Question

The temperature distribution across a wall 0.3 m thick at a certain instant of time is T(x) = a+bx+cx², where T is in degrees Celsius and x is in meters, a=200⁰C, b=-200⁰C/m, and c=30⁰C/m². The wall has a thermal conductivity of 1 W/m⁰C.

a) On a unit surface area basis, determine the rate of heat transfer in the wall on the left face.
b) On a unit surface area basis, determine the rate of heat transfer of the wall on the right face.

Answer 1

The rate of heat transfer at the left face of the wall is 200 W/m², while at the right face it is 182 W/m², computed using Fourier's law of heat conduction and the given temperature distribution function.

Step-by-step explanation:

To determine the rate of heat transfer through the wall at the left face (x = 0) and the right face (x = 0.3 m), we can use Fourier's law of heat conduction. This law states:

Q/t = -k * A *(dT/dx)

where Q/t is the rate of heat transfer in watts (W), k is the thermal conductivity of the material, A is the cross-sectional area through which heat is being transferred (which is unit area in this case), and dT/dx is the temperature gradient at the position x.

Given that T(x) = 200 - 200x + 30x², the derivative of T with respect to x, dT/dx, is:

dT/dx = -200 + 60x

(a) The rate of heat transfer at the left face (x = 0) is:

dT/dx (left face) = -200 + 60(0) = -200 °C/m

Q/t (left face) = -1 W/m°C * (-200 °C/m)

Q/t (left face) = 200 W/m²

(b) The rate of heat transfer at the right face (x = 0.3 m) is:

dT/dx (right face) = -200 + 60(0.3) = -182 °C/m

Q/t (right face) = -1 W/m°C * (-182 °C/m)

Q/t (right face) = 182 W/m²