ME 150 – Heat and Mass Transfer
Chap. 14: Internal Convection
Forced Convection in Internal Geometries No free flow condition (u∞ and T∞) exists in this case: Flow is determined by two boundary layers
r0
x
xfd, d,hx
at entrance: homogeneous velocity profile after hydrodynamic entrance length: fully developed velocity profile (no changes with x anymore) Prof. Nico Hotz
1
ME 150 – Heat and Mass Transfer
Chap. 14.1: Pipe Flow
Pipe Flow – Hagen-Poiseuille Equation The Hagen–Poiseuille equation describes a fluid flowing through a long cylindrical pipe due to a pressure drop. The assumptions of the equation are that the flow is laminar and incompressible and the flow is through a constant circular cross-section that is substantially longer than its diameter. This leads to the following velocity profile u(r):
(
u (r ) = r02 − r 2 ⎛ r 2 ⎞ u (r ) = umax ⋅ ⎜⎜1 − 2 ⎟⎟ ⎝ r0 ⎠
) (4p ⋅ −µ p⋅ l ) 1
2
=−
with
Prof. Nico Hotz
dp 1 ⋅ ⋅ r02 − r 2 dx 4 ⋅ µ
(
umax
)
dp r02 =− ⋅ dx 4 ⋅ µ
2
ME 150 – Heat and Mass Transfer
Chap. 14.1: Pipe Flow
Definition of a mean velocity in the pipe: V = u m ⋅ A
Mean velocity: um =
1 ∫ u (r ) ⋅ dA AA
Using the result for u(r): um = −
2 0
r u dp ⋅ = max 8 ⋅ µ dx 2
→
Prof. Nico Hotz
⎡ ⎛ r ⎞ 2 ⎤ u (r ) = 2 ⋅ um ⋅ ⎢1 − ⎜⎜ ⎟⎟ ⎥ ⎢⎣ ⎝ r0 ⎠ ⎥⎦ 3
ME 150 – Heat and Mass Transfer
Chap. 14.1: Pipe Flow
Mass flow rate the entrance region: u = u(x,r) m = ρ ⋅ u m ⋅ A = ρ ⋅ ∫ u( x, r ) ⋅ dA A
Using the pipe geometry: r0
m um = = ρ⋅A
2π ⋅ ρ ⋅ ∫ u (x, r ) ⋅ r ⋅ dr 0
ρ ⋅ π ⋅ r02
2 r0 = 2 ⋅ ∫ u ( x, r ) ⋅ r ⋅ dr r0 0
Using mass conservation (incompressible fluid) : um ≠ f(x)
Turbulence for ReD > ReD,krit
Re D =
ρ ⋅ um ⋅ D µ
Prof. Nico Hotz
Re D,krit ≈ 2300
4
ME 150 – Heat and Mass Transfer
Chap. 14.1: Pipe Flow
Hydrodynamic entrance length (semi-empirical):
Laminar flow:
⎛ X fd ,h ⎞ ⎜⎜ ⎟⎟ ≈ 0.05 ⋅ Re D ⎝ D ⎠lam
Turbulent flow:
⎛ X fd ,h ⎞ ⎟⎟ ≤ 60 10 ≤ ⎜⎜ ⎝ D ⎠turb
Moody Friction Factor: For laminar flow: Use results from Hagen-Poiseuille equation: −
dp ρ ⋅ D = f ⋅ ⋅ um2 dx 2
f ≡
− (dp dx )⋅ D 64 = ρ⋅ um2 / 2 Re D
Prof. Nico Hotz
(for laminar flow)
5
ME 150 – Heat and Mass Transfer
Chap. 14.1: Pipe Flow
Moody Diagram
Prof. Nico Hotz
6
ME 150 – Heat and Mass Transfer
Chap. 14.2: Pipe Flow – Thermal Considerations
Thermal Considerations Considering pipe flow with heat transfer between fluid and wall:
Thermal entrance length (laminar):
Prof. Nico Hotz
⎛ X fd ,t ⎜⎜ ⎝ D
⎞ ⎟⎟ ≈ 0.05 ⋅ Re D ⋅ Pr ⎠lam
7
ME 150 – Heat and Mass Transfer
Chap. 14.2: Pipe Flow – Thermal Considerations
Definition of mean temperature Tm using the flow enthalpy: E conv = ∫ ρ⋅ u (r ) ⋅ c p ⋅ T ( x, r ) ⋅ dA ≡ m ⋅ c p ⋅ Tm ( x) A
∫ ρ ⋅u(r ) ⋅ c
p
⋅ T ( x, r ) ⋅ dA
solved for Tm:
Tm (x ) =
Mass flow rate expressed by um :
2 0 Tm ( x) = u (r ) ⋅ T ( x, r ) ⋅ r ⋅dr 2 ∫ um ⋅ r0 0
A
m ⋅ c p r
