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Chapter 12: Problem 85

Consider an opaque plate that is well insulated on the edges and it is heated at the bottom with an electric heater. The plate has an emissivity of \(0.67\), and is situated in an ambient surrounding temperature of \(7^{\circ} \mathrm{C}\) where the natural convection heat transfer coefficient is \(7 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). To maintain a surface temperature of \(80^{\circ} \mathrm{C}\), the electric heater supplies \(1000 \mathrm{~W} / \mathrm{m}^{2}\) of uniform heat flux to the plate. Determine the radiosity of the plate under these conditions.

Short Answer

Expert verified
Answer: To find the radiosity of the plate, follow these steps: 1. Calculate the heat transfer by convection (q_conv) using the formula: q_conv = h * (T_s - T_inf) 2. Apply the energy balance equation: q_heater = q_conv + q_rad 3. Find the emissive power (E_b): q_rad = ε * σ * (T_s^4 - T_inf^4) 4. Calculate the radiosity (J): J = E_b + (q_conv / emissivity) By plugging in the given values and performing calculations according to the solution steps above, the radiosity of the plate can be determined.

Step by step solution

01

Calculate the heat transfer by convection

Calculate the heat transfer by convection between the plate and ambient surroundings using the natural convection heat transfer coefficient: \(q_{conv} = h \cdot (T_s - T_\infty)\) where \(h = 7 \: \mathrm{W/m^2 \cdot K}\) (given natural convection heat transfer coefficient) \(T_s = 80^{\circ} \mathrm{C}\) (desired surface temperature) \(T_\infty = 7^{\circ} \mathrm{C}\) (ambient surrounding temperature) Plugging in the values, we will find the heat transfer by convection.
02

Apply the energy balance equation

Use the energy balance equation to find the relationship between heat losses from the surface by convection and radiation. Let \(q_{rad}\) be the heat transfer by radiation. \(q_{heater} = q_{conv} + q_{rad}\) where \(q_{heater} = 1000 \: \mathrm{W/m^2}\) (heat supplied by the heater) Use the calculated value of \(q_{conv}\) from the previous step and plug it into the equation to find the heat transfer by radiation, \(q_{rad}\).
03

Find the emissive power

To find the emissive power for the given plate, we need to use the relationship: \(q_{rad} = \epsilon \cdot \sigma \cdot (T^4_s - T^4_\infty)\) where \(\epsilon = 0.67\) (emissivity of the plate) \(\sigma = 5.67 \times 10^{-8} \: \mathrm{W/m^2 \cdot K^4}\) (Stefan-Boltzmann constant) We know \(q_{rad}\) from the energy balance equation. We can rearrange the equation to directly calculate the emissive power \(E_b = \sigma \cdot T^4_s\).
04

Calculate the radiosity

Finally, we will calculate the radiosity using the following equation: \(J = E_b + \cfrac{q_{conv}}{emissivity} = \sigma \cdot T_s^4 + \cfrac{q_{conv}}{0.67}\) Plug in the values of emissive power \(E_b\) and the calculated heat transfer by convection, \(q_{conv}\), for the given system, to find the radiosity of the plate.

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Most popular questions from this chapter

Consider a black spherical ball, with a diameter of \(25 \mathrm{~cm}\), is being suspended in air. Determine the surface temperature of the ball that should be maintained in order to heat \(10 \mathrm{~kg}\) of air from 20 to \(30^{\circ} \mathrm{C}\) in the duration of 5 minutes.

Daylight and incandescent light may be approximated as a blackbody at the effective surface temperatures of \(5800 \mathrm{~K}\) and \(2800 \mathrm{~K}\), respectively. Determine the wavelength at maximum emission of radiation for each of the lighting sources.

A long metal bar \(\left(c_{p}=450 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \rho=\right.\) \(7900 \mathrm{~kg} / \mathrm{m}^{3}\) ) is being conveyed through a water bath to be quenched. The metal bar has a cross section of \(30 \mathrm{~mm} \times 15 \mathrm{~mm}\), and it enters the water bath at \(700^{\circ} \mathrm{C}\). During the quenching process, \(500 \mathrm{~kW}\) of heat is released from the bar in the water bath. In order to prevent thermal burn on people handling the metal bar, it must exit the water bath at a temperature below \(45^{\circ} \mathrm{C}\). A radiometer is placed normal to and at a distance of \(1 \mathrm{~m}\) from the bar to monitor the exit temperature. The radiometer receives radiation from a target area of \(1 \mathrm{~cm}^{2}\) of the bar surface. Irradiation signal detected by the radiometer is used to control the speed of the bar being conveyed through the water bath so that the exit temperature is safe for handing. If the radiometer detects an irradiation of \(0.015 \mathrm{~W} / \mathrm{m}^{2}\), determine the speed of the bar bar can be approximated as a blackbody.

When the earth is closest to the sun, we have winter in the northern hemisphere. Explain why. Also explain why we have summer in the northern hemisphere when the earth is farthest away from the sun.

Solar radiation is incident on a semi-transparent body at a rate of \(500 \mathrm{~W} / \mathrm{m}^{2}\). If \(150 \mathrm{~W} / \mathrm{m}^{2}\) of this incident radiation is reflected back and \(225 \mathrm{~W} / \mathrm{m}^{2}\) is transmitted across the body, the absorptivity of the body is (a) 0 (b) \(0.25\) (c) \(0.30\) (d) \(0.45\) (e) 1
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