Consider a 20 cm thick granite wall with a thermal conductivity of 2.79 W/m·K. The temperature of the left surface is held constant at 50°C, whereas the right face is exposed to a flow of 22°C air with a convection heat transfer coefficient of 15 W/m2·K. Neglecting heat transfer by radiation, find the right wall surface temperature and the heat flux through the wall.

Respuesta :

Answer:

The right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

Explanation:

Thickness of the wall is  L=  20cm = 0.2m

Thermal conductivity of the wall is  K = 2.79 W/m·K

Temperature at the left side surface is T₁ =  50°C

Temperature of the air is T = 22°C

Convection heat transfer coefficient is  h = 15 W/m2·K

Heat conduction process through wall is equal to the heat convection process so

[tex]Q_{conduction} = Q_{convection}[/tex]

Expression for the heat conduction process is

[tex]Q_{conduction} = \frac{K(T_1 - T)}{L}[/tex]

Expression for the heat convection process is

[tex]Q_{convection} = h(T_2 - T)[/tex]

Substitute the expressions of conduction and convection in equation above

[tex]Q_{conduction} = Q_{convection}[/tex]

[tex]\frac{K(T_1 - T_2)}{L} = h(T_2 - T)[/tex]

Substitute the values in above equation

[tex]\frac{2.79(50- T_2)}{0.2} = 15(T_2 - 22)\\\\T_2 = 35.5^\circC[/tex]

Now heat flux through the wall can be calculated as

[tex]q_{flux} = Q_{conduction} \\\\q_{flux} = \frac{K(T_1 - T_2)}{L}\\\\q_{flux} = \frac{2.79(50 - 35.5)}{0.2}\\\\q_{flux} = 202.3W/m^2[/tex]

Thus, the right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²