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Chapter 1: Problem 124

Consider a house in Atlanta, Georgia, that is maintained at \(22^{\circ} \mathrm{C}\) and has a total of \(20 \mathrm{~m}^{2}\) of window area. The windows are double-door type with wood frames and metal spacers and have a \(U\)-factor of \(2.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (see Prob. 1-125 for the definition of \(U\)-factor). The winter average temperature of Atlanta is \(11.3^{\circ} \mathrm{C}\). Determine the average rate of heat loss through the windows in winter.

Short Answer

Expert verified
Answer: The average rate of heat loss through the windows during winter in the given house is 535 W.

Step by step solution

01

Understand the U-factor and the formula for heat loss

The U-factor (also called thermal transmittance) represents the rate at which heat is transferred through a building component (in this case, windows). The U-factor is expressed in W/m²·K. The formula for heat loss through the windows can be given as: Heat Loss = U-factor × Window Area × Temperature Difference Now that we know the U-factor, we can plug in the given values for window area and temperature difference and calculate the average rate of heat loss through the windows in winter.
02

Calculate the temperature difference

The temperature difference between the inside and outside of the house is crucial for calculating heat loss. Given the inside temperature of the house C_{in} = 22°C and the winter average outdoor temperature, C_{out} = 11.3°C. We can calculate the temperature difference (∆T) as: ∆T = C_{in} - C_{out} ∆T = 22°C - 11.3°C ∆T = 10.7°C
03

Calculate the average rate of heat loss through the windows in winter

Now that we have the temperature difference (∆T), the U-factor (given), and the window area (given), we can plug in the values into the Heat Loss formula: Heat Loss = U-factor × Window Area × Temperature Difference Heat Loss = (2.5 W/m²·K) × (20 m²) × (10.7 °C) Heat Loss = 535 W The average rate of heat loss through the windows in winter is 535 W.

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Most popular questions from this chapter

A 40-cm-long, 0.4-cm-diameter electric resistance wire submerged in water is used to determine the convection heat transfer coefficient in water during boiling at \(1 \mathrm{~atm}\) pressure. The surface temperature of the wire is measured to be \(114^{\circ} \mathrm{C}\) when a wattmeter indicates the electric power consumption to be \(7.6 \mathrm{~kW}\). The heat transfer coefficient is (a) \(108 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\) (b) \(13.3 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(68.1 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\) (d) \(0.76 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(256 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\)

Hot air at \(80^{\circ} \mathrm{C}\) is blown over a 2-m \(\times 4\) - \(\mathrm{m}\) flat surface at \(30^{\circ} \mathrm{C}\). If the average convection heat transfer coefficient is \(55 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat transfer from the air to the plate, in \(\mathrm{kW}\).

Air at \(20^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) blows over a pond. The surface temperature of the pond is at \(40^{\circ} \mathrm{C}\). Determine the heat flux between the surface of the pond and the air.

Steady heat conduction occurs through a \(0.3\)-m-thick \(9 \mathrm{~m} \times 3 \mathrm{~m}\) composite wall at a rate of \(1.2 \mathrm{~kW}\). If the inner and outer surface temperatures of the wall are \(15^{\circ} \mathrm{C}\) and \(7^{\circ} \mathrm{C}\), the effective thermal conductivity of the wall is (a) \(0.61 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (b) \(0.83 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (c) \(1.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (d) \(2.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (e) \(5.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\)

Solve this system of two equations with two unknowns using EES: $$ \begin{aligned} x^{3}-y^{2} &=10.5 \\ 3 x y+y &=4.6 \end{aligned} $$
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