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Consider a 20 cm thick granite wall with a thermalconductivity of 2.79 W/m·K. The temperature of the leftsurface is held constant at 50°C, whereas the right face isexposed to a flow of 22°C air with a convection heat transfercoefficient of 15 W/m2·K. Neglecting heat transfer by radiation, find the right wall surface temperature and the heat fluxthrough the wall.

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Answer:

Q= 231.88 W/m²

Step-by-step explanation:

Given that

L= 20 cm

K= 2.79 W/m·K

T₁=50°C

T₂= 22°C

h=15 W/m₂·K

The total thermal resistance


R=(L)/(KA)+(1)/(hA)

Now by putting the values


R=(L)/(KA)+(1)/(hA)


R=(0.2)/(2.79A)+(1)/(15A)

R= 0.138 K/W ( for heat flux A= 1 m²)

Heat flux Q= ΔT/R

Q= (50-22)/0.138 = 32/0.138 W

Q= 231.88 W/m²

User Gerrit Brink
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