Thermodynamics Questions related to each other Steam is the

Steam is the working fluid in an ideal Rankine cycle. Steam enters the turbine at 12 MPa, 600 degree C and a mass flow rate of 5 times 10^5 kg/hr. The condenser operates at 8 kPa and cooling water enters the condenser at 20 degree C and exits at 35 degree C. For this cycle, determine: the net work output of the cycle, MW the cycle thermal efficiency the mass flow rate of condenser cooling water, kg/hr the quality of the exhaust steam 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. Reconsider the cycle in problem #3 adding a single stage of re-heat. Assume that the first turbine expands the steam to a pressure of 6 bar. The steam is then reheated to 500 degree C before expanding in the second turbine to the condenser pressure of 8 kPa. Using the same isentropic efficiencies as in problem 3, answer the same questions.

Solution

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=43%

mass flow rate of condenser cooling water = (2130.5- 173.80)*5*105/4.19*15 = 155.66 *105 kg /hr=4323.88 kg/s

thank you