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Chapter 1: Problem 154

Heat is lost through a brick wall \((k=0.72 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), which is \(4 \mathrm{~m}\) long, \(3 \mathrm{~m}\) wide, and \(25 \mathrm{~cm}\) thick at a rate of \(500 \mathrm{~W}\). If the inner surface of the wall is at \(22^{\circ} \mathrm{C}\), the temperature at the midplane of the wall is (a) \(0^{\circ} \mathrm{C}\) (b) \(7.5^{\circ} \mathrm{C}\) (c) \(11.0^{\circ} \mathrm{C}\) (d) \(14.8^{\circ} \mathrm{C}\) (e) \(22^{\circ} \mathrm{C}\)

Short Answer

Expert verified
#tag_title# Step 2: Calculate the surface area of the wall #tag_content# To find the surface area, multiply the length and the width of the wall: \(A = L \times W = 4 \mathrm{~m} \times 3 \mathrm{~m} = 12 \mathrm{~m^2}\) #tag_title# Step 3: Rearrange the heat transfer formula to solve for ΔT #tag_content# We have the values for \(k\), \(A\), and \(d\), so we can rearrange the heat transfer formula to solve for the temperature difference, \(\Delta T\): \(\Delta T = \dfrac{Q \times d}{k \times A}\) #tag_title# Step 4: Substitute the known values and calculate ΔT #tag_content# Plug in the known values into the expression for \(\Delta T\), and then calculate it: \(\Delta T = \dfrac{500 \mathrm{~W} \times 0.25 \mathrm{~m}}{0.72 \mathrm{W/m\cdot K} \times 12 \mathrm{~m^2}} \approx 14.58 \mathrm{K}\) #tag_title# Step 5: Calculate the midplane temperature #tag_content# Now, we need to find the temperature at the midplane. Since we know the inner surface temperature (\(T_{inner}\)) and the complete temperature difference across the wall (\(\Delta T\)), we can assume that half of the temperature difference occurs across half the thickness of the wall. So we can find the midplane temperature by subtracting half of \(\Delta T\) from \(T_{inner}\): \(T_{midplane} = T_{inner} - \dfrac{\Delta T}{2} \approx 22^{\circ} \mathrm{C} - \dfrac{14.58 \mathrm{K}}{2} \approx 14.71^{\circ} \mathrm{C}\) So the temperature at the midplane of the brick wall is approximately \(14.71^{\circ} \mathrm{C}\).

Step by step solution

01

List the given values

We are given the following information about the brick wall and heat transfer: - Thermal conductivity, \(k = 0.72 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) - Length, \(L = 4 \mathrm{~m}\) - Width, \(W = 3 \mathrm{~m}\) - Thickness, \(d = 25 \mathrm{~cm} = 0.25 \mathrm{~m}\) (converted to meters) - Rate of heat loss, \(Q = 500 \mathrm{~W}\) - Inner surface temperature, \(T_{inner} = 22^{\circ} \mathrm{C}\)

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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 property that measures a material's ability to conduct heat. It represents how easily heat can pass through a material. The higher the thermal conductivity of a substance, the more efficient it is at transferring heat. It is denoted by the symbol \( k \) and is measured in watts per meter Kelvin \( (W/m \cdot K) \). For example, metals typically have high thermal conductivity, making them good heat conductors, while materials like brick, with a lower \( k \) value, are less efficient but still used in construction due to their thermal mass properties.

In the exercise, we use the thermal conductivity \( k = 0.72 W/m \cdot K \) for calculations. This value is crucial to determine the rate of heat transfer through the brick wall. The choice of materials with specific thermal conductivities can impact heating and cooling requirements for buildings retrospectively.
Steady-State Heat Conduction
Steady-state heat conduction occurs when the temperature distribution in a material does not change over time; that is, the amount of heat entering a section of the material is equal to the amount of heat exiting the same section. There is no net accumulation of heat within the material, and the temperature gradient remains constant.

