- Two constant pressure operations (4-1) and (2-3) and
- Two friction less adiabatics (1-2) and (3-4).
These operations are discussed below :
1. Operation (4-1) : In this operation 1 kg of boiling water at temperature T1 is heated to form wet steam of dryness fraction x1. Thus heat is absorbed at constant temperature T1 and pressure p1 during this operation.
2. Operation (1-2) : During this operation steam is expanded isentropically to temperature
T2 and pressure p2. The point ‘2’ represents the condition of steam after expansion.
3. Operation (2-3) : During this operation heat is rejected at constant pressure p2 and
temperature T2. As the steam is exhausted it becomes wetter and cooled from 2 to 3.
4. Operation (3-4) : In this operation the wet steam at ‘3’ is compressed isentropically till
the steam regains its original state of temperature T1 and pressure p1. Thus cycle is completed.
Also read:- single acting reciprocating pump
Efficience of carnot cycle
Heat supplied at constant temperature T1 [operation (4-1)] = T1 (s1 – s4) or T1 (s2 – s3).
Heat rejected at constant temperature T2 (operation 2-3) = T2 (s2 – s3).
Since there is no exchange of heat during isentropic operations (1-2) and (3-4)
Net work done = Heat supplied – Heat rejected
= T1 (s2 – s3) – T2 (s2 – s3)
= (T1 – T2) (s2 – s3).