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Chapter 3: Problem 154

A 3-m-diameter spherical tank containing some radioactive material is buried in the ground \((k=1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The distance between the top surface of the tank and the ground surface is \(4 \mathrm{~m}\). If the surface temperatures of the tank and the ground are \(140^{\circ} \mathrm{C}\) and \(15^{\circ} \mathrm{C}\), respectively, determine the rate of heat transfer from the tank.

Short Answer

Expert verified
Answer: The rate of heat transfer from the tank to the ground is approximately 665.7 W.

Step by step solution

01

Identify the dimensions of the tank and ground

The diameter of the spherical tank is given as 3 meters. The radius of the tank can be calculated as half of the diameter, which is 1.5 meters. The distance between the top surface of the tank and the ground surface is given as 4 meters.
02

Calculate the conductive thermal resistance

The heat transfer through the ground can be modeled as a conductive thermal resistance. The conductive thermal resistance R_cond can be calculated as: \(R_\mathrm{cond} = \dfrac{r_\mathrm{outer}}{4 \pi \cdot k \cdot r_\mathrm{inner}}\) where r_inner is the inner radius of the tank, r_outer is the outer radius of the ground covering the tank, and k is the thermal conductivity of the ground. In this case, r_inner = 1.5 m, r_outer = 1.5 m + 4 m = 5.5 m, and k = 1.4 W/(m·K). Substituting these values, we get: \(R_\mathrm{cond} = \dfrac{5.5}{4 \pi \cdot 1.4 \cdot 1.5} = 0.1879 \, \mathrm{K} / \mathrm{W}\)
03

Determine temperature difference

The temperature difference between the surface of the tank (T_tank) and the ground (T_ground) is given as: \(\Delta T = T_\mathrm{tank} - T_\mathrm{ground}\) Substituting the given values, we get: \(\Delta T = 140 - 15 = 125^{\circ} \mathrm{C}\)
04

Calculate the rate of heat transfer

The rate of heat transfer (Q) can be determined using the overall thermal resistance (R_cond) and the temperature difference. The formula is as follows: \(Q = \dfrac{\Delta T}{R_\mathrm{cond}}\) Substituting the values, we get: \(Q = \dfrac{125}{0.1879} = 665.7 \, \mathrm{W}\) Therefore, the rate of heat transfer from the tank to the ground is approximately 665.7 W.

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Most popular questions from this chapter

Chilled water enters a thin-shelled 5-cm-diameter, 150-mlong pipe at \(7^{\circ} \mathrm{C}\) at a rate of \(0.98 \mathrm{~kg} / \mathrm{s}\) and leaves at \(8^{\circ} \mathrm{C}\). The pipe is exposed to ambient air at \(30^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the pipe is to be insulated with glass wool insulation \((k=0.05 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) in order to decrease the temperature rise of water to \(0.25^{\circ} \mathrm{C}\), determine the required thickness of the insulation.

Computer memory chips are mounted on a finned metallic mount to protect them from overheating. A \(152 \mathrm{MB}\) memory chip dissipates \(5 \mathrm{~W}\) of heat to air at \(25^{\circ} \mathrm{C}\). If the temperature of this chip is to not exceed \(50^{\circ} \mathrm{C}\), the overall heat transfer coefficient- area product of the finned metal mount must be at least (a) \(0.2 \mathrm{~W} /{ }^{\circ} \mathrm{C}\) (b) \(0.3 \mathrm{~W} /{ }^{\circ} \mathrm{C}\) (c) \(0.4 \mathrm{~W} /{ }^{\circ} \mathrm{C}\) (d) \(0.5 \mathrm{~W} /{ }^{\circ} \mathrm{C}\) (e) \(0.6 \mathrm{~W} /{ }^{\circ} \mathrm{C}\)

A 10-m-long 5-cm-outer-radius cylindrical steam pipe is covered with \(3-\mathrm{cm}\) thick cylindrical insulation with a thermal conductivity of \(0.05 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). If the rate of heat loss from the pipe is \(1000 \mathrm{~W}\), the temperature drop across the insulation is (a) \(163^{\circ} \mathrm{C}\) (b) \(600^{\circ} \mathrm{C}\) (c) \(48^{\circ} \mathrm{C}\) (d) \(79^{\circ} \mathrm{C}\) (e) \(251^{\circ} \mathrm{C}\)

A 1-cm-diameter, 30-cm-long fin made of aluminum \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is attached to a surface at \(80^{\circ} \mathrm{C}\). The surface is exposed to ambient air at \(22^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(11 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the fin can be assumed to bery long, its efficiency is (a) \(0.60\) (b) \(0.67\) (c) \(0.72\) (d) \(0.77\) (e) \(0.88\)

A room at \(20^{\circ} \mathrm{C}\) air temperature is loosing heat to the outdoor air at \(0^{\circ} \mathrm{C}\) at a rate of \(1000 \mathrm{~W}\) through a \(2.5\)-m-high and 4 -m-long wall. Now the wall is insulated with \(2-\mathrm{cm}\) thick insulation with a conductivity of \(0.02 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Determine the rate of heat loss from the room through this wall after insulation. Assume the heat transfer coefficients on the inner and outer surface of the wall, the room air temperature, and the outdoor air temperature to remain unchanged. Also, disregard radiation. (a) \(20 \mathrm{~W}\) (b) \(561 \mathrm{~W}\) (c) \(388 \mathrm{~W}\) (d) \(167 \mathrm{~W}\) (e) \(200 \mathrm{~W}\)
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