B) A Carnot heat engine receives heat at 1200 K and rejects...

b) A Carnot heat engine receives heat at 1200 K and rejects the waste heat to the environment at 25°C. The entire work output of the heat engine is used to drive a Carnot refrigerator that removes heat from the cooled space at -10°C at a rate of 300 kJ/min and rejects it to the same environment at 25°C. i) Draw the diagrams that show the components and flow of the process for the heat engine and the refrigerator. ii) Determine the rate of heat supplied to the heat engine (in kW). iii) Determine the total rate of heat rejection to the environment (in kW).

Transcript

00:01 Question, we have a carnot engine which is receiving at heat temperature t1 equals to 1200 kelvin and resetting the waste heat to the environment at a temperature t2 equals to 25 degrees centigrade that is equals to 298 kelvin, 298 kelvin.

00:21 The entire work output from the heat engine is used to derive a carnot refrigerator that removes heat from the colder space that is temperature equals to minus 10 degrees centigrade, this become 263 kelvin and rejecting the heat at the same environment that is 298 kelvin.

00:40 So at the rate which is given as suppose q3 which is equals to 300 kilojoule per minute and this will be equals to 5 kilojoules per second.

00:54 So for the part first we have to draw the diagram that shows the component and flow of of the process for the engine and the refrigerator.

01:02 So suppose this is our heat engine and it is receiving heat from here, this is temperature t1 and rejecting the heat at the temperature t2.

01:13 Okay, now we have a refrigerator.

01:17 This work output has been transferred to a refrigerator.

01:21 So suppose this is refrigerator and this is heat engine and this receive the work output and reject taking the heat from the temperature t3 and rejecting the heat at the same environment that is again t4.

01:36 So this t2 and t4 both are at the same temperature.

01:40 So heat is flowing from here and work output is in this direction and heat is rejecting in this direction and the refrigerator is taking heat from here.

01:50 So this value is q3 which is given and the refrigerator is rejecting heat in this direction.

01:56 So suppose this is q1, this is q2 and this is q3 and this is a little.

02:00 This is q4.

02:02 So this is the diagram that shows all the component.

02:05 Now for the second part, we can, we have to determine the rate of the heat supplied to the heat engine that is q1.

02:15 So we have to determine for the first part, q1 in the kilowatt.

02:19 So now for the refrigerator, first of all, we can write that refrigerator cop, it is equals to t lower or we can write that t lower, which is a lower.

02:30 Is actually t3 divided by t4 minus t3 this will be equals to q3 divided by work done w so we have all the values so we get t3 which is 263 divided by t4 which is 298 minus 263 this will be equals to q3 which is 5 kilojou per second divided by w so from here we get work work input to the refrigerator w that is equals to 0 .665 kilo so, suppose this is our equation number 1.

03:04 So now for the refrigerator, we can write that q4, it will be equals to q3 plus w.

03:10 So q3 is 5 plus w is 0 .265.

03:14 So from here we get q4 that is equal to 5 .265 kilojoule per second.

03:21 So suppose this is our equation number 2.

03:24 Now for the heat engine, we can write that the efficiency, this will be equal to efficiency, see this will be equals to t1 minus t2 divided by t1, this will be equals to work output w divided by heat input q1.

03:39 So from here after substituting values, so t1 which is 1 ,200 kelvin, 2 which is 298 kelvin divided by t1 which is 1, which is 1 ,200 kelvin that is equals to work output which is from the equation 1 is 0 .665 divided by heat input q1...

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