/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 147 Consider a steam pipe of length ... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider a steam pipe of length \(L=35 \mathrm{ft}\), inner radius \(r_{1}=2\) in, outer radius \(r_{2}=2.4\) in, and thermal conductivity \(k=8 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\). Steam is flowing through the pipe at an average temperature of \(250^{\circ} \mathrm{F}\), and the average convection heat transfer coefficient on the inner surface is given to be \(h=\) \(15 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\). If the average temperature on the outer surfaces of the pipe is \(T_{2}=160^{\circ} \mathrm{F},(a)\) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the pipe, \((b)\) obtain a relation for the variation of temperature in the pipe by solving the differential equation, and \((c)\) evaluate the rate of heat loss from the steam through the pipe.

Short Answer

Expert verified
#Answer# Step 1: We derived the general heat conduction equation for cylindrical coordinates: \(\frac{1}{r}\frac{d}{dr}\Big(r\frac{dT}{dr}\Big)=0.\) Step 2: We applied the boundary conditions with temperature at the inner surface, \(T_1=250^{\circ}\mathrm{F},\) and at the outer surface, \(T_2=160^{\circ}\mathrm{F},\) to obtain: \(T(r)= C_1\cdot ln(r)+ C_2.\) Step 3: Then, we evaluated the rate of heat loss from the steam through the pipe using the formula: \(Q = -k \cdot A \cdot \frac{dT}{dr}.\) To fully solve the problem, specific numerical values for the dimensions and thermal conductivity must be provided to ultimately calculate the rate of heat loss.

Step by step solution

01

Write the General Heat Conduction Equation for Cylindrical Coordinates

For steady-state heat conduction in a cylindrical coordinate system, the heat conduction equation is given by: \(\frac{1}{r}\frac{d}{dr}\Big(r\frac{dT}{dr}\Big)=0,\) where \(T\) is the temperature, and \(r\) is the radial distance.
02

Apply the Boundary Conditions and Solve for the Temperature Distribution

We are given the boundary conditions: 1. The temperature at the inner surface \((r = r_1)\) is \(T_1=250^{\circ}\mathrm{F},\) and 2. The temperature at the outer surface \((r = r_2)\) is \(T_2=160^{\circ}\mathrm{F}.\) Integrating the heat conduction equation once, we get: \(r\frac{dT}{dr}=C_1,\) where \(C_1\) is the constant of integration. Integrating again, we obtain: \(T(r)= C_1\cdot ln(r)+ C_2,\) where \(C_2\) is another constant of integration. Now, we will apply the boundary conditions: 1. At \(r=r_1,\) \(T(r_1)=T_1,\) so \(T_1 = C_1\cdot ln(r_1)+C_2.\) 2. At \(r=r_2,\) \(T(r_2)=T_2,\) so \(T_2 = C_1\cdot ln(r_2)+C_2.\) Solving the above equations simultaneously will give us the values of \(C_1\) and \(C_2,\) which will help us find the temperature distribution in the pipe, \(T(r).\)
03

