Question

Two thick walls made of the same material are separated by a vacuum gap of thickness L. A cylinder, made of the same material as the walls, and of diameter D runs between the walls. All surfaces are highly polished (their emissivity is small). The walls are at temperatures Tbi and Tbz at locations far from the cylinder. The thermal conductivity of the walls and the rod is k.
(a) Draw the thermal resistance network.

(b) Derive an expression for the shape factor, S, associated with conduction between Tb, and Tba Use: S = k(Tb,1-Tb,2)

(c) Determine the value of the shape factor for D = 0.01 m, L = 0.5 m, and k = 23 W/mK.

Answer 1

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.