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
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In the one-dimensional case
Јl+^ = 0. (4,1871
Dr2 rdr
The solution of Eq. (4.187) is
With boundary conditions
T = Tj when r = rx
T = T2 when r = r2 The logarithmic temperature distribution is
In—
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 —
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