Question

The heat flux that is applied to the left face of a plane wall is q" = 20 w/m2. the wall is of thickness l = 10 mm and of thermal conductivity k = 12 w/m · k. if the surface temperatures of the wall are measured to be 50°c on the left side and 30°c on the right side, do steady-state conditions exist?

Answer 1

Using Fourier's law of heat conduction and the given data, the calculated rate of heat transfer exceeds the applied heat flux, which suggests that steady-state conditions do not exist for the wall in question.

Step-by-step explanation:

To determine if steady-state conditions exist for the wall with an applied heat flux to its left face (q" = 20 W/m2), a thickness of l = 10 mm, and thermal conductivity k = 12 W/m·K, we can use Fourier's law of heat conduction. The law states that the rate of heat transfer (Q/t) through a material is proportional to the thermal conductivity (k), the area (A) through which the heat is being transferred, and the temperature difference across the material (T2 – T1), and inversely proportional to the thickness (d) of the material.

Using the given information and assuming a steady-state operation, Q/t = kA(T2 – T1)/d. Given the temperature measurements are 50°C on the left side and 30°C on the right side, the temperature difference is 20°C. The rate of heat transfer can be calculated as follows: Q/t = (12 W/m·K) · (1 m2) · (20 K) / (0.01 m), which simplifies to Q/t = 24000 W/m2.

However, this calculated heat transfer rate is significantly greater than the applied heat flux (20 W/m2), indicating that steady-state conditions do not exist or there is a measurement or recording error.

Answer 2

given flux is very small then the above calculated value so it is not is steady state

Step-by-step explanation:

As we know that heat flux is given in steady state by the formula

`Q = (k(T_1 - T_2))/(L)`

here we know that

`T_1 = 50^o C`

`T_2 = 30 ^o C`

k = 12 W/m k

L = 10 mm

now we have

`Q = (12(50 - 30))/(0.01)`

`Q = 24000 w/m^2`

since given flux is very small then the above calculated value so it is not is steady state