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In the United States, building insulation is specified by the \(R\)-value (thermal resistance in $\mathrm{h} \cdot \mathrm{ft}^{2}+{ }^{\circ} \mathrm{F} /$ Btu units). A homeowner decides to save on the cost of heating the home by adding additional insulation in the attic. If the total \(R\)-value is increased from 15 to 25 , the homeowner can expect the heat loss through the ceiling to be reduced by (a) \(25 \%\) (b) \(40 \%\) (c) \(50 \%\) (d) \(60 \%\) (e) \(75 \%\)

Short Answer

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Answer: 40%

Step by step solution

01

Recall the relationship between R-value and heat loss

The \(R\)-value measures the thermal resistance of a material, which means that the higher the \(R\)-value, the better the insulation and the lower the heat loss. The heat loss is inversely proportional to the \(R\)-value: \(Q \propto \frac{1}{R}\), where \(Q\) is the heat loss and \(R\) is the \(R\)-value.
02

Calculate the initial heat loss

Let's denote the initial heat loss as \(Q_1\) and the final heat loss as \(Q_2\). Based on the initial \(R\)-value, we have \(Q_1 \propto \frac{1}{R_1}\), where \(R_1 = 15\).
03

Calculate the final heat loss

Similarly, based on the final \(R\)-value, we have \(Q_2 \propto \frac{1}{R_2}\), where \(R_2 = 25\).
04

Calculate the reduction in heat loss

To find the reduction in heat loss, we can compare the final heat loss (\(Q_2\)) to the initial heat loss (\(Q_1\)). The reduction in heat loss can be calculated as follows: Reduction in heat loss \(= \frac{Q_1 - Q_2}{Q_1} \times 100\%\) Since \(Q_1 \propto \frac{1}{R_1}\) and \(Q_2 \propto \frac{1}{R_2}\), we can rewrite the reduction in heat loss as: Reduction in heat loss \(= \frac{\frac{1}{R_1} - \frac{1}{R_2}}{\frac{1}{R_1}} \times 100\% = \frac{1 - \frac{R_1}{R_2}}{1} \times 100\%\) Now, we can plug in the given \(R\)-values (15 and 25) and calculate the reduction in heat loss: Reduction in heat loss \(= \frac{1 - \frac{15}{25}}{1} \times 100\% = \frac{10}{25} \times 100\% = 0.4 \times 100\% = 40\%\) So, the heat loss through the ceiling will be reduced by (b) \(40\%\).

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Most popular questions from this chapter

Hot liquid is flowing in a steel pipe with an inner diameter of $D_{1}=22 \mathrm{~mm}\( and an outer diameter of \)D_{2}=27 \mathrm{~mm}$. The inner surface of the pipe is coated with a thin fluorinated ethylene propylene (FEP) lining. The thermal conductivity of the pipe wall is $15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$. The pipe outer surface is subjected to a uniform flux of \(1200 \mathrm{~W} / \mathrm{m}^{2}\) for a length of \(1 \mathrm{~m}\). The hot liquid flowing inside the pipe has a mean temperature of $180^{\circ} \mathrm{C}\( and a convection heat transfer coefficient of \)50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. The interface between the FEP lining and the steel surface has a thermal contact conductance of $1500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Determine the temperatures at the lining and at the pipe outer surface for the pipe length subjected to the uniform heat flux. What is the total thermal resistance between the two temperatures? The ASME Code for Process Piping (ASME B31.3-2014, A.323) recommends a maximum temperature for FEP lining to be \(204^{\circ} \mathrm{C}\). Does the FEP lining comply with the recommendation of the code?

Two very long, slender rods of the same diameter and length are given. One rod (Rod 1) is made of aluminum and has a thermal conductivity $k_{1}=200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$, but the thermal conductivity of Rod \(2, k_{2}\), is not known. To determine the thermal conductivity of Rod 2 , both rods at one end are thermally attached to a metal surface which is maintained at a constant temperature \(T_{b}\). Both rods are losing heat by convection, with a convection heat transfer coefficient \(h\) into the ambient air at \(T_{\infty}\). The surface temperature of each rod is measured at various distances from the hot base surface. The measurements reveal that the temperature of the aluminum rod (Rod 1) at \(x_{1}=40 \mathrm{~cm}\) from the base is the same as that of the rod of unknown thermal conductivity (Rod 2) at \(x_{2}=20 \mathrm{~cm}\) from the base. Determine the thermal conductivity \(k_{2}\) of the second rod \((\mathrm{W} / \mathrm{m} \cdot \mathrm{K})\).

Consider a \(1.5\)-m-high electric hot-water heater that has a diameter of $40 \mathrm{~cm}\( and maintains the hot water at \)60^{\circ} \mathrm{C}$. The tank is located in a small room whose average temperature is $27^{\circ} \mathrm{C}$, and the heat transfer coefficients on the inner and outer surfaces of the heater are 50 and $12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, respectively. The tank is placed in another \(46-\mathrm{cm}\)-diameter sheet metal tank of negligible thickness, and the space between the two tanks is filled with foam insulation $(k=0.03 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})$. The thermal resistances of the water tank and the outer thin sheet metal shell are very small and can be neglected. The price of electricity is \(\$ 0.08 / \mathrm{kWh}\), and the homeowner pays \(\$ 280\) a year for water heating. Determine the fraction of the hot-water energy cost of this household that is due to the heat loss from the tank. Hot-water tank insulation kits consisting of 3 -cm-thick fiberglass insulation \((k=0.035 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) large enough to wrap the entire tank are available in the market for about \(\$ 30\). If such an insulation is installed on this water tank by the homeowner himself, how long will it take for this additional insulation to pay for itself?

The walls of a food storage facility are made of a 2 -cm-thick layer of wood \((k=0.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) in contact with a 5 -cm- thick layer of polyurethane foam $(k=0.03 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\(. If the temperature of the surface of the wood is \)-10^{\circ} \mathrm{C}$ and the temperature of the surface of the polyurethane foam is \(20^{\circ} \mathrm{C}\), the temperature of the surface where the two layers are in contact is (a) \(-7^{\circ} \mathrm{C}\) (b) \(-2^{\circ} \mathrm{C}\) (c) \(3^{\circ} \mathrm{C}\) (d) \(8^{\circ} \mathrm{C}\) (e) \(11^{\circ} \mathrm{C}\)

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