Introduction to material balances and basics of mass conservation One of the most basic principles applied in the design and analysis of chemical engineering processes is the law of conservation. This is applied to the following conserved quantities: Mass (material balances) Energy (first law of thermodynamic - energy balances) Linear momentum fluid mechanics (fluid mechanics) Material balances provide a tool for engineers in many fields to grasp and to quantify the factors that determine the behaviour of both man-made and natural systems. Chemical processes are complex and typically involve many interacting factors in non-linear relationships. In this respect, the material balance techniques used by engineers to model industrial processes can be used as both an educational and a predictive tool to deal with complex processes. The principles involved to solve material balance calculations are those you probably learned in secondary school such as: -
The law of conversation Physical quantities Chemical reactions and stoichiometry The ideal gas law The first law of thermodynamics Simple algebra and a little calculus Computer calculations with spreadsheets
Conservation of mass Consider a simple boiler involving two components, A and B, shown below (this could be desalinisation, for example with A being water and B being salt):
Here FT1 is the total mass flowrate of stream 1. The symbols FA1 and FB1 are the mass flowrates of A and B, respectively in stream 1. These have units mass per time.
Mass conservation equation Over a duration of time t1 to t2 = t1+ Δt, material flows into and out of a process unit. Mass conservation demands that: mass in boiler at t2 = mass in boiler t1 + mass flowing in boiler over Δt - mass flowing out boiler over Δt
we can re-arrange it to: mass in boiler at t2 - mass in boiler t1 = mass flowing in boiler over Δt - mass flowing out boiler over Δt
or we can write down this mathematically: M(t2) – M(t1) = FT1 Δt – (FT2+FT3) Δt where M is the total mass of all material in the boiler and F is the flowrate of the material. The relationship between F and M is defined as: F = M/t We can rearrange the equation: M(t1+ Δt) – M(t1) = [FT1-(FT2+FT3)] Δt or: �
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to calculate the accumulation of mass in the boiler:
which gives:
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which means:
Rate of accumulation = rate of input - rate of output At steady state (process variables do not change with time), the rate of accumulation is zero, therefore:
Rate of input = rate of output Component balances In a process where only physical changes occur (in the absence of chemical reaction), the same material balance equation can be applied directly to individual components:
FA1 = FA2 + FA3 FB1 = FB2 + FB3 Note that the component balances are not independent of the overall material balance since they can be added to give: FA1+FB1 = (FA2+FB2)+(FA3+FB3) FT1 = FT2 + FT3 If n is the number of the components, n+1 equations can be generated in total. From these equations, the number of independent equations is n. The dependent equation can be used to check the final answer. When chemical reactions are present, atomic or mole balances can be useful for the analysis of steady state processes. We look into material balance with chemical reactions later. Independent equations: equations are independent if none of them can be derived from the others. For example, not one of the set of equations can be obtained by adding or subtracting multiples of the others. Example: Try to obtain the numerical vale of x and y in the following examples. If it is not possible, why not? (i) x + y = 1 (ii) x + y = 1; x – y = 2 (iii) x + y = 1; x – y = 2; y = 3 (iv) x + y = 1; 2x + 2y = 2
Example: Modeling desalinator Desalination of water is an important process in many parts of the world. Seawater Drinking water
H2O, % 96.5 100
Salt, % 3.5 0
Analyse a desalination plant producing a pure product of water and waste stream of 27% brine. The plant is processing 100 kg/min of seawater. We always take a basis. Step one: construct a process flow diagram
Note that the streams have been labelled and a control volume (the dashed box ) has been constructed over the process of interest. In this simple system, this detail may seem unnecessary, but it is recommended practice. Step two: define variables Stream 1:
FT1, Fw1 = FT1.xw1, FS1=FT1.xS1
Stream 2:
FT2, Fw2 = FT2.xw2, FS2=FT2.xS2
Stream 3:
FT3, Fw3 = FT3.xw3, FS3=FT3.xS3
Unknowns:
FT2, FT3
Step three: apply overall material balance FT1 = FT2 + FT3 Step four: apply component material balance Water:
FW1 = FW2 + FW3 FT1.xW1= FT2.xW2+ FT3.xW3
Salt:
FS1 = FS2 + FS3
(Dependent)
FT1.xS1= FT2.xS2+ FT3.xS3
Step five: solve for water production rate 1)
FT1 = FT2 + FT3 100 = FT2 + FT3
2)
FT1.xW1= FT2.xW2+ FT3.xW3 100(96.5/100) = FT2(73/100) + FT3 96.5 = 0.73 FT2 + FT3
3)
FT3 = 96.5 – 0.73(100-FT3)
FT3 = 87 kg/min
If desired, a component balance on the water can be used to check the result. This is where the dependent equation can be used to check the final result: FT1.xS1= FT2.xS2+ FT3.xS3 100 x 0.035 = 13 x 0.27 + 87 x 0 3.5 = 3.5