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Chapter 9: Problem 132

Consider a flat-plate solar collector placed horizontally on the flat roof of a house. The collector is \(1.5 \mathrm{~m}\) wide and \(4.5 \mathrm{~m}\) long, and the average temperature of the exposed surface of the collector is \(42^{\circ} \mathrm{C}\). Determine the rate of heat loss from the collector by natural convection during a calm day when the ambient air temperature is \(8^{\circ} \mathrm{C}\). Also, determine the heat loss by radiation by taking the emissivity of the collector surface to be \(0.85\) and the effective sky temperature to be \(-15^{\circ} \mathrm{C}\).

Short Answer

Expert verified
In this exercise, we have found the rates of heat loss from a flat-plate solar collector by natural convection and radiation. The natural convection heat loss is 1241.39 W, and the radiation heat loss is 450.17 W.

Step by step solution

01

Calculate the surface area of the collector

We are provided the dimensions of the collector, which is \(1.5 \mathrm{~m}\) wide and \(4.5 \mathrm{~m}\) long. The area of the collector can be calculated using the formula: \(A = width \times length\) \(A = 1.5 \mathrm{~m} \times 4.5 \mathrm{~m} = 6.75 \mathrm{m^2}\)
02

Calculate the temperature difference for natural convection

We are provided the surface temperature, \(\mathrm{T_s}\), and the ambient temperature, \(\mathrm{T_{ambient}}\). The temperature difference can be calculated as follows: \(\Delta T = \mathrm{T_s - T_{ambient}}\) \(\Delta T = 42^{\circ} \mathrm{C} - 8^{\circ} \mathrm{C} = 34^{\circ} \mathrm{C}\)
03

Calculate the heat transfer coefficient for natural convection

To find the heat transfer coefficient (\(h\)), we can use the correlation for horizontal flat plates with temperatures warmer than ambient air, which states: \(h = 1.31 (\Delta T)^{1/3}\) Using the temperature difference calculated in Step 2: \(h = 1.31 (34)^{1/3} = 5.43 \mathrm{~W/m^2K}\)
04

Calculate the natural convection heat loss

Now, we can find the natural convection heat loss using the formula mentioned in the analysis: \(Q_{conv} = h\cdot A\cdot(\mathrm{T_s - T_{ambient}})\) \(Q_{conv} = 5.43 \mathrm{W/m^2K} \times 6.75 \mathrm{m^2} \times 34 \mathrm{~K} = 1241.39 \mathrm{~W}\)
05

Calculate the temperature difference for radiation heat loss

We are provided the surface temperature, \(\mathrm{T_s}\), and the effective sky temperature, \(\mathrm{T_{sky}}\). The temperature difference can be calculated as follows: \(\Delta T' = T_s - T_{sky} = 42^{\circ} \mathrm{C} - (-15^{\circ} \mathrm{C}) = 57^{\circ} \mathrm{C}\)
06

Calculate the radiation heat loss

Now, we can find the radiation heat loss using the formula mentioned in the analysis: \(Q_{rad} = \epsilon\cdot\sigma\cdot A\cdot(T_s^4 - T_{sky}^4)\) We are given the emissivity (\(\epsilon = 0.85\)) and will use the Stefan-Boltzmann constant (\(\sigma = 5.67 × 10^{-8} \mathrm{~W/m^2K^4}\)): \(Q_{rad} = 0.85 \times 5.67\times10^{-8} \mathrm{~W/m^2K^4} \times 6.75\mathrm{m^2} \times [(273 + 42)^4 - (273 - 15)^4] \mathrm{K^4} = 450.17 \mathrm{~W}\) The rate of heat loss from the collector by natural convection is \(1241.39 \mathrm{~W}\), and the heat loss by radiation is \(450.17 \mathrm{~W}\).

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Most popular questions from this chapter

The primary driving force for natural convection is (a) shear stress forces (b) buoyancy forces (c) pressure forces (d) surface tension forces (e) none of them

A hot fluid \(\left(k_{\text {fluid }}=0.72 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) is flowing as a laminar fully-developed flow inside a pipe with an inner diameter of \(35 \mathrm{~mm}\) and a wall thickness of \(5 \mathrm{~mm}\). The pipe is \(10 \mathrm{~m}\) long and the outer surface is exposed to air at \(10^{\circ} \mathrm{C}\). The average temperature difference between the hot fluid and the pipe inner surface is \(\Delta T_{\text {avg }}=10^{\circ} \mathrm{C}\), and the inner and outer surface temperatures are constant. Determine the outer surface temperature of the pipe. Evaluate the air properties at \(50^{\circ} \mathrm{C}\). Is this a good assumption?

Consider a \(15-\mathrm{cm} \times 20\)-cm printed circuit board \((\mathrm{PCB})\) that has electronic components on one side. The board is placed in a room at \(20^{\circ} \mathrm{C}\). The heat loss from the back surface of the board is negligible. If the circuit board is dissipating \(8 \mathrm{~W}\) of power in steady operation, determine the average temperature of the hot surface of the board, assuming the board is \((a)\) vertical, \((b)\) horizontal with hot surface facing up, and (c) horizontal with hot surface facing down. Take the emissivity of the surface of the board to be \(0.8\) and assume the surrounding surfaces to be at the same temperature as the air in the room. Evaluate air properties at a film temperature of \(32.5^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure. Is this a good assumption?

A vertical \(1.5\)-m-high and \(3.0\)-m-wide enclosure consists of two surfaces separated by a \(0.4-\mathrm{m}\) air gap at atmospheric pressure. If the surface temperatures across the air gap are measured to be \(280 \mathrm{~K}\) and \(336 \mathrm{~K}\) and the surface emissivities to be \(0.15\) and \(0.90\), determine the fraction of heat transferred through the enclosure by radiation. Answer: \(0.30\)

A \(50-\mathrm{cm} \times 50-\mathrm{cm}\) circuit board that contains 121 square chips on one side is to be cooled by combined natural convection and radiation by mounting it on a vertical surface in a room at \(25^{\circ} \mathrm{C}\). Each chip dissipates \(0.18 \mathrm{~W}\) of power, and the emissivity of the chip surfaces is 0.7. Assuming the heat transfer from the back side of the circuit board to be negligible, and the temperature of the surrounding surfaces to be the same as the air temperature of the room, determine the surface temperature of the chips. Evaluate air properties at a film temperature of \(30^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure. Is this a good assumption?
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