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Thermal efficiency of a coal powered power station

  • Thread starter coffeem
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  • #1
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[SOLVED] Thermal efficiency of a coal powered power station

Homework Statement



A company plans to build a coal powered plant for eletricity generation. The coal is to be burnt ina heat engine designed to maintain a hot resovoir temp Th, and a cold resovoir of 120 degs celcius. What is the minimum Th for the company to break even, if domestic users are charged 15p kWh? Company buys coal at 6p kWh. 10% of electrical energy is lost in distribution system. Ignore capital costs.


Homework Equations



n = 1 - Tc/Th

The Attempt at a Solution



Well n = 0.9*6/15 = 0.36

So rearranging the equation i get:

Th = Tc/0.64

So remebering to change the units to Kelvins:

Th = 392.4 K

Problem is the anwer should be: 703K so where am I going wrong? thanks
 

Answers and Replies

  • #2
alphysicist
Homework Helper
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Hi coffeem,

Homework Statement



A company plans to build a coal powered plant for eletricity generation. The coal is to be burnt ina heat engine designed to maintain a hot resovoir temp Th, and a cold resovoir of 120 degs celcius. What is the minimum Th for the company to break even, if domestic users are charged 15p kWh? Company buys coal at 6p kWh. 10% of electrical energy is lost in distribution system. Ignore capital costs.


Homework Equations



n = 1 - Tc/Th

The Attempt at a Solution



Well n = 0.9*6/15 = 0.36
I don't think this is right. The overall minimum efficiency (combined effect of the heat engine efficiency and distribution efficiency) is 6/15.
 
  • #3
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Hi coffeem,



I don't think this is right. The overall minimum efficiency (combined effect of the heat engine efficiency and distribution efficiency) is 6/15.

Hi thanks for taking your time to help. Im not 100& clear on what you mean. I agree that the efficiency should by 6/15 but as it said 10% is lost am I not correct in multiplying by 0.9 to account for this?

Also when I use that value as the efficiency level I still fail to get the required answer.
 
  • #4
alphysicist
Homework Helper
2,238
1
Yes, multiplying by 0.9 accounts for the efficiency of the distribution system, because since 10% is lost it is 90% efficient.

But in this problem there are three efficiencies to keep track of: the efficiency of the distribution system (which is 0.9), the efficiency of the heat engine (which you call n), and the efficiency of the total process (distribution system + heat engine) which is 6/15.

You did get the equation that the separate efficiencies here can be multiplied to get the total efficiency, but your equation (n=(0.9)(6/15)) has the terms in the wrong place. What do you get?
 
  • #5
91
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Oh ok. I get it that if i divide my 0.9 it works. Thanks.
 

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