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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路K) and the 11 outer wall surface temperatures. | T2 (掳C) | Q (0.75 W/m路K) (W) | Q (1 W/m路K) (W) | Q (1.25 W/m路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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Key Concepts

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

Thermal Conductivity
Thermal conductivity is a measure of a material's ability to conduct heat. This property is vital in determining how quickly heat can transfer through a material. In this exercise, we looked at a concrete wall with varying thermal conductivity values of 0.75, 1, and 1.25 W/m路K. Higher values of thermal conductivity indicate that heat moves more easily through the material. For our concrete wall, as the thermal conductivity increases, the rate of heat transfer from the warm interior to the cold exterior also increases. This is because more heat will be conducted per unit time for a higher thermal conductivity value. This principle can be applied to numerous materials, guiding decisions in construction, insulation, and thermal management.
Concrete Wall Heat Loss
Heat loss through a concrete wall is determined by how much heat moves from one side of the wall to the other. The key factors affecting this are the wall's surface area, thickness, the temperature difference across the wall, and the material鈥檚 thermal conductivity. The formula to calculate heat loss (\(\dot{Q}\)) through a flat wall is given by: \(\dot{Q} = k \cdot \frac{A(T_1 - T_2)}{L}\). Here, \(k\) is the thermal conductivity, \(A\) is the surface area, \(T_1 - T_2\) is the temperature difference, and \(L\) is the thickness of the wall.
By using this formula in our exercise, we computed the heat loss for different external temperatures and for different thermal conductivities. Large temperature differences lead to higher heat loss since \(T_1 - T_2\) is a critical component in the formula. It is important to manage heat loss to reduce energy consumption and maintain comfortable indoor temperatures.
Temperature Gradient
The temperature gradient is the rate at which temperature changes in relation to distance through the material. It is expressed as \(\frac{\Delta T}{L}\), where \(\Delta T\) is the temperature difference and \(L\) is the material thickness. In our concrete wall exercise, the interior side is at a higher temperature compared to the exterior.
Understanding the temperature gradient helps in predicting the heat flow direction and magnitude. A steeper gradient indicates faster heat flow. Temperature gradients are crucial in engineering applications to predict heat transfer effectively, allowing for better design of structures to minimize unnecessary heat loss and maintain energy efficiency.
Graphical Data Analysis
Graphical data analysis in this context involves plotting the computed heat loss against the varying external temperatures on a graph. This helps in visualizing the relationship between outside temperature, heat loss, and thermal conductivity. For the exercise, plotting all three thermal conductivity values on the same graph allowed us to clearly see how each affects the heat loss.
The graph shows how heat loss reduces as the external temperature (\(T_2\)) increases towards the internal temperature (\(T_1\)). By observing the slope and curve of each line, you can determine the efficiency of the concrete wall as an insulator at different conditions. This kind of analysis aids in making informed decisions, such as choosing the right materials for specific environmental conditions to ensure optimal insulation performance.

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Most popular questions from this chapter

Using the parametric table and plot features of \(\mathrm{EES}\), determine the squares of the number from 1 to 100 in increments of 10 in tabular form, and plot the results.

Consider a person whose exposed surface area is \(1.7 \mathrm{~m}^{2}\), emissivity is \(0.5\), and surface temperature is \(32^{\circ} \mathrm{C}\). Determine the rate of heat loss from that person by radiation in a large room having walls at a temperature of \((a) 300 \mathrm{~K}\) and (b) \(280 \mathrm{~K}\). Answers: (a) \(26.7 \mathrm{~W}\), (b) \(121 \mathrm{~W}\)

A person standing in a room loses heat to the air in the room by convection and to the surrounding surfaces by radiation. Both the air in the room and the surrounding surfaces are at \(20^{\circ} \mathrm{C}\). The exposed surface of the person is \(1.5 \mathrm{~m}^{2}\) and has an average temperature of \(32^{\circ} \mathrm{C}\), and an emissivity of \(0.90\). If the rates of heat transfer from the person by convection and by radiation are equal, the combined heat transfer coefficient is (a) \(0.008 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (b) \(3.0 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(5.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (d) \(8.3 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(10.9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

A soldering iron has a cylindrical tip of \(2.5 \mathrm{~mm}\) in diameter and \(20 \mathrm{~mm}\) in length. With age and usage, the tip has oxidized and has an emissivity of \(0.80\). Assuming that the average convection heat transfer coefficient over the soldering iron tip is \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and the surrounding air temperature is \(20^{\circ} \mathrm{C}\), determine the power required to maintain the tip at \(400^{\circ} \mathrm{C}\).

Consider a person standing in a room at \(18^{\circ} \mathrm{C}\). Determine the total rate of heat transfer from this person if the exposed surface area and the skin temperature of the person are \(1.7 \mathrm{~m}^{2}\) and \(32^{\circ} \mathrm{C}\), respectively, and the convection heat transfer coefficient is \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Take the emissivity of the skin and the clothes to be \(0.9\), and assume the temperature of the inner surfaces of the room to be the same as the air temperature.
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