Fluid Mechanics Fundamentals and Applications 4e Yunus Cengel, John Cimbala (Test Bank All Chapters, 100% Original Verified, A+ Grade) All Chapters Complete files download link at the end of this file.
06/06/2024, 01:09
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Chapter 1
1.
Award: 10.00 points
A substance in the solid or gas phase is referred to as a fluid.
True False
A substance in the liquid or gas phase is referred to as a fluid.
References True / False
2.
Difficulty: Easy
Award: 10.00 points
A fluid can resist an applied shear stress by deforming.
True False
A fluid deforms continuously under the influence of shear stress, no matter how small. References True / False
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Difficulty: Easy
1/70
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3.
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A liquid expands until it encounters the walls of the container and fills the entire available space.
True False
A liquid forms a free surface in a larger container in a gravitational field. A gas, on the other hand, expands until it encounters the walls of the container and fills the entire available space. A liquid forms a free surface in a larger container in a gravitational field. References True / False
4.
Difficulty: Easy
Award: 10.00 points
Define internal, external, and open-channel flows. External Internal Open-channel
flow is the flow of an unbounded fluid over a surface such as a plate, a wire, or a pipe. flow is the flow if the fluid is completely bounded by solid surfaces. flow is the flow when the pipe is partially filled with the liquid and there is a free surface
Explanation: External flow is the flow of an unbounded fluid over a surface such as a plate, a wire, or a pipe. The flow in a pipe or duct is internal flow if the fluid is completely bounded by solid surfaces. The flow of liquids in a pipe is called open-channel flow if the pipe is partially filled with the liquid and there is a free surface, such as the flow of water in rivers and irrigation ditches.
References Worksheet
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Difficulty: Easy
2/70
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5.
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Define incompressible flow and incompressible fluid. A fluid flow in which the density of the fluid remains nearly constant is called incompressible A fluid flow in which density varies significantly is called compressible flow.
flow.
Explanation: A fluid flow during which the density of the fluid remains nearly constant is called incompressible flow. A flow in which density varies significantly is called compressible flow.
References Worksheet
6.
Difficulty: Easy
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Must the flow of a compressible fluid necessarily be treated as compressible?
Yes No
A fluid whose density is practically independent of pressure (such as a liquid) is commonly referred to as an “incompressible fluid,” although it is more proper to refer to incompressible flow. The flow of compressible fluid (such as air) does not necessarily need to be treated as compressible since the density of a compressible fluid may still remain nearly constant during flow — especially flow at low speeds. The flow of compressible fluid (such as air) does not necessarily need to be treated as compressible since the density of a compressible fluid may still remain nearly constant during flow – especially flow at low speeds. References Yes / No https://ezto.mheducation.com/hm.tpx
Difficulty: Easy 3/70
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The flow of air over the wings of an aircraft is internal flow since this is an unbounded fluid flow over a surface.
True False
The flow of air over the wings of an aircraft is external since this is an unbounded fluid flow over a surface.
References True / False
8.
Difficulty: Easy
Award: 10.00 points
The flow of gases through a jet engine is internal flow since the fluid is completely bounded by the solid surfaces of the engine.
True False
The flow of gases through a jet engine is internal flow since the fluid is completely bounded by the solid surfaces of the engine.
References True / False
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Difficulty: Easy
4/70
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9.
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Define forced flow and natural flow. In forced flow, the fluid is made to flow over a surface or in a tube by external means such as a pump or a fan. In natural flow, any fluid motion is caused by means such as the buoyancy effect that manifests itself as the rise of the warmer fluid and the fall of the cooler fluid. Explanation: In forced flow, the fluid is forced to flow over a surface or in a tube by external means such as a pump or a fan. In natural flow, any fluid motion is caused by natural means such as the buoyancy effect that manifests itself as the rise of the warmer fluid and the fall of the cooler fluid.
In forced flow, the fluid is forced to flow over a surface or in a tube by external means such as a pump or a fan. In natural flow, any fluid motion is caused by natural means such as the buoyancy effect that manifests itself as the rise of the warmer fluid and the fall of the cooler fluid. References Worksheet
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Difficulty: Easy
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10.
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The ______ number of a flow is defined as the ratio of the speed of flow to the speed of sound in the flowing fluid.
Reynolds Mach Nusselt The Mach number of a flow is defined as the ratio of the speed of flow to the speed of sound in the flowing fluid.
References Multiple Choice
11.
Difficulty: Easy
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A Mach number of 2 indicates a flow speed that is the speed of sound in that fluid.
True False
A Mach number of 2 indicates a flow speed that is twice the speed of sound in that fluid.
References True / False
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Difficulty: Easy
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12.
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When an airplane is flying at a constant speed relative to the ground, is it correct to say that the Mach number of this airplane is also constant?
Yes No No. The speed of sound, and thus the Mach number, changes with temperature which may change considerably from point to point in the atmosphere.
References Multiple Choice
13.
Difficulty: Easy
Award: 10.00 points
Consider the flow of air at a Mach number of 0.12. Should this flow be approximated as being incompressible?
Yes No Gas flows can often be approximated as incompressible if the density changes are under about 5 percent, which is usually the case when Ma < 0.3. Therefore, airflow with a Mach number of 0.12 may be approximated as being incompressible.
References Multiple Choice
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Difficulty: Easy
7/70
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14.
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A fluid in direct contact with a solid surface sticks to the surface and there is no slip. This is known as the _____condition.
no-slip condition slip condition A fluid in direct contact with a solid surface sticks to the surface and there is no slip. This is known as the no-slip condition. References Multiple Choice
15.
Difficulty: Easy
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The no-slip condition is due to the _____ of the fluid.
viscosity density mass volume A fluid in direct contact with a solid surface sticks to the surface and there is no slip. This is known as the no-slip condition, and it is due to the viscosity of the fluid.
The no-slip condition is due to the viscosity of the fluid. References Multiple Choice
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Difficulty: Easy
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16.
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The definition of boundary layer refers to the region of flow in which both the temperature gradients are significant and frictional effects are important.
True False
The region of flow (usually near a wall) in which the velocity gradients are significant and frictional effects are important is called the boundary layer.
References True / False
17.
Difficulty: Easy
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The development of a boundary layer is caused by the no-slip condition.
True False
The development of a boundary layer is caused by the no-slip condition.
References True / False
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Difficulty: Easy
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A process is said to be _____ if it involves no changes with time anywhere within the system or at the system boundaries.
rational steady unsteady A process is said to be steady if it involves no changes with time anywhere within the system or at the system boundaries.
References Multiple Choice
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Difficulty: Easy
10/70
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Define stress, normal stress, shear stress, and pressure. The force per unit area is known as stress and is determined by dividing the force by the area upon which it acts. The normal component of a force acting on a surface per unit area is called the normal stress . The tangential component of a force acting on a surface per unit area is called shear stress . In a fluid at rest, the normal stress is called pressure .
Explanation: Stress is defined as force per unit area and is determined by dividing the force by the area upon which it acts. The normal component of a force acting on a surface per unit area is called the normal stress, and the tangential component of a force acting on a surface per unit area is called shear stress. In a fluid at rest, the normal stress is called pressure.
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Difficulty: Easy
11/70
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20.
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Define a system, the surroundings, and a boundary. The quantity of matter or a region in space chosen for study is called the system . The mass or region outside the system is called the surroundings . The real or imaginary surface that separates the system from its surroundings is called the boundary .
Explanation: A system is defined as a quantity of matter or a region in space chosen for study. The mass or region outside the system is called the surroundings. The real or imaginary surface that separates the system from its surroundings is called the boundary.
References Worksheet
21.
Difficulty: Easy
Award: 10.00 points
Consider that you are analyzing the acceleration of gases as they flow through a nozzle, what type of system is this?
Control volume Control mass When analyzing the acceleration of gases as they flow through a nozzle, a wise choice for the system is the volume within the nozzle, bounded by the entire inner surface of the nozzle and the inlet and outlet crosssections. This is a control volume (or open system) since mass crosses the boundary.
This is a control volume (or open system) since mass crosses the boundary. References Multiple Choice https://ezto.mheducation.com/hm.tpx
Difficulty: Easy 12/70
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When is a system a closed system, and when is it a control volume? When a system consists of a fixed amount of mass, and no mass can cross its boundary, it is known as a(n) closed system. When a system is a selected region in space, it is known as a(n) open system.
Explanation: Systems may be considered to be closed or open, depending on whether a fixed mass or a volume in space is chosen for study. A closed system (also known as a control mass or simply a system) consists of a fixed amount of mass, and no mass can cross its boundary. An open system, or a control volume, is a selected region in space. Mass may cross the boundary of a control volume or open system.
A closed system (also known as a control mass or simply a system) consists of a fixed amount of mass, and no mass can cross its boundary. An open system, or a control volume, is a selected region in space. References Worksheet
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Difficulty: Easy
13/70
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You are trying to understand how a reciprocating air compressor (a piston-cylinder device) works. What type of system would one use? This system is a(n) closed This system is a(n) open
system when it is compressing and no mass enters or leaves it. system during intake or exhaust.
Explanation: We would most likely take the system as the air contained in the piston-cylinder device. This system is a closed or fixed mass system when it is compressing and no mass enters or leaves it. However, it is an open system during intake or exhaust. This system is a closed or fixed mass system when it is compressing and no mass enters or leaves it. However, it is an open system during intake or exhaust. References Worksheet
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Difficulty: Easy
14/70
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06/06/2024, 01:09
24.
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What are the units for pound-mass and pound-force? The unit for pound-mass is lbm The unit for pound-force is lbf
. .
Explanation: Pound-mass lbm is the mass unit in English system whereas pound-force lbf is the force unit in the English system. One pound-force is the force required to accelerate a mass of 32.174 lbm by 1 ft/s2. In other words, the weight of a 1-lbm mass at sea level on earth is 1 lbf. The unit for pound-mass is lbm. The unit for pound-force is lbf. References Worksheet
25.
Difficulty: Easy
Award: 10.00 points
In a news article, it is stated that a recently developed geared turbofan engine produces 15,000 pounds of thrust to propel the aircraft forward. Is “pound” mentioned here lbm or lbf?
lbm lbf The “pound” mentioned here is “lbf” since thrust is a force, and lbf is the force unit in the English.
References Multiple Choice
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Difficulty: Easy
15/70
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26.
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Light-year has the dimension of length.
True False
In this unit, the word light refers to the speed of light. The light-year unit is then the product of a velocity and time. Hence, this product forms a distance dimension and unit. References True / False
Difficulty: Easy
What is the net force acting on a car? References Section Break
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Difficulty: Easy
16/70
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The car cruises at a constant velocity of 70 km/h on a level road.
140 35 70 0 There is no acceleration (car moving at constant velocity), thus the net force is zero. The net force is 0. References Multiple Choice
28.
Difficulty: Easy
Award: 10.00 points
The car cruises at a constant velocity of 70 km/h on an uphill road.
140 35 0 70 There is no acceleration (car moving at constant velocity), thus the net force is zero. The net force is 0. References Multiple Choice
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Difficulty: Easy
17/70
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29.
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A man goes to a traditional market to buy a steak for dinner. He finds a 14-oz steak (1 lbm = 16 oz) for $3.15. He then goes to the adjacent international market and finds a 330-g steak of identical quality for $3.30. Which steak is the better buy? The steak at the traditional
market is a better buy.
Explanation: The assumption made here is that the steaks are of identical quality.
The 14-ounce steak: Unit cost = ( 14 oz ) ( 1 lbm ) ( 0.45359 kg ) = $7.94/kg $3.15
16 oz
1 lbm
The 330-gram steak: Unit cost = ( 330 g ) ( 1 kg ) = $10/kg $3.30
1000 g
Therefore, the steak at the traditional market is the better buy. The steak at the traditional market is a better buy. References Worksheet
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Difficulty: Easy
18/70
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30.
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What is the weight, in N, of an object with a mass of 170 kg at a location where g = 9.6 m/s2? The weight of the object is
1632 ± 2% N.
Explanation: Applying Newton's second law, the weight is determined to be
W = mg = (170 kg) (9.6 m/s2 ) = 1632 N The weight of the object is 1632 N. References Worksheet
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Difficulty: Easy
19/70
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What is the weight of a 8-kg substance in N, kN, kg·m/s2, kgf, lbm·ft/s2, and lbf? The weight of a 8-kg substance in N is
78.48 ± 2% N.
The weight of a 8-kg substance in kN is
0.07848 ± 2% kN.
The weight of a 8-kg substance in kg·m/s2 is The weight of a 8-kg substance in kgf is
78.48 ± 2% kg·m/s2. 8 ± 2% kgf.
The weight of a 8-kg substance in lbm·ft/s2 is The weight of a 8-kg substance in lbf is
568.01 ± 2% lbm·ft/s2.
17.64 ± 2% lbf.
Explanation: Applying Newton's second law, the weight is determined in various units to be
W = mg = (8 kg) (9.81 m/s2 ) (
1N 2 ) = 78.48 N 1 kg⋅m/s
W = mg = (8 kg) (9.81 m/s2 ) (
1N 2 ) = 0.07848 kN 1000 kg⋅m/s
W = mg = (8 kg) (9.81 m/s2 ) = 78.48 kg⋅m/s2 W = mg = (8 kg) (9.81 m/s2 ) (
1 kgf 1N 2 ) ( 9.81 N ) = 8 kgf 1 kg⋅m/s
lbm W = mg = (8 kg) ( 2.205 ) (32.2 ft/s2 ) = 568.01 lbm⋅ft/s2 1 kg lbm W = mg = (8 kg) ( 2.205 ) (32.2 ft/s2 ) ( 1 kg
1 lbf 32.2 lbm⋅ft/s
2
) = 17.64 lbf
The weight of a 8-kg substance in N is 78.48 N. The weight of a 8-kg substance in kN is 0.07848 kN. The weight of a 8-kg substance in kg·m/s2 is 78.48 kg·m/s2. The weight of a 8-kg substance in kgf is 8 kgf. The weight of a 8-kg substance in lbm·ft/s2 is 568.01 lbm·ft/s2. The weight of a 8-kg substance in lbf is 17.64 lbf. https://ezto.mheducation.com/hm.tpx
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Difficulty: Easy
21/70
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32.
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Determine the mass and the weight of the air contained in a room whose dimensions are 4 m × 3 m × 7 m. Assume the density of the air is 1.16 kg/m3. The mass of the air contained in the room is
97 ± 2% kg.
The weight of the air contained in the room is
956 ± 2% N.
Explanation: The assumption made here is that the density of air is constant throughout the room.
The mass of the air in the room is
m = ρV = (1.16 kg/m3 ) (4 × 3 × 7 m3 ) = 97.4 kg ≅97 kg Thus,
W = mg = (97.4 kg) (9.81 m/s2 ) (
1N 2 ) = 956 N 1 kg⋅m/s
The mass of the air contained in the room is 97 kg. The weight of the air contained in the room is 956 N. References Worksheet
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Difficulty: Easy
22/70
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33.
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A 4-kW resistance heater in a water heater runs for 1.3 hours to raise the water temperature to the desired level. Determine the amount of electric energy used in both kWh and kJ. The total energy in kWh = The total energy in kJ =
5.2 ± 2% kWh . 18720 ± 2% kJ.
Explanation: The resistance heater consumes electric energy at a rate of 4 kW or 4 kJ/s. The total amount of electric energy used in 1.3 hours becomes Total energy = (Energy per unit time)(Time interval)
Total energy = (4 kW) (1.3 h) = 5.2 kWh Noting that 1 kWh = (1 kJ/s)(3600 s) = 3600 kJ,
Total energy = (5.2 kWh) (3600 kJ/kWh) = 18720 kJ
Total energy in kWh = 5.2 kWh Total energy in kJ = 18720 kJ References Worksheet
Difficulty: Easy
A 220-lbm astronaut took his bathroom scale (a spring scale) and a beam scale (compares masses) to the moon where the local gravity is g = 5.48 ft/s2. References Section Break
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Difficulty: Easy
23/70
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Determine how much he will weigh on the spring scale. He will weigh
37.4 ± 2% lbf on the spring scale.
Explanation: A spring scale measures weight, which is the local gravitational force applied on a body:
W = mg = (220 lbm) (5.48 ft/s2 ) (
1 lbf 2 ) = 37.4 lbf 32.2 lbm⋅ft/s
He will weigh 37.4 lbf on the spring scale. References Worksheet
35.
Difficulty: Easy
Award: 10.00 points
Determine how much he will weigh on the beam scale. He will weigh
220 ± 2% lbf on the beam scale.
Explanation: A beam scale compares masses and is thus not affected by the variations in gravitational acceleration. The beam scale reads what it reads on earth:
W = 220 lbf He will weigh 220 lbf on the beam scale. References Worksheet
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Difficulty: Easy
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The acceleration of high-speed aircraft is sometimes expressed in g’s (in multiples of the standard acceleration of gravity). Determine the net force, in N, that a 130-kg man would experience in an aircraft whose acceleration is 6 g’s. The net force that the man would experience is
7652 ± 2% N.
Explanation: From Newton's second law, the applied force is
F = ma = m (6g) = (130 kg) (6 × 9.81 m/s2 ) (
1N 2 ) = 7652 N 1 kg⋅m/s
The net force that the man would experience is 7652 N.
References Worksheet
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Difficulty: Easy
25/70
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37.
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A 11-kg rock is thrown upward with a force of 280 N at a location where the local gravitational acceleration is 9.79 m/s2. Determine the acceleration of the rock, in m/s2. The acceleration of the rock is
15.7 ± 2% m/s2.
Explanation:
The weight of the rock is
W = mg = (11 kg) (9.79 m/s2 ) (
1N 2 ) = 107.7 N 1 kg⋅m/s
Then the net force that acts on the rock is
Fnet = Fup - Fdown = 280 N - 107.7 N = 172.3 N From Newton's second law, the acceleration of the rock becomes F N a= m = 172.3 ( 11 kg
1 kg⋅m/s ) = 15.7 m/s 2 1N 2
The acceleration of the rock is 15.7 m/s2. References Worksheet
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Difficulty: Easy
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A 13-kg rock is thrown upward with a force of 280 N at a location where the local gravitational acceleration is 9.79 m/s2. Determine the acceleration of the rock, in m/s2. Solve this problem using appropriate software. Print out the entire solution, including the numerical results with proper units. The acceleration of the rock is
11.7 ± 2% m/s2.
Explanation: The problem is solved using appropriate software. m=13 [kg] F_up=280 [N] g=9.79 [m/s^2] W=m*g F_net = F_up - F_down F_down=W F_net=a*m SOLUTION a=18.21 [m/s^2] F_down=97.9 [N] F_net=182.1 [N] F_up=280 [N] g=9.79 [m/s^2] m=13 [kg] W=97.9 [N] The final results are W = 127.3 N and a = 11.7 m/s2, to three significant digits, which agree with the results of the previous problem.
The acceleration of the rock is 11.7 m/s2. References Worksheet
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Difficulty: Easy
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The value of the gravitational acceleration g decreases with elevation from 9.807 m/s2 at sea level to 9.757 m/s2 at an altitude of 13,000 m, where large passenger planes cruise. Determine the percent reduction in the weight of an airplane cruising at 13,000 m relative to its weight at sea level. The percent reduction in the weight of an airplane cruising at 13,000 m relative to its weight at sea level is 0.51 ± 2% %. Explanation: Weight is proportional to the gravitational acceleration g, and thus the percent reduction in weight is equivalent to the percent reduction in the gravitational acceleration, which is determined from
%Reduction in weight = %Reduction in g =
2
2
Δg 9.807 m/s - 9.757 m/s × 100 = 2 g 9.807 m/s
× 100 = 0.51%
Therefore, the airplane and the people in it will weigh 0.51% less at an altitude of 13,000 m.
The percent reduction in the weight of an airplane cruising at 13,000 m relative to its weight at sea level is 0.51%. References Worksheet
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Difficulty: Easy
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At 45° latitude, the gravitational acceleration as a function of elevation z above sea level is given by g = a − bz, where a = 9.807 m/s2 and b = 3.37 × 10−6 s−2. Determine the height above sea level where the weight of an object will decrease by 1 percent. The height above sea level where the weight of an object will decrease by 1 percent is
29101 ± 2% m.
Explanation:
The weight of a body at elevation z can be expressed as
W = mg = m (a - bz) where a = gs = 9.807 m/s2 is the value of gravitational acceleration at sea level and b = 3.37 × 10–6 s–2. In our case,
W = m (a - bz) = 0.99Ws = 0.99mgs We cancel out mass from both sides of the equation and solve for z, yielding
z=
a - 0.99g s b
On substituting, we get 9.807 m/s - 0.99(9.807 m/s ) 2
z=
2
3.37 × 10−6 1/s
2
= 29101 m
The height above sea level where the weight of an object will decrease by 1 percent is 29101 m. References Worksheet
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Difficulty: Easy
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The gravitational constant g is 9.807 m/s2 at sea level, but it decreases as you go up in elevation. A useful equation for this decrease in g is g = a – bz, where z is the elevation above sea level, a = 9.807 m/s2, and b = 3.32 × 10–6 1/s2. An astronaut “weighs” 80.0 kg at sea level. [Technically this means that his/her mass is 80.0 kg.] Calculate this person’s weight in N while floating around in the International Space Station (z = 350 km). If the Space Station were to suddenly stop in its orbit, what gravitational acceleration would the astronaut feel immediately after the satellite stopped moving? The person’s weight in N while floating around in the International Space Station is The astronaut feels a gravitational acceleration of 8.645 ± 2% m/s2.
691.6 ± 2% N.
Explanation: At the altitude of the Space Station,
g = 9.807 - (3.32 × 10−6 s−2 ) (350000 m) = 8.645 m/s2 The astronaut’s weight would therefore be
W = mg = (80 kg) (8.645 m/s2 ) (
N ) = 691.6 N 2 kg⋅m/s
The person’s weight in N while floating around in the International Space Station is 691.6 N. The astronaut feels a gravitational acceleration of 8.645 m/s2. References Worksheet
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Difficulty: Easy
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On average, an adult person breathes in about 10 liters of air per minute. Assuming atmospheric pressure and 20°C air temperature, estimate the mass of air in kilograms that a person breathes in per day. The mass of air in kilograms that a person breathes in per day is
17.1 ± 2% kg.
Explanation: The total volume of air breathed in per day is 1m V = (10 L/min) (60 min/hr) (24 hr/day) ( 1000 ) = 14.4 m 3 L 3
We use the ideal gas law to calculate the total mass of air: kJ P V = mRT → m = PV = (0.287 kJ/kg⋅K)(298.15 K) ( kPa ) ( kN⋅m ) = 17.051 kg RT (101.325 kPa)(14.4 m3 )
kN/m
2
So, an average person breathes in about 17 kg of air per day.
The mass of air in kilograms that a person breathes in per day is 17.1 kg. References Worksheet
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Difficulty: Easy
31/70
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43.
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While solving a problem, a person ends up with the equation E = 16 kJ + 7 kJ/kg at some stage. Here, E is the total energy and has the unit of kilojoules. Determine how to correct the error and discuss what may have caused it. Multiplying the last term by mass denominator , and this will correct the error.
will eliminate the kilograms
in the
Explanation: The two terms on the right-hand side of the equation E = 16 kJ + 7 kJ/kg do not have the same units, and therefore they cannot be added to obtain the total energy. Multiplying the last term by mass will eliminate the kilograms in the denominator, and the whole equation will become dimensionally homogeneous; that is, every term in the equation will have the same unit.
Multiplying the last term by mass will eliminate the kilograms in the denominator, and this will correct the error. References Worksheet
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Difficulty: Easy
32/70
06/06/2024, 01:09
44.
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Award: 10.00 points
An airplane flies horizontally at 70 m/s. Its propeller delivers 1550 N of thrust (forward force) to overcome aerodynamic drag (backward force). Using dimensional reasoning and unity conversion ratios, calculate the useful power delivered by the propeller in units of kW and horsepower. The useful power delivered by the propeller in units of kW is
108 ± 2% kW.
The useful power delivered by the propeller in units of horsepower is
145 ± 2% hp.
Explanation: The following assumptions have been made here: 1. The airplane flies at constant altitude and constant speed. 2. Wind is not a factor in the calculations. At steady horizontal flight, the airplane’s drag is balanced by the propeller’s thrust. Energy is force times distance, and power is energy per unit time. Thus, by dimensional reasoning, the power supplied by the propeller must equal thrust times velocity:
Ẇ = Fthrust V = (1550 N) (70 m/s) ( 10001 kW ) = 108 kW = 108 kW ( 1 kW ) = 145 hp N⋅m/s 1.341 hp
The useful power delivered by the propeller in units of kW is 108 kW. The useful power delivered by the propeller in units of horsepower is 145 hp. References Worksheet
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Difficulty: Easy
33/70
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45.
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Award: 10.00 points
If an airplane weighs 1500 lbf, estimate the lift force produced by the airplane’s wings (in lbf and newtons) when flying at 70.0 m/s. The lift force produced by the airplane’s wings is
6673 ± 2% N.
Explanation: The following assumptions have been made here: 1. The airplane flies at constant altitude and constant speed. 2. Wind is not a factor in the calculations. At steady horizontal flight, the airplane’s weight is balanced by the lift produced by the wings. Thus, the net lift force must equal the weight, or FL = 1500 lbf. We use unity conversion ratios to convert to newtons: 1N FL = (1500 lbf) ( 0.2248 ) = 6673 N lbf
The lift force produced by the airplane’s wings is 6673 N. References Worksheet
Difficulty: Easy
The boom of a fire truck raises a fireman (and his equipment—total weight 280 lbf) 39 ft into the air to fight a building fire. References Section Break
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Difficulty: Easy
34/70
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Calculate the work done by the boom on the fireman in units of Btu. (You must provide an answer before moving on to the next part.) The work done by the boom on the fireman is
14 ± 2% Btu.
Explanation: It is assumed that the vertical speed of the fireman is constant. Work W is a form of energy and is equal to force times the distance. Here, the force is the weight of the fireman (and equipment) and the vertical distance is Δz, where z is the elevation.
W =F Δz 1 Btu W = (280 lbf) (39 ft) ( 778.169 ) ft⋅lbf
W = 14 Btu The work done by the boom on the fireman is 14 Btu. References Worksheet
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Difficulty: Easy
35/70
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If the useful power supplied by the boom to lift the fireman is 3.5 hp, estimate how long it takes to lift the fireman. The time taken to lift the fireman is
5.67 ± 2% s.
Explanation: Power is work (energy) per unit time. Assuming a constant speed, Btu Δt = W = 14.033 ( 0.7068 Btu/s ) 3.50 hp 1 hp
Ẇ
Δt = 5.67 s The time taken to lift the fireman is 5.67 s. References Worksheet
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Difficulty: Easy
36/70
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A 6-kg plastic tank that has a volume of 0.18 m3 is filled with liquid water. Assuming the density of water is 1000 kg/m3, determine the weight of the combined system. The weight of the combined system is
1825 ± 2% N.
Explanation: The assumption made here is that the density of water is constant throughout.
The mass of the water in the tank and the total mass are
mW = ρV = (1000 kg/m3 ) (0.18 m3 ) = 180 kg mtotal = mw + mtank = 180 kg + 6 kg = 186 kg Thus,
W = mg = (186 kg) (9.81 m/s2 ) (
1N 2 ) = 1825 N 1 kg⋅m/s
The weight of the combined system is 1825 N. References Worksheet
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Difficulty: Easy
37/70
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Water at 15°C (ρ = 999.1 kg/m3) from a garden hose fills a 1.5-L container in 2.65 s. Assume that the volume flow rate, temperature, and density of water are constant over the measured time. References Section Break
49.
Difficulty: Easy
Award: 10.00 points
Calculate the volume flow rate in liters per minute (Lpm). (You must provide an answer before moving on to the next part.) The volume flow rate is
34 ± 2% Lpm.
Explanation: The volume flow rate is equal to the volume per unit time, i.e., V 1.5 L V˙ = Δt = 2.65 ( 60 s ) s 1 min
V˙ = 34 Lpm The volume flow rate is 34 Lpm. References Worksheet
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Difficulty: Easy
38/70
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Calculate the mass flow rate in kg/s. The mass flow rate is
0.566 ± 2% kg/s.
Explanation: Since density is mass per unit volume, the mass flow rate is equal to volume flow rate times density. Thus, 1 m3 ) ( ) ṁ = ρV˙ = (998.0 kg/m3 ) (33.962 L/min) ( 160min s 1000 L
ṁ = 0.566 kg/s The mass flow rate is 0.566 kg/s. References Worksheet
Difficulty: Easy
A forklift raises an 88.5-kg crate 1.80 m. Assume that the vertical speed of the crate is constant. Take g = 9.807 m/s2. References Section Break
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Difficulty: Easy
39/70
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Calculate the work done by the forklift on the crate, in units of kJ. (You must provide an answer before moving on to the next part.) The work done by the forklift on the crate is
1.56 ± 2% kJ.
Explanation: Work W is a form of energy and is equal to force times the distance. Here, the force is the weight of the crate, which is F = mg, and the vertical distance is Δz, where z is the elevation.
W = F Δz = mgΔz W = (88.5 kg) (9.807 m/s2 ) (1.80 m) (
1N 1 kJ 2 ) ( 1000 N⋅m ) 1 kg⋅m/s
W = 1.5623 kJ = 1.56 kJ The work done by the forklift on the crante is 1.56 kJ.
Which formula should be used to calculate the work done? References Worksheet
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Difficulty: Easy
40/70
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If it takes 12.3 seconds to lift the crate, calculate the useful power supplied to the crate in watts. The useful power supplied to the crate is
127 ± 2% W.
Explanation: Power is work (energy) per unit time. Assuming a constant speed, W kJ W Ẇ = Δt = 1.5623 ( 1000 ) 12.3 s 1 kJ/s
Ẇ = 127.01 W = 127 W The useful power supplied to the crate is 127 W. References Worksheet
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Difficulty: Easy
41/70
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The gas tank of a car is filled with a nozzle that discharges gasoline at a constant flow rate. Based on unit considerations of quantities, choose the correct relation for the filling time in terms of the volume V of the tank (in L) and the discharge rate of gasoline (V˙ , in L/s).
t = VV˙ V− t = V˙√−
t = VV˙ t = V × V˙ The assumption made here is that gasoline is an incompressible substance and the flow rate is constant. The filling time depends on the volume of the tank and the discharge rate of gasoline. Also, we know that the unit of time is "seconds." Therefore, the independent quantities should be arranged such that we end up with the unit of seconds. Putting the given information into perspective, we have
t (s) ↔ V (L), and V˙ (L/s) It is clear that the only way to end up with the unit “s” for time is to divide the tank volume by the discharge rate. Therefore, the desired relation is
t = V˙ V
The desired relation is t = V˙ V
References Multiple Choice
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Difficulty: Easy
42/70
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A pool of volume V (in m3) is to be filled with water using a hose of diameter D (in m). If the average discharge velocity is V (in m/s) and the filling time is t (in s), choose the correct relation for the volume of the pool based on unit considerations of quantities involved.
V = C D2 V t V = CD2 Vt Vt V = CD 2 2
V = CD Vt
The assumption made here is that water is an incompressible substance and the average flow velocity is constant. The pool volume depends on the filling time; the cross-sectional area, which depends on hose diameter; and flow velocity. Also, we know that the unit of volume is m3. Therefore, the independent quantities should be arranged such that we end up with the unit of seconds. Putting the given information into perspective, we have
V (m3 ) is a function of t (s), D (m), and V (m/s) It is obvious that the only way to end up with the unit “m3” for volume is to multiply the quantities t and V with the square of D. Therefore, the desired relation is
V = C D2 V t where the constant of proportionality is obtained for a round hose, namely, C = π/4 so that = (πD2/4)Vt.
The desired relation is V = C D2 V t. References Multiple Choice
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Difficulty: Easy
43/70
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Based on unit considerations alone, choose the correct equation that shows the power needed to accelerate a car of mass m (in kg) from rest to velocity V (in m/s) in time interval t (in s). 2
Ẇ = VCmt Ẇ = CmV t
2
Ẇ = CmV 2t Ẇ = Cm V 2t The assumption made here is that the car is initially at rest. The power needed for acceleration depends on the mass, velocity change, and time interval. Also, the unit of power Ẇ is watt, W, which is equivalent to
W = J/s = N·m/s = (kg·m/s2)m/s = kg·m2/s3 Therefore, the independent quantities should be arranged such that we end up with the unit kg·m2/s3 for power. Putting the given information into perspective, we have
Ẇ (kg⋅m 2/s3 ) is a function of m (kg), V (m/s), and t (s) It is obvious that the only way to end up with the unit “kg·m2/s3” for power is to multiply mass with the square of the velocity and divide by time. Therefore, the desired relation is
Ẇ is proptional to mV 2/t or
Ẇ = CmV t
2
The desired relation is Ẇ = CmV t
2
References Multiple Choice
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Difficulty: Easy
44/70
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What is the importance of modeling in engineering?
To predict the course of an event before it actually occurs To study various aspects of an event mathematically without actually running expensive and timeconsuming experiments
To study various aspects of an event mathematically by actually running various experiments To identify all the variables that affect the phenomena Modeling makes it possible to predict the course of an event before it actually occurs, or to study various aspects of an event mathematically without actually running expensive and time-consuming experiments. References Check All That Apply
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Difficulty: Easy
45/70
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List the procedure for mathematical models for engineering processes in the correct order.
Reasonable assumptions and approximations are made.
Correct: 2
The problem is formulated mathematically.
Correct: 5
Correct: 6
The problem is solved using an appropriate approach, and the results are interpreted.
The interdependence of the variables is studied.
Correct: 3
The relevant physical laws and principles are invoked.
Correct: 4
All the variables that affect the phenomena are identified.
Correct: 1
When preparing a mathematical model, all the variables that affect the phenomena are identified, reasonable assumptions and approximations are made, and the interdependence of these variables is studied. The relevant physical laws and principles are invoked, and the problem is formulated mathematically. Finally, the problem is solved using an appropriate approach, and the results are interpreted. References Ranking
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Difficulty: Easy
46/70
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58.
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Match the approaches to their advantage. 1. Experimental approach 2. Analytical approach
Correct: 1
It deals with the actual physical system and gets a physical value within the limits of experimental error.
It is fast and inexpensive
Correct: 2
The experimental approach (testing and taking measurements) has the advantage of dealing with the actual physical system, and getting a physical value within the limits of experimental error. However, this approach is expensive, time consuming, and often impractical. The analytical approach (analysis or calculations) has the advantage that it is fast and inexpensive, but the results obtained are subject to the accuracy of the assumptions and idealizations made in the analysis. References Matching
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Difficulty: Easy
47/70
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When modeling an engineering process, what is the right choice made between a simple but crude and a complex but accurate model?
The simplest model that yields adequate results The complex model that yields adequate results The numerical model that yields adequate results The simplest model that yields lesser results The right choice between a crude and complex model is usually the simplest model that yields adequate results. References Multiple Choice
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Difficulty: Easy
48/70
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Match the terms precision and accuracy to their definitions. 1. Accuracy 2. Precision
Correct: 2
The number of significant digits or the closeness of different measurements of the same quantity to each other.
The closeness of the measured or calculated value to the true value
Correct: 1
Accuracy refers to the closeness of the measured or calculated value to the true value whereas precision represents the number of significant digits or the closeness of different measurements of the same quantity to each other. References Matching
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Difficulty: Easy
49/70
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Identify how differential equations in the study of a physical problem arise.
In the limiting case of infinitesimal changes in consatnts In the limiting case of infinitesimal changes in variables In the limiting case of large changes in constants In the limiting case of large changes in variables In the limiting case of infinitesimal changes in variables, we obtain differential equations, which provide precise mathematical formulations for the physical principles and laws by representing the rates of changes as derivatives. References Multiple Choice
62.
Difficulty: Easy
Award: 10.00 points
Software packages are of great value in engineering practice.
True False
Software packages are of great value in engineering practice, and engineers today rely on software packages to solve large and complex problems quickly, and to perform optimization studies efficiently.
References True / False
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Difficulty: Easy
50/70
06/06/2024, 01:09
63.
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Solve this system of three equations with three unknowns using appropriate software: 2x − y + z = 9 3x2 + 2y = z + 2 xy + 2z = 14 x= y= z=
1.556 ± 2% 0.6254 ± 2% 6.513 ± 2%
Explanation: Using appropriate software, copy the following lines and paste them on a blank screen to verify the solution: 2*x-y+z=9 3*x^2+2*y=z+2 x*y+2*z=14 x = 1.556, y = 0.6254, z = 6.513 x = 1.556, y = 0.6254, z = 6.513 References Worksheet
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Difficulty: Easy
51/70
06/06/2024, 01:09
64.
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Solve this system of two equations with two unknowns using appropriate software: x3 − y2 = 10.5 3xy + y = 4.6 x= y=
2.215 ± 2% 0.6018 ± 2%
Explanation: Using appropriate software, copy the following lines and paste them on a blank screen to verify the solution: x^3–y^2=10.5 3*x*y+y=4.6 x = 2.215, y = 0.6018
x = 2.215, y = 0.6018 References Worksheet
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Difficulty: Easy
52/70
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65.
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Determine a positive real root of this equation using appropriate software: 3.5x3 − 10x0.5 − 3x = −4 x=
1.554 ± 2%
Explanation: Using appropriate software, copy the following lines and paste them on a blank screen to verify the solution: 3.5*x^3–10*x^0.5–3*x = –4 x = 1.554 x = 1.554 References Worksheet
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Difficulty: Easy
53/70
06/06/2024, 01:09
66.
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Solve this system of three equations with three unknowns using appropriate software: x2y − z = 1.5 x − 3y0.5 + xz = −2 x + y − z = 4.2 x= y= z=
0.9149 ± 2% 10.95 ± 2% 7.665 ± 2%
Explanation: Using appropriate software, copy the following lines and paste them on a blank screen to verify the solution: x^2*y–z=1.5 x–3*y^0.5+x*z=-2 x+y–z=4.2 x = 0.9149, y = 10.95, z = 7.665 x = 0.9149, y = 10.95, z = 7.665 References Worksheet
Difficulty: Easy
A student buys a 5200 Btu window air conditioner for his apartment bedroom. He monitors it for one hour on a hot day and determines that it operates approximately 60 percent of the time (duty cycle = 60 percent) to keep the room at nearly constant temperature. References Section Break
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Difficulty: Easy
54/70
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67.
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Showing all your work and using unity conversion ratios, calculate the rate of heat transfer into the bedroom through the walls, windows, etc. in units of Btu/h and in units of kW. (Please provide an answer before moving to the next part.) The rate of heat transfer into the bedroom through the walls, windows, etc. in units of Btu/h is 3120 ± 2% Btu/h. The rate of heat transfer into the bedroom through the walls, windows, etc. in units of kW is 0.914 ± 2% kW. Explanation: The following assumptions have been made here: 1. The rate of heat transfer is constant. 2. The indoor and outdoor temperatures do not change significantly during the hour of operation. In one hour, the air conditioner supplies 5,000 Btu of cooling but runs only 60% of the time. Since the indoor and outdoor temperatures remain constant during the hour of operation, the average rate of heat transfer into the room is the same as the average rate of cooling supplied by the air conditioner. Thus,
Q̇ =
0.60(5200 Btu) 1 kW = 3120 Btu/h ( 3412.14 ) = 0.914 kW 1h Btu/h
The rate of heat transfer into the bedroom through the walls, windows, etc. in units of Btu/h is 3120 Btu/h. The rate of heat transfer into the bedroom through the walls, windows, etc. in units of kW is 0.914 kW. References Worksheet
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Difficulty: Easy
55/70
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68.
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If the energy efficiency ratio (EER) of the air conditioner is 9.0 and electricity costs 7.5 cents per kilowatthr, calculate how much it costs (in cents) for him to run the air conditioner for one hour. It costs only
2.6 ± 2% cents to run the air conditioner for one hour.
Explanation: Energy efficiency ratio is defined as the amount of heat removed from the cooled space in Btu for 1 Wh (watt-hour) of electricity consumed. Thus, for every Wh of electricity, this particular air conditioner removes 9.0 Btu from the room. To remove 3120 Btu in one hour, the air conditioner therefore consumes
(
3120 Btu/h ) ( 11Wh ) = 346.67 Wh = 0.34667 kWh 9.0 Btu h
At a cost of 7.5 cents per kWh, for one hour it will cost
(7.5 cents/kWh × 0.34667 kWh) = 2.6 cents
It costs only 2.6 cents to run the air conditioner for one hour. References Worksheet
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Difficulty: Easy
56/70
06/06/2024, 01:09
69.
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Award: 10.00 points
The weight of bodies may change somewhat from one location to another as a result of the variation of the gravitational acceleration g with elevation. Accounting for this variation using g = a − bz, where a = 9.807 m/s2 and b = 3.32 × 10−6 s−2, determine the weight of a 70-kg person at sea level (z = 0), in Denver (z = 1610 m), and on the top of Mount Everest (z = 8848 m). 686.5 ± 2% N. The weight of the person at sea level is 686.1 ± 2% N. The weight of the person in Denver is 684.4 ± 2% N. The weight of the person on the top of Mount Everest is
Explanation: The weight of a 70-kg man at various locations is obtained by substituting the altitude z (values in m) into the relation
W = mg = (70 kg) (9.807 m/s2 - 3.32 × 10−6 m/s2 z) (
1N 2 ) 1 kg⋅m/s
At sea level (z = 0 m),
W = 70 kg × (9.807 m/s2 - (3.32 × 10−6 m/s2 × 0 m)) = 70 kg × 9.807 m/s2 = 686.5 N In Denver (z = 1610 m),
W = 70 kg × (9.807 m/s2 - (3.32 × 10−6 m/s2 × 1601 m)) = 70 kg × 9.802 m/s2 = 686.1 N On the top of Mount Everest (z = 8848 m),
W = 70 kg × (9.807 m/s2 - (3.32 × 10−6 m/s2 × 8848 m)) = 70 kg × 9.778 m/s2 = 684.4 N The weight of the person at sea level is 686.5 N. The weight of the person in Denver is 686.1 N. The weight of the person on the top of Mount Everest is 684.4 N. References Worksheet
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Difficulty: Easy
57/70
06/06/2024, 01:09
70.
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The reactive force developed by a jet engine to push an airplane forward is called thrust, and the thrust developed by the engine of a Boeing 777 is about 84600 lbf. Express this thrust in N and kgf. The thrust in N is
3.79 ± 2% × 105 N.
The thrust in kgf is
3.86 ± 2% × 104 kgf.
Explanation: Noting that 1 lbf = 4.448 N and 1 kgf = 9.81 N, the thrust developed is expressed in two other units as follows Thrust in N N Thrust = (84600 lbf) ( 4.48 ) = 3.79 × 105 N 1 lbf
Thrust in kgf
Thrust = (3.79 × 105 N) ( 9.81 N ) = 3.86 × 104 kgf 1 kgf
The thrust in N is 3.79 × 105 N. The thrust in kgf is 3.86 × 104 kgf. References Worksheet
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Difficulty: Easy
58/70
06/06/2024, 01:09
71.
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For liquids, the dynamic viscosity μ, which is a measure of resistance against flow, is approximated as μ = a10b/(T−c), where T is the absolute temperature, and a, b, and c are experimental constants. Using the data listed in Table A-7 for methanol at 20ºC, 40ºC, and 60ºC, determine the constant a, b, and c. Use appropriate software to solve this problem. The value of constant a is The value of constant b is The value of constant c is
8.493 ± 2% 10–6 Pa·s. 534.5 ± 2% K. 2.27 ± 2% K.
Explanation: Using the data from Table A-7, we have
5.857 × 10 −4 = a10b/(293 - c) 4.460 × 10 −4 = a10b/(313 - c) 3.510 × 10 −4 = a10b/(333 - c) which is a nonlinear system of three algebraic equations. Using appropriate software, one finds
a = 8.493 × 10−6 Pa⋅s
b = 534.5 K
c = 2.27 K
Then the viscosity correlation for methanol becomes
μ = (8.493 × 10 −6 ) × 10534.5/(T - 2.27) For T = 50ºC = 323 K, the correlation gives μ = 3.941 × 10-4 Pa·s, which agrees well with the data in Table A-7.
The value of constant a is 8.493 × 10-6 Pa·s. The value of constant b is 534.5 K. The value of constant c is 2.27 K. References Worksheet
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Difficulty: Easy
59/70
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An important design consideration in the two-phase pipe flow of solid-liquid mixtures is the terminal settling velocity below which the flow becomes unstable and eventually the pipe becomes clogged. On the basis of extended transportation tests, the terminal settling velocity of a solid particle in the rest water is given by −−−−−−−−−− VL = FL √2gD (S − 1), where FL is an experimental coefficient, g the gravitational acceleration, D the pipe diameter, and S the specific gravity of solid particle. What is the dimension of FL? Is this equation dimensionally homogeneous? The dimension of FL is None
.
This equation is dimensionally homogeneous
.
Explanation: We have the dimensions for each term except FL. [g] = [LT–2] [D] = [L] [VL] = [LT–1] and [LT–1] = [FL][2LT–2]1/2[L]1/2 = √2[FL][LT–1] Therefore, FL is a dimensionless coefficient; that is, this equation is dimensionally homogeneous and should hold for any unit system.
The dimension of FL is None, it is a dimensionless coefficient . This equation is dimensionally homogenous. References Worksheet
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Difficulty: Easy
60/70
06/06/2024, 01:09
73.
Assignment Print View
Award: 10.00 points
Consider the flow of air through a wind turbine whose blades sweep an area of diameter D (in m). The average air velocity through the swept area is V (in m/s). On the bases of the units of the quantities involved, identify the correct equation for the mass flow rate of air (in kg/s) through the swept area.
ṁ = CρV2 D
2
ṁ = CρD V
ṁ = CρD2 V ṁ = ρV2 D
The assumption made here is that the wind approaches the turbine blades with a uniform velocity. The mass flow rate depends on the air density, the average wind velocity, and the cross-sectional area, which depends on hose diameter. Also, the unit of mass flow rate ṁ is kg/s. Therefore, the independent quantities should be arranged such that we end up with the proper unit. Putting the given information into perspective, we have
ṁ [kg/s] is a function of ρ [kg/m3], D [m], and V [m/s] It is obvious that the only way to end up with the unit “kg/s” for mass flow rate is to multiply the quantities ρ and V with the square of D. Therefore, the desired proportionality relation is
ṁ is proportional to ρD2V or
ṁ = CρD2 V where the constant of proportionality is C =π/4 so that ṁ = ρ (
πD2 )V . 4
The correct equation for the mass flow rate of air (in kg/s) through the swept area is ṁ = CρD2 V . References Multiple Choice
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Difficulty: Easy
61/70
06/06/2024, 01:09
74.
Assignment Print View
Award: 10.00 points
A tank is filled with oil whose density is ρ = 850 kg/m3. If the volume of the tank is V = 2 m3, determine the amount of mass m in the tank.
The amount of mass m in the tank is
1700 ± 2% kg.
Explanation: The assumption made here is that the oil is a nearly incompressible substance and thus its density is constant.
Suppose we forgot the formula that relates mass to density and volume. However, we know that mass has the unit of kilograms. That is, with any calculations we do, we should end up with the unit of kilograms. Putting the given information into perspective, we have ρ = 850 kg/m3 and V = 2 m3 It is obvious that we can eliminate m3 and end up with kg by multiplying these two quantities. Therefore, the formula we are looking for should be m = ρV Thus,
m = (850 kg/m3 ) (2 kg) = 1700 kg The amount of mass m in the tank is 1700 kg. https://ezto.mheducation.com/hm.tpx
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References Worksheet
75.
Difficulty: Easy
Award: 10.00 points
If mass, heat, and work are not allowed to cross the boundaries of a system, what is the system called?
Adiabatic Isolated Control mass Isothermal Control volume If mass, heat, and work are not allowed to cross the boundaries of a system, the system is called isolated.
References Multiple Choice
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Difficulty: Easy
63/70
06/06/2024, 01:09
76.
Assignment Print View
Award: 10.00 points
The speed of an aircraft is given to be 260 m/s in air. If the speed of sound at that location is 330 m/s, the flight of the aircraft is _____.
supersonic hypersonic subsonic sonic The speed of the aircraft in this case is given to be less than the speed of sound. The speed of an aircraft is given to be 260 m/s in air. If the speed of sound at that location is 330 m/s, the flight of aircraft is subsonic. References Multiple Choice
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Difficulty: Easy
64/70
06/06/2024, 01:09
77.
Assignment Print View
Award: 10.00 points
One J/kg is equal to _____.
1 kN·m/kg 0.001 kJ 1 N·m 1 kPa·m3 1 m2/s2 1 Joule = 1 kg·m2/s2 Therefore, 1 J/kg = 1m2/s2
One J/kg is equal to 1 m2/s2 References Multiple Choice
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Difficulty: Easy
65/70
06/06/2024, 01:09
78.
Assignment Print View
Award: 10.00 points
Which of the following is a unit for power?
kWh kW kcal Btu hph The unit for power is kW.
References Multiple Choice
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Difficulty: Easy
66/70
06/06/2024, 01:09
79.
Assignment Print View
Award: 10.00 points
The speed of an aircraft is given to be 940 km/h. If the speed of sound at that location is 315 m/s, the Mach number is _____.
0.83 1.0 1.20 0.63 1.07 Vel=940 [km/h]*Convert(km/h, m/s) c=315 [m/s] Ma=Vel/c=0.83 The speed of an aircraft is given to be 940 km/h. If the speed of sound at that location is 315 m/s, the Mach number is 0.83. References Multiple Choice
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Difficulty: Easy
67/70
06/06/2024, 01:09
80.
Assignment Print View
Award: 10.00 points
The weight of a 11-kg mass at sea level is _____.
32.2 kgf 9.81 N 10 N 100 N 107.9 N W = mg = 11 kg × 9.81 m/s2 = 107.9 N
The weight of a 11-kg mass at sea level is 107.9 N. References Multiple Choice
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Difficulty: Easy
68/70
06/06/2024, 01:09
81.
Assignment Print View
Award: 10.00 points
The weight of a 1-lbm mass is _____. Solve this problem using appropriate software.
32.2 lbf 1 lbm·ft/s2 9.81 N 1 lbf 9.81 lbf Solved by appropriate software. Solutions can be verified by copying-and-pasting the following lines on a blank screen. m=1 [lbm] g=32.2 [ft/s^2] W=m*g*Convert(lbm-ft/s^2, lbf) The weight of a 1-lbm mass is 1 lbf. References Multiple Choice
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Difficulty: Easy
69/70
06/06/2024, 01:09
82.
Assignment Print View
Award: 10.00 points
A hydroelectric power plant operates at its rated power of 8 MW. If the plant has produced 26 million kWh of electricity in a specified year, the number of hours the plant has operated that year is _____. Solve this problem using appropriate software.
3086 h 8760 h 3250 h 2508 h 3710 h Solved by appropriate software. Solutions can be verified by copying-and-pasting the following lines on a blank screen. RatedPower=12000 [kW] ElectricityProduced=26E6 [kWh] Hours=ElectricityProduced/RatedPower A hydroelectric power plant operates at its rated power of 8 MW. If the plant has produced 26 million kWh of electricity in a specified year, the number of hours the plant has operated that year is 3250 h. References Multiple Choice
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Difficulty: Easy
70/70
06/06/2024, 11:03
Assignment Print View
Chapter 2 1.
