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Temperature Distribution Through a Plane Wall

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Details

For one-dimensional conduction in a plane wall, temperature is a function of the x coordinate only and heat is transferred exlucively in this direction. Shown in the following figure, a plane wall separates two fluids of different temperatures. Heat transfer occurs by convection from the hot fluid at T_{∞, 1} to one surface of the wall at T_{s, 1}, by conduction through the wall, and by convection from the other surface of the wall at T_{s, 2} to the cold fluid at T_{∞, 2}.

First consider the conditions within the wall. The temperature distribution is determined, from which the conduction heat transfer rate can be obtained.

The temperature distribution in the wall can be determined by solving the heat equation with the proper boundary conditions. For steady-state conditions with no distributed source or sink of energy within the wall, the appropriate form of the heat equation is Eq12 from the Heat Diffusion Equation lesson:

(Eq1)

d dx ( k dT dx )   =  0

Hence, from Eq2 of the Fourier's Law, The Conduction Rate Equation lesson, it follows that, for one-dimensional, steady-state conduction in a plane wall with no heat generation, the heat flux is a constant, independent of x. If the thermal conductivity of the wall material is assumed to be constant, the equation may be integrated twice to obtain the general solution:

(Eq2)

T(x) = C1x + C2

To obtain the constants of integration, C₁ and C₂, boundary conditions must be introduced. Boundary conditions are chosen at x = 0 and x = L, in which case:

T(0) = T_s,1

and

T(L) = T_s,2

Applying the condition at x = 0 to the general solution, it follows that:

T_s,1 = C₂

Similarly, at x = L:

T_s,2 = C₁L + C₂ = C₁L + T_s,1

in which case:

Ts,2 − Ts,1 L

=  C1

Substituting into the general solution, the temperature distribution is then:

(Eq3)

T(x) = (Ts,2 − Ts,1) x L   + Ts,1

From this result it is evident that, for one-dimensional, steady-state conduction in a plane wall with no heat generation and constant thermal conductivity, the temperature varies linearly with x.

Now that the temperature distribution is obtained, Fourier's law, Eq1 of the Fourier's Law, The Conduction Rate Equation lesson, may be used to determine the conduction heat transfer rate. That is:

(Eq4)

qx  =  −kA dT dx   =   kA L (Ts,1 − Ts,2)

Note that A is the area of the wall normal to the direction of heat transfer and, for the plane wall, it is a constant independent of x. The heat flux is then:

(Eq5)

qx"  =   qx A   =   k L (Ts,1 − Ts,2)

Eq4 and Eq5 indicate that both the heat rate q_x and heat flux q"_x are constants, independent of x.

In the foregoing paragraphs the standard approach has been used for solving conduction problems. That is, the general solution for the temperature distribution is first obtained by solving the appropriate form of the heat equation. The boundary conditions are then applied to obtain the particular solution, which is used with Fourier's law to determine the heat transfer rate. Note that the surface temperature at x = 0 and x = L were denoted as boundary conditions, even though it is the fluid temperature, and not the surface temperatures, that are typically known. However, since adjoining fluid and surface temperatures are easily related through a surface energy balance (see The Surface Energy Balance lesson), it is a simple matter to express Eq3, Eq4, and Eq5 in terms of fluid temperatures rather than surface temperatures. Alternatively, equivalent results could be obtained by using the surface energy balances as boundary conditions of the third kind in evaluating the constants of Eq2.