A steam power plant operates on the reheat Rankine cycle. Steam enters the high-

Question

A steam power plant operates on the reheat Rankine cycle. Steam enters the high-pressure turbine at 12.5 MPa and 550 degree C at a rate of 7.7 kg/s and leaves at 2 MPa. Steam is then reheated at constant pressure to 450 degree C before it expands in the low-pressure turbine. The isentropic efficiencies of the turbine and the pump arc 85 percent and 90 percent, respectively. Steam leaves the condenser as a saturated liquid. If the moisture content of the steam at the exit of the turbine is not to exceed 5 percent, determine: the condenser pressure, the net power output, and the thermal efficiency.

Explanation / Answer

1.Assume steady state condition exists,2.Kinetic and potential energy changes are neglible

From the steam tables (Tables A-4, A-5, and A-6),

1.

P3=12.5MPa

T3= 550oC

h3=3476.5KJ/Kg

s3=6.6317 KJ/Kg.K

P4=2MPa

T4=450oC

h4s=2948.1KJ/Kg

isentropic efficiency = h3-h4/h3-h4s  

h4={isentropic efficency*h3-h4s}/h3

=3027KJ/Kg

P5=2MPa

T5=450oC

h5= 3358 KJ/Kg

s5=7.2815KJ/Kg.K

P6= ?

x6=0.95

h6=?

s6=ss

Isentropic efficiency = h5-h6/h5-h6s

h6= {isentropic efficiency*h5-h6s}/h5

h6=2463 KJ/Kg

P6 = 9.73 kPa

2.

h1=

hf @ 9.73 kPa = 189.57 kJ/kg

v1=

v f @ 10 kPa = 0.001010 m3/kg

wp.in=v1(P2-P1)/Efficiency=14.02KJ/Kg

h2=h1+wp.in=189.57 14.02= 203.59 kJ/kg

qin =(h3 - h2)+(h5 -h4)= 3476.5 - 203.59 +3358.2 -2463.3 = 3603.8 kJ/kg

qout= h6-h1 =2463-189.57 =2273.7 kJ/kg

Wnet = m(qin - qout )= (7.7 kg/s)(3603.8 - 2273.7)kJ/kg = 10,242 kW

3.Thermal Efficiency

= 1-qout/qin

= 1- 2273.7/ 3603.8

= 0.369

= 36.9%

1) condesner pressure

h6=2463 KJ/Kg

P6 = 9.73 kPa

2)net power output

Wnet =10,242 kW

3)Thermal Efficiency

= 36.9%

2.

h1=

hf @ 9.73 kPa = 189.57 kJ/kg

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