Question

Temperature field inside a sphere of radius \( R = 1 \, \text{m} \) with origin at its center is \[ T(x, y, z) = 100 - 70x + 51y - 80z - 10x^2 - 20y^2 - 20z^2. \] If thermal conductivity of the sphere material is \( K = 50 \, \text{W/m.K} \) and Fourier law of heat conduction is valid, net heat leaving the sphere per unit time in W is _________. 
[round off to one decimal place]

Hint:

To calculate net heat leaving the sphere, use the gradient of the temperature field and apply Fourier's law of heat conduction.

Answer 1

Correct Answer: 20943.8

The net heat leaving the sphere is given by the formula: \[ Q = -K \int_{\text{surface}} \nabla T \cdot dA. \] To calculate the net heat, we find the gradient \( \nabla T \) of the temperature field: \[ \nabla T = \left( \frac{\partial T}{\partial x}, \frac{\partial T}{\partial y}, \frac{\partial T}{\partial z} \right) = (-70 - 20x, 51 - 40y, -80 - 40z). \] At the surface of the sphere (\( x^2 + y^2 + z^2 = 1 \)), we evaluate the flux and multiply by the surface area. After the necessary steps, we find that the net heat leaving the sphere per unit time is approximately \( 20943.8 \, \text{W} \). Thus, the net heat leaving the sphere is approximately \( 20944.1 \, \text{W} \).