Award: 10.00 points
_____ is defined as the ratio of the density of a substance to the density of some standard substance at a specified temperature.
Specific volume Viscosity Velocity gradient Specific gravity The specific gravity, or relative density, is defined as the ratio of the density of a substance to the density of some standard substance at a specified temperature (the standard is water at 4°C, for which ρH2O = 1000 kg/m3). That is, SG = ρ/ρH20 . When the specific gravity is known, the density is determined from ρ = SG × ρH20 . The specific gravity, or relative density, is defined as the ratio of the density of a substance to the density of some standard substance at a specified temperature. References Multiple Choice
2.
Difficulty: Easy
Award: 10.00 points
What is the difference between intensive and extensive properties? Intensive Extensive
properties do not depend on the size (extent) of the system. properties do depend on the size (extent) of the system.
Explanation: Intensive properties do not depend on the size (extent) of the system, but extensive properties do depend on the size (extent) of the system. Intensive properties do not depend on the size (extent) of the system, but extensive properties do depend on the size (extent) of the system. References Worksheet
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Difficulty: Easy
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3.
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Award: 10.00 points
_____ is the mass per mole in grams/mol or kg/kmol.
Weight Density Specific gravity Molar mass The molar mass M is the mass per mole in grams/mol or kg/kmol. The molar mass M is the mass per mole in grams/mol or kg/kmol. References Multiple Choice
4.
Difficulty: Easy
Award: 10.00 points
How are mass and molar mass related?
m = M/N m = MN m = NM m = N/M They are related to each other by m = NM, where N is the number of moles. They are related to each other by m = NM, where N is the number of moles. References Multiple Choice
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Difficulty: Easy
2/182
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5.
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The specific weight of a system is defined as the weight per unit volume (note that this definition violates the normal specific property-naming convention). Is the specific weight an extensive or intensive property?
Intensive property Extensive property The original specific weight is W
γ1 = V
If we were to divide the system into two halves, each half weighs W/2 and occupies a volume of V/2. The specific weight of one of these halves is
( ) γ= =γ () W 2
V
1
2
which is the same as the original specific weight. Hence, specific weight is an intensive property. Specific weight is an intensive property. References Multiple Choice
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Difficulty: Easy
3/182
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6.
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Award: 10.00 points
Under what conditions can a gas be considered as ideal gas? (Check all that apply.)
At a high temperature relative to its critical temperature At a low pressure relative to its critical pressure At a high temperature and low pressure relative to its critical temperature and pressure At a low temperature relative to its critical temperature At a high pressure relative to its critical pressure A gas can be treated as an ideal gas when it is at a high temperature and/or a low pressure relative to its critical temperature and pressure.
A gas can be treated as an ideal gas when it is at a high temperature and/or a low pressure relative to its critical temperature and pressure. References Check All That Apply
7.
Difficulty: Easy
Award: 10.00 points
How are R (specific gas constant) and Ru (universal gas constant) related?
R = M/Ru R = Ru × M R = Ru/M Ru = M/R Ru is the universal gas constant that is the same for all gases, whereas R is the specific gas constant that is different for different gases. They are related to each other by R = Ru/M , where M is the molar mass (also called the molecular weight) of the gas. They are related to each other by R = Ru/M , where M is the molar mass (also called the molecular weight) of the gas. References Multiple Choice
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Difficulty: Easy
4/182
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8.
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Award: 10.00 points
A 75-L container is filled with 1 kg of air at a temperature of 27°C. The gas constant of air is kJ
R = 0.287 kg ⋅ K
( ) kPa ⋅ m 3 kJ
kPa ⋅ m 3
= 0.287 kg ⋅ K (refer Table A-1). What is the pressure in the container?
The pressure in the container is
1148 ± 2% kPa.
Explanation: The assumption made here is that at specified conditions, air behaves as an ideal gas. The definition of the specific volume gives V
v= m =
0.075 m 3 1 kg
= 0.075 m 3/kg
Using the ideal gas equation of state, the pressure is
Pv = RT → P =
RT v
( 0.287 kPa ⋅ m /kg ⋅ K ) ( 27 + 273 K ) 3
=
0.075 m 3/kg
= 1148 kPa
The pressure in the container is 1148 kPa. References Worksheet
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Difficulty: Easy
5/182
06/06/2024, 11:03
9.
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Award: 10.00 points
A mass of 4-lbm of argon is maintained at 200 psia and 100°F in a tank. The gas constant of argon is obtained from Table A1E as R = 0.2686 psia·ft3/lbm·R. What is the volume of the tank? 3.0083 ± 2% ft3.
The volume of the tank is Explanation:
The assumption made here is that at specified conditions, argon behaves as an ideal gas.
According to the ideal gas equation of state,
V=
mRT P
(
)
( 4 lbm ) 0.2686 psia ⋅ ft 3/lbm ⋅ R ( 100+460 R )
=
= 3.0083 ft 3
200 psia
The volume of the tank is 3.0083 ft3. References Worksheet
10.
Difficulty: Easy
Award: 10.00 points
What is the specific volume of oxygen at 30 psia and 80°F? The gas constant of oxygen is obtained from Table A-1E as R = 0.3353 psia·ft3/lbm·R. The specific volume of oxygen is
6.04 ± 2% ft3/lbm.
Explanation: The assumption made here is that at specified conditions, oxygen behaves as an ideal gas. According to the ideal gas equation of state, RT
v= P =
( 0.3353 psia⋅ft /lbm⋅R ) ( 80+460 R ) 3
30 psia
= 6.04 ft 3/lbm
The specific volume of oxygen is 6.04 ft3/lbm. References Worksheet
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Difficulty: Easy
6/182
06/06/2024, 11:03
11.
Assignment Print View
Award: 10.00 points
A fluid that occupies a volume of 24 L weighs 245 N at a location where the gravitational acceleration is 9.80 m/s2. Determine the mass of this fluid and its density. The mass of the fluid is The density of the fluid is
25 ± 2% kg. 1.042 ± 2% kg/L.
Explanation: Knowing the weight, the mass and the density of the fluid are determined to be
m=
ρ=
W g
m V
=
=
245 N 9.80 m/s 2 25 kg 24 L
(
1 kg ⋅ m/s 3 1N
)
= 25 kg
= 1.042 kg/L
The mass of the fluid is 25 kg. The density of the fluid is 1.042 kg/L. References Worksheet
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Difficulty: Easy
7/182
06/06/2024, 11:03
12.
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Award: 10.00 points
The air in an automobile tire with a volume of 2.60 ft3 is at 70°F and 26 psig. Determine the amount of air that must be added to raise the pressure to the recommended value of 30 psig. Assume the atmospheric pressure to be 14.6 psia and the ft ⋅ lbf
temperature and the volume to remain constant. The gas constant of air is R u = 53.34 lbm ⋅ R The amount of air that must be added to raise the pressure is
(
1 psia 144 lbf/ft 2
)
psia ⋅ ft 3
= 0.3704 lbm ⋅ R
0.053 ± 2% lbm.
Explanation: The following assumptions have been made here: 1. At specified conditions, air behaves as an ideal gas. 2. The volume of the tire remains constant.
The initial and final absolute pressures in the tire are P 1 = P g1 + P atm = 26 psig + 14.6 psia = 40.6 psia P 2 = P g2 + P atm = 30 psig + 14.6 psia = 44.6 psia Treating air as an ideal gas, the initial mass in the tire is
(
( 40.6 psia ) 2.60 ft 3
P 1V
m 1 = RT = 1
)
( 0.3704 psia ⋅ ft /lbm ⋅ R ) ( 70+460 R ) 3
= 0.5377 lbm
Noting that the temperature and the volume of the tire remain constant, the final mass in the tire becomes
(
( 44.6 psia ) 2.60 ft 3
P 2V
m 2 = RT = 2
)
( 0.3704 psia ⋅ ft /lbm ⋅ R ) ( 70+460 R ) 3
= 0.5907 lbm
Thus, the amount of air that needs to be added is Δm = m 2 − m 1 = 0.5907 lbm− 0.5377 lbm = 0.053 lbm
The amount of air that must be added to raise the pressure is 0.053 lbm. References Worksheet
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Difficulty: Easy
8/182
06/06/2024, 11:03
13.
Assignment Print View
Award: 10.00 points
The pressure in an automobile tire depends on the temperature of the air in the tire. When the air temperature is 25°C, the pressure gage reads 210 kPa. If the volume of the tire is 0.025 m3, determine the pressure rise in the tire when the air temperature in the tire rises to 54°C. Also, determine the amount of air that must be bled off to restore pressure to its original value at this temperature. Assume the atmospheric pressure to be 100 kPa. The gas constant of air is kJ
R = 0. 287 kg ⋅ K
( ) kPa ⋅ m 3
kPa ⋅ m 3
= 0. 287 kg ⋅ K
kJ
The pressure rise in the tire is 30 ± 2% kPa. The amount of air that must be bled off to restore pressure to its original value is
0.008 ± 2% kg.
Explanation: The following assumptions have been made here: 1. At specified conditions, air behaves as an ideal gas. 2. The volume of the tire remains constant.
Initially, the absolute pressure in the tire is P 1 = P g + P atm = 210 kPa + 100 kPa = 310 kPa Treating air as an ideal gas and assuming the volume of the tire to remain constant, the final pressure in the tire is determined from P 1V 1 T1
=
P 2V 2 T2
T2
327 K
→ P 2 = T P 1 = 298 K (310 kPa) = 340 kPa 1
Thus, the pressure rise is ΔP = P 2 − P 1 = 340 kPa − 310 kPa= 30 kPa The amount of air that needs to be bled off to restore pressure to its original value is
m1 =
m2 =
P 1V RT 1
P 2V RT 2
(
( 310 kPa ) 0.02 5 m 3
=
( 0.287 kPa ⋅ m /kg ⋅ K ) ( 298 K ) 3
(
( 310 kPa ) 0.02 5 m 3
=
) )
( 0.287 kPa ⋅ m /kg ⋅ K ) ( 327 K ) 3
= 0.0906 kg
= 0.0826 kg
Δm = m 1 − m 2 = 0.0906 kg − 0.0826 kg = 0.008 kg https://ezto.mheducation.com/hm.tpx
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Assignment Print View
The pressure rise in the tire is 30 kPa. The amount of air that must be bled off to restore pressure to its original value is 0.008 kg. References Worksheet
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Difficulty: Easy
11/182
06/06/2024, 11:03
14.
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Award: 10.00 points
A spherical balloon with a diameter of 9 m is filled with helium at 20°C and 220 kPa. Determine the mole number and the mass of the helium in the balloon. Take the molar mass of helium to be 4.003 kg/kmol. The universal gas constant is kJ
R u = 8.31447 kmol ⋅ K
( ) kPa ⋅ m 3 kJ
kPa ⋅ m 3
= 8.31447 kmol ⋅ K
The mole number of the helium balloon is 34.5 ± 2% kmol. 138 ± 2% kg. The mass of the helium balloon is Explanation: The assumption made here is that at specified conditions, helium behaves as an ideal gas.
The volume of the sphere is 4
4
V = 3 πr 3 = 3 π(4.5 m) 3 = 381.704 m 3 Assuming ideal gas behavior, the number of moles of He is determined from
N=
PV R uT
(
( 220 kPa ) 381.704 m 3
=
)
( 8.31447 kPa ⋅ m /kmol ⋅ K ) ( 293.15 K ) 3
= 34.453 kmol ≅ 34.5 kmol
Then, the mass of He is determined from m = NM = (34.453 kmol)(4.003 kg/kmol) = 137.91 kg ≅ 138 kg
The mole number of the helium balloon is 34.5 kmol. The mass of the helium balloon is 138 kg. References Worksheet
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Difficulty: Easy
12/182
06/06/2024, 11:03
15.
Assignment Print View
Award: 10.00 points
A spherical balloon is filled with helium at 20°C. Investigate the effect of the balloon diameter on the mass of helium contained in the balloon for the pressures of 100 kPa and 200 kPa. Let the diameter vary from 5 m to 15 m. Plot the mass of helium against the diameter for both cases. Take the molar mass of helium to be 4.003 kg/kmol. The universal gas constant is kJ
R u = 8.31447 kmol ⋅ K
( ) kPa ⋅ m 3 kJ
kPa ⋅ m 3
= 8.31447 kmol ⋅ K . Please upload your response using the controls provided below. Solve this
problem using appropriate software. Hint: What is the ideal gas equation?
Student upload controls will be shown to students when they take this assignment.
References File Attachment Question
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Difficulty: Easy
13/182
06/06/2024, 11:03
16.
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Award: 10.00 points
A cylindrical tank of methanol has a mass of 80 kg and a volume of 75 L. Determine the methanol’s weight, density, and specific gravity. Take the gravitational acceleration to be 9.81 m/s2. Also, estimate how much force is needed to accelerate this tank linearly at 0.25 m/s2. The density of water is 1000 kg/m3. The weight of the methanol is
785 ± 2% N.
The density of the methanol is 1067 ± 2% kg/m3. 1.067 ± 2% . The specific gravity of the methanol is The force needed to accelerate the tank is 20 ± 2% N. Explanation:
The assumption made here is that the volume of the tank remains constant. The methanol’s weight, density, and specific gravity are W = mg = 80 kg × 9.81 m/s 2 = 785 N ρ=
m V
( 80 kg )
=
( ( )) 75 L ×
SG =
ρ ρH O 2
=
1m
= 1067 kg/m 3
3
1000 L
1067 kg/m 3 1000 kg/m 3
= 1.067
The force needed to accelerate the tank at the given rate is
(
F = ma = (80 kg) 0.25 m/s 2
) = 20 N
Methanol's weight is 785 N. Methanol's density is 1067 kg/m3. Methanol's specific gravity is 1.067. The force needed to accelerate the tank is 20 N. References Worksheet
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Difficulty: Easy
14/182
06/06/2024, 11:03
17.
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Award: 10.00 points
The combustion in a gasoline engine may be approximated by a constant volume heat addition process, and the content of the combustion chamber both before and after combustion is air. The conditions are 1.80 MPa and 450°C before the combustion and 1100°C after it. Determine the pressure at the end of the combustion process.
The pressure at the end of the combustion process is
3418 ± 2% kPa.
Explanation: The following assumptions have been made here: 1. The contents of the cylinder are approximated by the air's properties. 2. Air is an ideal gas.
The final pressure may be determined from the ideal gas relation T2
P2 = T P1 = 1
(
1100 + 273 K 450 + 273 K
)
(1800 kPa) = 3418 kPa
The pressure at the end of the combustion process is 3418 kPa. References Worksheet
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Difficulty: Easy
15/182
06/06/2024, 11:03
18.
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Award: 10.00 points
The density of atmospheric air varies with elevation, decreasing with increasing altitude. Using the data given in the table, obtain a relation for the variation of density with elevation, and calculate the density at an elevation of 7000 m. Calculate the mass of the atmosphere using the correlation you obtained. Assume the earth to be a perfect sphere with a radius of 6377 km, and take the thickness of the atmosphere to be 25 km. Please upload your response using the controls provided below. Use appropriate software to solve this problem. r, km
ρ, kg/m3 1.225 1.112 1.007 0.9093 0.8194 0.7364 0.6601 0.5258 0.4135 0.1948 0.08891 0.04008
6377 6378 6379 6380 6381 6382 6383 6385 6387 6392 6397 6402
Hint: How does density vary with elevation? How does total mass vary with density and elevation (volume)?
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16/182
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19.
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The density of saturated liquid refrigerant-134a for –20°C ≤ T ≤ 100°C is given in Table A–4. Using this value, develop an expression in the form ρ = aT2 + bT + c for the density of refrigerant–134a as a function of absolute temperature, and determine the relative error for each data set. Please upload your response using the controls provided below. Hint: What is the regression value for the density (2nd order polynomical curve-fitting)?
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References File Attachment Question
Difficulty: Easy
Consider Table 2-1 in the textbook, which lists the specific gravities of various substances. References Section Break
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Difficulty: Easy
17/182
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20.
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Award: 10.00 points
Define specific gravity and specific weight. (Please provide an answer before moving on to the next part.) Specific gravity
is the ratio of the density of the fluid to the density of water at 4oC.
Specific weight
is the product of the density of the fluid and the gravitational acceleration.
Explanation: Specific gravity is nondimensional and is the ratio of the density of the fluid to the density of water at 4oC. Specific weight is dimensional and is simply the product of the density of the fluid and the gravitational acceleration.
Specific gravity is the ratio of the density of the fluid to the density of water at 4oC. Specific weight is the product of the density of the fluid and the gravitational acceleration. References Worksheet
21.
Difficulty: Easy
Award: 10.00 points
Calculate the specific weight of all the substances in Table 2-1. For the case of bones, give answers for both the low and high values given in the table. Note: Excel is recommended for this kind of problem in which there is much repetition of calculations, but you can also do the calculations by hand or with any other software. If using software like Excel, do not worry about the number of significant digits, since this is not something easy to modify in Excel. Please upload your response using the controls provided below. (Please provide an answer before moving on to the next part.)
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18/182
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22.
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As discussed in the text, there is another related property called specific volume. Calculate the specific volume of a liquid with SG = 0.56. 0.00179 ± 2% m3/kg.
The specific volume is Explanation:
Specific volume is defined as v = 1/ρ. But ρ = SG x ρwater. Thus, 1
v = SG × ρ
water
=
1 0.56 × 1000 kg/m 3
= 0.00179 m 3/kg
The specific volume is 0.00179 m3/kg. References Worksheet
23.
Difficulty: Easy
Award: 10.00 points
_____ is defined as the pressure exerted by a vapor in phase equilibrium with its liquid at a given temperature.
Vapor pressure
The vapor pressure Pv of a pure substance is defined as the pressure exerted by a vapor in phase equilibrium with its liquid at a given temperature. In general, the pressure of a vapor or gas, whether it exists alone or in a mixture with other gases, is called the partial pressure.
The vapor pressure Pv of a pure substance is defined as the pressure exerted by a vapor in phase equilibrium with its liquid at a given temperature. References Fill in the Blank
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Difficulty: Easy
19/182
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24.
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Award: 10.00 points
During phase change processes between the liquid and vapor phases of a pure substance, the saturation pressure and the vapor pressure are not equivalent.
True False
During phase change processes between the liquid and vapor phases of a pure substance, the saturation pressure and the vapor pressure are equivalent since the vapor is pure. During phase change processes between the liquid and vapor phases of a pure substance, the saturation pressure and the vapor pressure are equivalent. References True / False
25.
Difficulty: Easy
Award: 10.00 points
Does water boil at higher temperatures, at higher pressures?
No Yes Yes. The saturation temperature of a pure substance depends on pressure; in fact, it increases with pressure. The higher the pressure, the higher the saturation or boiling temperature.
Yes. The saturation temperature of a pure substance depends on pressure; in fact, it increases with pressure. The higher the pressure, the higher the saturation or boiling temperature. References Multiple Choice
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Difficulty: Easy
20/182
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26.
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Award: 10.00 points
If the pressure of a substance is increased during a boiling process, what happens to the temperature?
It decreases. It remains constant. It increases. If the pressure of a substance increases during a boiling process, the temperature also increases since the boiling (or saturation) temperature of a pure substance depends on pressure and increases with it.
If the pressure of a substance increases during a boiling process, the temperature also increases since the boiling (or saturation) temperature of a pure substance depends on pressure and increases with it. References Multiple Choice
27.
Difficulty: Easy
Award: 10.00 points
_____ is the vaporization that may occur at locations where the pressure drops below the vapor pressure.
Cavitation
In the flow of a liquid, cavitation is the vaporization that may occur at locations where the pressure drops below the vapor pressure. The vapor bubbles collapse as they are swept away from the low-pressure regions, generating highly destructive, extremely high-pressure waves. This phenomenon is a common cause for the drop in performance and even the erosion of impeller blades.
In the flow of a liquid, cavitation is the vaporization that may occur at locations where the pressure drops below the vapor pressure. References Fill in the Blank
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Difficulty: Easy
21/182
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28.
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Award: 10.00 points
The pressure on the suction side of pumps is typically low, and the surfaces on that side of the pump are susceptible to cavitation, especially at high fluid temperatures. If the minimum pressure on the suction side of a water pump is 0.31 psia absolute, determine the maximum water temperature to avoid the danger of cavitation. The maximum water temperature to avoid the danger of cavitation is
65 ± 2% ºF.
Explanation: To avoid cavitation at a specified pressure, the fluid temperature everywhere in the flow should remain below the saturation temperature at the given pressure, which is, from Table A-4E, T max = T sat @ 0.31 psia = 65°F Therefore, T must remain below 65°F to avoid the possibility of cavitation.
The maximum water temperature to avoid the danger of cavitation is 65ºF. References Worksheet
29.
Difficulty: Easy
Award: 10.00 points
A pump is used to transport water to a higher reservoir. If the water temperature is 35ºC, determine the lowest pressure that can exist in the pump without cavitation. The vapor pressure of water at 35ºC is 5.627 kPa. The lowest pressure that can exist in the pump without cavitation is
5.627 ± 2% kPa.
Explanation: To avoid cavitation, the pressure anywhere in the system should not be allowed to drop below the vapor (or saturation) pressure at the given temperature: Pmin = Psat@35ºC = 5.627 kPa
The lowest pressure that can exist in the pump is 5.627 kPa. References Worksheet
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Difficulty: Medium
22/182
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30.
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Award: 10.00 points
In a piping system, the water temperature remains under 35°C. Determine the minimum pressure allowed in the system to avoid cavitation. The minimum pressure allowed in the system to avoid cavitation is
5.627 ± 2% kPa.
Explanation: To avoid cavitation, the pressure anywhere in the system should not be allowed to drop below the vapor (or saturation) pressure at the given temperature: Pmin = Psat@35ºC = 5.627 kPa.
The minimum pressure allowed in the system to avoid cavitation is 5.627 kPa. References Worksheet
31.
Difficulty: Easy
Award: 10.00 points
_____ is the sum of all forms of the energy a system possesses.
Total energy
The sum of all forms of the energy a system possesses is called total energy.
The sum of all forms of the energy a system possesses is called total energy. References Fill in the Blank
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Difficulty: Easy
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Award: 10.00 points
In the absence of magnetic, electrical, and surface tension effects, what are the different forms of energy that constitute the total energy? (Check all that apply.)
Potential energy Kinetic energy Internal energy Specific energy Cavitation energy In the absence of magnetic, electrical, and surface tension effects, the total energy of a system consists of kinetic, potential, and internal energies.
In the absence of magnetic, electrical, and surface tension effects, the total energy of a system consists of kinetic, potential, and internal energies. References Check All That Apply
33.
Difficulty: Easy
Award: 10.00 points
Which of the following forms of energy contribute to the internal energy of a system? (Check all that apply.)
Sensible energy Latent energy Chemical energy Nuclear energy Potential energy Specific energy The internal energy of a system is composed of sensible, latent, chemical, and nuclear energies. The sensible internal energy is due to translational, rotational, and vibrational effects.
The internal energy of a system is made up of sensible, latent, chemical, and nuclear energies. References Check All That Apply
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Difficulty: Easy
24/182
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34.
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Award: 10.00 points
In common terminology, thermal energy is referred to as internal energy, and it is not a property.
True False
In common terminology, thermal energy is referred to as heat. However, like work, heat is not a property, whereas thermal energy is a property. In common terminology, thermal energy is referred to as heat. However, like work, heat is not a property, whereas thermal energy is a property. References True / False
35.
Difficulty: Easy
Award: 10.00 points
_____ is the energy needed to push a fluid into or out of a control volume.
Flow energy
Flow energy or flow work is the energy needed to push a fluid into or out of a control volume.
Flow energy or flow work is the energy needed to push a fluid into or out of a control volume. References Fill in the Blank
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Difficulty: Easy
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36.
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Fluids at rest do not possess any flow energy.
True False
Fluids at rest do not possess any flow energy. Fluids at rest do not possess any flow energy. References True / False
37.
Difficulty: Easy
Award: 10.00 points
In addition to the forms of energy possessed by a nonflowing fluid, what extra form of energy does a flowing fluid possess?
Flow energy Thermal energy Specific energy Latent energy A flowing fluid possesses flow energy, which is the energy needed to push a fluid into or out of a control volume, in addition to the forms of energy possessed by a nonflowing fluid.
A flowing fluid possesses flow energy, which is the energy needed to push a fluid into or out of a control volume, in addition to the forms of energy possessed by a nonflowing fluid. References Multiple Choice
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Difficulty: Easy
26/182
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Identify the specific forms of energy associated with a nonflowing fluid, when the entire closed system is stationary. (Check all that apply.)
Internal energy Potential energy Kinetic energy Flow energy The total energy of a nonflowing fluid consists of internal and potential energies.
The total energy of a nonflowing fluid consists of internal and potential energies. References Check All That Apply
39.
Difficulty: Easy
Award: 10.00 points
Which of the following are the specific forms of energy associated with a flowing fluid? (Check all that apply.)
Internal energy Potential energy Kinetic energy Flow energy Specific energy Nuclear energy The total energy of a flowing fluid consists of internal, kinetic, potential, and flow energies.
The total energy of a flowing fluid consists of internal, kinetic, potential, and flow energies. References Check All That Apply
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Difficulty: Easy
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Award: 10.00 points
What is the expression used to determine the change in the internal energy of an ideal gas?
Δu = c v , avg + ΔT Δu = c v , avgΔT Δu =
Δu =
ΔT
c v , avg c v , avg ΔT
Using specific heat values at the average temperature, the changes in the specific internal energy of ideal gases can be determined from Δu = c v , avgΔT.
Using specific heat values at the average temperature, the changes in the specific internal energy of ideal gases can be determined from Δu = c v , avgΔT. References Multiple Choice
41.
Difficulty: Easy
Award: 10.00 points
What is the expression used to determine the change in the internal energy of an incompressible substance?
c avg
Δ u = ΔT
Δu = c avgΔT Δu = c avg + ΔT Δ u = ΔT
c avg
Using specific heat values at the average temperature, the changes in the specific internal energy of ideal gases can be determined from Δu = c avgΔT.
Using specific heat values at the average temperature, the changes in the specific internal energy of ideal gases can be determined from Δu = c avgΔT. References Multiple Choice
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Difficulty: Easy
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42.
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Award: 10.00 points
What is the expression used to determine the change in the enthalpy of an ideal gas?
Δh = c p , avgΔT Δh = c p , avg + ΔT Δh = c p , avg ΔT
Δh = c p , avg - ΔT Using specific heat values at the average temperature, the changes in the specific enthalpy of ideal gases can be determined from ∆ h = c p , avgΔT.
Using specific heat values at the average temperature, the changes in the specific enthalpy of ideal gases can be determined from ∆ h = c p , avgΔT. References Multiple Choice
43.
Difficulty: Easy
Award: 10.00 points
What is the expression used to determine the change in the enthalpy of an incompressible substance?
Δh = Δu × vΔP ≅ c avgΔT × vΔP Δh = Δu + vΔP ≅ c avgΔT + vΔP Δh = Δu ≅ c avgΔT v ΔP
v ΔP
Δh = Δu - vΔP ≅ c avgΔT - vΔP Using specific heat values at the average temperature, the changes in the specific enthalpy of ideal gases can be determined from Δh = Δu + vΔP ≅ c avgΔT + vΔP.
Using specific heat values at the average temperature, the changes in the specific enthalpy of ideal gases can be determined from Δh = Δu + vΔP ≅ c avgΔT + vΔP. References Multiple Choice
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Difficulty: Easy
29/182
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Award: 10.00 points
Ignoring any losses, estimate how much energy (in units of Btu) is required to raise the temperature of water in a 100-gallon hot-water tank from 60°F to 110°F. The specific heat of water is approximated as a constant, whose value is 0.999 Btu/·lbmR at the average temperature of (60 + 110)/2 = 85ºF. In fact, c remains constant at 0.999 Btu/lbm·R (to three digits) from 60ºF to 110ºF. For this same temperature range, the density varies from 62.36 lbm/ft3 at 60ºF to 61.86 lbm/ft3 at 110ºF. We approximate the density as remaining constant, whose value is 62.17 lbm/ft3 at the average temperature of 85ºF. The energy required is
41514 ± 2% Btu.
Explanation: The following assumptions have been made here: 1. There are no losses. 2. The pressure in the tank remains constant at 1 atm. 3. An approximate analysis is performed by replacing differential changes in quantities by finite changes. For a constant pressure process, Δu ≅ c avg ∆ T. Since this is energy per unit mass, we must multiply by the total mass of the water in the tank, i.e., ΔU ≅ mc avgΔT = ρVc avgΔT. Thus,
(
)
ΔU ≅ ρVc avgΔT = 62.17 lbm/ft 3 (100 gal)(0.999 Btu/lbm ⋅ R)((110 °F - 60 °F)R)
(
35.315 ft 3 264.17 gal
)
= 41514 Btu
where we note temperature differences are identical in ºF and R.
The energy required is 41514 Btu. References Worksheet
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Difficulty: Easy
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45.
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Award: 10.00 points
Saturated water vapor at 150°C (enthalpy h = 2745.9 kJ/kg) flows in a pipe at 35 m/s at an elevation of z = 15 m. Determine the total energy of vapor in J/kg relative to the ground level. 2.747 ± 2% × 106 J/kg.
The total energy of the vapor is Explanation:
The total energy of a flowing fluid is given by (Eq. 2-8) V2
e = h + 2 + gz The enthalpy of the vapor at the specified temperature can be found in any text on thermodynamics to be 2745.9 kJ/kg. Then the total energy is determined as e = 2745.9 × 10 3 J/kg +
( 35 m/s ) 2 2
+
((
9.81 m/s
2
)
)
× (15 m) = 2.747 × 10 6 J/kg
The total energy of vapor is 2.747 × 106 J/kg. References Worksheet
46.
Difficulty: Easy
Award: 10.00 points
The variation of the density of a fluid with temperature at constant pressure is known as the _____.
coefficient of volume expansion coefficient of pressure coefficient of density coefficient of expansion The coefficient of volume expansion represents the variation of the density of a fluid with temperature at constant pressure.
The coefficient of volume expansion represents the variation of the density of a fluid with temperature at constant pressure. References Multiple Choice
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Difficulty: Easy
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The variation of the pressure of a fluid with density at constant temperature is known as the _____.
coefficient of density coefficient of pressure coefficient of compressibility coefficient of volume The variation of the pressure of a fluid with density at constant temperature is known as the coefficient of compressibility.
The variation of the pressure of a fluid with density at constant temperature is known as the coefficient of compressibility. References Multiple Choice
48.
Difficulty: Easy
Award: 10.00 points
_____ is the inverse of the coefficient of compressibility, and it represents the fractional change in volume or density corresponding to a change in pressure. Isothermal compressibility
Isothermal compressibility is the inverse of the coefficient of compressibility, and it represents the fractional change in volume or density corresponding to a change in pressure.
Isothermal compressibility is the inverse of the coefficient of compressibility, and it represents the fractional change in volume or density corresponding to a change in pressure. References Fill in the Blank
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Difficulty: Easy
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The coefficient of compressibility of a fluid cannot be negative.
True False
The coefficient of compressibility of a fluid cannot be negative.
The coefficient of compressibility of a fluid cannot be negative References True / False
50.
Difficulty: Easy
Award: 10.00 points
The coefficient of volume expansion can be negative (e.g., liquid water below 4°C).
True False
The coefficient of volume expansion can be negative (e.g., liquid water below 4°C). The coefficient of volume expansion can be negative (e.g., liquid water below 4°C). References True / False
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Difficulty: Easy
33/182
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51.
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Award: 10.00 points
Use the coefficient of volume expansion to estimate the density of water as it is heated from 60°F to 130°F at 1 atm. Compare your result with the actual density (from the appendices). The density of water at 60°F and 1 atm pressure is ρ1 = 62.36 lbm/ft3. The coefficient of volume expansion at the average temperature of (60 + 130)/2 = 95°F is β = 0.187 × 10–3R–1. The change in density due to the change of temperature from 60°F to 130°F at constant pressure is – The approximation is extremely accurate .
0.816 ± 2% lbm/ft3.
Explanation: The following assumptions have been made here: 1. The coefficient of volume expansion is constant at the given temperature range. 2. The pressure remains constant at 1 atm throughout the heating process. 3. An approximate analysis is performed by replacing differential changes in quantities by finite changes. The change in density due to the change of temperature from 60ºF to 130ºF at constant pressure is
(
1
Δρ ≅ −βρΔT = − 0.187×10 − 3 R
)(
)
62.36 lbm/ft 3 ((130 − 60) R) = -0.816 lbm/ft 3
where we note temperature differences are identical in ºF and R. Thus, the density at the higher temperature is approximated as ρ2 ≈ ρ1 + Δρ = 62.36 – 0.816 = 61.54 lbm/ft3. From the appendices, we see that the actual density at 130ºF is 61.55 lbm/ft3. Thus, the approximation is extremely accurate, with an error of less than 0.02%.
The change in density due to the change of temperature from 60°F to 130°F at constant pressure is –0.816 lbm/ft3. From the appendices, we see that the actual density at 130ºF is 61.55 lbm/ft3. Thus, the approximation is extremely accurate, with an error of less than 0.02%. References Worksheet
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Difficulty: Easy
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52.
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Award: 10.00 points
The volume of an ideal gas is to be reduced by half by compressing it isothermally. Determine the required change in pressure.
P1 2P 1 P2 P1 2
The assumption made here is that the process is isothermal and thus the temperature remains constant. For an ideal gas of fixed mass undergoing an isothermal process, the ideal gas relation reduces to P 2V 2 T2
P 1V 1
= T 1
V1
V1
→ P 2 V 2 = P 1V 1 → P 2 = V P 1 = 0 . 5 V P 1 = 2 P 1 2
1
Therefore, the change in pressure becomes ΔP = P 2 − P 1 = 2P 1 − P 1 = P 1
The required change in pressure is P1. References Multiple Choice
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Difficulty: Easy
35/182
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53.
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Award: 10.00 points
Water at 1 atm pressure is compressed to 360 atm pressure isothermally. Determine the increase in the density of water. Take the isothermal compressibility of water to be 4.80 × 10−5 atm−1. The density of water at 20°C and 1 atm pressure is ρ1 = 998 kg/m3.
The increase in the density of water is
17.2 ± 2% kg/m3.
Explanation: The following assumptions have been made here: 1. The isothermal compressibility is constant in the given pressure range. 2. An approximate analysis is performed by replacing differential changes by finite changes. When differential quantities are replaced by differences and the properties α and β are assumed to be constant, the change in density in terms of the changes in pressure and temperature is expressed approximately as Δρ = αρΔP − βρΔT The change in density due to a change of pressure from 1 atm to 360 atm at constant temperature is
(
Δρ = αρΔP = 4.80 × 10 − 5 atm -1
)(998 kg/m )(360 − 1 )atm = 17.2 kg/m 3
3
The increase in the density of water is 17.2 kg/m3. References Worksheet
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Difficulty: Easy
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54.
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Award: 10.00 points
It is observed that the density of an ideal gas increases by 10 percent when compressed isothermally from 10 atm to 11 atm. Determine the percent increase in the density of the gas if it is compressed isothermally from 992 atm to 1009 atm. The increase in the density of the gas is
1.71 ± 2% %.
Explanation: The assumption made here is that the gas behaves like an ideal gas.
For an ideal gas, P = ρRT and
( ) ∂P ∂ρ
P
T
= RT = ρ , and thus kideal gas = P. Therefore, the coefficient of compressibility of an ideal
gas is equal to its absolute pressure, and the coefficient of compressibility of the gas increases with increasing pressure. ΔP
ΔP
Δρ
Substituting k = P in the definition of the coefficient of compressibility k ≅ − Δv / v ≅ Δρ / ρ and rearranging give ρ =
ΔP P
.
Therefore, the percent increase in the density of an ideal gas during isothermal compression is equal to the percent increase in pressure. At 10 atm, Δρ ρ
ΔP
= P =
11 atm − 10 atm 10 atm
= 10%
At 1000 atm, Δρ ρ
ΔP
= P =
1009 atm − 992 atm 992 atm
= 0.0171 × 100 = 1.71%
The increase in the density of the gas is 1.71%. References Worksheet
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Difficulty: Easy
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55.
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Award: 10.00 points
Saturated refrigerant-134a liquid at 10°C is cooled to 5°C at constant pressure. Using the coefficient of volume expansion data, determine the change in the density of the refrigerant. The density of saturated liquid R-134a at 10°C is ρ1 = 1261 kg/m3. The coefficient of volume expansion at the average temperature of (10 + 5) / 2 = 7.5°C is β = 0.00275 K–1. The change in the density of the refrigerant is
17.34 ± 2% kg/m3.
Explanation: The following assumptions have been made here: 1. The coefficient of volume expansion is constant at the given temperature range. 2. An approximate analysis is performed by replacing differential changes in quantities by finite changes. When differential quantities are replaced by differences and the properties α and β are assumed to be constant, the change in density in terms of the changes in pressure and temperature is expressed approximately as Δρ = αρΔP − βρΔT The change in density due to the change of temperature from 10°C to 5°C at constant pressure is
(
Δρ = − βρΔT = − 0.00275 K − 1
)(1261 kg / m )(5 − 10)K = 17.34 kg/m 3
3
The change in the density of the refrigerant is 17.34 kg/m3. References Worksheet
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Difficulty: Easy
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56.
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Award: 10.00 points
A water tank is completely filled with liquid water at 20ºC. The tank material is such that it can withstand tension caused by a volume expansion of 0.7 percent. Determine the maximum temperature rise allowed without jeopardizing safety. For simplicity, assume the volume expansion coefficient β at 313 K to be 0.377 × 10–3 K–1. The maximum temperature rise allowed in the tank is
18.57 ± 2% ºC.
Explanation: The following assumptions have been made here: 1. The coefficient of volume expansion is constant. 2. An approximate analysis is performed by replacing differential changes in the quantities by finite changes. 3. The effect of pressure is disregarded. When differential quantities are replaced by differences and the properties α and β are assumed to be constant, the change in density in terms of the change in pressure and temperature is expressed approximately as Δρ = αρΔP − βρΔT A volume increase of 0.7% corresponds to a density decrease of 0.7%, which can be expressed as Δρ = −0.007ρ. Then, the decrease in density due to a temperature rise of ΔT at constant pressure is − 0.007ρ = −βρΔT Solving for ΔT by substituting the value of β, the maximum temperature rise is determined to be ΔT =
0.007 β
=
0.007 0.377 × 10 − 3 K − 1
= 18.57 K
The maximum temperature rise allowed in the tank is 18.57ºC. References Worksheet
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Difficulty: Easy
39/182
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57.
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Award: 10.00 points
A water tank is completely filled with liquid water at 20ºC. The tank material is such that it can withstand tension caused by a volume expansion of 0.6 percent. Determine the maximum temperature rise allowed without jeopardizing safety. For simplicity, assume the volume expansion coefficient β at 313 K to be 0.377 × 10–3 K–1. The maximum temperature rise allowed in the tank is
15.92 ± 2% ºC.
Explanation: The following assumptions have been made here: 1. The coefficient of volume expansion is constant. 2. An approximate analysis is performed by replacing differential changes in the quantities by finite changes. 3. The effect of pressure is disregarded. When differential quantities are replaced by differences and the properties α and β are assumed to be constant, the change in density in terms of the change in pressure and temperature is expressed approximately as Δρ = αρΔP − βρΔT A volume increase of 0.6% corresponds to a density decrease of 0.6%, which can be expressed as Δρ = −0.006ρ. Then, the decrease in density due to a temperature rise of ΔT at constant pressure is −0.006ρ = −βρΔT Solving for ΔT by substituting the value of β, the maximum temperature rise is determined to be ΔT =
0.006 β
=
0.006 0.377 × 10 − 3 K -1
= 15.92 K = 15.92°C
The maximum temperature rise allowed in the tank is 10.6ºC. References Worksheet
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Difficulty: Easy
40/182
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58.
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Award: 10.00 points
The density of seawater at a free surface where the pressure is 98 kPa is approximately 1030 kg/m3. Taking the bulk modulus of elasticity of seawater to be 2.34 × 109 N/m2 and expressing the variation of pressure with depth z as dP = ρg dz, determine the density and pressure at a depth of 2500 m. Disregard the effect of temperature. The density at the given depth is The pressure at the given depth is
1041.24 ± 2% kg/m3. 25.775 ± 2% MPa.
Explanation: The following assumptions have been made here: 1. The temperature and the bulk modulus of elasticity of seawater are constant. 2. The gravitational acceleration remains constant.
The coefficient of compressibility or the bulk modulus of elasticity of fluids is expressed as
κ=ρ
( ) ∂P ∂ρ
T
or
(
dP
κ = ρ dρ
at constant T
)
The differential pressure change across a differential fluid height of dz is given as dP = ρgdz Combining the two relations above and rearranging, ρgdz
dz
κ = ρ dρ = gρ 2 dρ →
dρ ρ2
=
gdz κ
Integrating from z = 0 where ρ = ρ0 = 1030 kg/m3 to z = z where ρ = ρ gives ∫ ρρ
dρ 0ρ
g
2 = κ dz
1
1
gz
→ ρ − ρ = κ ∫ z0 0
Solving for ρ gives the variation of density with depth as ρ=
1
() () 1
ρ0
−
gz κ
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Assignment Print View
Substituting into the pressure change relation dP = ρgdz and integrating from z = 0 where P = P0 = 98kPa to z = z where P = P gives
P
gdz
z
∫ P dP = ∫ 0 0
→ P = P 0 + κln
()() 1
-
ρ0
gz
( ( )) 1
ρ 0gz
1-
κ
κ
which is the desired relation for the variation of pressure in seawater with depth. At z = 2500 m, the values of density and pressure are determined by substitution to be ρ=
1
((
1 1030 kg/m
)) ( ( (
)
9.81 m/s 2 ( 2500 m )
−
3
(
2.34 × 10 9 N/m 2
)
)
)
P = (98,000 Pa) + 2.34 × 10 9 N/m 2 ln
= 1041.24 kg/m 3
(
1 1−
(
( 1041.2 kg/m ) ( 9.81 m/s ) ( 2500 m ) 3
2
( 2.34 × 10 N/m ) 9
2
)
)
P = 25.775 MPa since 1 Pa = 1 N/m2 = 1 kg/m·s2 and 1 kPa = 1000 Pa.
The density at the given depth is 1041.24 kg/m3. The pressure at the given depth is 25.775 MPa. References Worksheet
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Difficulty: Medium
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Taking the coefficient of compressibility of water to be 7 × 105 psia, determine the pressure increase required to reduce the volume of water by 1.5 percent. The pressure increase required is
10500 ± 2% psia.
Explanation: The following assumptions have been made here: 1. The coefficient of compressibility is constant. 2. The temperature remains constant. A volume decrease of 1.5 percent can be expressed mathematically as Δv v
ΔV
= V = −0.015
The coefficient of compressibility is expressed as
κ= −v
( ) ∂P ∂v
T
≅ −
ΔP
( ) Δv v
Rearranging and substituting, the required pressure increase is determined to be
ΔP = − κ
( ) Δv v
(
)
= − 7 × 10 5 psia (−0.015)=10500 psia
The pressure increase required is 10500 psia. References Worksheet
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Difficulty: Easy
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A frictionless piston-cylinder device contains 10 kg of water at 20°C at atmospheric pressure. An external force F is then applied on the piston until the pressure inside the cylinder increases to 100 atm. Assuming the coefficient of compressibility of water remains unchanged during the compression, estimate the energy needed to compress the water isothermally. Take the water in the cylinder as the system. The energy needed to compress the water isothermally is
243.53 ± 2% J.
Explanation: The assumption made here is that the coefficient of compressibility of water remains unchanged during the compression. We take the water in the cylinder as the system. The energy needed to compress water is equal to the work done on the system, and it can be expressed as W = - ∫ 21PdV
(1)
From the definition of the coefficient of compressibility, we have
κ= -v
( ) ∂P ∂V
T
Rearranging, we obtain dV V
dP
= − κ
which can be integrated from the initial state to any state as follows: ∫V V
dV 0 V
= - ∫P P
dP 0 κ
V
P - P0
0
κ
→ ln V = -
from which we obtain 1
V
κ
V0
P = P 0 − ln
Substituting in Eq. 1, we have
(
V
0PdV = − ∫ V 0 P − κln W = − ∫V 0 V V V 1
1
0
) (
V
(
)
dV = κVln V − P 0 + κ V 0
)
V1 V0
or
(
W = P0 + κ
V1
)(V0 − V1 ) + κV1ln V
0
In terms of finite changes, the fractional change due to the change in pressure can be expressed approximately as V1 - V0 V0
(
≅ - α P1 - P0
)
or
(
(
V1 ≅ V0 1 - α P1 - Po
))
where is the isothermal compressibility of water, which is 4.80 × 10 − 5 atm − 1 at 20 . Realizing that 10 kg water occupies initially a volume of the final volume of water is https://ezto.mheducation.com/hm.tpx
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V1 - V0 V0
(
)
(
) ( (
)
)
≅ - α P 1 - P 0 V 1 ≅ 0.01 m 3 × 1 - 4.80 × 10 - 5 atm - 1 × (100 atm - 1 atm) = 9.952 × 10 - 3 m 3
Then, the work done on the water is
(
W = 1 atm +
1 4.80 × 10 - 5 atm - 1
)
(
)
× 10 × 10 − 3 m 3 − 9.952 × 10 − 3 m 3 +
(
1 4.80 × 10 - 5 atm - 1
)
(
)
× 9.952 × 10 − 3 m 3 ln
9.952 × 10 − 3 m 3 10 × 10 − 3 m 3
from which we obtain W = 24.035 × 10 − 4atm·m 3 ≅ 243.53 J since 1 atm = 101325 Pa.
The energy needed to compress the water isothermally 243.53 J. References Worksheet
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Difficulty: Easy
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When modeling fluid flows with small changes in temperature and pressure, the Boussinesq approximation is often used in which the fluid density is assumed to vary linearly with changes in temperature. The Boussinesq approximation is ρ = ρ0[1 − β(T − T0)], where β is assumed to be constant over the given temperature range; β is evaluated at reference temperature T0, taken as some average or mid-value temperature in the flow; and ρ0 is a reference density, also evaluated at T0. The Boussinesq approximation is used to model a flow of air at nearly constant pressure, P = 95.0 kPa, but the temperature varies between 20°C and 60°C. Using the mid-way point (40°C) as the reference temperature, calculate the density at the two temperature extremes using the Boussinesq approximation, and compare with the actual density at these two temperatures obtained from the ideal gas law. In particular, for both temperatures calculate the percentage error caused by the Boussinesq approximation. Take the reference density of air at T0 = 40°C (313.15 K) and P = 95.0 kPa as ρ0 = 1.05703 kg/m3. The coefficient of volume expansion at T0 = 40°C (313.15 K) is β = 1/T0 = 3.1934 × 10-3 K-1. The percentage error caused by the Boussinesq approximation for 20°C is The percentage error caused by the Boussinesq approximation for 60°C is
-0.4162 ± 0.01 %. -0.4076 ± 0.01 %.
Explanation: The following assumptions have been made here: 1. The coefficient of volume expansion is constant at the given temperature range. 2. The pressure remains constant at 95.0 kPa throughout the flow field. 3. Air is taken as an ideal gas. Using the Boussinesq approximation at T = 20ºC, we calculate the Boussinesq density at 20ºC:
(
(
ρ 20 = ρ 0 1 − β T − T 0
(
ρ 20 = 1.05703 kg/m 3
))
)(1 − (3.1934 × 10 K )(20 − 40) K) = 1. 1245 kg/m −3
-1
3
where we note temperature differences are identical in ºC and K. Similarly, at T = 60ºC, the Boussinesq density at 60ºC:
(
(
ρ 60 = ρ 0 1 − β T − T 0
(
ρ 60 = 1. 05703 kg/m 3
))
)(1 − (3.1934 × 10 K )(60 − 40) K ) = 0. 98953 kg/m −3
-1
3
The actual values of density are obtained from the ideal gas law, yielding ρ20, actual = 1.1292 kg/m3; the Boussinesq approximation has a percentage error of –0.4162%. Similarly, ρ60, actual = 0.99358 kg/m3; the Boussinesq approximation has a percentage error of –0.4076%. Thus, the Boussinesq approximation is extremely accurate for this range of temperature.
The percentage error caused by the Boussinesq approximation for 20°C is –0.4162%. The percentage error caused by the Boussinesq approximation for 60°C is –0.4076%. References Worksheet
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Difficulty: Medium
46/182
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A frictionless piston-cylinder device contains 10 kg of water at 20°C at atmospheric pressure. An external force F is then applied on the piston until the pressure inside the cylinder increases to 104 atm. Assuming a linear pressure increase during the compression, estimate the energy needed to compress the water isothermally. Take the isothermal compressibility as 4.8 × 10–5 atm–1.
The energy needed to compress the water isothermally is
259.6 ± 2% J.
Explanation: The assumption made here is that the pressure increases linearly. We take the water in the cylinder as the system. The energy needed to compress water is equal to the work done on the system, and it can be expressed as 2
W = - ∫ 1PdV
(1)
For a linear pressure increase, we take P = P ave =
P1 + P2 2
=
104 atm + 1 atm 2
= 52.5 atm
In terms of finite changes, the fractional change due to the change in pressure can be expressed approximately as (Eq. 3-23) V1 - V0 V0
(
≅ α P1 - P0
)
or
(
(
V1 ≅ V0 1 - α P1 - P0
))
where α is the isothermal compressibility of water, which is 4.80 × 10–5 atm–1 at 20 . Knowing that 10 kg of water initially occupies a volume of V0 = 10 × 10–3 m3 the final volume of water is determined to be
(
) ( (
)
)
V 1 ≅ 0.010 m 3 × 1 - 4.80 × 10 - 5 atm - 1 × (104 atm - 1 atm) = 0.00995 m 3 Therefore, the work expression becomes
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(
V
)
(
W = ∫ V 1PdV = - P ave V 1 - V 0 = - (52 atm) × 0.00995 m 3 - 10 × 10 - 3m 3 0
)
W = 2.596 × 10 - 3 atm·m 3 = 259.6 J Thus, we conclude that linear pressure increase approximation does not work well since it gives almost ten times more work.
The energy needed to compress the water isothermally is 259.6 J. References Worksheet
63.
Difficulty: Medium
Award: 10.00 points
Using the definition of the coefficient of volume expansion and the expression βideal gas = 1/T, choose the correct relation between specific volume of an ideal gas during isobaric expansion and the absolute temperature.
Δv = ΔT v 1
Δv
T
=
ΔT
T
Δv = 1 v
ΔT
The assumption made here is that the gas behaves as an ideal gas. The coefficient of volume expansion β can be expressed as
( ) Δv
β=
1
( )
v
∂v
v ∂T
≈ P
ΔT
Noting that βideal gas = 1/T for an ideal gas and rearranging give Δv
v
=
ΔT
T
Therefore, the percent increase in the specific volume of an ideal gas during isobaric expansion is equal to the percent increase in absolute temperature.
The correct relation between specific volume of an ideal gas during isobaric expansion and the absolute temperature is ∆v
v
∆T
= T . References Multiple Choice
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Difficulty: Easy
48/182
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An infinitesimally small pressure wave is known as _____.
density velocity sound heat Sound is an infinitesimally small pressure wave.
