Answer:
engine B is more efficient.
Explanation:
We know that Carnot cycle is an ideal cycle for all working heat engine.In Carnot cycle there are four processes in which two are constant temperature processes and others two are isentropic process.
We also kn ow that the efficiency of Carnot cycle given as follows
Here temperature should be in Kelvin.
For engine A
For engine B
So from above we can say that engine B is more efficient.
Answer:
Explanation:
First, we will find actual properties at given inlet and outlet states by the use of steam tables:
AT INLET:
At 4MPa and 350°C, from the superheated table:
h₁ = 3093.3 KJ/kg
s₁ = 6.5843 KJ/kg.K
AT OUTLET:
At P₂ = 125 KPa and steam is saturated in vapor state:
h₂ = = 2684.9 KJ/kg
Now, for the isentropic enthalpy, we have:
P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K
Since s₂ is less than and greater than
at 125 KPa. Therefore, the steam is in a saturated mixture state. So:
Now, we will find (enthalpy at the outlet for the isentropic process):
Now, the isentropic efficiency of the turbine can be given as follows: