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The \(700 \mathrm{~m}^{2}\) ceiling of a building has a thermal resistance of \(0.52 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). The rate at which heat is lost through this ceiling on a cold winter day when the ambient temperature is \(-10^{\circ} \mathrm{C}\) and the interior is at \(20^{\circ} \mathrm{C}\) is (a) \(23.1 \mathrm{~kW}\) (b) \(40.4 \mathrm{~kW}\) (c) \(55.6 \mathrm{~kW}\) (d) \(68.1 \mathrm{~kW}\) (e) \(88.6 \mathrm{~kW}\)

Short Answer

Expert verified
Answer: (b) 40.4 kW

Step by step solution

01

Write down the given variables

We are given: - Area of the ceiling: \(A = 700 \mathrm{~m}^{2}\) - Thermal resistance: \(R = 0.52 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) - Interior temperature: \(T_{interior} = 20^{\circ} \mathrm{C}\) - Exterior (ambient) temperature: \(T_{exterior} = -10^{\circ} \mathrm{C}\)
02

Calculate the temperature difference

We can find the temperature difference between the interior and exterior as: \(\Delta T = T_{interior} - T_{exterior} = 20^{\circ} \mathrm{C} - (-10^{\circ} \mathrm{C}) = 30^{\circ} \mathrm{C}\)
03

Find the rate of heat loss

Using the formula \(Q = \frac{A \cdot \Delta T}{R}\), we can find the rate of heat loss: \(Q = \frac{700 \cdot 30}{0.52} = \frac{21000}{0.52} \approx 40384.6 \mathrm{~W}\)
04

Convert the result to kilowatts

To convert the rate of heat loss to kilowatts, divide by 1000: \(Q = \frac{40384.6}{1000} \approx 40.4 \mathrm{~kW}\)
05

Choose the correct answer

The calculated rate of heat loss is approximately \(40.4 \mathrm{~kW}\), which corresponds to option (b). So the correct answer is (b) \(40.4 \mathrm{~kW}\).

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Most popular questions from this chapter

For simplicity, approximate the forearm as a one-dimensional cylinder and ignore the presence of bones. The metabolic heat generation rate \(\left(\dot{e}_{\mathrm{m}}\right)\) and perfusion rate \((\dot{p})\) are both constant throughout the muscle. The blood density and specific heat are \(\rho_{b}\) and \(c_{b}\), respectively. The core body temperate \(\left(T_{c}\right)\) and the arterial blood temperature \(\left(T_{a}\right)\) are both assumed to be the same and constant. The muscle and the skin/fat layer thermal conductivities are \(k_{m}\) and \(k_{s f}\), respectively. The skin has an emissivity of \(\varepsilon\), and the forearm is subjected to an air environment with a temperature of \(T_{\infty 0}\), a convection heat transfer coefficient of \(h_{\text {com }}\), and a radiation heat transfer coefficient of \(h_{\text {rad }}\). Assuming blood properties and thermal conductivities are all constant, \((a)\) write the bioheat transfer equation in radial coordinates. The boundary conditions for the forearm are specified constant temperature at the outer surface of the muscle \(\left(T_{i}\right)\) and temperature symmetry at the centerline of the forearm. (b) Solve the differential equation and apply the boundary conditions to develop an expression for the temperature distribution in the forearm. (c) Determine the temperature at the outer surface of the muscle \(\left(T_{i}\right)\) and the maximum temperature in the forearm $\left(T_{\max }\right)$ for the following conditions: $$ \begin{aligned} &r_{m}=0.05 \mathrm{~m}, t_{s f}=0.003 \mathrm{~m}, \dot{e}_{\mathrm{m}}=700 \mathrm{~W} / \mathrm{m}^{3}, \dot{p}=0.00051 / \mathrm{s}, \\ &T_{a}=37^{\circ} \mathrm{C}, T_{\text {co }}=T_{\text {surr }}=24^{\circ} \mathrm{C}, \varepsilon=0.95 \\ &\rho_{b}=1000 \mathrm{~kg} / \mathrm{m}^{3}, c_{b}=3600 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k_{m}=0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \\ &k_{s f}=0.3 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, h_{\text {conv }}=2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}, h_{\text {rad }}=5.9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \end{aligned} $$

A 1-m-inner-diameter liquid-oxygen storage tank at a hospital keeps the liquid oxygen at \(90 \mathrm{~K}\). The tank consists of a \(0.5\)-cm-thick aluminum \((k=170 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) shell whose exterior is covered with a 10 -cm-thick layer of insulation \((k=0.02\) $\mathrm{W} / \mathrm{m} \cdot \mathrm{K})$. The insulation is exposed to the ambient air at \(20^{\circ} \mathrm{C}\), and the heat transfer coefficient on the exterior side of the insulation is \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The rate at which the liquid oxygen gains heat is (a) \(141 \mathrm{~W}\) (b) \(176 \mathrm{~W}\) (c) \(181 \mathrm{~W}\) (d) \(201 \mathrm{~W}\) (e) \(221 \mathrm{~W}\)

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.

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?

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