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Chapter 1: Problem 142

A thin metal plate is insulated on the back and exposed to solar radiation on the front surface. The exposed surface of the plate has an absorptivity of \(0.7\) for solar radiation. If solar radiation is incident on the plate at a rate of \(550 \mathrm{~W} / \mathrm{m}^{2}\) and the surrounding air temperature is \(10^{\circ} \mathrm{C}\), determine the surface temperature of the plate when the heat loss by convection equals the solar energy absorbed by the plate. Take the convection heat transfer coefficient to be \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and disregard any heat loss by radiation.

Short Answer

Expert verified
Answer: The surface temperature of the plate is 25.4°C.

Step by step solution

01

Calculate the absorbed solar energy

To calculate the solar energy absorbed by the surface, we need to multiply the incident solar radiation (\(550 W/m^2\)) by the absorptivity (\(0.7\)), which will give us the absorbed energy rate per unit area: $$E_\text{absorbed} = 0.7 \cdot 550 W/m^2 = 385 W/m^2$$
02

Equation for the heat loss due to convection

The heat loss due to convection can be calculated using the following equation: $$q_\text{conv} = h_\text{conv} A (T_\text{plate} - T_\text{air})$$ In our case, we are given the convection heat transfer coefficient \(h_\text{conv} = 25 W/m^2 \cdot K\) and the air temperature \(T_\text{air} = 10^\circ C\). The terms q_conv and A cancel out, giving us: $$q_\text{loss} = 25 (T_\text{plate} - 10)$$
03

Equate the absorbed solar energy and the convection heat loss, then solve for the surface temperature of the plate

Now, we equate the absorbed solar energy and the convection heat loss: $$E_\text{absorbed} = q_\text{loss}$$ $$385 W/m^2 = 25 W/m^2 \cdot K \cdot (T_\text{plate} - 10^\circ\textbf{ C})$$ Now we solve for \(T_\text{plate}\):$$T_\text{plate} - 10^\circ \textbf{ C} = \frac{385}{25}$$ $$T_\text{plate} = 10^\circ \textbf{ C} + \frac{385}{25}$$ $$T_\text{plate} = 25.4^\circ \textbf{ C}$$ So the surface temperature of the plate when the heat loss by convection equals the solar energy absorbed by the plate is \(25.4^\circ C\).

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Most popular questions from this chapter

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: \(2945 \mathrm{~W}\)

What is a blackbody? How do real bodies differ from blackbodies?

Consider a person standing in a room at \(18^{\circ} \mathrm{C}\). Determine the total rate of heat transfer from this person if the exposed surface area and the skin temperature of the person are \(1.7 \mathrm{~m}^{2}\) and \(32^{\circ} \mathrm{C}\), respectively, and the convection heat transfer coefficient is \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Take the emissivity of the skin and the clothes to be \(0.9\), and assume the temperature of the inner surfaces of the room to be the same as the air temperature.

A 10 -cm-high and 20-cm-wide circuit board houses on its surface 100 closely spaced chips, each generating heat at a rate of \(0.12 \mathrm{~W}\) and transferring it by convection and radiation to the surrounding medium at \(40^{\circ} \mathrm{C}\). Heat transfer from the back surface of the board is negligible. If the combined convection and radiation heat transfer coefficient on the surface of the board is \(22 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), the average surface temperature of the chips is (a) \(41^{\circ} \mathrm{C}\) (b) \(54^{\circ} \mathrm{C}\) (c) \(67^{\circ} \mathrm{C}\) (d) \(76^{\circ} \mathrm{C}\) (e) \(82^{\circ} \mathrm{C}\)

Heat treatment is common in processing of semiconductor material. A 200-mm- diameter silicon wafer with thickness of \(725 \mu \mathrm{m}\) is being heat treated in a vacuum chamber by infrared heater. The surrounding walls of the chamber have a uniform temperature of \(310 \mathrm{~K}\). The infrared heater provides an incident radiation flux of \(200 \mathrm{~kW} / \mathrm{m}^{2}\) on the upper surface of the wafer, and the emissivity and absorptivity of the wafer surface are \(0.70\). Using a pyrometer, the lower surface temperature of the wafer is measured to be \(1000 \mathrm{~K}\). Assuming there is no radiation exchange between the lower surface of the wafer and the surroundings, determine the upper surface temperature of the wafer. (Note: A pyrometer is a non-contacting device that intercepts and measures thermal radiation. This device can be used to determine the temperature of an object's surface.)
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