A steam power plant operates on the reheat Rankine cycle Ste

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.

Solution

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= 550^oC

h₃=3476.5KJ/Kg

s₃=6.6317 KJ/Kg.K

P₄=2MPa

T₄=450^oC

h_4s=2948.1KJ/Kg

isentropic efficiency = h₃-h₄/h₃-h_4s  

h₄={isentropic efficency*h₃-h_4s}/h₃

=3027KJ/Kg

P₅=2MPa

T₅=450^oC

h₅= 3358 KJ/Kg

s₅=7.2815KJ/Kg.K

P₆= ?

x₆=0.95

h₆=?

s₆=s_s

Isentropic efficiency = h₅-h₆/h₅-h_6s

h₆= {isentropic efficiency*h₅-h_6s}/h₅

h₆=2463 KJ/Kg

P6 = 9.73 kPa

2.

h₁=

h_{f @ 9.73 kPa} = 189.57 kJ/kg

^v1=

v _{f @ 10 kPa} = 0.001010 m³/kg

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

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

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

q_out= h_6-h₁ =2463-189.57 =2273.7 kJ/kg

W_net = m(q_in - q_out )= (7.7 kg/s)(3603.8 - 2273.7)kJ/kg = 10,242 kW

3.Thermal Efficiency

= 1-q_out/q_in

= 1- 2273.7/ 3603.8

= 0.369

= 36.9%

1) condesner pressure

h₆=2463 KJ/Kg

P6 = 9.73 kPa

2)net power output

W_net =10,242 kW

3)Thermal Efficiency

= 36.9%

2. h1=

hf @ 9.73 kPa = 189.57 kJ/kg