Finding Qdot for a Thermodynamic Rankine Cycle

[jaredogden]

Homework Statement

Water is the working fluid in a Rankine cycle. Superheated vapor enters the turbine at 10 MPa, 480 °C condenser pressure = 6 kPa η_turbine = 80% η_pump = 70% determine a.) Qdot of the steam generator. b.) η_th c.) Qdot from the condenser to the cooling water.

Homework Equations

x2 = (s2 - sf)/(sg-sf)
W_turbine = Qdot/mdot = (h₁ - h₂
q_in = Qdot/mdot = (h₂ - h₃ )
W_pump = Wdot_pump/mdot = (h₄ - h₃)
q_in = Qdot/mdot = (h₁ - h₄)

The Attempt at a Solution

s1 = s2 x3 = 0
T2 = T3 P4 = P1
P2 = P3 h3 = hf
s3 = s4

From the superheated water vapor table at 10 MPa and 480°C:
h1 = 3321.4 kJ/kg s1 = 6.5282 kJ/kg*K

From the saturated water mixture table at 6 kPa:
hf = 151.53 kJ/kg sg = 8.3304 kJ/kg*K
hg = 2567.4 kJ/kg sf = 0.5210 kJ/kg*K
hfg = 2415.9 kJ/kg vf = 1.0064x10^-3 m³/kg


x2 = (6.5282 - 0.5210) [kJ/kg*K]/(8.3304 - 0.5210)[kJ/kg*K]
x2 = 0.7692

h2 = 151.53 kJ/kg + 0.7692(2415.9 kJ/kg)
h2 = 2009.84 kJ/kg
h3 = 151.53 kJ/kg
h4 = 151.53 kJ/kg + 1.0064x10^-3 m³/kg(10 - 0.006)[MPa]
h4 = 161.59 kJ/kg

 

[Valkarie]

Wturbine = (h1 - h2)/0.8Wturbine = 678.7 kJ/kgqin = (h2 - h3)qin = 1858.31 kJ/kgWpump = (h4 - h3)/0.7Wpump = 10.06 kJ/kgQdot = mdot(h1 - h4)Qdot = mdot(3321.4 - 161.59)Qdot = mdot3159.81 kJ/kgm dot = Wturbine + Wpump/qinmdot = 678.7 + 10.06/1858.31mdot = 0.361 kg/sQdot = 0.361(3159.81)Qdot = 1139.6 kWηth = Wturbine/qinηth = 678.7/1858.31ηth = 0.365Qdot from the condenser to the cooling waterQdot = mdot(h3 - h4)Qdot = 0.361(151.53 - 161.59)Qdot = -5.46 kW