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A wall has inner and outer surface temperatures of 16C and 6C respectively. The interior and exterior air temperatures are 20 and 5C respectively. The inner and outer convection heat transfer coefficients are 5 and 20 W/m^2-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 :

cjmejiab

To develop this problem we will apply the concept related to heat transfer defined as the product between the transfer coefficient and the temperature difference between two spaces, that is,

[tex]q=h(T_s-T_0)[/tex]

Here,

h = Heat transfer coefficient

[tex]T_{s,0}[/tex]= Temperature at each point

  • Interior air and inner wall:

[tex]q = 5(20-16)[/tex]

[tex]q = 20 W/m^2[/tex]

  • Outer wall to exterior air:

[tex]q = 20(6-5)[/tex]

[tex]q = 20 W/m^2[/tex]

  • Inner wall to interior air:

[tex]q = 5(16-20)[/tex]

[tex]q = -20 W/m^2[/tex]

We can see that the magnitude of the heat fluxes in the three states are the same (The negative sign only indicates the change of direction) so the wall is in steady state conditions

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