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Chapter 1: Problem 72

A 5-cm-external-diameter, 10-m-long hot-water pipe at \(80^{\circ} \mathrm{C}\) is losing heat to the surrounding air at \(5^{\circ} \mathrm{C}\) by natural convection with a heat transfer coefficient of \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the rate of heat loss from the pipe by natural convection. Answer: \(2945 \mathrm{~W}\)

Short Answer

Expert verified
Answer: The rate of heat loss from the pipe by natural convection is approximately 2945 W.

Step by step solution

01

Calculate the surface area of the pipe

The pipe is of cylindrical shape, so its surface area can be calculated using the formula \(A=2 \pi r L\), where \(r\) is the radius and \(L\) is the length of the pipe. Since the diameter is given, we can find the radius by dividing the diameter by two. So \(r = \frac{5}{2} \mathrm{~cm} = 0.025 \mathrm{~m}\). Now we can find the surface area of the pipe: \(A=2 \pi (0.025 \mathrm{~m})(10 \mathrm{~m}) \approx 1.571 \mathrm{~m}^2\).
02

Find the temperature difference

The temperature difference between the pipe and the surrounding air can be found by subtracting the surrounding air temperature (\(T_\mathrm{air}=5^{\circ} \mathrm{C}\)) from the pipe temperature (\(T_\mathrm{pipe}=80^{\circ} \mathrm{C}\)). So, the temperature difference is \(ΔT = T_\mathrm{pipe} - T_\mathrm{air} = 80^{\circ} \mathrm{C} - 5^{\circ} \mathrm{C} = 75 \mathrm{~K}\).
03

Calculate the rate of heat loss

Now we can use the formula for heat transfer by natural convection, which is \(q=h \cdot A \cdot ΔT\), where \(q\) represents the rate of heat loss, \(h\) is the heat transfer coefficient, \(A\) is the surface area of the pipe, and \(ΔT\) is the temperature difference. Plugging in the given and calculated values, we get: \(q = (25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}) \cdot (1.571 \mathrm{~m}^2) \cdot (75 \mathrm{~K}) \approx 2945 \mathrm{~W}\). The rate of heat loss from the pipe by natural convection is approximately \(2945 \mathrm{~W}\).

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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) \(256 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\)

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.

A solid plate, with a thickness of \(15 \mathrm{~cm}\) and a thermal conductivity of \(80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), is being cooled at the upper surface by air. The air temperature is \(10^{\circ} \mathrm{C}\), while the temperatures at the upper and lower surfaces of the plate are 50 and \(60^{\circ} \mathrm{C}\), respectively. Determine the convection heat transfer coefficient of air at the upper surface and discuss whether the value is reasonable or not for force convection of air.

A hollow spherical iron container with outer diameter \(20 \mathrm{~cm}\) and thickness \(0.2 \mathrm{~cm}\) is filled with iced water at \(0^{\circ} \mathrm{C}\). If the outer surface temperature is \(5^{\circ} \mathrm{C}\), determine the approximate rate of heat loss from the sphere, in \(\mathrm{kW}\), and the rate at which ice melts in the container. The heat of fusion of water is \(333.7 \mathrm{~kJ} / \mathrm{kg}\).

How does forced convection differ from natural convection?
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