A 1000 - W iron is left on the iron board with its base exposed to ambient air a

Question

A 1000 - W iron is left on the iron board with its base exposed to ambient air at 20 degree C. The base plate of the iron has a thickness of L = 0.5 cm, base area of A = 150 cm2, and thermal conductivity of k = 18 W/m degree C. The inner surface of the base plate is subjected to uniform heat flux generated by the resistance heaters inside. The outer surface of the base plate whose emissivity is epsilon = 0.7, loses heat by convection to ambient air at T infinity = 22 degree C with an average heat transfer coefficient of h = 30 W/m2 degree C as well as by radiation to the surrounding surfaces at an average temperature of Tsurr = 290 K. Disregarding any heat loss through the upper part of the iron. (a) express the differential equation and the boundary conditions for steady one - dimensional heat conduction through the plate, (b) obtain a relation for the temperature of the outer surface of the plate by solving the differential equation, and (c) evaluate the outer surface temperature.

Explanation / Answer

(a) Energy balance on a thin slice of length dx gives,

-kA (dT/dx)x - (-kA (dT/dx)x+dx) = 0

This gives d2T/dx2 = 0.

Boundary conditions will be,

(i) At x=0, qwA = -kA(dT/dx)

(ii) At x = L, -kA (dT/dx) = hA(T-T) + A(T4 - Tsurr4)

(b) Solving the differential equation yields, T = C1x + C2.

Using first boundary condition, we get, C1 = -qw/k

Using second boundary condition, we get, qwA = hA(TL - T) + A(TL4 - Tsurr4)

This is the expression for outer wall temperature TL.

(c) We have qwA = 1000 Watts.

Putting values, 1000 = 30*(150*10-4)(TL - (22+273)) + 0.7*(5.67*10-8)*(150*10-4)*(TL4 - 2904)

Solving this gives, TL = 465 K = 192 deg C.

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