The figure below shows the schematic diagram of an ideal vapor power plant in wh

Question

The figure below shows the schematic diagram of an ideal vapor power plant in which steam with a mass flow rate of 50 kg/s steadily circulates through the four components shown. The steam enters the turbine at 3 MPa and 360 0C and is condensed in the condenser at a pressure of 70 KPa. Draw a (T-S) diagram and Determine:

The power developed by the turbine

The power input to the pump

The thermal efficiency of the ideal cycle

The mass flow rate of condenser cooling water if the water experiences a temperature increase from 15 to 25 0F

Redo the problem using an isentropic efficiency of 0.8 for the turbine and 0.75 for the pump.

     Boiler                                                                                                    Wout

Explanation / Answer

For stem properties, see http://www.irc.wisc.edu/properties/

At 3 MPa, 360 deg C, we get h1 = 3140 kJ / kg and s1 = 6780 J / kg-K

At s2 = s1 = 6780 J / kg-K and P2 = 70 kPa, we get h2 = 2410 kJ / kg

At P3 = 70 kPa and quality = 0 (sat.liquid) we get h3 = 377 kJ / kg and s1 = 1190 j /kg - K

At P4 = 3 MPa and s4 = s3 = 1190 j / kg-K, we get h4 = 379 kJ / kg

a)

Power developed by turbine = m [(h1 - h2)] = 50*(730) = 36500 kW

b)

Pump work = m(h4 - h3) = 100 kW

c) Thermal eff = Net Turb output / [m(h1 - h4)]

= (36500-100) / [50*(3140-379)]

= 0.2637 or 26.37 %

d)

Temperature rise = 15 to 25 F = 263.7 K to 269.3 K = 5.6 K

m_water*Cp*5.6 = m_steam*(h2 - h3)

m_water*4.18*5.6 = 50*(2410 - 377)

m_water = 4342.5 kg/s

e)

Turb power = 0.8*36500 = 29200 kW

Pump power = 100 / 0.75 = 133.3 kW

Eff = (29200 - 133.3) / [50*(3140-379)] = 0.21 or 21 %

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.