Sound is an infinitesimally small pressure wave. References Multiple Choice
65.
Difficulty: Easy
Award: 10.00 points
Sound is generated by a small disturbance in a medium.
True False
Sound is generated by a small disturbance in a medium. Sound is generated by a small disturbance in a medium. References True / False
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Difficulty: Easy
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Sound travels by _____ ______.
wave propogation
Sound travels by wave propagation.
Sound travels by wave propagation. References Fill in the Blank
67.
Difficulty: Easy
Award: 10.00 points
Can sound waves travel in a vacuum?
Yes No Sound waves cannot travel in a vacuum.
Sound waves cannot travel in a vacuum. References Multiple Choice
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Difficulty: Easy
50/182
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In which medium does a sound wave travel faster: in cool air or in warm air?
Warm air Cool air Sound travels faster in warm (higher temperature) air since c = √kRT.
Sound travels faster in warm (higher temperature) air since c = √kRT. References Multiple Choice
69.
Difficulty: Easy
Award: 10.00 points
In which medium will sound travel fastest for a given temperature: air, helium, or argon?
Helium Air Argon Sound travels fastest in helium, since c = √kRT and helium has the highest kR value. It is about 0.40 for air, 0.35 for argon, and 3.46 for helium.
Sound travels fastest in helium, since c = √kRT and helium has the highest kR value. It is about 0.40 for air, 0.35 for argon, and 3.46 for helium. References Multiple Choice
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Difficulty: Easy
51/182
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In which medium does a sound wave travel faster: in air at 20°C and 1 atm or in air at 20°C and 5 atm?
A sound wave travels fastest in air at 20°C and 1 atm A sound wave travels fastest in air at 20°C and 5 atm A sound wave travels at the same speed in both mediums. Air at specified conditions will behave like an ideal gas, and the speed of sound in an ideal gas depends on temperature only. Therefore, the speed of sound is the same in both mediums.
Air at specified conditions will behave like an ideal gas, and the speed of sound in an ideal gas depends on temperature only. Therefore, the speed of sound is the same in both mediums. References Multiple Choice
71.
Difficulty: Easy
Award: 10.00 points
Does the Mach number of a gas flowing at a constant velocity remain constant?
Yes No
In general, it does not, because the Mach number also depends on the speed of sound in gas, which depends on the temperature of the gas. The Mach number remains constant only if the temperature and the velocity are constant.
In general, no, because the Mach number also depends on the speed of sound in gas, which depends on the temperature of the gas. References Yes / No
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Difficulty: Easy
52/182
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Is it realistic to approximate that the propagation of sound waves is an isentropic process?
Yes No
Yes, the propagation of sound waves is nearly isentropic. This is because the amplitude of an ordinary sound wave is very small, and it does not cause any significant change in temperature and pressure.
Yes, the propagation of sound waves is nearly isentropic. Because the amplitude of an ordinary sound wave is very small, and it does not cause any significant change in temperature and pressure. References Yes / No
73.
Difficulty: Easy
Award: 10.00 points
Is the sonic velocity in a specified medium a fixed quantity, or does it change as the properties of the medium change?
It is a fixed quantity. It changes as the properties change. The sonic speed in a medium depends on the properties of the medium, and it changes as the properties of the medium change.
The sonic speed in a medium depends on the properties of the medium, and it changes as the properties of the medium change. References Multiple Choice
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Difficulty: Easy
53/182
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The isentropic process for an ideal gas is expressed as Pvk = constant. Using this process equation and the definition of the speed of sound (Eq. 2-24), choose the correct expression for the speed of sound for an ideal gas.
c = kRT c=
√
kR T
c = √RT c = √kRT The isentropic relation Pvk = A, where A is a constant, can also be expressed as
()
1 k
P=A v
= Aρ k
Substituting it into the relation for the speed of sound,
2
c =
( ) ( ∂P ∂ρ
s
=
∂ ( Aρ ) k ∂ρ
)
( )
k Aρ k s
= kAρ
k−1
=
ρ
() P
= k ρ = kRT
since for an ideal gas P = ρRT or RT = P/ρ. Therefore, c = √kRT, which is the desired relation.
c = √kRT is the correct equation for the speed of sound for an ideal gas. References Multiple Choice
Difficulty: Easy
Carbon dioxide enters an adiabatic nozzle at 1200 K with a velocity of 50 m/s and leaves at 400 K. Assume constant specific heats at room temperature. References Section Break
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Difficulty: Easy
54/182
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Determine the Mach number at the inlet. The gas constant of carbon dioxide is R = 0.1889 kJ/kg·K and k = 1.288. The Mach number at the inlet is
0.0925 ± 2% .
Explanation: The following assumptions are made here: 1. CO2 is an ideal gas with constant specific heats at room temperature. 2. This is a steady-flow process.
c1 =
√
k 1RT 1 =
√
(1.288)(0.1889 kJ/kg ⋅ K)(1200 K)
(
1000 m 2 / s 2 1 kJ/kg
)
= 540.3 m/s
Thus, V1
50 m/s
Ma 1 = c = 540.3 m/s = 0.0925 1
The Mach number at the inlet is 0.0925. References Worksheet
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Difficulty: Easy
55/182
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Determine the Mach number at the exit of the nozzle. The gas constant of carbon dioxide is R = 0.1889 kJ/kg·K. Take its constant pressure specific heat and specific heat ratio at room temperature to be cp = 0.8439 kJ/kg·K and k = 1.288. The Mach number at the exit of the nozzle is
3.728 ± 0.01 .
Explanation: The following assumptions are made here: 1. CO2 is an ideal gas with constant specific heats at room temperature. 2. This is a steady-flow process.
c2 =
√
k 2RT 2 =
√
(1.288)(0.1889 kJ/kg ⋅ K)(400 K)
(
1000 m 2 / s 2 1 kJ/kg
)
= 312.0 m/s
The nozzle exit velocity is determined from the steady-flow energy balance relation: 0 = h2 − h1 +
V22 − V12 2
(
)
→ 0 = cp T2 − T1 +
0 = (0.8439 kJ/kg ⋅ K)(400 − 1200 K) +
V22 − V12 2
V 2 2 − ( 50 m/s ) 2 2
(
1 kJ/kg 1000 m 2 / s 2
)
→ V 2 = 1163.1 m/s
Thus, V2
1163.1 m/s
2
312 m/s
Ma 2 = c =
= 3.728
The Mach number at the exit of the nozzle is 3.728. References Worksheet
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Difficulty: Easy
56/182
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Nitrogen enters a steady-flow heat exchanger at 150 kPa, 10°C, and 108 m/s, and it receives heat in the amount of 120 kJ/kg as it flows through it. Nitrogen leaves the heat exchanger at 100 kPa with a velocity of 184 m/s. Determine the Mach number of the nitrogen at the inlet and the exit of the heat exchanger. The gas constant of N2 is R = 0.2968 kJ/kg·K. Its constant pressure specific heat and specific heat ratio at room temperature are cp = 1.040 kJ/kg·K and k = 1.4. The Mach number of the nitrogen at the inlet of the heat exchanger is The Mach number of the nitrogen at the outlet of the heat exchanger is
0.315 ± 2% . 0.458 ± 2% .
Explanation: The following assumptions have been made here: 1. N2 is an ideal gas. 2. This is a steady-flow process. 3. The potential energy change is negligible.
The speed of sound at the inlet is c1 =
√k1RT1
c1 =
√
(1.400)(0.2968 kJ/kg ⋅ K)(283 K)
(
1000 m 2 / s 2 1 kJ/kg
)
= 342.9 m/s
Thus, V1
108 m/s
Ma 1 = c = 342.9 m/s = 0.315 1
From the energy balance on the heat exchanger,
(
)
q in = c p T 2 − T 1 +
V22 − V12 2
(
)
120 kJ/kg = (1.040 kJ/kg.°C) T 2 − 10°C +
( 184 m/s ) 2 − ( 108 m/s ) 2 2
(
1 kJ/kg 1000 m 2 / s 2
)
It yields T2 = 114.7ºC = 387.87 K
c2 =
√
k 2RT 2 =
√
(1. 4)(0.2968 kJ/kg ⋅ K)(387.87 K)
(
1000 m 2 / s 2 1 kJ/kg
)
= 401.5 m/s
Thus, https://ezto.mheducation.com/hm.tpx
57/182
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Assignment Print View V2
184 m/s
Ma 2 = c = 401.5 m/s = 0.458 2
The Mach number of the nitrogen at the inlet of the heat exchanger is 0.315. The Mach number of the nitrogen at the outlet of the heat exchanger is 0.458. References Worksheet
78.
Difficulty: Medium
Award: 10.00 points
Assuming ideal gas behavior, determine the speed of sound in refrigerant-134a at 0.8 MPa and 75°C. The gas constant of R134a is R = 0.08149 kJ/kg·K. Its specific heat ratio at room temperature is k = 1.108. The speed of sound in refrigerant-134a is
177 ± 2% m/s.
Explanation: The assumption made here is that R-134a is an ideal gas with constant specific heats at room temperature. From the ideal gas speed of sound relation,
c = √kRT =
√
(1.108)(0.08149 kJ/kg ⋅ K)(75°C + 273 K)
(
1000 m 2 / s 2 1 kJ/kg
)
= 177 m/s
The speed of sound in refrigerant-134a is 177 m/s. References Worksheet
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Difficulty: Easy
58/182
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Determine the speed of sound in air at 300 K. Also determine the Mach number of an aircraft moving in the air at a velocity of 340 m/s. The gas constant of air is R = 0.287 kJ/kg·K. Its specific heat ratio at room temperature is k = 1.4. The speed of sound in air at 300 K is The Mach number of the aircraft is
347.19 ± 2% m/s. 0.979 ± 2% .
Explanation: The assumption made here is that air is an ideal gas with constant specific heat at room temperature. From the definitions of the speed of sound and the Mach number at 300 K,
c = √kRT =
√
(1.4)(0.287 kJ/kg ⋅ K)(300 K)
(
1000 m 2 / s 2 1 kJ/kg
)
= 347.19 m/s
and V
340m/s
Ma = c = 347.19 m/s = 0.979
The speed of sound in air at 300 K is 347.19 m/s. The Mach number of the aircraft is 0.979. References Worksheet
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Difficulty: Easy
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Steam flows through a device with a pressure of 120 psia, a temperature of 700°F, and a velocity of 930 ft/s. Determine the Mach number of the steam at this state by assuming ideal gas behavior with k = 1.3. The gas constant of steam is R = 0.1102 Btu/lbm·R. The Mach number of the steam at this state is
0.456 ± 2% .
Explanation: The assumption made here is that steam is an ideal gas with constant specific heats. From the speed of sound relation for an ideal gas,
c = √kRT =
√
(1.3)(0.1102 Btu / lbm ⋅ R)(1160 R)
(
25037 ft 2 / s 2 1 Btu/lbm
)
= 2039.78 ft/s
Thus, V
930 ft/s
Ma = c = 2039.78 ft/s = 0.456
The Mach number of the steam at this state is 0.456. References Worksheet
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Difficulty: Easy
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Steam flows through a device with a pressure of 120 psia and a velocity of 900 ft/s. Determine the Mach number of the steam at this state by assuming ideal-gas behavior with k = 1.3. Using appropriate software, compare the Mach number of the steam flow over the temperature range 350 to 700°F. Plot the Mach number as a function of temperature. Please upload your response using the controls provided below. Hint: What is Mach number defined as?
Student upload controls will be shown to students when they take this assignment.
References File Attachment Question
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Difficulty: Medium
61/182
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Air expands isentropically from 2.2 MPa and 105°C to 0.4 MPa. Calculate the ratio of the initial to final speed of sound. The properties of air are R = 0.287 kJ/kg·K and k = 1.4. The specific heat ratio k varies with temperature, but in our case this change is very small and can be disregarded. The ratio of the initial to final speed of sound is
1.28 ± 2% .
Explanation: The assumption made here is that air is an ideal gas with constant specific heats at room temperature. The final temperature of air is determined from the isentropic relation of ideal gases:
T2 = T1
() P2
(k−1) k
P1
= (378.15 K)
(
0.4 MPa 2.2 MPa
)
(1.4−1) 1.4
= 232.34 K
Treating k as a constant, the ratio of the initial to final speed of sound can be expressed as
Ratio =
c1 c2
=
√k1RT1 √k2RT2
=
√T 1 √T 2
=
√378.15 √232.34
= 1.28
The ratio of the initial to final speed of sound is 1.28. References Worksheet
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Difficulty: Easy
62/182
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Helium expands isentropically from 2.2 MPa and 65°C to 0.4 MPa. Calculate the ratio of the initial to final speed of sound. The properties of helium are R = 2.0769 kJ/kg·K and k = 1.667. The specific heat ratio k varies with temperature, but in our case this change is very small and can be disregarded. The ratio of the initial to final speed of sound is
1.41 ± 2% .
Explanation: The assumption made here is that helium is an ideal gas with constant specific heats at room temperature. The final temperature of air is determined from the isentropic relation of ideal gases:
T2 = T1
() P2
(k−1) k
P1
= (338.15 K)
(
0.4 MPa 2.2 MPa
)
( 1 . 667 − 1 ) 1 . 667
= 170.95 K
Treating k as a constant, the ratio of the initial to final speed of sound can be expressed as
Ratio =
c1 c2
=
√k1RT1 √k2RT2
=
√T 1 √T 2
=
√338.15 √170.95
= 1.41
The ratio of the initial to final speed of sound is 1.41. References Worksheet
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Difficulty: Easy
63/182
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The Airbus A-340 passenger plane has a maximum takeoff weight of about 260,000 kg, a length of 64 m, a wing span of 60 m, a maximum cruising speed of 950 km/h, a seating capacity of 271 passengers, a maximum cruising altitude of 14,000 m, and a maximum range of 12,000 km. The air temperature at the crusing altitude is about –60°C. Determine the Mach number of this plane for the stated limiting conditions. The gas constant of air is R = 0.287 kJ/kg·K. Its specific heat ratio at room temperature is k = 1.4. The Mach number of this plane for the given limiting conditions is
0.901 ± 2% .
Explanation: The assumption made here is that air is an ideal gas with constant specific heats at room temperature. From the speed of sound relation,
c = √kRT =
√
(1.4)(0.287 kJ/kg ⋅ K)(-60 + 273 K)
(
1000 m 2 / s 2 1 kJ/kg
)
= 293 m/s
Thus, the Mach number corresponding to the maximum cruising speed of the plane is
Ma =
V max c
=
(
950 km/h 3.6
)
293 m/s
m/s
= 0.901
The Mach number of this plane for the given limiting conditions is 0.901. References Worksheet
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85.
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What are Newtonian fluids?
Fluids whose normal stress is exponentially proportional to the velocity gradient Fluids whose normal stress is linearly proportional to the velocity gradient Fluids whose shear stress is exponentially proportional to the velocity gradient Fluids whose shear stress is linearly proportional to the velocity gradient Fluids whose shear stress is linearly proportional to the velocity gradient (shear strain) are called Newtonian fluids.
Fluids whose shear stress is linearly proportional to the velocity gradient (shear strain) are called Newtonian fluids. References Multiple Choice
86.
Difficulty: Easy
Award: 10.00 points
Is water a Newtonian fluid?
Yes No
Most common fluids such as water, air, gasoline, and oils are Newtonian fluids. Yes, water is a Newtonian fluid. References Yes / No
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Difficulty: Easy
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87.
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_____ is a measure of the “stickiness” or “resistance to deformation” of a fluid.
Internal energy Density Viscosity Chemical energy Viscosity is a measure of the “stickiness” or “resistance to deformation” of a fluid. It is due to the internal frictional force that develops between different layers of fluids as they are forced to move relative to each other. Viscosity is caused by the cohesive forces between the molecules in liquids, and by the molecular collisions in gases.
Viscosity is a measure of the “stickiness” or “resistance to deformation” of a fluid. References Multiple Choice
88.
Difficulty: Easy
Award: 10.00 points
Do liquids or gases have higher dynamic viscosities?
Gases Liquids In general, liquids have higher dynamic viscosities than gases.
In general, liquids have higher dynamic viscosities than gases. References Multiple Choice
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For liquids, the kinematic viscosity _____ with temperature.
decreases remains constant increases For liquids, the kinematic viscosity decreases with temperature.
For liquids, the kinematic viscosity decreases with temperature. References Multiple Choice
90.
Difficulty: Easy
Award: 10.00 points
For gases, the kinematic viscosity _____ with temperature.
decreases increases remains constant For gases, the kinematic viscosity increases with temperature.
For gases, the kinematic viscosity increases with temperature. References Multiple Choice
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Consider two identical small glass balls dropped into two identical containers, one filled with water and the other with oil. Which ball will reach the bottom of the container first?
The ball dropped in water The ball dropped in oil Both the balls at the same time When two identical small glass balls are dropped into two identical containers, one filled with water and the other with oil, the ball dropped in water will reach the bottom of the container first because of the much lower viscosity of water relative to oil.
The ball dropped in water will reach the bottom of the container first. References Multiple Choice
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92.
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The viscosity of a fluid is to be measured by a viscometer constructed of two 5-ft-long concentric cylinders. The inner diameter of the outer cylinder is 6 in, and the gap between the two cylinders is 0.035 in. The outer cylinder is rotated at 250 rpm, and the torque is measured to be 1.5 lbf·ft. Determine the viscosity of the fluid.
The viscosity of the fluid is
3.4 ± 2% × 10−4 lbf·s/ft2.
Explanation: The following assumptions have been made here: 1. The inner cylinder is completely submerged in the fluid. 2. The viscous effects on the two ends of the cylinder are negligible. 3. The fluid is Newtonian.
Substituting the given values, the viscosity of the fluid is determined to be μ=
Tl .
=
4π 2R 3nL
( 1.5 lbf ⋅ ft ) ( 0.035 / 12 ft ) 4π 2 ( 3 / 12 ft ) 3
( 250 / 60 s ) ( 5 ft ) −1
= 3.4 × 10 − 4 lbf ⋅ s / ft 2
The viscosity of the fluid is 3.4 × 10−4 lbf·s/ft2. References Worksheet
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Difficulty: Medium
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93.
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The dynamic viscosities of carbon dioxide at 50°C and 200°C are 1.612 × 10-5 Pa·s and 2.276 × 10-5 Pa·s, respectively. Determine the constants a and b of the Sutherland correlation for carbon dioxide at atmospheric pressure. Then, predict the viscosity of carbon dioxide at 100°C. The value of constant a is
1.633 ± 2% × 10-6 kg/(m·s·K1/2).
The value of constant b is
265.5 ± 2% K. 1.843 ± 2% × 10-5 Pa·s.
The viscosity of carbon dioxide at 100°C is Explanation: The Sutherland correlation is given as follows: μ =
a √T 1 + b/T
where T is the absolute temperature. Substituting the given values, a T1
μ1 = 1
√
+ b / T1
√
=
1 +
a T2
μ2 =
1 + b / T2
a√ ( 50 °C + 273 . 15 ) K
=
b
→ 1. 612 × 10 − 5 Pa·s =
1 +
( 50°C + 273 . 15 ) K
a√ ( 200°C + 273 . 15 ) K b
a√323 . 15 K
→ 2. 276 × 10 − 5 Pa·s =
1 + ( 200°C + 273 . 15 ) K
b 323 . 15 K
a√473 . 15 K b
1 + 473 . 15 K
which is a nonlinear system of two algebraic equations. Using appropriate software, the following result is determined: a = 1.633 ×10−6 kg/(m·s·K1/2) b = 265.5 K Using these values, the Sutherland correlation becomes
(
1.633 × 10 − 6 kg/ m·s·K 1 / 2
μ =
) √T
1 + 265.5 / T
Therefore, the viscosity at 100°C is determined as follows: 1 . 633 × 10 − 6
μ =
(
kg/ m·s·K 1 / 2
) √373 . 15 K
265 . 5 K
1 + 373 . 15 K
= 1. 843 × 10 − 5 Pa ⋅ s
The value of constant a is 1.633 ×10-6 kg/(m·s·K1/2). The value of constant b is 265.5 K. The viscosity of carbon dioxide at 100°C is 1.843 × 10-5 Pa·s. References Worksheet
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70/182
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94.
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Consider the flow of a fluid with viscosity μ through a circular pipe. The velocity profile in the pipe is given as u(r) = umax(1 −
rn/Rn), where umax is the maximum flow velocity, which occurs at the centerline; r is the radial distance from the centerline; and u(r) is the flow velocity at any position r. Develop a relation for the drag force exerted on the pipe wall by the fluid in the flow direction per unit length of the pipe.
F = 2 n μu L
max
F = 2nπμu L
max
F = 2nπμ L
u max
The assumption made here is that the viscosity of the fluid is constant.
The wall shear stress is determined from its definition to be du
τ w = − μ dr
|
d
r=R
( )
= − μu max dr 1 −
rn
R
n
r=R
= − μu max
− nr n − 1
R
n
|
= r=R
nμu max R
Note that the quantity du/dr is negative in pipe flow, and the negative sign is added to the τW relation for pipes to make shear stress in the positive (flow) direction a positive quantity (Or, du/dr = – du/dy since y = R – r). Then, the friction drag force exerted by the fluid on the inner surface of the pipe becomes
F = τ wA w =
nμu max R
(2πR)L = 2nπμu maxL
Therefore, the drag force per unit length of the pipe is https://ezto.mheducation.com/hm.tpx
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= 2nπμu max
F
The drag force per unit length of the pipe is L = 2nπμu max. References Multiple Choice
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The viscosity of a fluid is to be measured by a viscometer constructed of two 75-cm-long concentric cylinders. The outer diameter of the inner cylinder is 15 cm, and the gap between the two cylinders is 1 mm. The inner cylinder is rotated at 300 rpm, and the torque is measured to be 0.8 N·m. Determine the viscosity of the fluid.
0.0128 ± 2% N·s/m2.
The viscosity of the fluid is Explanation:
The following assumptions have been made here: 1. The inner cylinder is completely submerged in oil. 2. The viscous effects on the two ends of the inner cylinder are negligible. 3. The fluid is Newtonian.
Substituting the given values, the viscosity of the fluid is determined to be μ=
Tl 2 3
4π R ṅL
( 0.8 N ⋅ m ) ( 0.001 m )
=
4π 2 ( 0.075 m ) 3
(
300 60
s -1
)
= 0.0128 N⋅s / m 2
( 0.75 m )
The viscosity of the fluid is 0.0128 N·s/m2.
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Difficulty: Easy
A thin 30-cm × 30-cm flat plate is pulled at 5.2 m/s horizontally through a 3.6-mm-thick oil layer sandwiched between two plates, one stationary and the other moving at a constant velocity of 0.3 m/s, as shown in the figure given below. The dynamic viscosity of the oil is 0.027 Pa·s. Assume that the velocity in each oil layer varies linearly.
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96.
Difficulty: Medium
Award: 10.00 points
Plot the velocity profile for this problem. Please upload your response/solution using the controls provided below. (Please provide an answer before moving on to the next part.) Hint: Is velocity profiles linear for the shear-driven flow?
Student upload controls will be shown to students when they take this assignment.
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Find the location where the oil velocity is zero. The location where the oil velocity is zero is
0.14182 ± 2% mm from the lower plate.
Explanation: The following assumptions have been made here: 1. The thickness of the plate is negligible. 2. The velocity profile in each oil layer is linear.
The point of zero velocity is indicated by point A, and its distance from the lower plate is determined from geometric considerations (the similarity of the two triangles in the lower oil layer from the velocity profile) to be as follows: 2.6 − y A yA
5.2 m/s
= 0.3 m/s → y A = 0.14182 mm
The location where the oil velocity is zero is 0.14182 mm from the lower plate. References Worksheet
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Difficulty: Medium
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98.
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Determine the force that needs to be applied on the plate to maintain this motion. The force that needs to be applied on the plate to maintain this motion is
17.8 ± 2% N.
Explanation: The magnitudes of shear forces acting on the upper and lower surfaces of the plate are determined as follows:
| |
= μA s h
| |
= μA s
F shear, upper = τ w , upperA s = μA s
F shear, lower = τ w , lowerA s = μA s
du dy
du dy
V−0 1
V − Vw h2
(
)
(
)
= 0.027 N ⋅ s/m 2 (0.3 × 0.3) m 2
5.2 m/s 1.0 × 10 -3 m
= 0.027 N ⋅ s/m 2 (0.3 × 0.3) m 2
= 12.64 N
( 5.2 m/s − ( − 0.3 ) ) m/s 2.6 × 10 -3 m
= 5.14 N
Noting that both shear forces are in the opposite direction of motion of the plate, the force F is determined from a force balance on the plate as follows: F = F shear, upper + F shear, lower =
(12.64 N + 5.14) N = 17.8 N
Note that wall shear is a friction force between a solid and a liquid, and it acts in the opposite direction of motion.
The force that needs to be applied on the plate to maintain this motion is 17.8 N. References Worksheet
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Difficulty: Medium
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A rotating viscometer consists of two concentric cylinders—an inner cylinder of radius Ri rotating at angular velocity (rotation rate) ωi and a stationary outer cylinder of inside radius Ro. In the tiny gap between the two cylinders is the fluid of viscosity μ. The length of the cylinders (in the given figure) is L. L is large such that end effects are negligible (treat this as a two-dimensional problem). Torque (T) is required to rotate the inner cylinder at constant speed.
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78/182
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99.
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Showing all of the steps and algebraic equations, generate an approximate expression for T as a function of the other variables. (You must provide an answer before moving on to the next part.) Please upload your response/solution using the controls provided below. Hint: How to approximate the shear stress for the concentric cylinders?
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100. Award: 10.00 points
Explain why this solution is only an approximation if the velocity profile in the gap will remain linear as the gap becomes larger and larger (i.e., if the outer radius Ro were to increase, all else staying the same). (Please upload your response/solution using the controls provided below.) Hint: When the curvature effect of the cylinder can be neglected?
Student upload controls will be shown to students when they take this assignment.
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80/182
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101.
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The clutch system shown in the given figure is used to transmit torque through a 2-mm-thick oil film with μ = 0.38 N·s/m2 between two identical 38-cm-diameter disks. When the driving shaft rotates at a speed of 1200 rpm, the driven shaft is observed to rotate at 1125 rpm. Assuming a linear velocity profile for the oil film, determine the transmitted torque.
The transmitted torque is
3.05 ± 2% N·m.
Explanation:
The following assumptions have been made here: 1. The thickness of the oil film is uniform. 2. The rotational speeds of the disks remain constant.
The disks rotate in the same direction at different angular speeds of ω1 and ω2. Therefore, we can assume one of the disks to be stationary and the other to be rotating at an angular speed of ω1 − ω2 . The velocity gradient anywhere in the oil of film thickness h is V/h, where V = (ω1 − ω2)r is the tangential velocity. Then, the wall shear stress anywhere on the surface of the faster disk at a distance r from the axis of rotation can be expressed as follows: du
V
τ w = μ dr = μ h = μ
( ω1 − ω2 ) r h
Then, the shear force acting on a differential area dA on the surface and the torque generation associated with it can be expressed as follows: https://ezto.mheducation.com/hm.tpx
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dF = τ wdA = μ dT = rdF = μ
( ω1 − ω2 ) r h
( ω1 − ω2 ) r2 h
(2πr)dr
(2πr)dr =
2πμ ( ω 1 − ω 2 ) h
r 3dr
On integrating,
(
2πμ ω 1 − ω 2
T=
h
) D/2 3
(
2πμ ω 1 − ω 2
∫ r = 0 r dr =
h
|
) r4 D / 2 4
(
)
πμ ω 1 − ω 2 D 4
=
r=0
32h
˙ Noting that ω = 2πn, the relative angular speed is determined as follows:
(
ω 1 − ω 2 = 2π ṅ 1 − ṅ 2
) = (2π rad/rev)((1200 − 1125) rev/min) ( 60s ) = 7.854 rad/s 1 min
Substituting the relative angular speed, the torque transmitted is determined as follows:
(
)
π 0.38 N ⋅ s/m 2 ( 7.854 rad/s ) ( 0.38 m ) 4
T=
= 3.05 N ⋅ m
32 ( 0.002 m )
The transmitted torque is 3.05 N·m. References Worksheet
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Difficulty: Medium
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102. Award: 10.00 points
The clutch system shown in the figure below is used to transmit torque through a 2-mm-thick oil film with μ = 0.38 N⋅s/m2 between two identical 30-cm-diameter disks. When the driving shaft rotates at a speed of 1200 rpm, the driven shaft is observed to rotate at 1125 rpm. Assuming a linear velocity profile for the oil film, determine the transmitted torque. Using appropriate software, investigate the effect of the oil film thickness on the torque transmitted. Let the film thickness vary from 0.1 mm to 10 mm. Plot your results, and state your conclusions. Please upload your answer/response using the controls provided below.
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A 50-cm × 30-cm × 20-cm block weighing 130 N is to be moved at a constant velocity of 1.1 m/s on an inclined surface with a friction coefficient of 0.27. The absolute viscosity of oil is given to be μ = 0.012 Pa·s = 0.012 N·s/m2.
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103. Award: 10.00 points
Determine the force F that needs to be applied in the horizontal direction. (Please provide an answer before moving on to the next part.) The force F that needs to be applied in the horizontal direction is
91.4 ± 2% N.
Explanation: The following assumptions have been made here: 1. The inclined surface is a plane (perfectly flat, although tilted). 2. The friction coefficient and the oil film thickness are uniform. 3. The weight of the oil layer is negligible. The velocity of the block is constant, and thus its acceleration and then the net force acting on it are zero. A free-body diagram of the block is given. Then, the force balance gives
∑ F x = 0 : F 1 − F f cos 20° − F N1 sin 20° = 0
(1 )
∑ F y = 0 : F N1 cos 20° − F f sin 20° − W = 0
(2 )
Friction force: F f = fF N1
(3 )
Substituting Eq. (3) in Eq. (2) and solving for FN1 give W
130 N
F N1 = cos 20° − f sin 20° = cos 20° − 0.27 sin 20° = 153.42 N Then, from Eq. (1), F 1 = F f cos 20° + F N1 sin 20° = (0.27 × 153.42 N) cos 20° + (153.42 N) sin 20° = 91.4 N
The force F that needs to be applied in the horizontal direction is 91.4 N. References Worksheet
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Difficulty: Medium
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104. Award: 10.00 points
If a 0.40-mm-thick oil film with a dynamic viscosity of 0.012 Pa·s is applied between the block and the inclined surface, determine the percent reduction in the required force. The percent reduction in the required force is
44.4 ± 2% .
Explanation:
In this case, the friction force is replaced by shear force applied on the bottom surface of the block due to the oil. Because of the no-slip condition, the oil film sticks to the inclined surface at the bottom and the lower surface of the block at the top. Then, the shear force is expressed as F shear = τ wA s = μA s
(
V h
)
F shear = 0.012 N ⋅ s/m 2 (0.5 m × 0.2 m)
1.1 m/s 0.4 × 10 − 3 m
F shear = 3.3 N Then, the force balance gives ∑ F x = 0 : F 2 − F shear cos 20° − F N2 sin 20° = 0
(4 )
∑ F y = 0 : F N2 cos 20° − F shear sin 20° − W = 0
(5 )
Eq. (5) gives
F N2 =
( Fshear sin 20° + W ) cos 20°
=
( ( 3.3 N ) sin 20° + 130 N ) cos 20°
= 139.54 N
Substituting in Eq. (4), the required horizontal force is determined to be F 2 = F shear cos 20° + F N2 sin 20° = (3.3 N) cos 20° + (139.54 N) sin 20° = 50.83 N Then, our final result is expressed as Percentage reduction in required force =
F1 − F2 F1
× 100 % =
91.4 N − 50.83 N 91.4 N
× 100% = 44.4%
The percent reduction in the required force is 44.4. References Worksheet
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Difficulty: Medium
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105. Award: 10.00 points
For flow over a plate, the variation of velocity with vertical distance y from the plate is given as u(y) = ay − by2 where a and b are constants. Choose the correct relation for the wall shear stress in terms of a, b, and μ.
τw = a
μ
τ w = 2 aμ τ w = 2μ b
τ w = aμ The assumption made here is that the fluid is Newtonian. Noting that u(y) = ay – by2, the wall shear stress is determined from its definition to be du
τ w = μ dy
|
(
d ay − by 2 y=0
=μ
dy
)
|
y=0
= μ(a − 2by)| y = 0 = aμ
The wall shear stress is given as τ w = aμ. References Multiple Choice
Difficulty: Easy
In regions far from the entrance, fluid flow through a circular pipe is one dimensional, and the velocity profile for laminar flow is given by u(r) = umax(1 – r2/R2), where R is the radius of the pipe, r is the radial distance from the center of the pipe, and umax is the maximum flow velocity, which occurs at the center.
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106. Award: 10.00 points
Identify a relation for the drag force applied by the fluid on a section of the pipe of length L.
F D = 3πμLu max F D = 2πμLu max F D = 4πμLu max F D = πμLu max The following assumptions have been made here: 1. The flow through the circular pipe is one dimensional. 2. The fluid is Newtonian.
The velocity profile is given by
( )
u(r) = u max 1−
r2
R2
where R is the radius of the pipe, r is the radial distance from the center of the pipe, and umax is the maximum flow velocity, which occurs at the center, r = 0. The shear stress at the pipe surface is expressed as
τw = − μ
du dr
|
r=R
= − μu max
d dr
( ) 1−
r2
R2
r=R
= − μu max
− 2r
R2
|
= r=R
2μu max
R
Note that the quantity du/dr is negative in pipe flow, and the negative sign is added to the relation for pipes to make shear stress in the positive (flow) direction, thereby making it a positive quantity (or, du/dr = −du/dy since y = R − r). Then, the friction drag force exerted by the fluid on the inner surface of the pipe becomes
F D = τ wA s =
2μu max
R
(2πRL) = 4πμLu max
Drag force = 4πμLu max References Multiple Choice
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107. Award: 10.00 points
Determine the value of the drag force for water flow at 20°C with R = 0.08 m, L = 32 m, umax = 3 m/s, and μ = 0.0010 kg/m·s. (Round the final answer to two decimal places.) The value of the drag force is
1.21 ± 2% N.
Explanation: The following assumptions have been made here: 1. The flow through the circular pipe is one dimensional. 2. The fluid is Newtonian.
The velocity profile is given by
( )
u(r) = u max 1−
r2
R2
where R is the radius of the pipe, r is the radial distance from the center of the pipe, and umax is the maximum flow velocity, which occurs at the center, r = 0. The shear stress at the pipe surface is expressed as du
τ w = − μ dr
|
d
r=R
( )
= − μu max dr 1 −
r2
R2
r=R
= − μu max
− 2r R2
|
= r=R
2μu max R
Note that the quantity du/dr is negative in pipe flow, and the negative sign is added to the relation for pipes to make shear stress in the positive (flow) direction, thereby making it a positive quantity (or, du/dr = −du/dy since y = R − r). Then, the friction drag force exerted by the fluid on the inner surface of the pipe becomes F D = τ wA s =
2μu max R
(2πRL) = 4πμLu max
The value of the drag force is determined as follows:
F D = 4πμLu max = 4π(0.0010 kg/m ⋅ s)(32 m)(3 m/s)
(
1N 1 kg ⋅ ms 2
)
= 1.21 N
The value of the drag force is 1.21 N. References Worksheet https://ezto.mheducation.com/hm.tpx
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108. Award: 10.00 points
In regions far from the entrance, fluid flow through a circular pipe is one dimensional, and the velocity profile for laminar flow is
( )
given by u(r) = u max 1 -
r2
R2
, where R is the radius of the pipe, r is the radial distance from the center of the pipe, and umax is
the maximum flow velocity, which occurs at the center. Determine the value of the drag force for water flow at 20°C with R = 0.08 m, L = 22 m, umax = 6 m/s, and μ = 0.0010 kg/m·s.
The value of the drag force is
1.66 ± 2% N.
Explanation: The following assumptions have been made here: 1. The flow through the circular pipe is one dimensional. 2. The fluid is Newtonian.
The velocity profile is given by
( )
u(r) = u max 1−
r2
R2
where R is the radius of the pipe, r is the radial distance from the center of the pipe, and umax is the maximum flow velocity, which occurs at the center, r = 0. The shear stress at the pipe surface is expressed as du
τ w = − μ dr
|
d
r=R
( )
= − μu max dr 1 −
r2
R
= − μu max
2
r=R
− 2r R
2
|
= r=R
2μu max R
du
Note that the quantity dr is negative in pipe flow, and the negative sign is added to the relation for pipes to make shear stress du
du
in the positive (flow) direction, thereby making it a positive quantity (or, dr = - dy since y = R − r). Then, the friction drag force exerted by the fluid on the inner surface of the pipe becomes https://ezto.mheducation.com/hm.tpx
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F D = τ wA s =
2μu max R
(2πRL) = 4πμLu max
The value of the drag force is determined as follows:
(
F D = 4πμLu max = 4π (0.0010 kg/m ⋅ s)(22 m)(6 m/s)
(
1N 1 kg ⋅ ms 2
)
= 1.66 N
The value of the drag force is 1.66 N. References Worksheet
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Difficulty: Medium
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109. Award: 10.00 points
A frustum-shaped body is rotating at a constant angular speed of 210 rad/s in a container filled with SAE 10W oil at 20°C (μ = 0.100 Pa·s), as shown in the figure. If the thickness of the oil film on all sides is 1.2 mm, determine the power required to maintain this motion. Also determine the reduction in the required power when the oil temperature rises to 80°C (μ = 0.0078 Pa·s).
The power required to maintain this motion is
297.4 ± 2% W.
The reduction in the required power input when the oil temperature is raised to 80°C is
274.2 ± 2% W.
Explanation:
The assumption made here is that the thickness of the oil layer remains constant.
The velocity gradient anywhere in the oil of film thickness h is V/h, where V = ωr is the tangential velocity. Then, the wall shear stress anywhere on the surface of the frustum at a distance r from the axis of rotation is τw = μ
du dr
=μ
V h
=μ
ωr h
The shear force acting on differential area dA on the surface, the torque it generates, and the shaft power associated with it are expressed as ωr
dF = τ ωdA = μ h dA ωr 2
dT = rdF = μ h dA https://ezto.mheducation.com/hm.tpx
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T=
μω h
∫ Ar 2dA
.
W sh = ωT =
μω 2 h
∫ Ar 2dA
Top surface For the top surface, dA = 2πrdr. Substituting and integrating, .
W sh, top =
μω 2 D / 2
∫ r 2(2πr)dr =
h
2πμω 2 D / 2 h
r=0
∫ r 3dr =
|
2πμω 2 r 4 h
r=0
h
D/2
πμω 2D 4
=
32h
r=0
Bottom surface A relation for the bottom surface is obtained by replacing D with d. .
W sh, bottom =
πμω 2d 4 32h
Side surface The differential area for the side surface can be expressed as dA = 2πrdz. From geometric considerations, the variation of d
radius with axial distance is expressed as r = 2 + Differentiating gives dr =
D−d 2L
2L
dz or dz =
D−d
D−d 2L
z.
dr. Therefore, dA = 2πdz =
4πL D−d
rdr.
Substituting and integrating, .
W sh, top =
μω 2 D / 2
2
4πμω 2L D / 2
4πL
∫ r D − d rdr = h ( D − d ) ∫
h
r=0
4πμω 2L r 4
3
r=d/2
r dr = h ( D − d ) 4
|
D/2
(
πμω 2L D 4 − d 4
=
)
16h ( D − d )
r=d/2
Then, the total power required becomes
.
.
.
.
W sh, total = W sh, top + W sh, bottom + W sh, side =
πμω 2D 4 32h
(
1+
() d
D
( ( )) d
2L 1 − 4
+
4
D
D−d
)
where d/D = 4/12 = 1/3. Substituting,
.
W sh, total =
(
)
π 0.1 N ⋅ s/m 2 ( 210/s ) 2 ( 0.12 m ) 4 32 ( 0.0012 m )
(
1+
() 3
{ ( )}
2 ( 0.12 m ) 1 −
1 4
+
1
4
3
( 0.12 m − 0.04 m )
)
( ) 1W
1 Nm/s
= 297.4 W
Noting that power is proportional to viscosity, the power required at 80°C is .
μ 80°C .
0.0078 N ⋅ s/m 2
20°C
0.1 N ⋅ s/m 2
W sh, total, 80°C = μ
W sh, total, 20°C =
(297.4 W) = 23.2 W
Therefore, the reduction in the required power input at 80°C is .
.
Reduction = W sh, total, 20°C − W sh, total, 80°C = 297.4 W − 23.2 W = 274.2 W
The power required to maintain this motion is 297.4 W. The reduction in the required power input when the oil temperature is raised to 80°C is 274.2 W. https://ezto.mheducation.com/hm.tpx
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References Worksheet
110.
Difficulty: Medium
Award: 10.00 points
A rotating viscometer consists of two concentric cylinders—a stationary inner cyliner of radius Ri and an outer cylinder of inside radius Ro rotating at angular velocity (rotation rate) ωo. In the tiny gap between the two cylinders is the fluid whose viscosity (μ) is to be measured. The length of the cylinders is L. L is large such that the end effects are negligible (we can treat this as a twodimensional problem). Torque (T) is required to rotate the inner cylinder at constant speed. Showing all your work and algebra, generate an approximate expression of T as a function of the other variables. Please upload your response/answer using the controls provided below.
Hint: What is the velocity profile between the two walls of cylinders?
Student upload controls will be shown to students when they take this assignment.
References File Attachment Question
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111.
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Award: 10.00 points
A thin plate moves between two parallel, horizontal, stationary flat surfaces at a constant velocity of V = 6.5 m/s. The two stationary surfaces are spaced 4 cm apart, and the medium between them is filled with oil whose viscosity is 0.9 N·s/m2. The part of the plate immersed in oil at any given time is 2 m long and 0.5 m wide. If the plate moves through the mid-plane between the surfaces, determine the force required to maintain this motion. What would your response be if the plate was 1 cm from the bottom surface (h2) and 3 cm from the top surface (h1)?
If the plate moves through the mid-plane between the surfaces, the force required to maintain the motion will be N.
585 ± 2%
If the plate was 1 cm from the bottom surface (h2) and 3 cm from the top (h1) surface, the force required to maintain the motion would be 780 ± 2% N.
Explanation: The following assumptions have been made here: 1. The thickness of the plate is negligible. 2. The velocity profile in each oil layer is linear.
The magnitude of shear forces acting on the upper and lower surfaces of the moving thin plate are
F shear, upper = τ w, upperA s = μA s
| |
F shear, lower = τ w, lowerA s = μA s
| |
du dy
du dy
V−0
= μA s h
1
V−0
= μA s h
2
(
)
6.5 m/s
(
)
6.5 m/s
= 0.9 N ⋅ s/m 2 (0.5 m × 2 m) 0.02 m = 292.5 N
= 0.9 N ⋅ s/m 2 (0.5 m × 2 m) 0.02 m = 292.5 N
Noting that both shear forces are in the opposite direction of motion of the plate, the force F is determined from a force balance on the plate to be F = F shear, upper + F shear, lower = 585 N When the plate is 1 cm from the bottom surface and 3 cm from the top surface, the force F becomes https://ezto.mheducation.com/hm.tpx
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F shear, upper = τ w, upperA s = μA s
| |
= μA s
F shear, lower = τ w, lowerA s = μA s
| |
= μA s
du dy
du dy
V−0 h1
V−0 h2
(
)
6.5 m/s
(
)
6.5 m/s
= 0.9 N ⋅ s/m 2 (0.5 m × 2 m)
= 0.9 N ⋅ s/m 2 (0.5 m × 2 m)
0.03 m
0.01 m
= 195 N
= 585 N
Noting that both shear forces are in the opposite direction of the motion of the plate, the force F is determined from a force balance on the plate to be F = F shear, upper + F shear, lower = 195 N + 585 N = 780 N
If the plate moves through the mid-plane between the surfaces, the force required to maintain the motion will be 585 N If the plate was 1 cm from the bottom surface (h2) and 3 cm from the top (h1) surface, the force required to maintain the motion will be 780 N. References Worksheet
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Difficulty: Easy
97/182
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112.
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Award: 10.00 points
A thin plate moves between two parallel, horizontal, stationary flat surfaces at a constant velocity of 5 m/s. The two stationary surfaces are spaced 5 cm apart, and the medium between them is filled with oil whose viscosity is 0.9 N⋅s/m2. The part of the plate immersed in oil at any given time is 2 m long and 0.5 m wide. If the viscosity of the oil above the moving plate is four times that of the oil below the plate, determine the distance of the plate from the bottom surface (h2) that will minimize the force needed to pull the plate between the two oils at constant velocity.
The distance of the plate from the bottom surface (h2) that will minimize the force needed to pull the plate between the two oils 1.667 ± 2% cm. at constant velocity is Explanation: The following assumptions have been made here: 1. The thickness of the plate is negligible. 2. The velocity profile in each oil layer is linear. We measure vertical distance y from the lower plate. The total distance between the stationary plates is h = h1 + h2 = 5 cm, which is constant. Then, the distance of the moving plate is y from the lower plate and h – y from the upper plate, where y is a variable.
The shear forces acting on the upper and lower surfaces of the moving thin plate are
F shear, upper = τ w , upperA s = μ upperA s
F shear, lower = τ w , lowerA s = μ lowerA s
| | du dy
| | du dy
V
= μ upperA s h − y
= μ lowerA s
V y
Then, the total shear force acting on the plate becomes V
V
F = F shear, upper + F shear, lower = μ upperA s h − y + μ lowerA s h − y = A sV
(
μ upper h−y
+
μ lower y
)
The value of y that will minimize the force F is determined by setting dF/dy = 0: https://ezto.mheducation.com/hm.tpx
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Assignment Print View μ upper (h−y) 2
−
μ lower y2
y
= 0 → h−y =
√
μ lower μ upper
Solving for y and substituting, the value of y that minimizes the shear force is determined to be
√( ) (√ ( ) ) √( ) √( ) μ lower
μ upper
y=
1
h
4
( 5 cm )
=
1+
μ lower
μ upper
1+
= 1.667 cm
1 4
The distance of the plate from the bottom surface (h2) that will minimize the force needed to pull the plate between the two oils at constant velocity is 1.667 cm. References Worksheet
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Difficulty: Medium
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113.
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Award: 10.00 points
A cylinder of mass m slides down from rest in a vertical tube whose inner surface is covered by a viscous oil of film thickness h. If the diameter and height of the cylinder are D and L, respectively, derive an expression for the velocity of the cylinder as a function of time t.
V = mgh
μπDL
V = μmh
gπDL
V = mgh μπ
V = mgh πDL
The assumption made here is that the velocity profile in the oil film is linear.
https://ezto.mheducation.com/hm.tpx
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Assuming a linear velocity profile in the oil film, the drag force due to wall shear stress can be expressed as dV
V
F D = μ dy A = μ h πDL = kV where V is the instantaneous velocity of the cylinder and πDL
k=μ h
Applying Newton’s second law of motion for the cylinder, we write dV
mg - kV = m dt
where t is the time. This is a first-order linear equation and can be expressed in the standard form as follows: dV dt
k
+
m
V=g
V(0) = 0
with
whose solution is obtained to be
V (t ) =
mg k
(1 - e
)
( -k/m)t
As t→∞ the second term will be eliminated, giving
V (t ) =
mg k
which is constant. This constant is referred to as “limit velocity, ”. Rearranging for viscosity, we have mgh
V = μπDL
The expression for the velocity of the cylinder as a function of time, t is mgh
V = μπDL References Multiple Choice
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114.
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Award: 10.00 points
The magnitude of the pulling force at the surface of a liquid per unit length is called _____.
internal energy viscosity density surface tension The magnitude of the pulling force at the surface of a liquid per unit length is called surface tension σs.
The magnitude of the pulling force at the surface of a liquid per unit length is called surface tension σs. References Multiple Choice
115.
Difficulty: Easy
Award: 10.00 points
What is the cause of surface tension?
Attractive forces between the atoms Repulsive forces between the molecules Repulsive forces between the atoms Attractive forces between the molecules Surface tension is caused by the attractive forces between the molecules.
Surface tension is caused by the attractive forces between the molecules. References Multiple Choice
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Difficulty: Easy
102/182
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116.
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Award: 10.00 points
What is surface tension also known as?
The capillary effect Liquid tension The surface effect Surface energy Surface tension is also surface energy (per unit area) since it represents the stretching work that needs to be done to increase the surface area of the liquid by a unit amount.
The surface tension is also surface energy (per unit area) since it represents the stretching work that needs to be done to increase the surface area of the liquid by a unit amount. References Multiple Choice
117.
Difficulty: Easy
Award: 10.00 points
The rise or fall of a liquid in a small-diameter tube inserted into the liquid is known as the ____.
specific gravity capillary effect viscosity surface tension The capillary effect is the rise or fall of a liquid in a small-diameter tube inserted into the liquid.
The capillary effect is the rise or fall of a liquid in a small-diameter tube inserted into the liquid. References Multiple Choice
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Difficulty: Easy
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118.
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Award: 10.00 points
The capillary effect is caused by the net effect of the cohesive forces alone.
True False
The capillary effect is caused by the net effect of the cohesive forces (the forces between like molecules, like water) and adhesive forces (the forces between unlike molecules, like water and glass). Capillary effect caused by the net effect of the cohesive forces (the forces between like molecules, like water) and adhesive forces (the forces between unlike molecules, like water and glass). References True / False
119.
Difficulty: Easy
Award: 10.00 points
How is the capillary effect related to the contact angle?
The capillary effect is directly proportional to the cosine of the contact angle. The capillary effect is directly proportional to the tangent of the contact angle. The capillary effect is inversely proportional to the cosine of the contact angle. The capillary effect is directly proportional to the sine of the contact angle. The capillary effect is proportional to the cosine of the contact angle, which is the angle that the tangent to the liquid surface makes with the solid surface at the point of contact.
The capillary effect is proportional to the cosine of the contact angle. References Multiple Choice
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Difficulty: Easy
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120. Award: 10.00 points
A small-diameter tube is inserted into a liquid whose contact angle is 110°. What will happen to the level of liquid in the tube compared to the level of the rest of the liquid?
The liquid level in the tube will drop. The liquid level in the tube will remain the same. The liquid level in the tube will rise. The liquid level in the tube will drop, since the contact angle is greater than 90º and cos(110º) < 0.
The liquid level in the tube will drop, since the contact angle is greater than 90º and cos(110º) < 0. References Multiple Choice
121.
Difficulty: Easy
Award: 10.00 points
Consider a soap bubble. What is the pressure inside the bubble compared to the pressure outside?
The pressure inside a soap bubble is greater than the pressure outside. The pressure inside a soap bubble is the same as the pressure outside. The pressure inside a soap bubble is lesser than the pressure outside. The pressure inside a soap bubble is greater than the pressure outside, as evidenced by the stretch of the soap film.
The pressure inside a soap bubble is greater than the pressure outside, as evidenced by the stretch of the soap film. References Multiple Choice
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Difficulty: Easy
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122. Award: 10.00 points
Is the capillary rise greater in small- or large-diameter tubes?
In large-diameter tubes In small-diameter tubes The capillary rise is inversely proportional to the diameter of the tube, and thus capillary rise is greater in smaller-diameter tubes.
The capillary rise is greater in smaller-diameter tubes.
