A furnace wall is made up of three layers of thicknesses 250 mm, 100 mm and 150 mm with thermal conductivities of 1.65, k and 9.2 W/m°C respectively. The inside is exposed to gases... A furnace wall is made up of three layers of thicknesses 250 mm, 100 mm and 150 mm with thermal conductivities of 1.65, k and 9.2 W/m°C respectively. The inside is exposed to gases at 1250°C with a convection coefficient of 25 W/m² °C and the inside surface is at 1100°C, the outside surface is exposed to air at 25°C with convection coefficient of 12 W/m²°C. Determine: (i) The unknown thermal conductivity 'k'; (ii) The overall heat transfer coefficient; (iii) All surface temperatures. Read more

Steps to Solve

1. Calculate the heat flux (q) through the first layer

Since the inside surface temperature ($T_1$) is given as 1100°C and the inside gas temperature ($T_g$) is 1250°C, we can find the heat flux using the convection coefficient ($h_i$) on the inside: $ q = h_i (T_g - T_1) $ $ q = 25 \frac{W}{m^2°C} (1250°C - 1100°C) = 25 \frac{W}{m^2°C} \cdot 150°C = 3750 \frac{W}{m^2} $

2. Determine the temperature at the outer surface of the third layer ($T_4$)

We know the heat flux ($q$) and the convection coefficient ($h_o$) and ambient temperature ($T_a$) on the outside. We can find the outer surface temperature ($T_4$) using: $ q = h_o (T_4 - T_a) $ Rearrange to solve for $T_4$: $ T_4 = \frac{q}{h_o} + T_a $ $ T_4 = \frac{3750 \frac{W}{m^2}}{12 \frac{W}{m^2 °C}} + 25°C = 312.5°C + 25°C = 337.5°C $

3. Calculate the temperature at the interface between the second and third layers ($T_3$)

We can calculate the temperature drop across the third layer using Fourier's Law:

$ q = -k_3 \frac{T_4 - T_3}{L_3} $ where $k_3$ is the thermal conductivity of the third layer and $L_3$ is its thickness. Rearrange to solve for $T_3$: $ T_3 = T_4 + \frac{q \cdot L_3}{k_3} $ $ T_3 = 337.5°C + \frac{3750 \frac{W}{m^2} \cdot 0.15 m}{9.2 \frac{W}{m°C}} = 337.5°C + 61.03°C = 398.53°C $

4. Calculate the temperature at the interface between the first and second layers ($T_2$)

We can also calculate the temperature drop across the first layer using Fourier's Law:

$ q = -k_1 \frac{T_2 - T_1}{L_1} $ where $k_1$ is the thermal conductivity of the first layer and $L_1$ is its thickness. Rearrange to solve for $T_2$: $ T_2 = T_1 - \frac{q \cdot L_1}{k_1} $ $ T_2 = 1100°C - \frac{3750 \frac{W}{m^2} \cdot 0.25 m}{1.65 \frac{W}{m°C}} = 1100°C - 568.18°C = 531.82°C $

5. Calculate the unknown thermal conductivity 'k' of the second layer

Using Fourier's Law for the second layer:

$ q = -k_2 \frac{T_3 - T_2}{L_2} $ where $k_2$ is the thermal conductivity of the second layer and $L_2$ is its thickness. $k_2 = k$

Rearrange to solve for $k$: $ k = - \frac{q \cdot L_2}{T_3 - T_2} $ $ k = - \frac{3750 \frac{W}{m^2} \cdot 0.1 m}{398.53°C - 531.82°C} = - \frac{375 \frac{W}{m}}{-133.29°C} = 2.817 \frac{W}{m°C} $

6. Calculate the overall heat transfer coefficient (U) $ U = \frac{1}{\frac{1}{h_i} + \frac{L_1}{k_1} + \frac{L_2}{k_2} + \frac{L_3}{k_3} + \frac{1}{h_o}} $ $ U = \frac{1}{\frac{1}{25} + \frac{0.25}{1.65} + \frac{0.1}{2.817} + \frac{0.15}{9.2} + \frac{1}{12}} $ $ U = \frac{1}{0.04 + 0.1515 + 0.0355 + 0.0163 + 0.0833} = \frac{1}{0.3266} = 3.06 \frac{W}{m^2°C} $

7. Summarize all surface temperatures $ T_1 = 1100°C $ (Given) $ T_2 = 531.82°C $ $ T_3 = 398.53°C $ $ T_4 = 337.5°C $