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A plane brick wall \((k=0.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is \(10 \mathrm{~cm}\) thick. The thermal resistance of this wall per unit of wall area is (a) \(0.143 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) (b) \(0.250 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) (c) \(0.327 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) (d) \(0.448 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) (e) \(0.524 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\)

Short Answer

Expert verified
Answer: (a) 0.143 m²K/W

Step by step solution

01

Convert wall thickness to meters

As the wall thickness is given in centimeters, we should convert it to meters: $$ 10\,\text{cm} \times \frac{1\,\text{m}}{100\,\text{cm}} = 0.1\,\text{m} $$
02

Use the formula to find thermal resistance per unit area

We will use the formula R/A = L/k. In our case, L = 0.1 m and k = 0.7 W/mK. $$ \frac{R}{A} = \frac{0.1\,\text{m}}{0.7\,\text{W/mK}} = 0.1428... \,\text{m}^2 \cdot \text{K/W} $$
03

Compare the result with the options

Our result, 0.1428 m²K/W, is closest to option (a) 0.143 m²K/W. So the correct answer is: (a) \(0.143 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\)

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Key Concepts

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

Heat Transfer
Heat transfer is the process of energy moving from one body or substance to another due to a temperature difference. This fundamental physical concept is crucial in many aspects of science and engineering. There are three primary modes through which heat can be transferred: conduction, convection, and radiation. Each operates differently in various scenarios.
  • **Conduction** involves heat transfer through a material without the movement of the material itself. It happens when atoms or molecules within a substance exchange energy with neighboring particles.
  • **Convection** occurs in fluids (liquids and gases) and is due to the movement of the fluid itself. It often involves the bulk movement of molecules within fluids.
  • **Radiation** doesn't require a medium and involves heat transfer through electromagnetic waves.
Understanding heat transfer is vital for designing thermal systems, like insulation, heating, and cooling technologies. It also helps scientists and engineers develop better methods to conserve energy and improve efficiency in industrial processes.
Conduction
Conduction describes the way heat is transferred through solid materials by direct contact of their particles. This process primarily occurs in metals and ionic solids, where atoms are densely packed, allowing easy transfer of vibrational energy.
In conduction:
  • The transfer of heat happens from the "hotter" region to the "cooler" region within the material.
  • The rate of heat transfer depends on the thermal conductivity of the material, denoted as **k**.
  • The higher the thermal conductivity, the more efficiently heat can pass through a material.
For example, a brick wall has a specific thermal conductivity, which determines how well it can conduct heat. Using the formula for thermal resistance, \( R = \frac{L}{k} \), where \( L \) is the thickness of the wall, we can calculate how much resistance the wall provides against heat transfer. This understanding can inform choices about building materials to optimize for energy efficiency.
Plane Wall Theory
The plane wall theory simplifies the analysis of heat conduction in flat surfaces, such as walls or slabs, where the thickness is small compared to its other dimensions. This theory assumes that heat transfer through the wall occurs in one dimension, making calculations more straightforward.
Key ideas of plane wall theory:
  • Assumes a steady state, so the temperature doesn't change with time.
  • Considers conduction as the primary mode of heat transfer, with effects like convection and radiation being negligible at the wall's surfaces.
  • Typically uses Fourier’s Law of Heat Conduction, which links the thermal conductivity, temperature gradient, and heat transfer rate.
Applying plane wall theory helps engineers calculate thermal resistance, which is a measure of a material's resistance to conductive heat flow. This is critical for designing effective insulation in building walls, ensuring comfort inside while minimizing energy loss. The derived formula for thermal resistance per unit area from the plane wall theory, \( \frac{R}{A} = \frac{L}{k} \), is used to find the correct answer in our given exercise, highlighting its practical utility.

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Most popular questions from this chapter

The boiling temperature of nitrogen at atmospheric pressure at sea level ( 1 atm pressure) is \(-196^{\circ} \mathrm{C}\). Therefore, nitrogen is commonly used in low-temperature scientific studies since the temperature of liquid nitrogen in a tank open to the atmosphere will remain constant at \(-196^{\circ} \mathrm{C}\) until it is depleted. Any heat transfer to the tank will result in the evaporation of some liquid nitrogen, which has a heat of vaporization of \(198 \mathrm{~kJ} / \mathrm{kg}\) and a density of \(810 \mathrm{~kg} / \mathrm{m}^{3}\) at 1 atm. Consider a 3-m-diameter spherical tank that is initially filled with liquid nitrogen at 1 atm and \(-196^{\circ} \mathrm{C}\). The tank is exposed to ambient air at \(15^{\circ} \mathrm{C}\), with a combined convection and radiation heat transfer coefficient of \(35 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The temperature of the thin-shelled spherical tank is observed to be almost the same as the temperature of the nitrogen inside. Determine the rate of evaporation of the liquid nitrogen in the tank as a result of the heat transfer from the ambient air if the tank is \((a)\) not insulated, \((b)\) insulated with 5 -cm-thick fiberglass insulation \((k=0.035 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), and (c) insulated with 2 -cm-thick superinsulation which has an effective thermal conductivity of \(0.00005 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

A cylindrical pin fin of diameter \(0.6 \mathrm{~cm}\) and length of \(3 \mathrm{~cm}\) with negligible heat loss from the tip has an efficiency of 0.7. The effectiveness of this fin is (a) \(0.3\) (b) \(0.7\) (c) 2 (d) 8 (e) 14

The \(700 \mathrm{~m}^{2}\) ceiling of a building has a thermal resistance of \(0.52 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). The rate at which heat is lost through this ceiling on a cold winter day when the ambient temperature is \(-10^{\circ} \mathrm{C}\) and the interior is at \(20^{\circ} \mathrm{C}\) is (a) \(23.1 \mathrm{~kW} \quad\) (b) \(40.4 \mathrm{~kW}\) (c) \(55.6 \mathrm{~kW}\) (d) \(68.1 \mathrm{~kW}\) (e) \(88.6 \mathrm{~kW}\)

What is the reason for the widespread use of fins on surfaces?

A 50 -m-long section of a steam pipe whose outer (€) diameter is \(10 \mathrm{~cm}\) passes through an open space at \(15^{\circ} \mathrm{C}\). The average temperature of the outer surface of the pipe is measured to be \(150^{\circ} \mathrm{C}\). If the combined heat transfer coefficient on the outer surface of the pipe is \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine (a) the rate of heat loss from the steam pipe; \((b)\) the annual cost of this energy lost if steam is generated in a natural gas furnace that has an efficiency of 75 percent and the price of natural gas is $$\$ 0.52 /$$ therm ( 1 therm \(=105,500 \mathrm{~kJ})\); and \((c)\) the thickness of fiberglass insulation \((k=0.035 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) needed in order to save 90 percent of the heat lost. Assume the pipe temperature to remain constant at \(150^{\circ} \mathrm{C}\).

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