Question

The overall heat transfer coefficient (the \(U\)-value) of a wall under winter design conditions is \(U=2.25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Now a layer of \(100-\mathrm{mm}\) face brick is added to the outside, leaving a 20 -mm airspace between the wall and the bricks. Determine the new \(U\)-value of the wall. Also, determine the rate of heat transfer through a \(3-\mathrm{m}\)-high, 7-m-long section of the wall after modification when the indoor and outdoor temperatures are \(22^{\circ} \mathrm{C}\) and $-25^{\circ} \mathrm{C}$, respectively.

Short answer

Answer: The new U-value of the wall is 0.731 W/m²⋅K and the rate of heat transfer through the modified wall section is 720.729 W.

Step-by-step solution

(Step 1: Determine the thermal resistance of the original wall)

To find the original thermal resistance of the wall, we use the formula: \(R = \frac{1}{U}\), where \(R\) is the thermal resistance and \(U\) is the overall heat transfer coefficient. Given, \(U = 2.25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), we can calculate the thermal resistance of the original wall. \(R_\text{original} = \frac{1}{2.25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}} = 0.444~\mathrm{m}^{2} \cdot \mathrm{K} / \mathrm{W}\).

(Step 2: Determine the thermal resistance of the face brick layer and the airspace)

We're given that the added layer of face brick is 100 mm, and the airspace is 20 mm. To calculate the thermal resistance of both the face brick layer and the airspace, we will need their respective thermal conductivities. Commonly, the thermal conductivity of face brick(\(k_\text{brick}\)) is approximately \(0.81~\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\), and for the airspace (\(k_\text{air}\)), it is \(0.025~\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\). Now, we can calculate the thermal resistance using the formula: \( R_{i} = \frac{d_{i}}{k_{i}} \), where \(R_{i}\) is the thermal resistance of the ith layer, \(d_{i}\) is the layer thickness, and \(k_{i}\) is the layer thermal conductivity. \( R_\text{brick} = \frac{0.1 \mathrm{~m}}{0.81 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}} = 0.1235~\mathrm{m}^{2}\cdot \mathrm{K} / \mathrm{W} \) \( R_\text{air} = \frac{0.02 \mathrm{~m}}{0.025 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}} = 0.8~\mathrm{m}^{2}\cdot \mathrm{K} / \mathrm{W} \).

(Step 3: Calculate the new U-value of the wall)

After adding the face brick layer and the airspace, the overall thermal resistance of the modified wall is the sum of the thermal resistances of the original wall, face brick layer, and the airspace. Then, we can calculate the new U-value using the formula: \(U_\text{new} = \frac{1}{R_\text{total}}\). \( R_\text{total} = R_\text{original} + R_\text{brick} + R_\text{air} = 0.444 + 0.1235 + 0.8 = 1.3675~\mathrm{m}^{2} \cdot \mathrm{K} / \mathrm{W} \) \( U_\text{new} = \frac{1}{1.3675~\mathrm{m}^{2} \cdot \mathrm{K} / \mathrm{W}} = 0.731~\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K} \).

(Step 4: Calculate the rate of heat transfer through the modified wall)

Now that we have the new U-value, we can calculate the rate of heat transfer through the modified wall using the formula: \(Q = U_\text{new} \cdot A \cdot \Delta T\), where \(Q\) is the rate of heat transfer, \(A\) is the area of the wall section, and \(\Delta T\) is the temperature difference. We're given that the wall section is 3 m high and 7 m long, so its area is: \(A = 3\,\mathrm{m} \times 7\,\mathrm{m} = 21\,\mathrm{m}^2\) Also, we're given that the indoor temperature is \(22^{\circ}\mathrm{C}\), and the outdoor temperature is \(-25^{\circ}\mathrm{C}\). Therefore, the temperature difference is: \(\Delta T = 22^{\circ}\mathrm{C} - (-25^{\circ}\mathrm{C})= 47\,\mathrm{K}\) Now, calculating the rate of heat transfer: \(Q = 0.731~\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K} \times 21\,\mathrm{m}^2 \times 47\,\mathrm{K} = 720.729\,\mathrm{W}\). So, the new U-value of the wall is \(0.731 \,\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and the rate of heat transfer through the modified wall section is \(720.729 \,\mathrm{W}\).