Answer:
a. [tex]\frac{-kdT(0)}{dx} =q_{0}[/tex]=5000W/m^2
b.833.3(0.006-x)+112
c. 117 deg C
Explanation:
Consider the base plate of an 800-W household iron with a thickness of L 5 0.6 cm, base area of A 5 160 cm2, and thermal conductivity of k 5 60 W/m·K. The inner surface of the base plate is subjected to uniform heat flux generated by the resistance heaters inside. When steady operating conditions are reached, the outer surface temperature of the plate is measured to be 112°C. Disregarding any heat loss through the upper part of the iron,
Assumption
Heat conduction is steady state and unidimensional 2. thermal conductivity is constant. Heat supplied is not in the plate
4. we disregard heat loss
Heat flux=heat/area
[tex]\alpha[/tex]/A=800W/160*10^-4
with direction to the surface been in the x direction,
the mathematical expression will be
[tex]\frac{d^2T}{dx^2}[/tex]=0..............1
and [tex]\frac{-kdT(0)}{dx} =q_{0}[/tex]=5000W/m^2
from fourier law, for conductivity
T(L)=T2=112C
b. integrating equation 1 twice we have\dT/dx=c1
T(x)=C1x+C2
C1 and C2 are arbitrary constant
at x=0 the boundary conditions become
-kC1=qo
C1=-(qo/k)
at x=L
=T(L)=C1L+C2=T2
C2=T2-cL1
C2=T2+qoL/k
Juxtaposing C1 and C2 into the general equation , we have
T(x)=-qo/k+T2+qoL/k=qo(L-k)/k+T2
50000*(0.006-x)/60+112
833.3(0.006-x)+112
c. inner surface plate temperature is
T(0)=833.33(0.006-0)+112 ( using the derivation in answer b)
117 deg C
