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Chapter 1: Problem 94

Consider a 20-cm thick granite wall with a thermal conductivity of \(2.79 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The temperature of the left surface is held constant at \(50^{\circ} \mathrm{C}\), whereas the right face is exposed to a flow of \(22^{\circ} \mathrm{C}\) air with a convection heat transfer coefficient of \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Neglecting heat transfer by radiation, find the right wall surface temperature and the heat flux through the wall.

Short Answer

Expert verified
Answer: The right wall surface temperature is approximately 38.42°C, and the heat flux through the wall is approximately 20.94 W/m².

Step by step solution

01

List the given information and variables to find

We are given the following information: - Wall thickness: \(d = 20 \mathrm{cm} = 0.2 \mathrm{m}\) - Thermal conductivity: \(k = 2.79 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) - Temperature of left surface: \(T_{1} = 50^{\circ} \mathrm{C}\) - Temperature of air: \(T_{3} = 22^{\circ} \mathrm{C}\) - Convection heat transfer coefficient: \(h = 15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) We need to find: - Right wall surface temperature: \(T_{2}\) - Heat flux through the wall: \(q\)
02

Use one-dimensional heat conduction equation

Use the one-dimensional heat conduction equation to find the heat flux through the wall due to conduction: \(q_{cond} = k \frac{T_{1} - T_{2}}{d}\)
03

Use convection heat transfer equation

Use the convection heat transfer equation to find the heat transfer rate due to convection: \(q_{conv} = h (T_{2} - T_{3})\)
04

Equate heat transfer rates and solve for \(T_{2}\)

Since there is no heat transfer by radiation, the heat transfer rate due to conduction and convection are equal: \(q_{cond} = q_{conv}\) Substitute the equations from steps 2 and 3, and rearrange to solve for \(T_{2}\): \(k \frac{T_{1} - T_{2}}{d} = h (T_{2} - T_{3})\) \(k (T_{1} - T_{2}) = hd (T_{2} - T_{3})\) \((k + hd) T_{2} = kT_{1} + hdT_{3}\) \(T_{2} = \frac{kT_{1} + hdT_{3}}{k + hd}\) Plug in the given values: \(T_{2} = \frac{(2.79)(50) + (15)(0.2)(22)}{2.79 + (15)(0.2)}\) \(T_{2} ≈ 38.42^{\circ} \mathrm{C}\)
05

Find the heat flux through the wall

Now that we have found the right wall surface temperature, we can use the equation from step 2 to find the heat flux through the wall: \(q = k \frac{T_{1} - T_{2}}{d}\) \(q = 2.79 \frac{50 - 38.42}{0.2}\) \(q ≈ 20.94 \mathrm{~W} / \mathrm{m}^{2}\) The right wall surface temperature is approximately \(38.42^{\circ} \mathrm{C}\), and the heat flux through the wall is approximately \(20.94 \mathrm{~W} / \mathrm{m}^{2}\).

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Most popular questions from this chapter

Consider steady heat transfer between two large parallel plates at constant temperatures of \(T_{1}=290 \mathrm{~K}\) and \(T_{2}=150 \mathrm{~K}\) that are \(L=2 \mathrm{~cm}\) apart. Assuming the surfaces to be black (emissivity \(\varepsilon=1\) ), determine the rate of heat transfer between the plates per unit surface area assuming the gap between the plates is (a) filled with atmospheric air, \((b)\) evacuated, \((c)\) filled with fiberglass insulation, and \((d)\) filled with superinsulation having an apparent thermal conductivity of \(0.00015 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

Can all three modes of heat transfer occur simultaneously (in parallel) in a medium?

A concrete wall with a surface area of \(20 \mathrm{~m}^{2}\) and a thickness of \(0.30 \mathrm{~m}\) separates conditioned room air from ambient air. The temperature of the inner surface of the wall \(\left(T_{1}\right)\) is maintained at \(25^{\circ} \mathrm{C}\). (a) Determine the heat loss \(\dot{Q}(\mathrm{~W})\) through the concrete wall for three thermal conductivity values of \((0.75,1\), and \(1.25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and outer wall surface temperatures of \(T_{2}=-15,-10,-5,0,5,10,15,20,25,30\), and \(38^{\circ} \mathrm{C}\) (a total of 11 data points for each thermal conductivity value). Tabulate the results for all three cases in one table. Also provide a computer generated graph [Heat loss, \(\dot{Q}(\mathrm{~W})\) vs. Outside wall temperature, \(\left.T_{2}\left({ }^{\circ} \mathrm{C}\right)\right]\) for the display of your results. The results for all three cases should be plotted on the same graph. (b) Discuss your results for the three cases.

Consider a flat-plate solar collector placed on the roof of a house. The temperatures at the inner and outer surfaces of the glass cover are measured to be \(33^{\circ} \mathrm{C}\) and \(31^{\circ} \mathrm{C}\), respectively. The glass cover has a surface area of \(2.5 \mathrm{~m}^{2}\), a thickness of \(0.6 \mathrm{~cm}\), and a thermal conductivity of \(0.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Heat is lost from the outer surface of the cover by convection and radiation with a convection heat transfer coefficient of \(10 \mathrm{~W} /\) \(\mathrm{m}^{2} \cdot \mathrm{K}\) and an ambient temperature of \(15^{\circ} \mathrm{C}\). Determine the fraction of heat lost from the glass cover by radiation.

Consider heat transfer through a windowless wall of a house on a winter day. Discuss the parameters that affect the rate of heat conduction through the wall.
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