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Chapter 2: Problem 172

The temperatures at the inner and outer surfaces of a \(15-\mathrm{cm}\)-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
Answer: The correct expression for the steady, one-dimensional temperature variation in the wall is T(x) = -0.8x + 40.

Step by step solution

01

Set up the general equation for temperature distribution.

First, we set up the general equation for the steady-state, one-dimensional temperature distribution, which is: $$ T(x) = Ax + B $$ Where \(T(x)\) is the temperature at a position \(x\) from the inner surface, \(A\) is a linear coefficient (rate of temperature change) and \(B\) is the constant term (base temperature).
02

Use boundary conditions to solve for A and B.

We are given the temperatures at the inner and outer surfaces of the wall, which are \(T(0) = 40\)°C and \(T(15) = 28\)°C, respectively. We will use these values as boundary conditions in our general equation to find the values of \(A\) and \(B\). At the inner surface \((x = 0)\), the temperature is \(40\)°C. Plug these values into the equation: $$ 40 = A \cdot 0 + B \Rightarrow B = 40 $$ At the outer surface \((x = 15)\), the temperature is \(28\)°C. Plug these values into the equation, replacing \(B\) with \(40\): $$ 28 = A \cdot 15 + 40 \Rightarrow A \cdot 15 = -12 \Rightarrow A = -\frac{12}{15} = -0.8$$
03

Plug the values of A and B into the equation.

Now that we have the values of \(A\) and \(B\), plug them into the equation \(T(x) = Ax + B\): $$ T(x) = -0.8x + 40 $$
04

Compare the resulting equation to the given options.

We compare the resulting equation with the options given in the exercise: (a) \(T(x)=28 x+40\) (Incorrect) (b) \(T(x)=-40 x+28\) (Incorrect) (c) \(T(x)=40 x+28\) (Incorrect) (d) \(T(x)=-80 x+40\) (Incorrect) (e) \(T(x)=40 x-80\) (Incorrect) None of the given options match our resulting equation \(T(x) = -0.8x + 40\). There might be an error in the options provided. The correct expression for the steady, one-dimensional temperature variation in the wall should be: $$ T(x) = -0.8x + 40 $$

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Most popular questions from this chapter

Consider a medium in which the heat conduction equation is given in its simplest form as $$ \frac{1}{r} \frac{d}{d r}\left(r k \frac{d T}{d r}\right)+\dot{e}_{\text {gen }}=0 $$ (a) Is heat transfer steady or transient? (b) Is heat transfer one-, two-, or three-dimensional? (c) Is there heat generation in the medium? (d) Is the thermal conductivity of the medium constant or variable?

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.

A solar heat flux \(\dot{q}_{s}\) is incident on a sidewalk whose thermal conductivity is \(k\), solar absorptivity is \(\alpha_{s^{+}}\)and convective heat transfer coefficient is \(h\). Taking the positive \(x\)-direction to be toward the sky and disregarding radiation exchange with the surrounding surfaces, the correct boundary condition for this sidewalk surface is (a) \(-k \frac{d T}{d x}=\alpha_{s} \dot{q}_{s}\) (b) \(-k \frac{d T}{d x}=h\left(T-T_{\infty}\right)\) (c) \(-k \frac{d T}{d x}=h\left(T-T_{o c}\right)-\alpha_{s} \dot{q}_{s}\) (d) \(h\left(T-T_{\infty}\right)=\alpha_{s} \dot{q}_{s}\) (e) None of them

A large plane wall has a thickness \(L=50 \mathrm{~cm}\) and thermal conductivity \(k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). On the left surface \((x=0)\), it is subjected to a uniform heat flux \(\dot{q}_{0}\) while the surface temperature \(T_{0}\) is constant. On the right surface, it experiences convection and radiation heat transfer while the surface temperature is \(T_{L}=225^{\circ} \mathrm{C}\) and the surrounding temperature is \(25^{\circ} \mathrm{C}\). The emissivity and the convection heat transfer coefficient on the right surface are \(0.7\) and $15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, respectively. (a) Show that the variation of temperature in the wall can be expressed as $T(x)=\left(\dot{q}_{0} / k\right)(L-x)+T_{L},(b)\( calculate the heat flux \)\dot{q}_{0}$ on the left face of the wall, and (c) determine the temperature of the left surface of the wall at \(x=0\).

Heat is generated in a long \(0.3-\mathrm{cm}\)-diameter cylindrical electric heater at a rate of \(150 \mathrm{~W} / \mathrm{m}^{3}\). The heat flux at the surface of the heater in steady operation is (a) \(42.7 \mathrm{~W} / \mathrm{cm}^{2}\) (b) \(159 \mathrm{~W} / \mathrm{cm}^{2}\) (c) \(150 \mathrm{~W} / \mathrm{cm}^{2}\) (d) \(10.6 \mathrm{~W} / \mathrm{cm}^{2}\) (e) \(11.3 \mathrm{~W} / \mathrm{cm}^{2}\)
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