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ȠCarnot = Wnet/q1 =q1 – q2/ q1 = 1 – q2/q1

= 1 – (RT2 In v3/v4)/ (RT1 In v2/v1)

We are now applying adiabatic law to process 2 – 3 and 4 – 1 and getting

P2 u2r = p3 u3r

P2 u2 u2r-1 = p3 u3u3 r-1

RT1 u2r-1 = RT2 u3 r-1

T1 / T2 = ( u3 / u2) r-1

Similarly for process 4 – 1

T1 / T2 = ( u3 / u2) r-1

Equating equations (7.4) and (7.5) we get

u3 / u2 = u4 / u1

u2 / u1= u3 / u4

Thus equation (7.3) becomes

Where T1 = source temperature and T2 = Sink temperature

 From above equations it is clear that Carnot cycle depends on source and sinks temperature. Here the fluid as working medium doesn’t play any part. When you will need more efficiency in this process, you will simply increase T1. T2 cannot be decreased at any cost since atmosphere is taken down to sink.

Practical hindrances to perform this cycle:

There are certain points in the Carnot cycle process that will prove to be quite impractical:

  • The interchangeable cylinder head is impossible in real time experiments.
  • A complete frictionless movement of the piston in the cylinder is also impossible.
  • You will find isothermal expansion rather slow and adiabatic expansion very quick.
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