Convective heat transfer depending on Tm:
qʹ′ʹ′ = h ⋅ (Ts − Tm )
Note: For pipe flow, um und Tm are used instead of u∞ and T∞ Prof. Nico Hotz
8
ME 150 – Heat and Mass Transfer
Chap. 14.2: Pipe Flow – Thermal Considerations
Is the temperature profile stationary / fully developed ? The absolute temperature changes along x: Tm = Tm ( x)
and
dTm ≠ 0! dx
Introduction of a dimensionless temperature: Ts − T (r , x) θ ( r , x) = Ts − Tm
The ‘relative shape‘ of the temperature profile is constant: ∂θ (r , x) ∂ ⎛ T − T (r , x) ⎞ ⎟⎟ = 0 = ⎜⎜ s ∂x ∂x ⎝ Ts − Tm ⎠
Prof. Nico Hotz
i.e. θ = θ (r ) ≠ f ( x)
9
ME 150 – Heat and Mass Transfer
Chap. 14.3: Pipe Flow – Energy Balance
Energy Balance for volume element Goal: Temperature profile of Tm(x) for known boundary conditions and h value
Assumptions for derivation: - Pipe flow, steady-state: m = const. - ΔKE = ΔPE = 0 - no heat conduction in axial direction Prof. Nico Hotz
10
ME 150 – Heat and Mass Transfer
Chap. 14.3: Pipe Flow – Energy Balance
First Law for open system for volume element (δW = 0): Entering heat
+
Entering enthalpy flux
-
Leaving enthalpy flux
=0
d (cv ⋅ Tm + p ⋅ v ) ⎤ ⎡ dq s + m ⋅ (cv ⋅ Tm + p ⋅ v) − ⎢m ⋅ (cv ⋅ Tm + p ⋅ v ) + m dx ⎥ = 0 dx ⎣ ⎦ → dq s = m ⋅ d (cv ⋅ Tm + p ⋅ v ) = m ⋅ c p ⋅ dTm
After integration over entire pipe length: First Law for pipe Heat transfer through pipe wall:
qs = m ⋅ c p ⋅ (Tm ,out − Tm ,in )
Valid for compressible and incompressible fluids
Prof. Nico Hotz
11
ME 150 – Heat and Mass Transfer
Chap. 14.3: Pipe Flow – Energy Balance
Using heat flux through the outer surface of the control volume (P = inner pipe perimeter) dq s = qʹ′ʹ′ ⋅ P ⋅ dx = m ⋅ c p ⋅ dTm →
dTm P P = ⋅ qʹ′ʹ′ = ⋅ h ⋅ (Ts − Tm ) dx m ⋅ c p m ⋅ c p
ODE for the development of Tm along x - direction 2 specific cases are of interest: • Heat flux is constant • Inner wall temperature is constant
Prof. Nico Hotz
12
ME 150 – Heat and Mass Transfer
Chap. 14.3.1: Pipe Flow – Constant Heat Flux
Solution for constant heat flux through pipe wall ODE to be solved: dTm P = ⋅ qʹ′sʹ′ dx m ⋅ c p
Boundary condition: Tm(0) = Tm,in Tm ( x) = Tm,in +
qʹ′sʹ′ ⋅ P ⋅x m ⋅ cp
Tm increases linearly Entrance length: h high, (Ts - Tm) small
No quantitative result for h possible yet !
Prof. Nico Hotz
13
ME 150 – Heat and Mass Transfer
Chap. 14.3.2: Pipe Flow – Constant Wall Temp.
Solution for constant wall temperature ODE to be solved: formal analogy to transient conduction (0-D) dTm P = ⋅ h ⋅ (Ts − Tm ) dx m ⋅ c p
Homogeneous ODE by using new variable ΔT = Ts - Tm −
d (ΔT ) P = ⋅ h ⋅ΔT dx m ⋅ c p
Separation of variables, integration over entire pipe length ΔTout
L ⎞ d (ΔT ) P ⋅ L ⎛ 1 ⎜ ⎟ = − h ⋅ dx ∫ΔT ΔT ∫ ⎜ ⎟ m ⋅ c L p 0 ⎝ ⎠ in
Prof. Nico Hotz
14
ME 150 – Heat and Mass Transfer
Chap. 14.3.2: Pipe Flow – Constant Wall Temp.
