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Integrated Concepts(a) A large electrical power facility produces 1600 MW of “waste heat,” which is dissipated to the environment in cooling towers by warming air flowing through the towers by$5.00^{\circ} {C}$ . What is the necessary flow rate of air in$m^{3} / s$? (b) Is your result consistent with the large cooling towers used by many large electrical power plants?

Part $(a) :$ The flow rate is $3.4 \times 10^{5} \mathrm{m}^{3} / \mathrm{s}$Part $(\mathrm{b}) :$ The result is very high and not consistent with the large cooling tower used by the electrical power plants.

Physics 101 Mechanics

Chapter 14

Heat and Heat Transfer Methods

Thermal Properties of Matter

Rutgers, The State University of New Jersey

University of Washington

Hope College

McMaster University

Lectures

02:20

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06:13

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03:55

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01:31

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20.5. A certain nuclear-po…

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05:22

You're the environmen…

to solve this question. We are going to use the formula which relates heat with mass specific heat on difference in the temperature, since we want, uh, per second values that his volume per second on also the heat is not given to us. But the power is given to us in terms of what which is Q by t. So we'll divide both sides by t. Also, we can replace mass bye density times volume so we can find the volume rate. Now we know that Cube I t is given a 1600 megawatt that is 1600 times 10 power, six masses replaced with Roe and be that his density and we volume. So let's keep really worried Vitus separate because that's the quantity we want to calculate. Density times C. Times T is what we need. Now let's substitute density times specific It times Delta T density is 1.269 for I guess that's the average density killed in terms of kilograms per meter cube, specifically it off a gas. It's 7 21 June's per kilogram. Remember to use proper units on dhe. The difference in the temperature as the question say's is five decree Celcius. So let's calculate this. So 1.269 signs 7 21 times five on We're dividing this number by 1600 May the words. So the answer here will be three point forward 97 So if you calculate this very mighty, you will get three point for nine seven times 10 Power five Mater Cube The seconds. So let's keep only two significant figures That will be 3.5 times 10 power, five meter cube for seconds. And in the part be question asks if your result is consistent with the large cooling towers used by many, largely to be power plants. Well, it is not. Result is very high. If you compare the values off air flow rate by the ceiling fan, it is in some feet cubed per second per hour. So using many friends off that capacity or even higher capacity cannot produce this high value. The result is too high. Thank you for listening

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