the efficiency of a heat engine is given by:

efficiency = net work output divide by total heat input
= heat input minus heat output divide by heat input

there is an upper limit to the efficiency given by:

maximum effiecency = 1 - low temp divide by high temp

a heat pump is a heat engine woerking in reverse. Its coefficient performance is given by:

coefficeint of performance = heat supplied to high temp divide by work done

questions:

1. what is the maximum possible efficiency of a stean engine working with a steam at a high temp of 127 degree celsius (400 K) and a low temp, at which it condenses, of 77 degrees celsius (350 K)?

2. how do the temperatures used need to be changed in order to increase the efficiency?

3. what would the low temp need to be to have a 100% efficient engine?

4. a power station has a pratical effiecincy of 33% and a maximum theoretical efficiency of 68%. Sugest reasons these figures are so different.

5. a coventional power station (coal fired) has an electrical output of 1200 MW. Its actual effiecincy is 36%.
a) how much heat is provided by the burning coal each second?
b) how much heat is wasted by the power station each second?
c) state 2 ways in which this wasted heat can be removed from the power station

1. what is the maximum possible efficiency of a steam engine working with a steam at a high temp of 127 degree Celsius (400 K) and a low temp, at which it condenses, of 77 degrees Celsius (350 K)?


From what you wrote down, the maximum efficiency is given by

Max = 1 - \frac {T_{low}}{T_{high}} = 1 - \frac {350\,K}{400\,K}



2. how do the temperatures used need to be changed in order to increase the efficiency?


Should be pretty obvious. Try changing the temperatures in the above equation.



3. what would the low temp need to be to have a 100% efficient engine?


100%= 1 = 1 - \frac {LowT}{400\,K}

Solve for LowT



4. a power station has a practical efficiency of 33% and a maximum theoretical efficiency of 68%. Suggest reasons these figures are so different.


The first thing that you would think of is heat loss from friction. Can you come up with other possibilities as to why the engine isn't 100% efficient?



5. a conventional power station (coal fired) has an electrical output of 1200 MW. Its actual efficiency is 36%.
a) how much heat is provided by the burning coal each second?
b) how much heat is wasted by the power station each second?
c) state 2 ways in which this wasted heat can be removed from the power station


Output = 1200 \times 10^6\,watts

Actual efficiency is 36% so

a. Total \, input \, heat = \frac {1200 \times 10^6\,watts}{0.36}

You can leave this as watts or you can convert it to calories or BTUs or whatever.

b. Just subtract the output heat from the input heat.

c. How could you cool off the power station if you had to do it?

Thank you for your help it helped me alto but for the first question I got the answer 125%... im not sure about this.. can you please tell me if this is right

125%? You can't get over 100% efficiency. That would be a "perpetual motion machine" and perpetual motion is impossible.

I think you goofed on your decimal point:

1-\frac {350}{400} = 1-0.875 = 0.125\,(12.5%)

125%? You can't get over 100% efficiency. That would be a "perpetual motion machine" and perpetual motion is impossible.

I think you goofed on your decimal point:

1-\frac {350}{400} = 1-0.875 = 0.125\,(12.5%)

thank you i now understand... i also have more questions i have posted with the title more on gas laws and heat engines. Please would you be able to help me on those questions as well