Question

Consider a 20-cm-thick granite wall with a thermal conductivity of $2.79 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$. The temperature of the left surface is held constant at \(50^{\circ} \mathrm{C}\), whereas the right face is exposed to a flow of \(22^{\circ} \mathrm{C}\) air with a convection heat transfer coefficient of \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Neglecting heat transfer by radiation, find the right wall surface temperature and the heat flux through the wall.

Short answer

The right wall surface temperature is 35.50°C, and the heat flux through the wall is 202.31 W, neglecting heat transfer by radiation.

Step-by-step solution

Determine the conduction resistance

To find the conduction resistance \(R_{cond}\) through the granite wall, we can use the formula for the resistance of a planar wall: \(R_{cond} = \dfrac{L}{kA}\) where: \(R_{cond}\) = Conduction resistance [\(\mathrm{K / W}\)] \(L\) = Thickness of the wall = 0.2 m (converting 20 cm to meters) \(k\) = Thermal conductivity of the granite = 2.79 \(\mathrm{W / m \cdot K}\) \(A\) = Area of the wall We will assume the wall has an area of 1 \(\mathrm{m^2}\) for simplicity. \(R_{cond} = \dfrac{0.2 \mathrm{m}}{2.79 \mathrm{\dfrac{W}{m \cdot K}} \cdot 1 \mathrm{m^2}} = 0.0717 \mathrm{\dfrac{K}{W}}\)

Determine the convection resistance

To find the convection resistance \(R_{conv}\) at the right face of the wall, we can use the formula: \(R_{conv} = \dfrac{1}{hA}\) where: \(R_{conv}\) = Convection resistance [\(\mathrm{K / W}\)] \(h\) = Convection heat transfer coefficient = 15 \(\mathrm{\dfrac{W}{m^2 \cdot K}}\) \(A\) = Area of the wall Using the same assumption of area of 1 \(\mathrm{m^2}\): \(R_{conv} = \dfrac{1}{15 \mathrm{\dfrac{W}{m^2 \cdot K}} \cdot 1 \mathrm{m^2}} = 0.0667 \mathrm{\dfrac{K}{W}}\)

Find the total thermal resistance

The total thermal resistance \(R_{tot}\) is the sum of the conduction resistance and the convection resistance: \(R_{tot} = R_{cond} + R_{conv} = 0.0717 + 0.0667 = 0.1384 \mathrm{\dfrac{K}{W}}\)

Calculate the heat flux

We can use the total thermal resistance and the temperature difference across the wall to find the heat flux \(q\): \(q = \dfrac{T_{left} - T_{air}}{R_{tot}}\) where: \(q\) = Heat flux through the wall [\(\mathrm{W}\)] \(T_{left}\) = Left surface temperature = \(50^{\circ} \mathrm{C}\) \(T_{air}\) = Air temperature = \(22^{\circ} \mathrm{C}\) \(q = \dfrac{50 - 22}{0.1384} = 202.31 \mathrm{W}\)

Find the right wall surface temperature

We can use the heat flux \(q\) and the convection resistance \(R_{conv}\) to find the right wall surface temperature \(T_{right}\): \(q = (T_{right} - T_{air}) / R_{conv}\) Solving for \(T_{right}\), we get: \(T_{right} = T_{air} + qR_{conv} = 22 + 202.31 \cdot 0.0667 = 35.50^{\circ} \mathrm{C}\)

Final Results

The right wall surface temperature is \(35.50^{\circ} \mathrm{C}\) and the heat flux through the wall is \(202.31 \mathrm{W}\), neglecting heat transfer by radiation.