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Chapter 3: Problem 202

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) \(251^{\circ} \mathrm{C}\)

Short Answer

Expert verified
Answer: (c) 48°C

Step by step solution

01

Identifying the given parameters

From the exercise, we have the following parameters: - Length of the pipe, L = 10m, - Outer radius of the pipe, r1 = 5cm = 0.05m, - Thickness of the insulation, t = 3cm = 0.03m, - Thermal conductivity of the insulation, k = 0.05 W/m.K, - Rate of heat loss from the pipe, Q = 1000 W. With this information, we can now proceed to find the temperature drop across the insulation.
02

Calculate the inner and outer radii of the insulation

The inner radius of the insulation (r2) is equal to the outer radius of the pipe (r1), which is 0.05m. The outer radius of the insulation (r3) will be the sum of the inner radius and the thickness of the insulation: r3 = r2 + t = 0.05m + 0.03m = 0.08m Now we have the inner and outer radii of the insulation, r2 = 0.05m and r3 = 0.08m.
03

Use the formula for heat transfer through a cylindrical wall

The formula for the rate of heat transfer through a cylindrical wall is: Q = 2 * pi * L * (T1 - T2) * k / ln(r3/r2) where Q is the rate of heat transfer, L is the length of the pipe, T1 is the temperature at the inner radius (r2), T2 is the temperature at the outer radius (r3), k is the thermal conductivity, r2 is the inner radius of the insulation, and r3 is the outer radius of the insulation. We are given Q, L, k, r2, and r3, and we need to find the temperature drop, which is (T1 - T2).
04

Rearrange the equation and solve for the temperature drop

Rearranging the formula to find (T1 - T2): (T1 - T2) = Q * ln(r3/r2) / (2 * pi * L * k) Now plug in the given values: (T1 - T2) = (1000 W) * ln(0.08m/0.05m) / [2 * pi * (10m) * (0.05 W/m.K)] (T1 - T2) ≈ 48.5 °C
05

Identify the correct answer among the given options

The calculated temperature drop across the insulation is approximately 48.5°C, so the correct option among the given choices is: (c) 48°C

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Most popular questions from this chapter

A \(2.2\)-mm-diameter and 10-m-long electric wire is tightly wrapped with a \(1-m m\)-thick plastic cover whose thermal conductivity is \(k=0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Electrical measurements indicate that a current of \(13 \mathrm{~A}\) passes through the wire and there is a voltage drop of \(8 \mathrm{~V}\) along the wire. If the insulated wire is exposed to a medium at \(T_{\infty}=30^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(h=24 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the temperature at the interface of the wire and the plastic cover in steady operation. Also determine if doubling the thickness of the plastic cover will increase or decrease this interface temperature.

Circular cooling fins of diameter \(D=1 \mathrm{~mm}\) and length \(L=25.4 \mathrm{~mm}\), made of copper \((k=400 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), are used to enhance heat transfer from a surface that is maintained at temperature \(T_{s 1}=132^{\circ} \mathrm{C}\). Each rod has one end attached to this surface \((x=0)\), while the opposite end \((x=L)\) is joined to a second surface, which is maintained at \(T_{s 2}=0^{\circ} \mathrm{C}\). The air flowing between the surfaces and the rods is also at \(T_{\infty}=0^{\circ} \mathrm{C}\), and the convection coefficient is \(h=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Express the function \(\theta(x)=T(x)-T_{\infty}\) along a fin, and calculate the temperature at \(x=L / 2\). (b) Determine the rate of heat transferred from the hot surface through each fin and the fin effectiveness. Is the use of fins justified? Why? (c) What is the total rate of heat transfer from a \(10-\mathrm{cm}\) by 10 -cm section of the wall, which has 625 uniformly distributed fins? Assume the same convection coefficient for the fin and for the unfinned wall surface.

Consider an insulated pipe exposed to the atmosphere. Will the critical radius of insulation be greater on calm days or on windy days? Why?

Superheated steam at an average temperature \(200^{\circ} \mathrm{C}\) is transported through a steel pipe \(\left(k=50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, D_{o}=8.0 \mathrm{~cm}\right.\), \(D_{i}=6.0 \mathrm{~cm}\), and \(L=20.0 \mathrm{~m}\) ). The pipe is insulated with a 4-cm thick layer of gypsum plaster \((k=0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The insulated pipe is placed horizontally inside a warehouse where the average air temperature is \(10^{\circ} \mathrm{C}\). The steam and the air heat transfer coefficients are estimated to be 800 and \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Calculate \((a)\) the daily rate of heat transfer from the superheated steam, and \((b)\) the temperature on the outside surface of the gypsum plaster insulation.

Determine the winter \(R\)-value and the \(U\)-factor of a masonry wall that consists of the following layers: \(100-\mathrm{mm}\) face bricks, 100 -mm common bricks, \(25-\mathrm{mm}\) urethane rigid foam insulation, and \(13-\mathrm{mm}\) gypsum wallboard.
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