2 Steam is the working fluid in an ideal Rankine cycle Steam

2.         Steam is the working fluid in an ideal Rankine cycle. Steam enters the turbine at 12 MPa, 600^oC and a mass flow rate of 5x10⁵ kg/hr. The condenser operates at 8 kPa and cooling water enters the condenser at 20^oC and exits at 35^oC . For this cycle, determine:

                                                                                                a.         the net work output of the cycle, MW

                                                                                                b.         the cycle thermal efficiency

                                                                                                c.         the mass flow rate of condenser cooling water, kg/hr

                                                                                                d.         the quality of the exhaust steam

3.         Reconsider the cycle in problem #2, but now account for the non-isentropic operation of the turbine and pumps by assuming isentropic efficiencies of 85% for both. Repeat the calculations in problem 2.

Solution

2 Ans:

Mass flow rate of steam = 5*10⁵ kg/hr,steam enters the turbine at 12 MPa, 600^oC

at 12 Mpa,Saturated Steam Temperature= 324.7 ^oC,so steam entering the turbine is at superheated state

from steam tables,

Enthalpy of steam entering the turbine = 3609.02 KJ/kg

Entropy of steam entering the turbine = 6.8097 kJ/(kg·K)

for isentropic expansion,entropy of steam leaving the turbine = 6.8097 kJ/(kg·K),let quality of exhaust steam = x

at  8 kPa,S_f = 0.5924  kJ/(kg·K),S_fg = 7.6372  kJ/(kg·K),h_f = 173.80 KJ/Kg , h_fg= 2403.58 KJ/Kg

so ,quality of exhaust steam = (6.8097-0.5924)/(7.6372) = 0.81 =x

enthalpy of exhaust steam = 173.80+(0.81*2403.58) = 2130.5 KJ/Kg

turbine work output = (3609.02- 2130.5)*5*10⁵/3600000 = 205.375 MW,

enthalpy of liquid leaving condenser = 173.80 KJ/Kg,enthalpy of liquid entering bolier = ((12000-8)*0.00100847)+ (173.8) = 12.09+173.8=185.9 KJ/Kg

heat input = (3609.02 - 185.9) *5*10⁵/3600000= 475.43 MW

net work output = 205.375 - (12.09*5*10⁵/3600000) = 205.375-1.68 = 203.7 MW

cycle thermal efficiency = net work output / heat input = 203.7/475.43 = 0.43

mass flow rate of condenser cooling water = (2130.5- 173.80)*5*10⁵/4.19*15 = 155.66 *10⁵ kg /hr

3 Ans:

similar to 2nd Ans but after calculating enthalpy of steam leaving turbine and pump,use that calculated enthalpy to find actual enthalpy by using below formulas

turbine efficiency = (inlet enthalpy - actual exit enthalpy)/(inlet enthalpy - isentropic exit enthalpy) = 0.85

Pump efficency =( inlet enthalpy - isentropic exit enthalpy) / (inlet enthalpy - actual exit enthalpy) = 0.85