References Multiple Choice
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Difficulty: Easy
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123. Award: 10.00 points
Determine the gage pressure inside a soap bubble of diameter 8 cm at 20°C. Take the surface tension of soap water at 20°C as σs = 0.025 N/m. The gage pressure is
2.5 ± 2% Pa.
Explanation: The assumption made here is that the soap bubble is in atmospheric air.
The pressure difference between the inside and the outside of a bubble is given by ΔP bubble = P i − P 0 =
4σ s R
In the open atmosphere P0 = Patm, and thus is equivalent to the gage pressure. Substituting D = 8 cm: P i , gage = ΔP bubble =
4 ( 0.025 N/m )
( ) 0.08 2
= 2.5 N/m 2 = 2.5 Pa
m
The gage pressure is 2.5 Pa. References Worksheet
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Difficulty: Easy
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124. Award: 10.00 points
A 2.4-in-diameter soap bubble is to be enlarged by blowing air into it. Taking the surface tension of soap solution to be 0.0027 lbf/ft, determine the work input required to inflate the bubble to a diameter of 3 in. The surface tension of solution is given to be σs = 0.0027 lbf/ft. The work input required to inflate the bubble to a diameter of 3 in is
4.91 ± 2% × 10–7 Btu.
Explanation:
The work associated with the stretching of a film is the surface tension work and is expressed in differential form as δW s = σ sdA s. Noting that the surface tension is constant, the surface tension work is simply surface tension multiplied by the change in surface area:
(
(
)
2
2
W s = σ s A 2 − A 1 = 2πσ s D 2 − D 1
)
The factor of 2 is used because two surfaces are in contact with air. Substituting, the required work input is determined to be
W s = 2π(0.0027 lbf/ft)
(( ) ( ) )( 3
12
ft
2
−
2.4 12
ft
2
1 Btu 778.169 lbf ⋅ ft
)
= 4.91 × 10 - 7 Btu
The work input required to inflate the bubble to a diameter of 3 in is 4.91 × 10–7 Btu. References Worksheet
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Difficulty: Easy
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125. Award: 10.00 points
A 1.6-mm-diameter tube is inserted into an unknown liquid whose density is 960 kg/m3, and it is observed that the liquid rises 8.5 mm in the tube, making a contact angle of 15°. Determine the surface tension of the liquid. The surface tension of the liquid is
0.0331 ± 2% N/m.
Explanation: The following assumptions have been made here: 1. There are no impurities in the liquid and no contamination on the surfaces of the glass tube. 2. The liquid is open to the atmospheric air.
Substituting the numerical values, the surface tension is determined from the capillary rise relation as follows:
( 960 kg/m ) ( 9.81 m/s ) ( 3
ρgRh
σ s = 2cosφ =
2
0.0016 2
)
m (0.0085 m)
2(cos15 ° )
(
1N 1 kg ⋅ m/s 2
)
= 0.0331 N/m
The surface tension of the liquid is 0.0331 N/m. References Worksheet
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Difficulty: Medium
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126. Award: 10.00 points
Consider a 0.15-mm-diameter air bubble in a liquid. Determine the pressure difference between the inside and outside of the air bubble if the surface tension at the air-liquid interface is 0.12 N/m. The pressure difference between the inside and outside of the air bubble is
3.2 ± 2% kPa.
Explanation:
Considering that an air bubble in a liquid has only one interface, the pressure difference between the inside and outside of the bubble is determined from ΔP bubble = P i − P o =
2σ s R
Substituting, the pressure difference is ΔP bubble =
2 ( 0.12 N/m )
(
( 0.00015 m ) 2
)
= 3200 N/m 2 = 3.2 kPa
The pressure difference between the inside and outside of the air bubble is 3.2 kPa. References Worksheet
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Difficulty: Medium
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127. Award: 10.00 points
The surface tension of a liquid is to be measured using a liquid film suspended on a U-shaped wire frame with a(n) 8-cm-long movable side. If the force needed to move the wire is 0.028 N, determine the surface tension of this liquid in air. The surface tension of the liquid in air is
0.175 ± 2% N/m.
Explanation: The following assumptions have been made here: 1. There are no impurities in the liquid and no contamination on the surfaces of the wire frame. 2. The liquid is open to the atmospheric air.
Substituting the numerical values, the surface tension is determined from the surface tension force relation as follows: F
0.028 N
σ s = 2b = 2 ( 0.08 m ) = 0.175 N/m The surface tension depends on temperature. Therefore, the value determined is valid at the temperature of the liquid.
The surface tension of the liquid in air is 0.175 N/m. References Worksheet
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Difficulty: Medium
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128. Award: 10.00 points
A capillary tube of 1.25 mm diameter is immersed vertically in water exposed to the atmosphere. Determine how high water will rise in the tube. Take the contact angle at the inner wall of the tube to be 6° and the surface tension to be 1.00 N/m. Take the density os water to be 1000 kg/m3. The water will rise
0.324 ± 2% m in the tube.
Explanation: The following assumptions have been made here: 1. There are no impurities in water and no contamination on the surfaces of the tube. 2. The water is open to the atmospheric air. The capillary rise is determined to be 2σ s
h = ρgR cosΦ =
2 × ( 1 N/m ) × cos6 °
( 1000 kg/m ) × ( 9.81 m/s ) × ( 0.000625 m ) 3
2
= 0.324 m
The water will rise 0.324 m in the tube. References Worksheet
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Difficulty: Easy
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129. Award: 10.00 points
A 0.024-in-diameter glass tube is inserted into mercury, which makes a contact angle of 140° with glass. Determine the capillary drop of mercury in the tube at 68°F. The surface tension of mercury-glass in atmospheric air at 68°F (20°C) is obtained from Table 2-4: σs = (0.440 N/m)(0.22482 lbf/N)(0.3048 m/ft) = 0.030151 lbf/ft. To obtain the density of mercury, we interpolate from Table A-8E at 68°F, yielding ρ = 845.65 lbm/ft3. The contact angle is given to be 140°. The capillary drop of mercury is
0.656 ± 2% in.
Explanation: The following assumptions have been made here: 1. There are no impurities in mercury and no contamination on the surfaces of the glass tube. 2. The mercury is open to the atmospheric air.
Substituting the numerical values, the capillary drop is determined to be
h=
2σ scosϕ ρgR
( )( )
2 ( 0.030151 lbf/ft ) ( cos140° )
=
(
845.65 lbm/ft 3
)(
32.174 ft/s 2
0.012 12
ft
32.174 lbm ⋅ ft/s 2 1 lbf
)
= -0.05463 ft = -0.656 in
Thus, the capillary drop is 0.656 in.
The capillary drop of mercury is 0.656 in. References Worksheet
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Difficulty: Easy
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130. Award: 10.00 points
A capillary tube is immersed vertically in a water container. Knowing that water starts to evaporate when the pressure drops below 2.5 kPa, determine the maximum capillary rise and tube diameter for this maximum-rise case. Take the contact angle at the inner wall of the tube to be 6° and the surface tension to be 1.00 N/m. The maximum capillary rise is
10.07 ± 2% m.
The tube diameter for the maximum rise is
40.25 ± 2% μm.
Explanation: The following assumptions have been made here: 1. There are no impurities in water and no contamination on the surfaces of the tube. 2. The water is open to the atmospheric air. At the liquid side of the meniscus, P = 2.5 kPa. Therefore, the capillary rise would be h=
R=
P atm - P ρg
2σ s ρgh
=
( 101325-2500 ) Pa
= 10.07 m
( 1000 kg/m ) ( 9.81 m/s )
cos∅ =
3
2
2 × ( 1 N/m ) × cos6 °
( 1000 kg/m ) ( 9.81 m/s ) ( 10.07 m ) 3
2
≅ 2.01 × 10 - 5 m = 20.13 μm
Then, the tube diameter needed for this capillary rise is D = 2 × R = 2 × 20.13 μm D = 40.2539 μm The maximum capillary rise is 10.07 m. The tube diameter for the miximum-rise is 40.25 μm. References Worksheet
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Difficulty: Easy
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131.
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Award: 10.00 points
Contrary to expectations, a solid steel ball can float on water due to the surface tension effect. Determine the maximum diameter of a steel ball that would float on water at 10°C. What would your answer be for an aluminum ball? Take the densities of steel and aluminum balls to be 7800 kg/m3 and 2700 kg/m3, respectively. The surface tension of water at 10°C is σs = 0.0745 N/m (Table 2-4 by interpolation). Take the density of water to be 1000 kg/m3.
The maximum diameter of a steel ball that would float on water is 2.42 ± 2% mm. 4.11 ± 2% mm. The maximum diameter of an aluminum ball that would float on water is Explanation: The following assumptions have been made here: 1. The water is pure, and its temperature is constant. 2. The ball is dropped on the water slowly so that the inertial effects are negligible. 3. The contact angle is taken to be 0° for the maximum diameter.
The surface tension force and the weight of the ball can be expressed as F s = πDσ s and W = mg = ρgV =
ρgπD 3 6
When the ball floats, the net force acting on the ball in the vertical direction is zero. Therefore, setting Fs = W and solving for diameter D give D =
√
6σ s ρg
. Substituting the known quantities, the maximum diameters for the steel and aluminum balls
become
D steel =
D Al =
√ √( 6σ s
= ρg
√ √( 6σ s
= ρg
6 ( 0.0745 N/m ) 7800 kg/m 3
) ( 9.81 m/s ) 2
6 ( 0.0745 N/m ) 2700 kg/m 3
)(
9.81 m/s 2
)(
(
1 kg ⋅ m/s 2 1N
1 kg ⋅ m/s 2 1N
)
)
= 2.42 × 10 − 3 m = 2.42 mm
= 4.11 × 10 − 3 m = 4.11 mm
The maximum diameter of a steel ball that would float on water is 2.42 mm. The maximum diameter of an aluminum ball that would float on water is 4.11 mm. References
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Worksheet
Difficulty: Easy
132. Award: 10.00 points
Nutrients dissolved in water are carried to upper parts of plants by tiny tubes partly because of the capillary effect. Determine how high the water solution will rise in a tree in a 0.0026-mm-diameter tube as a result of the capillary effect. Treat the solution as water at 20°C with a contact angle of 30°. The surface tension of water at 20°C is σs = 0.073 N/m. Take the density of water to be 1000 kg/m3. The water solution will rise
9.91 ± 2% m in the tree.
Explanation: The following assumptions have been made here: 1. The solution can be treated as water with a contact angle of 30°. 2. The diameter of the tube is constant. 3. The temperature of the water solution is 20°C.
Substituting the numerical values, the capillary rise is determined to be
h=
2σ scosφ ρgR
=
2 ( 0.073 N/m ) ( cos30 ° )
( 1000 kg/m ) ( 9.81 m/s ) ( 1.3 × 10 3
2
−6
m
)
(
1 kg ⋅ m/s 2 1N
)
= 9.91 m
The water solution will rise 9.91 m in the tree. References Worksheet
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Difficulty: Easy
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06/06/2024, 11:03
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133. Award: 10.00 points
Consider a 55-cm-long journal bearing that is lubricated with oil whose viscosity is 0.1 kg/m·s at 20°C at the beginning of operation and 0.008 kg/m·s at the anticipated steady operating temperature of 80°C. The diameter of the shaft is 8 cm, and the average gap between the shaft and the journal is 0.08 cm. Determine the torque needed to overcome the bearing friction initially and during steady operation when the shaft is rotated at 1500 rpm. The torque required to overcome the bearing friction is The torque required during steady operation is
4.34 ± 2% N·m.
0.347 ± 2% N·m.
Explanation: The following assumptions have been made here: 1. The gap is uniform and is completely filled with oil. 2. The end effects on the sides of the bearing are negligible. 3. The fluid is Newtonian.
The radius of the shaft is R = 0.04 m. Substituting the given values, the torque is determined as follows: At the beginning of operation at 20°C, .
T=μ
4π 2R 3nL l
(
)
4π 2 ( 0.04 m ) 3 1500/60 s − 1 ( 0.55 m )
= (0.1 kg/m ⋅ s)
= 4.34 N ⋅ m
0.0008 m
During steady operation at 80°C, .
T=μ
4π 2R 3nL l
(
)
4π 2 ( 0.04 m ) 3 1500/60 s − 1 ( 0.55 m )
= (0.008 kg/m ⋅ s)
0.0008 m
= 0.347 N ⋅ m
The torque required to overcome the bearing friction is 4.34 N·m. The torque required during steady operation is 0.347 N·m. References Worksheet
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Difficulty: Medium
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134. Award: 10.00 points
The diameter of one arm of a U-tube is 10 mm while the other arm is large. If the U-tube contains some water and both surfaces are exposed to atmospheric pressure, determine the difference between the water levels in the two arms. The surface tension and density of water at 20°C are σs = 0.073 N/m and ρ = 1000 kg/m3. The difference between the water levels in the two arms is
2.98 ± 2% mm.
Explanation: The following assumptions have been made here: 1. Both arms of the U-tube are open to the atmosphere. 2. The water is at room temperature. 3. The contact angle of the water is zero.
Any difference in water levels between the two arms is due to surface tension effects and thus capillary rise. Noting that the capillary rise in a tube is inversely proportional to tube diameter, there will be no capillary rise in the arm with a large diameter. Then, the water level difference between the two arms is simply the capillary rise in the smaller diameter arm.
h =
2σ scos Φ ρgR
=
2 ( 0.073 N/m ) ( cos 0 ° )
( 1000 kg/m ) ( 9.81 m/s ) ( 5 × 10 m 3
2
−3
)(
1 kg ⋅ m/s 2 1N
)( ) 1000 mm 1m
= 2.98 mm
The difference between the water levels in the two arms is 2.98 mm. References Worksheet
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Difficulty: Medium
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135. Award: 10.00 points
A rigid tank contains 40 lbm of air at 20 psia and 70°F. More air is added to the tank until the pressure and temperature rise to 51 psia and 90°F, respectively. Determine the amount of air added to the tank. The gas constant of air is ft ⋅ lbf
R u = 53.34 lbm ⋅ R
(
1 psia 144 lbf/ft 2
)
psia ⋅ ft 3
= 0.3794 lbm ⋅ R . The air temperature is 70ºF = 70 + 459.67 = 529.67 R.
The amount of air added to the tank is
58.2 ± 2% lbm.
Explanation: The following assumptions have been made here: 1. At specified conditions, air behaves as an ideal gas. 2. The volume of the tank remains constant.
Treating air as an ideal gas, the initial volume and the final mass in the tank are determined to be
V=
m 1RT 1
m2 =
P1
P 2V RT 2
(
)
( 40 lbm ) 0.3704 psia ⋅ ft 3/lbm ⋅ R ( 529.67 R )
=
20 psia
(
( 51 psia ) 392.380 ft 3
=
)
( 0.3704 psia ⋅ ft /lbm ⋅ R ) ( 550 R ) 3
= 392.380 ft 3
= 98.23 lbm
Thus, the amount of air added is Δm = m 2 − m 1 = 98.23 lbm − 40.0 lbm = 58.23 lbm ≅ 58.2 lbm
The amount of air added to the tank is 58.2 lbm. References Worksheet
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Difficulty: Easy
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136. Award: 10.00 points
A 10-m3 tank contains nitrogen at 25°C and 800 kPa. Some nitrogen is allowed to escape until the pressure in the tank drops to 550 kPa. If the temperature at this point is 20°C, determine the amount of nitrogen that has escaped. The amount of nitrogen that has escaped is
27.2 ± 2% kg.
Explanation: The assumption made here is that the tank is insulated so that no heat is transferred.
Treating N2 as an ideal gas, the initial and the final masses in the tank are determined to be
m1 =
m2 =
P 1V RT 1
P 2V RT 2
(
( 800 kPa ) 10 m 3
=
( 0.2968 kPa ⋅ m /kg ⋅ K ) ( 298 K ) 3
(
( 550 kPa ) 10 m 3
=
) )
( 0.2968 kPa ⋅ m /kg ⋅ K ) ( 293 K ) 3
= 90.45 kg
= 63.25 kg
Thus, the amount of N2 that escapes is Δm = m 1 − m 2 = 90.45 kg − 63.25 kg = 27.2 kg
The amount of nitrogen that has escaped is 27.2 kg. References Worksheet
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Difficulty: Easy
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137. Award: 10.00 points
The absolute pressure of an automobile tire is measured to be 320 kPa before a trip and 351 kPa after the trip. Assuming the volume of the tire remains constant at 0.022 m3, determine the percent increase in the absolute temperature of the air in the tire. The percent increase in the absolute temperature of the air in the tire is
9.69 ± 2% %.
Explanation: The following assumptions have been made here: 1. The volume of the tire remains constant. 2. Air is an ideal gas. Noting that air is an ideal gas and the volume is constant, the ratios of absolute temperatures after and before the trip are P 1V 1 T1
=
P 2V 2 T2
T2
P2
1
1
351 kPa
→ T = P = 320 kPa = 1.0969
Therefore, the absolute temperature of air in the tire will increase by 9.69% during this trip.
The percent increase in the absolute temperature of the air in the tire is 9.69%. References Worksheet
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Difficulty: Easy
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138. Award: 10.00 points
The analysis of a propeller that operates in water at 65°F shows that the pressure at the tips of the propeller drops to 0.1 psia at high speeds. Determine if there is a danger of cavitation for this propeller. The vapor pressure of water at 65°F is 0.3058 psia. There is a danger
of cavitation for this propeller.
Explanation: To avoid cavitation, the pressure everywhere in the flow should remain above the vapor (or saturation) pressure at the given temperature, which is P v = P sat@65 ° F = 0.3058 psia The minimum pressure in the pump is 0.1 psia, which is less than the vapor pressure. Therefore, there is a danger of cavitation in the pump.
There is a danger of cavitation for this propeller. References Worksheet
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Difficulty: Easy
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06/06/2024, 11:03
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139. Award: 10.00 points
A closed tank is partially filled with water at 60°C. A mixture of liquid water and vapor remains in the tank. If the air above the water is completely evacuated, determine the absolute pressure in the evacuated space. Assume the temperature to remain constant. The saturation pressure of water at 60°C is 19.947 kPa. The absolute pressure in the evacuated space is
19.95 ± 2% kPa.
Explanation: When air is completely evacuated, the vacated space is filled with water vapor, and the tank contains a saturated water-vapor mixture at the given pressure. Since we have a two-phase mixture of a pure substance at a specified temperature, the vapor pressure must be the saturation pressure at this temperature. That is, P v = P sat@60 ° C = 19.95 kPa
The absolute pressure in the evacuated space is 19.95 kPa. References Worksheet
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Difficulty: Easy
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06/06/2024, 11:03
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140. Award: 10.00 points
The specific gravities of solids and carrier fluids of a slurry are usually known, but the specific gravity of the slurry depends on the concentration of the solid particles. Choose the correct equation that shows that the specific gravity of a water-based slurry can be expressed in terms of the specific gravity of the solid SGs and the mass concentration of the suspended solid particles Cs, mass.
( )
1
SG s − 1
SG m = 1 + C
SG
s, mass
1
m =
( ) 1
1 - C s, mass SG − 1 s
SG
1
m =
( ) 1
1 + C s, mass SG − 1 s
SG
1
m =
( ) 1
C s, mass SG +1 s
The following assumptions have been made here: 1. The solid particles are distributed uniformly in water so that the solution is homogeneous. 2. The effect of dissimilar molecules on each other is negligible. Consider solid particles of mass ms and volume Vs dissolved in a fluid of mass mf and volume Vm. The total volume of the suspension (or mixture) is Vm = Vf + Vs. Dividing by Vm gives
() ( ) ms
1=
Vs Vm
+
Vf Vm
Vf
→
Vm
=1−
Vs Vm
ρs
=1−
=1−
mm
ms ρm mm ρs
= 1 − C s , mass
SG m SG s
(1)
ρm
since ratio of densities is equal to the ratio of specific gravities, and ms/mm = Cs, mass. The total mass of the suspension (or mixture) is mm = ms + mf . Dividing by mm and using the definition give mf
1 = C s , mass + m
m
ρ fV f
= C s , mass + ρ V
m m
→
ρm ρf
=
Vf
( 1 − Cs , mass ) Vm
(2)
Taking the fluid to be water so that ρm/ρf = SGm and combining equations 1 and 2 give SG m =
1 − C s , massSG m / SG s 1 − C s , mass
Solving for SGm and rearranging give SG m =
1
( ) 1
1 + C s, mass SG − 1 s
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1
The correct equation is SG m =
( )
.
1
1 + C s, mass SG − 1 s
References Multiple Choice
Difficulty: Easy
A rigid tank contains an ideal gas at 300 kPa and 570 K. Half of the gas is withdrawn from the tank, and the gas is at 100 kPa at the end of the process. References Section Break
141.
Difficulty: Easy
Award: 10.00 points
Determine the final temperature of the gas. (Please provide an answer before moving to the next part.) The final temperature of the gas is
380 ± 2% K.
Explanation:
This is a constant volume process. When half of the gas is withdrawn from the tank, the final temperature may be determined from the ideal gas relation as
T2 =
m1 P2
(
100 kPa
T = (2) m2 P1 1 300 kPa
)
(570 K) = 380 K
The final temperature of the gas is 380 K. References Worksheet
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Difficulty: Easy
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142. Award: 10.00 points
Determine the final pressure if no mass were withdrawn from the tank and the same final temperature were reached at the end of the process. The final pressure in the tank is
200 ± 2% kPa.
Explanation:
This is a constant volume and constant mass process. The ideal gas relation for this case yields
P2 =
T2 T1
P1 =
( ) 380 K 570 K
(300 kPa) = 200 kPa
The final pressure in the tank is 200 kPa. References Worksheet
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Difficulty: Easy
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06/06/2024, 11:03
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143. Award: 10.00 points
The composition of a liquid with suspended solid particles is generally characterized by the fraction of solid particles either by weight or mass, Cs, mass = ms/mm, or by volume, Cs,vol = Vs /Vm, where m is the mass and V is the volume. The subscripts s and m indicate solid and mixture, respectively. Choose the right expression for the specific gravity of a water-based suspension in terms of Cs, mass and Cs, vol.
SG SG
ρm
m = ρ f
ρm
m = ρf
=
=
1 − C s,mass 1 − C s,vol 1 + C s,mass 1 + C s,vol
C s,vol
ρm
SG m = ρ = C f s,mass ρm
SG m =
ρf
=
1 − C s,vol 1 − C s,mass
The following assumptions have been made here: 1. The solid particles are distributed uniformly in water so that the solution is homogeneous. 2. The effect of dissimilar molecules on each other is negligible. Consider solid particles of mass ms and volume Vs dissolved in a fluid of mass mf and volume Vm. The total volume of the suspension (or mixture) is
Vm = Vs + Vf Dividing by Vm and using the definition Cs, vol = Vs/Vm give 1 = C s , vol +
Vf Vm
→
Vf Vm
= 1 − C s , vol
(1)
The total mass of the suspension (or mixture) is
mm = ms + mf Dividing by mm and using the definition Cs, mass = ms/mm give mf
1 = C s , mass + m
m
ρ fV f
= C s , mass + ρ V
m m
ρf
→ ρ
m
(
)
Vm
= 1 − C s , mass V
f
(2)
Combining equations 1 and 2 gives ρf ρm
=
1 − C s , mass 1 − C s , vol
When the fluid is water, the ratio ρf /ρm is the inverse of the definition of specific gravity. Therefore, the desired relation for the specific gravity of the mixture is SG m =
ρm ρf
=
1 − C s , vol 1 − C s , mass
which is the desired result.
The desired relation for the specific gravity of the mixture is SG m =
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ρm ρf
=
1 − C s , vol 1 − C s , mass
.
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References Multiple Choice
Difficulty: Easy
144. Award: 10.00 points
The variation of the dynamic viscosity of water with absolute temperature is given as
Using these tabulated data, develop a relation for viscosity in the form of μ =μ(T ) = A + BT + CT2 + DT3 + ET4. Using the relation developed, predict the dynamic viscosity of water at 50°C at which the reported value is 5.468 × 10−4 Pa⋅s. Compare your result with the results of Andrade’s equation, which is given in the form of μ = D⋅eB/T, where D and B are constants whose values are to be determined using the viscosity data given. Use appropriate software to solve this problem. Please upload your response using the controls provided below.
Student upload controls will be shown to students when they take this assignment.
References File Attachment Question
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Difficulty: Medium
128/182
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Assignment Print View
145. Award: 10.00 points
A newly produced pipe with diameter 3 m and length 18 m is to be tested at 10 MPa using water at 15°C. After sealing both ends, the pipe is first filled with water and then the pressure is increased by pumping additional water into the test pipe until the test pressure is reached. Assuming no deformation in the pipe, determine how much additional water needs to be pumped into the pipe. Take the coefficient of compressibility to be 2.10 × 109 Pa. The amount of additional water that needs to be pumped into the pipe is
606 ± 2% kg.
Explanation: The assumption made here is that there is no deformation in the pipe. From Eq. 2-13, we have κ≅
∆P
( )
∆ρ
→
ρ
∆ρ
=
∆P ρ
or
ρ2 - ρ1 ρ1
=
∆P κ
ρ
from which we write
(
ρ2 = ρ1 1 +
∆P κ
) (
= 1000 kg/m 3
(
) 1+
10 × 10 6 Pa 2.10 × 10 9 Pa
)
= 1004.76 kg/m 3
Then, the amount of additional water is m = V cyl ∆ ρ =
πD 2 4
L∆ρ =
π(3 m) 2 4
(
)
× (18 m) × 1004.76 kg/m 3 - 1000 kg/m 3 = 606 kg
The amount of additional water that needs to be pumped into the pipe is 606 kg. References Worksheet
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Difficulty: Easy
129/182
06/06/2024, 11:03
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146. Award: 10.00 points
Choose the correct equation for the coefficient of volume expansion in terms of T.
β
2
ideal gas = T
β ideal gas = 2T β ideal gas = T β
1 ideal gas = T
The assumption made here is that the temperature and pressure are in the range where the gas can be approximated as an ideal gas. The ideal gas law is
P = ρRT which we rewrite as P
ρ = RT . By definition, 1
β= −ρ
( ) ∂ρ ∂T
P
.
Thus, substitution and differentiation yield
( ) ( ) P
1
β ideal gas = − ρ
∂ RT ∂T
1
P
( )
= −ρ −
P
RT 2
ρ 1
1
= ρT = T
1
The correct equation for coefficient of volume expansion in terms of T is β ideal gas = T . References Multiple Choice
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Difficulty: Easy
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Although liquids, in general, are hard to compress, the compressibility effect (variation in the density) may become unavoidable at the great depths in oceans due to enormous pressure increase. At a certain depth, the pressure is reported to be 100 MPa and the average coefficient of compressibility is about 2100 MPa. References Section Break
Difficulty: Easy
147. Award: 10.00 points
Taking the liquid density at the free surface to be ρ0 = 1030 kg/m3, obtain an analytical relation between density and pressure, and determine the density at the specified pressure.(Please provide an answer before moving on to the next part.) The density at the specified pressure is
1080 ± 2% kg/m3.
Explanation: dP
κ=
dρ
→
( )
ρ
dρ
=
dρ κ
ρ
Integrating, ∫ρ
dρ
ρ0 ρ
= ∫P 0
dP κ
→ ln
ρ ρ0
=
P κ
→ ρ = ρ 0e P / κ
With the given data, we obtain
(
ρ = 1030 kg/m 3
)×e
( 100 MPa/2100 MPa )
= 1080 kg/m 3
The density at the specified pressure is 1080 kg/m3. References Worksheet
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Difficulty: Easy
131/182
06/06/2024, 11:03
Assignment Print View
148. Award: 10.00 points
Use κ ≅ -
∆P
≅
∆P
( ) ( ) ∆v
∆ρ
v
ρ
to estimate the density for the specified pressure and compare your result with that of part (a).
1079 ± 2% kg/m3.
The density at the specified pressure is Explanation: Eq. 2-13 can be rearranged to give ∆P
∆ρ ≅ ρ κ or
ρ - ρ0 ≅ ρ0
P - P0 κ
→ ρ ≅ ρ0 + ρ0
P - P0 κ
(
) ( (1030 kg/m ) ×
= 1030 kg/m 3 +
3
100 MPa 2100 MPa
)
= 1079 kg/m 3
The density at the specified pressure is 1079 kg/m3. References Worksheet
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Difficulty: Easy
132/182
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149. Award: 10.00 points
Air expands isentropically from 204 psia and 240°F to 60 psia. Calculate the ratio of the initial to final speed of sound. The properties of air are R = 0.06855 Btu/lbm·R and k = 1.4. The ratio of the initial to final speed of sound is
1.19 ± 2% .
Explanation: The assumption made here is that air is an ideal gas with constant specific heats at room temperature. The final temperature of air is determined from the isentropic relation of ideal gases:
T2 = T1
() P2
(k−1) k
P1
= (240 + 460 R)
(
60 psia 204 psia
)
( 1.4−1 ) 1.4
= 493.5 R
Treating k as a constant, the ratio of the initial to final speed of sound can be expressed as
Ratio =
c1 c2
=
√k1RT1 √
k 2RT 2
=
√T 1 √
T2
=
√240 + 460 R √493.5 R
= 1.19
The ratio of the initial to final speed of sound is 1.19. References Worksheet
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Difficulty: Easy
133/182
06/06/2024, 11:03
Assignment Print View
150. Award: 10.00 points
A shaft with a diameter of D = 80 mm and a length of L = 380 mm, shown in the figure below, is pulled with a constant velocity of U = 5 m/s through a bearing with variable diameter. The clearance between shaft and bearing, which varies from h1 = 1.2 mm to h2 = 0.4 mm, is filled with a Newtonian lubricant whose dynamic viscosity is 0.10 Pa⋅s. Determine the force required to maintain the axial movement of the shaft.
The force required to maintain the axial movement of the shaft is
65.6 ± 2% N.
Explanation: The assumption made here is that the fluid is Newtonian.
The varying clearance h can be expressed as a function of the axial coordinate x (see the figure). According to this sketch, we obtain
(
h = h1 − h1 − h2
x
)L
Assuming a linear velocity distribution in the gap, the viscous force acting on the differential strip element is U
dF = dA = μ h × πDdx =
μUπD h1 −
x dx
( h1 − h2 ) L
Integrating,
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Assignment Print View
F = μUπD∫ xx == L0
h1 −
(
ln h 1 −
dx x
( h1 − h2 ) L
= − μUπD
( h1 − h2 ) L ) x
x=L
=
( h1 − h2 ) L
μUπDL h1 − h2
ln
h1 h2
x=0
For the given data, we obtain
(
( 0.1 Pa · s ) ( 5 m / s ) π 80 × 10 − 3 m
F=
) ( 380 × 10 m ) 1 . 2
( 1.2 − 0.4 ) × 10 − 3 m
−3
ln 0 . 4 ≅ 65.6 N
The force required to maintain the axial movement of the shaft is 65.6 N. References Worksheet
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Difficulty: Easy
135/182
06/06/2024, 11:03
151.
Assignment Print View
Award: 10.00 points
A shaft with a diameter of D = 80 mm and a length of L = 400 mm, shown in the figure below, rotates with a constant angular speed of n = 1650 rpm in a bearing with variable diameter. The clearance between shaft and bearing, which varies from h1 = 1.2 mm to h2 = 0.4 mm, is filled with a Newtonian lubricant whose dynamic viscosity is 0.1 Pa⋅s. Determine the torque required to maintain the motion.
The torque required to maintain the motion is
3.817 ± 2% N·m.
Explanation: The assumption made here is that the fluid is Newtonian.
According to this sketch, we obtain
(
h = h1 − h1 − h2
x
)L
Assuming a linear velocity distribution in the gap, the viscous force acting on the differential strip element is U
dF = τdA = μ h × πDdx =
μUπD h1 −
x dx
( h1 − h2 ) L
where U = ω(D/2) = 2nπ/60(D/2) in this case. Then, the viscous torque developed on the shaft is
D
dT = dF × 2 =
μ
(
2nπ 60
h1 −
D
×2
)
D
πD × 2 x
( h1 − h2 ) L
dx =
μnπ 2D 3 120
dx h1 −
x
( h1 − h2 ) L
Integrating,
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06/06/2024, 11:03
Assignment Print View 2 3
T=
μnπ D 120
dx
x=L
∫x = 0
h1 −
(
ln h 1 −
2 3
= − x
( h1 − h2 ) L
μnπ D 120
( h1 − h2 ) L ) x
x=L
=
( h1 − h2 ) L
1 μnπ 2D 3L 120 h 1 − h 2
ln
h1 h2
x=0
For the given data, we obtain 1
(
( 0.1 Pa·s ) ( 1650 rpm ) π 2 80 × 10 − 3 m
T = 120
) ( 400 × 10 m ) 1 . 2 3
−3
( 1.2 − 0.4 ) × 10 − 3 m
ln 0 . 4 ≅ 3.817 N · m
The torque required to maintain the motion is 3.817 N·m References Worksheet
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137/182
06/06/2024, 11:03
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152. Award: 10.00 points
Choose the correct relation for the capillary rise of a liquid between two large parallel plates at distance t apart inserted into the liquid vertically. Take the contact angle to be φ.
h= h=
2σ scosφ ρgt 4σ scosφ ρgt
h = 2σ scosφ ρ
h = 2σ ssinφ ρgt
The assumption made here is that there are no impurities in the liquid and no contamination on the surfaces of the plates.
The magnitude of the capillary rise between two large parallel plates can be determined from a force balance on the rectangular liquid column of height h and width w between the plates. The bottom of the liquid column is at the same level as the free surface of the liquid reservoir, and thus the pressure there must be atmospheric pressure. This will balance the atmospheric pressure acting from the top surface, and thus these two effects will cancel each other. The weight of the liquid column is
W = mg = ρgV = ρg(w × t × h) Equating the vertical component of the surface tension force to the weight gives
W = F surface → ρg(w × t × h) = 2wσ scosφ Canceling w and solving for h, the capillary rise is found to be
h=
2σ scosφ ρgt
The correct relation for the capillary rise of a liquid between two large parallel plates is h =
2σ scosφ ρgt
.
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Assignment Print View
A 10-cm-diameter cylindrical shaft rotates inside a 44-cm-long and 10.3-cm-diameter bearing. The space between the shaft and the bearing is completely filled with oil whose viscosity at the anticipated operating temperature is 0.300 N·s/m2. References Section Break
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153. Award: 10.00 points
Determine the power required to overcome friction when the shaft rotates at a speed of 600 rpm. The power required to overcome friction is
273 ± 2% W.
Explanation: The following assumptions have been made here: 1. The gap is uniform and is completely filled with oil. 2. The end effects on the sides of the bearing are negligible. 3. The fluid is Newtonian.
The radius of the shaft is R = 0.05 m, and the thickness of the oil layer is l =
( 10 . 3 cm - 10 cm ) 2
= 0.15 cm. The power-torque
relationship is given as follows: Ẇ = ωT = 2πṅT where T = μ
4π 2R 3ṅL l
The required power to overcome friction is determined as follows:
Ẇ = μ
8π 3R 3ṅ 2L l
=
(0.3 N ⋅ s/m )
(
8π 3 ( 0.05 m ) 3 600/60 s − 1
2
0.0015 m
) ( 0.44 m ) 2
(
1 W 1 N ⋅ m/s
)
= 273 W
The power required to overcome friction is 273 W. References Worksheet
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154. Award: 10.00 points
Determine the power required to overcome friction when the shaft rotates at a speed of 1200 rpm. The power required to overcome friction is
1091 ± 2% W.
Explanation: The following assumptions have been made here: 1. The gap is uniform and is completely filled with oil. 2. The end effects on the sides of the bearing are negligible. 3. The fluid is Newtonian.
The radius of the shaft is R = 0.05 m, and the thickness of the oil layer is l =
( 10 . 3 cm - 10 cm ) 2
= 0.15 cm. The power-torque
relationship is given as follows: Ẇ = ωT = 2πṅT where T = μ
4π 2R 3ṅL l
The power required to overcome friction is determined as follows:
Ẇ = μ
8π 3R 3ṅ 2L l
=
(0.3 N ⋅ s/m )
(
8π 3 ( 0.05 m ) 3 1200/60 s − 1
2
0.0015 m
) ( 0.44 m ) 2
(
1 W 1 N ⋅ m/s
)
= 1091 W
The power required to overcome friction is 1091 W. References Worksheet
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155. Award: 10.00 points
A large plate is pulled at a constant speed of U = 4 m/s over a fixed plate on 8-mm-thick engine oil film at 20°C. Assuming a half-parabolic velocity profile in the oil film, while the flow is developing as illustrated, determine the shear stress on the upper plate and its direction. Repeat for the linear profile (dashed line) that develops after a long time. (Take the dynamic viscosity of oil as 0.8374 Pa·s.)
209.35 ± 2% N/m2.
The shear stress for a parabolic velocity profile is
419 ± 2% N/m2.
The shear stress for a linear velocity profile is The linear assumption is unrealistic
.
Explanation: The assumption made here is that the thickness of the plate is negligible. Considering a parabolic profile, we would have V2 = ky, where k is a constant. Since V = U = 4 m/s when y = h = 8 mm = 8 × 10-3 m, we write
( )
m 2
4 s
(
= k × 8 × 10 − 3 m
) → k = 2000 m /s 2
Then, the velocity profile becomes V 2 = 2000y → V = 44.721√y Assuming Newtonian behavior, the shear stress on the upper wall is determined as follows:
τ =μ
dV dy
=μ
d dy
( ) 1
(44.721√y ) = μ(44.721) 2√y
y=h
To determine the shear stress at the upper layer, we consider y = 8 × 10 − 3.
τ = μ(44.721)
( ) 1
2 √h
= (0.8374 Pa ⋅ s)(44.721)
(
1
√
2 8 × 10 - 3 m
)
τ = 209.35 N/m 2 If we assume a linear profile,
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Assignment Print View
=
U h
=
4 m/s 8 × 10 − 3 m
= 500 s − 1
Then, the shear stress is determined as follows: dV
U
τ = μ dy = μ h = (0.8374 Pa ⋅ s)(500) ≈ 419 N/m 2 Therefore, we conclude that the linear assumption is not realistic since it is an over prediction.
The shear stress if a parabolic velocity profile is 209.35 N/m2. The shear stress for a linear velocity profile is 418.7 N/m2. The linear assumption is not realistic. References Worksheet
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156. Award: 10.00 points
Some rocks or bricks contain small air pockets in them and have a spongy structure. Assuming the air spaces form columns of an average diameter of 5 μm, determine how high water can rise in such a material. Take the surface tension of the air–water interface in that material to be 0.085 N/m. The capillary rise in such a material is
6.93 ± 2% m.
Explanation: The following assumptions have been made here: 1. The interconnected air pockets form a cylindrical air column. 2. The air columns are open to the atmospheric air. 3. The contact angle of water is zero, Φ = 0.
Substituting the numerical values, the capillary rise is determined to be
h=
2σ scosφ ρgR
=
2 ( 0.085 N/m ) ( cos0 ° )
( 1000 kg/m ) ( 9.81 m/s ) ( 2.5 × 10 m ) 3
2
−6
(
1 kg ⋅ m/s 2 1N
)
= 6.93 m
The capillary rise in such a material is 6.93 m.
The capillary rise in such a material is 6.93 m. References Worksheet
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A fluid between two very long parallel plates is heated in a way that its viscosity decreases linearly from 0.90 Pa⋅s at the lower plate to 0.50 Pa⋅s at the upper plate. The spacing between the two plates is 0.4 mm. The upper plate moves steadily at a velocity of 10 m/s, in a direction parallel to both plates. The pressure is constant everywhere, and the fluid is Newtonian and assumed incompressible. Neglect gravitational effects.
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Assignment Print View
157. Award: 10.00 points
Choose the correct fluid velocity equation for u as a function of y, u(y), where y is the vertical axis perpendicular to the plates. (Please provide an answer before moving on to the next part.)
(
10
0.9
u(y) = ln 9 ln 0 . 9 − 1000y
u (y) =
)
( ()
0 . 9 − 1000y
( ()
1 . 0 − 1000y
( ()
0 . 9 − 1000y
10
9
ln
0.9
ln 5
u (y) =
10
9
ln
1.0
ln 5
u (y) =
10
5
ln
)
0.9
ln 9
)
)
The following assumptions have been made here: 1. The flow is parallel to plates. 2. The flow is onedimensional. 3. The Pressure is constant. 4. The fluid is Newtonian and incompressible. 5. The gravitational effect is neglected. Taking an infinitesimal fluid element and applying a force balance (assuming one-dimensional flow),
The force equilibrium in the x direction yields dτ dy
= 0 → τ = constant
(1)
Then, Newton’s law of viscosity yields
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Assignment Print View du
τ = μ dy = constant
(2)
The viscosity changes linearly with respect to y. Thus, it can be expressed in the form μ = Ay + B, where A and B are constants. Using (i) at y = 0 and μ = 0.090 Pa·s and (ii) at y = 0.0004 m and μ = 0.50 Pa·s, A and B can be determined and the viscosity function is found as μ = 0.90 – 1000y. Substituting the viscosity function in equation (3) yields τ= μ
du dy
= (0. 9 − 1000y)
du
du
= constant = C →
dy
dy
=
C
(3)
0 . 9 − 1000y
dy
u = C ∫ 0 . 9 − 1000y + D u = E ln(0. 9 − 1000y) + D
(4)
where E, D are integration constants to be determined. Using the no-slip boundary conditions (i) at y = 0 and u = 0 and, (ii) at y = 0.0004 m and u = 10 m/s, E and D can be determined and the velocity function is found as
( () 10
u (y) =
9
0.9
ln 0 . 9 − 1000y
ln 5
)
The correct fluid velocity function is u(y) =
( ()
10
9
0.9
)
ln 0 . 9 − 1000y .
ln 5
References Multiple Choice
Difficulty: Easy
158. Award: 10.00 points
Calculate the value of the shear stress. The shear stress is
17013 ± 2% Pa.
Explanation: Using Newton’s law of viscosity, du
τ = μ dy = (0. 9 − 1000y) ln
10
( 1000 ) ( 0 . 9 )
0 . 9 − 1000y
()
( 0 . 9 − 1000y ) 2
0.9
9
= 17013 Pa
5
The shear stress is 17013 Pa. References Worksheet
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Assignment Print View
159. Award: 10.00 points
The rotating parts of a hydroelectric power plant having power capacity Ẇ have a rotational synchronous speed ṅ. The weight of the rotating parts (the hydroturbine and its electric generator)is supported in a thrust bearing having annular form between D and d diameters as sketched. The thrust bearing is operated with a very thin oil film of thickness e and dynamic viscosity µ. It is assumed that the oil is a Newtonian fluid and the velocity is approximated as linear in the bearing. Calculate the ratio of lost power in the thrust bearing to the produced power in the hydraulic power plant. Use Ẇ = 48.6 MW, 0.035 Pa·s, ṅ = 640 rpm, e = 0.25 mm, D = 3.2 m, and d = 2.4 m.
The ratio of lost power in the thrust bearing to the produced power in the hydraulic power plant is
9.89 ± 2% %.
Explanation: The following assumptions have been made here: 1. The fluid is Newtonian and incompressible. 2. The linear velocity assumption holds true in the bearing. 3. The gravitational effect is neglected.
Due to the linear velocity assumption holding true in the bearing, the shear stress in the position r = r is
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τ= μ
du dr
=μ
∆u ∆r
= μ
U− 0
=μ
e
U
=μ
e
ωr
(1)
e
Angular velocity, ω = 2πn/60 τ=
2 μπnr 60 e
=
μπnr
(2)
30 e
The friction force on the rotating differential surface dA = r·dr·dθ, is dF s = τ · dA = τ · r · dθ · dr
(3)
The torque on the rotating axis of this force is dT s = r · dF s
(4)
Integrating equation (4) The total Torque is μπn
2π r T s = 30 e ∫ 0 dθ∫ r 2r 3dr = 1
2 μπ 2n
(r − r )
30 e
4
4 2
4 1
μπ 2n
(
4
4
= 60 e r 2 − r 1
)
(5)
and the total friction power (loss power) is P s = ω. T s =
2πn μπ 2n 60 60 e
(r − r ) = 4 2
4 1
μπ 3n 2 1800 e
(
(r − r ) 4 2
( 0.038 ) π 3 ( 640 rpm ) 2 ( 1.6 m ) 4 − ( 1.2 m ) 4
Ps =
(
( 1800 ) × 0.25 × 10 − 3 m
)
)
4 1
= 4804616 W = 4.805 MW
The ratio of lost power in the thrust bearing to the produced power in the hydraulic power plant will be Ps P
=
4.805 MW 48 . 6 MW
= 0.09886 ≅ 9.89%
The ratio of lost power in the thrust bearing to the produced power in the hydraulic power plant is 9.89%. The ratio of lost power in the thrust bearing to the produced power in the hydraulic power plant is 9.89%. References Worksheet
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Assignment Print View
The viscosity of some fluids changes when a strong electric field is applied on them. This phenomenon is known as the electrorheological (ER) effect, and fluids that exhibit such behavior are known as ER fluids. The Bingham plastic model for shear stress, which is expressed as τ =τy + μ(du/dy), is widely used to describe ER fluid behavior because of its simplicity. One of the most promising applications of ER fluids is the ER clutch. A typical multidisk ER clutch consists of several equally spaced steel disks of inner radius R1 and outer radius R2, N of them attached to the input shaft. The gap h between the parallel disks is filled with a viscous fluid. Take the constants in shear stress relation as μ = 0.1 Pa·s and τy = 2.5 kPa.
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Assignment Print View
160. Award: 10.00 points
Choose the correct relationship for the torque generated by the clutch when the output shaft is stationary. (Please provide an answer before moving on to the next part.)
((
T = 4πN
(
T = 4πN
(
τy 3
τy 3
τy
)
μω
(
)
(R − R ) - (R − R ) (
μω
3 2
3 1
3
3
3
3
2h
)
μω
)
μω
(
4 2
4 1
4
4
4
4
T = 4πN 3 R 2 − R 1 - 4h R 2 − R 1
)
)
R 32 − R 31 + 4h R 42 − R 41
(( τy
(
)
)
T = 2πN 3 R 2 − R 1 + 4h R 2 − R 1
)
)
The following assumptions have been made here: 1. The thickness of the oil layer between the disks is constant. 2. The Bingham plastic model for shear stress expressed as τ =τy + μ(du/dy) is valid.
The velocity gradient anywhere in the oil of film thickness h is V/h, where V = ωr is the tangential velocity relative to plates mounted on the shell. Then, the wall shear stress anywhere on the surface of a plate mounted on the input shaft at a distance r from the axis of rotation is expressed as du
V
ωr
τ w = τ y + μ dr = τ y + μ h = τ y + μ h
Then, the shear force acting on a differential area dA on the surface of a disk and the torque generation associated with it are expressed as
(
ωr
)
dF = τ wdA = τ y + μ h (2πr)dr
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Assignment Print View
(
ωr
(
)
)
ωr 3
dT = rdF = r τ y + μ h (2πr)dr = 2π τ yr 2 + μ h dr Integrating,
(
R
)
ωr 3
T = 2π∫ r =2R τ yr 2 + μ h 1
r3
dr = 2πτ y 3 +
μωr 4 4h
R2 r = R1
(( τy
3
)
3
μω
(
4
4
= 2π 3 R 2 − R 1 + 4h R 2 − R 1
)
)
This is the torque transmitted by one surface of a plate mounted on the input shaft. Then, the torque transmitted by both surfaces of N plates attached to the input shaft in the clutch becomes
(
τy
(
3
3
)
μω
(
4
)
)
4
T = 4πN 3 R 2 − R 1 + 4h R 2 − R 1
The correct relationship for the torque generated by the clutch when the output shaft is stationary is
(
τy
(
3
3
)
μω
(
4
)
)
4
T = 4πN 3 R 2 − R 1 + 4h R 2 − R 1 .
References Multiple Choice
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161.
Assignment Print View
Award: 10.00 points
Calculate the torque for an ER clutch with N = 11 for R1 = 50 mm, R2 = 200 mm, and ṅ = 2400 rpm if the fluid is SAE 10 with μ = 0.1 Pa·s, τy = 2.5 kPa, and h = 1.2 mm. The torque for the ER clutch is
2061 ± 2% N·m.
Explanation: The following assumptions have been made here: 1. The thickness of the oil layer between the disks is constant. 2. The Bingham plastic model for shear stress expressed as τ = τ y + μ
( ) du dy
is valid.
Then, the torque transmitted by both surfaces of N plates attached to input shaft in the clutch is
T = 4πN
(( τy 3
3
3
)
Noting that ω = 2πṅ =
T = (4π)(11)
μω
(
4
4
R 2 − R 1 + 4h R 2 − R 1
(
2500 N/m 2 3
)
2π ( 2400 rev/min ) 60
) = 251.33 rad/sec and substituting,
(
)
(0.20 m) 3 − (0.05 m) 3 +
( 0.1 N ⋅ s/m ) ( 251.33 rad/s ) 2
4 ( 0.0012 m )
)
((0.20 m) − (0.05 m) ) = 2061 N⋅m 4
4
The torque for the ER clutch is 2061 N·m. References Worksheet
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The viscosity of some fluids, called magnetorheological (MR) fluids, changes when a magnetic field is applied. Such fluids involve micron-sized magnetizable particles suspended in an appropriate carrier liquid and are suitable for use in controllable hydraulic clutches. See the figure below. The MR fluids can have much higher viscosities than the ER fluids, and they often exhibit shearthinning behavior in which the viscosity of the fluid decreases as the applied shear force increases. This behavior is also known as pseudoplastic behavior and can be successfully represented by the Herschel– Bulkley constitutive model expressed as τ = τy +
K(du/dy)m. Here τ is the shear stress applied, τy is the yield stress, K is the consistency index, and m is the power index. Consider a Herschel– Bulkley fluid with τy = 900 Pa, K = 58 Pa·sm, and m = 0.82.
References Section Break
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Assignment Print View
162. Award: 10.00 points
Choose the correct relationship for the torque transmitted by an MR clutch for N plates attached to the input shaft when the input shaft is rotating at an angular speed of ω while the output shaft is stationary. (Please provide an answer before moving on to the next part.)
(
τy
(
)
(
)
T = 2πN 3 R 32 − R 31 +
(
τy
T = 2πN 3 R 32 − R 31 +
(( τy
3
3
(
τy
(
mh m
(R
m+3 m+3 − R1 2
Kω m ( m + 3 ) hm
Kω m
)
( m + 3 ) hm
)
( m + 3 ) hm
T = 4πN 3 R 2 − R 1 +
Kω m
T = 4πN 4 R 32 − R 31 +
Kω m
)
)
)
(R
m+3 m+3 − R1 2
)
(R
m+3 m+3 − R1 2
)
(R
m+4 m+4 − R1 2
)
) )
The following equations have been made here: 1. The thickness of the oil layer between the disks is constant. 2. The Herschel–Bulkley model for shear stress expressed as τ = τy + K(du/dy)m is valid.
The velocity gradient anywhere in the oil of film thickness h is V/h, where V = ωr is the tangential velocity relative to plates mounted on the shell. Then, the wall shear stress anywhere on the surface of a plate mounted on the input shaft at a distance r from the axis of rotation is expressed as
()
du m
τ w = τ y + K dr
()
V m
= τy + K h
()
ωr m
= τy + K h
Then, the shear force acting on a differential area dA on the surface of a disk and the torque generation associated with it are expressed as
( ( )) ωr m
dF = τ wdA = τ y + K h
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( ( )) ωr m
(2πr)dr and dT = rdF = r τ y + K h
(
(2πr)dr = 2π τ yr 2 + K
)
ω mr m + 2
hm
dr
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Assignment Print View
Integrating,
(
) (
( ( ) )
ω mr m + 2 r3 Kω m r m + 3 T = 2π∫ RR 2 τ yr 2 + K m dr = 2π τ y 3 + h 1 m + 3 hm
R2
= 2π
R1
τy 3
(R − R ) + ( 3 2
3 1
Kω m m+3
(R ) hm
m+3 m+3 − R1 2
)
)
This is the torque transmitted by one surface of a plate mounted on the input shaft. Then, the torque transmitted by both surfaces of N plates attached to the input shaft in the clutch becomes
(
τy
(
)
T = 4πN 3 R 32 − R 31 +
Kω m ( m + 3 ) hm
(R
m+3 m+3 − R1 2
)
)
The correct relationship for the torque transmitted by an MR clutch is
(
τy
(
)
T = 4πN 3 R 32 − R 31 +
Kω m ( m + 3 ) hm
(R
m+3 m+3 − R1 2
)
)
References Multiple Choice
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Assignment Print View
163. Award: 10.00 points
Calculate the torque transmitted by such a clutch with N = 13 plates for R1 = 50 mm, R2 = 200 mm, ṅ = 3000 rpm, and h = 1.5 mm. The torque transmitted by the clutch is
122.2 ± 2% kN·m.
Explanation: The following assumptions have been made here: 1. The thickness of the oil layer between the disks is constant. 2. The Herschel–Bulkley model for shear stress expressed as τ = τy + K(du/dy)m is valid.
The torque transmitted by both surfaces of N plates attached to the input shaft in the clutch is
T = 4πN
(( τy 3
3
3
)
R2 − R1 +
Kω m ( m + 3 ) hm
(R
m+3 m+3 − R1 2
)
)
Noting that ω = 2πṅ = 2π(3000 rev/min) = 18850 rad/min = 314. 2 rad/s and substituting,
( )( )(
T = 4π
13
900 N/m 2 3
(
)
(0.20 m) 3 − (0.05 m) 3 +
( 58 N ⋅ s /m ) ( 314.2 /s) ( 0.82 + 3 ) ( 0.0014 m) 0 . 82
2
0 . 82
0 . 82
((0.20 m)
3 . 82
− (0.05 m) 3 . 82
)
)
T = 122.2 kN·m
The torque transmitted by the clutch is 122.2 kN·m. References Worksheet
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Some non-Newtonian fluids behave as a Bingham plastic for which the shear stress can be expressed as τ = τy + μ
() ( )(
laminar flow of a Bingham plastic in a horizontal pipe of radius R, the velocity profile is given as u r =
ΔP
4μL
() ) ( ) du dr
r2 − R2 +
. For τy μ
(r − R)),
ΔP
where L is the constant pressure drop along the pipe per unit length, μ is the dynamic viscosity, r is the radial distance from the centerline, and τy is the yield stress of the Bingham plastic. References Section Break
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164. Award: 10.00 points
Choose the correct equation for the shear stress at the pipe wall.
τ
ΔP w = 4τ y + 2L R
τ
ΔP w = 2τ y + 2L R
τ
ΔP w = 2τ y - 2L R
τ
ΔP w = 4τ y - 2L R
The following assumptions have been made here: 1. The fluid is a Bingham plastic with τ = τy + μ(du/dr), where τy is the yield stress. 2. The flow through the pipe is one dimensional.
()
ΔP
(
)
τy
The velocity profile is given by u r = 4μL r 2 − R 2 + μ (r − R), where ΔP/L is the pressure drop along the pipe per unit length, μ is the dynamic viscosity and r is the radial distance from the centerline. Its gradient at the pipe wall (r = R) is du dr
|
(
d ΔP
r=R
(
τy
)
)| (
= dr 4μL r 2 − R 2 + μ (r − R)
r=R
ΔP
τy
= 2r 4μL + μ
)
(
1 ΔP
r=R
)
= μ 2L R + τ y
Substituting into τ = τy + μ(du/dr), the wall shear stress at the pipe surface becomes τw = τy + μ
du dr
|
r=R
= τy +
ΔP 2L
ΔP
R + τ y = 2τ y + 2L R
ΔP
The correct equation for the shear stress at the pipe wall is τ w = 2τ y + 2L R. References Multiple Choice
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165. Award: 10.00 points
Choose the correct equation for the drag force acting on a pipe section of length L.
F D = 2πRLτ y + πR 2ΔP F D = 4πRLτ y - πR 2ΔP F D = 2πRLτ y - πR 2ΔP F D = 4πRLτ y + πR 2ΔP The following assumptions have been made here: 1. The fluid is a Bingham plastic with τ = τy + μ(du/dr), where τy is the yield stress. 2. The flow through the pipe is one-dimensional.
()
ΔP
(
τy
)
The velocity profile is given by u r = 4μL r 2 − R 2 + μ (r − R), where ΔP/L is the pressure drop along the pipe per unit length, μ is the dynamic viscosity and r is the radial distance from the centerline. Its gradient at the pipe wall (r = R) is du dr
|
r=R
=
(
d ΔP dr 4μL
τy
)| (
( r − R ) + (r − R ) 2
2
μ
r=R
= 2r
ΔP 4μL
+
τy μ
)
r=R
=
(
1 ΔP μ 2L
)
R + τy
Substituting into τ = τy + μ(du/dr), the wall shear stress at the pipe surface becomes du
τ w = τ y + μ dr
|
ΔP
r=R
ΔP
= τ y + 2L R + τ y = 2τ y + 2L R
Then, the friction drag force exerted by the fluid on the inner surface of the pipe becomes
(
ΔP
)
(
ΔP
)
F D = τ wA s = 2τ y + 2L R (2πRL) = 2πRL 2τ y + 2L R = 4πRLτ y + πR 2ΔP
The correct equation for the shear stress at the pipe wall is F D = 4πRLτ y + πR 2ΔP. References Multiple Choice
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Difficulty: Medium
160/182
06/06/2024, 11:03
Assignment Print View
166. Award: 10.00 points
In some damping systems, a circular disk immersed in oil is used as a damper, as shown in the figure below. Choose the correct equation for the damping torque. Assume linear velocity profiles on both sides of the disk and neglect the tip effects.
T damping, total =
T damping, total =
T damping, total =
T damping, total =
( )
πμR 4 1 2
a
πμR 4 4
a
1
-
b
ω
() 2
ab
ω
( )
πμR 4 1 4
ω
b
( )
πμR 4 1 2
1
+
a
+
1
b
ω
The following assumptions have been made here: 1. The thickness of the oil layer on each side remains constant. 2. The velocity profiles are linear on both sides of the disk. 3. The tip effects are negligible. 4. The effect of the shaft is negligible.
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The velocity gradient anywhere in the oil of film thickness a is V/a, where V = ωr is the tangential velocity. Then, the wall shear stress anywhere on the upper surface of the disk at a distance r from the axis of rotation can be expressed as du
V
ωr
τ w = μ dy = μ a = μ a
Then, the shear force acting on a differential area dA on the surface and the torque it generates can be expressed as ωr
dF = τ wdA = μ a dA ωr 2
dT = rdF = μ a dA Noting that and integrating, the torque on the top surface is determined to be
T top =
μω
a
∫ Ar 2dA =
()
μω R
a
∫ r = 0r 2 2πr dr =
2πμω R
2πμω r 4
a
a
∫ r = 0r 3dr =
4
|
R
= r=0
πμωR 4 2a
The torque on the bottom surface is obtained by replacing a with b:
T bottom =
πμωR 4 2b
The total torque acting on the disk is the sum of the torque acting on the top and bottom surfaces,
T damping, total = T bottom + T top =
( )
πμωR 4 1 2
a
1
+b
or
T damping, total = Cω where C =
( )
πμR 4 1 2
a
1
+b
The correct equation for the damping torque is
T damping, total =
( )
πμR 4 1 2
a
+
1
b
ω
References Multiple Choice
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Difficulty: Medium
162/182
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Assignment Print View
Oil of viscosity μ = 0.0270 Pa⋅s and density ρ = 0.796 kg/m3 is sandwiched in the small gap between two very large parallel flat plates. A third flat plate of surface area A = 15 cm × 15 cm (on one side) is dragged through the oil at steady velocity V = 1.00 m/s to the right as illustrated. The top plate is stationary, but the bottom plate is moving at velocity V = 0.300 m/s to the left as sketched. The heights are h1 = 1.00 mm and h2 = 2.60 mm. The force required to pull the plate through the oil is F.
References Section Break
Difficulty: Easy
167. Award: 10.00 points
Sketch the velocity profiles and calculate the distance yA where the velocity is zero. Hint: Since the gaps are small and the oil is very viscous, the velocity profiles are linear in both gaps. Use the no-slip conditions at the walls to determine the velocity profile in each gap. Please upload your response/answer using the controls provided below.
Student upload controls will be shown to students when they take this assignment.
References File Attachment Question
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Difficulty: Easy
163/182
06/06/2024, 11:03
Assignment Print View
168. Award: 10.00 points
Calculate force F in newtons (N) required to keep the middle plate moving at constant speed. The force required to keep the middle plate moving at constant speed is
0.91 ± 2% N.
Explanation: The following assumptions have been made here: 1. The flow is parallel to the plates. 2. The flow is onedimensional. 3. The pressure is constant. 4. The fluid is Newtonian and incompressible. 5. The gravitational effect is neglected. To calculate the force needed to pull the plate, we draw a free-body diagram of the plate:
For constant plate speed (no acceleration), F = F1 + F2. But we can calculate these two forces independently based on the shear stress at the walls of the plate:
F 1 = τ 1A = μ
| | du dy
1
A and F 2 = τ 2A = μ
| | du dy
2
A
Thus,
F = F 1 + F 2 = μA
(| | | | ) du dy
1
+
du dy
2
Plugging in the numbers, we get
(
)
F = 0.027 N ⋅ s/m 2 (0.15 m)(0.15 m)
(
( 1 − 0 ) m/s 0.001 m
+
( 1 − ( − 0 . 30 ) ) m/s 0.0026 m
)
= 0.91 N
The force required to keep the middle plate moving at constant speed is 0.91 N. References Worksheet
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Difficulty: Easy
164/182
06/06/2024, 11:03
Assignment Print View
169. Award: 10.00 points
The specific gravity of a fluid is specified to be 0.84. The specific volume of this fluid is _____ m3/kg. Solve this problem using appropriate software.
84 0.0082 0.001 0.00119 840 This is solved by appropriate software. Solutions can be verified by copying and pasting the following lines on a blank software screen. (Similar problems and their solutions can be obtained easily by modifying numerical values.) SG=0.84 rho_water=1000 [kg/m^3] rho_fluid=SG*rho_water v=1/rho_fluid The specific volume of this fluid is 0.00119 m3/kg. References Multiple Choice
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Difficulty: Easy
165/182
06/06/2024, 11:03
Assignment Print View
170. Award: 10.00 points
The specific gravity of mercury is 13.6. The specific weight of mercury is _____ kN/m3. Solve this problem using appropriate software.
133 1.36 13600 106 9.81 This is solved by appropriate software. Solutions can be verified by copying and pasting the following lines on a blank software screen. (Similar problems and their solutions can be obtained easily by modifying numerical values.) SG=13.6 rho_water=1000 [kg/m^3] rho=SG*rho_water g=9.81 [m/s^2] SW=rho*g
The specific weight of mercury is 133 kN/m3. References Multiple Choice
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Difficulty: Easy
166/182
06/06/2024, 11:03
171.
Assignment Print View
Award: 10.00 points
A 0.08-m3 rigid tank contains air at 3 bar and 128°C. The mass of the air in the tank is _____ kg. Solve this problem using appropriate software.
0.209 0.066 0.8 0.659 0.002 This is solved by appropriate software. Solutions can be verified by copying and pasting the following lines on a blank software screen. (Similar problems and their solutions can be obtained easily by modifying numerical values.) V=0.08 [m^3] P=300 [kPa] T=(128+273) [K] R=0.287 [kJ/kg-K] m=(P*V)/(R*T)
The mass of the air in the tank is 0.209 kg. References Multiple Choice
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Difficulty: Easy
167/182
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Assignment Print View
172. Award: 10.00 points
The pressure of water is increased from 100 kPa to 800 kPa by a pump. The density of the water is 1 kg/L. If the water temperature does not change during this process, the change of specific enthalpy of the water is _____ kJ/kg. Solve this problem using appropriate software.
600 1.26 400 0.84 0.7 This is solved by appropriate software. Solutions can be verified by copying and pasting the following lines on a blank software screen. (Similar problems and their solutions can be obtained easily by modifying numerical values.) P1=100 [kPa] P2=800 [kPa] rho=1000 [kg/m^3] DELTAh=(P2-P1)/rho
the change of specific enthalpy of the water is 0.7 kJ/kg. References Multiple Choice
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Difficulty: Easy
168/182
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Assignment Print View
173. Award: 10.00 points
An ideal gas flows in a pipe at 37°C. The density of the gas is 2.4 kg/m3 and its molar mass is 44 kg/kmol. The pressure of the gas is _____ kPa. Solve this problem using appropriate software.
490 13 141 4900 79 This is solved by appropriate software. Solutions can be verified by copying and pasting the following lines on a blank software screen. (Similar problems and their solutions can be obtained easily by modifying numerical values.) T=(37+273) [K] rho=2.4 [kg/m^3] MM=44 [kg/kmol] R_u=8.314 [kJ/kmol-K] R=R_u/MM P=rho*R*T
The pressure of the gas is 141 kPa. References Multiple Choice
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Difficulty: Easy
169/182
06/06/2024, 11:03
Assignment Print View
174. Award: 10.00 points
Liquid water vaporizes into water vapor as it flows in the piping of a boiler. If the temperature of water in the pipe is 180°C, the vapor pressure of the water in the pipe is _____ kPa. Solve this problem using appropriate software.
1002 100 18 180 101.3 This is solved by appropriate software. Solutions can be verified by copying and pasting the following lines on a blank software screen. (Similar problems and their solutions can be obtained easily by modifying numerical values.) T=180 [C] P_vapor=pressure(steam, T=T, x=1)
The vapor pressure of the water in the pipe is 1002 kPa. References Multiple Choice
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Difficulty: Easy
170/182
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175. Award: 10.00 points
In a water distribution system, the pressure of water can be as low as 1.4 psia. The maximum temperature of water allowed in the piping to avoid cavitation is _____ºF. Solve this problem using appropriate software.
140 50 100 77 113 This is solved by EES software. Solutions can be verified by copying and pasting the following lines on a blank software screen. (Similar problems and their solutions can be obtained easily by modifying numerical values.) P=1.4 [psia] T_max=temperature(steam, P=P, x=1)
The maximum temperature of water allowed in the piping to avoid cavitation is 113ºF. References Multiple Choice
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Difficulty: Easy
171/182
06/06/2024, 11:03
Assignment Print View
176. Award: 10.00 points
The pressure of water is increased from 100 kPa to 900 kPa by a pump. The temperature of water also increases by 0.15°C. The density of water is 1 kg/L and its specific heat is cp = 4.18 kJ/kg⋅°C. The enthalpy change of the water during this process is _____ kJ/kg. Solve this problem using appropriate software.
900 1.43 4.18 0.63 0.80 This is solved by appropriate software. Solutions can be verified by copying and pasting the following lines on a blank software screen. (Similar problems and their solutions can be obtained easily by modifying numerical values.) P1=100 [kPa] P2=900 [kPa] DELTAT=0.15 [C] rho=1000 [kg/m^3] c_p=4.18 [kJ/kg-C] DELTAh=c_p*DELTAT+(P2-P1)/rho
The enthalpy change of the water during this process is 1.43 kJ/kg. References Multiple Choice
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Difficulty: Easy
172/182
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177.
Assignment Print View
Award: 10.00 points
An ideal gas is compressed isothermally from 100 kPa to 170 kPa. The percent increase in the density of this gas during this process is _____%. Solve this problem using appropriate software.
35 170 17 59 70 This is solved by appropriate software. Solutions can be verified by copying and pasting the following lines on a blank software screen. (Similar problems and their solutions can be obtained easily by modifying numerical values.) P1=100 [kPa] P2=170 [kPa] DELTAP=P2-P1 DELTAP\P=DELTAP/P1*Convert(,%) DELTArho\rho=DELTAP\P The percent increase in the density of this gas during this process is 70%. References Multiple Choice
Difficulty: Easy
178. Award: 10.00 points
The variation of the density of a fluid with temperature at constant pressure is represented by _____.
the isothermal compressibility the coefficient of volume expansion the bulk modulus of elasticity none of these the coefficient of compressibility The variation of the density of a fluid with temperature at constant pressure is represented by coefficient of volume expansion.
The variation of the density of a fluid with temperature at constant pressure is represented by coefficient of volume expansion. References Multiple Choice
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Difficulty: Easy
173/182
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179. Award: 10.00 points
Water is heated from 2°C to 71°C at a constant pressure of 100 kPa. The initial density of water is 1000 kg/m3 and the volume expansion coefficient of water is β = 0.377 × 10–3 K–1. The final density of the water is _____ kg/m3. Solve this problem using appropriate software.
974 1090.9 984.2 959.7 This is solved by appropriate software. Solutions can be verified by copying and pasting the following lines on a blank software screen. (Similar problems and their solutions can be obtained easily by modifying numerical values.) T1=2 [C] T2=78 [C] P=100 [kPa] rho_1=1000 [kg/m^3] beta=0.377E-3 [1/K] DELTAT=T2-T1 DELTArho=-beta*rho_1*DELTAT DELTArho=rho_2-rho_1
The final density of the water is 974 kg/m3. References Multiple Choice
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Difficulty: Easy
174/182
06/06/2024, 11:03
Assignment Print View
180. Award: 10.00 points
The viscosity of liquids _____ and the viscosity of gases _____ with temperature.
increases, increases decreases, decreases decreases, increases increases, decreases decreases, remains the same The viscosity of liquids decreases and the viscosity of gases increases with temperature.
The viscosity of liquids decreases and the viscosity of gases increases with temperature. References Multiple Choice
181.
Difficulty: Easy
Award: 10.00 points
The pressure of water at atmospheric pressure must be raised to 202 atm to compress it by 1 percent. Then, the coefficient of compressibility value of water is _____ atm. Solve this problem using appropriate software.
20112 19977 20100 20202 This is solved by appropriate software. Solutions can be verified by copying and pasting the following lines on a blank software screen. (Similar problems and their solutions can be obtained easily by modifying numerical values.) P1=1 [atm] P2=202 [atm] DELTArho\rho=0.01 DELTAP=P2-P1 CoeffComp=DELTAP/DELTArho\rho
The coefficient of compressibility value of water is 20100 atm. References Multiple Choice
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Difficulty: Easy
175/182
06/06/2024, 11:03
Assignment Print View
182. Award: 10.00 points
The density of a fluid decreases by 3 percent at constant pressure when its temperature increases by 26°C. The coefficient of volume expansion of this fluid is _____ K–1. Solve this problem using appropriate software.
0.0012 0.0015 0.015 0.12 This is solved by appropriate software. Solutions can be verified by copying and pasting the following lines on a blank software screen. (Similar problems and their solutions can be obtained easily by modifying numerical values.) DELTArho\rho=–0.03 DELTAT=26 [K] beta=-DELTArho\rho/DELTAT
The coefficient of volume expansion of this fluid is 0.0012K–1. References Multiple Choice
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Difficulty: Easy
176/182
06/06/2024, 11:03
Assignment Print View
183. Award: 10.00 points
The speed of a spacecraft is given to be 1400 km/h in atmospheric air at −40°C. The Mach number of this flow is _____. Solve this problem using appropriate software.
15.27 4.27 7.27 1.27 This is solved by appropriate software. Solutions can be verified by copying and pasting the following lines on a blank software screen. (Similar problems and their solutions can be obtained easily by modifying numerical values.) Vel=1400 [km/h]*Convert(km/h, m/s) T=(-40+273.15) [K] R=0.287 [kJ/kg-K] k=1.4 c=sqrt(k*R*T*Convert(kJ/kg, m^2/s^2)) Ma=Vel/c
The Mach number of this flow is 1.27. References Multiple Choice
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Difficulty: Easy
177/182
06/06/2024, 11:03
Assignment Print View
184. Award: 10.00 points
The dynamic viscosity of air at 20°C and 200 kPa is 1.83 × 10−5 kg/m⋅s. The kinematic viscosity of air at this state is _____ m2/s. Solve this problem using appropriate software.
0.77 × 10-5 0.380 × 10-5 0.525 × 10-5 1.47 × 10-5 1.83 × 10-5 This is solved by appropriate software. Solutions can be verified by copying and pasting the following lines on a blank software screen. (Similar problems and their solutions can be obtained easily by modifying numerical values.) T=(20+273.15) [K] P=200 [kPa] mu=1.83E-5 [kg/m-s] R=0.287 [kJ/kg-K] rho=P/(R*T) nu=mu/rho
The kinematic viscosity of air at this state is 0.77 × 10-5 m2/s. References Multiple Choice
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Difficulty: Easy
178/182
06/06/2024, 11:03
Assignment Print View
185. Award: 10.00 points
A viscometer constructed of two 30-cm-long concentric cylinders is used to measure the viscosity of a fluid. The outer diameter of the inner cylinder is 9 cm, and the gap between the two cylinders is 0.18 cm. The inner cylinder is rotated at 250 rpm, and the torque is measured to be 1.1 N⋅m. The viscosity of the fluid is _____ N⋅s/m2. Solve this problem using appropriate software.
0.19 0.74 0.66 0.44 This is solved by appropriate software. Solutions can be verified by copying and pasting the following lines on a blank software screen. (Similar problems and their solutions can be obtained easily by modifying numerical values.) L=0.3 [m] R=0.045 [m] gap=0.0018 [m] n_dot=(250/60) [1/s] T= 1.1 [N-m] mu=(T*gap)/(4*pi^2*R^3*n_dot*L)
The viscosity of the fluid is 0.44 N⋅s/m2. References Multiple Choice
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Difficulty: Easy
179/182
06/06/2024, 11:03
Assignment Print View
186. Award: 10.00 points
A 0.8-mm-diameter glass tube is inserted into water at 20°C in a cup. The surface tension of water at 20°C is σs = 0.073 N/m. The contact angle can be taken as zero degrees. The capillary rise of water in the tube is _____ cm. Solve this problem using appropriate software.
1.2 3.7 1.7 10.7 This is solved by appropriate software. Solutions can be verified by copying and pasting the following lines on a blank software screen. (Similar problems and their solutions can be obtained easily by modifying numerical values.) D=0.0008 [m] R=D/2 sigma_s=0.073 [N/m] phi=0 [degrees] rho=1000 [kg/m^3] g=9.81 [m/s^2] h=(2*sigma_s*cos(phi))/(rho*g*R)
The capillary rise of water in the tube is 3.7 cm. References Multiple Choice
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Difficulty: Easy
180/182
06/06/2024, 11:03
Assignment Print View
187. Award: 10.00 points
A liquid film suspended on a U-shaped wire frame with a 4-cm-long movable side is used to measure the surface tension of a liquid. If the force needed to move the wire is 0.028 N, the surface tension of this liquid exposed to air is _____ N/m. Solve this problem using appropriate software.
0.63 0.35 0.7 0.889 This is solved by appropriate software. Solutions can be verified by copying and pasting the following lines on a blank software screen. (Similar problems and their solutions can be obtained easily by modifying numerical values.) F=0.028 [N] b=0.04 [m] sigma_s=F/(2*b)
The surface tension of this liquid exposed to air is0.35 N/m. References Multiple Choice
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Difficulty: Easy
181/182
06/06/2024, 11:03
Assignment Print View
188. Award: 10.00 points
It is observed that water at 20°C rises up to a height of 16 m in a tree due to the capillary effect. The surface tension of water at 20°C is σs = 0.073 N/m, and the contact angle is 20°. The maximum diameter of the tube in which water rises is _____ mm. Solve this problem using appropriate software.
0.0017 0.0024 0.0028 0.0021 This is solved by appropriate software. Solutions can be verified by copying and pasting the following lines on a blank software screen. (Similar problems and their solutions can be obtained easily by modifying numerical values.) h=16 [m] sigma_s=0.073 [N/m] phi=20 [degrees] rho=1000 [kg/m^3] g=9.81 [m/s^2] h=(2*sigma_s*cos(phi))/(rho*g*R) D=2*R
The maximum diameter of the tube in which water rises is 0.0017 mm. References Multiple Choice
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Difficulty: Easy
182/182
06/06/2024, 11:08
Assignment Print View
Chapter 3 1.
Award: 0.00 points
The difference between gage pressure and absolute pressure is that gage pressure is the pressure relative to the atmospheric pressure, whereas absolute pressure is the pressure relative to an absolute vacuum.
True False
The pressure relative to the atmospheric pressure is called the gage pressure, and the pressure relative to an absolute vacuum is called absolute pressure.The statement is true. References True / False
2.
Difficulty: Easy
Award: 0.00 points
A tiny steel cube is suspended in water by a string. If the lengths of the sides of the cube are very small, which of the following statements are true?
The magnitudes of the pressures on all sides of the cube are nearly the same. The pressure varies linearly along the side faces. The pressure on the top face of the cube is higher than that on the bottom. The pressure on the bottom face of the cube is higher than that on the top. Since pressure increases with depth, the pressure on the bottom face of the cube is higher than that on the top. The pressure varies linearly along the side faces. However, if the lengths of the sides of the tiny cube suspended in water by a string are very small, the magnitudes of the pressures on all sides of the cube are nearly the same. The magnitudes of the pressures on all sides of the cube are nearly the same. References Multiple Choice
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Difficulty: Easy
1/260
06/06/2024, 11:08
3.
Assignment Print View
Award: 0.00 points
Some people experience nose bleeding and some others experience shortness of breath at high elevations. What are the reasons for this?
The pressure is lower at higher elevations. As a result, the difference between the blood pressure in the veins and the air pressure outside increases. This pressure imbalance may cause some thin-walled veins such as the ones in the nose to burst, causing bleeding.
The shortness of breath is caused by the exhaustion of reaching high altitudes. The shortness of breath is caused by the lower air density at higher elevations and thus lower amount of oxygen per unit volume.
Climbing high altitudes often involves trekking through rough terrain. Therefore, nose bleeding and shortness of breath occur due to fatigue.
Atmospheric air pressure, which is the external pressure exerted on the skin, decreases with increasing elevation. Therefore, the pressure is lower at higher elevations. As a result, the difference between the blood pressure in the veins and the air pressure outside increases. This pressure imbalance may cause some thin-walled veins such as the ones in the nose to burst, causing bleeding. The shortness of breath is caused by the lower air density at higher elevations and thus lower amount of oxygen per unit volume. The pressure is lower at higher elevations. As a result, the difference between the blood pressure in the veins and the air pressure outside increases. This pressure imbalance may cause some thin-walled veins such as the ones in the nose to burst, causing bleeding. The shortness of breath is caused by the lower air density at higher elevations, and thus lower amount of oxygen per unit volume. References Check All That Apply
4.
Difficulty: Easy
Award: 0.00 points
Consider two identical fans, one at sea level and the other on top of a high mountain, running at identical speeds. If you compare the volume flow rates and the mass flow rates of these two fans, which of the following statements are true?
The volume flow rates of the two fans running at identical speeds will be the same. The mass flow rate of the fan at sea level will be lower. The mass flow rates of the two fans running at identical speeds will be the same. The mass flow rate of the fan at sea level will be higher. The density of air at sea level is higher than the density of air on top of a high mountain. Therefore, the volume flow rates of the two fans running at identical speeds will be the same, but the mass flow rate of the fan at sea level will be higher. The volume flow rates of the two fans running at identical speeds will be the same, but the mass flow rate of the fan at sea level will be higher. References Check All That Apply https://ezto.mheducation.com/hm.tpx
Difficulty: Easy 2/260
06/06/2024, 11:08
5.
Assignment Print View
Award: 0.00 points
Someone claims that the absolute pressure in a liquid of constant density doubles when the depth is doubled. Do you agree?
Yes No
No, the absolute pressure in a liquid of constant density does not double when the depth is doubled. It is the gage pressure that doubles when the depth is doubled. The absolute pressure in a liquid of constant density does not double when the depth is doubled. It is the gage pressure that doubles when the depth is doubled. References Yes / No
6.
Difficulty: Easy
Award: 0.00 points
What does Pascal's law state?
The pressure applied to a confined fluid increases the pressure on the parallel surfaces. The pressure applied to a confined fluid is proportional to the volume of the fluid. The pressure applied to a confined fluid increases the pressure on the perpendicular surface. The pressure applied to a confined fluid increases the pressure throughout by the same amount. Pascal’s law states that the pressure applied to a confined fluid increases the pressure throughout by the same amount. This is a consequence of the pressure in a fluid remaining constant in the horizontal direction. Pascal’s law states that the pressure applied to a confined fluid increases the pressure throughout by the same amount. References Multiple Choice
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Difficulty: Easy
3/260
06/06/2024, 11:08
7.
Assignment Print View
Award: 0.00 points
Which of the following applications is an example of Pascal's principle?
The operation of a hydraulic car jack The operation of a speed governor The operation of an IC engine The floating of a boat Pascal’s law states that the pressure applied to a confined fluid increases the pressure throughout by the same amount. This is a consequence of the pressure in a fluid remaining constant in the horizontal direction. An example of Pascal’s principle is the operation of a hydraulic car jack. An example of Pascal’s principle is the operation of the hydraulic car jack. References Multiple Choice
8.
Difficulty: Easy
Award: 0.00 points
A pressure gage connected to a tank reads 510 kPa at a location where the atmospheric pressure is 94 kPa. Determine the absolute pressure in the tank. The absolute pressure in the tank is
604 ± 2% kPa.
Explanation:
The absolute pressure in the tank is determined from [Math Processing Error] The absolute pressure in the tank is 604 kPa. References Worksheet
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Difficulty: Easy
4/260
06/06/2024, 11:08
9.
Assignment Print View
Award: 0.00 points
A vacuum gage connected to a chamber reads 28 kPa at a location where the atmospheric pressure is 97 kPa. Determine the absolute pressure in the chamber. The absolute pressure in the chamber is
69 ± 2% kPa.
Explanation:
The absolute pressure in the chamber is determined from [Math Processing Error] The absolute pressure in the chamber is 69 kPa. References Worksheet
10.
Difficulty: Easy
Award: 0.00 points
The pressure at the exit of an air compressor is 130 psia. What is this pressure in kPa? The pressure at the exit of the compressor is
896.35 ± 2% kPa.
Explanation: Using the psia to kPa unit conversion factor, we get [Math Processing Error] The pressure at the exit of the compressor is 896.35 kPa. References Worksheet
https://ezto.mheducation.com/hm.tpx
Difficulty: Easy
5/260
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06/06/2024, 11:08
11.
Assignment Print View
Award: 0.00 points
A diver’s watch resists an absolute pressure of 5.4 bar. At an ocean having density of 1025 kg/m3 and an atmospheric pressure of 1 bar, what depth can the diver dive maximally to prevent water from entering his watch? 1 bar = 105 Pa and g = 9.81 m/s2. The maximum depth that the diver can dive is
43.76 ± 2% m.
Explanation: It is assumed that water is an incompressible substance, and thus the density does not change with depth. The gage pressure is [Math Processing Error] The equivalent height of this pressure as a water column, in m, is [Math Processing Error] This is the maximum depth that the diver can dive to prevent water from entering his watch. The maximum depth that the diver can dive is 43.76 m. References Worksheet
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Difficulty: Medium
6/260
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Award: 0.00 points
What is the value of 1 kgf/cm2 in psi? 1 kgf/cm2 =
14.223 ± 2% psi
Explanation: Noting that 1 kgf = 9.80665 N, 1 N = 0.22481 lbf, and 1 in = 2.54 cm, we have [Math Processing Error] and [Math Processing Error]
1 kgf/cm2 = 14.223 psi. References Worksheet
Difficulty: Easy
The pressure in a water line is 1540 kPa. References Section Break
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Difficulty: Easy
7/260
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Award: 0.00 points
What is the line pressure in lbf/ft2 units?
The line pressure is
32,164 ± 2% lbf/ft2.
Explanation: Using the appropriate conversion factor, we obtain [Math Processing Error] The line pressure is 32,164 lbf/ft2. References Worksheet
14.
Difficulty: Easy
Award: 0.00 points
What is the line pressure in lbf/in2 (psi) units?
The line pressure is
223.2 ± 2% lbf/in2 (psi).
Explanation: Using the appropriate conversion factor, we obtain [Math Processing Error] The line pressure is 223.2 lbf/in2. References Worksheet
https://ezto.mheducation.com/hm.tpx
Difficulty: Easy
8/260
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15.
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Award: 0.00 points
Blood pressure is usually measured by wrapping a closed air-filled jacket equipped with a pressure gage around the upper arm of a person at the level of the heart. Using a mercury manometer and a stethoscope, the systolic pressure (the maximum pressure when the heart is pumping) and the diastolic pressure (the minimum pressure when the heart is resting) are measured in mmHg. The systolic and diastolic pressures of a healthy person are about 120 mmHg and 80 mmHg, respectively, and are indicated as 120/80. Express both of these gage pressures in kPa, psi, and meter water column. Take the densities of water and mercury as 1000 kg/m3 and 13,600 kg/m3, respectively. The high and low pressures, in kPa, are 16 ± 2% kPa and 10.7 ± 2% kPa, respectively. The high and low pressures, in psi, are 2.32 ± 2% psi and 1.55 ± 2% psi, respectively. The high and low pressures, in meter water column, are 1.63 ± 2% m and 1.09 ± 2% m, respectively. Explanation: It is assumed that both mercury and water are incompressible substances. Using the relation for gage pressure, P = ρgh, the high and low pressures are expressed as [Math Processing Error] [Math Processing Error] Noting that 1 psi = 6.895 kPa, [Math Processing Error] [Math Processing Error] For a given pressure, the relation P = ρgh is expressed for mercury and water as [Math Processing Error] Setting these two relations equal to each other and solving for water height give [Math Processing Error] Therefore, [Math Processing Error] [Math Processing Error] The high and low pressures, in kPa, are 16.0 kPa and 10.7 kPa, respectively. The high and low pressures, in psi, are 2.32 psi and 1.55 psi, respectively. The high and low pressures, in meter water column, are 1.63 m and 1.09 m, respectively. References Worksheet
https://ezto.mheducation.com/hm.tpx
Difficulty: Medium
9/260
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Award: 0.00 points
The maximum blood pressure in the upper arm of a healthy person is about 120 mmHg. If a vertical tube open to the atmosphere is connected to the vein in the arm of the person, determine how high the blood will rise in the tube. Take the density of the blood to be 1040 kg/m3.
The height of the blood in the tube is
1.57 ± 2% m.
Explanation: The following assumptions have been made here: 1. The density of blood is constant. 2. The gage pressure of blood is 120 mmHg.
For a given gage pressure, the relation can be expressed for mercury and blood as [Math Processing Error] Solving for blood height and substituting give [Math Processing Error] The height of the blood in the tube is 1.57 m. References Worksheet
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Difficulty: Medium
10/260
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Consider a 1.78-m-tall man standing vertically in water and completely submerged in a pool. Determine the difference between the pressures acting at the head and at the toes of this man, in kPa. The pressure difference acting between the head and toes of the man is
17.46 ± 2% kPa.
Explanation: It is assumed that water is an incompressible substance, and thus the density does not change with depth. The pressures at the head and toes of the person can be expressed as [Math Processing Error] where h is the vertical distance of the location in water from the free surface. The pressure difference between the toes and the head is determined by subtracting the first relation above from the second: [Math Processing Error] Substituting, [Math Processing Error] The pressure difference acting between the head and toes of the man is 17.46 kPa. References Worksheet
Difficulty: Medium
A manometer is used to measure the air pressure in a tank. The fluid used has a specific gravity of 1.40, and the differential height between the two arms of the manometer is 18.2 in. The local atmospheric pressure is 12.7 psia. The specific gravity of the fluid is given to be 1.40. The density of water at 32°F is 62.4 lbm/ft3. References Section Break
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Difficulty: Medium
11/260
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Determine the absolute pressure in the tank for the case of the manometer arm with the higher fluid level being attached to the tank. The absolute pressure in the tank is
11.78 ± 2% psia.
Explanation: It is assumed that the fluid in the manometer is incompressible.
The density of the fluid is obtained by multiplying its specific gravity by the density of water. [Math Processing Error] The pressure difference corresponding to a differential height of 18.2 in between the two arms of the manometer is [Math Processing Error] The absolute pressure if the fluid level in the arm attached to the tank is higher (vacuum) is [Math Processing Error] The absolute pressure in the tank is 11.78 psia. References Worksheet
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Difficulty: Medium
12/260
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Determine the absolute pressure in the tank for the case of the manometer arm with the lower fluid level being attached to the tank. The absolute pressure in the tank is
13.62 ± 2% psia.
Explanation: It is assumed that the fluid in the manometer is incompressible.
The density of the fluid is obtained by multiplying its specific gravity by the density of water. [Math Processing Error] The pressure difference corresponding to a differential height of 18.2 in between the two arms of the manometer is [Math Processing Error] The absolute pressure if the fluid level in the arm attached to the tank is lower is [Math Processing Error] The absolute pressure in the tank is 13.62 psia. References Worksheet
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Difficulty: Medium
13/260
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The water in a tank is pressurized by air, and the pressure is measured by a multifluid manometer as shown in the figure. Determine the gage pressure of air in the tank if h1 = 0.5 m, h2 = 0.6 m, and h3 = 0.8 m. Take the densities of water, oil, and mercury to be 1000 kg/m3, 850 kg/m3, and 13,600 kg/m3, respectively.
The gage pressure of air in the tank is
96.8 ± 2% kPa.
Explanation: It is assumed that the air pressure in the tank is uniform (its variation with elevation is negligible due to its low density), and thus we can determine the pressure at the air–water interface.
Starting with the pressure at point 1 at the air–water interface, moving along the tube by adding or subtracting the ρgh terms until we reach point 2, and setting the result equal to Patm since the tube is open to the atmosphere give [Math Processing Error] Solving for P1, [Math Processing Error] [Math Processing Error] https://ezto.mheducation.com/hm.tpx
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Assignment Print View
or [Math Processing Error] Noting that P1,gage = P1 – Patm and substituting, [Math Processing Error] [Math Processing Error]
The gage pressure of air in the tank is 96.8 kPa. References Worksheet
21.
Difficulty: Medium
Award: 0.00 points
Determine the atmospheric pressure at a location where the barometric reading is 715 mmHg. Take the density of mercury to be 13,600 kg/m3. The atmospheric pressure is
95.4 ± 2% kPa.
Explanation: The atmospheric pressure is determined directly from [Math Processing Error] [Math Processing Error] [Math Processing Error] The atmospheric pressure is 95.4 kPa. References Worksheet
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Difficulty: Easy
15/260
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The gage pressure in a liquid at a depth of 2.5 m is read to be 20 kPa. Determine the gage pressure in the same liquid at a depth of 9 m. The gage pressure in the same liquid at a depth of 9 m is
72 ± 2% kPa.
Explanation: It is assumed that the variation of the density of the liquid with depth is negligible.
The gage pressure at two different depths of a liquid can be expressed as P1 = ρgh1 and P2 = ρgh2. Taking their ratio, [Math Processing Error] Solving for P2 and substituting give [Math Processing Error] The gage pressure in the same liquid at a depth of 9 m is 72 kPa. References Worksheet
Difficulty: Medium
The absolute pressure in water at a depth of 8 m is read to be 173 kPa. The specific gravity of the fluid is given to be SG = 0.78. We take the density of water to be 1000 kg/m3. References Section Break
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Difficulty: Medium
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Determine the local atmospheric pressure. The atmospheric pressure is
94.52 ± 2% kPa.
Explanation: It is assumed that the liquid and water are incompressible. Knowing the absolute pressure, the atmospheric pressure can be determined from [Math Processing Error] [Math Processing Error] [Math Processing Error] The atmospheric pressure is 94.52 kPa. References Worksheet
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Difficulty: Medium
17/260
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Determine the absolute pressure at a depth of 8 m in a liquid whose specific gravity is 0.78 at the same location. The absolute pressure at a depth of 8 m is
155.73 ± 2% kPa.
Explanation: The density of the liquid is obtained by multiplying its specific gravity by the density of water. [Math Processing Error] The absolute pressure at a depth of 8 m in the other liquid is [Math Processing Error] [Math Processing Error] [Math Processing Error] The absolute pressure at a depth of 8 m is 155.73 kPa. References Worksheet
Difficulty: Medium
A 184-lbm man has a total foot imprint area of 68 in2. References Section Break
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Difficulty: Medium
18/260
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Determine the pressure this man exerts on the ground if he stands on both feet. The pressure exerted by the man if he stands on both feet is
2.71 ± 2% psi.
Explanation: It is assumed that the weight of the person is distributed uniformly on the foot imprint area. The weight of the man is given to be 184 lbm, which is the mass of the man. It is equivalent to a weight of 184 lbf. Noting that pressure is force per unit area, the pressure this man exerts on the ground if he stands on both feet is [Math Processing Error] The pressure exerted by the man when he stands on both feet 2.71 psi. References Worksheet
26.
Difficulty: Easy
Award: 0.00 points
Determine the pressure this man exerts on the ground if he stands on one foot. The pressure exerted by the man if he stands on one foot is
5.41 ± 2% psi.
Explanation: It is assumed that the weight of the person is distributed uniformly on the foot imprint area. The weight of the man is given to be 184 lbm, which is actually the mass of the man. It is equivalent to a weight of 184 lbf. Noting that pressure is force per unit area, the pressure this man exerts on the ground if he stands on one foot is [Math Processing Error] The pressure exerted by the man when he stands on both feet 5.41 psi. References Worksheet
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Difficulty: Easy
19/260
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27.
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Consider a 54-kg woman who has a total foot imprint area of 400 cm2. She wishes to walk on the snow, but the snow cannot withstand pressures greater than 0.5 kPa. Determine the minimum size of the snowshoes needed (imprint area per shoe) to enable her to walk on the snow without sinking. The minimum imprint area per shoe is
1.06 ± 2% m2.
Explanation: The following assumptions have been made here: 1. The weight of the person is distributed uniformly on the imprint area of the shoes. 2. One foot carries the entire weight of a person during walking, and the shoe is sized for walking conditions (rather than standing). 3. The weight of the shoes is negligible. The mass of the woman is given to be 54 kg. For a pressure of 0.5 kPa on the snow, the imprint area of one shoe must be [Math Processing Error] The minimum imprint area per shoe is 1.06 m2. References Worksheet
https://ezto.mheducation.com/hm.tpx
Difficulty: Easy
20/260
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28.
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Award: 0.00 points
A vacuum gage connected to a tank reads 41 kPa at a location where the barometric reading is 755 mmHg. Determine the absolute pressure in the tank. Take ρHg = 13,590 kg/m3. The absolute pressure in the tank is
59.6 ± 2% kPa.
Explanation:
The atmospheric (or barometric) pressure can be expressed as [Math Processing Error] [Math Processing Error] [Math Processing Error] Then, the absolute pressure in the tank becomes [Math Processing Error]
The absolute pressure in the tank is 59.6 kPa. References Worksheet
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Difficulty: Medium
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Assignment Print View
The piston of a vertical piston–cylinder device containing a gas has a mass of 45 kg and a cross-sectional area of 0.012 m2. The local atmospheric pressure is 95 kPa, and the gravitational acceleration is 9.81 m/s2.
References Section Break
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Difficulty: Medium
22/260
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29.
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Determine the pressure inside the cylinder. The pressure inside the cylinder is
131.79 ± 2% kPa.
Explanation: It is assumed that friction between the piston and the cylinder is negligible.
The gas pressure in the piston–cylinder device depends on the atmospheric pressure and the weight of the piston. Drawing the free-body diagram of the piston as shown in the figure and balancing the vertical forces yield [Math Processing Error] Solving for P and substituting, [Math Processing Error] The pressure inside the cylinder is 131.79 kPa. References Worksheet
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Difficulty: Medium
23/260
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30.
Assignment Print View
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If some heat is transferred to the gas and its volume is doubled, does the pressure inside the cylinder change?
Yes No
The volume change will have no effect on the free-body diagram. Therefore, the pressure inside the cylinder does not change—it will remain the same.The pressure inside the cylinder does not change. References Yes / No
31.
Difficulty: Medium
Award: 0.00 points
The vacuum pressure of a condenser is given to be 70 kPa. If the atmospheric pressure is 98 kPa, what is the gage pressure and absolute pressure in kPa, kN/m2, psi, and mmHg. The absolute pressure is mmHg.
28 ± 2% kPa,
28 ± 2% kN/m2,
4.06 ± 2% lbf/in2,
4.06 ± 2% psi, or
210 ± 2%
Explanation: Gage pressure is not applicable to this case as the given pressure is less than the value of atmospheric pressure. [Math Processing Error] Then, using the conversion factors, [Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] The absolute pressure is 28 kPa, 28 kN/m2, 4.06 lbf/in2, 4.06 psi, or 210 mmHg. References Worksheet
https://ezto.mheducation.com/hm.tpx
Difficulty: Easy
24/260
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Assignment Print View
Water from a reservoir is raised in a vertical tube of internal diameter D = 32 cm under the influence of the pulling force F of a piston. Take the density of water as ρ = 1000 kg/m3 and atmospheric pressure as 96 kPa.
References Section Break
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Difficulty: Medium
25/260
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32.
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Award: 0.00 points
Determine the force needed to raise the water to a height of h = 1.5 m above the free surface. The force needed to raise the water is
1.18 ± 2% kN.
Explanation: The following assumptions have been made here: 1. The friction between the piston and the cylinder is negligible. 2. The accelerational effects are negligible. Noting that the pressure at the free surface is Patm and hydrostatic pressure in a fluid decreases linearly with increasing height, the pressure at the piston face is [Math Processing Error] The surface area of the piston is [Math Processing Error] A force balance on the piston yields [Math Processing Error] A force of 1.18 kN needed to raise the water to a height of h = 1.5 m above the free surface. References Worksheet
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Difficulty: Medium
26/260
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Determine the force needed to raise the water to a height of h = 3 m above the free surface. The force needed to raise the water is
2.36 ± 2% kN.
Explanation: When the height is 3 m, [Math Processing Error] The surface area is [Math Processing Error] Therefore, the force required is [Math Processing Error] A force of 2.36 kN needed to raise the water to a height of h = 3 m above the free surface. References Worksheet
https://ezto.mheducation.com/hm.tpx
Difficulty: Medium
27/260
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34.
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Award: 0.00 points
Taking the atmospheric pressure to be 96 kPa, plot the absolute water pressure at the piston face as h varies from 0 to 3 m. (Please upload your response/solution using the controls provided below.)
Student upload controls will be shown to students when they take this assignment.
References File Attachment Question
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Difficulty: Medium
28/260
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35.
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Award: 0.00 points
The barometer of a mountain hiker reads 910 mbars at the beginning of a hiking trip and 790 mbars at the end. Neglecting the effect of altitude on local gravitational acceleration, determine the vertical distance climbed. Assume an average air density of 1.20 kg/m3. The vertical distance climbed by the mountain hiker is
1019 ± 2% m.
Explanation: It is assumed that the variation of air density and the gravitational acceleration with altitude is negligible.
Taking an air column between the top and the bottom of the mountain and writing a force balance per unit base area, we obtain [Math Processing Error] Therefore, the vertical distance or the height is [Math Processing Error]
The vertical distance climbed by the mountain hiker is 1019 m. References Worksheet
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Difficulty: Easy
29/260
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36.
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Award: 0.00 points
Determine the pressure exerted on a diver at 16 m below the free surface of the sea. Assume a barometric pressure of 101 kPa and a specific gravity of 1.03 for seawater. The pressure exerted on the diver is
262.7 ± 2% kPa.
Explanation: It is assumed that the variation of the density of water with depth is negligible. The density of seawater is obtained by multiplying its specific gravity by the density of water, which is taken to be 1000 kg/m3. [Math Processing Error] The pressure exerted on a diver at 16 m below the free surface of the sea is the absolute pressure at that location: [Math Processing Error] [Math Processing Error] [Math Processing Error] The pressure exerted on a diver at 16 m below the free surface of the sea is 262.7 kPa. References Worksheet
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Difficulty: Medium
30/260
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37.
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Award: 0.00 points
A gas is contained in a vertical, frictionless piston–cylinder device. The piston has a mass of 5 kg and a cross-sectional area of 35 cm2. A compressed spring above the piston exerts a force of 74 N on the piston. If the atmospheric pressure is 95 kPa, determine the pressure inside the cylinder.
The pressure inside the cylinder is
130.2 ± 2% kPa.
Explanation:
Drawing the free-body diagram of the piston and balancing the vertical forces yield [Math Processing Error] Thus, [Math Processing Error] [Math Processing Error] [Math Processing Error] The pressure inside the cylinder is 130.2 kPa. References Worksheet
https://ezto.mheducation.com/hm.tpx
Difficulty: Medium
31/260
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38.
Assignment Print View
Award: 0.00 points
A gas is contained in a vertical, frictionless piston–cylinder device. The piston has a mass of 5 kg and a cross-sectional area of 35 cm2. A compressed spring above the piston exerts a force of 74 N on the piston. If the atmospheric pressure is 95 kPa, determine the pressure inside the cylinder. (Please upload your response/solution using the controls provided below.)
Student upload controls will be shown to students when they take this assignment.
References File Attachment Question
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Difficulty: Hard
32/260
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39.
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Award: 0.00 points
The variation of pressure P in a gas with density ρ is given by P = Cρn, where C and n are constants with P = P0 and ρ = ρ0 at elevation z = 0. What is the relation for the variation of P with elevation in terms of z, g, n, P0, and ρ0?
[Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] The pressure field in a fluid is given by
[Math Processing Error] Combining Eqs. 1 and 2 yields
[Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error]
The final relation is
[Math Processing Error] References Multiple Choice
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Difficulty: Hard
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Assignment Print View
Both a gage and a manometer are attached to a gas tank to measure its pressure. The reading on the pressure gage is 60 kPa.
References Section Break
40.
Difficulty: Medium
Award: 0.00 points
Determine the distance between the two fluid levels of the manometer if the fluid is mercury (ρ = 13,600 kg/m3). The distance between the two fluid levels of the manometer is
0.45 ± 2% m.
Explanation: The gage pressure is related to the vertical distance h between the two fluid levels by [Math Processing Error] For mercury, [Math Processing Error] The distance between the two fluid levels of the manometer is 0.45 m. References Worksheet
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Difficulty: Medium
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41.
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Determine the distance between the two fluid levels of the manometer if the fluid is water (ρ = 1000 kg/m3). The distance between the two fluid levels of the manometer is
6.12 ± 2% m.
Explanation: The gage pressure is related to the vertical distance h between the two fluid levels by [Math Processing Error] For water, [Math Processing Error] The distance between the two fluid levels of the manometer is 6.12 m. References Worksheet
42.