Using average h value: L
1 hL = ∫ h ⋅ dx L0
ln
ΔTout P⋅L =− ⋅ hL ΔTin m ⋅ cp
Rearranging and using absolute temperatures: ⎛ P ⋅ L ⎞ ΔTout Ts − Tm,out ⎜ = = exp − ⋅ hL ⎟ ⎜ ⎟ ΔTin Ts − Tm,in ⎝ m ⋅ c p ⎠
Mean fluid temperature at any location x: ⎛ P ⋅ x ⎞ Ts − Tm ( x) ⎜ = exp − ⋅ hx ⎟ ⎜ m ⋅ c ⎟ Ts − Tm,in p ⎝ ⎠
After thermal entrance length, h is constant. Therefore, for short entrance lengths:
Prof. Nico Hotz
hx = hL = h
15
ME 150 – Heat and Mass Transfer
Chap. 14.3.2: Pipe Flow – Constant Wall Temp.
Total heat transfer calculated with averaged values: qs = P ⋅ L ⋅ h ⋅ ΔT
Average temperature difference: L
1 ΔT = ⋅ ∫ ΔT ( x) ⋅ dx L 0 where :
ΔT ( x) = ΔTin ⋅ e −γ ⋅ x
Prof. Nico Hotz
and
γ=
h⋅P m ⋅ c p
16
ME 150 – Heat and Mass Transfer
Chap. 14.3.2: Pipe Flow – Constant Wall Temp.
Using function for ΔT(x): ΔT = −
ΔTin −γ ⋅x L ΔT ⋅e = − in ⋅ e −γ ⋅L −1 0 γ ⋅L γ ⋅L
(
Substituting with ΔT at outlet → ΔTin ⋅ e −γ ⋅L = ΔTout
)
rearranging →
Log Mean Temperature Difference: (for known inlet and outlet temperatures)
Prof. Nico Hotz
⎛ ΔTin ⎞ ⎟⎟ ⎝ ΔTout ⎠
γ ⋅ L = ln⎜⎜
ΔT =
ΔTout − ΔTin ΔT ⎞ ln⎛⎜ out ⎟ Δ T in ⎠ ⎝
17
ME 150 – Heat and Mass Transfer
Chap. 14.3.2: Pipe Flow – Constant Wall Temp.
If instead of the wall temperature the outer surrounding fluid temperature is known:
qs = U L ⋅ L ⋅ ΔT = U L ⋅ L ⋅
ΔTout − ΔTin ΔT ⎞ ln⎛⎜ out ⎟ Δ T in ⎠ ⎝
The convective heat transfer is substituted by a total heat transfer coefficient (h → UL): linear tube U-value !
Prof. Nico Hotz
18
ME 150 – Heat and Mass Transfer
Chap. 14.3.3: Pipe Flow – Heat Transfer Coeff.
Heat transfer coefficients for laminar pipe flows
We need to solve for the temperature gradient on the wall surface: ∂T ∂ T α ∂ ⎛ ∂ T ⎞ ⎟⎟ u⋅ +v⋅ = ⋅ r ⋅ ⎜⎜ ∂x ∂ r r ∂ r ⎝ ∂ r ⎠
If u and v are known: ODE can be solved for laminar pipe flow in the fully developed region: v=0
and
∂u =0 ∂x
Prof. Nico Hotz
⎡ ⎛ r ⎞ 2 ⎤ u (r ) = 2 ⋅ um ⋅ ⎢1 − ⎜⎜ ⎟⎟ ⎥ ⎢⎣ ⎝ r0 ⎠ ⎥⎦
19
ME 150 – Heat and Mass Transfer
Chap. 14.3.3: Pipe Flow – Heat Transfer Coeff.
Heat conduction in axial direction is negligible compared to convection: assumption already used for Tm(x)
∂ 2T =0 2 ∂x
Using solution for u(r): Conduction radial
1 d ⎛ dT ⋅ ⎜ r ⋅ r dr ⎝ dr
Convection axial 2 ⎡ ⎤ ⎛ ⎞ 2 ⋅ u dT r ⎞ m ⎛ m ⎞ ⎜ ⎟ ⋅ ⎢1 − ⎜⎜ ⎟⎟ ⎥ ⎟ = α ⎝ dx ⎠ ⎢ ⎝ r0 ⎠ ⎥ ⎠ ⎣ ⎦
For case: qʹ′sʹ′ known = constant , the temperature gradient
dTm is known = constant dx
Prof. Nico Hotz
20
ME 150 – Heat and Mass Transfer
Chap. 14.3.3: Pipe Flow – Heat Transfer Coeff.
