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Two concentric spheres with diameters of \(5 \mathrm{~cm}\) and \(10 \mathrm{~cm}\) are having the surface temperatures maintained at \(100^{\circ} \mathrm{C}\) and \(200^{\circ} \mathrm{C}\), respectively. The enclosure between the two concentric spherical surfaces is filled with nitrogen gas at \(1 \mathrm{~atm}\). Determine the rate of heat transfer through the enclosure.

Short Answer

Expert verified
Answer: The rate of heat transfer through the enclosure is approximately 65 Watts.

Step by step solution

01

Determine the radii of the spheres and their temperature difference

We are given the diameters of the spheres as 5 cm and 10 cm. We will find their radii by dividing the diameters in half: \(r_1 = \dfrac{5}{2} \mathrm{~cm} = 2.5 \mathrm{~cm}\) \(r_2 = \dfrac{10}{2} \mathrm{~cm} = 5 \mathrm{~cm}\) The surface temperatures of the two spheres are given as \(T_1 = 100^{\circ}\mathrm{C}\) and \(T_2 = 200^{\circ}\mathrm{C}\). We will find the temperature difference as: \(\Delta T = T_2 - T_1 = 200^{\circ}\mathrm{C} - 100^{\circ}\mathrm{C} = 100^{\circ}\mathrm{C}\)
02

Find the thermal conductivity of nitrogen gas

Next, we need to find the thermal conductivity of nitrogen gas. At 1 atm and room temperature, the thermal conductivity of nitrogen gas is approximately \(k = 0.026 \mathrm{~W/m \cdot K}\).
03

Calculate the rate of heat transfer through the enclosure

Now we will use the formula for the rate of heat transfer through a spherical shell: \(q = 4 \pi k \dfrac{r_1 r_2}{r_2 - r_1} \Delta T\) We will convert the radii to meters (\(1 \mathrm{~cm} = 0.01 \mathrm{~m}\)) and plug in the values to find the heat transfer rate: \(q = 4 \pi (0.026 \mathrm{~W/m \cdot K}) \dfrac{(2.5 \cdot 0.01)(5 \cdot 0.01)}{(5 \cdot 0.01) - (2.5 \cdot 0.01)} (100^{\circ}\mathrm{C})\) \(q = 4 \pi (0.026) \dfrac{0.0125}{0.025} (100)\) \(q = 4 \pi (0.026) (0.5) (100)\) \(q \approx 65 \mathrm{~W}\) The rate of heat transfer through the enclosure is approximately 65 Watts.

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Key Concepts

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

Concentric Spheres
Concentric spheres describe a situation where one sphere is located inside another, with both sharing the same center but having different radii. In this scenario, the inner sphere has a diameter of 5 cm, while the outer one has a diameter of 10 cm.

These spheres can contain various mediums between them, such as gases or liquids, often used in thermal experiments or engineering to study heat transfer.

The difference in the diameters helps define the geometry and the surface area over which heat transfer occurs, crucial for calculations involving thermal processes.
Thermal Conductivity
Thermal conductivity is a material property that indicates its ability to conduct heat. It is represented by the symbol \( k \) and is measured in units of \( \, W/m \cdot K \, \). This value helps us understand how easily heat can pass through a material, the higher the thermal conductivity, the more efficient the material is at allowing heat flow.

In our exercise, we use the thermal conductivity of nitrogen gas, which fills the enclosure between the two concentric spheres. Nitrogen at standard atmospheric conditions has a low thermal conductivity of approximately 0.026 \( \, W/m \cdot K \).

This low value means nitrogen does not conduct heat very effectively, acting more as an insulator in thermal systems.
Spherical Shell
A spherical shell is the space or layer of material that exists between two concentric spheres. In the exercise, this shell is filled with nitrogen gas.

The shell’s thickness is the radial distance between the two spheres, calculated by subtracting the radius of the inner sphere from the radius of the outer sphere. The heat transfer process through this shell can be determined using heat transfer equations that account for the geometry of spheres.

The expression used in such cases is \( q = 4 \pi k \cdot \frac{r_1 r_2}{r_2 - r_1} \cdot \Delta T \), which accounts for the shape and size of the shell when computing the rate of heat transfer.
Nitrogen Gas
Nitrogen gas is a diatomic molecule (â‚‚) and is a major component of Earth's atmosphere, accounting for about 78%. Known for its inertness and low reactivity, nitrogen gas is commonly used in various industrial applications.

In the given problem, nitrogen fills the space between the concentric spheres. At 1 atmosphere (atm) pressure, nitrogen gas is a poor conductor of heat, with a thermal conductivity of 0.026 \( \, W/m \cdot K \).

This property makes it act as an insulating layer, minimizing heat loss in systems where applied, such as in thermal insulation or controlled heating environments.

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Most popular questions from this chapter

Consider a 3-m-high rectangular enclosure consisting of two surfaces separated by a \(0.1-\mathrm{m}\) air gap at \(1 \mathrm{~atm}\). If the surface temperatures across the air gap are \(30^{\circ} \mathrm{C}\) and \(-10^{\circ} \mathrm{C}\), determine the ratio of the heat transfer rate for the horizontal orientation (with hotter surface at the bottom) to that for vertical orientation.

A spherical tank \((k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) with an inner diameter of \(3 \mathrm{~m}\) and a wall thickness of \(10 \mathrm{~mm}\) is used for storing hot liquid. The hot liquid inside the tank causes the inner surface temperature to be as high as \(100^{\circ} \mathrm{C}\). To prevent thermal burns on the skin of the people working near the vicinity of the tank, the tank is covered with a \(7-\mathrm{cm}\) thick layer of insulation \((k=0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and the outer surface is painted to give an emissivity of \(0.35\). The tank is located in a surrounding with air at \(16^{\circ} \mathrm{C}\). Determine whether or not the insulation layer is sufficient to keep the outer surface temperature below \(45^{\circ} \mathrm{C}\) to prevent thermal burn hazards. Discuss ways to further decrease the outer surface temperature. Evaluate the air properties at \(30^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure. Is this a good assumption?

Consider a 2-m-high electric hot-water heater that has a diameter of \(40 \mathrm{~cm}\) and maintains the hot water at \(60^{\circ} \mathrm{C}\). The tank is located in a small room at \(20^{\circ} \mathrm{C}\) whose walls and ceiling are at about the same temperature. The tank is placed in a 44-cm-diameter sheet metal shell of negligible thickness, and the space between the tank and the shell is filled with foam insulation. The average temperature and emissivity of the outer surface of the shell are \(40^{\circ} \mathrm{C}\) and \(0.7\), respectively. The price of electricity is \(\$ 0.08 / \mathrm{kWh}\). Hot-water tank insulation kits large enough to wrap the entire tank are available on the market for about \(\$ 60\). If such an insulation is installed on this water tank by the home owner himself, how long will it take for this additional insulation to pay for itself? Disregard any heat loss from the top and bottom surfaces, and assume the insulation to reduce the heat losses by 80 percent.

Consider a heat sink with optimum fin spacing. Explain how heat transfer from this heat sink will be affected by (a) removing some of the fins on the heat sink and (b) doubling the number of fins on the heat sink by reducing the fin spacing. The base area of the heat sink remains unchanged at all times.

Consider laminar natural convection from a vertical hot-plate. Will the heat flux be higher at the top or at the bottom of the plate? Why?

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