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Classical Thermodynamics

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Heat transfer


Classical Thermodynamics
Heat transfer

1. Fourier's Law: Conduction

1.1. Definitions

We have seen that this difference ΔT12 betwenn two points respectively a temperature T2 and T2 can be expressed by the Fourier's law which states that the flow ?J(the thermal energy that passes per unit time) through the surface A of a cube across the length d is :br> ΔJ = K (T1 - T2)A/d; where K is the thermal conductivity of the cube. Let's write:
ΔJ = K (T1 - T2)A/d = Q = κ A T12; where: κ = K/d.

  1. K is the thermal conductivity,
  2. κ is the conductance,
  3. Q the heat flux, and
  4. R = d/KA is the thermal resistance .
Q = (K A /d)ΔT12 = κ A ΔT12 = (1/R) ΔT12
Q = ΔT12/R where R = d/KA
Q in Watt (W), ΔT12 in oC or oK, R in oC/W, and K in W/oC . m

1.2 Application 1: Fourier's Law for a wall

1.2.1. One wall

For this case here, the surface at left is A and d = L. Let's write:

R = L/KA and Q = K A ΔT12/L
We have an analogy with electrical circuis, where Q is the electrical current I, &Delata;T is analog to the potential electrical difference, and the electrical resistance is equivalent to the thermal resistance. As with electrical connections in series, we add the thermal resistances Σri.

1.2.2. Two or more walls

If another wall, with K2 as the thermal conductivity coefficient, is juxtaposed with the first, we write:
Rtotal = R1st + R2nd = (1/A) [L1/K1 + L2/K2]
and :
Q = ΔT13/Rtotal
Because T13 = T12 + T23 = (T1 - T2 )+ (T2 - T3)

For more than two walls, we have:
Rtotal = Σ Ri = (1/A) Σ[Li/Ki]
Q = ΔTinitial-final/Rtotal

1.3. Application 2: Fourier's Law for a tube

Let's consider a tube full of material of thermal conductivity K. In this case the heat flux is directed laterally across the surface S = 2πrL. The thermal resistance is equal to R. R varies. How to calculate R?

1.4. Thermal resistance formula

If dr is the differential resistance, we can write:
dR = dr/KA = dr/K 2πrL = (1/K 2πL)dr/r → R
= (1/K 2πL) ∫rd/r = (1/K 2πL) Ln (r)[r1 → r2]

R = [Ln (r2/r1)]/K 2πL

2. Fourier's Law: Convection

In the case of convection, we take the thermal conductivity for the conduction is replaced by the thermal convection h. The related thermal resistance is given by:
R = 1/hA
A is the surface where the convection occurs from.

3. Conduction & Convection

The rule is to add the thermal resistances to obtain the total thermal resistance Rtotal = Σ Rconduction + Rconvection
In the case of walls:
Rtotal = (1/A) Σ(Li /Ki)
In the case of tubes:
Rtotal = Σ [Ln(ri+1/ri)/2πKiLi]
In the general case, we have:
Rtotal = (1/A) Σ(Li /Ki)
+ (1/A)Σ(1/hi)

4. Heat exchanger:

4. 1. Differential Temperature Logarithm Mean: DTLM

We have Q = ΔT/R where the thermal resistance is equal to R for a static system; where there is no current of fluid. In the case in which we have a heat exchanger of two currents: 1 cold current and 1 hot current that will heat the colder; we express the heat flux Q as:
Q = U A ΔTm
where K/L is replaced by U.
U is called the global heat exchange coefficient and ΔTm the differential temperature logarithm Mean (DTLM).