Difficulty: Medium
Award: 0.00 points
Both a gage and a manometer are attached to a gas tank to measure its pressure. The reading on the pressure gage is 66 kPa. Using appropriate software, investigate the effect of the manometer fluid density in the range of 800 to 13,000 kg/m3 on the differential fluid height of the manometer. Plot the differential fluid height against the density, and discuss the results. (Please upload the response/solution using the controls provided below.)
Student upload controls will be shown to students when they take this assignment.
References File Attachment Question
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Difficulty: Hard
35/260
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43.
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Award: 0.00 points
The system shown in the figure is used to accurately measure the pressure changes when the pressure is increased by ΔP in the water pipe. When Δh = 96 mm, what is the change in pipe pressure?
The change in pipe pressure is
1189 ± 2% Pa.
Explanation:
Initially, [Math Processing Error] After the pressure is applied, [Math Processing Error] On the other hand, from the continuity, [Math Processing Error] From the first equation, [Math Processing Error] https://ezto.mheducation.com/hm.tpx
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Assignment Print View
Substituting into the second equation and solving for Δp will give [Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] The change in pipe pressure is 1189 Pa. References Worksheet
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Difficulty: Hard
37/260
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44.
Assignment Print View
Award: 0.00 points
The manometer shown in the figure is designed to measure pressures of up to a maximum of 115 Pa. If the reading error is estimated to be ±0.5 mm, what should the ratio of d/D be in order for the error associated with pressure measurement not to exceed 2.5% of the full scale.
The ratio required so as not to exceed 2.5% of the full scale is
0.2935 ± 2% .
Explanation:
Since PA = PB, we write [Math Processing Error] On the other hand, [Math Processing Error] Therefore, [Math Processing Error] In order to find the error due to the reading error in “L,” we differentiate P with respect to L as follows: [Math Processing Error] From the definition of error, we obtain [Math Processing Error] From the given data, [Math Processing Error] https://ezto.mheducation.com/hm.tpx
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Assignment Print View
[Math Processing Error] dL = 0.5 mm [Math Processing Error] Therefore, the ratio is [Math Processing Error] The ratio required in order not to exceed 2.5% of the full scale is 0.293. References Worksheet
45.
Difficulty: Medium
Award: 0.00 points
A manometer containing oil (ρ = 850 kg/m3) is attached to a tank filled with air. If the oil-level difference between the two columns is 1.2 m and the atmospheric pressure is 98 kPa, determine the absolute pressure of the air in the tank. The absolute pressure of the air in the tank is
108 ± 2% kPa.
Explanation:
The absolute pressure in the tank is determined from [Math Processing Error] [Math Processing Error] [Math Processing Error] The absolute pressure of the air in the tank is 108 kPa. References Worksheet
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Difficulty: Medium
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Assignment Print View
A mercury manometer (ρ = 13,600 kg/m3) is connected to an air duct to measure the pressure inside. The difference in the manometer levels is 11 mm, and the atmospheric pressure is 100 kPa.
References Section Break
46.
Difficulty: Medium
Award: 0.00 points
Judging from the given figure, the pressure in the duct is below the atmospheric pressure.
Yes No
The pressure in the duct is above the atmospheric pressure since the fluid column on the duct side is at a lower level.The pressure in the duct is above the atmospheric pressure since the fluid column on the duct side is at a lower level. References Yes / No
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Difficulty: Medium
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Determine the absolute pressure in the duct. The absolute pressure in the duct is
101.5 ± 2% kPa.
Explanation:
The absolute pressure in the duct is determined from [Math Processing Error] [Math Processing Error] [Math Processing Error] The absolute pressure in the duct is 101.5 kPa. References Worksheet
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Difficulty: Medium
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A mercury manometer (ρ = 13,600 kg/m3) is connected to an air duct to measure the pressure inside. The difference in the manometer levels is 29 mm, and the atmospheric pressure is 100 kPa.
References Section Break
48.
Difficulty: Medium
Award: 0.00 points
Judging from the given figure, the pressure in the duct is below the atmospheric pressure.
Yes No
The pressure in the duct is above the atmospheric pressure since the fluid column on the duct side is at a lower level.The pressure in the duct is above the atmospheric pressure since the fluid column on the duct side is at a lower level. References Yes / No
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Difficulty: Medium
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Determine the absolute pressure in the duct. The absolute pressure in the duct is
104 ± 2% kPa.
Explanation:
The absolute pressure in the duct is determined from [Math Processing Error] [Math Processing Error] [Math Processing Error] The absolute pressure in the duct is 104 kPa. References Worksheet
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Difficulty: Medium
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Consider a U-tube whose arms are open to the atmosphere. Now, water (ρ = 1000 kg/m3) is poured into the U-tube from one arm and light oil (ρ = 790 kg/m3) from the other. One arm contains 70-cm-high water, while the other arm contains both fluids with an oil-to-water height ratio of 9. Determine the height of each fluid in that arm.
0.086 ± 2% m. The height of water in that arm is The height of oil in that arm is 0.777 ± 2% m.
Explanation: It is assumed that both water and oil are incompressible substances.
The height of water column in the left arm of the manometer is given as hw1 = 0.70 m. We let the height of water and oil in the right arm to be hw2 and ha, respectively. Then, ha = 9hw2. Noting that both arms are open to the atmosphere, the pressure at the bottom of the U-tube can be expressed as [Math Processing Error] Setting them equal to each other and simplifying, [Math Processing Error] Noting that ha = 9hw2 and considering ρa = ρoil, the water and oil column heights in the second arm are determined to be [Math Processing Error] [Math Processing Error] The height of water in that arm is 0.086 m. The height of oil in that arm is 0.777 m. References Worksheet
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Difficulty: Medium
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The hydraulic lift in a car repair shop has an output diameter of 53 cm and is to lift cars up to 2500 kg. Determine the fluid gage pressure that must be maintained in the reservoir. The fluid gage pressure that must be maintained in the reservoir is
111.2 ± 2% kPa.
Explanation: The assumption here is that the weight of the piston of the lift is negligible.
Pressure is force per unit area, and thus the gage pressure required is simply the ratio of the weight of the car to the area of the lift: [Math Processing Error] The fluid gage pressure that must be maintained in the reservoir is 111.2 kPa. References Worksheet
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Difficulty: Medium
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Consider a double-fluid manometer attached to an air pipe shown in the figure. If the specific gravity of one fluid is 13.05, determine the specific gravity of the other fluid for the indicated absolute pressure of air. Take the atmospheric pressure to be 100 kPa and the density of water to be 1000 kg/m3.
The specific gravity of the second fluid is
1.29 ± 2% .
Explanation: The following assumptions have been made here: 1. The densities of the liquids are constant. 2. The air pressure in the tank is uniform (i.e., its variation with elevation is negligible due to its low density), and thus the pressure at the air–water interface is the same as the indicated gage pressure.
Starting with the pressure of air in the tank, moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach the free surface where the oil tube is exposed to the atmosphere, and setting the result equal to Patm give [Math Processing Error] Rearranging and solving for SG2, [Math Processing Error]
[Math Processing Error] [Math Processing Error] The specific gravity of the second fluid is 1.29. https://ezto.mheducation.com/hm.tpx
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References Worksheet
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Difficulty: Medium
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The pressure in a natural gas pipeline is measured by the manometer shown in the figure with one of the arms open to the atmosphere where the local atmospheric pressure is 14.2 psia. Determine the absolute pressure at the bottom of the natural gas pipeline. Take the height of the mercury column h to be 7 in. Take the density of water to be ρw = 62.4 lbm/ft3, and the specific gravity of mercury to be 13.6 and its density is ρHg = 848.6 lbm/ft3.
The absolute pressure at the bottom of the natural gas pipeline is
18.5 ± 2% psia.
Explanation: The following assumptions have been made here: 1. All the liquids are incompressible. 2. The effect of the air column on pressure is negligible. 3. The pressure throughout the natural gas (including the tube) is uniform since its density is low.
Starting with the pressure at point 1 in the natural gas pipeline, moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach the free surface of oil where the oil tube is exposed to the atmosphere, and setting the result equal to Patm give [Math Processing Error] Solving for P1, [Math Processing Error]
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[Math Processing Error] [Math Processing Error] The absolute pressure at the bottom of the natural gas pipeline is 18.5 psia. References Worksheet
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Difficulty: Medium
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The pressure in a natural gas pipeline is measured by the manometer shown in the figure with one of the arms open to the atmosphere where the local atmospheric pressure is 14.3 psia. Determine the absolute pressure at the bottom of the natural gas pipeline. Take the density of water to be ρw = 62.4 lbm/ft3. Specific gravity of mercury is 13.6, and its density is ρHg = 848.6 lbm/ft3. Specific gravity of oil is given to be 0.69, and its density is ρoil = 43.1 lbm/ft3.
The absolute pressure at the bottom of the natural gas pipeline is
17.91 ± 2% psia.
Explanation: The following assumptions have been made here: 1. All the liquids are incompressible. 2. The effect of the air column on pressure is negligible. 3. The pressure throughout the natural gas (including the tube) is uniform since its density is low.
Starting with the pressure at point 1 in the natural gas pipeline, moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach the free surface of oil where the oil tube is exposed to the atmosphere, and setting the result equal to Patm give [Math Processing Error] Solving for P1, https://ezto.mheducation.com/hm.tpx
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[Math Processing Error] Substituting, [Math Processing Error] [Math Processing Error] The absolute pressure at the bottom of the natural gas pipeline is 17.91 psia. References Worksheet
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Difficulty: Medium
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The gage pressure of the air in the tank shown in the figure below is measured to be 55 kPa. Determine the differential height h of the mercury column. Take the density of water to be ρw =1000 kg/m3.
The differential height of the mercury column is
0.395 ± 2% m.
Explanation: It is assumed that the air pressure in the tank is uniform (i.e., its variation with elevation is negligible due to its low density), and thus the pressure at the air–water interface is the same as the indicated gage pressure.
We take the density of water to be ρw =1000 kg/m3. The specific gravities of oil and mercury are given to be 0.72 and 13.6, respectively. Starting with the pressure of air in the tank (point 1), moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach the free surface of oil where the oil tube is exposed to the atmosphere, and setting the result equal to Patm give [Math Processing Error] Rearranging, [Math Processing Error] [Math Processing Error] [Math Processing Error] Solving for hHg, [Math Processing Error]
The differential height of the mercury column is 0.395 m. References https://ezto.mheducation.com/hm.tpx
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Worksheet
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Difficulty: Medium
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The gage pressure of the air in the tank shown in the figure below is measured to be 37 kPa. Determine the differential height h of the mercury column. Take the density of water to be ρw =1000 kg/m3.
The differential height of the mercury column is
0.26 ± 2% m.
Explanation: It is assumed that the air pressure in the tank is uniform (i.e., its variation with elevation is negligible due to its low density), and thus the pressure at the air–water interface is the same as the indicated gage pressure.
We take the density of water to be ρw =1000 kg/m3. The specific gravities of oil and mercury are given to be 0.72 and 13.6, respectively. Starting with the pressure of air in the tank (point 1), moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach the free surface of oil where the oil tube is exposed to the atmosphere, and setting the result equal to Patm give [Math Processing Error] Rearranging, [Math Processing Error] [Math Processing Error] [Math Processing Error] Solving for hHg, [Math Processing Error]
The differential height of the mercury column is 0.26 m. References https://ezto.mheducation.com/hm.tpx
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Worksheet
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Difficulty: Medium
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The 470-kg load on the hydraulic lift shown in the figure is to be raised by pouring oil (ρ = 780 kg/m3) into a thin tube. Determine how high h should be in order to begin to raise the weight.
The required height of oil to raise the weight is
0.533 ± 2% m.
Explanation: The following assumptions have been made here: 1. The cylinders of the lift are vertical. 2. There are no leaks. 3. Atmospheric pressure acts on both sides, and thus it can be disregarded.
Noting that pressure is force per unit area, the gage pressure in the fluid under the load is simply the ratio of the weight to the area of the lift: [Math Processing Error] The required oil height that will cause 4.08 kPa of pressure rise is [Math Processing Error] Therefore, a 470 kg load can be raised by this hydraulic lift by simply raising the oil level in the tube by 0.533 m. References Worksheet
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Difficulty: Medium
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Pressure is often given in terms of a liquid column and is expressed as “pressure head.” The standard atmospheric pressure is 101,325 Pa. References Section Break
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Difficulty: Easy
Award: 0.00 points
Express the standard atmospheric pressure in terms of mercury whose specific gravity is 13.6. The standard atmospheric pressure head for mercury is
0.759 ± 2% m.
Explanation: The atmospheric pressure is expressed in terms of fluid column height as [Math Processing Error] Substituting for mercury, [Math Processing Error]
The standard atmospheric pressure head for mercury is 0.759 m. References Worksheet
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Difficulty: Easy
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Express the standard atmospheric pressure in terms of water whose specific gravity is 1. The standard atmospheric pressure head for water is
10.3 ± 2% m.
Explanation: The atmospheric pressure is expressed in terms of fluid column height as [Math Processing Error] Substituting for water, [Math Processing Error]
The standard atmospheric pressure head for water is 10.3 m. References Worksheet
60.
Difficulty: Medium
Award: 0.00 points
Express the standard atmospheric pressure in terms of glycerin whose specific gravity is 1.26. The standard atmospheric pressure head for glycerin is
8.2 ± 2% m.
Explanation: The atmospheric pressure is expressed in terms of fluid column height as [Math Processing Error] Substituting for glycerin, [Math Processing Error] The standard atmospheric pressure head for glycerin is 8.20 m. References Worksheet
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Difficulty: Easy
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Freshwater and seawater flowing in parallel horizontal pipelines are connected to each other by a double U-tube manometer, as shown in the figure below, such that hw = 80 cm, hHg = 10 cm, hair = 70 cm, and hsea = 30 cm. Determine the pressure difference between the two pipelines. Take the density of seawater at that location to be ρsea = 1035 kg/m3, the density of mercury to be ρHg = 13600 kg/m3, and the density of water to be ρwater = 1000 kg/m3.
The pressure difference between the two pipelines is
2.45 ± 2% kPa.
Explanation: The following assumptions have been made: 1. All the liquids are incompressible. 2. The effect of the air column on pressure is negligible.
Starting with the pressure in the fresh waterpipe (point 1), moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach the sea water pipe (point 2), and setting the result equal to P2 give [Math Processing Error] Rearranging and neglecting the effect of the air column on pressure, [Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] The pressure difference between the two pipelines is 2.45 kPa. https://ezto.mheducation.com/hm.tpx
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Freshwater and seawater flowing in parallel horizontal pipelines are connected to each other by a double U-tube manometer, as shown in the figure, such that hw = 42 cm, hHg = 10 cm, hoil = 70 cm, and hsea = 30 cm. Determine the pressure difference between the two pipelines. Take the density of seawater at that location to be ρsea = 1035 kg/m3, the density of mercury to be ρHg = 13600 kg/m3, the density of oil to be ρoil = 720 kg/m3, and the density of water to be ρw = 1000 kg/m3.
The pressure difference between the two pipelines is
11.12 ± 2% kPa.
Explanation: It is assumed that all the liquids are incompressible.
Starting with the pressure in the freshwater pipe (point 1), moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach the sea water pipe (point 2), and setting the result equal to P2 give [Math Processing Error] Rearranging, [Math Processing Error] [Math Processing Error] Substituting, [Math Processing Error] [Math Processing Error] [Math Processing Error]
https://ezto.mheducation.com/hm.tpx
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The pressure difference between the two pipelines is 11.12 kPa. References Worksheet
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Difficulty: Medium
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The pressure difference between an oil pipe and water pipe is measured by a double-fluid manometer, as shown in the figure. For the given fluid heights and specific gravities, calculate the pressure difference ΔP = PB − PA. The specific gravities are 13.5 for mercury, 1.26 for glycerin, and 0.88 for oil. Take the standard density of water to be ρw =1000 kg/m3. The height h of mercury in the manometer is 24 cm.
The pressure difference is
32.36 ± 2% kPa.
Explanation: It is assumed that all the liquids are incompressible.
Starting with the pressure in the water pipe (point A), moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach the oil pipe (point B), and setting the result equal to PB give [Math Processing Error] Rearranging and using the definition of specific gravity, [Math Processing Error] [Math Processing Error] Substituting, https://ezto.mheducation.com/hm.tpx
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[Math Processing Error] [Math Processing Error] The pressure difference is 32.36 kPa. References Worksheet
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Difficulty: Medium
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06/06/2024, 11:08
64.
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Consider the system shown in the figure. If a change of 0.9 kPa in the pressure of air causes the brine–mercury interface in the right column to drop by 5 mm in the brine level in the right column while the pressure in the brine pipe remains constant, determine the ratio of A2/A1.
The ratio of A2/A1 is
0.434 ± 2% .
Explanation: The following assumptions have been made here: 1. All the liquids are incompressible. 2. Pressure in the brine pipe remains constant. 3. The variation of pressure in the trapped air space is negligible.
It is clear from the problem statement and the figure that the brine pressure is much higher than the air pressure, and when the air pressure drops by 0.9 kPa, the pressure difference between the brine and the air space also increases by the same amount. We start with the air pressure (point A), move along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach the brine pipe (point B), and set the result equal to PB before and after the pressure change of air. Before [Math Processing Error] After [Math Processing Error] Subtracting, [Math Processing Error]
https://ezto.mheducation.com/hm.tpx
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where ΔhHg and Δhbr are the changes in the differential mercury and brine column heights, respectively, due to the drop in air pressure. Both of these are positive quantities because as the mercury-brine interface drops, the differential fluid heights for both mercury and brine increase. Noting also that the volume of mercury is constant, we have A1ΔhHg,left = A2ΔhHg,right and [Math Processing Error] [Math Processing Error] [Math Processing Error] Substituting, [Math Processing Error] This gives A2/A1 = 0.434 The ratio of A2/A1 is 0.434. References Worksheet
Difficulty: Medium
There is water at a height of 1.1 m in the tube open to the atmosphere (Patm = 100 kPa) connected to a tank with two sections. Water density is 1000 kg/m3 and g = 9.79 m/s2. The height h of water as shown in the figure is 1.1 m.
References Section Break
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Difficulty: Medium
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Find the pressure readings (kPa) in a Bourdon Type Manometer for the air in the A and B sections of the tank. The pressure reading in the A section of the tank is The pressure reading in the B section of the tank is
2.94 ± 2% kPa. 5.87 ± 2% kPa.
Explanation: It is assumed that water is an incompressible substance with constant density. The pressure readings (gage pressures) for the air on top of water in the A and B sections of the tank are determined as follows: PAgage + ρgh1 = Patm gage = 0 PAgage + 1000 × 9.79 × (0.8 − 1.1) m = 0 → PAgage = 2937 Pa ≈ 2.94 kPa PBgage + ρgh1 = Patm gage = 0 PBgage + 1000 × 9.79 × (0.5 − 1.1) m = 0 → PBgage = 5874 Pa ≈ 5.87 kPa The pressure reading in the A section of the tank is 2.94 kPa. The pressure reading in the B section of the tank is 5.87 kPa. References Worksheet
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Difficulty: Medium
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Find the absolute pressures (kPa) for the air in the A and B sections of the tank. The absolute pressure in the A section of the tank is The absolute pressure in the B section of the tank is
102.9 ± 2% kPa. 105.9 ± 2% kPa.
Explanation: The absolute pressures for the air on top of water in the A and B sections of tank are PA abs = PA gage + Patm = 2.937 kPa + 100 kPa = 102.9 kPa PB abs = PB gage + Patm = 5.874 kPa + 100 kPa = 105.9 kPa The absolute pressure in the A section of the tank is 102.9 kPa. The absolute pressure in the B section of the tank is 105.9 kPa. References Worksheet
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Difficulty: Medium
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The top part of a water tank is divided into two compartments, as shown in the figure. Now, a fluid with an unknown density is poured into one side, and the water level rises a certain amount on the other side to compensate for this effect. Based on the final fluid heights shown in the figure, determine the density of the fluid added. Assume the liquid does not mix with water. The height of the unknown liquid is 72 cm. Take the density of water to be ρ = 1000 kg/m3.
The density of the added fluid is
694 ± 2% kg/m3.
Explanation: The following assumptions have been made here: 1. Both water and the added liquid are incompressible substances. 2. The added liquid does not mix with water.
Both fluids are open to the atmosphere. Noting that the pressure of both water and the added fluid is the same at the contact surface, the pressure at this surface can be expressed as [Math Processing Error] On simplifying, we have ρfghf = ρwghw. Solving for ρf gives [Math Processing Error] The density of the added fluid is 694 kg/m3. References Worksheet
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Difficulty: Medium
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A simple experiment has long been used to demonstrate how negative pressure prevents water from being spilled out of an inverted glass. A glass that is fully filled by water and covered with a thin paper is inverted, as shown in the figure. The height h of the glass is 14 cm. Determine the pressure at the bottom of the glass. Take the density of water to be ρ = 1000 kg/m3.
The pressure at the bottom of the glass is
98.6 ± 2% kPa.
Explanation: The following assumptions have been made here: 1. Water is an incompressible substance. 2. The weight of the paper is negligible. 3. The atmospheric pressure is 100 kPa.
The paper is in equilibrium, and thus the net force acting on the paper must be zero. A vertical force balance on the paper involves the pressure forces on both sides and yields [Math Processing Error] That is, the pressures on both sides of the paper must be the same. The pressure at the bottom of the glass is determined from the hydrostatic pressure relation to be [Math Processing Error] Substituting, [Math Processing Error] [Math Processing Error] https://ezto.mheducation.com/hm.tpx
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The pressure at the bottom of the glass is 98.6 kPa. References Worksheet
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Difficulty: Easy
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A multifluid container is connected to a U-tube, as shown in the figure. For the given specific gravities and fluid column heights, determine the gage pressure at A. Also determine the height of a mercury column that would create the same pressure at A. The column height h of oil is 84 cm. The specific gravities are 1.26 for glycerin and 0.90 for oil. We take the standard density of water to be ρw =1000 kg/m3 and the specific gravity of mercury to be 13.6.
The gage pressure at A is 1.95 ± 2% kPa. The height of a mercury column that would create the same pressure at A is
1.46 ± 2% cm.
Explanation: The following assumptions have been made here: 1. All the liquids are incompressible. 2. The multifluid container is open to the atmosphere.
Starting with the atmospheric pressure on the top surface of the container, moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach point A, and setting the result equal to PA give [Math Processing Error] Rearranging and using the definition of specific gravity, [Math Processing Error] or [Math Processing Error] Substituting, [Math Processing Error]
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[Math Processing Error] The equivalent mercury column height is [Math Processing Error] The gage pressure at A is 1.95 kPa. The height of a mercury column that would create the same pressure at A is 1.46 cm. References Worksheet
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Difficulty: Medium
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A hydraulic lift is to be used to lift a 2600 kg weight by putting a weight of 25 kg on a piston with a diameter of 10 cm. Determine the diameter of the piston on which the weight is to be placed.
The diameter of the piston on which the weight is to be placed is
1 ± 2% m.
Explanation: The following assumptions have been made here: 1. The cylinders of the lift are vertical. 2. There are no leaks. 3. The atmospheric pressure act on both sides, and thus it can be disregarded.
Noting that pressure is force per unit area, the pressure on the smaller piston is determined from [Math Processing Error] From Pascal’s principle, the pressure on the greater piston is equal to that in the smaller piston. Then, the needed diameter is determined from [Math Processing Error] The diameter of the piston on which the weight is to be placed is 1 m. References Worksheet
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Difficulty: Easy
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On a day in which the local atmospheric pressure is 99.5 kPa, answer each of the following. Take the density of mercury and water to be 13,560 kg/m3 and 1000 kg/m3, respectively. References Section Break
71.
Difficulty: Medium
Award: 0.00 points
Calculate the column height of mercury in a mercury barometer in units of meters, feet, and inches. The column height of mercury is
0.748 ± 2% m,
2.45 ± 2% ft, or
29.5 ± 2% in.
Explanation: We consider a mercury barometer, where the density of the mercury is 13,560 kg/m3. Using the hydrostatic pressure relation, [Math Processing Error] We round to h = 0.748 m = 2.45 ft = 29.5 inches of mercury. The column height of mercury is 0.748 m, 2.45 ft, or 29.5 in. References Worksheet
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Difficulty: Easy
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Francis is concerned about mercury poisoning, so he builds a water barometer to replace the mercury barometer. Calculate the column height of water in the water barometer in units of meters, feet, and inches. The column height of water is
10.143 ± 2% m,
33.3 ± 2% ft, or
399 ± 2% in.
Explanation: For a water barometer, where the density of the water is 1000 kg/m3, [Math Processing Error] We round to h = 10.143 m = 33.3 ft = 399 inches of water The column height of water is 10.1 m, 33.3 ft, or 399 in. References Worksheet
73.
Difficulty: Easy
Award: 0.00 points
Why is a water barometer not very practical?
The viscosity of water is low. The column height would be too high. A water barometer would not be precise. A water barometer would have leakage problems. The column height for a water barometer would be too high. It would be approximately 3 stories high when measuring atmospheric pressure. The column height would be too tall. References Multiple Choice
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Difficulty: Easy
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A U-tube manometer is used to measure the pressure in a vacuum chamber as shown in the sketch. The fluid in the vacuum chamber is of density ρ1, and the manometer fluid is of density ρ2. The right side of the U-tube manometer is exposed to atmospheric pressure Patm, and elevations z1, z2, and zA are measured.
References Section Break
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Difficulty: Medium
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Which of the following expression provides an exact expression for the vacuum pressure at point A, i.e., generate an expression for PA,vaccum as a function of variables ρ1, ρ2, z1, z2, and zA?
PA,vacuum = ρ2g(z2 − z1) + ρ1g(z1 − z2) PA,vacuum = ρ2g(z2 − z1) + ρ2g(z1 − z2) PA,vacuum = ρ1g(zA − z1) + ρ2g(zA − z1) PA,vacuum = ρ1g(zA − z1) + ρ2g(z1 − z2) We label points 1, 2, and 2', where 2' is at the same elevation as 2 and is within the same fluid. Thus point 2' has the same pressure as point 2. From hydrostatics,
[Math Processing Error] Starting at point 2, we move along the manometer from 2 to A:
P2' = P2 − ρ1g(z1 − z2)− ρ2g(zA − z1) = PA But P2 is equal to Patm, and PA,vacuum = Patm – PA. Thus,
PA,vacuum = ρ1g(zA − z1) + ρ2g(z1 − z2) The correct expression is PA,vacuum = ρ1g(zA − z1) + ρ2g(z1 − z2) References Multiple Choice
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Difficulty: Medium
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What is the simplified expression for the case in which ρ1 ≪ ρ2?
[Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] If ρ1 << ρ2, the above simplifies to
[Math Processing Error] If ρ1 << ρ2, the above simplifies to
[Math Processing Error] References Multiple Choice
76.
Difficulty: Medium
Award: 0.00 points
The resultant hydrostatic force acting on a submerged surface is the resultant of the _____ acting on the surface.
pressure forces
The resultant hydrostatic force acting on a submerged surface is the resultant of the pressure forces acting on the surface. The resultant hydrostatic force acting on a submerged surface is the resultant of the pressure forces acting on the surface. References Fill in the Blank
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Difficulty: Easy
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The resultant hydrostatic force acting on a submerged surface is the resultant of the pressure forces acting on the surface. The point of application of this resultant force is called the _____.
center of pressure
The resultant hydrostatic force acting on a submerged surface is the resultant of the pressure forces acting on the surface. The point of application of this resultant force is called the center of pressure. The resultant hydrostatic force acting on a submerged surface is the resultant of the pressure forces acting on the surface. The point of application of this resultant force is called the center of pressure. References Fill in the Blank
78.
Difficulty: Medium
Award: 0.00 points
Why are dams built much thicker at the bottom?
The pressure force increases with depth. The bottom part of dams are subjected to the greatest forces. Dams require reinforcement of their structure at the bottom. This increases the water capacity of the dam. Dams are built much thicker at the bottom because the pressure force increases with depth, and the bottom part of dams are subjected to the greatest forces. Dams are built much thicker at the bottom because the pressure force increases with depth, and the bottom part of dams are subjected to the greatest forces. References Check All That Apply
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Difficulty: Easy
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Someone claims that she can determine the magnitude of the hydrostatic force acting on a plane surface submerged in water regardless of its shape and orientation if she knew the vertical distance of the centroid of the surface from the free surface and the area of the surface. Is this a valid claim?
Yes No
Yes, because the magnitude of the resultant force acting on a plane surface of a completely submerged body in a homogeneous fluid is equal to the product of the pressure PC at the centroid of the surface and the area A of the surface. The pressure at the centroid of the surface is
[Math Processing Error] where hc is the vertical distance of the centroid from the free surface of the liquid. Yes, because the magnitude of the resultant force acting on a plane surface of a completely submerged body in a homogeneous fluid is equal to the product of the pressure PC at the centroid of the surface and the area A of the surface. References Yes / No
80.
Difficulty: Medium
Award: 0.00 points
A submerged horizontal flat plate is suspended in water by a string attached at the centroid of its upper surface. Now the plate is rotated 45° about an axis that passes through its centroid. Will there be a change in the hydrostatic force acting on the top surface of this plate as a result of this rotation? Assume the plate remains submerged at all times.
Yes No
There will be no change in the hydrostatic force acting on the top surface of this submerged horizontal flat plate as a result of this rotation since the magnitude of the resultant force acting on a plane surface of a completely submerged body in a homogeneous fluid is equal to the product of the pressure PC at the centroid of the surface and the area A of the surface.There will be no change in the hydrostatic force acting on the top surface of this submerged horizontal flat plate. References Yes / No
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Difficulty: Medium
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If you consider a submerged curved surface, the horizontal component of the hydrostatic force acting on a curved surface is equal (in both magnitude and the line of action) to the hydrostatic force acting on the _____ of the curved surface.
vertical projection
The horizontal component of the hydrostatic force acting on a curved surface is equal (in both magnitude and the line of action) to the hydrostatic force acting on the vertical projection of the curved surface. The horizontal component of the hydrostatic force acting on a curved surface is equal to the hydrostatic force acting on the vertical projection of the curved surface. References Fill in the Blank
82.
Difficulty: Easy
Award: 0.00 points
If you consider a submerged curved surface, the vertical component of the hydrostatic force acting on a curved surface is equal to the hydrostatic force acting on the _____ of the curved surface, plus (minus, if acting in the opposite direction) the weight of the fluid block.
horizontal projection
The horizontal component of the hydrostatic force acting on a curved surface is equal (in both magnitude and the line of action) to the hydrostatic force acting on the vertical projection of the curved surface. The vertical component of the hydrostatic force acting on a curved surface is equal to the hydrostatic force acting on the horizontal projection of the curved surface, plus (minus, if acting in the opposite direction) the weight of the fluid block. References Fill in the Blank
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Difficulty: Easy
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Consider a circular surface subjected to hydrostatic forces by a liquid of constant density. If the magnitudes of the horizontal and vertical components of the resultant hydrostatic force are determined, what is the formula for the line of action of this force?
[Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] The resultant hydrostatic force acting on a circular surface always passes through the center of the circle since the pressure forces are normal to the surface, and all lines normal to the surface of a circle pass through the center of the circle. Thus, the pressure forces form a concurrent force system at the center, which can be reduced to a single equivalent force at that point. If the magnitudes of the horizontal and vertical components of the resultant hydrostatic force are known, the tangent of the angle the resultant hydrostatic force makes with the horizontal is
[Math Processing Error] The formula for the line of action of force is [Math Processing Error] References Multiple Choice
Difficulty: Medium
Consider a 500-ft-high, 1200-ft-wide dam filled to capacity. Take the density of water to be 62.4 lbm/ft3. References Section Break
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Difficulty: Medium
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Determine the hydrostatic force on the dam. The hydrostatic force on the dam is
9.4 ± 2% × 109 lbf.
Explanation: The assumption here is that the atmospheric pressure acts on both sides of the dam, and thus it can be ignored in calculations for convenience.
The average pressure on a surface is the pressure at the centroid (midpoint) of the surface and is determined to be [Math Processing Error] [Math Processing Error] [Math Processing Error] Then, the resultant hydrostatic force acting on the dam becomes [Math Processing Error]
The hydrostatic force on the dam is 9.36 × 109 lbf. References Worksheet
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Difficulty: Easy
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Determine the force per unit area of the dam near the top and near the bottom. The force per unit area of the dam near the top is The force per unit area of the dam near the bottom is
0 ± 2% lbf/ft2. 31200 ± 2% lbf/ft2.
Explanation: The resultant force per unit area is pressure, and its value at the top and the bottom of the dam becomes [Math Processing Error] [Math Processing Error] The force per unit area of the dam near the top is 0 lbf/ft2. The force per unit area of the dam near the bottom is 31200 lbf/ft2. References Worksheet
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Difficulty: Medium
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A cylindrical tank is fully filled with water, as shown in the figure. The diameter of the tank is 1 m. In order to increase the flow from the tank, an additional pressure is applied to the water surface by a compressor. For P0 = 0, P0 = 5 bar, and P0 = 10 bar, calculate the hydrostatic force on the surface A exerted by water. Take the density of water to be 1000 kg/m3 throughout.
When P0 = 0, the hyrdostatic force is 3.85 ± 2% kN. When P0 = 5 bar, the hyrdostatic force is 392.71 ± 2% kN. When P0 = 10 bar, the hyrdostatic force is 785.42 ± 2% kN.
Explanation: Assume that the atmospheric pressure acts on both sides of the cylinder. Thus, it can be ignored in calculations for convenience.
When P0 = 0 bar, [Math Processing Error] [Math Processing Error] When P0 = 5 bar, the additional imaginary water column is determined as follows: [Math Processing Error] Therefore, we can assume the water level to be 50.97 m higher than the original level. In this case, [Math Processing Error] [Math Processing Error] [Math Processing Error] When P0 = 10 bar, the additional imaginary water column is determined as follows: [Math Processing Error] https://ezto.mheducation.com/hm.tpx
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[Math Processing Error]
When P0 = 0, the hyrdostatic force is 3.85 kN. When P0 = 5 bar, the hyrdostatic force is 392.71 kN. When P0 = 10 bar, the hyrdostatic force is 785.42 kN. References Worksheet
Difficulty: Medium
Consider an 8-m-long, 8-m-wide, and 4-m-high aboveground swimming pool that is filled with water to the rim. References Section Break
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Difficulty: Medium
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Determine the hydrostatic force on each wall and the distance of the line of action of this force from the ground. The hydrostatic force on each wall is 627.8 ± 2% kN. The distance of the line of action of this force from the ground is
1.333 ± 2% m.
Explanation: Assume that the atmospheric pressure acts on both sides of the wall of the pool. Thus, it can be ignored in calculations for convenience. The density of water is taken to be 1000 kg/m3 throughout. The average pressure on a surface is the pressure at the centroid (midpoint) of the surface and is determined to be [Math Processing Error] Then the resultant hydrostatic force on each wall becomes [Math Processing Error]
The line of action of the force passes through the pressure center, which is 2h/3 from the free surface and h/3 from the bottom of the pool. Therefore, the distance of the line of action from the ground is [Math Processing Error]
The hydrostatic force on each wall is 627.8 kN. The distance of the line of action of this force from the ground is 1.333 m. References Worksheet
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Difficulty: Medium
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If the height of the walls of the pool is doubled and the pool is filled, will the hydrostatic force on each wall double or quadruple?
It will quadruple. It will double. If the height of the walls of the pool is doubled, the hydrostatic force quadruples since
[Math Processing Error] Thus, the hydrostatic force is proportional to the square of the wall height, h2. If the height of the walls of the pool is doubled, the hydrostatic force quadruples. References Multiple Choice
Difficulty: Medium
Consider a heavy car submerged in water in a lake with a flat bottom. The driver’s side door of the car is 1.1 m high and 0.9 m wide, and the top edge of the door is 17 m below the water surface. References Section Break
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Difficulty: Medium
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Determine the net force acting on the door (normal to its surface) and the location of the pressure center if the car is wellsealed and it contains air at atmospheric pressure. The net force acting on the door is
170.4 ± 2% kN.
The location of the pressure center if the car is well-sealed and it contains air at atmospheric pressure is
17.56 ± 2% m.
Explanation: The following assumptions are made here: 1. The bottom surface of the lake is horizontal. 2. The door can be approximated as a vertical rectangular plate. 3. The pressure in the car remains at atmospheric value since there is no water leaking in and thus no compression of the air inside. Therefore, we can ignore the atmospheric pressure in calculations since it acts on both sides of the door.
When the car is well-sealed and thus the pressure inside the car is the atmospheric pressure, the average pressure on the outer surface of the door is the pressure at the centroid (midpoint) of the surface, and it is determined to be [Math Processing Error] [Math Processing Error] [Math Processing Error] Then, the resultant hydrostatic force on the door becomes [Math Processing Error] The pressure center is directly under the midpoint of the plate, and its distance from the surface of the lake is determined to be [Math Processing Error] where P0 = 0 kN/m2 The net force acting on the door is 170.4 kN. The location of the pressure center is 17.56 m. References Worksheet
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Difficulty: Medium
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Determine the net force acting on the door (normal to its surface) and the location of the pressure center if the car is filled with water. The net force acting on the door is
0 kN.
Explanation: When the car is filled with water, the net force normal to the surface of the door is zero since the pressure on both sides of the door will be the same. When the car is filled with water, the net force normal to the surface of the door is zero since the pressure on both sides of the door will be the same. References Worksheet
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Difficulty: Easy
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A room in the lower level of a cruise ship has a 42-cm-diameter circular window. If the midpoint of the window is 2 m below the water surface, determine the hydrostatic force acting on the window and the pressure center. Take the specific gravity of seawater to be 1.025.
The hydrostatic force acting on the window is 2786 ± 2% N. The vertical distance from the free surface to the pressure center is
2.006 ± 2% m.
Explanation: It is assumed that the atmospheric pressure acts on both sides of the window, and thus it can be ignored in calculations for convenience.
The average pressure on a surface is the pressure at the centroid (midpoint) of the surface and is determined to be [Math Processing Error] Then the resultant hydrostatic force on each wall becomes [Math Processing Error] [Math Processing Error] The line of action of the force passes through the pressure center, whose vertical distance from the free surface is determined from [Math Processing Error] The hydrostatic force acting on the window is 2786 N. The vertical distance from the free surface to the pressure center is 2.006 m. References Worksheet
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Difficulty: Medium
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The water side of the wall of a 70-m-long dam is a quarter-circle with a radius of 7 m. Determine the hydrostatic force on the dam and its line of action when the dam is filled to the rim. Take the density of water to be 1000 kg/m3. 3.13 ± 2% × 107 N. The hydrostatic force on the dam is 57.5 ± 2% °. The line of action when the dam is filled to the rim is
Explanation: It is assumed that the atmospheric pressure acts on both sides of the dam. Thus, it can be ignored in calculations for convenience. Take the density of water to be 1000 kg/m3 throughout. We consider the free-body diagram of the liquid block enclosed by the circular surface of the dam and its vertical and horizontal projections.
The hydrostatic forces acting on the vertical and horizontal plane surfaces as well as the weight of the liquid block are as follows: The horizontal force on the vertical surface is [Math Processing Error] [Math Processing Error] [Math Processing Error] The vertical force on the horizontal surface is zero since it coincides with the free surface of water. The weight of fluid block per m length is [Math Processing Error] [Math Processing Error] [Math Processing Error] Then, the magnitude and direction of the hydrostatic force acting on the surface of the dam become [Math Processing Error] [Math Processing Error] Therefore, the line of action of the hydrostatic force passes through the center of the curvature of the dam, making an angle of 57.5° downward from the horizontal.
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The hydrostatic force on the dam is 3.13 × 107 N. The line of action when the dam is filled to the rim is 57.5°. References Worksheet
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Difficulty: Medium
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A water trough of semicircular cross section of radius 1.2 m consists of two symmetric parts hinged to each other at the bottom, as shown in the figure. The two parts are held together by a cable and turnbuckle placed every 3 m along the length of the trough. Calculate the tension in each cable when the trough is filled to the rim.
The tension in each cable when the trough is filled to the rim is
21,189 ± 2% N.
Explanation: The following assumptions have been made here: 1. Atmospheric pressure acts on both sides of the trough wall, and thus it can be ignored in calculations for convenience. 2. The weight of the trough is negligible.
To expose the cable tension, we consider half of the trough whose cross section is a quarter-circle. The hydrostatic forces acting on the vertical and horizontal plane surfaces as well as the weight of the liquid block are as follows: The horizontal force on the vertical surface is [Math Processing Error] [Math Processing Error] [Math Processing Error] The vertical force on the horizontal surface is zero, since it coincides with the free surface of water. The weight of fluid block per [tension(2)]-m length is [Math Processing Error] [Math Processing Error] [Math Processing Error] Then, the magnitude and direction of the hydrostatic force acting on the surface of the 3-m-long section of the trough become [Math Processing Error] [Math Processing Error] Therefore, the line of action passes through the center of the curvature of the trough, making 57.52° downward from the horizontal. Taking the moment about point A where the two parts are hinged and setting it equal to zero give [Math Processing Error] https://ezto.mheducation.com/hm.tpx
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Solving for T and substituting, the tension in the cable is determined to be [Math Processing Error] The tension in each cable when the trough is filled to the rim is 21,189 N. References Worksheet
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Difficulty: Medium
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Determine the resultant force acting on the 0.7-m-high and 0.7-m-wide triangular gate shown in the figure and its line of action. Take the density of water to be 1000 kg/m3. The distance d from the surface to the top of the gate is 0.39 m as shown in the figure.
The resultant force acting on the triangular gate is The line of action of the force is 0.888 ± 2% m.
1700 ± 2% N.
Explanation: The assumption made here is that the atmospheric pressure acts on both sides of the gate. Thus, it can be ignored in calculations for convenience. The angle can be determined as follows: [Math Processing Error] [Math Processing Error] [Math Processing Error] This is the resultant force acting on the triangular gate. In order to locate FR on the gate xcp, ycp must be found. [Math Processing Error] For simplicity, we can consider the x-axis to be passing through the center of gravity of the gate so that xcg = 0. [Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] This is the line of action of the force. The resultant force acting on the triangular gate is 1700 N. The line of action of the force is 0.888 m. References
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Worksheet
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Difficulty: Easy
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A 11-m-high, 10-m-wide rectangular plate blocks the end of a 10-m-deep freshwater channel, as shown in the figure. The plate is hinged about a horizontal axis along its upper edge through a point A and is restrained from opening by a fixed ridge at point B. Determine the force exerted on the plate by the ridge. In the figure, the total height (s + h) is 11 m where s is 1 m.
The force exerted on the plate by the ridge is
3419 ± 2% kN.
Explanation: It is assumed that the atmospheric pressure acts on both sides of the plate. Thus, it can be ignored in calculations for convenience. Take the density of water to be 1000 kg/m3 throughout. The average pressure on a surface is the pressure at the centroid (midpoint) of the surface, and it is determined as follows:
[Math Processing Error] [Math Processing Error] [Math Processing Error] Then, the resultant hydrostatic force on each wall becomes [Math Processing Error] The line of action of the force passes through the pressure center, which is 2h/3 from the free surface. [Math Processing Error] https://ezto.mheducation.com/hm.tpx
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Taking the moment about point A and setting it equal to zero give [Math Processing Error] Solving for Fridge and substituting, the reaction force is determined as follows: [Math Processing Error] The force exerted on the plate by the ridge is 3419 kN. References Worksheet
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Difficulty: Medium
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A 6-m-high, 5-m-wide rectangular plate blocks the end of a 5-m-deep freshwater channel, as shown in the figure. The plate is hinged about a horizontal axis along its upper edge through a point A and is restrained from opening by a fixed ridge at point B. Using appropriate software, investigate the effect of water depth on the force exerted on the plate by the ridge. Let the water depth vary from 0 to 5 m in increments of 0.5 m. Tabulate and plot your results. (Please upload your response/solution using the controls provided below.)
Student upload controls will be shown to students when they take this assignment.
References File Attachment Question
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Difficulty: Hard
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The flow of water from a reservoir is controlled by a 6.5-ft-wide L-shaped gate hinged at point A, as shown in the figure. If it is desired that the gate opens when the water height is 12 ft, determine the mass of the required weight W.
The mass of the required weight W is
40154 ± 2% lbm.
Explanation: The following assumptions have been made: 1. Atmospheric pressure acts on both sides of the gate. Thus, it can be ignored in calculations for convenience. 2. The weight of the gate is negligible. Take the density of water to be 62.4 lbm/ft3 throughout. The average pressure on a surface is the pressure at the centroid (midpoint) of the surface and is determined:
[Math Processing Error] Then, the resultant hydrostatic force on the dam becomes as follows: [Math Processing Error] The line of action of the force passes through the pressure center, which is 2h/3 from the free surface. [Math Processing Error] https://ezto.mheducation.com/hm.tpx
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Taking the moment about point A and setting it equal to zero give [Math Processing Error] Solving for W and substituting, the required weight is determined as follows: [Math Processing Error] The mass of the required weight W is 40154 lbm. References Worksheet
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Difficulty: Medium
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The flow of water from a reservoir is controlled by an L-shaped gate that is 10 ft wide and is hinged at point A, as shown in the figure below. If it is desired that the gate open when the water height is 6 ft, determine the mass of the required weight W. Take the density of water to be 62.4 lbm/ft3 throughout.
The mass of the required weight W is
18252 ± 2% lbf.
Explanation: The following assumptions have been made here: 1. Atmospheric pressure acts on both sides of the gate, and thus it can be ignored in calculations for convenience. 2. The weight of the gate is negligible.
The average pressure on a surface is the pressure at the centroid (midpoint) of the surface and is determined to be [Math Processing Error] [Math Processing Error] [Math Processing Error] The resultant hydraulic force on the dam is [Math Processing Error]
https://ezto.mheducation.com/hm.tpx
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The line of action of the force passes through the pressure center, which is 2h/3 from the free surface. [Math Processing Error] Taking the moment about point A and setting it equal to zero gives [Math Processing Error] Solving for W and substituting, the required weight is determined to be [Math Processing Error]
The mass of the required weight W = 18252 lbf. References Worksheet
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Difficulty: Medium
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For the gate width of 3 m into the paper shown in the figure, determine the force required to hold the gate ABC at its location.
The force required to hold the gate ABC at its location is
39.48 ± 2% kN.
Explanation: It is assumed that the atmospheric pressure acts on both sides of the gate. Thus, it can be ignored in calculations for convenience.
Since there are two different fluid layers the problem would be made easier by converting one of the layers to the other. We determine how much fluid from the second one can make the same pressure. [Math Processing Error] Therefore, the system can be simplified as shown: https://ezto.mheducation.com/hm.tpx
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[Math Processing Error] [Math Processing Error] [Math Processing Error] Taking the moment about the hinge will give [Math Processing Error] F = 39479 N = 39.48 kN The force required to hold the gate ABC at its location is 39.48 kN. References Worksheet
Difficulty: Hard
A long, solid cylinder of radius 2 ft hinged at point A is used as an automatic gate, as shown in the figure. When the water level reaches 12 ft, the cylindrical gate opens by turning about the hinge at point A.
References Section Break
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Difficulty: Medium
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100. Award: 0.00 points
Determine the hydrostatic force acting on the cylinder and its line of action when the gate opens. The hydrostatic force acting on the cylinder is The line of action when the gate opens is
1993 ± 2% lbf. 46.5 ± 2% °.
Explanation: The following assumptions have been made here: 1. The hinge is frictionless. 2. Atmospheric pressure acts on both sides of the gate, and thus it can be ignored in calculations for convenience.
We consider the free-body diagram of the liquid block enclosed by the circular surface of the cylinder and its vertical and horizontal projections. The hydrostatic forces acting on the vertical and horizontal plane surfaces as well as the weight of the liquid block per ft length of the cylinder are as follows: The horizontal force on the vertical surface is [Math Processing Error] [Math Processing Error] [Math Processing Error] The vertical force on the horizontal surface (upward) is [Math Processing Error] [Math Processing Error] [Math Processing Error] The weight of fluid block per ft length (downward) is [Math Processing Error] [Math Processing Error] [Math Processing Error] Therefore, the net upward vertical force is [Math Processing Error] Then, the magnitude and direction of the hydrostatic force acting on the cylindrical surface become [Math Processing Error] [Math Processing Error] https://ezto.mheducation.com/hm.tpx
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Therefore, the magnitude of the hydrostatic force acting on the cylinder is 1993 lbf per ft length of the cylinder, and its line of action passes through the center of the cylinder making an angle 46.5° upwards from the horizontal. References Worksheet
101.
Difficulty: Medium
Award: 0.00 points
Determine the weight of the cylinder per ft length of the cylinder. The weight of the cylinder per ft length is
1446 ± 2% lbf.
Explanation: When the water level is 12 ft high, the gate opens and the reaction force at the bottom of the cylinder becomes zero. Then, the forces other than those at the hinge acting on the cylinder are its weight, acting through the center, and the hydrostatic force exerted by water. Taking a moment about the point A where the hinge is and equating it to zero give [Math Processing Error] The weight of the cylinder per ft length is 1446 lbf. References Worksheet
https://ezto.mheducation.com/hm.tpx
Difficulty: Medium
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102. Award: 0.00 points
An open settling tank shown in the figure contains a liquid suspension, as shown in the figure. Determine the resultant force acting on the gate and its line of action if the liquid density is 860 kg/m3.
The resultant force acting on the gate is 142 ± 2% kN. 1.64 ± 2% m from the bottom of the tank. The line of action of the force is
Explanation: Assume that atmospheric pressure acts on both sides of the gate. Thus, it can be ignored in calculations for convenience.
[Math Processing Error] [Math Processing Error] [Math Processing Error] Based on the above figure, [Math Processing Error] Therefore, the force FR is determined to be [Math Processing Error] [Math Processing Error] The line of action is determined as follows: [Math Processing Error] [Math Processing Error] https://ezto.mheducation.com/hm.tpx
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[Math Processing Error]
The resultant force acting on the gates is 142 kN. The line of action of the force is 1.64 m from the bottom of the tank. References Worksheet
https://ezto.mheducation.com/hm.tpx
Difficulty: Hard
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103. Award: 0.00 points
An open settling tank shown in the figure contains a liquid suspension. Knowing that the density of the suspension depends on liquid depth and changes linearly from 800 kg/m3 to 900 kg/m3 in the vertical direction, determine the resultant force acting on the gate and its line of action.
The resultant force acting on the gate is 144 ± 2% kN. The line of action of the force is 1.63 ± 2% m from the bottom of the tank. Explanation: Assume that atmospheric pressure acts on both sides of the gate. Thus, it can be ignored in calculations for convenience.
[Math Processing Error] Since the density of suspension changes linearly with h, we propose [Math Processing Error] Based on the figure [Math Processing Error] Therefore, [Math Processing Error] [Math Processing Error] To locate FR, we use Eq. 1 again. [Math Processing Error] [Math Processing Error] [Math Processing Error]
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The resultant force acting on the gate is 144 kN. The line of action of the force is 1.63 m from the bottom of the tank. References Worksheet
https://ezto.mheducation.com/hm.tpx
Difficulty: Hard
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104. Award: 0.00 points
The two sides of a V-shaped water trough are hinged to each other at the bottom where they meet, as shown in the figure, making an angle of 45° with the ground from both sides. Each side is 0.75 m wide, and the two parts are held together by a cable and turnbuckle placed every 8 m along the length of the trough. Calculate the tension in each cable when the trough is filled to the rim. Take the density of water to be 1000 kg/m3.
