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Chapter 2: Problem 154

A pipe is used for transporting boiling water in which the inner surface is at \(100^{\circ} \mathrm{C}\). The pipe is situated in a surrounding where the ambient temperature is \(20^{\circ} \mathrm{C}\) and the convection heat transfer coefficient is \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The pipe has a wall thickness of \(3 \mathrm{~mm}\) and an inner diameter of \(25 \mathrm{~mm}\), and it has a variable thermal conductivity given as \(k(T)=k_{0}(1+\beta T)\), where \(k_{0}=1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \beta=0.003 \mathrm{~K}^{-1}\) and \(T\) is in \(\mathrm{K}\). Determine the outer surface temperature of the pipe.

Short Answer

Expert verified
Answer: The outer surface temperature of the pipe is approximately \(66.23 ^{\circ} \mathrm{C}\).

Step by step solution

01

Convert temperatures to Kelvin

First, we must convert the inner surface and ambient temperatures to Kelvin. \(T_{i} = 100^{\circ} \mathrm{C} + 273.15 = 373.15 \mathrm{K}\) \(T_{\infty} = 20^{\circ} \mathrm{C} + 273.15 = 293.15 \mathrm{K}\)
02

Calculate the heat transfer through the pipe using convection

Using the convection heat transfer formula, we can find the heat transfer, \(q\): \(q = hA(T_{i} - T_{\infty})\) Here, \(A = \pi dL\), where \(d = 25 \times 10^{-3} \mathrm{m}\) is the pipe's inner diameter and \(L\) is the length of the pipe. \(q = 50 \cdot \pi(25 \times 10^{-3})L (373.15 - 293.15)\)
03

Calculate the thermal conductivity at the average temperature

First, we need to find the average temperature between the inner surface and ambient temperatures: \(T_{avg} = \frac{T_{i} + T_{\infty}}{2} = 333.15 \mathrm{K}\) Now, we can calculate the thermal conductivity, \(k(T)\), at the average temperature: \(k(T_{avg}) = k_{0}(1 + \beta T_{avg}) = 1.5(1 + 0.003 \cdot 333.15)\)
04

Calculate the heat transfer through the pipe using conduction

Using the conduction heat transfer formula, we can find the heat transfer, \(q\): \(q = kA\frac{T_{i} - T_{o}}{t}\)
05

Equate the convection and conduction heat transfer equations and solve for the outer surface temperature

Now we set the convection and conduction heat transfer equations equal to each other and solve for \(T_{o}\): \(50 \cdot \pi(25 \times 10^{-3})L (373.15 - 293.15) = 1.5(1 + 0.003 \cdot 333.15) \cdot \pi(25 \times 10^{-3})L \frac{T_{i} - T_{o}}{3 \times 10^{-3}}\) Solve for \(T_{o}\): \(T_{o} = T_{i} - \frac{50(373.15 - 293.15)}{1.5(1 + 0.003 \cdot 333.15) \cdot \frac{1}{3 \times 10^{-3}}} = 339.38 \mathrm{K}\)
06

Convert the outer surface temperature back to Celsius

Our final step is to convert \(T_{o}\) back to Celsius: \(T_{o} = 339.38 \mathrm{K} - 273.15 = 66.23 ^{\circ} \mathrm{C}\) The outer surface temperature of the pipe is approximately \(66.23 ^{\circ} \mathrm{C}\).

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Most popular questions from this chapter

A stainless steel spherical container, with \(k=\) \(15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is used for storing chemicals undergoing exothermic reaction. The reaction provides a uniform heat flux of \(60 \mathrm{~kW} / \mathrm{m}^{2}\) to the container's inner surface. The container has an inner radius of \(50 \mathrm{~cm}\) and a wall thickness of \(5 \mathrm{~cm}\) and is situated in a surrounding with an ambient temperature of \(23^{\circ} \mathrm{C}\). The container's outer surface is subjected to convection heat transfer with a coefficient of \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). For safety reasons to prevent thermal burn to individuals working around the container, it is necessary to keep the container's outer surface temperature below \(50^{\circ} \mathrm{C}\). Determine the variation of temperature in the container wall and the temperatures of the inner and outer surfaces of the container. Is the outer surface temperature of the container safe to prevent thermal burn?

A spherical container, with an inner radius \(r_{1}=1 \mathrm{~m}\) and an outer radius \(r_{2}=1.05 \mathrm{~m}\), has its inner surface subjected to a uniform heat flux of \(\dot{q}_{1}=7 \mathrm{~kW} / \mathrm{m}^{2}\). The outer surface of the container has a temperature \(T_{2}=25^{\circ} \mathrm{C}\), and the container wall thermal conductivity is \(k=1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Show that the variation of temperature in the container wall can be expressed as \(T(r)=\left(\dot{q}_{1} r_{1}^{2} / k\right)\left(1 / r-1 / r_{2}\right)+T_{2}\) and determine the temperature of the inner surface of the container at \(r=r_{1}\).

The conduction equation boundary condition for an adiabatic surface with direction \(n\) being normal to the surface is (a) \(T=0\) (b) \(d T / d n=0\) (c) \(d^{2} T / d n^{2}=0\) (d) \(d^{3} T / d n^{3}=0\) (e) \(-k d T / d n=1\)

Water flows through a pipe at an average temperature of \(T_{\infty}=90^{\circ} \mathrm{C}\). The inner and outer radii of the pipe are \(r_{1}=\) \(6 \mathrm{~cm}\) and \(r_{2}=6.5 \mathrm{~cm}\), respectively. The outer surface of the pipe is wrapped with a thin electric heater that consumes \(400 \mathrm{~W}\) per \(\mathrm{m}\) length of the pipe. The exposed surface of the heater is heavily insulated so that the entire heat generated in the heater is transferred to the pipe. Heat is transferred from the inner surface of the pipe to the water by convection with a heat transfer coefficient of \(h=85 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming constant thermal conductivity and one-dimensional heat transfer, express the mathematical formulation (the differential equation and the boundary conditions) of the heat conduction in the pipe during steady operation. Do not solve.

Consider a cylindrical shell of length \(L\), inner radius \(r_{1}\), and outer radius \(r_{2}\) whose thermal conductivity varies in a specified temperature range as \(k(T)=k_{0}\left(1+\beta T^{2}\right)\) where \(k_{0}\) and \(\beta\) are two specified constants. The inner surface of the shell is maintained at a constant temperature of \(T_{1}\) while the outer surface is maintained at \(T_{2}\). Assuming steady one-dimensional heat transfer, obtain a relation for the heat transfer rate through the shell.
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