/*! 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 vertical slab of Wood's metal is joined to a substrate on one surface and is melted as it is uniformly irradiated by a laser source on the opposite surface. The metal is initially at its fusion temperature of \(T_{f}=72^{\circ} \mathrm{C}\), and the melt runs off by gravity as soon as it is formed. The absorptivity of the metal to the laser radiation is \(\alpha_{1}=0.4\), and its latent heat of fusion is \(h_{s f}=33 \mathrm{~kJ} / \mathrm{kg}\). (a) Neglecting heat transfer from the irradiated surface by convection or radiation exchange with the surroundings, determine the instantaneous rate of melting in \(\mathrm{kg} / \mathrm{s} \cdot \mathrm{m}^{2}\) if the laser irradiation is \(5 \mathrm{~kW} / \mathrm{m}^{2}\). How much material is removed if irradiation is maintained for a period of \(2 \mathrm{~s}\) ? (b) Allowing for convection to ambient air, with \(T_{\infty}=20^{\circ} \mathrm{C}\) and \(h=15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and radiation exchange with large surroundings \((\varepsilon=0.4\), \(T_{\text {sur }}=20^{\circ} \mathrm{C}\) ), determine the instantaneous rate of melting during irradiation.

The heat flux through a wood slab \(50 \mathrm{~mm}\) thick, whose inner and outer surface temperatures are 40 and \(20^{\circ} \mathrm{C}\), respectively, has been determined to be \(40 \mathrm{~W} / \mathrm{m}^{2}\). What is the thermal conductivity of the wood?

An aluminum plate \(4 \mathrm{~mm}\) thick is mounted in a horizontal position, and its bottom surface is well insulated. A special, thin coating is applied to the top surface such that it absorbs \(80 \%\) of any incident solar radiation, while having an emissivity of \(0.25\). The density \(\rho\) and specific heat \(c\) of aluminum are known to be \(2700 \mathrm{~kg} / \mathrm{m}^{3}\) and \(900 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively. (a) Consider conditions for which the plate is at a temperature of \(25^{\circ} \mathrm{C}\) and its top surface is suddenly exposed to ambient air at \(T_{\infty}=20^{\circ} \mathrm{C}\) and to solar radiation that provides an incident flux of \(900 \mathrm{~W} / \mathrm{m}^{2}\). The convection heat transfer coefficient between the surface and the air is \(h=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What is the initial rate of change of the plate temperature? (b) What will be the equilibrium temperature of the plate when steady-state conditions are reached? (c) The surface radiative properties depend on the specific nature of the applied coating. Compute and plot the steady-state temperature as a function of the emissivity for \(0.05 \leq \varepsilon \leq 1\), with all other conditions remaining as prescribed. Repeat your calculations for values of \(\alpha_{S}=0.5\) and \(1.0\), and plot the results with those obtained for \(\alpha_{S}=0.8\). If the intent is to maximize the plate temperature, what is the most desirable combination of the plate emissivity and its absorptivity to solar radiation?

Pressurized water \(\left(p_{\text {in }}=10\right.\) bar, \(\left.T_{\text {in }}=110^{\circ} \mathrm{C}\right)\) enters the bottom of an \(L=10\)-m-long vertical tube of diameter \(D=100 \mathrm{~mm}\) at a mass flow rate of \(\dot{m}=1.5 \mathrm{~kg} / \mathrm{s}\). The tube is located inside a combustion chamber, resulting in heat transfer to the tube. Superheated steam exits the top of the tube at \(p_{\text {out }}=7\) bar, \(T_{\text {out }}=600^{\circ} \mathrm{C}\). Determine the change in the rate at which the following quantities enter and exit the tube: (a) the combined thermal and flow work, (b) the mechanical energy, and (c) the total energy of the water. Also, (d) determine the heat transfer rate, \(q\). Hint: Relevant properties may be obtained from a thermodynamics text.

Under conditions for which the same room temperature is maintained by a heating or cooling system, it is not uncommon for a person to feel chilled in the winter but comfortable in the summer. Provide a plausible explanation for this situation (with supporting calculations) by considering a room whose air temperature is maintained at \(20^{\circ} \mathrm{C}\) throughout the year, while the walls of the room are nominally at \(27^{\circ} \mathrm{C}\) and \(14^{\circ} \mathrm{C}\) in the summer and winter, respectively. The exposed surface of a person in the room may be assumed to be at a temperature of \(32^{\circ} \mathrm{C}\) throughout the year and to have an emissivity of \(0.90\). The coefficient associated with heat transfer by natural convection between the person and the room air is approximately \(2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

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