# The Ideal Reheat Rankine Cycle

One of the problems in increasing the efficiency of the ideal Rankine cycle is the moisture content of the steam at the final stages of the expansion. A two-stage turbine with reheat in between, is the solution to the excessive moisture problem in turbines and, is commonly used in actual steam power plants. The sole purpose of the reheat cycle is to reduce the moisture content of the steam at the final stages of the expansion process.

The difference between the simple ideal Rankine cycle and the ideal reheat cycle is that the expansion process occurs in two stages. In the first stage, the high-pressure stage, the steam expands isentropically to an intermediate pressure and is sent back to the boiler, where it is reheated at constant pressure, to the first inlet temperature. In the second stage, the low-pressure stage, the steam expands isentropically to the condenser’s pressure. The total heat input and the total turbine work output for a reheat cycle become:

qin = qprimary + qreheat =  (h3 – h2) + (h5 – h4)

and:

wturb,out = wturb,high-P + wturb,low-P = (h3 – h4) + (h5 – h6)

wturb,out = wturb,high- + wturb,low-p = (h3 – h4) + (h5 – h6)

A single reheat process in a modern power plant improves the cycle efficiency by 4 to 5 percent by increasing the average temperature at which heat is transferred to the steam. More reheat processes can be used but, due to the added cost and complexity to modify the cycle, it is not practical to use more than two reheat processes.

## Example

A steam power plant operates on the ideal reheat Rankine cycle. Steam enters the high-pressure turbine at 8 MPa and 500°C and leaves at 3 MPa. Steam is then reheated at constant pressure to 500°C before it expands to 20 kPa in the low-pressure turbine. Determine the turbine work output, in kJ/kg, and the thermal efficiency of the cycle. This example was solved using Engineering Equation Solver, or just EES, but steam charts can be used also.

```"State 1"
P1=20 [kPa]
x1=0
h1=Enthalpy(Steam;x=x1;P=P1)
s1=Entropy(Steam;x=x1;P=P1)
"State 2"
P2=8000 [kPa]
s2=s1
h2=Enthalpy(Steam;s=s2;P=P2)
"State 3"
P3=8000 [kPa]
T3=500 [C]
h3=Enthalpy(Steam;T=T3;P=P3)
s3=Entropy(Steam;T=T3;P=P3)
"State 4"
P4=3000 [kPa]
s4=s3
h4=Enthalpy(Steam;s=s4;P=P4)
"State 5"
P5=3000 [kPa]
T5=500 [C]
h5=Enthalpy(Steam;T=T5;P=P5)
s5=Entropy(Steam;T=T5;P=P5)
"State 6"
P6=20 [kPa]
s6=s5
h6=Enthalpy(Steam;s=s6;P=P6)
"Turbine Work Output"
w_turb=(h3-h4)+(h5-h6)
"Thermal efficiency of the cycle"
q_in=(h3-h2)+(h5-h4)
eta_therm=w_turb/q_in``` The turbine output is 1367 kJ/kg and the thermal efficiency of the cycle is 0,39 or 39%.