Question about Convective Heat Transfer Coefficient?

[LaReina]

Homework Statement

The cross section of a wall is shown in the figure below. To reduce the heat flow, an air gap (still air) of 6cm is created between the brick wall layers of 14cm thick, each. The thermal conductivities of the brick wall and the still air are: K_air = 0.032 W/m.K and K_brick = 0.9 W/m.K, respectively.

Calculate thermal resistance.

This is only part of the question as I have no problem with the rest. The convective heat transfer coefficient has not been given thus making it very difficult for me to even begin.

Homework Equations

R=1/hA
R=L/kA
Q=T₁-T₂/R_total

The Attempt at a Solution

As no convective heat transfer coefficient has been given, I can't even begin to calculate the thermal resistance. It's probably something very simple but I just can't get my head around it.

 

[Chestermiller]

Hi LaRiena. Welcome to Physics Forums.

They are only asking for the thermal resistance of the wall, so the convective heat transfer resistance on either side of the wall is not relevant to the question. Since the air within the wall is still, convective heat transfer within the air compartment is negligible. This is a straightforward conduction problem.

Chet

 

[LaReina]

Hi LaRiena. Welcome to Physics Forums.

They are only asking for the thermal resistance of the wall, so the convective heat transfer resistance on either side of the wall is not relevant to the question. Since the air within the wall is still, convective heat transfer within the air compartment is negligible. This is a straightforward conduction problem.

Chet

Hi Chet
Thank you for your response.

Correct me if I'm wrong but since the convective heat transfer coefficient is negligent would I use R=L/k instead to find the thermal resistance?

So R_total=((0.14/0.9) x 2) + 0.06/0.032
R_total=2.1861 °C/W

 

[Chestermiller]

Yes. This looks correct to me.

When you are talking about the possibility of convective heat transfer within the air pocket, natural convection there would increase the rate of heat transfer and reduce the heat transfer resistance of the wall.

Chet

 

[LaReina]

Thank you so much! :)