What's the expression of the DTLM?
In the expression Q = U A ΔT, Q and U are constant; but A and Δ vary. Let's write the expression as follows:
(Q /U ) [1/ΔT]= A and calculate the mean of the two side of this expression, we have:
(Q /U ) <(1/ΔT)>= <A>
(Q /U ) [1/(ΔT1 - ΔT2)] ∫(dΔT/ΔT)= (1/A) ∫ A dA
[ΔT: ΔT1 → ΔT2]
(1/A) ∫ A dA = A and
<(1/ΔT)>= [1/(ΔT1 - ΔT2)] ∫(dΔT/ΔT) = [1/(ΔT1 - ΔT2)] Ln (ΔT)= [(Ln (ΔT1/ΔT2))/(ΔT1 - Δ2)]
We find:
ΔTm = [ΔT1 - ΔT2]/[Ln (ΔT1/ΔT2]

4.2. Co-Current: 1 shell pass & 1 tube pass

The system is counter-current; that is the directions of the two currents are the same.
ΔT1 = Thi - Tci
ΔT2 = Tho - Tco
ΔTm = [ΔT1 - ΔT2]/[Ln(ΔT1/ΔT2)]

4.3. Counter-Current: 1 shell pass & 1 tube pass

The system is counter-current; that is the directions of the two currents are opposite.
ΔT1 = Tho - Tci
ΔT2 = Thi - Tco
ΔTmcc = [ΔT1 - ΔT2]/[Ln(ΔT1/ΔT2)]

4.4. Multipasses shell and tube

In this case, We define the DTLM as follows:
DTLM = F (ΔTm)cc.

F is a correction factor defined as the value of a R-function of P, that is FP(P); geven by charts.
Here are the four steps:
1. calculate:ΔTmcc (counter current 1 pass shell and 1 pass tube)
2. Calculate the two defined ratios P and R as follows:
P = ΔT(tube)/ΔT(in-in)
R = ΔT(shell)/ΔT(tube)

ΔT(tube) = |T(tube-in) - T(tube-out)|
ΔT(shell) = |T(shell-in) - T(shell-out)|
ΔT(in-in) = |T(tube-in) - T(shell-in)| (or the maximum temperature difference)
Expressed in absolute values.

3. find the factor F in the charts.
4.Calculate the related DTLM

DTLM = F (ΔTm)cc.

F is a correction factor defined as the value of a R-function of P, that is FP(P); geven by charts.

Example :

1-2 Heat Exchager ( 1 pass shell and 2 passes tube)that cools water going through tubes
ΔT(tube) = T(tube-in) - T(tube-out) = 120 - 60 = 60 oC
ΔT(shell) = T(shell-out) - T(shell-in) = 60 - 10 = 50 oC
ΔT(in-in) = T(tube-in) - T(shell-in)
= 120 - 10 = 110oC
ΔT1 = 120 - 60 = 60
ΔT2 = 60 - 10 = 50
ΔTmcc = (60 - 50)/Ln(60/50) = 55.0
P = ΔT(tube)/ΔT(in-in) = 60/110 = 0.55
R = ΔT(shell)/ΔT(tube) = 50/60 = 0.83
The chart gives: F = 0.95
ΔTm = F. ΔTmcc = 0.95 x 55.0 = 52.25 oC.

4.5. Energy conservation

The shell:
Let's assume a fluid of thermal convection h circulate from the left to the right across a shell. Its rate D is the quantity of fluid, per unit time, crossing the surface S1 = π r12. At the entrance, the fluid has a temperature Thi and at the exit, it becomes less hot and has a temperature Tco (Thi > Tco). If the specific heat at constant volume of the fluid is cv, and the its mass is m contained in the part volume V = S1 L; then the energy loss of this fluid is: m cv ΔTfluid-shell ; where ΔThc = Thi - Tco is the temperature difference.
This difference in temperature occurs when the fluid touches the wall of the tube of thermal conductivity kfluid-tube and heated it.
The tube:
If the temperature of the fluid passes from Tci inside the tube, that is before crossing the tube and Tho the temperature outside the tube, the difference ΔTfluid-tube = Tci - Tho.
Energy conservation
The loss of energy for the fluid in the shell is retreived by the fluid in the shell; that is
Qfluid-tube = Qfluid-shell or
mfluid-tube cfluid-tube ΔTfluid-tube = mfluid-shell cfluid-shell ΔTfluid-shell
As the rate D = dm/dt; we have:

[D c ΔT]fluid-tube = [D c ΔT]fluids-hell = U A ΔTm





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