# One-Dimensional Steady-State Heat Conduction Infinite Plate

A simple case of heat conduction is a plate of finite thickness but infinite in other directions. If the temperature is constant around the plate, the mate­rial is assumed to have a constant thermal conductivity. In this case the linear temperature distribution and the heat flow through the plate is easy to deter­mine from Fourier’s law (Eq. (4.154)).

In a case similar to Fig. 4.23 the heat conduction equation (Eq. (4.180)) becomes

Ј1=0. (4.182)

Dx2

In steady-state conditions the right side of Eq. (4.180) is zero, and no heat genera­tion takes place; the thermal conductivity in the one-dimensional case is constant. The solution of Eq. (4.182) is

T=ClX + C2, (4.183)

With boundary conditions

T = Tj when x = Xj T = T-, when x = x2

This gives the linear temperature distribution

T=T1 + (T2-Tl)^ (4.184)

Substituting the above equation into Eq. (4.154) gives the heat flow through the plate:

Q = ~AS = = aIjaT2- (4-185)

OX Xj —Xy ISX

Axial-Symmetric Case

For the axial-symmetric case the equation is

3 HEAT AND MASS TRANSFER

 113 In the one-dimensional case

Јl+^ = 0. (4,1871

Dr2 rdr

The solution of Eq. (4.187) is

T= Q lnr + C2, (4.188)

With boundary conditions

T = Tj when r = rx

T = T2 when r = r2 The logarithmic temperature distribution is

In—

T= Tj + tT^-T,)— . (4.189)

TOC o "1-5" h z “ In—

R-1

Thus the heat flow per a unit of length is

= —A 27rr~j^~ = 2irT-lr-2 . (4.190)

Dr. r-,

In —