Answer:
a. -k {(dT(L)/dx} = h{T(L) - T(s)}
b. T(x) = 90 - 131.09(x)
c. 9045W
Step-by-step explanation:
Given detail
Thickness of wall L = 0.4 m
Left side temperature of the wall T1 = 90°C,
Thermal conductivity k = 2.3 W/m.k
Right side temperature of the wall T(infinity) = T(s) = 25°C
Heat transfer coefficient h = 24W/m2.K
(a) Differential equation and the boundary conditions for steady one-dimensional heat conduction through the wall,
Qwall = -kA {(dT(0)/dx}
{d²T /dx²} = 0
T(0) = 90°C
-k {(dT(L)/dx} = h{T(L) - T(s)}
(b) relation for the variation of temperature in the wall by solving the differential equation.
T(x) = xC1 + C2
T(0) = 0*C1 + C2 = 90°C
T(0) = C2 = 90°C
-kC1 = hLC1 + hC2 -T(s)h
-C1 (k+hL) = h (C2 - T(s))
- C1 = {h(C2 - T(s)) / (k + hL)}
C1 = - { 24(90 - 25) / (2.3 + 24*0.4) }
C1 = - 131.09°C
T(x) = x*(-131.09) + 90
T(x) = 90 - 131.09(x)
(c) rate of heat transfer through the wall
Qwall = - 2.3 * 30 * (-131.09)
Qwall = 9045W ans