USING ENGINEERING EQUATION SOLVER 1 Using EES An ideal Ranki

USING ENGINEERING EQUATION SOLVER

1.    Using EES. An ideal Rankine power cycle with super-heat (turbine inlet state is super-heated!!!) uses solar energy for its heat input and refrigerant 134a as the working fluid. The fluid enters the pump as saturated liquid at 7 bars and is pumped to 14 bars. The turbine inlet is at a temperature of 140 C and the mass flow rate is 1200 kg/hr. Determine:

the net work in kJ/kg Ans. 20.4 kJ/kg

the thermal efficiency Ans. 0.071

the area of the solar collectors (in m²) if the collector picks up 630 W/m². Ans. 125.5 m².

Include an EES generated T-s diagram which includes the dome, relevant isobars and states.

2.      Using EES! The turbine inlet conditions in a Rankine steam power cycle with super-heat are 30 bars and 600 C. The turbine exhaust pressure is 0.1 bar. The turbine is adiabatic with an isentropic efficiency of 88% and the pump has an efficiency of 78%. The steam-generator/boiler and condenser are iso-baric. Determine the thermal efficiency of the cycle. Include an EES generated T-s diagram which includes the dome, relevant isobars and states. Ans. 33%

Hints using EES: Greek letters are typed in as any other variable (e.g. , ). For air IGB problems make sure to type in “air” as your working fluid. For constant specific heat problems, use the built-in function “Cp()” and “Cv()”. Also, under function info, make sure to select “Ideal gas”. Finally, if you want to convert from temperature in C to K for example, use the converttemp() built-in function.

Solution

1) From the properties of R134a, we get following values of specific enthalpy

Enthalpy at pump inlet = h₃ = 196.342 KJ/kg

Enthalpy at pump outlet = h₄ = 284.942 KJ/kg

Enthalpy at turbine inlet = h1 = 562.696 KJ/kg

Enthalpy at turbine outlet = h2 = 451.6896 KJ/kg

Therefore, Net work done = Turbine work - pump work = (h₁-h₂) - (h₄-h₃)

Substituting values, net work done = 21.4 KJ/kg (answer may vary depending upon values of enthalpy)

Thermal efficiency = (Net work ) / (Heat supplied)

Heat supplied = h₁ - h₄ = 301.408 KJ/kg

Therefore, Thermal efficiency = 0.073

Area of solar collector = (Heat supplied * mass flowrate)/( Heat picked up by collector) = 125.8 m²

2) From steam table , we get following values of enthalpy

enthalpy at turbine inlet = h₁ = 3682.3 KJ/kg

isentropic enthalpy at turbine outlet = h_2a = 2651.9 KJ/kg

enthalpy at pump inlet = h₃ = 192.6 KJ/kg

isentropic enthalpy at pump outlet = h_2a = 1008.4 KJ/kg

Turbine efficiency = ( h₁ - h₂)/( h₁ - h_2a)

substituting values , we get h₂= 2857.98 KJ/kg

Pump efficiency = ( h_4a- h₃)/(h₄ - h₃)

substituting values , we get h₄= 828.924 KJ/kg

Thermal efficiency of cycle = 1 - ((h₂ - h₃)/(h₁- h₄)) = 0.33 = 33%