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Calculating the thermodynamic efficency of an electrical energy power station

  • Thread starter higheye
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  • #1
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I have to calculate the thermodynamic efficiency of a power station. I have to assume the cooling towers maintain 70 degrees C and hence the amount of electrical energy (in J) from one mole of methan. I've been given the enthalpy of combustion of Methane. I know the thermo efficiency is (sum of net useful power + and useful thermal outputs) / total fuel input.....(am i right?) i just cant figure out how to get the electrical energy!! I have figured out how many moles of water the combustion of methane heats, and figured out the energy needed to take that one mole to steam at 500 degrees C (a question earlier i think it has relevance) but thats all i got, i dont know how to calculate the electrical energy output from just this information!! Is there something in the question missing, or am i missing something?? Any help is greatly appreciated as i've been up all night doing this thanks
 

Answers and Replies

  • #2
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It is really difficult to understand your problem. Could you describe it exactly as it is given to you, with all the data, and show what you have already done?
 
  • #3
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This is the 4th question of the sheet, it directly asks;

Once the power station ia ready, calculate its thrmodynamic efficiency, assuming cooling towers maintain a temperatue of 70 degrees C and hence the amount of electrical energy (in J) from one mole of methane.

Thats the question as it is on paper.

The whole sheet is on a power station starting up from cold. Initally the boiler contains 23,000 kg of water at a temp of 10 degree C.

I was then asked to calculate the energy need to raise water to boiling point
The energy required to vaporise all water to steam
Energy required to raise temp of steam to 500 degrees C

It told me to add these 3 results then, (from which I got 119 GJ), and stated this is the power stations (rough estimate) amount of energy needed to prepare it for electrical generation.
My next question is to calculate the mass of methane that must be burnt to prepare the power station for generation. I calculated this, and now on the 5th question that I wrote at the begining.

Data I been given: Molar heat capacity of water - 75.29 J K-1 Mol-1
'' '' '' steam - 36.54 J K-1 Mol-1
Enthalpy of Vapourisation of water - 40.656 kJ Mol-1
Enthalpy of combustion if methane - -890 kJ Mol-1
Gibbs free energy for the combustion of methane - -845 kJ Mol-1
 
  • #4
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I am lead to believe that 500 C is the "high" temperature, and 70 C is the "low" temperature. This is enough to determine the Carnot cycle efficiency.
 
  • #5
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Yeah i picked up on this difference in temperature but didn't know what to do with it and what the loss of 430 C ment, i'll research the carnot cycle now thanks
 
  • #6
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efficiency = work done by system / energy put in

am i right in calculating the work done is the energy used to transfer 890 kJ from the methane, heating up the water to boiling then vapourising then heating steam to 500 C?

Is the above equation the same as: 1 - (the absolute temperature of cold resevoir/the absolute temperature of hot resevoir)
 
  • #7
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efficiency = work done by system / energy put in

am i right in calculating the work done is the energy used to transfer 890 kJ from the methane, heating up the water to boiling then vapourising then heating steam to 500 C?
No, this is the work done by the heat engine, resulting from the heat it consumes. In your case the work done is spinning the generator's shaft. You can't calculate this directly, rather, you compute the efficiency in some other way and since you know the input heat, you can then estimate the work done.

Is the above equation the same as: 1 - (the absolute temperature of cold resevoir/the absolute temperature of hot resevoir)
In the case of the Carnot cycle, yes. Note that a real power plant does not use the Carnot cycle, the Carnot cycle is only the the upper bound theoretical estimate.
 
  • #8
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ah so i can take the second equation like this: 1 - ( 70 / 500 ) = 0.86

then I can re-arrange the the top one: work done = 0.86 * 890 kJ?

this come out to work done = 765.4 kJ

Is this sensible? i really dont know?
 
  • #9
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ah so i can take the second equation like this: 1 - ( 70 / 500 ) = 0.86
You need to use the absolute temperature.

then I can re-arrange the the top one: work done = 0.86 * 890 kJ?

this come out to work done = 765.4 kJ
Yes, that's the principle (but the numbers are wrong). Note that what you get is mechanical energy. There is also some loss of energy when it is converted to electrical energy, but the loss is negligible. I leave it to you to find out the efficiency of power plant electrical generators to prove that.
 
  • #10
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I re-calculated the mechanical energy and resulted in 356 kJ, I have been researching and to no avail have I been able to progress with working out how much electrical energy is produced. Maybe im looking for the wrong thing? I been looking for equations or conversions for mechanical enrergy into electrical energy through generators. Am i at least on the right path? Maybe I need sleep
 
  • #11
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I re-calculated the mechanical energy and resulted in 356 kJ
This seems incorrect. Please show your calculations.

I have been researching and to no avail have I been able to progress with working out how much electrical energy is produced. Maybe im looking for the wrong thing? I been looking for equations or conversions for mechanical enrergy into electrical energy through generators. Am i at least on the right path? Maybe I need sleep
You merely need to find some facts on the current electrical generators. For example: http://www.energy.siemens.com/hq/en/power-generation/generators/sgen-4000w.htm
 
  • #12
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I mistaskenly used the upper temperature as 575.15 K instead of 773.15 K. I have repremanded this and my new results is 498.4 kJ
 
  • #13
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1 - ( 343.15 / 773.15 )
 
  • #14
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This result seems correct.
 
  • #15
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Is there a way to find out how much electrical energy can be generated form 498.4 kJ of mechanical? Im seriously out of practice with area of physics sorry
 
  • #16
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is the carnot efficiency, at 56%, not the answer?

why has it given me the 1 mole information and electrical energy information?
 
  • #17
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The efficiency of good electrical generators is close to 100% (see the page I linked earlier). So you can assume 100% mechanical energy is converted into electrical. Now, given the energy content of one mole of the fuel, you can find out the resultant mechanical energy (via efficiency) and thus the electrical energy.
 

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