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

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 -5°C. The entire work output of the heat engine is used to drive a heat pump that extracts heat from the ambient air at -5°C and transfer it to a heated space at 27°C. a) Draw the Schematic Diagram of the Combined System. b) Determine the maximum rate of heat supplied to the heated space, in kJ/min. c) Determine the net rate of heat transfer with the ambient air, in kJ/min.

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 at 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 and we know that the sense the heat engine is a car note cycle is a is a is based on the carnote cycle so we can say that the efficiency will be equals to one minus t lower divided by t higher and simply will plug the values one minus here t lower is 27 degrees celsius if you convert this thing into kelvin will get 300 kelvin and here will get 100 one hundred one thousand one hundred seventy three kelvin so one 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 divided by minus 5 degrees 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 dot so again cop of refrigerator can also be written in this way also ql dot of this time is for a refrigerator divided by the rate of work input so this is a rate of work input and this thing you 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 starts to be nearly 4 ,982 kilojoule per minute.

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 a total heat of the heat, the total rate of heat rejection to the ambient.

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

04:55 This is qh, this is sorry, this is ql for heat engine and this is for qh for refrigerators.

05:13 So we have to find the sum of this and this.

05:19 So first of all, we'll be calculating this ql dot for heat engine.

05:24 So we'll be applying the first law of thermodynamics here...

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