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I am trying to complete the following question. But feel it is not correct, so need some help.

Given that heat transfer via convection off a planar wall is given by:

enter image description here

Write out the combined thermal resistance for the following system (only thermal conduction and convection present)

enter image description here

My attempt:

Combined Thermal Resistance:

$$R1 = \frac{L1}{K1A1}$$ $$R2 = \frac{L2}{K2A2}$$

$$RTotal = R1 + R2$$ $$RTotal = \frac{L1}{K1A1} + \frac{L2}{K2A2}$$

$$RTotal = \frac{L1}{K1A} + \frac{L2}{K2A}$$

$$Q= \frac{\delta T}{ER}$$

Is what i have done so far correct?

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Conductive

The Conductive heat transfer is given by:

$$\dot{Q} = \frac{k}{L} A \Delta T$$

where:

  • $k$ = is the heat conductivity of the material in this case aluminimum ($\frac{kCal}{m°C}$)
  • $L$ is the thickness of the wall
  • A is the total exchange surface
  • $\Delta T$ the temperature difference

Convective heat transfer

Convective heat transfer is when a solid surface and a fluid (liquid or gas) exchange heat. The total rate of exchanged heat is:

$$\dot{Q} = h_c A \Delta T$$

where:

  • $h_c$ = heat transfer coefficient ($\frac{kCal}{m^2h°C}$)
  • A is the total exchange surface
  • $\Delta T$ the temperature difference

Total thermal resistance:

The total thermal resistance in your example will be given by

$$R = \frac{1}{h_{1}A} + \frac{L_1}{k_1 A}+\frac{L_2}{k_2 A}+ \frac{1}{h_{2}A}$$

where:

  • R is the thermal resistance
  • $h_{1}$ is the convective coefficient on the left side of the problem
  • $h_{2}$ is the convective coefficient on the right side of the problem
  • $k_i$ is the heat conductivity coefficient of the material
  • $L_i$ is the thickness for each wall
  • $A$ is the area of the surface.

When you calculate R, then you can use it the following way:

$$ \dot{Q} = \frac{T_{\infty 1}-T_{\infty 2}}{R}$$

where:

  • $T_{\infty 1}$ is the operating temperature inside of the box
  • $T_{\infty 2}$ is the operating temperature outside of the box
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