10—Partial Differential Equations 251position.
x= 0 x=a
y=b
y=b−a T(x,b−a)≈
4
π
T 0
∑∞
`=01
2 `+ 1
e−(2`+1)πsin
(2`+ 1)πx
a
For= 0, the exponential factor ise−π = 0. 043 , and for = 1this factor ise−^3 π = 0. 00008.
This means that measured from theT 0 end, within the very short distance equal to the width, the
temperature has dropped 95% of the way down to its limiting value of zero. The temperature in the
rod is quite uniform until you are very close to the heated end.
The Heat Flow into the Box
All the preceding analysis and discussion was intended to make this problem and its solution sound
oh-so-plausible. There’s more, and it isn’t pretty.
The temperature on one of the four sides was given as different from the temperatures on the
other three sides. What will the heat flow into the region be? That is, what power must you supply to
maintain the temperatureT 0 on the single wall?
At the beginning of this chapter, Eq. (10.1), you have the equation for the power through anareaA, but that equation assumed that the temperature gradient∂T/∂xis the same all over the area
A. If it isn’t, you simply turn it into a density.
∆P=−κ∆A
∂T
∂x
, and then
∆P
∆A
→
dP
dA
=−κ
∂T
∂x
(10.28)
Equivalently, just use the vector form from Eq. (10.6),H~ =−κ∇T. In Eq. (10.22) the temperature is
T 0 alongy=b, and the power density (energy/(time.area)) flowing in the+ydirection is−κ∂T/∂y,
so the power density flowingintothis area has the reversed sign,
+κ∂T/∂y (10.29)
The total power flow is the integral of this over the area of the top face.
LetLbe the length of this long rectangular rod, its extent in thez-direction. The element of
area along the surface aty=bis thendA=Ldx, and the power flow into this face is
∫a0Ldxκ
∂T
∂y
∣
∣∣
∣
y=bThe temperature function is the solution Eq. (10.26), so differentiate that equation with respect toy.
∫a0Ldxκ
4
π
T 0
∑∞
`=0[(2`+ 1)π/a]
2 `+ 1
cosh(
(2`+ 1)πy/a
)
sinh(
(2`+ 1)πb/a
)sin(2`+ 1)πx
a
at y=b
=
4 LκT 0
a
∫a0dx
∑∞
`=0sin