1. Fourier's Law: Conduction
1.1. Definitions
We have seen that this difference ΔT_{12} betwenn two points respectively a temperature T_{2}
and T_{2} 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 (T_{1}  T_{2})A/d; where K is the thermal conductivity of the cube. Let's write:
ΔJ = K (T_{1}  T_{2})A/d = Q = κ A T_{12}; where: κ = K/d.
 K is the thermal conductivity,
 κ is the conductance,
 Q the heat flux, and
 R = d/KA is the thermal resistance .
Thus:
Q = (K A /d)ΔT_{12} = κ A ΔT_{12} = (1/R) ΔT_{12}
Q = ΔT_{12}/R
where R = d/KA
Q in Watt (W), ΔT_{12} in ^{o}C or ^{o}K, R in ^{o}C/W, and
K in W/^{o}C . 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 ΔT_{12}/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 Σr_{i}.

1.2.2. Two or more walls
If another wall, with K_{2} as the thermal conductivity coefficient, is juxtaposed with
the first, we write:
R_{total} = R_{1st} + R_{2nd} = (1/A) [L_{1}/K_{1} + L_{2}/K_{2}]
and :
Q = ΔT_{13}/R_{total}
Because T_{13} = T_{12} + T_{23} = (T_{1}  T_{2} )+ (T_{2}  T_{3})
For more than two walls, we have:
R_{total} = Σ R_{i}
= (1/A) Σ[L_{i}/K_{i}]
and
Q = ΔT_{initialfinal}/R_{total}


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)[r_{1} → r_{2}]
R = [Ln (r_{2}/r_{1})]/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 R_{total} = Σ R_{conduction} + R_{convection}
In the case of walls:
R_{total} = (1/A) Σ(L_{i} /K_{i})
In the case of tubes:
R_{total} = Σ [Ln(r_{i+1}/r_{i})/2πK_{i}L_{i}]
In the general case, we have:
R_{total} = (1/A) Σ(L_{i} /K_{i}) + (1/A)Σ(1/h_{i})


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 ΔT_{m}
where K/L is replaced by U.
U is called the global heat exchange coefficient and ΔT_{m} 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/(ΔT_{1}  ΔT_{2})] ∫(dΔT/ΔT)= (1/A) ∫ A dA
[ΔT: ΔT_{1} → ΔT_{2}]
(1/A) ∫ A dA = A and
<(1/ΔT)>= [1/(ΔT_{1}  ΔT_{2})] ∫(dΔT/ΔT)
= [1/(ΔT_{1}  ΔT_{2})] Ln (ΔT)=
[(Ln (ΔT_{1}/ΔT_{2}))/(ΔT_{1}  Δ_{2})]
We find:
ΔT_{m} = [ΔT_{1}  ΔT_{2}]/[Ln (ΔT_{1}/ΔT_{2}]
4.2. CoCurrent: 1 shell pass & 1 tube pass

The system is countercurrent; that is the directions of the two currents are the same.
ΔT_{1} = T_{hi}  T_{ci}
ΔT_{2} = T_{ho}  T_{co}
ΔT_{m} = [ΔT_{1}  ΔT_{2}]/[Ln(ΔT_{1}/ΔT_{2})]

4.3. CounterCurrent: 1 shell pass & 1 tube pass
The system is countercurrent; that is the directions of the two currents are opposite.
ΔT_{1} = T_{ho}  T_{ci}
ΔT_{2} = T_{hi}  T_{co}
ΔT_{m}_{cc} = [ΔT_{1}  ΔT_{2}]/[Ln(ΔT_{1}/ΔT_{2})]


4.4. Multipasses shell and tube
In this case, We define the DTLM as follows:
DTLM = F (ΔT_{m})_{cc}.
F is a correction factor defined as the value of
a Rfunction of P, that is F_{P}(P); geven by charts.
Here are the four steps:
1. calculate:ΔT_{m}_{cc}
(counter current 1 pass shell and 1 pass tube)
2.
Calculate the two defined ratios P and R as follows:
P = ΔT(tube)/ΔT(inin)
and
R = ΔT(shell)/ΔT(tube)
Where:
ΔT(tube) = T(tubein)  T(tubeout)
ΔT(shell) = T(shellin)  T(shellout)
ΔT(inin) = T(tubein)  T(shellin) (or the maximum temperature difference)
Expressed in absolute values.
3. find the factor F in the charts.
4.Calculate the related DTLM
DTLM = F (ΔT_{m})_{cc}.
F is a correction factor defined as the value of
a Rfunction of P, that is F_{P}(P); geven by charts.

Example :

12 Heat Exchager ( 1 pass shell and 2 passes tube)that cools water going through tubes
ΔT(tube) = T(tubein)  T(tubeout) = 120  60 = 60 ^{o}C
ΔT(shell) = T(shellout)  T(shellin) = 60  10 = 50 ^{o}C
ΔT(inin) = T(tubein)  T(shellin) = 120  10 = 110^{o}C
Then:
ΔT_{1} = 120  60 = 60
ΔT_{2} = 60  10 = 50
ΔT_{m}_{cc} = (60  50)/Ln(60/50) = 55.0
P = ΔT(tube)/ΔT(inin) = 60/110 = 0.55
R = ΔT(shell)/ΔT(tube) = 50/60 = 0.83
The chart gives: F = 0.95
ΔT_{m} = F. ΔT_{m}_{cc}
= 0.95 x 55.0 = 52.25 ^{o}C.

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 S_{1} = π r_{1}^{2}.
At the entrance, the fluid has a temperature T_{hi} and at the exit, it becomes less hot and has a
temperature T_{co} (T_{hi} > T_{co}). If the specific heat at constant volume
of the fluid is c_{v}, and the its mass is m contained in the part volume V = S_{1} L; then
the energy loss of this fluid is: m c_{v} ΔT_{fluidshell} ; where ΔT_{hc} = T_{hi}  T_{co}
is the temperature difference.
This difference in temperature occurs when the fluid touches the wall of the tube of thermal
conductivity k_{fluidtube} and heated it.
The tube:
If the temperature of the fluid passes from T_{ci} inside the
tube, that is before crossing the tube and T_{ho} the temperature outside the tube, the
difference ΔT_{fluidtube} = T_{ci}  T_{ho}.
Energy conservation
The loss of energy for the fluid in the shell is retreived by the fluid in the shell; that is
Q_{fluidtube} = Q_{fluidshell} or
m_{fluidtube} c_{fluidtube} ΔT_{fluidtube} = m_{fluidshell} c_{fluidshell} ΔT_{fluidshell}
As the rate D = dm/dt; we have:
[D c ΔT]_{fluidtube} = [D c ΔT]_{fluidshell} = U A ΔT_{m}