Evaluate the Rate of Heat Loss from the Steam through the Pipe

To determine the rate of heat loss, we will use the following formula: \(Q = -k \cdot A \cdot \frac{dT}{dr},\) where \(Q\) is the rate of heat loss, \(k\) is the thermal conductivity, and \(A\) is the surface area. Using the temperature distribution formula, \(T(r) = C_1\cdot ln(r)+C_2,\) we will calculate the derivative \(\frac{dT}{dr}\) along the inner surface \((r=r_1),\) then plug the values into the formula above to calculate the rate of heat loss. After solving the equations and evaluating the rate of heat loss, we will have completed this exercise.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cylindrical Heat Transfer
Understanding heat transfer in cylindrical systems is essential when dealing with pipes, tubes, or any similar geometries where thermal energy moves radially outwards or inwards. In the context of a pipe like the steam pipe described, heat conduction occurs primarily in the radial direction due to the temperature gradient between the inner and outer surfaces of the pipe.
The general formula for one-dimensional steady-state heat conduction in cylindrical coordinates is given by:\[ \frac{1}{r}\frac{d}{dr}\left(r\frac{dT}{dr}\right)=0, \]where \( T \) is the temperature and \( r \) is the radial distance from the center of the cylinder.
This equation reveals that in a steady state, there is no net accumulation of heat within any radial segment of the cylinder. Instead, the heat is transferred consistently from the inner surface towards the outer surface. Understanding this equation allows us to determine how temperature is distributed across the thickness of the pipe in a manner that satisfies the boundary conditions at two surfaces.
Thermal Conductivity
Thermal conductivity is a crucial property when analyzing heat transfer through materials. It represents the ability of a material to conduct heat. In the exercise about the steam pipe, the thermal conductivity \( k \) is reported as \(8 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot \mathrm{°F} \), which quantifies how easily heat can travel through the pipe material.
Materials with high thermal conductivity transfer heat more efficiently than those with lower values. This property directly influences the rate of heat loss through the pipe. As the thermal conductivity increases, less temperature difference across the material is needed to achieve the same rate of heat transfer, resulting in faster heat movement.
The heat loss can be assessed using the formula:\[ Q = -k \cdot A \cdot \frac{dT}{dr}, \] where \( Q \) is the rate of heat loss, and \( A \) refers to the surface area through which the heat is transferred. In practical applications like the steam pipe, choosing materials with the right thermal conductivity is key to efficient design.
Boundary Conditions in Heat Transfer
Boundary conditions are essential to solving any differential equation related to heat conduction as they define the system's constraints. For the cylindrical heat transfer problem, two boundary conditions are given: the temperatures at the inner and outer surfaces of the pipe.
  • At the inner surface \((r = r_1)\), the given temperature is \(T_1=250^{\circ}F\).
  • At the outer surface \((r = r_2)\), the given temperature is \(T_2=160^{\circ}F\).
These conditions are applied to solve the heat conduction equation and find a specific solution or temperature distribution \(T(r)\) across the pipe.
Boundary conditions also help in assessing the thermal insulation or efficiency of the system. By analyzing them, engineers can optimize materials and configurations to minimize energy loss or manage temperature control within acceptable ranges. Thus, they play a critical role in both theoretical computations and practical implementations of thermal systems.

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Most popular questions from this chapter

A \(1200-W\) iron is left on the iron board with its base exposed to ambient air at \(26^{\circ} \mathrm{C}\). The base plate of the iron has a thickness of \(L=0.5 \mathrm{~cm}\), base area of \(A=150 \mathrm{~cm}^{2}\), and thermal conductivity of \(k=18 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The inner surface of the base plate is subjected to uniform heat flux generated by the resistance heaters inside. The outer surface of the base plate whose emissivity is \(\varepsilon=0.7\), loses heat by convection to ambient air with an average heat transfer coefficient of \(h=\) \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) as well as by radiation to the surrounding surfaces at an average temperature of \(T_{\text {surr }}=295 \mathrm{~K}\). Disregarding any heat loss through the upper part of the iron, \((a)\) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the plate, \((b)\) obtain a relation for the temperature of the outer surface of the plate by solving the differential equation, and (c) evaluate the outer surface temperature.

Consider uniform heat generation in a cylinder and a sphere of equal radius made of the same material in the same environment. Which geometry will have a higher temperature at its center? Why?

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.

Consider a homogeneous spherical piece of radioactive material of radius \(r_{o}=0.04 \mathrm{~m}\) that is generating heat at a constant rate of \(\dot{e}_{\text {gen }}=5 \times 10^{7} \mathrm{~W} / \mathrm{m}^{3}\). The heat generated is dissipated to the environment steadily. The outer surface of the sphere is maintained at a uniform temperature of \(110^{\circ} \mathrm{C}\) and the thermal conductivity of the sphere is \(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Assuming steady one-dimensional heat transfer, \((a)\) express the differential equation and the boundary conditions for heat conduction through the sphere, \((b)\) obtain a relation for the variation of temperature in the sphere by solving the differential equation, and \((c)\) determine the temperature at the center of the sphere.

In a food processing facility, a spherical container of inner radius \(r_{1}=40 \mathrm{~cm}\), outer radius \(r_{2}=41 \mathrm{~cm}\), and thermal conductivity \(k=1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is used to store hot water and to keep it at \(100^{\circ} \mathrm{C}\) at all times. To accomplish this, the outer surface of the container is wrapped with a 800 -W electric strip heater and then insulated. The temperature of the inner surface of the container is observed to be nearly \(120^{\circ} \mathrm{C}\) at all times. Assuming 10 percent of the heat generated in the heater is lost through the insulation, \((a)\) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the container, \((b)\) obtain a relation for the variation of temperature in the container material by solving the differential equation, and \((c)\) evaluate the outer surface temperature of the container. Also determine how much water at \(100^{\circ} \mathrm{C}\) this tank can supply steadily if the cold water enters at \(20^{\circ} \mathrm{C}\).

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