Question

Thermo Final Exam open book open notes 2hrs 1. (10pts) A power plant can be described by an ideal Rankine cycle without any irreversibility's. he boiler produces superheated steam at a pressure of 200 bars and temperature of 600c (1). The steam expands adiabatically (and isentropicly) thru a turbine exiting as a saturated vapor/liquid mixture at pressure of 0.1bar (2). The mixture passes thru a condenser (P-0.1bar) until it is all saturated liquid (3). It is then pumped adiabatically to the boiler pressure of 200 bar(4) at which point heat is added in the boiler to produce saturated steam to begin the cycle again(1). The total flow rate of steam and/or water thru the plant is 2x10 kg/hr. a) Find the quality (mass fraction of vapor) of the vapor/liquid mixture exiting the turbine b) Find the thermal efficiency and total power output of the turbine You may use h-h+(p-p3)vs, where v refers to specific volume c)(extra credit-5pts)If the turbine is only 80% efficient what effect will this have on the work output of the turbine and the plant efficiency assuming the heat input from the boiler is constant?

Answer 1

The T-S diagram for the Rankine cycle is drawn below showing various states

0-O) MPa 3 1.

Here 1-2 $\rightarrow$ isentropic expansion in turbine

2-3$\rightarrow$constant pressure heat rejection in consdensor

3-4$\rightarrow$isentropic compression in pump

4-1$\rightarrow$constant pressure heat addition in boiler

From the steam tables, find out various properties of each state,

State 1, 20 MPa, 600 deg C(superheated table referred)

specific enthalpy , h₁ = 3539 KJ/Kg

specific entropy, s₁ = 6.5075 KJ/Kg.K

For state 2 and 3 (at same pressure of 0.01 MPa)

h_f= 191.81 KJ/Kg, h_fg = 2392.1 KJ/Kg

s_f = 0.64920 KJ/Kg.K, s_fg = 7.4996 KJ/Kg.K

specific volume at state 3, v₃ = 0.00101027 m³/Kg

1)

From the graph we find that entropy of state 1 and 2 are same

i.e s₁ = s₂ = s_f2 + xs_fg2, x is the dryness fraction at state 2

6.5075 = 0.64920 + x7.4996

x = (6.5075-0.64920)/7.4996 = 0.78115

Hence dryness fraction at exit of turbine(quality of vapour) = x₂ = 0.78115

2)

using x₂, we get h₂ = h_f2 + x₂*h_fg2

h₂ = 191.81+0.78115*2392.1 = 2060.375 KJ/Kg

h₃ = h_f3 = 191.81 KJ/Kg

we have for a pump specific work = vdP, v is the specific volume and dP is the pressure difference,

Here we have, h₄ - h₃ = v₃*(P₄ - P₃ ) =0.00101027*(20-0.01)*1000 = 20.1953 KJ/Kg

therefore , h₄ = h₃ + 20.1953 KJ/Kg = 191.81 + 20.1953 = 212.005 KJ/Kg

we have efficiency of the plant,

Wturbine - Wpump(hi - h2) - (h4 h3) Qinput Worknet tur 02e h1 -h4 heatinput при

substituting values we get

(3539 - 2060.375) - (212.005-191.81) 1478.625 20.195 = 0.4384 3539 - 212.005 3326.995

Hence efficiency of the plant = 43.84 %

mass flow rate of water = 2*10⁶ Kg/hr = 555.55 Kg/s

Total power output = W_turbine-W_pump = m{(h₁ - h₂) - (h₄ - h₃ ) =555.55*{(3539-2060.375)-(212.005-191.81) =

810230.7865 W = 810.23 KW

3)

When there is inefficiency of turbine, the graph changes as the turbine is no longer isentropic

600 C -O) MPa 2

Here, state 2s represents the ideal state of exit and state 2 represents the actual state of exit

we have from above h_2s = 2060.375 KJ/Kg

we have isentropic efficeincy of turbine(being a work producing device)

hi - h2 actual hi - h2s VVtheoretical

we have h₁ = 3539 KJ/Kg.K

we can calculate h₂ (actual exit state of turbine) as,

h2 = h1-lT(hi-ha) = 3539-0.8*(3539-2060.375) = 2356.1 KJ/Kg. K

New power output = W_turbine-W_pump = m{(h₁ - h₂) - (h₄ - h₃ )}= 555.55*{(3539-2356.1)-(212.005-191.81)} = 645940.7628 W = 645.94KW

new efficiency =

Wturbine - Wpump(hi - h2) - (h4 h3) Qinput Worknet tur 02e h1 -h4 heatinput при

substituting values we get

(3539- 2356.1) (212.005- 191.81) 1182.9 20.195 0.349.5 3539 212.005 3326.995

hence new efficiency = 34.95 %