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Chapter 2: Problem 150

Consider a water pipe of length \(L=17 \mathrm{~m}\), inner radius \(r_{1}=15 \mathrm{~cm}\), outer radius \(r_{2}=20 \mathrm{~cm}\), and thermal conductivity \(k=14 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Heat is generated in the pipe material uniformly by a \(25-\mathrm{kW}\) electric resistance heater. The inner and outer surfaces of the pipe are at \(T_{1}=60^{\circ} \mathrm{C}\) and \(T_{2}=80^{\circ} \mathrm{C}\), respectively. Obtain a general relation for temperature distribution inside the pipe under steady conditions and determine the temperature at the center plane of the pipe.

Short Answer

Expert verified
Answer: The temperature at the center plane of the water pipe is 70.12°C.

Step by step solution

01

1. Identify the steady-state heat equation

The steady-state heat equation is given by: \(-k\frac{d^{2}T}{dr^{2}} = q\) Where: \(k\) is the thermal conductivity, \(T\) is the temperature, \(r\) is the radial distance from the pipe's centerline, and \(q\) is the heat generation rate per unit volume.
02

2. Calculate the heat generation rate per unit volume

The heat generation rate per unit volume, \(q\), in the entire pipe wall is given by: \(q = \frac{P}{V} = \frac{P}{\pi (r_{2}^2 - r_{1}^2) L}\) We are given the power \(P = 25,000 \mathrm{~W}\), \(r_{1} = 15 \mathrm{~cm}\), \(r_{2} = 20\mathrm{~cm}\), and \(L = 17 \mathrm{~m}\). Converting radii to meters: \(r_{1} = 0.15 \mathrm{~m}\) and \(r_{2} = 0.2 \mathrm{~m}\). Now, calculate the heat generation rate per unit volume: \(q = \frac{25,000}{\pi (0.2^2 - 0.15^2)(17)} = 26384.4635 \mathrm{~W/m^3}\)
03

3. Solving the heat equation

Inserting the calculated value of \(q\) into the steady-state heat equation, we have: \(-14 \frac{d^{2}T}{dr^{2}} = 26384.4635\) Now we integrate it twice with respect to \(r\) to obtain the expression for temperature distribution: \(\frac{d^{2}T}{dr^{2}} = -1884.61\) Integrate once: \(\frac{dT}{dr} = -1884.61 r + C_{1}\) Integrate again: \(T(r) = -942.305 r^{2} + C_{1} r + C_{2}\)
04

4. Apply boundary conditions

We have two boundary conditions: \(T_1 = T(r_1) = 60^{\circ}\)C and \(T_2 = T(r_2) = 80^{\circ}\)C. Apply the first boundary condition to find \(C_1\) and \(C_2\): \(60 = -942.305 (0.15)^2 + C_{1}(0.15) + C_{2}\) Apply the second boundary condition to find \(C_1\) and \(C_2\): \(80 = -942.305 (0.2)^2 + C_{1}(0.2) + C_{2}\) Solve this system of linear equations to find \(C_{1} = 5184.37\) and \(C_{2} = 156.52\).
05

5. Complete the temperature distribution expression

Now that we have the values for \(C_1\) and \(C_2\), we can write the complete expression for the temperature distribution inside the pipe: \(T(r) = -942.305 r^{2} + 5184.37 r + 156.52\)
06

6. Find the temperature at the center plane of the pipe

We want to find the temperature at the radial distance \(r =\) halfway between \(r_1\) and \(r_2\). So, let's use \(r = (r_1 + r_2) / 2 = (0.15 + 0.20) / 2 = 0.175 \mathrm{~m}\). Evaluate the temperature distribution expression at \(r = 0.175\): \(T(0.175) = -942.305 (0.175)^2 + 5184.37 (0.175) + 156.52 = 70.12^{\circ}\mathrm{C}\) The temperature at the center plane of the pipe is \(70.12^{\circ}\mathrm{C}\).

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Most popular questions from this chapter

A large steel plate having a thickness of \(L=4\) in, thermal conductivity of \(k=7.2 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\), and an emissivity of \(\varepsilon=0.7\) is lying on the ground. The exposed surface of the plate at \(x=L\) is known to exchange heat by convection with the ambient air at \(T_{\infty}=90^{\circ} \mathrm{F}\) with an average heat transfer coefficient of \(h=12 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\) as well as by radiation with the open sky with an equivalent sky temperature of \(T_{\text {sky }}=480 \mathrm{R}\). Also, the temperature of the upper surface of the plate is measured to be \(80^{\circ} \mathrm{F}\). Assuming steady onedimensional heat transfer, \((a)\) express the differential equation and the boundary conditions for heat conduction through the plate, \((b)\) obtain a relation for the variation of temperature in the plate by solving the differential equation, and \((c)\) determine the value of the lower surface temperature of the plate at \(x=0\).

Heat is generated in a \(3-\mathrm{cm}\)-diameter spherical radioactive material uniformly at a rate of \(15 \mathrm{~W} / \mathrm{cm}^{3}\). Heat is dissipated to the surrounding medium at \(25^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(120 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The surface temperature of the material in steady operation is (a) \(56^{\circ} \mathrm{C}\) (b) \(84^{\circ} \mathrm{C}\) (c) \(494^{\circ} \mathrm{C}\) (d) \(650^{\circ} \mathrm{C}\) (e) \(108^{\circ} \mathrm{C}\)

Consider a large plane wall of thickness \(L=0.4 \mathrm{~m}\), thermal conductivity \(k=1.8 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and surface area \(A=\) \(30 \mathrm{~m}^{2}\). The left side of the wall is maintained at a constant temperature of \(T_{1}=90^{\circ} \mathrm{C}\) while the right side loses heat by convection to the surrounding air at \(T_{\infty}=25^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(h=24 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming constant thermal conductivity and no heat generation in the wall, \((a)\) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the wall, \((b)\) obtain a relation for the variation of temperature in the wall by solving the differential equation, and \((c)\) evaluate the rate of heat transfer through the wall. Answer: (c) \(7389 \mathrm{~W}\)

A spherical container, with an inner radius of \(1 \mathrm{~m}\) and a wall thickness of \(5 \mathrm{~mm}\), has its inner surface subjected to a uniform heat flux of \(7 \mathrm{~kW} / \mathrm{m}^{2}\). The outer surface of the container is maintained at \(20^{\circ} \mathrm{C}\). The container wall is made of a material with a thermal conductivity given as \(k(T)=k_{0}(1+\beta T)\), where \(k_{0}=1.33 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \beta=0.0023 \mathrm{~K}^{-1}\), and \(T\) is in \(\mathrm{K}\). Determine the temperature drop across the container wall thickness.

Consider a large plane wall of thickness \(L=0.8 \mathrm{ft}\) and thermal conductivity \(k=1.2 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\). The wall is covered with a material that has an emissivity of \(\varepsilon=0.80\) and a solar absorptivity of \(\alpha=0.60\). The inner surface of the wall is maintained at \(T_{1}=520 \mathrm{R}\) at all times, while the outer surface is exposed to solar radiation that is incident at a rate of \(\dot{q}_{\text {solar }}=300 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}\). The outer surface is also losing heat by radiation to deep space at \(0 \mathrm{~K}\). Determine the temperature of the outer surface of the wall and the rate of heat transfer through the wall when steady operating conditions are reached.
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