/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 A concrete wall, which has a sur... [FREE SOLUTION] | 91Ó°ÊÓ

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A concrete wall, which has a surface area of \(20 \mathrm{~m}^{2}\) and is \(0.30 \mathrm{~m}\) thick, separates conditioned room air from ambient air. The temperature of the inner surface of the wall is maintained at \(25^{\circ} \mathrm{C}\), and the thermal conductivity of the concrete is \(1 \mathrm{~W} / \mathrm{m}=\mathrm{K}\). (a) Determine the heat loss through the wall for outer surface temperatures ranging from \(-15^{\circ} \mathrm{C}\) to \(38^{\circ} \mathrm{C}\), which correspond to winter and summer extremes, respectively. Display your results graphically. (b) On your graph, also plot the heat loss as a function of the outer surface temperature for wall materials having thermal conductivities of \(0.75\) and \(1.25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Explain the family of curves you have obtained.

Short Answer

Expert verified
The heat loss rate (Q) can be calculated for various outer surface temperatures (T2) ranging from -15°C to 38°C using the formula \(Q = kA\frac{(T_2 - T_1)}{t}\), where k is the thermal conductivity of the wall material, A is the surface area, t is the thickness, T1 is the inner surface temperature, and T2 is the outer surface temperature. The graph of heat loss as a function of the outer surface temperature for the given thermal conductivities (1 W/m·K, 0.75 W/m·K, and 1.25 W/m·K) exhibits different slopes based on the varying thermal conductivities. In general, higher thermal conductivities result in greater heat transfer, causing higher heat loss rates.

Step by step solution

01

Identify Known Values

The known values in this exercise are: - Surface area (A) = 20 m² - Wall thickness (t) = 0.30 m - Inner surface temperature (T1) = 25°C - Thermal conductivity of concrete (k) = 1 W/m·K We will solve part (a) first and then move on to part (b).
02

Evaluate Heat Loss Rate for Given Outer Surface Temperatures

We will calculate the heat loss rate for outer surface temperatures ranging from -15°C to 38°C using the formula: \(Q = kA\frac{(T_2 - T_1)}{t}\). For each outer surface temperature (T2) in the given range, we will solve for Q (heat loss rate) using the values provided in Step 1.
03

Graph Heat Loss vs. Outer Surface Temperature

Next, plot a graph with the OUTER SURFACE TEMPERATURE on the x-axis, and HEAT LOSS RATE on the y-axis, using the heat loss rate (Q) obtained for each outer surface temperature of the concrete wall in Step 2.
04

Evaluate Heat Loss Rate for Other Thermal Conductivities

Now for part (b), we will calculate the heat loss rate for the other two thermal conductivities: - k = 0.75 W/m·K - k = 1.25 W/m·K For each outer surface temperature (T2) in the given range (-15°C to 38°C), we will solve for Q (heat loss rate) using the values provided in Step 1 and the new values of the thermal conductivities provided in this step.
05

Add Curves to Graph for Other Thermal Conductivities

Using the heat loss rate (Q) calculated for different thermal conductivities in Step 4, plot curves on the graph from Step 3, representing the heat loss as a function of the outer surface temperature for each thermal conductivity separately.
06

Analyze the Family of Curves

Finally, analyze the obtained family of curves on the graph. Notice how the curves exhibit different slopes and variations in heat loss as a function of the outer surface temperature based on the thermal conductivity of the wall materials. The curve with the highest slope corresponds to the greater thermal conductivity (1.25 W/m·K) and the curve with the lowest slope corresponds to the lower thermal conductivity (0.75 W/m·K). Thus, the wall with greater thermal conductivity will transfer more heat through it, and vice versa.

