Consider a house that is maintained at a constant temperature of \(22^{\circ} \mathrm{C}\). One of the walls of the house has three singlepane glass windows that are \(1.5 \mathrm{~m}\) high and \(1.8 \mathrm{~m}\) long. The glass \((k=0.78 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is \(0.5 \mathrm{~cm}\) thick, and the heat transfer coefficient on the inner surface of the glass is \(8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Now winds at \(35 \mathrm{~km} / \mathrm{h}\) start to blow parallel to the surface of this wall. If the air temperature outside is \(-2^{\circ} \mathrm{C}\), determine the rate of heat loss through the windows of this wall. Assume radiation heat transfer to be negligible. Evaluate the air properties at a film temperature of \(5^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\).

Step by step solution

01

Calculate the Area of the Windows

First, we will calculate the area of each window, which will be used later in our calculations. The dimensions given are 1.5 m high and 1.8 m long, so the area of one window is: $$ A = (1.5\,\mathrm{m})(1.8\,\mathrm{m}) = 2.7\,\mathrm{m}^2 $$ There are three windows, so the total area of the windows is: $$ A_{total} = 3 \times A = 3 \times 2.7\,\mathrm{m}^2 = 8.1\,\mathrm{m}^2 $$

02

Calculate Thermal Resistance of Glass and Convective Heat Transfer Coefficients

Now that we have the area of the windows, we will calculate the combined thermal resistance of conduction through the glass and convection at the inner surface of the glass. First, the thermal resistance of conduction through the glass can be expressed as: $$ R_{cond} = \frac{L}{kA} = \frac{0.005\,\mathrm{m}}{0.78\,\mathrm{W/m\,K}\times 8.1\,\mathrm{m}^2} $$ Next, compute the thermal resistance of convection at the inner surface of the glass: $$ R_{conv} = \frac{1}{hA} = \frac{1}{8\,\mathrm{W/m^2\,K} \times 8.1\,\mathrm{m}^2} $$ Then, find the total combined thermal resistance by adding these two values: $$ R_{total} = R_{cond} + R_{conv} $$

03

Calculate Temperature Difference

The temperature difference between the inside and outside of the house is needed to find the rate of heat loss. The inside temperature is 22°C, and the outside temperature is -2°C. The temperature difference is: $$ \Delta T = T_{inside} - T_{outside} = 22^{\circ}\mathrm{C} - (-2^{\circ}\mathrm{C}) = 24^{\circ}\mathrm{C} $$

04

Calculate Rate of Heat Loss

Now that we have the combined thermal resistance and temperature difference, we can determine the rate of heat loss through the windows using the following equation: $$ Q = \frac{\Delta T}{R_{total}} $$ Plug in the values calculated in previous steps and find the rate of heat loss through the windows.

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