Question

Problem 3. (30') Water is the working fluid in an ideal Rankine cycle with regeneration. All steam enters the first turbine stage at 12 MPa, 520 °C and expands to 1 MPa where some of the steam is extracted and diverted to the open feedwater heat exchanger operating at 1 MPa. The remaining steam keeps expanding through the second turbine stage and exits at 6 kPa, then goes through a heat exchange to cool to saturated liquid water. After that, the pump increases the pressure back to 1 MPa. Then it goes to the feedwater heat exchanger. Satureated liquid water leaves the feedwater heat exchanger then pumped back to 12 MPa. Find: (1) Sketch the process on a T-s diagram. (6') (2) Thermal efficiency. (22') (3) Mass flow rate of the total steam if net power output is 330 MW. (2')

Answer 1

Part a

T-S diagram of the process

a

p=12MPa p= 1 MPa 2 p=6kPa 4 3

Part b

Determine the properties from steam table at every state

Initial conditions

State 1

P1 = 12 MPa

T1 = 520 °C

Enthalpy h1 = 3404 kJ/kg

Entropy S1 = 6.5584 kJ/kg-K

State 2 at P2 = 1 MPa

This is an isentropic process

Entropy S2 = S1 = 6.5584 kJ/kg-K

Enthalpy h2 = 2823.3 kJ/kg

State 3

P3 = 6 kPa

S3 = S1 = 6.5584 kJ/kg-K

Enthalpy h3 = 2058.2 kJ/kg

State 4 is saturated liquid

P4 = 6 kPa

Enthalpy h4 = 151.53 kJ/kg

Entropy S4 = 0.5210 kJ/kg-K

Specific volume v4 = 0.0010064 m3/kg

State 5

P5 = P2 = 1 MPa

S5 = S4 = 0.5210 kJ/kg-K

Enthalpy Balance

h5 = h4 + v4 (P5 - P4)

= 151.53 kJ/kg + 0.0010064 m3/kg * (1000 - 6) kPa

= 152.53 kJ/kg

State 6 is saturated liquid

P6 = 1 MPa

Enthalpy h6 = 762.81 kJ/kg

Entropy S6 = 2.1387 kJ/kg-K

Specific volume v6 = 0.0011273 m3/kg

State 7

P7 = P1 = 12 MPa

Entropy S7 = S6 = 2.1387 kJ/kg-K

Enthalpy Balance

h7 = h6 + v6 (P7 - P6)

= 762.81 kJ/kg + 0.0011273 m3/kg * (12000 - 1000) kPa

= 775.21 kJ/kg

At state 2

Enthalpy Balance, quality of steam

x =( h6 - h5) /( h2 - h5)

= (762.81 - 152.53) /(2823.3 - 152.53)

= 0.2285

Specific work done by turbine

Wt/m1 =( h1 - h2) + (1- x) (h2 - h3)

=( 3404 - 2823.3) + (1-0.2285)(2823.3 - 2058.2)

= 1170.97 kJ/kg

Specific work done by pump

Wp/m1 =( h7 - h6) + (1- x) (h5 - h4)

=( 775.21 - 762.81) + (1-0.2285)(152.53 - 151.53)

= 13.17 kJ/kg

Heat in For steam generator

Q/m1 = h1 - h7

= 3404 - 775.21

= 2628.79 kJ/kg

Thermal efficiency

= (work done by turbine - work done by pump) / heat inlet to steam generator

= (1170.97 - 13.17)kJ/kg / (2628.79 kJ/kg)

= 0.4404

= 44.04%

Part C

Mass flow rate of total steam

= net power output / (work done by turbine - work done by pump)

= 330 MW x 1000 kW/MW / (1170.97 - 13.17)kJ/kg

= 285.02 kg/s