Steam is expanded from 800 F and 250 psia to 5 psia through an 85% efficient turbine. What is the enthalpy of the steam?

ANSWER: Since T=800 F > Tcritical, the initial steam is superheated. Enthalpy and entropy at this stage is obtained by:

PHASE: Superheated Steam

UNITS: US

Given T800 P250

Find H, H=1423.4082 (T=800 P=250)

S, S=1.7405 (T=800 P=250)

If the expansion is isentropic (100% efficient), then the enthalpy at the end of the expansion can be found:

PHASE: Water-Steam Mix

UNITS: US

Given P5 S1.7405

Find H, H=1066.5264 (P=5 S=1.7405)

The energy extracted by the ideal turbine is:

H1-H2=1423.4082-1066.5264=356.8818 Btu/lbm

The energy extracted by the real turbine is:

85% (H1-H2)=0.85*356.8818=303.3495 Btu/lbm

The final enthalpy is:

H1-303.3495=1120.0587 Btu/lbm

A boiler feed pump increases the pressure of 15 psia 100 F water to 180 psia. What is the final water temperature if the pump has an isentropic efficiency of 80%?

ANSWER:

PHASE: Compressed Water

UNITS: US

Given T100 P15

Find H, H=68.0360 (T=100 P=15)

S, S=0.1295 (T=100 P=15)

If the compression is isentropic (100% efficient), then the enthalpy at the end of compression can be found:

PHASE: Compressed Water

UNITS: US

Given P180 S0.1295

Find H, H=68.5101 (P=180 S=0.1295)

The energy added to the water by the ideal pump is:

H2-H1=68.5101-68.0360=0.4741 Btu/lbm

Since the pump is not 100% efficient, not all of the 0.4741 Btu/lbm enthalpy increase goes into raising the pressure. Therefore more energy must be added to the water:

(H2-H1)/80% =0.4741/0.8=0.5926 Btu/lbm

The final enthalpy is:

H1+0.5926=68.6286 Btu/lbm

The final temperature can be found:

PHASE: Compressed Water

UNITS: US

Given P180 H68.6286

Find T, T=100.1576 (P=180 H=68.6286)

End of Advanced Problems.