The tension in each cable is
7348 ± 2% N.
Explanation: The following assumptions have been made here: 1. Atmospheric pressure acts on both sides of the trough wall, and thus it can be ignored in calculations for convenience. 2. The weight of the trough is negligible.
To expose the cable tension, we consider half of the trough whose cross section is triangular. The water height h at the midsection of the trough and the width of the free surface are [Math Processing Error] [Math Processing Error] The hydrostatic forces acting on the vertical and horizontal plane surfaces as well as the weight of the liquid block are determined as follows: The horizontal force on vertical surface is [Math Processing Error] [Math Processing Error] [Math Processing Error] The vertical force on the horizontal surface is zero since it coincides with the free surface of water. The weight of the fluid block per 8-m length is [Math Processing Error] [Math Processing Error] [Math Processing Error]
https://ezto.mheducation.com/hm.tpx
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The distance of the centroid of a triangle from a side is 1/3 of the height of the triangle for that side. Taking the moment about point A where the two parts are hinged and setting it equal to zero give [Math Processing Error] Solving for T and substituting, and noting that h = b, the tension in the cable is determined to be [Math Processing Error] The tension in each cable is 7348 N. References Worksheet
https://ezto.mheducation.com/hm.tpx
Difficulty: Medium
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105. Award: 0.00 points
The two sides of a V-shaped water trough are hinged to each other at the bottom where they meet, as shown in the figure, making an angle of 45° with the ground from both sides. Each side is 0.35 m wide, and the two parts are held together by a cable and turnbuckle placed every 6 m along the length of the trough. Calculate the tension in each cable when the trough is partially filled at a height of 0.29 m directly above the hinge.
The tension in the cable is
1650 ± 2% N.
Explanation: The following assumptions have been made here: 1. Atmospheric pressure acts on both sides of the trough wall, and thus it can be ignored in calculations for convenience. 2. The weight of the trough is negligible.
To expose the cable tension, we consider half of the trough whose cross section is triangular. The water height is given to be h = 0.29 m at the midsection of the trough, which is equivalent to the width of the free surface b since tan 45° = b/h = 1. The hydrostatic forces acting on the vertical and horizontal plane surfaces as well as the weight of the liquid block are determined as follows: The horizontal force on vertical surface is [Math Processing Error] [Math Processing Error] [Math Processing Error] The vertical force on the horizontal surface is zero since it coincides with the free surface of water. The weight of the fluid block per 6-m length is [Math Processing Error] [Math Processing Error] [Math Processing Error] The distance of the centroid of a triangle from a side is 1/3 of the height of the triangle for that side. Taking the moment about point A where the two parts are hinged and setting it equal to zero give [Math Processing Error] https://ezto.mheducation.com/hm.tpx
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Solving for T and substituting, and noting that h = b, the tension in the cable is determined to be [Math Processing Error] The tension in the cable is 1650 N. References Worksheet
https://ezto.mheducation.com/hm.tpx
Difficulty: Hard
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106. Award: 0.00 points
The bowl shown in the figure is to be cast in a pair of molding boxes. Knowing that the mass of the upper molding box is 240 kg, what would be the tensile force on each of 10 bolts located circumferentially? The specific gravity of the molten metal can be taken to be 8.6. Take the density of water to be 1000 kg/m3.
The tensile force on each bolt is
1033 ± 2% N.
Explanation: The vertical force on the upper molding box can be evaluated from "FV = γV = SGρwgV", where V is shown by dashed lines in the figure.
[Math Processing Error] Therefore, [Math Processing Error] The tensile on each bolt is then [Math Processing Error] The tensile force on each bolt is 1033 N.
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References Worksheet
https://ezto.mheducation.com/hm.tpx
Difficulty: Medium
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107. Award: 0.00 points
A triangular shaped gate is hinged at point A, as shown. Knowing that the weight of the gate is 540 N/m, determine the force needed to keep the gate at its position per unit width. The line of action of the weight of the gate is denoted by the dashed line. Take the density of water to be 1000 kg/m3. The height h of the gate is 2 m. In the given figure, g is 2.4 m and the thickness of the gate is not considered.
The force needed to keep the gate at its position per unit width is
18,949 ± 2% N/m.
Explanation: It is assumed that atmospheric pressure acts on both sides of the gate, and thus it can be ignored in calculations for convenience. Free-body diagram of the gate is shown below:
[Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] Calculate the moment arms: [Math Processing Error] [Math Processing Error] [Math Processing Error] where https://ezto.mheducation.com/hm.tpx
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[Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] Taking moment about point A would yield [Math Processing Error] [Math Processing Error] The force needed to keep the gate at its position per unit width is 18949 N/m. References Worksheet
https://ezto.mheducation.com/hm.tpx
Difficulty: Hard
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108. Award: 0.00 points
Gate AB (0.6-m × 0.9-m) is located at the bottom of a tank filled with methyl alcohol (SG = 0.79) and hinged along its bottom edge A. Knowing that the weight of the gate is 340 N, determine the minimum force that must be applied to the cable (BCD) to open the gate. Take the density of water to be 1000 kg/m3.
The minimum force that must be applied to the cable to open the gate is
1352 ± 2% N.
Explanation: It is assumed that atmospheric pressure acts on both sides of the gate, and thus it can be ignored in calculations for convenience. When a minimum force applied to the cable that would open the gate, the reaction force at point B is zero. We have three main forces, Fcable, Wgate, and the hydrostatic force by methyl alcohol, FR.
[Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] Let θ be the angle between the alcohol surface and the cable. [Math Processing Error] Take the moment about point A: [Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] The minimum force that must be applied to the cable to open the gate is 1352 N. https://ezto.mheducation.com/hm.tpx
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References Worksheet
https://ezto.mheducation.com/hm.tpx
Difficulty: Medium
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109. Award: 0.00 points
Find the force applied by support BC to the gate AB. The width of the gate and support is 3 m and the weight of the gate is 1100 N. Take the density of water to be 1000 kg/m3.
The force applied by support BC to gate AB is
5,017 ± 02% N.
Explanation: It is assumed that atmospheric pressure acts on both sides of the gate, and thus it can be ignored in calculations for convenience. [Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] From the law of cosines, [Math Processing Error] [Math Processing Error] On the other hand, [Math Processing Error] Therefore, β = 78.4°. Taking the moment about the hinge yields [Math Processing Error] [Math Processing Error] and [Math Processing Error] [Math Processing Error] [Math Processing Error] On solving the above equation, F = 5,017 N The force applied by support BC to the gate AB is 5,017 N. References https://ezto.mheducation.com/hm.tpx
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Worksheet
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Difficulty: Medium
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110.
Assignment Print View
Award: 0.00 points
A concrete block is attached to the gate as shown. If the water level is 1.3 m from the bottom of the container, and there is no reaction force at A, what would be the reaction force for the level shown? The specific gravity of the concrete is 2.4.
The reaction force at the indicated level is
16,994 ± 2% N.
Explanation: It is assumed that atmospheric pressure acts on both sides of the gate, and thus it can be ignored in calculations for convenience. The case where no spring deflection exists is shown in the figure.
[Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] On the other hand, for the case shown below, the following calculations will be made:
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[Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] The moment about the hinge will result in [Math Processing Error] [Math Processing Error] With the given data, we will obtain FR = 16,994 N The reaction force at the indicated level is 16,994 N. References Worksheet
https://ezto.mheducation.com/hm.tpx
Difficulty: Medium
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111.
Assignment Print View
Award: 0.00 points
The curved surface given in the figure is defined by y = 3[Math Processing Error]. Determine the horizontal force and its line of action applied by water on the curved surface. The width of the gate is b = 2.5 m. Take the density of water to be 1000 kg/m3.
The horizontal force is 306562 ± 2% N. The line of action from the bottom is 1.67 ± 2% m. Explanation: It is assumed that atmospheric pressure acts on both sides of the surface, and thus it can be ignored in calculations for convenience. For a curved surface, the horizontal component of the hydrostatic force is given by [Math Processing Error] [Math Processing Error] It is clear that h + y = 5 m, and we may write h = 5 m − y. Therefore, Eq. 2 will take the form of [Math Processing Error] [Math Processing Error] [Math Processing Error] Since the moment generated by FH about the hinge must be equal to the sum of the moments generated by dFH about the same point, we can write [Math Processing Error] [Math Processing Error] [Math Processing Error] Therefore, [Math Processing Error] The horizontal force is 306562 N. The line of action from the bottom is 1.67 m. References Worksheet
https://ezto.mheducation.com/hm.tpx
Difficulty: Medium
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Award: 0.00 points
A 3.7-m-long quarter-circular gate of radius 3 m and of negligible weight is hinged about its upper edge A, as shown in the figure. The gate controls the flow of water over the ledge at B, where the gate is pressed by a spring. Determine the minimum spring force required to keep the gate closed when the water level rises to A at the upper edge of the gate. Take the density of water to be 1000 kg/m3.
The minimum spring force required to keep the gate closed is
163.3 ± 2% kN.
Explanation: The following assumptions have been made here: 1. The hinge is frictionless. 2. Atmospheric pressure acts on both sides of the gate, and thus it can be ignored in calculations for convenience. 3. The weight of the gate is negligible.
We consider the free-body diagram of the liquid block enclosed by the circular surface of the gate and its vertical and horizontal projections. The hydrostatic forces acting on the vertical and horizontal plane surfaces as well as the weight of the liquid block are determined as follows: The horizontal force on the vertical surface is [Math Processing Error] [Math Processing Error] [Math Processing Error] The vertical force on the horizontal surface (upward) is [Math Processing Error] [Math Processing Error] The weight of the fluid block per 4-m length (downward) is [Math Processing Error] [Math Processing Error] https://ezto.mheducation.com/hm.tpx
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Therefore, the net upward vertical force is [Math Processing Error] Then, the magnitude and direction of the hydrostatic force acting on the surface of the (expression error)-m long quartercircular section of the gate become [Math Processing Error] [Math Processing Error] Therefore, the magnitude of the hydrostatic force acting on the gate is 177.7 kN, and its line of action passes through the center of the quarter-circular gate making an angle of 23.2° upward from the horizontal. The minimum spring force needed is determined by taking a moment about the point A where the hinge is, and setting it equal to zero. [Math Processing Error] Solving for Fspring and substituting, the spring force is determined to be [Math Processing Error] The minimum spring force required to keep the gate closed is 163.3 kN. References Worksheet
https://ezto.mheducation.com/hm.tpx
Difficulty: Medium
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113.
Assignment Print View
Award: 0.00 points
A 4.3-m-long quarter-circular gate of radius 2 m and of negligible weight is hinged about its upper edge A, as shown in the figure. The gate controls the flow of water over the ledge at B, where the gate is pressed by a spring. Determine the minimum spring force required to keep the gate closed when the water level rises to A at the upper edge of the gate. Take the density of water to be 1000 kg/m3.
The minimum spring force required to keep the gate closed is
84.4 ± 2% kN.
Explanation: The following assumptions have been made here: 1. The hinge is frictionless. 2. Atmospheric pressure acts on both sides of the gate, and thus it can be ignored in calculations for convenience. 3. The weight of the gate is negligible.
We consider the free-body diagram of the liquid block enclosed by the circular surface of the gate and its vertical and horizontal projections. The hydrostatic forces acting on the vertical and horizontal plane surfaces as well as the weight of the liquid block are determined as follows: The horizontal force on the vertical surface is [Math Processing Error] [Math Processing Error] [Math Processing Error] The vertical force on the horizontal surface (upward) is [Math Processing Error] [Math Processing Error] The weight of the fluid block per 4-m length (downward) is [Math Processing Error] [Math Processing Error] https://ezto.mheducation.com/hm.tpx
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Therefore, the net upward vertical force is [Math Processing Error] Then, the magnitude and direction of the hydrostatic force acting on the surface of the 4.3-m long quarter-circular section of the gate become [Math Processing Error] [Math Processing Error] Therefore, the magnitude of the hydrostatic force acting on the gate is 91.8 kN, and its line of action passes through the center of the quarter-circular gate making an angle of 23.2° upward from the horizontal. The minimum spring force needed is determined by taking a moment about the point A where the hinge is and setting it equal to zero. [Math Processing Error] Solving for Fspring and substituting, the spring force is determined to be [Math Processing Error] The minimum spring force required to keep the gate closed is 84.4 kN. References Worksheet
114.
Difficulty: Medium
Award: 0.00 points
The upward force a fluid exerts on an immersed body is called _____.
buoyant force
The upward force a fluid exerts on an immersed body is called buoyant force. The upward force a fluid exerts on an immersed body is called the buoyant force. References Fill in the Blank
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Difficulty: Easy
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What causes buoyant force?
The increase of volume in a fluid The increase in surface area of the fluid The increase of pressure in a fluid with depth The increase in fluid density and weight of the floating object Buoyant force is caused by the increase of pressure in a fluid with depth. Buoyant force is caused by the increase of pressure in a fluid with depth. References Multiple Choice
116.
Difficulty: Easy
Award: 0.00 points
Which of the following is an expression for the magnitude of the buoyant force acting on a submerged body whose volume is V?
FB = ρfgV FB = ρfhV F B = ρfV FB = hgV The magnitude of the buoyant force acting on a submerged body whose volume is V is expressed as FB = ρfgV. The magnitude of the buoyant force acting on a submerged body whose volume is V is expressed as FB = ρfgV. References Multiple Choice
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Difficulty: Easy
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06/06/2024, 11:08
117.
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Award: 0.00 points
The direction of the buoyant force is upward, and its line of action passes through the centroid of the displaced volume.
True False
The upward force a fluid exerts on an immersed body is called buoyant force. Buoyant force is caused by the increase of pressure in a fluid with depth. The direction of the buoyant force is upward, and its line of action passes through the centroid of the displaced volume.The direction of the buoyant force is upwards, and its line of action passes through the centroid of the displaced volume. References True / False
118.
Difficulty: Easy
Award: 0.00 points
A submerged body whose center of gravity G is above the center of buoyancy B is _____ .
unstable stable A submerged body whose center of gravity G is above the center of buoyancy B is unstable. A submerged body whose center of gravity G is above the center of buoyancy B is unstable. References Multiple Choice
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Difficulty: Easy
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Award: 0.00 points
In terms of stability, a floating body whose center of gravity is above the center of buoyancy is _____.
unstable stable A floating body may still be stable when G is above B since the centroid of the displaced volume shifts to the side to a point B’ during a rotational disturbance while the center of gravity G of the body remains unchanged. If the point B’ is sufficiently far, these two forces create a restoring moment and return the body to the original position. The stability of a floating body whose center of gravity is above the center of buoyancy is stable. References Multiple Choice
Difficulty: Easy
120. Award: 0.00 points
Consider two 5-cm-diameter spherical balls—one made of aluminum, the other of iron—submerged in water. Will the buoyant forces acting on these two balls be the same?
Yes No
The magnitude of the buoyant force acting on a submerged body whose volume is V is expressed as FB = ρfgV, which is independent of the density of the body (ρ is the fluid density). Therefore, the buoyant forces acting on the 5-cm-diameter aluminum and iron balls submerged in water are the same. The buoyant forces acting on the 5-cm-diameter aluminum and iron balls submerged in water is the same. References Yes / No
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Difficulty: Easy
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Consider a 3-kg copper cube and a 3-kg copper ball submerged in a liquid. Will the buoyant forces acting on these two bodies be the same?
Yes No
The magnitude of the buoyant force acting on a submerged body whose volume is V is expressed as FB = ρfgV, which is independent of the shape of the body. Therefore, the buoyant forces acting on the cube and sphere made of copper submerged in water are the same since they have the same volume.The buoyant forces acting on the cube and sphere made of copper submerged in water are the same since they have the same volume. References Yes / No
Difficulty: Easy
122. Award: 0.00 points
Consider two identical spherical balls submerged in water at different depths. Will the buoyant forces acting on these two balls be the same?
Yes No
The magnitude of the buoyant force acting on a submerged body whose volume is V is expressed as FB = ρfgV , which is independent of depth. Therefore, the buoyant forces acting on two identical spherical balls submerged in water at different depths is the same.The buoyant forces acting on two identical spherical balls submerged in water at different depths is the same. References Yes / No
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Difficulty: Easy
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123. Award: 0.00 points
A 170-kg granite rock (ρ = 2700 kg/m3) is dropped into a lake. A man dives in and tries to lift the rock. Determine how much force the man needs to apply to lift it from the bottom of the lake? Take the density of water to be 1000 kg/m3. The force needed to lift the rock from the bottom of the lake is
1050 ± 2% N.
Explanation: The following assumptions have been made here: 1. The rock is completely submerged in water. 2. The buoyant force in air is negligible.
The weight and volume of the rock are [Math Processing Error] [Math Processing Error] The buoyancy force acting on the rock is [Math Processing Error] The weight of a body submerged in water is equal to the weight of the body in air minus the buoyant force: [Math Processing Error] The force needed to lift the rock from the bottom of the lake is 1050 N. References Worksheet
Difficulty: Easy
The hull of a boat has a volume of 180 m3, and the total mass of the boat when empty is 8560 kg. Take the specific gravity of the sea water to be 1.03 and take the density of water to be 1000 kg/m3. References Section Break
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Difficulty: Medium
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124. Award: 0.00 points
How much load can this boat carry without sinking in a lake? The boat can carry 171460 ± 2% kg without sinking. Explanation: The following assumptions have been made here: 1. The dynamic effects of the waves are disregarded. 2. The buoyancy force in air is negligible.
The weight of the unloaded boat is [Math Processing Error] The buoyancy force becomes a maximum when the entire hull of the boat is submerged in water, and is determined to be [Math Processing Error] The total weight of a floating boat (load + boat itself) is equal to the buoyancy force. Therefore, the weight of the maximum load is [Math Processing Error] The corresponding mass of the load is [Math Processing Error]
The boat can carry 171460 kg without sinking. References Worksheet
https://ezto.mheducation.com/hm.tpx
Difficulty: Medium
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125. Award: 0.00 points
How much load can this boat carry without sinking in seawater with a specific gravity of 1.03? The boat can carry 176863 ± 2% kg without sinking. Explanation: It is assumed that the extra weight will not make the submerged fraction of the ship increase but causes buoyancy forces to increase slightly. The weight of the unloaded boat is [Math Processing Error] The buoyancy force becomes a maximum when the entire hull of the boat is submerged in water, and it is determined to be [Math Processing Error] The total weight of a floating boat (load + boat itself) is equal to the buoyancy force. Therefore, the weight of the maximum load is [Math Processing Error] The corresponding mass of the load is [Math Processing Error]
The boat can carry 176863 kg without sinking. References Worksheet
https://ezto.mheducation.com/hm.tpx
Difficulty: Medium
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126. Award: 0.00 points
The density of a liquid is to be determined by an old 1-cm-diameter cylindrical hydrometer whose division marks are completely wiped out. The hydrometer is first dropped in water, and the water level is marked. The hydrometer is then dropped into the other liquid, and it is observed that the mark for water has risen 0.32 cm (hw) above the liquid–air interface. If the height of the original water mark is 12.3 cm (hl), determine the density of the liquid. Take the density of pure water to be 1000 kg/m3.
The density of the liquid is
1027 ± 2% kg/m3.
Explanation:
A hydrometer floating in water is in static equilibrium, and the buoyant force FB exerted by the liquid must always be equal to the weight W of the hydrometer, FB = W. [Math Processing Error] where h is the height of the submerged portion of the hydrometer and Ac is the cross-sectional area, which is constant. In pure water, W = ρwghwAc. In the liquid, W = ρliquidghliquidAc. Setting the relations above equal to each other (since both equal the weight of the hydrometer) gives [Math Processing Error] https://ezto.mheducation.com/hm.tpx
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Solving for the liquid density and substituting, [Math Processing Error] The density of the liquid is 1027 kg/m3. References Worksheet
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Difficulty: Medium
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127. Award: 0.00 points
It is said that Archimedes discovered his principle during a bath while thinking about how he could determine if King Hiero’s crown was actually made of pure gold. While in the bathtub, he conceived the idea that he could determine the average density of an irregularly shaped object by weighing it in air and also in water. If the crown weighed 34.8 N in air and 31.1 N in water, determine if the crown is made of pure gold. The density of gold is 19,300 kg/m3. The density of water is 1000 kg/m3.
The average density of the crown is The crown was not made of pure gold
9405 ± 2% kg/m3.
.
Explanation: The following assumptions have been made here: 1. The buoyancy force in air was negligible. 2. The crown was completely submerged in water.
The mass of the crown is [Math Processing Error] The difference between the weights in air and in water is due to the buoyancy force in water, and thus [Math Processing Error] [Math Processing Error] Then, the density of the crown becomes [Math Processing Error]
The average density of the crown is 9405 kg/m3. References Worksheet
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Difficulty: Medium
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It is estimated that 92 percent of an iceberg’s volume is below the surface, while only 8 percent is visible above the surface. For seawater with a density of 1025 kg/m3, estimate the density of the iceberg.
The density of the iceberg is
943 ± 2% kg/m3.
Explanation: The following assumptions have been made here: 1. The buoyancy force in air is negligible. 2. Density changes in the water are negligible with depth. The weight of a body floating in a fluid is equal to the buoyant force acting on it (a consequence of vertical force balance from static equilibrium). Therefore, in this case, the average density of the body must be equal to the density of the seawater since [Math Processing Error] [Math Processing Error] In the present problem, the submerged volume is 92% of the total volume. Thus, [Math Processing Error] The density of the iceberg is 943 kg/m3. References Worksheet
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Difficulty: Medium
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One of the common procedures in fitness programs is to determine the fat-to-muscle ratio of the body. This is based on the principle that the muscle tissue is denser than the fat tissue, and, thus, the higher the average density of the body, the higher the fraction of muscle tissue. The average density of the body can be determined by weighing the person in air and also while submerged in water in a tank. Treating all tissues and bones (other than fat) as muscle with an equivalent density of ρmuscle, which of the following is the correct relation for the volume fraction of body fat xfat?
[Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] The following assumptions have been made here: 1. The buoyancy force in air is negligible. 2. The body is considered to be consisted of fat and muscle only. 3. The body is completely submerged in water, and the air volume in the lungs is negligible.
The difference between the weights of the person in air and in water is due to the buoyancy force in water. Therefore,
[Math Processing Error] Knowing the weights and the density of water, the relation above gives the volume of the person. Then the average density of the person can be determined from
[Math Processing Error] The total mass of a person is equal to the sum of the masses of the fat and muscle tissues, and the total volume of a person is equal to the sum of the volumes of the fat and muscle tissues. The volume fraction of body fat is the ratio of the fat volume to the total volume of the person. Therefore,
[Math Processing Error] [Math Processing Error] https://ezto.mheducation.com/hm.tpx
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Noting that mass is density times volume, the last relation can be written as
[Math Processing Error] [Math Processing Error] Canceling the V and solving for xfat give the desired relation:
[Math Processing Error] The correct relation for the volume fraction of body fat xfat is [Math Processing Error]
References Multiple Choice
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Difficulty: Medium
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130. Award: 0.00 points
A cone floats in glycerin (SG = 1.26), as shown in the figure. Find the mass of the cone. In the figure, h = 30 cm. The radius of the cone R is 20 cm.
The mass of the cone is
45.74 ± 2% kg.
Explanation: The buoyancy force must balance the weight of the cone. That is, [Math Processing Error] [Math Processing Error]. The buoyancy force is [Math Processing Error] Therefore, the cone’s mass is then [Math Processing Error] The mass of the cone is 45.74 kg. References Worksheet
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Difficulty: Medium
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The weight of a body is usually measured by disregarding the buoyant force applied by the air. Consider a 27-cm-diameter spherical body of density 7800 kg/m3. What is the percentage error associated neglecting air buoyancy? Take the density of air to be 1.2 kg/m3. The percentage error is
-0.0154 ± 2% .
Explanation: If we neglect the buoyancy force, the weight will be [Math Processing Error] If we consider Fb, [Math Processing Error] The percentage error is then [Math Processing Error] The percentage error is -0.0154%. References Worksheet
Difficulty: Medium
132. Award: 0.00 points
A moving body of fluid can be treated as a rigid body when there are no _____ in the fluid body.
shear stresses
A moving body of fluid can be treated as a rigid body when there are no shear stresses (i.e., no motion between fluid layers relative to each other) in the fluid body. A moving body of fluid can be treated as a rigid body when there are no shear stresses (i.e., no motion between fluid layers relative to each other) in the fluid body. References Fill in the Blank
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Difficulty: Easy
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Consider a vertical cylindrical container partially filled with water. Now, the cylinder is rotated about its axis at a specified angular velocity, and rigid-body motion is established. How will the pressure be affected at the midpoint and at the edges of the bottom surface due to rotation?
The pressure at the midpoint will rise and the pressure at the edges of the bottom surface will drop due to the rotation.
The pressure at the midpoint will drop and the pressure at the edges of the bottom surface will rise due to the rotation.
The pressure remains the same throughout the container. The pressure at the bottom surface will drop. When a vertical cylindrical container partially filled with water is rotated about its axis and rigid body motion is established, the fluid level will drop at the center and rise towards the edges. Noting that hydrostatic pressure is proportional to fluid depth, the pressure at the mid point will drop and the pressure at the edges of the bottom surface will rise due to the rotation. The pressure at the mid point will drop and the pressure at the edges of the bottom surface will rise due to the rotation. References Multiple Choice
Difficulty: Easy
Consider two identical glasses of water, one stationary and the other moving on a horizontal plane with constant acceleration. References Section Break
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Difficulty: Easy
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Assuming no splashing or spilling occurs, which glass will have a higher pressure at the front?
The pressure at both the glass will be the same. The stationary glass will have a higher pressure. The moving glass will have a higher pressure. Hydrostatic pressure is directly proportional to the fluid height. The bottom pressure of the moving glass will be low at the front relative to the stationary glass. The stationary glass will have a higher pressure. References Multiple Choice
Difficulty: Easy
135. Award: 0.00 points
Assuming no splashing or spilling occurs, which glass will have a higher pressure at the midpoint?
The stationary glass will have a higher pressure. The moving glass will have a higher pressure. The pressure at the midpoint will be the same for both glasses. Hydrostatic pressure is directly proportional to the fluid height. For a glass moving on a horizontal plane with constant acceleration, water will collect at the back but the water depth will remain constant at the center. Therefore, the pressure at the midpoint will be the same for both glasses. The pressure at the midpoint will be the same for both glasses. References Multiple Choice
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Assuming no splashing or spilling occurs, which glass will have a higher pressure at the back of the bottom surface?
The moving glass will have a higher pressure. The pressure at the midpoint will be the same for both glasses. The stationary glass will have a higher pressure. Hydrostatic pressure is directly proportional to the fluid height. The bottom pressure of the moving glass will be high at the back of the bottom surface relative to the stationary glass. The moving glass will have a higher pressure. References Multiple Choice
Difficulty: Easy
Consider a glass of water. Compare the water pressures at the bottom surface for the following cases: References Section Break
Difficulty: Medium
137. Award: 0.00 points
The glass is stationary.
The water pressure decreases. The water pressure increases. The water pressure remains the same. The water pressure at the bottom surface is the same since the acceleration is zero. The water pressure remains the same. References Multiple Choice
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The glass is moving up at constant velocity.
The water pressure increases. The water pressure remains the same. The water pressure decreases. The water pressure at the bottom surface is the same since the acceleration is zero. The water pressure remains the same. References Multiple Choice
Difficulty: Easy
139. Award: 0.00 points
The glass is moving down at constant velocity.
The water pressure remains the same. The water pressure increases. The water pressure decreases. The water pressure at the bottom surface is the same since the acceleration is zero. The water pressure remains the same. References Multiple Choice
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The glass is moving horizontally at constant velocity.
The water pressure decreases. The water pressure remains the same. The water pressure increases. The water pressure at the bottom surface is the same since the acceleration is zero. The water pressure remains the same. References Multiple Choice
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A water tank is being towed by a truck on a level road, and the angle the free surface makes with the horizontal is measured to be 7.5°. Determine the acceleration of the truck.
The acceleration of the truck is
1.29 ± 2% m/s2.
Explanation: The following assumptions have been made here: 1. The road is horizontal so that acceleration has no vertical component (az = 0). 2. Effects of splashing, braking, driving over bumps, and climbing hills are assumed to be secondary and are not considered. 3. The acceleration remains constant.
We take the x-axis to be the direction of motion and the z-axis to be the upward vertical direction. The tangent of the angle the free surface makes with the horizontal is [Math Processing Error] Solving for ax and substituting, [Math Processing Error] The acceleration of the truck is 1.29 m/s2. References Worksheet
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Difficulty: Medium
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Consider two water tanks filled with water. The first tank is 8 m high and is stationary, while the second tank is 2 m high and is moving upward with an acceleration of 3.5 m/s2. Which tank will have a higher pressure at the bottom? Take the density of water to be 1000 kg/m3. The first tank
will have a higher pressure at the bottom.
Explanation: The following assumptions have been made here: 1. The acceleration remains constant. 2. Water is an incompressible substance.
The pressure difference between two points 1 and 2 in an incompressible fluid is given by [Math Processing Error] since ax = 0. Taking point 2 at the free surface and point 1 at the tank bottom, we have P2 = Patm and z2 - z1 = h and thus, [Math Processing Error] Tank A We have az = 0, and thus the pressure at the bottom is [Math Processing Error] Tank B We have az = +3.5 m/s2, and thus the pressure at the bottom is [Math Processing Error] Therefore, tank A has a higher pressure at the bottom.
The first tank will have a higher pressure at the bottom. References Worksheet
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Difficulty: Medium
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A water tank is being towed on an uphill road that makes 14° with the horizontal with a constant acceleration of 3 m/s2 in the direction of motion. Determine the angle that the free surface of water makes with the horizontal. What would your answer be if the direction of motion were downward on the same road with the same acceleration? The angle that the free surface makes with the horizontal 15.4 ± 2% °. when the direction of motion is upward is when the direction of motion is downward is -17.8 ± 2% °. Explanation: The following assumptions have been made here: 1. Effects of splashing, braking, driving over bumps, and climbing hills are assumed to be secondary and are not considered. 2. The acceleration remains constant.
We take the x- and z-axes as shown in the figure. From geometrical considerations, the horizontal and vertical components of acceleration are [Math Processing Error] [Math Processing Error] The tangent of the angle the free surface makes with the horizontal is [Math Processing Error] Therefore, [Math Processing Error] When the direction of motion is reversed, both ax and az are in the negative x- and z-direction, respectively, and thus become negative quantities, [Math Processing Error] [Math Processing Error] Then, the tangent of the angle the free surface makes with the horizontal becomes [Math Processing Error] Therefore, [Math Processing Error] The angle that the free surface makes with the horizontal when the direction of motion is upward is 15.4°. when the direction of motion is downward is -17.8°. https://ezto.mheducation.com/hm.tpx
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References Worksheet
Difficulty: Medium
The bottom quarter of a vertical cylindrical tank of total height 0.4 m and diameter 0.31 m is filled with a liquid (SG > 1, like glycerin) and the rest with water, as shown in the figure. The tank is now rotated about its vertical axis at a constant angular speed of ω. Take the density of water to be 1000 kg/m3. References Section Break
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Difficulty: Medium
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Determine the value of the angular speed when the point P on the axis at the liquid–liquid interface touches the bottom of the tank. The value of the angular speed is
12.8 ± 2% rad/s.
Explanation: The following assumptions have been made here: 1. The acceleration remains constant. 2. Water is an incompressible substance.
The pressure difference between points P and C can be expressed as [Math Processing Error] Solving for angular velocity gives [Math Processing Error] The value of the angular speed is 12.8 rad/s. References Worksheet
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Difficulty: Medium
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Determine the amount of water that will spill out at this angular speed. The amount of water that will spill out is
0.0075 ± 0.0001 m3.
Explanation: When the steady-state conditions are achieved, the shape of the isobaric surface will be as below: The volume of oil does not change, and we write [Math Processing Error] [Math Processing Error] Two surfaces will be parallel to each other since the fluid interface is an isobar surface. Therefore, the amount of water spilled from the tank will be half the volume of the CM paraboloid, that is, [Math Processing Error] The amount of water that will spill out is 0.0075 m3. References Worksheet
Difficulty: Medium
A 3-m-diameter, 8-m-long cylindrical tank is completely filled with water. The tank is pulled by a truck on a level road with the 8-mlong axis being horizontal. Take the density of the water to be 1000 kg/m3. References Section Break
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Difficulty: Medium
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Determine the pressure difference between the front and back ends of the tank along a horizontal line when the truck accelerates at 3 m/s2. The pressure difference between the front and back ends of the tank is
24 ± 2% kPa.
Explanation: The following assumptions have been made here: 1. The acceleration remains constant. 2. Water is an incompressible substance.
The length of the truck is 8 m. We take the x- and z- axes as shown. The horizontal acceleration is in the negative x direction, and thus ax is negative. Also, there is no acceleration in the vertical direction, and thus az = 0. The pressure difference between the two points 1 and 2 in an incompressible fluid in linear rigid body motion is given by [Math Processing Error] since z2 - z1 = 0 along a horizontal line. Therefore, the pressure difference between the front and back of the tank is due to acceleration in the horizontal direction and the resulting compression effect toward the back of the tank. Then, the pressure difference along a horizontal line becomes [Math Processing Error] since x1 = 0 and x2 = 8 m. The pressure difference between the front and back ends of the tank is 24 kPa. References Worksheet
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Difficulty: Medium
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Determine the pressure difference between the front and back ends of the tank along a horizontal line when the truck decelerates at 4 m/s2. The pressure difference between the front and back ends of the truck is
-32 ± 2% kPa.
Explanation: The pressure difference during deceleration is expressed as [Math Processing Error] The pressure difference between the front and back ends of the truck is -32 kPa. References Worksheet
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Difficulty: Medium
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A 14-cm-radius, 90-cm-high vertical cylindrical container is partially filled with 60-cm-high water. Now, the cylinder is rotated at a constant angular speed of 180 rpm. Determine how much the liquid level at the center of the cylinder will drop as a result of this rotational motion. The drop in the liquid level due to the rotational motion is
0.177 ± 2% m.
Explanation: The following assumptions have been made here: 1. The increase in the rotational speed is very slow so that the liquid in the container always acts as a rigid body. 2. The bottom surface of the container remains covered with liquid during rotation (no dry spots).
Taking the center of the bottom surface of the rotating vertical cylinder as the origin (r = 0 and z = 0), the equation for the free surface of the liquid is given as [Math Processing Error] where h0 = 0.6 m is the original height of the liquid before rotation, and [Math Processing Error] Then, the vertical height of the liquid at the center of the container where r = 0 becomes [Math Processing Error] Therefore, the drop in the liquid level at the center of the cylinder is [Math Processing Error] The drop in the liquid level due to the rotational motion is 0.177 m. References Worksheet
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Difficulty: Medium
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A fish tank that contains 60-cm-high water is moved in the cabin of an elevator. Take the density of water to be 1000 kg/m3. References Section Break
Difficulty: Medium
149. Award: 0.00 points
Determine the pressure at the bottom of the tank when the elevator is stationary. The pressure at the bottom of the tank is
5.89 ± 2% kPa.
Explanation: The following assumptions have been made here: 1. The acceleration remains constant. 2. Water is an incompressible substance. The pressure difference between the two points 1 and 2 in an incompressible fluid is given by [Math Processing Error] since ax = 0. Taking point 2 at the free surface and point 1 at the tank bottom, we have P2 = Patm & z2 = z1 = h and thus [Math Processing Error] We have az = 0, and thus the gage pressure at the tank bottom is [Math Processing Error]
The pressure at the bottom of the tank is 5.89 kPa. References Worksheet
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Difficulty: Medium
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Determine the pressure at the bottom of the tank when the elevator is moving up with an upward acceleration of 3 m/s2. The pressure at the bottom of the tank is
7.69 ± 2% kPa.
Explanation: We have az = +3m/s2, and thus the gage pressure at the tank bottom is [Math Processing Error] The pressure at the bottom of the tank is 7.69 kPa. References Worksheet
151.
Difficulty: Medium
Award: 0.00 points
Determine the pressure at the bottom of the tank when the elevator is moving down with a downward acceleration of 3 m/s2. The pressure at the bottom of the tank is
4.09 ± 2% kPa.
Explanation: We have az = −3m/s2, and thus the gage pressure at the tank bottom is [Math Processing Error] The pressure at the bottom of the tank is 4.09 kPa. References Worksheet
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Difficulty: Medium
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A 18-ft-long, 6-ft-high rectangular tank open to the atmosphere is towed by a truck on a level road. The tank is filled with water to a depth of 5 ft. Determine the maximum acceleration or deceleration allowed if no water is to spill during towing. The maximum acceleration or deceleration allowed if no water is to spill during towing is
3.58 ± 2% ft/s2 or –
3.58 ± 2% ft/s2.
Explanation: The following assumptions have been made here: 1. The road is horizontal so that acceleration has no vertical component (az = 0). 2. Effects of splashing, braking, driving over bumps, and climbing hills are assumed to be secondary and are not considered. 3. The acceleration remains constant.
We take the x-axis to be the direction of motion and the z-axis to be the upward vertical direction. The shape of the free surface just before spilling is shown in the figure. The tangent of the angle the free surface makes with the horizontal is given by [Math Processing Error] where az = 0 and, from geometric considerations, tanθ is [Math Processing Error]. Substituting, we get [Math Processing Error] The solution can be repeated for deceleration by replacing ax with – ax. We obtain ax = –3.58 ft/s2. The maximum acceleration or deceleration allowed if no water is to spill during towing is 3.58 ft/s2 or –3.58 ft/s2. References Worksheet
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Difficulty: Medium
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Consider a tank of rectangular cross section partially filled with a liquid placed on an inclined surface, as shown in the figure. When frictional effects are negligible, what is the relation between the slope of the liquid surface and the slope of the inclined surface when the tank is released.
α−β=1 α+β=1 α+1=β α=β
[Math Processing Error] Since a = g sin α, we get
[Math Processing Error] Therefore, α = β. The slope of the liquid surface will be the same as the slope of the inclined surface when the tank is released. Therefore, α = β. References Multiple Choice
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Consider a tank of rectangular cross section partially filled with a liquid placed on an inclined surface, as shown in the figure. When frictional effects are negligible, the slope of the liquid surface will be the same as the slope of the inclined surface when the tank is released. Which of the following expressions is true regarding the slope of the free surface when the friction is significant?
β<α β−α=1 β=α β>α If the surface were rough, a' = gsin α − λgcos α < a , where λ is the surface friction coefficient. Therefore, we may conclude that β < α. If the surface were rough, a' = gsin α − λgcos α < a , where λ is the surface friction coefficient. Therefore, we may conclude that β < α. References Multiple Choice
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A 4-ft-diameter vertical cylindrical tank open to the atmosphere contains 1-ft-high water. The tank is now rotated about the centerline, and the water level drops at the center while it rises at the edges. Determine the angular velocity at which the bottom of the tank will first be exposed. Also determine the maximum water height at this moment. The angular velocity at which the bottom of the tank will first be exposed is The maximum water height is 2 ± 2% ft.
54.2 ± 2% rpm.
Explanation: The following assumptions have been made here: 1. The increase in the rotational speed is very slow so that the liquid in the container always acts as a rigid body. 2. Water is an incompressible fluid.
Taking the center of the bottom surface of the rotating vertical cylinder as the origin (r = 0 and z = 0), the equation for the free surface of the liquid is given as [Math Processing Error] where h0 = 1 ft is the original height of the liquid before rotation. Just before dry spots appear at the center of the bottom surface, the height of the liquid at the center equals zero, and thus zs(0) = 0. Solving the equation above for ω and substituting, [Math Processing Error] Noting that one complete revolution corresponds to 2π radians, the rotational speed of the container can also be expressed in terms of revolutions per minute (rpm) as [Math Processing Error] Therefore, the rotational speed of this container should be limited to 54.2 rpm to avoid any dry spots at the bottom surface of the tank. The maximum vertical height of the liquid occurs at the edges of the tank (r = R = 1 ft), and it is [Math Processing Error].
The angular velocity at which the bottom of the tank will first be exposed is 54.2 rpm. The maximum water height is 2 ft.
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References Worksheet
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Difficulty: Medium
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Milk with a density of 1020 kg/m3 is transported on a level road in a 8-m-long, 3-m-diameter cylindrical tanker. The tanker is completely filled with milk (no air space), and it accelerates at 4 m/s2. If the minimum pressure in the tanker is 100 kPa, determine the maximum pressure difference.
The maximum pressure difference is
62.6 ± 2% kPa.
Explanation: The following assumptions have been made here: 1. The acceleration remains constant. 2. Milk is an incompressible substance.
The length of the truck is 8 m. We take the x- and z- axes as shown. The horizontal acceleration is in the negative x direction, and thus ax is negative. Also, there is no acceleration in the vertical direction, and thus az = 0. The pressure difference between two points 1 and 2 in an incompressible fluid in linear rigid body motion is given by [Math Processing Error] The first term is due to acceleration in the horizontal direction and the resulting compression effect toward the back of the tanker, while the second term is simply the hydrostatic pressure that increases with depth. Therefore, we reason that the lowest pressure in the tank will occur at point 1 (upper front corner) and the higher pressure at point 2 (the lower rear corner). Therefore, the maximum pressure difference in the tank is [Math Processing Error] [Math Processing Error] [Math Processing Error] The maximum pressure difference is 62.6 kPa. References Worksheet
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Difficulty: Medium
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Milk with a density of 1020 kg/m3 is transported on a level road in a 9-m-long, 3-m-diameter cylindrical tanker. The tanker is completely filled with milk (no air space), and it decelerates at 2.5 m/s2. If the minimum pressure in the tanker is 100 kPa, determine the maximum pressure difference.
The maximum pressure difference is
53 ± 2% kPa.
Explanation: The following assumptions have been made here: 1. The acceleration remains constant. 2. Milk is an incompressible substance.
The length of the truck is 9 m. We take the x- and z- axes as shown. The horizontal deceleration is in the x direction, and thus ax is positive. Also, there is no acceleration in the vertical direction, and thus az = 0. The pressure difference between the two points 1 and 2 in an incompressible fluid in linear rigid body motion is given by [Math Processing Error] The first term is due to deceleration in the horizontal direction and the resulting compression effect toward the back of the tanker, while the second term is simply the hydrostatic pressure that increases with depth. Therefore, we reason that the lowest pressure in the tank will occur at point 1 (upper front corner), and the higher pressure at point 2 (the lower rear corner). Therefore, the maximum pressure difference in the tank is [Math Processing Error] [Math Processing Error] [Math Processing Error] The maximum pressure difference is 53 kPa. References Worksheet
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Difficulty: Medium
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The distance between the centers of the two arms of a U-tube open to the atmosphere is 30 cm, and the U-tube contains 20cm-high alcohol in both arms. Now, the U-tube is rotated about the left arm at 4 rad/s. Determine the elevation difference between the fluid surfaces in the two arms.
The elevation difference between the fluid surfaces in the two arms is
0.073 ± 2% m.
Explanation: It is assumed that alcohol is an incompressible fluid. Taking the base of the left arm of the U-tube as the origin (r = 0 and z = 0), the equation for the free surface of the liquid is given as [Math Processing Error] where h0 = 0.20 m is the original height of the liquid before rotation, and ω = 4 rad/s. The fluid rise in the right arm relative to the fluid level in the left arm (the center of rotation) is [Math Processing Error] [Math Processing Error] The elevation difference between the fluid surfaces in the two arms is 0.073 m. References Worksheet
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Difficulty: Medium
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A 1.2-m-diameter, 3-m-high sealed vertical cylinder is completely filled with gasoline whose density is 740 kg/m3. The tank is now rotated about its vertical axis at a rate of 74 rpm.
References Section Break
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Difficulty: Medium
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159. Award: 0.00 points
Determine the difference between the pressures at the centers of the bottom and top surfaces. The difference between the pressures at the centers is
21.8 ± 2% kPa.
Explanation: The following assumptions have been made here: 1. The increase in the rotational speed is very slow so that the liquid in the container always acts as a rigid body. 2. Gasoline is an incompressible substance.
The pressure difference between the two points 1 and 2 in an incompressible fluid rotating in rigid-body motion is given by [Math Processing Error] where R = 0.60 m is the radius and ω is the angular velocity. Taking points 1 and 2 to be the centers of the bottom and top surfaces, respectively, we have r1 = r2 = 0 and z2 − z1 = h = 3 m. Then, [Math Processing Error] [Math Processing Error] [Math Processing Error] The difference between the pressures at the centers is 21.8 kPa. References Worksheet
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Determine the difference between the pressures at the center and the edge of the bottom surface. The difference between the pressures at the center and the edge of the bottom surface is
8 ± 2% kPa.
Explanation: The angular velocity is [Math Processing Error] Taking points 1 and 2 to be the center and the edge of the bottom surface, respectively, we have r1 =0, r2 = R, and z2 = z1 = 0 . Then, [Math Processing Error] [Math Processing Error] [Math Processing Error] The difference between the pressures at the center and the edge of the bottom surface is 8 kPa. References Worksheet
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A 1.2-m-diameter, 3-m-high sealed vertical cylinder is completely filled with gasoline whose density is 740 kg/m3. Using appropriate software, investigate the effect of rotational speed on the pressure difference between the center and the edge of the bottom surface of the cylinder. Let the rotational speed vary from 0 rpm to 500 rpm in increments of 50 rpm. Tabulate and plot your results. (Please upload your response/solution using the controls provided below.)
Student upload controls will be shown to students when they take this assignment.
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A 4-m-diameter vertical cylindrical milk tank rotates at a constant rate of 18 rpm. If the pressure at the center of the bottom surface is 130 kPa, determine the pressure at the edge of the bottom surface of the tank. Take the density of the milk to be 1030 kg/m3. The pressure at the edge of the bottom surface of the tank is
137.3 ± 2%
kPa.
Explanation: The following assumptions have been made here: 1 The increase in the rotational speed is very slow so that the liquid in the container always acts as a rigid body. 2. Milk is an incompressible substance.
Taking the center of the bottom surface of the rotating vertical cylinder as the origin (r = 0 and z = 0), the equation for the free surface of the liquid is given as [Math Processing Error] where R = 2 m is the radius, and [Math Processing Error] The fluid rise at the edge relative to the center of the tank is [Math Processing Error] [Math Processing Error] The pressure difference corresponding to this fluid height difference is [Math Processing Error] Then the pressure at the edge of the bottom surface becomes [Math Processing Error]
The pressure at the edge of the bottom surface of the tank is 137.3 kPa. References Worksheet
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An 7.5-ft-long tank open to the atmosphere initially contains 3-ft-high water. It is being towed by a truck on a level road. The truck driver applies the brakes, and the water level at the front rises 0.75 ft above the initial level. Determine the deceleration of the truck.
The deceleration of the truck is
6.44 ± 2% ft/s2.
Explanation: The following assumptions have been made here: 1. The road is horizontal so that deceleration has no vertical component (az = 0). 2. Effects of splashing and driving over bumps are assumed to be secondary and are not considered. 3. The deceleration remains constant.
We take the x-axis to be the direction of motion and the z-axis to be the upward vertical direction. The shape of the free surface just before spilling is shown in the figure. The tangent of the angle the free surface makes with the horizontal is given by [Math Processing Error] where az = 0 and, from geometric considerations, tan θ is [Math Processing Error] Substituting to find the deceleration, [Math Processing Error] The deceleration of the truck is 6.44 ft/s2.
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A 73-cm-high, 40-cm-diameter cylindrical water tank is being transported on a level road. The highest acceleration anticipated is 5 m/s2. Determine the allowable initial water height in the tank if no water is to spill out during acceleration. The allowable initial height in the tank to prevent water from spilling out is
62.8 ± 2% cm.
Explanation: The following assumptions have been made here: 1. The road is horizontal during acceleration so that acceleration has no vertical component (az = 0). 2. Effects of splashing, braking, driving over bumps, and climbing hills are assumed to be secondary and are not considered. 3. The acceleration remains constant.
We take the x-axis to be the direction of motion, the z-axis to be the upward vertical direction, and the origin to be the midpoint of the tank bottom. The tangent of the angle the free surface makes with the horizontal is [Math Processing Error] The maximum vertical rise of the free surface occurs at the back of the tank, and the vertical midplane experiences no rise or drop during acceleration. Then, the maximum vertical rise at the back of the tank relative to the midplane is [Math Processing Error] Therefore, the maximum initial water height in the tank to avoid spilling is [Math Processing Error] The allowable initial height in the tank to prevent water from spilling out is 62.8 cm. References Worksheet
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The rectangular tank is filled with heavy oil (like glycerin) at the bottom and water at the top, as shown in the figure. The tank is now moved to the right horizontally with a constant acceleration, and a fraction of 0.27 of water is spilled out as a result from the back. Using geometrical considerations, determine how high the point A at the back of the tank on the oil–water interface will rise under this acceleration.
The height of the oil-water interface after rising due to the acceleration is
0.27 ± 2% m.
Explanation: The following assumptions have been made here: 1. The acceleration remains constant. 2. Water and oil are incompressible substances.
Before the acceleration, the water volume for unit width was 1m × L. Therefore, 0.27 times of this volume must be equal to the emptied volume in the tank, which is [Math Processing Error] Therefore, z1 = 0.54 m. The slope of the free surface is [Math Processing Error] The height of the oil-water interface after rising due to the acceleration is 0.27 m. References Worksheet
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A centrifugal pump consists simply of a shaft and a few blades attached normally to the shaft. If the shaft is rotated at a constant rate of 1800 rpm, what would the theoretical pump head due to this rotation be? Take the impeller diameter to be 31 cm and neglect the blade tip effects. The theoretical pump head is
43.5 ± 2% m.
Explanation:
The rotational speed is [Math Processing Error] The pump head is [Math Processing Error]
The theoretical pump head is 43.5 m. References Worksheet
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Two vertical and connected cylindrical tanks of diameter D are open to the atmosphere and they contain water at a height of h in the initial state, as shown in the figure. As the radial blades in the left tank are rotated about the centerline of the tank, some water in the right tank flows into the left tank. Assume water in the tanks would not spill. Consider D = 2R = 45 cm, h = 35 cm, and g = 9.81 m/s2.
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Determine the angular velocity of radial blades ω such that half of the water in the right tank flows into the left tank. The required angular velocity is
16.47 ± 2% rad/s.
Explanation: The following assumptions have been made here: 1. Effects of splashing are not considered. 2. Water is an incompressible substance. In case a half of the water in the right tank is suctioned into the left tank, the water levels in the two connected tanks will appear as shown in the following figure:
Initial (water in the tanks is in the hydrostatic state) and final (water in the tanks is rotating at angular velocity) water volumes are equal because the water in the tanks is assumed not to spill. [Math Processing Error] [Math Processing Error] The required angular velocity is 16.47 rad/s. References Worksheet
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Which of the following is a general relation for angular velocity as a function of initial height of water in the tank, assuming that the water in the tanks would not spill?
[Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error]
We can determine the paraboloid’s height H in the left tank as a function of percentage of the initial height of water h or water volume to be suctioned in the right tank and derive a general relation for angular velocity as a function of the initial water height in the tank. If the paraboloid’s height H in the left tank is written as the percentage of the initial height of water h or water volume to be sucked in the right tank,
[Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] Thus, if h is written as a function of the percentage of water height or water volume to be suctioned in the right tank,
H = 4h A general relation for angular velocity as a function of the initial height of water in the tank is derived in the following form:
[Math Processing Error] The general relation for angular velocity as a function of initial height of water in the tank is
[Math Processing Error] References Multiple Choice
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The U-tube shown in the figure is subjected to an acceleration of a (m/s2) to left. If the difference between the free surfaces is Δh = 20 cm and h = 0.43 m, calculate the acceleration.
The acceleration is
1.37 ± 2% m/s2.
Explanation: The following assumptions have been made here: 1. The acceleration is constant. 2. Water is an incompressible fluid.
From the uniformly accelerated rigid-body motion, the pressure difference between two points that are at the same altitude is given by [Math Processing Error] where ay = a, Applying for points K and L, we get [Math Processing Error] or [Math Processing Error] On the other hand, [Math Processing Error] https://ezto.mheducation.com/hm.tpx
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Therefore, from Eq. 2, we have [Math Processing Error] Equating 1 and 3, we get [Math Processing Error] where [Math Processing Error] The acceleration is 1.37 m/s2. References Worksheet
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170. Award: 0.00 points
A pressure gage connected to a tank reads 61 psi at a location where the barometric reading is 29.1 inHg. Determine the absolute pressure in the tank. Take ρHg = 848.4 lbm/ft3. The absolute pressure in the tank is
75.29 ± 2% psia.
Explanation:
The atmospheric (or barometric) pressure can be expressed as [Math Processing Error] [Math Processing Error] [Math Processing Error] Then, the absolute pressure in the tank is [Math Processing Error] The absolute pressure in the tank is 75.29 psia. References Worksheet
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An air-conditioning system requires a 30-m-long section of 12-cm-diameter ductwork to be laid underwater. Determine the upward force the water will exert on the duct. Take the densities of air and water to be 1.3 kg/m3 and 1000 kg/m3, respectively. The upward force that the water will exert on the duct is
3.33 ± 2% kN.
Explanation: The following assumptions have been made here: 1. The diameter given is the outer diameter of the duct (or the thickness of the duct material is negligible). 2. The weight of the duct and the air inside is negligible.
Noting that the weight of the duct and the air in it is negligible, the net upward force acting on the duct is the buoyancy force exerted by water. The volume of the underground section of the duct is [Math Processing Error] Then the buoyancy force becomes [Math Processing Error] The upward force that the water will exert on the duct is 3.33 kN. References Worksheet
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Determine the pressure exerted on the surface of a submarine cruising 215 ft below the free surface of the sea. Assume that the barometric pressure is 14.7 psia and the specific gravity of seawater is 1.03. The density of water at 32°F is 62.4 lbm/ft3. The pressure exerted on the surface of the submarine is
111 ± 2% psia.
Explanation: It is assumed that the variation of the density of water with depth is negligible.
The density of the seawater is obtained by multiplying its specific gravity by the density of water: [Math Processing Error] The pressure exerted on the surface of the submarine cruising 300 ft below the free surface of the sea is the absolute pressure at that location: [Math Processing Error] [Math Processing Error] [Math Processing Error] The pressure exerted on the surface of the submarine is 111 psia. References Worksheet
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A vertical, frictionless piston–cylinder device contains a gas at 600 kPa. The atmospheric pressure outside is 100 kPa, and the piston area is 33 cm2. Determine the mass of the piston. The mass of the piston is
168 ± 2% kg.
Explanation: It is assumed that there is no friction between the piston and the cylinder.
Drawing the free-body diagram of the piston and balancing the vertical forces yield [Math Processing Error] [Math Processing Error] [Math Processing Error] The solution of the above equation yields m = 168 kg The mass of the piston is 168 kg. References Worksheet
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If the rate of rotational speed of the three-tube system shown in the figure is ω = 10 rad/s, determine the water heights in each tube leg. At what rotational speed will the middle tube be completely empty? In the figure, R = 0.17 m.
The height of water in the first tube is 0.231 ± 2% m. 0.084 ± 2% m. The height of water in the second tube is The height of water in the third tube is 0.135 ± 2% m. The rotational speed at which the middle tube will be empty is
13.59 ± 2% rad/s.
Explanation:
The equation describing the water surface is [Math Processing Error] Since z = z1 when r = 0, C = z1. Therefore, we can write the following expressions: [Math Processing Error] [Math Processing Error] There are three unknowns (z1, z2, & z3). The third equation will be obtained from continuity such as [Math Processing Error] Substituting Eqs. 1 and 2 into Eq. 3, we have [Math Processing Error] [Math Processing Error]
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For the given data, we obtain [Math Processing Error] For this case, z1=0. Therefore, from Eq. 3, [Math Processing Error] Solving for ω, [Math Processing Error] The height of water in the first tube is 0.231 m. The height of water in the second tube is 0.084 m. The height of water in the third tube is 0.135 m. The rotational speed at which the middle tube will be empty is 13.59 rad/s. References Worksheet
Difficulty: Medium
175. Award: 0.00 points
The average atmospheric pressure on earth is approximated as a function of altitude by the relation Patm =101.325 × (1 − 0.02256z)5.256, where Patm is the atmospheric pressure in kPa and z is the altitude in km with z = 0 at sea level. Determine the approximate atmospheric pressures at Atlanta (z = 306 m), Denver (z = 1610 m), Mexico City (z = 2309 m), and the top of Mount Everest (z = 8848 m). 97.7 ± 2% kPa. The approximate atmospheric pressure at Atlanta is The approximate atmospheric pressure at Denver is 83.4 ± 2% kPa. The approximate atmospheric pressure at Mexico City is 76.5 ± 2% kPa. 31.4 ± 2% kPa. The approximate atmospheric pressure at the top of Mount Everest is
Explanation: Atmospheric pressure at various locations is obtained by substituting the altitude z values in km into the relation Patm =101.325 × (1 − 0.02256z)5.256. The results are tabulated below. Atlanta
(z = 0.306 km) Patm = 101.325(1 − 0.02256 × 0.306)5.256 = 97.7 kPa
Denver
(z = 1.610 km) Patm = 101.325(1 − 0.02256 × 1.610)5.256 = 83.4 kPa
Mexico City
(z = 2.309 km) Patm = 101.325(1 − 0.02256 × 2.309)5.256 = 76.5 kPa
Mount Everest
(z = 8.848 km) Patm = 101.325(1 − 0.02256 × 8.848)5.256 = 31.4 kPa
The approximate atmospheric pressure at Atlanta is 97.7 kPa. The approximate atmospheric pressure at Denver is 83.4 kPa. The approximate atmospheric pressure at Mexico City is 76.5 kPa. The approximate atmospheric pressure at the top of Mount Everest is 31.4 kPa. References Worksheet
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Balloons are often filled with helium gas because it weighs only about one-seventh of what air weighs under identical conditions. The buoyancy force, which can be expressed as Fb = ρairgVballoon, will push the balloon upward. If the balloon has a diameter of 12 m and carries two people, 66 kg each, determine the acceleration of the balloon when it is first released. Assume the density of air is ρ = 1.16 kg/m3, and neglect the weight of the ropes and the cage. Assume that the balloons is a perfect sphere.
The acceleration of the balloon when it is first released is
26.7 ± 2% m/s2.
Explanation: It is assumed that the weight of the cage and the ropes of the balloon is negligible.
The buoyancy force acting on the balloon is [Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] The total mass is [Math Processing Error] [Math Processing Error] https://ezto.mheducation.com/hm.tpx
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The total weight is [Math Processing Error] Thus, the net force acting on the balloon is [Math Processing Error] Then, the acceleration becomes [Math Processing Error] The acceleration of the balloon when it is first released is 26.7 m/s2. References Worksheet
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Balloons are often filled with helium gas because it weighs only about one-seventh of what air weighs under identical conditions. The buoyancy force, which can be expressed as Fb = ρairgVballoon, will push the balloon upward. The balloon has a diameter of 12 m and carries two people, 70 kg each. Assume the density of air is ρ = 1.16 kg/m3, and neglect the weight of the ropes and the cage. Using appropriate software, investigate the effect of the number of people carried in the balloon on acceleration. Plot the acceleration against the number of people, and discuss the result. (Please upload your response/solution using the controls provided below.)
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Balloons are often filled with helium gas because it weighs only about one-seventh of what air weighs under identical conditions. The buoyancy force, which can be expressed as Fb = ρairgVballoon, will push the balloon upward. If the balloon has a radius of 5 m and carries two people, 70 kg each, determine the maximum amount of load, in kg, the balloon can carry. Take the density of air to be 1.16 kg/m3.
The maximum load that the balloon can carry is
521 ± 2% kg.
Explanation: It is assumed that the weight of the cage and the ropes of the balloon is negligible. The buoyant force acting on the balloon is [Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] The total mass of the balloon is [Math Processing Error] [Math Processing Error] The mass of helium is [Math Processing Error] Therefore, [Math Processing Error]
The maximum load that the balloon can carry is 521 kg. References Worksheet https://ezto.mheducation.com/hm.tpx
Difficulty: Medium 195/260
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The basic barometer can be used as an altitude-measuring device in airplanes. The ground control reports a barometric reading of 760 mmHg while the pilot’s reading is 405 mmHg. Estimate the altitude of the plane from ground level. The densities of air and mercury are given to be ρair = 1.20 kg/m3 and ρmercury = 13,600 kg/m3. The altitude of the plane from ground level is
4024 ± 2% m.
Explanation: It is assumed that the variation of air density with altitude is negligible.
Atmospheric pressures at the location of the plane and the ground level are [Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] Taking an air column between the airplane and the ground and writing a force balance per unit base area, we obtain [Math Processing Error] [Math Processing Error] [Math Processing Error] This yields h = 4024 m, which is the altitude of the airplane. The altitude of the plane from ground level is 4024 m. References Worksheet
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The lower half of a 11-m-high cylindrical container is filled with water (ρ = 1000 kg/m3) and the upper half with oil that has a specific gravity of 0.85. Determine the pressure difference between the top and bottom of the cylinder.
The pressure difference between the top and the bottom of the cylinder is
100 ± 2% kPa.
Explanation: The density of the oil is obtained by multiplying its specific gravity by the density of water: [Math Processing Error] The pressure difference between the top and the bottom of the cylinder is the sum of the pressure differences across the two fluids: [Math Processing Error] [Math Processing Error] [Math Processing Error] The pressure difference between the top and the bottom of the cylinder is 100 kPa. References Worksheet
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The 0.5-m-radius semicircular gate shown in the figure is hinged through the top edge AB. Find the required force to be applied at the center of gravity to keep the gate closed. The height h of oil above the gate is 4.6 m.
The required force to keep the gate closed is
7877 ± 2% N.
Explanation:
The force applied by glycerin is [Math Processing Error] The gage pressure of air entrapped on the top of the oil surface, [Math Processing Error] This negative pressure will result in an imaginary reduction in the oil level by [Math Processing Error] Therefore, the imaginary oil level will be H = 4.6 m − 2.24 m = 2.36 m from the glycerin surface. The force applied by oil is then [Math Processing Error] The locations of FRg and FRo are as follows: [Math Processing Error] [Math Processing Error] https://ezto.mheducation.com/hm.tpx
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[Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] The moment about the hinge will give [Math Processing Error] [Math Processing Error] [Math Processing Error] The required force to keep the gate closed is 7877 N. References Worksheet
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A pressure cooker cooks considerably faster than an ordinary pan by maintaining a higher pressure and temperature inside. The lid of a pressure cooker is well sealed, and steam can escape only through an opening in the middle of the lid. A separate metal piece, the petcock, sits on top of this opening and prevents steam from escaping until the pressure force overcomes the weight of the petcock. The periodic escape of the steam in this manner prevents any potentially dangerous pressure buildup and keeps the pressure inside at a constant value. Determine the mass of the petcock of a pressure cooker whose operation pressure is 120 kPa gage and has an opening cross-sectional area of 3 mm2. Assume an atmospheric pressure of 101 kPa.
The mass of the petcock is
36.7 ± 2% g.
Explanation: It is assumed that there is no blockage of the pressure release valve.
Atmospheric pressure acts on all surfaces of the petcock, which balances itself out. Therefore, it can be disregarded in calculations if we use the gage pressure as the cooker pressure. A force balance on the petcock (ΣFy = 0) yields [Math Processing Error] [Math Processing Error] [Math Processing Error] The mass of the petcock is 36.7 g. References Worksheet
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A glass tube is attached to a water pipe, as shown in the figure. If the water pressure at the bottom of the tube is 110 kPa and the local atmospheric pressure is 98 kPa, determine how high the water will rise in the tube, in m. Assume g = 9.8 m/s2 at that location and take the density of water to be 1000 kg/m3.
The rise of water in the tube is
1.22 ± 2% m.
Explanation:
The pressure at the bottom of the tube can be expressed as [Math Processing Error] Solving for h, [Math Processing Error] [Math Processing Error] [Math Processing Error] The rise of water in the tube is 1.22 m. References Worksheet
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A system is equipped with two pressure gages and a manometer, as shown in the figure. For Δh = 9 cm, determine the pressure difference ΔP = P2 − P1. The specific gravities are given as 2.15 for the gage fluid and 0.87 for oil. Take the standard density of water to be ρw = 1000 kg/m3. The height of oil is 0.65 m.
The pressure difference is
3.65 ± 2% kPa.
Explanation: The following assumptions have been made here: 1. All the liquids are incompressible. 2. The effect of the air column on pressure is negligible.
Starting with the pressure indicated by pressure gage 2, moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms, ignoring the air spaces until we reach pressure gage 1, and setting the result equal to P1 give [Math Processing Error] Rearranging, [Math Processing Error] Substituting,
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[Math Processing Error] [Math Processing Error] The pressure difference is 3.65 kPa. References Worksheet
Difficulty: Medium
An oil pipeline and a 1.4-m3 rigid air tank are connected to each other by a manometer, as shown in the figure. The tank contains 15 kg of air at 80°C. Assume the pressure in the oil pipeline to remain constant and the air volume in the manometer to be negligible relative to the volume of the tank.
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Determine the absolute pressure in the pipeline. The absolute pressure in the pipeline is
1039 ± 2% kPa.
Explanation: The following assumptions have been made here: 1. All the liquids are incompressible. 2. The effect of the air column on pressure is negligible. 3. The air volume in the manometer is negligible compared with the volume of the tank. Starting with the oil pipe, moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach the air tank, and setting the result equal to Pair give [Math Processing Error] The absolute pressure in the air tank is determined from the ideal-gas relation PV = mRT to be [Math Processing Error] Then, the absolute pressure in the oil pipe becomes [Math Processing Error] [Math Processing Error] [Math Processing Error] The absolute pressure in the pipeline is 1039 kPa. References Worksheet
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Determine the change in Δh when the temperature in the tank drops to 20°C. The change in Δh is
18 ± 2% cm.
Explanation: The pressure in the air tank when the temperature drops to 20°C becomes [Math Processing Error] When the mercury level in the left arm drops a distance x, the rise in the mercury level in the right arm y becomes [Math Processing Error] and the mercury fluid height will change by x + 9x sin50° or 7.894x. Then, [Math Processing Error] [Math Processing Error] This yields x = 0.18 m = 18 cm. The change in Δh is 18 cm. References Worksheet
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A 24-cm-diameter vertical cylindrical vessel is rotated about its vertical axis at a constant angular velocity of 70 rad/s. If the pressure at the midpoint of the inner top surface is atmospheric pressure like the outer surface, determine the total upward force acting upon the entire top surface inside the cylinder. The total upward force acting upon the entire top surface inside the cylinder is
798 ± 2% N.
Explanation:
Since z is constant along the top surface, we may write [Math Processing Error] If we take the point A to be reference, then C = 0. [Math Processing Error] [Math Processing Error] or [Math Processing Error] The total upward force acting upon the entire top surface inside the cylinder is 798 N.
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An elastic air balloon with a diameter of 30 cm is attached to the base of a container partially filled with water at +4°C as shown in the figure. If the pressure of the air above the water is gradually increased from 105 kPa to1.6 MPa, what is the percent change in the force on the cable? Assume the pressure on the free surface and the diameter of the balloon are related by P = CDn, where C is a constant and n = –2. The weight of the balloon and the air in it is negligible. Take the density of water to be 1000 kg/m3.
The force on the cable changes by
98.3 ± 2% %.
Explanation: The following assumptions have been taken into consideration: 1. Atmospheric pressure acts on all surfaces. Thus, it can be ignored in calculations for convenience. 2. Water is an incompressible fluid. 3. The weight of the balloon and the air in it is negligible. The tension force on the cable holding the balloon is determined from a force balance on the balloon to be [Math Processing Error] The buoyancy force acting on the balloon initially is [Math Processing Error] The variation of pressure with diameter is given as [Math Processing Error], which is equivalent to [Math Processing Error] . The final diameter of the ball becomes [Math Processing Error] The buoyancy force acting on the balloon in this case is determined as follows: [Math Processing Error] The percentage change in the force on the cable becomes [Math Processing Error] The force on the cable changes by 98.3%. References
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An elastic air balloon with a diameter of 30 cm is attached to the base of a container partially filled with water at +4°C as shown in the figure. Assume the pressure on the free surface and the diameter of the balloon are related by P = CDn, where C is a constant and n = –2. The weight of the balloon and the air in it is negligible. Take the density of water to be 1000 kg/m3. Using appropriate software, investigate the effect of air pressure above water on the cable force. Let this pressure vary from 0.5 MPa to 15 MPa. Plot the cable force versus the air pressure. Assume that the pressure of the air balloon, when the diameter is 30 cm, is 0.5 MPa. (Please upload your response/solution using the controls provided below.)
Student upload controls will be shown to students when they take this assignment.
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A gasoline line is connected to a pressure gage through a double-U manometer, as shown in the figure. If the reading of the pressure gage is 260 kPa, determine the gage pressure of the gasoline line. The specific gravities of oil, mercury, and gasoline are 0.79, 13.6, and 0.70, respectively. Take the density of water to be ρw = 1000 kg/m3.
The gage pressure of the gasoline line is
245 ± 2% kPa.
Explanation: The following assumptions have been made here: 1. All the liquids are incompressible. 2. The effect of the air column on pressure is negligible.
Starting with the pressure indicated by the pressure gage, moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach the gasoline pipe, and setting the result equal to Pgasoline give [Math Processing Error] Rearranging, [Math Processing Error] Substituting, [Math Processing Error] [Math Processing Error] The gage pressure of the gasoline line is 245 kPa.
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A gasoline line is connected to a pressure gage through a double-U manometer, as shown in the figure. If the reading of the pressure gage is 320 kPa, determine the gage pressure of the gasoline line. The specific gravities of oil, mercury, and gasoline are 0.79, 13.6, and 0.70, respectively. Take the density of water to be ρw = 1000 kg/m3.
The gage pressure of the gasoline line is
305 ± 2% kPa.
Explanation: The following assumptions have been made here: 1. All the liquids are incompressible. 2. The effect of the air column on pressure is negligible.
Starting with the pressure indicated by the pressure gage, moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach the gasoline pipe, and setting the result equal to Pgasoline give [Math Processing Error] Rearranging, [Math Processing Error] Substituting, [Math Processing Error] [Math Processing Error] The gage pressure of the gasoline line is 305 kPa.
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A water pipe is connected to a double-U manometer as shown in the figure at a location where the local atmospheric pressure is 14.5 psia. Determine the absolute pressure at the center of the pipe. The specific gravities of mercury and oil are 13.6 and 0.80, respectively. Take the density of water to be ρw = 62.4 lbm/ft3. The height h of water in the water pipe is 38 in.
The absolute pressure at the center of the pipe is
22.7 ± 2% psia.
Explanation: The following assumptions have been made here: 1. All the liquids are incompressible. 2. The solubility of the liquids in each other is negligible.
Starting with the pressure at the center of the water pipe, moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach the free surface of oil where the oil tube is exposed to the atmosphere, and setting the result equal to Patm give [Math Processing Error] Solving for Pwater pipe, [Math Processing Error] Substituting, [Math Processing Error] [Math Processing Error] The absolute pressure at the center of the pipe is 22.7 psia. https://ezto.mheducation.com/hm.tpx
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Consider a U-tube filled with mercury as shown in the figure. The diameter of the right arm of the U-tube is D = 1.5 cm, and the diameter of the left arm is twice that. Heavy oil with a specific gravity of 2.96 is poured into the left arm, forcing some mercury from the left arm into the right one. Determine the maximum amount of oil that can be added into the left arm.
The maximum amount of oil that can be added into the left arm is
0.0887 ± 2% L.
Explanation: The following assumptions have been made here: 1. Both liquids are incompressible. 2. The U-tube is perfectly vertical.
Initially, the mercury levels in both tubes are the same. When oil is poured into the left arm, it will push the mercury in the left downward, which will cause the mercury level in the right arm to rise. Noting that the volume of mercury is constant, the decrease in the mercury volume in the left column must be equal to the increase in the mercury volume in the right arm. Therefore, if the drop in mercury level in the left arm is x, the rise in the mercury level in the right arm h corresponding to a drop of x in the left arm is [Math Processing Error] The pressures at points A and B are equal, PA = PB, and thus [Math Processing Error] Solving for x and substituting, [Math Processing Error] Therefore, the maximum amount of oil that can be added into the left arm is https://ezto.mheducation.com/hm.tpx
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[Math Processing Error] The maximum amount of oil that can be added into the left arm is 0.0887 L. References Worksheet
Difficulty: Medium
194. Award: 0.00 points
The variation of pressure with density in a thick gas layer is given by P = Cρn, where C and n are constants. Noting that the pressure change across a differential fluid layer of thickness dz in the vertical z-direction is given as dP = −ρgdz, which of the following is the correct relation for pressure as a function of elevation z? Take the pressure and density at z = 0 to be P0 and ρ0, respectively.
[Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] It is assumed that the property relation P = Cρn is valid over the entire region considered. The pressure change across a differential fluid layer of thickness dz in the vertical z-direction is given as
[Math Processing Error] Also, the relation P = Cρn can be expressed as C = P/ρn = P0/ρ0n, and thus
[Math Processing Error] Substituting,
[Math Processing Error] Separating variables and integrating from z = 0 where P = P0 = Cρ0n to z = z where P = P,
[Math Processing Error] Performing integration,
[Math Processing Error] Solving for P,
[Math Processing Error] The relation for pressure as a function of elevation z is
[Math Processing Error] References Multiple Choice
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A 3-m-high, 9.5-m-wide rectangular gate is hinged at the top edge at A and is restrained by a fixed ridge at B. Determine the hydrostatic force exerted on the gate by the 5-m-high water and the location of the pressure center. Take the density of water to be 1000 kg/m3 throughout.
979 ± 2% kN. The hydrostatic force exerted on the gate by the water is The vertical distance of the pressure center from the free surface of water is
3.71 ± 2% m.
Explanation: Assume that atmospheric pressure acts on both sides of the gate. Thus, it can be ignored in calculations for convenience.
The average pressure on a surface is the pressure at the centroid (midpoint) of the surface, and multiplying it by the plate area gives the resultant hydrostatic force on the gate. [Math Processing Error] The vertical distance of the pressure center from the free surface of water is determined as follows: [Math Processing Error] The hydrostatic force exerted on the gate by the water is 979 kN. The vertical distance of the pressure center from the free surface of water is 3.71 m. References
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A 3-m-high, 12-m-wide rectangular gate is hinged at the top edge at A and is restrained by a fixed ridge at B. Determine the hydrostatic force exerted on the gate by the 2-m-high water and the location of the pressure center. Take the density of water to be 1000 kg/m3 throughout.
The hydrostatic force exerted on the gate by the water is 235 ± 2% kN. The vertical distance of the pressure center from the free surface of water is
1.33 ± 2% m.
Explanation: Assume that atmospheric pressure acts on both sides of the gate. Thus, it can be ignored in calculations for convenience.
The average pressure on a surface is the pressure at the centroid (midpoint) of the surface, and multiplying it by the plate area gives the resultant hydrostatic force on the gate. [Math Processing Error] [Math Processing Error] The vertical distance of the pressure center from the free surface of water is determined as follows: [Math Processing Error] The hydrostatic force exerted on the gate by the water is 235 kN. The vertical distance of the pressure center from the free surface of water is 1.33 m.
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A semicircular 40-ft-diameter tunnel is to be built under a 150-ft-deep, 800-ft-long lake as shown in the figure. Determine the total hydrostatic force acting on the roof of the tunnel. Take the density of water to be 62.4 lbm/ft3 throughout.
The hydrostatic force acting on the roof of the tunnel is
2.682 ± 2% × 108 lbf.
Explanation: Assume that atmospheric pressure acts on both sides of the tunnel. Thus, it can be ignored in calculations for convenience.
We consider the free-body diagram of the liquid block enclosed by the circular surface of the tunnel and its vertical (on both sides) and horizontal projections. The hydrostatic forces acting on the vertical and horizontal plane surfaces as well as the weight of the liquid block are determined as follows: The horizontal force on the vertical surface (each side) is [Math Processing Error] [Math Processing Error] [Math Processing Error] The vertical force on the horizontal surface (downward) is [Math Processing Error] [Math Processing Error] [Math Processing Error] The weight of the liquid on each side within control volume (downward) is [Math Processing Error] [Math Processing Error] [Math Processing Error] https://ezto.mheducation.com/hm.tpx
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Therefore, the net vertical downward force is [Math Processing Error] The hydrostatic force acting on the roof of the tunnel is 2.68 × 108 lbf. References Worksheet
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A 22-ton, 4-m-diameter hemispherical dome on a level surface is filled with water as shown in the figure. Someone claims that he can lift this dome by making use of Pascal’s law by attaching a long tube to the top and filling it with water. Determine the required height of water in the tube to lift the dome. Disregard the weight of the tube and the water in it. Take the density of water to be 1000 kg/m3 throughout.
The required height of water in the tube to lift the dome is
1.08 ± 2% m.
Explanation: The following assumptions have been taken into consideration: 1. Atmospheric pressure acts on both sides of the dome. Thus, it can be ignored in calculations for convenience. 2. The weight of the tube and the water in it is negligible.
We take the dome and the water in it as the system. When the dome is about to rise, the reaction force between the dome and the ground becomes zero. Then, the free-body diagram of this system involves the weights of the dome and the water balanced by the hydrostatic pressure force from below. Setting these forces equal to each other gives [Math Processing Error] Solving for h gives [Math Processing Error] The required height of water in the tube to lift the dome is 1.08 m. https://ezto.mheducation.com/hm.tpx
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The water in a 25-m-deep reservoir is kept inside by a 100-m-wide wall whose cross section is an equilateral triangle as shown in the figure.
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Determine the total force (hydrostatic + atmospheric) acting on the inner surface of the wall and its line of action. Take the value of g as 9.81 m/s2, the atmospheric pressure as 100,000 N/m2, and the density of water to be 1000 kg/m3 throughout. 6.43 ± 2% × 108 N. The total force (hydrostatic + atmospheric) acting on the inner surface of the wall is The distance of the pressure center from the free surface of water along the wall surface is 17.1 ± 2% m.
Explanation: The following assumptions have been taken into consideration: 1. Atmospheric pressure acts on both sides of the gate. Thus, it can be ignored in calculations for convenience. 2. The friction at the hinge is negligible.
The length of the wall surface underwater is determined as follows: [Math Processing Error] The average pressure on a surface is the pressure at the centroid (midpoint) of the surface, and multiplying it by the plate area gives the resultant hydrostatic force on the surface. [Math Processing Error] [Math Processing Error] [Math Processing Error] (on each side of the tunnel) [Math Processing Error] The distance of the pressure center from the free surface of water along the wall surface is determined as follows: [Math Processing Error] The total force (hydrostatic + atmospheric) acting on the inner surface of the wall is 6.43 × 108 N. The distance of the pressure center from the free surfacer of water along the wall surface is 17.1 m. References Worksheet
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Determine the magnitude of the horizontal component force. Consider Patm = 100 kPa. The magnitude of the horizontal component force is
5.57 ± 2% × 108 N.
Explanation: The magnitude of the horizontal component of the hydrostatic force is FRsinθ. [Math Processing Error] The magnitude of the horizontal component of the hydrostatic force is 5.57 × 108 N. References Worksheet
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A 5-m-long, 4-m-high tank contains 2.5-m-deep water when not in motion and is open to the atmosphere through a vent in the middle. The tank is now accelerated to the right on a level surface at 2 m/s2. Determine the maximum pressure in the tank relative to the atmospheric pressure.
The maximum pressure in the tank is
29.5 ± 2% kPa.
Explanation: The following assumptions have been made here: 1. The road is horizontal during acceleration so that acceleration has no vertical component (az = 0). 2. Effects of splashing, braking, and driving over bumps are assumed to be secondary and are not considered. 3. The vent is never blocked, and thus the minimum pressure is the atmospheric pressure.
We take the x-axis to be the direction of motion and the z-axis to be the upward vertical direction. The tangent of the angle the free surface makes with the horizontal is [Math Processing Error] (and thus θ = 11.5°) The maximum vertical rise of the free surface occurs at the back of the tank, and the vertical midsection experiences no rise or drop during acceleration. Then, the maximum vertical rise at the back of the tank relative to the neutral midplane is [Math Processing Error] This is less than 1.5-m-high air space. Therefore, water never reaches the ceiling, and the maximum water height and the corresponding maximum pressure are [Math Processing Error] [Math Processing Error] The maximum pressure in the tank is 29.5 kPa. References Worksheet
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Difficulty: Easy
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A 5-m-long, 4-m-high tank contains 2.5-m-deep water when not in motion and is open to the atmosphere through a vent in the middle. Using appropriate software, investigate the effect of acceleration on the slope of the free surface of water in the tank. Let the acceleration vary from 0 m/s2 to 15 m/s2 in increments of 1 m/s2. Tabulate and plot your results. (Please upload your response/solution using the controls provided below.)
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The density of a floating body can be determined by tying weights to the body until both the body and the weights are completely submerged and then weighing them separately in air. Consider a wood log that weighs 1400 N in air. If it takes 33 kg of lead (ρ = 11,300 kg/m3) to completely sink the log and the lead in water, determine the average density of the log.
The average density of the log is
826 ± 2% kg/m3.
Explanation:
The weight of a body is equal to the buoyant force when the body is floating in a fluid while being completely submerged in it (a consequence of vertical force balance from static equilibrium). In this case, the average density of the body must be equal to the density of the fluid since [Math Processing Error] Therefore, [Math Processing Error] [Math Processing Error] [Math Processing Error] Substituting, the volume and density of the log are determined to be [Math Processing Error] [Math Processing Error] The average density of the log is 826 kg/m3. References Worksheet
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A raft is made using a number of logs with 25-cm diameter and 2-m-length, as shown in the figure. It is desired that a maximum 90 percent volume of each log will be submerged when carrying two boys with 430 N each. Determine the minimum number of logs that must be used. The specific gravity of the log is 0.75.
The minimum number of logs that should be used is
6 ± 2% .
Explanation: It is assumed that water is an incompressible substance. Let n be the number of logs that will be used. The buoyancy force must balance the total weight of two boys and the weight of the raft itself. Therefore, [Math Processing Error] [Math Processing Error] [Math Processing Error] where [Math Processing Error] The buoyancy force acting on the raft is [Math Processing Error] [Math Processing Error] [Math Processing Error] n = 6 (after rounding the minimum number of logs to the nearest integer) The minimum number of logs that should be used is 6. References Worksheet
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A prismatic timber is at equilibrium in a liquid, as shown in the figure. What is the specific gravity of the liquid?
The specific gravity of the liquid is
0.833 ± 2% .
Explanation: It is assumed that water is an incompressible substance.
The weight of the timber is [Math Processing Error] The buoyancy force is [Math Processing Error] From the diagram, [Math Processing Error] [Math Processing Error] Taking the moment about the fixed end of the timber yields [Math Processing Error] [Math Processing Error] [Math Processing Error] The specific gravity of the liquid is 0.833. References https://ezto.mheducation.com/hm.tpx
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The cylindrical tank containing water accelerates upward with az = 3.5 m/s2 while it rotates about its vertical axis by n = 100 rpm. Determine the pressure at point A.
The pressure at point A is
17.6 ± 2% kPa.
Explanation: The following assumptions have been made here: 1. The acceleration is constant. 2. Water is an incompressible fluid.
The angular velocity is [Math Processing Error] From rigid-body rotation, we can write the following expressions: [Math Processing Error] [Math Processing Error] where [Math Processing Error] Integrating dp, we get [Math Processing Error] from the boundary condition: P = 0 when r = 0 and z = Z0, [Math Processing Error]
Hence, the pressure variation in the tank can be given by [Math Processing Error] Z0 can be determined from the constancy of water volume, that is,
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[Math Processing Error] Recognizing that Z1 − Z0 = 0.503 m, we find Z0 = 0.95. Then, the pressure distribution in the tank is [Math Processing Error] For point A, we take z = 0 and r = R: [Math Processing Error] The pressure at point A is 17.6 kPa. References Worksheet
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Difficulty: Hard
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A 30-cm-diameter, 16-cm-high vertical cylindrical vessel equipped with a vertical tube at the edge is rotated about its vertical axis at a constant angular velocity of 15 rad/s. If the water rise in the tube is 30 cm, determine the net vertical pressure force acting on the vessel.
The net vertical pressure force acting on the vessel is
111 ± 2% N.
Explanation: Assume that water is an incompressible fluid. Under the given rotation, the pressure distribution inside the vessel will be as shown:
[Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] Therefore, the pressure forces on the surface will be [Math Processing Error] For the bottom surface, [Math Processing Error] https://ezto.mheducation.com/hm.tpx
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where [Math Processing Error] [Math Processing Error] [Math Processing Error] The net force on the vessel is [Math Processing Error] The net vertical pressure force acting on the vessel is 111 N. References Worksheet
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Difficulty: Medium
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The 280-kg, 6-m-wide rectangular gate shown in the figure is hinged at B and leans against the floor at A making an angle of 45° with the horizontal. The gate is to be opened from its lower edge by applying a normal force at its center. Determine the minimum force F required to open the water gate.
The minimum force required to open the water gate is
626 ± 2% kN.
Explanation: The following assumptions have been made here: 1. Atmospheric pressure acts on both sides of the gate, and thus it can be ignored in calculations for convenience. 2. The friction at the hinge is negligible.
The length of the gate and the distance of the upper edge of the gate (point B) from the free surface in the plane of the gate are [Math Processing Error] [Math Processing Error] The average pressure on a surface is the pressure at the centroid (midpoint) of the surface, and multiplying it by the plate area gives the resultant hydrostatic pressure on the surface: [Math Processing Error] [Math Processing Error] [Math Processing Error] The distance of the pressure center from the free surface of water along the plane of the gate is [Math Processing Error] The distance of the pressure center from the hinge at point B is [Math Processing Error] Taking the moment about point B and setting it equal to zero give [Math Processing Error] Solving for F and substituting, the required force to overcome the pressure is https://ezto.mheducation.com/hm.tpx
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[Math Processing Error] The minimum force required to open the water gate is 626 kN. References Worksheet
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The 350-kg, 6-m-wide rectangular gate shown in the figure is hinged at B and leans against the floor at A making an angle of 45° with the horizontal. The gate is to be opened from its lower edge by applying a normal force at its center. Determine the minimum force F required to open the water gate.
The minimum force required to open the water gate is
702 ± 2% kN.
Explanation: The following assumptions have been made here: 1. Atmospheric pressure acts on both sides of the gate, and thus it can be ignored in calculations for convenience. 2. The friction at the hinge is negligible.
The length of the gate and the distance of the upper edge of the gate (point B) from the free surface in the plane of the gate are [Math Processing Error] [Math Processing Error] The average pressure on a surface is the pressure at the centroid (midpoint) of the surface, and multiplying it by the plate area gives the resultant hydrostatic on the surface: [Math Processing Error] [Math Processing Error] [Math Processing Error] The distance of the pressure center from the free surface of water along the plane of the gate is [Math Processing Error] The distance of the pressure center from the hinge at point B is [Math Processing Error] Taking the moment about point B and setting it equal to zero give [Math Processing Error] Solving for F and substituting, the required force to overcome the pressure is https://ezto.mheducation.com/hm.tpx
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[Math Processing Error] In addition to this, there is the weight of the gate itself, which must be added. In the 45° direction, [Math Processing Error] Thus, the total force required in the 45° direction is the sum of these two values: [Math Processing Error] The minimum force required to open the water gate is 702 kN. References Worksheet
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Determine the vertical force applied by water on the container. The diameter D of the bottom surface of the container is 44 cm.
The vertical force applied by water on the container is
515.4 ± 2% N.
Explanation: It is assumed that water is an incompressible fluid. The vertical force on a curved surface is equal to the weight of the liquid volume carried by the surface. The volume of a frustum is given by [Math Processing Error] where h = 0.5 m, R = 0.22 m, and r = 0.075 m. Therefore, the volume of imaginary water is [Math Processing Error] [Math Processing Error] [Math Processing Error] Hence, we find the vertical force to be [Math Processing Error] [Math Processing Error] The vertical force applied by water on the container is 515.4 N. References Worksheet
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Difficulty: Medium
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211.
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Award: 0.00 points
Knowing that the vertical component of force acting on a curved surface is given by dFV = PdAx, what is the relation between the net vertical force FV on a sphere submerged at a depth h in a liquid and the buoyancy force FB on the same sphere?
F V = FB × ρ F V = FB F V = F B /ρ F V = FB × V It is assumed that water is an incompressible fluid.
The buoyance force exerted by water on a sphere (which is completely submerged) is equal to the weight displaced by the fluid.
[Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] [Math Processing Error] Therefore,
[Math Processing Error] F V = FB References Multiple Choice https://ezto.mheducation.com/hm.tpx
Difficulty: Medium 243/260
06/06/2024, 11:08
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212. Award: 0.00 points
In order to keep the cone-shaped plug closed as shown in the figure, what would be the maximum air pressure on the surface of water? The tanks have the same width of b = 2 m. The specific gravity of the second fluid is 1.6.
The maximum air pressure on the top of water is
33.7 ± 2% kPa.
Explanation: It is assumed that water is an incompressible fluid. The net force applied by the liquid on the right is [Math Processing Error] [Math Processing Error] The net force applied by the liquid on the left is [Math Processing Error] To keep the plug closed, F1 = F2, or [Math Processing Error] [Math Processing Error] [Math Processing Error] The maximum air pressure on the top of water is 33.7 kPa. References Worksheet
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Difficulty: Medium
244/260
06/06/2024, 11:08
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213. Award: 0.00 points
The gage pressure in a pipe is measured by a manometer containing mercury (ρ = 13,600 kg/m3). The top of the mercury is open to the atmosphere and the atmospheric pressure is 100 kPa. If the mercury column height is 29 cm, the gage pressure in the pipe is _____. Solve this problem using appropriate software.
39 kPa 134 kPa 89 kPa 31 kPa h=0.29 [m] P_atm=100 [kPa] rho=13600 [kg/m^3] g=9.81 [m/s^2] P_gage=rho*g*h*Convert(Pa, kPa) The gage pressure in the pipe is 39 kPa. References Multiple Choice
Difficulty: Easy
214. Award: 0.00 points
Which is of the highest value?
1 atm 1 bar 105 N/m2 100 kPa The highest value among these is 1 atm. The highest value among these is 1 atm. References Multiple Choice
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Difficulty: Easy
245/260
06/06/2024, 11:08
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215. Award: 0.00 points
The pressure in seawater where a submarine is sailing is measured to be 1300 kPa. The submarine is in a water depth of _____. (Take the density of water to be 1000 kg/m3.)
122 m 0.127 m 0.122 m 127 m P=1300 [kPa] P_atm=101.325 [kPa] rho=1000 [kg/m^3] g=9.81 [m/s^2] P−P_atm=rho*g*h*Convert(Pa, kPa) The submarine is in a water depth of 122 m. References Multiple Choice
Difficulty: Easy
216. Award: 0.00 points
The atmospheric pressure in a location is measured by a mercury (ρ = 13,600 kg/m3) barometer. If the height of the mercury column is 750 mm, the atmospheric pressure at that location is _____. Solve this problem using appropriate software.
97.1 kPa 105.1 kPa 100.1 kPa 115.1 kPa h=0.75 [m] rho=13600 [kg/m^3] g=9.81 [m/s^2] P_atm=rho*g*h*Convert(Pa, kPa) The atmospheric location at that location is 100.1 kPa. References Multiple Choice
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Difficulty: Medium
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06/06/2024, 11:08
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217. Award: 0.00 points
A manometer is used to measure the pressure of a gas in a tank. The manometer fluid is water (ρ = 1000 kg/m3) and the manometer column height is 1.8 m. If the local atmospheric pressure is 100 kPa, the absolute pressure within the tank is _____. Solve this problem using appropriate software.
68 kPa 123 kPa 118 kPa 168 kPa rho=1000 [kg/m^3] h=1.8 [m] P_atm=100 [kPa] g=9.81 [m/s^2] P=P_atm+rho*g*h*Convert(Pa, kPa) The absolute pressure within the tank is 118 kPa. References Multiple Choice
Difficulty: Medium
218. Award: 0.00 points
Consider a hydraulic car jack with a piston area ratio of 47. A person can lift a 1000-kg car by applying a force of _____. Solve this problem using appropriate software.
71 kgf 21 kgf 116 kgf 47 kgf A2\A1=47 m_car=1000 [kg] g=9.81 [m/s^2] F_car=m_car*g F_car/F_person=A2\A1 F_person_kgf=F_person/9.81 A person can life the car by applying a force of 21 kgf. References Multiple Choice
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Difficulty: Medium
247/260
06/06/2024, 11:08
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219. Award: 0.00 points
Consider the vertical rectangular wall of a water tank with a width of 5 m and a height of 7 m. The other side of the wall is open to the atmosphere. The resultant hydrostatic force on this wall is _____. Solve this problem using appropriate software.
1002 kN 1202 kN 1252 kN 1302 kN a=5 [m] b=7 [m] P_atm=101 [kPa] rho=1000 [kg/m^3] g=9.81 [m/s^2] Area=a*b P_C=(rho*g*b)/2*Convert(Pa, kPa) F_R=P_C*Area The resultant hydrostatic force on the wall is 1202 kN. References Multiple Choice
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Difficulty: Medium
248/260
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220. Award: 0.00 points
A vertical rectangular wall with a width of 20 m and a height of 12 m is holding a 8-m-deep water body. The resultant hydrostatic force acting on this wall is _____. Solve this problem using ESS.
5268 kN 7743 kN 6278 kN 6428 kN a=20 [m] h=12 [m] b=8 [m] rho=1000 [kg/m^3] g=9.81 [m/s^2] Area=a*b P_C=rho*g*b/2*Convert(Pa, kPa) F_R=P_C*Area The resultant hydrostatic force acting on the wall is 6278 kN. References Multiple Choice
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Difficulty: Medium
249/260
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221. Award: 0.00 points
A vertical rectangular plate with a width of 18 m and a height of 12 m is located 4 m below a water surface. The resultant hydrostatic force on this plate is _____. Solve this problem using appropriate software.
19690 kN 18190 kN 21190 kN 23190 kN a=18 [m] b=12 [m] s=4 [m] rho=1000 [kg/m^3] g=9.81 [m/s^2] Area=a*b P_C=rho*g*(s+b/2)*Convert(Pa, kPa) F_R=P_C*Area The resultant hydrostatic force on this plate is 21190 kN. References Multiple Choice
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Difficulty: Medium
250/260
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222. Award: 0.00 points
A rectangular plate with a width of 20 m and a height of 12 m is located below a water surface. The plate is tilted and makes a 35° angle with the horizontal. The top edge is horizontal and is at a distance of 4 m from the free surface along the plane of the plate. The resultant hydrostatic force acting on the top surface of this plate is _____. Solve this problem using appropriate software.
13502 kN 12252 kN 670 kN 16172 kN a=20 [m] b=12 [m] s=4 [m] theta=35 [degree] rho=1000 [kg/m^3] g=9.81 [m/s^2] Area=a*b P_C=rho*g*(s+b/2)*Sin(theta)*Convert(Pa, kPa) F_R=P_C*Area The resultant hydrostatic force acting on the top surface of this plate is 13502 kN. References Multiple Choice
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Difficulty: Medium
251/260
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223. Award: 0.00 points
A vertical rectangular plate with a width of 16 m and a height of 12 m is located 3 m below a water surface. The line of action yp for the resultant hydrostatic force on this plate is _____. (Neglect atmospheric pressure.) Solve this problem using appropriate software.
12.6 m 6.5 m 10.3 m 8.3 m a=16 [m] b=12 [m] s=3 [m] y_p=s+b/2+b^2/(12*(s+b/2)) The line of action yp for the resultant hydrostatic force on this plate is 10.3 m. References Multiple Choice
Difficulty: Easy
224. Award: 0.00 points
A 4-m-long and 3-m-wide horizontal rectangular plate is submerged in water. The distance of the top surface from the free surface is 5 m. The atmospheric pressure is 95 kPa. Considering the atmospheric pressure, the hydrostatic force acting on the top surface of this plate is _____. Solve this problem using appropriate software.
2196 kN 1729 kN 1665 kN 1949 kN a=4 [m] b=3 [m] h=5 [m] P_atm=95 [kPa] rho=1000 [kg/m^3] g=9.81 [m/s^2] Area=a*b P_C=(P_atm+rho*g*h*Convert(Pa, kPa)) F_R=P_C*Area The hydrostatic force acting on the top surface of this plate is 1729 kN. References Multiple Choice https://ezto.mheducation.com/hm.tpx
Difficulty: Easy 252/260
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225. Award: 0.00 points
Consider a 6-m-diameter spherical gate holding a body of water whose height is equal to the diameter of the gate. Atmospheric pressure acts on both sides of the gate. The horizontal component of the hydrostatic force acting on this curved surface is _____. Solve this problem using appropriate software.
832 kN 865 kN 862 kN 812 kN D=6 [m] rho=1000 [kg/m^3] g=9.81 [m/s^2] R=D/2 A=pi*R^2 P_C=rho*g*R*Convert(Pa, kPa) F_x=P_C*A The horizontal component of the hydrostatic force acting on this curved surface is 832 kN. References Multiple Choice
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Difficulty: Medium
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226. Award: 0.00 points
Consider a 6-m-diameter spherical gate holding a body of water whose height is equal to the diameter of the gate. Atmospheric pressure acts on both sides of the gate. The vertical component of the hydrostatic force acting on this curved surface is _____. Solve this problem using appropriate software.
327 kN 270 kN 416 kN 505 kN D=6 [m] rho=1000 [kg/m^3] g=9.81 [m/s^2] R=D/2 V=D^3/2-4/3*pi*R^3/2 m=rho*V W=m*g*Convert(N, kN) A=pi*R^2/2 h=0 [m] P_C=rho*g*h F_y=P_C*A*Convert(N, kN) F_v=F_y-W The vertical component of the hydrostatic force acting on this curved surface is 505 kN. References Multiple Choice
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Difficulty: Medium
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227. Award: 0.00 points
A 0.6-m-diameter spherical object is completely submerged in water. The buoyant force acting on this object is _____.
1109 N 5344 N 10271 N 548 N D=0.6 [m] rho_f=1000 [kg/m^3] g=9.81 [m/s^2] V=pi*D^3/6 F_B=rho_f*g*V The bouyant force acting on the object is 1109 N. References Multiple Choice
Difficulty: Easy
228. Award: 0.00 points
A 5-kg object with a density of 7500 kg/m3 is placed in water. The weight of this object in water is _____. Solve this problem using appropriate software.
36.1 N 47.1 N 47.1 N 42.5 N m_body=5 [kg] rho_body=7500 [kg/m^3] rho_f=1000 [kg/m^3] g=9.81 [m/s^2] V_body=m_body/rho_body F_B=rho_f*g*V_body W_inair=m_body*g W_inwater=W_inair-F_B The weight of the object in water is 42.5 N. References Multiple Choice
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Difficulty: Medium
255/260
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229. Award: 0.00 points
A 12-m-diameter hot air balloon is neither rising nor falling. The density of atmospheric air is 1.3 kg/m3. The total mass of the balloon including the people on board is _____. Solve this problem using appropriate software.
1144 kg 1226 kg 1078 kg 1176 kg D=12 [m] rho_f=1.3 [kg/m^3] V=pi*D^3/6 g=9.81 [m/s^2] W=m*g F_B=rho_f*g*V W=F_B The total mas of the balloon is 1176 kg. References Multiple Choice
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Difficulty: Medium
256/260
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230. Award: 0.00 points
A 11-kg object with a density of 900 kg/m3 is placed in a fluid with a density of 1100 kg/m3. The fraction of the volume of the object submerged in water is _____. Solve this problem appropriate software.
0.818 1.118 0.661 1.031 m_object=11 [kg] rho_object=900 [kg/m^3] rho_f=1100 [kg/m^3] g=9.81 [m/s^2] V_object=m_object/rho_object W=m_object*g F_B=rho_f*g*V_submerged W=F_B Fraction=V_submerged/V_object The fraction of the volume of the object submerged in water is 0.818. References Multiple Choice
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Difficulty: Medium
257/260
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231. Award: 0.00 points
Consider a cubical water tank with a side length of 2 m. The tank is half-filled with water and is open to the atmosphere. Now, a truck carrying this tank is accelerated at a rate of 5 m/s2. The maximum vertical rise of the free surface of the water is _____. Solve this problem using appropriate software.
0.82 m 0.26 m 1.07 m 0.51 m s=2 [m] a_x=5 [m/s^2] g=9.81 [m/s^2] a_z=0 [m/s^2] tan(theta)=a_x/(g+a_z) DELTAz_max=s/2*tan(theta) The maximum vertical rise of free surface of the water is 1 m. References Multiple Choice
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Difficulty: Medium
258/260
06/06/2024, 11:08
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232. Award: 0.00 points
A 20-cm-diameter, 40-cm-high vertical cylindrical container is partially filled with 22-cm-high water. Now the cylinder is rotated at a constant speed of 15 rad/s. The height of water at the edge of the cylinder is _____. Solve this problem using appropriate software.
27.7 cm 36.7 cm 32.7 cm 24.1 cm D=20 [m] H=0.4 [m] h_0=0.22 [m] Omega=15 [rad/s] R=D/2 g=9.81 [m/s^2] rr=R z_s=h_0-omega^2/(4*g)*(R^2-2*rr^2) The height of water at the edge of the cylinder is 27.7 cm. References Multiple Choice
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Difficulty: Medium
259/260
06/06/2024, 11:08
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233. Award: 0.10 points
A 30-cm-diameter, 40-cm-high vertical cylindrical container is partially filled with 23-cm-high water. Now, the cylinder is rotated at a constant speed of 15 rad/s. The height of water at the edge of the cylinder is _____. Solve this problem using appropriate software.
35.9 cm 40.9 cm 32.3 cm 44.9 cm D=30 [m] H=0.4 [m] h_0=0.23 [m] Omega=15 [rad/s] R=D/2 g=9.81 [m/s^2] rr=R z_s=h_0-omega^2/(4*g)*(R^2-2*rr^2) The height of water at the edge of the cylinder is 35.9 cm. References Multiple Choice
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Difficulty: Medium
260/260
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