Question

Consider a large plane wall of thickness L = 0.4 m, thermal conductivity k = 2.3 W/m·K, and surface area A = 30 m2. The left side of the wall is maintained at a constant temperature of T1 = 90°C, while the right side loses heat by convection to the surrounding air at T[infinity] = 25°C with a heat transfer coefficient of h = 24 W/m2·K. Assuming constant thermal conductivity and no heat generation in the wall, (a) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the wall, (b) obtain a relation for the variation of temperature in the wall by solving the differential equation, and (c) evaluate the rate of heat transfer through the wall. Answer: (c) 9045 W

Answer 1

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