Question

A proposed steam power plant design consists of an ideal Rankine cycle with reheat and regeneration. Steam enters Turbine 1 at P1 and T1 at the rate of m1 and exits at P2. A fraction (y') of the steam exiting Turbine 1 is diverted to an open feedwater heater while the remainder is reheated to T3 before entering Turbine 2. The condenser operates at P4. Saturated liquid exits the condenser and is fed to Pump 1. The outlet of Pump 1 is fed into the open feedwater heater. Saturated liquid exits the feedwater heater and is fed to Pump 2. All turbines and pumps are isentropic.

--Given Values--
m1 (kg/s) = 50
P1 (Bar) = 140
T1 (C) = 560
P2 (Bar) = 10
T3 (C) = 500
P4 (Bar) = 0.08

a) Determine the specific enthalpy (kJ/kg) at the inlet of turbine 1.
  
b) Determine the specific enthalpy (kJ/kg) at the exit of turbine 1.
  
c) Determine the specific enthalpy (kJ/kg) at the inlet of turbine 2 .
  
d) Determine the specific enthalpy (kJ/kg) at the exit of turbine 2.
  
e) Determine the specific enthalpy (kJ/kg) at the condenser exit.
  
f) Determine the specific enthalpy (kJ/kg) at the exit of the low pressure pump.
  
g) Determine the specific enthalpy (kJ/kg) at the exit of the feedwater heater.
  
h) Determine the specific enthalpy (kJ/kg) at the exit of the high pressure pump.
  
i) Determine the fraction (y') of flow diverted to the open feedwater heater.
  
j) Determine the power (MW) produced by turbine 1.
  
k) DeteRmine the power (MW) produced by turbine 2.
  
l) Determine the power (kW) required (a positive number) by the low pressure pump.
  
m) Determine the power (kW) required (a positive number) by the high pressure pump.
  
n) Determine the total rate of heat transfer (MW) supplied to the boiler.
  
o) Determine the thermal efficiency (%) of the power plant.
  

Answer 1

Consider below T-S diagram for further discussion.

To calculate properties at various points, steam table is used and wherever required, interpolation is performed.

(a) Specific enthalpy at point-1 is enthalpy at P1 = 140 bar and T1 = 560 deg C, specific enthalpy,

h1 = 3487.5 kJ/kg

(b) Specific enthalpy at exit of turbine-1 (i.e. Point -2) can be found by isentopic process 1-2 (as turbine is isentropic).

hence s2 = s1. (s1 = entropy at P1 & T1 = 6.5968 kJ / kg K)

hence, specific enthalpy at 2 = enthalpy at P2 = 10 bar and s2 = 6.5968 kJ / kg K

hence, h2 = 2782.5 kJ/kg

(c) specific enthalpy at inlet of turbine 2 (i.e. point -3) is enthalpy at P3 = P2 = 10 bar and T3 = 500 deg C.

hence, h3 = 3479.1 kJ/kg

(d) specific enthalpy at exit of turbine 2 (i.e. point-4) is enthalpy at P4 = 0.08 bar and s4 = s3 = 7.7641 kJ/ kg K.

hence, h4 = 2430.4 kJ/kg

(e) specific enthalpy at condenser exit (point-5) is hf at P4.

hence, h5 = 173.84 kJ/kg

(f) specific enthalpy at pump-1 exit (point-6) is enthalpy at s6 = s5 = 0.59249 kJ/ kg K and P6 = 10 bar.

hence h6 = 174.84 kJ/kg

(g) specific enthalpy at the exit of the feedwater heater (point-7) is hf at P7 = 10 bar (as it is mentioned in question that saturated liquid exits the feedwater heater.)

hence, h7 = 762.52 kJ/kg

(h) specific enthalpy at the exit of the high pressure pump (point-8) is enthalpy at s8 = s7 = 2.1381 kJ/ kg K and P8 = 140 bar

hence, h8 = 777.13 kJ/kg

(i) To determine y' we need to consider energy balance in feed water heater,

considering unit mass flow in the system,

m2 h2 + m6 h6 = m7 h7

but m2 = y' kg, m6 = 1-y' kg/s and m7 = m2+m6 = 1 kg/s

y' (2782.5) + (1-y') (174.84) = 1 (762.52)

hence. y' = 0.2254

(j) Power produced by Turbine - 1 :

P1 = m1 (h1 - h2) = 50 (3487.5 - 2782.5) = 35250 kW = 35.25 MW

(k) Power produced by Turbine - 2 :

P2 = m3 (h3 - h4) , where m3 = (1-y') m1 = (1-0.2254) 50 = 38.73 kg/s

P2 = 38.73 (3479.1 - 2430.4) = 40616.151 kW = 40.616 MW

(l) Power required by low pressure pump

W1 = m5 (h6 - h5) , where m5 = (1-y') m1 = 38.73 kg/s

W1 = 38.73 (174.84-173.84) = 38.73 kW

(m) Power required by high pressure pump

W2 = m7 (h8 - h7) , where m7 = m1 = 50 kg/s

W2 = 50 (777.13-762.52) = 730.5 kW

(n) Total heat supplied to boiler :

Q = m1 (h1 - h8 ) + m2 (h3 - h2)

where m1 = 50 kg/s and m2 = m3 = 38.73 kg/s

Q = 50 (3487.5 - 777.13) + 38.73 (3479.1 - 2782.5) = 162497.82 kW = 162.498 MW

(o) thermal efficiency (%) of the power plant = Total work gain / Total heat supplied

hence, Thermal Efficiency = 46.21 %