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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 fundamental property of materials that measures their ability to conduct heat. It is often denoted by the symbol \(k\) and is measured in units of \(W/\text{m} \cdot \text{K}\).
High thermal conductivity means heat passes through the material quickly, while low thermal conductivity means the material is more insulating.
For example, in our problem, concrete has a thermal conductivity of \(1 \text{ W/m} \cdot \text{K}\). This means that for every meter of concrete, 1 watt of heat is transferred per square meter for each degree of temperature difference across it.
Different materials have different thermal conductivities based on their molecular structure and physical properties.
  • Metals usually have high thermal conductivity because their electrons can move freely, carrying heat.
  • Insulating materials like wood or fiberglass have low thermal conductivity and resist heat flow.
This property is critical when selecting materials for building designs that focus on energy efficiency and the reduction of heat loss through walls.
Heat Loss Calculation
Calculating heat loss involves determining how much heat passes through a material over time. We use the formula: \[ Q = kA\frac{(T_2 - T_1)}{t} \] where:
  • \(Q\) is the heat loss rate.
  • \(k\) is the thermal conductivity of the material.
  • \(A\) is the surface area through which heat is being transferred.
  • \(T_1\) and \(T_2\) are the temperatures of the inner and outer surfaces, respectively.
  • \(t\) is the thickness of the wall.
In this example, the wall's thermal conductivity, size, and thickness combine with the temperature difference between the inner and outer surfaces to determine how much heat is lost.
The calculation demonstrates that the larger the temperature gradient or the higher the conductivity, the higher the rate of heat loss. By practicing these calculations, students can understand the influence of each factor and make appropriate decisions in real-world applications.
Temperature Gradient
A temperature gradient is the difference in temperature between two points divided by the distance separating them. It illustrates how rapidly temperature changes over a specific distance, such as the thickness of a wall.
In scenarios like this exercise, the temperature gradient is created between the inside and the outside air, with the inner wall surface at \(25^{\circ} \text{C}\) and the outer surface varying from \(-15^{\circ} \text{C}\) to \(38^{\circ} \text{C}\).
The resulting temperature gradient influences the rate of heat transfer:
  • A larger temperature gradient means more significant heat transfer.
  • If the outside temperature is much lower than inside, the gradient increases, meaning faster heat loss.
This concept is crucial in designing effective climate control in buildings. Understanding how temperature gradients function helps in creating systems that minimize unwanted heat loss or gain, crucial for energy efficiency and cost savings.

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Most popular questions from this chapter

A solar flux of \(700 \mathrm{~W} / \mathrm{m}^{2}\) is incident on a flat-plate solar collector used to heat water. The area of the collector is \(3 \mathrm{~m}^{2}\), and \(90 \%\) of the solar radiation passes through the cover glass and is absorbed by the absorber plate. The remaining \(10 \%\) is reflected away from the collector. Water flows through the tube passages on the back side of the absorber plate and is heated from an inlet temperature \(T_{i}\) to an outlet temperature \(T_{o}\). The cover glass, operating at a temperature of \(30^{\circ} \mathrm{C}\), has an emissivity of \(0.94\) and experiences radiation exchange with the sky at \(-10^{\circ} \mathrm{C}\). The convection coefficient between the cover glass and the ambient air at \(25^{\circ} \mathrm{C}\) is \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Perform an overall energy balance on the collector to obtain an expression for the rate at which useful heat is collected per unit area of the collector, \(q_{11}^{\prime \prime}\). Determine the value of \(q_{u r^{\prime \prime}}\). (b) Calculate the temperature rise of the water, \(T_{o}-T_{i}\), if the flow rate is \(0.01 \mathrm{~kg} / \mathrm{s}\). Assume the specific heat of the water to be \(4179 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). (c) The collector efficiency \(\eta\) is defined as the ratio of the useful heat collected to the rate at which solar energy is incident on the collector. What is the value of \(\eta\) ?

In one stage of an annealing process, 304 stainless steel sheet is taken from \(300 \mathrm{~K}\) to \(1250 \mathrm{~K}\) as it passes through an electrically heated oven at a speed of \(V_{s}=10 \mathrm{~mm} / \mathrm{s}\). The sheet thickness and width are \(t_{s}=8 \mathrm{~mm}\) and \(W_{s}=2 \mathrm{~m}\), respectively, while the height, width, and length of the oven are \(H_{o}=2 \mathrm{~m}\), \(W_{o}=2.4 \mathrm{~m}\), and \(L_{o}=25 \mathrm{~m}\), respectively. The top and four sides of the oven are exposed to ambient air and large surroundings, each at \(300 \mathrm{~K}\), and the corresponding surface temperature, convection coefficient, and emissivity are \(T_{s}=350 \mathrm{~K}, h=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and \(\varepsilon_{s}=0.8\). The bottom surface of the oven is also at \(350 \mathrm{~K}\) and rests on a \(0.5\)-m-thick concrete pad whose base is at \(300 \mathrm{~K}\). Estimate the required electric power input, \(P_{\text {elec }}\), to the oven.

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