In the ideal Rankine Cycle, pump and turbine operate isentropically, meaning their efficiency is 100% but, in the actual cycle, due to the irreversibilities, there are a few differences. The two main causes for irreversibilities are:
– heat loss to the surroundings
– the friction caused between the steam and the cycle components.
Heat is lost to the surroundings as the steam flows through the cycle components. To compensate the heat loss it is necessary to increase the heat transfer in the boiler.
Friction causes pressure to drop in a way that the pressure on the turbine’s entrance is a bit lower than the pressure right on the boiler’s exit. Lower pressure means less work output from the turbine. In the same way, the pump needs to operate at a higher pressure to compensate these pressure variations, consuming more power.
So, if the pump demands more power and the turbine produces less work then, the cycle’s efficiency will drop.
Pump isentropic efficiency
The efficiency is a comparison between the isentropic work and the actual work. For pumps, the isentropic work is the least work that a pump requires in order to produce a certain pressure rise.
Example
Q: A pump with an efficiency of 70%, raises a flow of water at a rate of 1,35 kg/s and 30 ºC, from 0,1 MPa to 4 MPa. What is the pump’s actual work?
S: to calculate the pump’s actual work, first, we need to calculate the actual enthalpy, h2a.
A: The actual pump work is 7,52 kW.
Turbine isentropic efficiency
For turbines, the isentropic work is the maximum output that a turbine can produce.
Example
Q: Steam enters a turbine at 15 MPa and 600 ºC and leaves at 10 kPa. The turbine’s isentropic efficiency is 87%. Determine the power output from the turbine in kJ/kg.
S: to calculate the turbine’s power output, we need to calculate the actual enthalpy, h2a.
A: The turbine’s actual work is 1277 kJ/kg.