# Thermal Resistance

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:

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

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?

# 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