Integrating twice: 2 ⋅ um ⎛ dTm ⎞ ⎛ r 2 dT r 4 ⎞ ⎟ + C1 r⋅ = ⎜ ⎟ ⋅ ⎜⎜ − 2 ⎟ dr α ⎝ dx ⎠ ⎝ 2 4 ⋅ r0 ⎠ 2 ⋅ um ⎛ dTm ⎞ ⎛ r 2 r 4 ⎞ ⎟ + C1 ⋅ ln(r ) + C2 T (r ) = ⎜ ⎟ ⋅ ⎜ − α ⎝ dx ⎠ ⎜⎝ 4 16 ⋅ r02 ⎟⎠
Boundary conditions: T(0) is finite / symmetry at x = 0 →
C1 = 0
T(r0) = Ts(x) is given (directly or through heat flux) T (r0 , x) = Ts ( x)
→
2 ⋅ um ⎛ dTm ⎞⎛ 3 ⋅ r02 ⎞ ⎟ C2 = Ts − ⎜ ⎟⎜ α ⎝ dx ⎠⎜⎝ 16 ⎟⎠
Prof. Nico Hotz
21
ME 150 – Heat and Mass Transfer
Chap. 14.3.3: Pipe Flow – Heat Transfer Coeff.
Entire temperature profile: 4 2 ⎡ 2 ⋅ um ⋅ r ⎛ dTm ⎞ 3 1 ⎛ r ⎞ 1 ⎛ r ⎞ ⎤ T (r ) = Ts − ⋅ ⎜ ⎟ ⋅ ⎢ + ⎜⎜ ⎟⎟ − ⎜⎜ ⎟⎟ ⎥ α ⎝ dx ⎠ ⎢⎣16 16 ⎝ r0 ⎠ 4 ⎝ r0 ⎠ ⎥⎦ 2 0
Definition of mean temperature Tm (see above): E conv ≡ ρ ⋅ um ⋅ c p ⋅ Tm (x )
Tm =
r0
2 u m ⋅ r0
2
∫ u (r ) ⋅ T (r ) ⋅ r ⋅dr
0
u(r) and T(r) are known, resulting in: 11 ⎛ um ⋅ r02 ⎞ dTm ⎟⎟ ⋅ Tm = Ts − ⋅ ⎜⎜ 48 ⎝ α ⎠ dx
Prof. Nico Hotz
22
ME 150 – Heat and Mass Transfer
Chap. 14.3.3: Pipe Flow – Heat Transfer Coeff.
48 Ts − Tm tan(ϕ ) = ⋅ 11 D
Temperatur
Temperature profile: Gradient is almost constant over half of cross-section
Temperature
Ts
ϕ
Tm D
-r0
0
+r0
Rohrdurchmesser
Pipe radius
Additional analysis for constant heat flux through pipe wall Axial temperature increase driven by heat flux: ρ ⋅ c p ⋅ um ⋅ π ⋅ r02 ⋅
dTm = qʹ′sʹ′ ⋅ 2π ⋅ r0 dx Prof. Nico Hotz
→
dTm 2 ⋅ qʹ′sʹ′ = dx ρ ⋅ c p ⋅ um ⋅ r0 23
ME 150 – Heat and Mass Transfer
Chap. 14.3.3: Pipe Flow – Heat Transfer Coeff.
Difference between wall and mean fluid temperature: f(h)
Tm − Ts = −
11 qʹ′sʹ′ ⋅ D qʹ′sʹ′ ⋅ = 48 k h
Using this result, we can calculate h: h=
48 k 11 D
→
Nu D =
h⋅ D = 4.36 k
Valid for laminar pipe flow and qʹ′sʹ′ = constant
Constant wall temperature (derivation not shown) Nu D = 3.66
Ts = const. Prof. Nico Hotz
24
ME 150 – Heat and Mass Transfer
Chap. 14.4: Pipe Flow – Summary
Summary for Laminar Pipe Flow ⎡ ⎛ r ⎞ 2 ⎤ u (r ) = 2 ⋅ um ⋅ ⎢1 − ⎜⎜ ⎟⎟ ⎥ ⎢⎣ ⎝ r0 ⎠ ⎥⎦
Velocity profile
qs = P ⋅ L ⋅ h ⋅ ΔT
Heat transfer through pipe wall Heat flux constant:
Wall temperature constant: ΔTout = ΔTin ⋅ e −γ ⋅L with γ =
qs = P ⋅ L ⋅ qʹ′ʹ′ qʹ′ʹ′ ΔT = h h=
48 k 11 D
→
ΔT =
Nu D = 4.36
Prof. Nico Hotz
h ⋅P m ⋅ c p
ΔTout − ΔTin ΔT ⎞ ln⎛⎜ out ΔTin ⎟⎠ ⎝
NuD = 3.66 25
ME 150 – Heat and Mass Transfer
Prof. Nico Hotz
26