A carnot heat engine receives heat from a reservoir at 900 c...

A Carnot heat engine receives heat from a reservoir at 900 C at a rate of 800 kJ/min and rejects the waste heat to the ambient air at 27 C. The work output is used to run devices in a well-insulated house nearby. If the house has only a Carnot air conditioner and a Carnot refrigerator. Portion of the work output of the heat engine is used to drive the refrigerator that keeps a refrigerated space at 0 C, while the other portion drives the air conditioner that keeps the house at 21 C. If the ambient temperature is 27 C, determine (a) the rate of heat removal from the refrigerated space, s(b) the total rate of heat rejection to the ambient air, (c) the air conditioner load, and (d) the power consumed by each device

Transcript

00:01 So here in this problem we are given that the carnot heat engine it received heat from a reservoir that is at 900 degrees celsius and the heat received the rate of heat received we are given as 800 kilojoule per minute and ambient condition we are given as the the lower temperature that is 27 degrees celsius if you convert this thing into kelvin we'll get 300 kelvin and we are given that the entire work output from the heat engine it is used by the refrigerator and the refrigerator space we have to maintain it at minus 5 degrees celsius and we have to calculate in the a part of the problem we have to calculate the maximum rate of heat removal from the refrigerated space that means we have to calculate this ql dot for refrigeration this thing if we see in the figure this thing we have to calculate so first of all to calculate this will be calculating the efficiency of the heat engine.

01:16 And we know that the sense the heat engine is a carnot cycle is based on the carnot cycle.

01:23 So we can say that the efficiency will be equals to 1 minus t lower divided by t higher.

01:30 And simply we'll plug the values 1 minus here t lower is 27 degrees celsius.

01:36 If you convert this thing into kelvin, we'll get 300 kelvin and here we'll get 100.

01:42 1 ,173 kelvin so 1 minus steel over is 300 kelvin and divide by 1 ,173 kelvin so we're going to solve this and we'll get the efficiency of the heat engine as 0 .744 now the maximum work output of this heat engine is determined from the definition of the thermal efficiency so we know that the work output the rate of work output so this is output is not input this is output and the same the work output is used by the refrigerator so it is equals to efficiency the thermal efficiency of the heat engine times qh we have all the values we can plug it here so this is 0 .744 times quh dot we have 800 kilojoule per minute so it comes out to be as nearly 595 .2 kilojoule per minute so this is the work output the rate of work output and the same work output is used by the refrigerator so we know that first of all it will be calculating the cop of the refrigerator so it is equals to 1 divided by th divided by tl minus 1 so you can plug all the values here 1 divided by th in this case is this 300 kelvin divide by minus 5 degree celsius that means it will be 268 kelvin minus 1 and it is equals to want to solve this and we get the cop as 8 .37 now we can calculate this ql and so again cop of refrigerator can also be written in this way also ql dot of this time is for a refrigerator divide by the rate of work input so this is the rate of work input and this thing we have to calculate cop we have already calculated it is 8 .37 is equals to ql dot for refrigerator we have to calculate and work input rate of work input is 595 .2 kilojoule per minute so from here the work this ql dot comes out to be as let me write here it comes out to be nearly 4 ,982 kilojoules j.

04:40 So this was the solution for the a part.

04:45 Next we'll be solving the b part of the problem and in the b part we have to calculate the total heat of the heat and the total rate of heat rejection to the ambient.

04:53 That means we have to calculate this heat...

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