2. Steam is the working fluid in an ideal Rankine cycle. Steam enters the turbin

Question

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

Explanation / Answer

2 Ans:

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

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,Sf = 0.5924  kJ/(kg·K),Sfg = 7.6372  kJ/(kg·K),hf = 173.80 KJ/Kg , hfg= 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

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