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Chapter 9: Problem 96

Two concentric spheres of diameters \(15 \mathrm{~cm}\) and \(25 \mathrm{~cm}\) are separated by air at \(1 \mathrm{~atm}\) pressure. The surface temperatures of the two spheres enclosing the air are \(T_{1}=350 \mathrm{~K}\) and \(T_{2}=\) \(275 \mathrm{~K}\), respectively. Determine the rate of heat transfer from the inner sphere to the outer sphere by natural convection.

Short Answer

Expert verified
Question: Determine the rate of heat transfer from the inner sphere to the outer sphere by natural convection, given the surface temperatures of the two concentric spheres with diameters of 15 cm and 25 cm are 350 K and 275 K, respectively, and the air pressure between them is 0.1 MPa. Answer: To calculate the rate of heat transfer from the inner sphere to the outer sphere by natural convection, follow these steps: 1. Calculate the radius of both spheres: r1 = 0.075 m, r2 = 0.125 m 2. Determine the mean film temperature and air properties at this temperature: T_film = 312.5 K 3. Calculate the Grashof (Gr) and Prandtl (Pr) numbers: Gr = (g * β * (T1 - T2) * (r2 - r1)^3) / ν^2, Pr = (μ * c_p) / k 4. Determine the Nusselt number: Nu = 0.154 * (Gr * Pr)^(1/5) 5. Calculate the convective heat transfer coefficient: h = (Nu * k) / (r2 - r1) 6. Determine the rate of heat transfer: Q = 4π * r1 * r2 * h * (T1 - T2) After calculating all the values, you will get the rate of heat transfer from the inner sphere to the outer sphere by natural convection.

Step by step solution

01

Calculate the radius of both spheres

Given the diameters, we can calculate the radii of the spheres: - Radius of sphere 1, \(r_1 = \frac{d_1}{2} = \frac{15\mathrm{~cm}}{2} = 7.5\mathrm{~cm} = 0.075\mathrm{~m}\) - Radius of sphere 2, \(r_2 = \frac{d_2}{2} = \frac{25\mathrm{~cm}}{2} = 12.5\mathrm{~cm} = 0.125\mathrm{~m}\)
02

Determine the mean film temperature and air properties at this temperature

To find the mean film temperature, we average the surface temperatures of the two spheres: $$ T_{film} = \frac{T_1 + T_2}{2} = \frac{350\mathrm{~K} + 275\mathrm{~K}}{2} = 312.5\mathrm{~K} $$ At this temperature, we can determine the properties of air using air property tables or correlations. The properties we'll need are the dynamic viscosity (\(\mu\)), thermal conductivity (\(k\)), thermal expansion coefficient (\(\beta\)), and specific heat (\(c_p\)).
03

Calculate the Grashof (Gr) and Prandtl (Pr) numbers

The Grashof number is a dimensionless number that characterizes the flow's buoyancy forces relative to viscous forces. The Prandtl number is a dimensionless number that characterizes the relative importance of momentum and thermal diffusion in convective heat transfer. $$Gr = \frac{g\beta(T_1 - T_2)(r_2 - r_1)^3}{\nu^2}$$ $$Pr = \frac{\mu c_p}{k}$$ where \(g\) is the acceleration due to gravity (\(9.81\mathrm{~m/s^2}\)), \(\nu\) is the kinematic viscosity (\(\nu = \frac{\mu}{\rho}\)), and \(\rho\) is the density of air.
04

Determine the Nusselt number

To find the Nusselt number (\(Nu\)), we use the correlation for natural convection between concentric spheres: $$Nu = 0.154 \times (Gr \times Pr)^{1/5}$$
05

Calculate the convective heat transfer coefficient

The convective heat transfer coefficient (\(h\)) can be calculated using the Nusselt number through: $$h = \frac{Nu \times k}{r_2 - r_1}$$
06

Determine the rate of heat transfer

Finally, we can calculate the rate of heat transfer (\(Q\)) using the convective heat transfer coefficient and the temperature difference between the surfaces of the spheres: $$Q = 4\pi r_1 r_2 h(T_1 - T_2)$$ After calculating all the values, we will get the rate of heat transfer from the inner sphere to the outer sphere by natural convection.

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Most popular questions from this chapter

Under what conditions does natural convection enhance forced convection, and under what conditions does it hurt forced convection?

In which mode of heat transfer is the convection heat transfer coefficient usually higher, natural convection or forced convection? Why?

Consider three similar double-pane windows with air gap widths of 5,10 , and \(20 \mathrm{~mm}\). For which case will the heat transfer through the window will be a minimum?

Two concentric cylinders of diameters \(D_{i}=30 \mathrm{~cm}\) and \(D_{o}=40 \mathrm{~cm}\) and length \(L=5 \mathrm{~m}\) are separated by air at \(1 \mathrm{~atm}\) pressure. Heat is generated within the inner cylinder uniformly at a rate of \(1100 \mathrm{~W} / \mathrm{m}^{3}\), and the inner surface temperature of the outer cylinder is \(300 \mathrm{~K}\). The steady-state outer surface temperature of the inner cylinder is (a) \(402 \mathrm{~K}\) (b) \(415 \mathrm{~K}\) (c) \(429 \mathrm{~K}\) (d) \(442 \mathrm{~K}\) (e) \(456 \mathrm{~K}\) (For air, use \(k=0.03095 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.7111, v=\) \(\left.2.306 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\right)\)

In a plant that manufactures canned aerosol paints, the cans are temperature- tested in water baths at \(55^{\circ} \mathrm{C}\) before they are shipped to ensure that they withstand temperatures up to \(55^{\circ} \mathrm{C}\) during transportation and shelving (as shown in Fig. P9-44 on the next page). The cans, moving on a conveyor, enter the open hot water bath, which is \(0.5 \mathrm{~m}\) deep, \(1 \mathrm{~m}\) wide, and \(3.5 \mathrm{~m}\) long, and move slowly in the hot water toward the other end. Some of the cans fail the test and explode in the water bath. The water container is made of sheet metal, and the entire container is at about the same temperature as the hot water. The emissivity of the outer surface of the container is 0.7. If the temperature of the surrounding air and surfaces is \(20^{\circ} \mathrm{C}\), determine the rate of heat loss from the four side surfaces of the container (disregard the top surface, which is open). The water is heated electrically by resistance heaters, and the cost of electricity is \(\$ 0.085 / \mathrm{kWh}\). If the plant operates \(24 \mathrm{~h}\) a day 365 days a year and thus \(8760 \mathrm{~h}\) a year, determine the annual cost of the heat losses from the container for this facility.
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