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Chapter 1: Problem 61

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.

Short Answer

Expert verified
Short Answer Question: Explain the relationship between the heat loss, outer wall surface temperature, and the thermal conductivity of a concrete wall based on the calculations and graph. Answer: The heat loss through a concrete wall increases with increasing thermal conductivity values and the difference in temperature between the inner and outer wall surfaces. As the thermal conductivity of the wall material increases, the amount of heat transferred through the wall also increases. Moreover, a larger temperature difference between the inner and outer wall surfaces results in more heat loss as well. By analyzing the graph and table, one can observe the mentioned relationships and comprehend the significance of thermal conductivity and temperature difference in heat transfer through a wall.

Step by step solution

01

Apply the formula for heat transfer through a flat wall

The formula to calculate the heat loss through the wall is: \(\dot{Q} = k \cdot \frac{A(T_1 - T_2)}{L}\) We'll use this formula to compute the heat loss for each thermal conductivity value and outer wall surface temperature.
02

Tabulate the results

Organize the data in the table for all three cases of thermal conductivity (0.75, 1, and 1.25 W/m·K) and the 11 outer wall surface temperatures. | T2 (°C) | Q (0.75 W/m·K) (W) | Q (1 W/m·K) (W) | Q (1.25 W/m·K) (W) | |---------|---------------------|----------------|--------------------| | -15 | - | - | - | | -10 | - | - | - | | -5 | - | - | - | | 0 | - | - | - | | 5 | - | - | - | | 10 | - | - | - | | 15 | - | - | - | | 20 | - | - | - | | 25 | - | - | - | | 30 | - | - | - | | 38 | - | - | - | Calculate the heat loss (Q) for each T2 value and replace the '-' signs in the table with the calculated values.
03

Plot the graph of heat loss vs. outside wall temperature

Create a graph depicting the heat loss (Q) on the y-axis and the outside wall temperature (T2) on the x-axis. Plot the results for all three thermal conductivity values on the same graph.
04

Discuss the results

Analyze the graph and the table to discuss the relationship between the heat loss, outer wall surface temperature, and the thermal conductivity values of the wall. Notice the differences between each case and interpret the implications of these differences.

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Most popular questions from this chapter

A solid plate, with a thickness of \(15 \mathrm{~cm}\) and a thermal conductivity of \(80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), is being cooled at the upper surface by air. The air temperature is \(10^{\circ} \mathrm{C}\), while the temperatures at the upper and lower surfaces of the plate are 50 and \(60^{\circ} \mathrm{C}\), respectively. Determine the convection heat transfer coefficient of air at the upper surface and discuss whether the value is reasonable or not for force convection of air.

What is a blackbody? How do real bodies differ from blackbodies?

The rate of heat loss through a unit surface area of a window per unit temperature difference between the indoors and the outdoors is called the \(U\)-factor. The value of the \(U\)-factor ranges from about \(1.25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (or \(0.22 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}\) ) for low-e coated, argon-filled, quadruple-pane windows to \(6.25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (or \(1.1 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}\) ) for a single-pane window with aluminum frames. Determine the range for the rate of heat loss through a \(1.2-\mathrm{m} \times 1.8-\mathrm{m}\) window of a house that is maintained at \(20^{\circ} \mathrm{C}\) when the outdoor air temperature is \(-8^{\circ} \mathrm{C}\).

A \(3-\mathrm{m}^{2}\) black surface at \(140^{\circ} \mathrm{C}\) is losing heat to the surrounding air at \(35^{\circ} \mathrm{C}\) by convection with a heat transfer coefficient of \(16 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and by radiation to the surrounding surfaces at \(15^{\circ} \mathrm{C}\). The total rate of heat loss from the surface is (a) \(5105 \mathrm{~W}\) (b) \(2940 \mathrm{~W}\) (c) \(3779 \mathrm{~W}\) (d) \(8819 \mathrm{~W}\) (e) \(5040 \mathrm{~W}\)

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.
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