Steam is the working fluid in an ideal saturated Rankine cyc

Steam is the working fluid in an ideal saturated Rankine cycle. Boiler pressure is 12 Mpa and condenser pressure is 8 kPa. For a steam flow rate of 5 times 10^5 kg/hr, determine: the network output of the cycle, MW the cycle thermal efficiency the quality of the exhaust steam

Solution

For problem from this you need to use Stream tables.

For Rankine cycle

For Boiler pressure 120 MPa P₂ = 120000kPa = 120 bar

For Condenser pressuer 80 kPa P₁ = .8 bar

At 80 kPa = .8 bar

h₁=h_f=391.7 kJ/kg

v₁=v_f= 0.001039 m³/kg

w_pin= v₁( P₂-P₁)

       =0.001039 ( 12000-80) = 124.5969 kJ/kg

h₂= h₁ +w_pin =391.7 +124.5969 =516.2969 kJ/kg

For P =120MPa = 120 bar

x₃=1 ( Dryness fraction is 1 at saturation condition)

h₃=2678.4 kJ/kg

s₃= 5.471 kJ/kg K

at P₄= 80kpa s₃₌ s_{4 ( entropy is constant})

x₄= (s_4-s_f )/s_{fg= ( 5.471-1.307)/ 6.048 = .6885}

h₄= h_f+ x_{4 .} h_f = 419.1 +( 06885* 2240.8)

h₄ = 1961.7730 kJ / kg

a)Work output = h₃ - h₄ = 2678.4- 1961.7730 = 716.6270 kJ / kg

Heat supplied Qin= h₃ - h₂ = 2678.4- 516.29 = 2162.11 kJ /kg

Heat Rejected Q out = h₄ - h₁ = 1961.7730 - 391.7 = 1570.07 kJ/kg

b) efficiency = 1-( Qout / Q in) = .27378 = 27.38 %

c) Quality of the stream at exhaust (x₄= 0.6885)

68.85% is the quality of the exhaust stream