This concept is essential for long-term analysis of heat transfer in structures, as it helps in the design phase to ensure that heating or cooling loads are consistent. It simplifies calculations since we can assume the heat transfer rate is unchanging. In our problem, we're inferring a steady-state condition to calculate the temperature at the midplane of the brick wall.
Fourier’s Law of Heat Conduction
Fourier’s law of heat conduction is a fundamental principle which quantifies the heat transfer through a material. It states that the heat transfer rate \( Q \) through a material is proportional to the negative gradient of the temperature and the area \( A \) perpendicular to the heat transfer direction, and inversely proportional to the thickness \( d \) of the material. Mathematically, it is expressed as:

\[ Q = -k A \frac{dT}{dx} \]
Where:\( Q \) is the heat transfer rate in watts (\( W \)),\( k \) is the thermal conductivity,\( A \) is the area through which heat is being transferred,\( \frac{dT}{dx} \) is the temperature gradient, and\( d \) is the thickness of the material.

Fourier's law helps us understand how different factors affect heat transfer. In practice, it means that if we have a material with a known thermal conductivity and area, and we can measure the temperature difference across it, we can calculate the rate at which heat flows through the material.
Temperature Distribution
Temperature distribution within a material under steady-state heat conduction tells us how temperature varies from one point to another. It's fundamental in evaluating and designing thermal systems, as uneven temperature distribution can lead to thermal stresses and inefficiencies.

With knowledge of the thermal conductivity and boundary conditions (like known surface temperatures), we can estimate the temperature at any point within the material. This assumes a linear temperature gradient between points when dealing with simple, homogeneous materials. In the provided exercise, determining the temperature at the midplane involves knowing the temperature at the surfaces and using the heat transfer rate to deduce the temperature distribution in the wall's interior.

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Most popular questions from this chapter

An ice skating rink is located in a building where the air is at \(T_{\text {air }}=20^{\circ} \mathrm{C}\) and the walls are at \(T_{w}=25^{\circ} \mathrm{C}\). The convection heat transfer coefficient between the ice and the surrounding air is \(h=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The emissivity of ice is \(\varepsilon=0.95\). The latent heat of fusion of ice is \(h_{i f}=333.7 \mathrm{~kJ} / \mathrm{kg}\) and its density is \(920 \mathrm{~kg} / \mathrm{m}^{3}\). (a) Calculate the refrigeration load of the system necessary to maintain the ice at \(T_{s}=0^{\circ} \mathrm{C}\) for an ice rink of \(12 \mathrm{~m}\) by \(40 \mathrm{~m}\). (b) How long would it take to melt \(\delta=3 \mathrm{~mm}\) of ice from the surface of the rink if no cooling is supplied and the surface is considered insulated on the back side?

A cylindrical resistor element on a circuit board dissipates \(1.2 \mathrm{~W}\) of power. The resistor is \(2 \mathrm{~cm}\) long, and has a diameter of \(0.4 \mathrm{~cm}\). Assuming heat to be transferred uniformly from all surfaces, determine \((a)\) the amount of heat this resistor dissipates during a 24-hour period, \((b)\) the heat flux, and \((c)\) the fraction of heat dissipated from the top and bottom surfaces.

The roof of a house consists of a 22-cm-thick (st) concrete slab \((k=2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) that is \(15 \mathrm{~m}\) wide and \(20 \mathrm{~m}\) long. The emissivity of the outer surface of the roof is \(0.9\), and the convection heat transfer coefficient on that surface is estimated to be \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The inner surface of the roof is maintained at \(15^{\circ} \mathrm{C}\). On a clear winter night, the ambient air is reported to be at \(10^{\circ} \mathrm{C}\) while the night sky temperature for radiation heat transfer is \(255 \mathrm{~K}\). Considering both radiation and convection heat transfer, determine the outer surface temperature and the rate of heat transfer through the roof. If the house is heated by a furnace burning natural gas with an efficiency of 85 percent, and the unit cost of natural gas is \(\$ 1.20\) / therm ( 1 therm \(=105,500 \mathrm{~kJ}\) of energy content), determine the money lost through the roof that night during a 14-hour period.

A 5-cm-external-diameter, 10-m-long hot-water pipe at \(80^{\circ} \mathrm{C}\) is losing heat to the surrounding air at \(5^{\circ} \mathrm{C}\) by natural convection with a heat transfer coefficient of \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the rate of heat loss from the pipe by natural convection. Answer: \(2945 \mathrm{~W}\)

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 \(0.00015 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).
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