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Chapter 2: Problem 165

The temperatures at the inner and outer surfaces of a 15 -cm-thick plane wall are measured to be \(40^{\circ} \mathrm{C}\) and \(28^{\circ} \mathrm{C}\), respectively. The expression for steady, one-dimensional variation of temperature in the wall is (a) \(T(x)=28 x+40\) (b) \(T(x)=-40 x+28\) (c) \(T(x)=40 x+28\) (d) \(T(x)=-80 x+40\) (e) \(T(x)=40 x-80\)

Short Answer

Expert verified
a) T(x) = 28x + 40 b) T(x) = -40x + 28 c) T(x) = 40x + 28 d) T(x) = -80x + 40 e) T(x) = 40x - 80 Answer: None of the given expressions satisfy both given temperature values at the inner and outer surfaces. There might be an error in the exercise or the given options.

Step by step solution

01

Identify given temperature values at wall surfaces

We are given the inner surface temperature \(T_1=40^{\circ} \mathrm{C}\) and the outer surface temperature \(T_2=28^{\circ} \mathrm{C}\). The wall has a thickness (\(x\)) of 15 cm.
02

Check which expression satisfies the given temperature values

We will plug in the distances \(x=0\) (inner surface) and \(x=15\) cm (outer surface) into each of the given expressions and see which expression satisfies the given temperature values. (a) \(T(x)=28x+40\) When \(x=0\), \(T(0)=28(0)+40=40^{\circ} \mathrm{C}\) (satisfies inner surface) When \(x=15\), \(T(15)=28(15)+40=460^{\circ} \mathrm{C}\) (does not satisfy outer surface) (b) \(T(x)=-40x+28\) When \(x=0\), \(T(0)=-40(0)+28=28^{\circ} \mathrm{C}\) (does not satisfy inner surface) (c) \(T(x)=40x+28\) When \(x=0\), \(T(0)=40(0)+28=28^{\circ} \mathrm{C}\) (does not satisfy inner surface) (d) \(T(x)=-80x+40\) When \(x=0\), \(T(0)=-80(0)+40=40^{\circ} \mathrm{C}\) (satisfies inner surface) When \(x=15\), \(T(15)=-80(15)+40=-1160^{\circ} \mathrm{C}\) (does not satisfy outer surface) (e) \(T(x)=40x-80\) When \(x=0\), \(T(0)=40(0)-80=-40^{\circ} \mathrm{C}\) (does not satisfy inner surface)
03

Select the correct expression

Based on our testing in Step 2, none of the given expressions satisfy both temperature values at the inner and outer surfaces. Therefore, there might be an error in the exercise or the given options. If the problem arises from the given options, it is crucial to inform the student that none of the given options are correct and to report the issue in the exercise.

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Key Concepts

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

Understanding the Temperature Gradient
The concept of a temperature gradient is essential when discussing heat transfer, particularly through materials like walls. A temperature gradient refers to the change in temperature with respect to distance. In our exercise, this is seen as the change from 40掳C at the inner surface of the wall to 28掳C at the outer surface, across a 15 cm thickness.
For heat to flow continuously, there must be a difference in temperature - this is our gradient.
  • The gradient is calculated by subtracting the lower temperature from the higher one and then dividing by the distance over which this change occurs. For instance, here it would be \(\frac{40 - 28}{15}\) 掳C/cm.
  • Since the gradient is negative, it confirms that temperature decreases as one moves from the inner surface to the outer surface.
This temperature difference drives heat transfer through the material.
Exploring Steady-State Conduction
In thermal conduction, steady-state refers to a condition where the temperature distribution does not change over time. This is a fundamental concept in thermal analysis and is particularly relevant in our exercise where we assume that the temperatures at the surfaces remain constant.
During steady-state conduction:
  • Heat transfer through the wall is constant, meaning the same amount of heat enters one side and exits the other.
  • The temperature profile remains linear for homogenous materials, indicating a consistent rate of heat loss or gain along the path.
In the exercise problem, if temperatures are constant at both ends over time, it indicates that the wall has reached a steady state. Understanding this helps identify why the given temperature equations in the exercise might not represent a realistic steady-state condition.
Applying Thermal Analysis to Problem-Solving
Thermal analysis involves using mathematical models to predict temperature distribution, heat flow, and other thermal properties in materials. In the exercise scenario, the goal of thermal analysis is to determine which mathematical expression correctly describes the temperature variation across the wall.
Given:
  • Inner temperature: 40掳C
  • Outer temperature: 28掳C
  • Wall thickness: 15 cm
When applying thermal analysis, we attempt to understand the realistic flow of heat. By plugging different distance values into potential equations, we try to match the calculated and given temperature values. In this problem, the expressions given don't satisfy both the inner and outer surface temperatures, suggesting a miscalculation or error in provided options.
Thermal analysis is critical in diagnosing issues like this, ensuring that we understand whether a condition is physically possible and consistent with the principles of steady-state conduction and temperature gradients.

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Most popular questions from this chapter

A spherical metal ball of radius \(r_{o}\) is heated in an oven to a temperature of \(T_{i}\) throughout and is then taken out of the oven and allowed to cool in ambient air at \(T_{\infty}\) by convection and radiation. The emissivity of the outer surface of the cylinder is \(\varepsilon\), and the temperature of the surrounding surfaces is \(T_{\text {surr }}\). The average convection heat transfer coefficient is estimated to be \(h\). Assuming variable thermal conductivity and transient one-dimensional heat transfer, express the mathematical formulation (the differential equation and the boundary and initial conditions) of this heat conduction problem. Do not solve.

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.

A pipe is used for transporting boiling water in which the inner surface is at \(100^{\circ} \mathrm{C}\). The pipe is situated in surroundings where the ambient temperature is \(10^{\circ} \mathrm{C}\) and the convection heat transfer coefficient is \(70 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The wall thickness of the pipe is \(3 \mathrm{~mm}\) and its inner diameter is \(30 \mathrm{~mm}\). The pipe wall has a variable thermal conductivity given as \(k(T)=k_{0}(1+\beta T)\), where \(k_{0}=1.23 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \beta=0.002 \mathrm{~K}^{-1}\), and \(T\) is in \(\mathrm{K}\). For safety reasons and to prevent thermal burn to workers, the outer surface temperature of the pipe should be kept below \(50^{\circ} \mathrm{C}\). Determine whether the outer surface temperature of the pipe is at a safe temperature so as to avoid thermal burn.

Consider a steam pipe of length \(L=30 \mathrm{ft}\), inner radius \(r_{1}=2\) in, outer radius \(r_{2}=2.4\) in, and thermal conductivity \(k=7.2 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\). Steam is flowing through the pipe at an average temperature of \(300^{\circ} \mathrm{F}\), and the average convection heat transfer coefficient on the inner surface is given to be \(h=12.5 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}\). If the average temperature on the outer surfaces of the pipe is \(T_{2}=175^{\circ} \mathrm{F},(a)\) 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.

Does heat generation in a solid violate the first law of thermodynamics, which states that energy cannot be created or destroyed? Explain.
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