Question

"Steam is in steady state flow through an ideal Rankine Cycle with a single pump and turbine. The steam enters the turbine, at State 1, as saturated vapor at 2 MPa. The condenser is at 10 kPa and discharges saturated liquid at State 3.

A) Find the cycle efficiency B) Find the back work ratio"

Please show all the work you can! Thank you in advance

Answer 1

The cycle squennce is like this.

1.Turbine-2. condenser-3. pump-4. boiler-1. turbine -

Now at state 1 : feed enters turbine as saturated vapor

P₁=2 Mpa=2*10⁶ Pa

Themodynamic property at this pressure

h₁= 2798.3 kJ/kg

s₁= 6.3390 kJ/kg K

State 2:

It is fixed P₂=10kPa . Here the process is adiabatic , process is reversible expansion through the turbine. So specific enrtropy is same.

So s₁=s₂

At 10 kPa from the literature

s_f=0.6492 kJ/kgK

s_fg = 7.4496 kJ/kgK

s_g=8.1488 kJ/kgK

h_f = 191.8 kJ/kg

h_fg = 2392.1 kJ/kg

So

find the fraction using above data

Now find enthalpy

h₂ =h_f+x₂h_fg

h₂ = 191.8 + 0.7433*2392.1 = 1969.92 kJ/kg

At state 3 : we have saturated liquid at 10 kPa

From the literature find thermodynamic property at 10 kPa for saturated liquid

So h₃ = 191.83 kJ/kg

v₃ = 0.00101 m³/ kg

State 4: boiler pressure P₄ , s₄ =s₃

h₄ = h₃ + W(work done) =h₃+ v₃ (P₄-P₃) = 191.83 +0.00101*(2000-10)=193.8399 kJ/kg

(a)

find the cycle efficiency

W_cycle =W_t - W_p

W_t = workdone in turbine

W_P = workdone in pump

So mass and enegy balance at turbine and pump

W_t /m =h₁-h₂

W_p/m =h₄-h₃

Where m is the mass flow rate

The rate of heat transfer to the working fluid as it passes through the boiler is determined using mass and energy rate balances as

Q/m =h₁-h₄

So thermal efficiency

So efficiency is 31.76%

(b)

Now back work ratio

BWR= W_P /W_t= (h₄-h₃) / (h₁-h₂) = 0.002423 =0.24%