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Chapter 3: Problem 227

A 25-cm-diameter, 2.4-m-long vertical cylinder containing ice at $0^{\circ} \mathrm{C}$ is buried right under the ground. The cylinder is thin-shelled and is made of a high-thermal-conductivity material. The surface temperature and the thermal conductivity of the ground are \(18^{\circ} \mathrm{C}\) and $0.85 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$, respectively. The rate of heat transfer to the cylinder is (a) \(37.2 \mathrm{~W}\) (b) \(63.2 \mathrm{~W}\) (c) \(158 \mathrm{~W}\) (d) \(480 \mathrm{~W}\) (e) \(1210 \mathrm{~W}\)

Short Answer

Expert verified
Answer: (b) 63.2 W

Step by step solution

01

Calculate the radius of the cylinder

First, we need to find the radius of the cylinder. Since the diameter is given as 25 cm, we can find the radius by dividing the diameter by 2: Radius (r) = Diameter/2 = 25 cm / 2 = 12.5 cm = 0.125 m
02

Calculate the cross-sectional area of the cylinder

Next, we need to calculate the cross-sectional area of the cylinder. The formula for the cross-sectional area of a circle is A = πr^2, where A is the area and r is the radius. Therefore, we have: A = π * (0.125 m)^2 = 0.0491 m²
03

Calculate the temperature difference

We are given two temperatures: the surface temperature of the ground, which is \(18^{\circ}\mathrm{C}\), and the temperature of the ice inside the cylinder, which is \(0^{\circ}\mathrm{C}\). The temperature difference (ΔT) between these two temperatures is: ΔT = 18°C - 0°C = 18°C
04

Apply the formula for rate of heat transfer

The formula for the rate of heat transfer (Q) through a cylindrical object is: Q = (2 * π * k * A * ΔT) / L where k is the thermal conductivity of the ground (\(0.85\mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\)), A is the cross-sectional area found in Step 2, ΔT is the temperature difference from Step 3, and L is the length of the cylinder (2.4 m). Plugging the values into the formula, we get: Q = (2 * π * 0.85 W/m·K * 0.0491 m² * 18°C) / 2.4 m Q = 63.243 W
05

Compare the calculated value with the options given

The calculated rate of heat transfer is 63.243 W. Comparing this value with the options given in the exercise, we can see that the closest answer is: (b) \(63.2 \mathrm{~W}\)

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Most popular questions from this chapter

A pipe is insulated such that the outer radius of the insulation is less than the critical radius. Now the insulation is taken off. Will the rate of heat transfer from the pipe increase or decrease for the same pipe surface temperature?

A plane brick wall \((k=0.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is $10 \mathrm{~cm}$ thick. The thermal resistance of this wall per unit of wall area is (a) \(0.143 \mathrm{~m}^{2}, \mathrm{~K} / \mathrm{W}\) (b) \(0.250 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) (c) \(0.327 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) (d) \(0.448 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) (e) \(0.524 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\)

Circular fins of uniform cross section, with diameter of \(10 \mathrm{~mm}\) and length of \(50 \mathrm{~mm}\), are attached to a wall with surface temperature of \(350^{\circ} \mathrm{C}\). The fins are made of material with thermal conductivity of \(240 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), they are exposed to an ambient air condition of \(25^{\circ} \mathrm{C}\), and the convection heat transfer coefficient is $250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Determine the heat transfer rate and plot the temperature variation of a single fin for the following boundary conditions: (a) Infinitely long fin (b) Adiabatic fin tip (c) Fin with tip temperature of \(250^{\circ} \mathrm{C}\) (d) Convection from the fin tip

Hot water at an average temperature of \(80^{\circ} \mathrm{C}\) and an average velocity of \(1.5 \mathrm{~m} / \mathrm{s}\) is flowing through a \(25-\mathrm{m}\) section of a pipe that has an outer diameter of $5 \mathrm{~cm}\(. The pipe extends \)2 \mathrm{~m}$ in the ambient air above the ground, dips into the ground $(k=1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\( ) vertically for \)3 \mathrm{~m}$, and continues horizontally at this depth for \(20 \mathrm{~m}\) more before it enters the next building.

Hot water $\left(c_{p}=4.179 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\( flows through an 80 -m-long PVC \)(k=0.092 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\( pipe whose inner diameter is \)2 \mathrm{~cm}$ and outer diameter is \(2.5 \mathrm{~cm}\) at a rate of $1 \mathrm{~kg} / \mathrm{s}\(, entering at \)40^{\circ} \mathrm{C}$. If the entire interior surface of this pipe is maintained at \(35^{\circ} \mathrm{C}\) and the entire exterior surface at \(20^{\circ} \mathrm{C}\), the outlet temperature of water is (a) \(35^{\circ} \mathrm{C}\) (b) \(36^{\circ} \mathrm{C}\) (c) \(37^{\circ} \mathrm{C}\) (d) \(38^{\circ} \mathrm{C}\) (e) \(39^{\circ} \mathrm{C}\)
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