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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.
Links of Previous Main Topic:-
- Statement of zeroth law of thermodynamics with explanation
- Heat and work introduction
- Open system and control volume
- Conversion of work into heat
- Introduction to carnot cycle
Links of Next Mechanical Engineering Topics:-
- Carnots theorem
- Absolute thermodynamic temperature scale or kelvin scale
- Absolute zero on thermodynamic temperature scale
- Clausius inequality entropy and irreversibility introduction
- Ideal gas or perfect gas
- Introduction about air standard cycles
- Properties of pure substances introduction
- Vapour compression refrigeration cycle introduction
- Basic fluid mechanics and properties of fluids introduction
