Answer :
Answer:
q = -
T
overall
Rth
Explanation:
First consider the plane wall where a direct application of Fourier’s law [Equation (1-1)]
may be made. Integration yields
q = − kA
-
x (T2 − T1) [2-1]
when the thermal conductivity is considered constant. The wall thickness is -
x, and T1
and T2 are the wall-face temperatures. If the thermal conductivity varies with temperature
according to some linear relation k = k0(1 + βT ), the resultant equation for the heat flow
is
q = −k0A
-
x -
(T2 − T1) + β
2 (T 2
2 − T 2
1 )
[2-2]
If more than one material is present, as in the multilayer wall shown in Figure 2-1, the
analysis would proceed as follows: The temperature gradients in the three materials are
shown, and the heat flow may be written
q = −kAA
T2 − T1
-
xA
= −kBA
T3 − T2
-
xB
= −kCA
T4 − T3
-
xC
Note that the heat flow must be the same through all sections.
Solving these three equations simultaneously, the heat flow is written
q = T1 − T4
-
xA/kAA + -
xB/kBA + -
xC/kCA [2-3]
At this point we retrace our development slightly to introduce a different conceptual view-
point for Fourier’s law. The heat-transfer rate may be considered as a flow, and the combina-
tion of thermal conductivity, thickness of material, and area as a resistance to this flow. The
temperature is the potential, or driving, function for the heat flow, and the Fourier equation
may be written
Heat flow = thermal potential difference
thermal resistance [2-4]
a relation quite like Ohm’s law in electric-circuit theory. In Equation (2-1) the thermal
resistance is -
x/kA, and in Equation (2-3) it is the sum of the three terms in the denominator.
We should expect this situation in Equation (2-3) because the three walls side by side act as
three thermal resistances in series. The equivalent electric circuit is shown in Figure 2-1b.
The electrical analogy may be used to solve more complex problems involving both
series and parallel thermal resistances. A typical problem and its analogous electric circuit
are shown in Figure 2-2. The one-dimensional heat-flow equation for this type of problem
may be written
q = -
T
overall
Rth
[2-5]
where the Rth are the thermal resistances of the various materials. The units for the thermal
resistance are ◦C/W or ◦F · h/Btu.
It is well to mention that in some systems, like that in Figure 2-2, two-dimensional
heat flow may result if the thermal conductivities of materials B, C, and D differ by an
appreciable amount. In these cases other techniques must be employed to effect a solution.
