Solve the problem with conditions if a wall has inner and outer surface temperatures of 16 and 6 C, respectively. The interior and exterior air temperatures are 20 and 5 C, respectively. The inner and outer convection heat transfer coefficients are 5 and 20 W/m2 K, respectively. Calculate the heat flux from the interior air to the wall, from the wall to the exterior air, and from the wall to the interior air. Is the wall under steady-state conditions?

Respuesta :

Answer:

Heat flux is 20 W/m^2

Explanation:

Heat flux (Q) is computed as

[tex] Q = h \, \Delta T [/tex]

where h is heat transfer coefficient and ΔT is the difference between body's temperature

From the interior air to the inner wall

[tex]Q = 5 \frac{W}{m^2 K} \, 4 K [/tex]

[tex]Q = 20 \frac{W}{m^2} [/tex]

From the the outer wall to the exterior air

[tex]Q = 20 \frac{W}{m^2 K} \, 1 K [/tex]

[tex]Q = 20 \frac{W}{m^2} [/tex]

The wall is under steady-state condition because heat flux is constant  

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