For this set of questions, give your answers correct to 2 decimal places. Solar

Question

For this set of questions, give your answers correct to 2 decimal places. Solar radiation with intensity of 928 W/m2 is incident on a solar thermal collector. The collector cover has transmission of 0.91 and the collector absorber has absorption of 0.96. How much solar radiation is absorbed by the absorber? Give your answer in W/m2 Answer: 890.88 The correct answer is: 810.70 f the collector has an area of 2.03 m2, what is the power input to the collector in kW? Answer: 1.81 The correct answer is: 1.65 Suppose that a propylene glycol solution is flowing through the collector at 0.005 kg/s and enters at 9.1°C. If the collector is operating with an overall efficiency of 49%, what is the temperature of the fluid leaving the collector? Assume that the specific heat capacity of the propylene glycol solution is 3.83475 kJ/kg C Answer: 37.10 The correct answer is: 57.24 Suppose that water is flowing through the collector at 0.0056 kg/s. The water is heated from 13.6°C to 38.7°C. Use a specific heat capacity of 4.199 kJ/kg C for water. What is the efficiency of the collector? Give your answer in %. Answer: 32.64

Explanation / Answer

1. Incident radiation= 928 W/m2

Collector has transmission of 0.91 and absorption of 0.96

so, total radiation absorbed=928 W/m2 * 0.91 * 0.96

=> 810.70 W/m2

2. If collector has an area of 2.03 m2,

then power output=> 810.70 W/m2 * 2.03 m2

=> 1645.7 W => 1.65 kW

3. Propylene glycol flows and absorbs the heat from the collector with an overall efficiency of 49%.

so, heat incident on collector * efficiency = heat gained by propylene glycol

=> 928 W/m2 * 2.03 m2 * 0.49 = 0.005 kg/s * 3.83475 kJ/kg°C * (Tf - 9.1°C)

where Tf = final temperature of ethylene glycol

on solving,

=> 0.92 kJ/s = 0.019 * (Tf- 9.1°C)

Tf-9.1°C = 48.14

=> Tf= 48.14 °C + 9.1°C => 57.14 °C

4. Now, similarly for water

let the efficiency be x.

then,

heat incident on collector*efficiency= heat absorbed by water

=> 928 W/m2 * 2.03 m2 * x = 0.0056 kg/s * 4.199 kJ/kg°C * (38.7-13.6)°C

=> 1.884 kJ/s * x = 0.590 kJ/s

=> x = 0.590/1.884 => 31.33%

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