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Chapter 3: Problem 195

A 1.4-m-diameter spherical steel tank filled with iced water at \(0^{\circ} \mathrm{C}\) is buried underground at a location where the thermal conductivity of the soil is \(k=0.55 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The distance between the tank center and the ground surface is \(2.4 \mathrm{~m}\). For ground surface temperature of \(18^{\circ} \mathrm{C}\), determine the rate of heat transfer to the iced water in the tank. What would your answer be if the soil temperature were \(18^{\circ} \mathrm{C}\) and the ground surface were insulated?

Short Answer

Expert verified
Answer: The rate of heat transfer to the iced water in the tank is approximately 33.29 W. 2. What is the rate of heat transfer to the iced water in the tank if the ground surface is insulated? Answer: If the ground surface is insulated and the soil temperature is 18°C, there will be no heat transfer to the iced water in the tank (0 W).

Step by step solution

01

Information and Formula

Given: - Spherical tank diameter: \(D = 1.4 m\) - Temperature of iced water: \(T_{i} = 0^{\circ} C\) - Distance between tank center and ground surface: \(H = 2.4 m\) - Thermal conductivity of soil: \(k = 0.55 \frac{\mathrm{W}}{\mathrm{m} \cdot \mathrm{K}}\) - Ground surface temperature: \(T_{s} = 18^{\circ} C\) We will use this formula for the rate of heat transfer: \(Q = \frac{k \cdot A \cdot \Delta T}{L}\), where: - \(Q\) – rate of heat transfer - \(k\) – thermal conductivity - \(A\) – area of the heat transfer surface - \(\Delta T\) – temperature difference between the surfaces - \(L\) – thickness between the surfaces
02

Calculate the Surface Area

Since the tank is spherical, the surface area can be found using the formula for the surface area of a sphere: $$A = 4 \pi r^2$$, where \(r\) is the radius of the sphere. Radius of the tank: \(r = \frac{D}{2} = \frac{1.4}{2} = 0.7 m\) Now we can find the surface area: \(A = 4 \pi (0.7)^2 = 6.16 m^2\)
03

Temperature Difference and Thickness

Now we need to find \(\Delta T\) and \(L\). Temperature difference: \(\Delta T = T_{s} - T_{i} = 18 - 0 = 18^{\circ} C\) Thickness between the surfaces: \(L = H - r = 2.4 - 0.7 = 1.7 m\)
04

Calculate the Rate of Heat Transfer

Using the formula for the rate of heat transfer, we can now determine the value of \(Q\): \(Q = \frac{0.55 \cdot 6.16 \cdot 18}{1.7} = 33.29 \mathrm{~W}\) So, the rate of heat transfer to the iced water in the tank is approximately \(33.29 \mathrm{W}\).
05

Insulated Ground Surface Case

If the ground surface is insulated, there will be no heat transfer from the insulated surface to the soil layer above the tank. Therefore, \(T_s\) is no longer the ground surface temperature but rather the soil temperature, which we are told is also \(18^{\circ} C\). In this case, the temperature difference between the surfaces becomes zero: \(\Delta T = T_s - T_i = 18 - 18 = 0^{\circ} C\) As a consequence, the rate of heat transfer will be zero: \(Q = \frac{0.55 \cdot 6.16 \cdot 0}{1.7} = 0 \mathrm{~W}\) So, if the soil temperature is \(18^{\circ} C\) and the ground surface is insulated, there will be no heat transfer to the iced water in the tank.

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Most popular questions from this chapter

Steam at \(450^{\circ} \mathrm{F}\) is flowing through a steel pipe \(\left(k=8.7 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right)\) whose inner and outer diameters are \(3.5\) in and \(4.0\) in, respectively, in an environment at \(55^{\circ} \mathrm{F}\). The pipe is insulated with 2 -in-thick fiberglass insulation \((k=\) \(\left.0.020 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right)\). If the heat transfer coefficients on the inside and the outside of the pipe are 30 and \(5 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\), respectively, determine the rate of heat loss from the steam per foot length of the pipe. What is the error involved in neglecting the thermal resistance of the steel pipe in calculations?

Using cylindrical samples of the same material, devise an experiment to determine the thermal contact resistance. Cylindrical samples are available at any length, and the thermal conductivity of the material is known.

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) \(11^{\circ} \mathrm{C}\)

Steam at \(320^{\circ} \mathrm{C}\) flows in a stainless steel pipe \((k=\) \(15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) ) whose inner and outer diameters are \(5 \mathrm{~cm}\) and \(5.5 \mathrm{~cm}\), respectively. The pipe is covered with \(3-\mathrm{cm}\)-thick glass wool insulation \((k=0.038 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). Heat is lost to the surroundings at \(5^{\circ} \mathrm{C}\) by natural convection and radiation, with a combined natural convection and radiation heat transfer coefficient of \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Taking the heat transfer coefficient inside the pipe to be \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat loss from the steam per unit length of the pipe. Also determine the temperature drops across the pipe shell and the insulation.

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