Thermodynamics Questions related to each other Steam is the working fluid in an

Question

Thermodynamics Questions related to each other 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.

Explanation / Answer

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

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.