1

8.16 Steam enters the turbine of a simple vapor power plant with a pressure of 12 MPa and a

temperature of 600oC and expands adiabatically to condenser pressure, p. Saturated liquid exits

the condenser at pressure p. The isentropic efficiency of both the turbine and the pump is 84%.

(a) For p = 30 kPa, determine the turbine exit quality and the cycle thermal efficiency.

(b) Plot the quantities of part (a) versus p ranging from 6 kPa to 100 kPa.

KNOWN: Water is the working fluid in a simple vapor power plant. Data are given at various

states in the cycle. The condenser pressure is p.

FIND: (a) For p = 30 kPa, determine the turbine exit quality and the cycle thermal efficiency,

(b) plot the quantities of part (a) versus p ranging from 6 kPa to 100 kPa.

SCHEMATIC AND GIVEN DATA:

ENGINEERING MODEL:

1. Each component of the cycle is analyzed as a control volume at steady state. The control

volumes are shown on the accompanying sketch by dashed lines.

2. Flow through the boiler and condenser occurs at constant pressure.

3. Stray heat transfer in the turbine, condenser, and pump is ignored.

4. Kinetic and potential energy effects are negligible.

5. Condensate exits the condenser as saturated liquid.

ANALYSIS: First fix each principal state with p2 = 30 kPa.

State 1: p1 = 12 MPa (120 bar), T1 = 600oC → h1 = 3608.3 kJ/kg, s1 = 6.8037 kJ/kg∙K

State 2s: p2s = p2 = 30 kPa (0.3 bar), s2s = s1 = 6. 8037 kJ/kg∙K → x2s = 0.8586,

h2s = 2295.0 kJ/kg

State 2: p2 = 30 kPa (0.3 bar), h2 = 2505.1 kJ/kg (see below)

tWTurbine

Cooling

water

2

Condenser

Pump

pW

3

4

1Boiler

inQ

p3 = p2 = 30 kPa (part (a))

x3 = 0 (saturated liquid)

p1 = 12 MPa

T1 = 600oC

p2 = 30 kPa (part (a))

p4 = p1 = 12 MPa

outQ

ht = 84%

hp = 84%

T

4

32

1

p

12 MPa

s

2s

T1 = 600oC

2

kg

kJ)0.22953.3608)(84.0(

kg

kJ3.3608)( 21t12

21

21t

s

s

hhhhhh

hhhh = 2505.1 kJ/kg

State 3: p3 = 30 kPa (0.3 bar), saturated liquid → h3 = hf3 = 289.23 kJ/kg,

v3 = vf3 = 0.0010223 m3/kg,

State 4: p4 = p1 = 12 MPa (120 bar), h4 = 303.80 kJ/kg (see below)

p

34334

34

343p

)()(

hh

pphh

hh

pp

vv

mN 1000

kJ 1

kPa 1m

N1000

84.0

kPa )3012000)(kg

m0010223.0(

kg

kJ23.289

2

3

4

h = 303.80 kJ/kg

(a) The turbine exit quality, x2, is

2fg

2f22

h

hhx

Substituting values from Table 3, hf2 = 289.23 kJ/kg and hfg2 = 2336.1 kJ/kg, gives

kJ/kg 2336.1

kJ/kg )23.289 .15052(2

x = 0.9485 (94.85%)

The thermal efficiency is

)(

)()(

/

//

41

3421

in

pt

hh

hhhh

mQ

mWmW

h

Substituting enthalpy values and solving yield

kJ/kg ).803033608.3(

kJ/kg )23.289.80303(kJ/kg )1.2505 3608.3(

h = 0.3294 (32.94%)

3

The data for the required plots are obtained using IT as follows:

IT Code IT Results for p2 = 30 kPa p1 = 12000 // kPa T1 = 600 // oC p2 = 30 // kPa eff_t = 0.84 eff_p = 0.84 p3 = p2 x3 = 0 p4 = p1 h1 = h_PT("Water/Steam", p1, T1) s1 = s_PT("Water/Steam", p1, T1) s2s = s1 p2s = p2 h2s = h_Ps("Water/Steam", p2s, s2s) h2 = h1 - eff_t*(h1 - h2s) x2 = x_hP("Water/Steam", h2, p2) h3 = hsat_Px("Water/Steam", p3, x3) v3 = vsat_Px("Water/Steam", p3, x3) h4 = h3 + (v3*(p4 - p3))/eff_p eta = ((h1 - h2) - (h4 - h3))/(h1 - h4)

eta 0.3295 h1 3608 h2 2505 h2s 2295 h3 289.9 h4 304.5 p2s 30 p3 30 p4 1.2E4 s1 6.803 s2s 6.803 v3 0.001022 x2 0.9486 eff_p 0.84 eff_t 0.84 p1 1.2E4 p2 30 T1 600 x3 0

IT results are consistent with the calculations in part (a).

Plots:

From the T-s diagram, we see that within the range of pressures considered the working fluid

is a liquid-vapor mixture (0 ≤ x ≤ 1) and as p2 increases, points 2s and 2 move to the right.

Thus, x2 increases as indicated in the plot above. Further as p2 increases, the average

temperature of heat rejection increases, lowering thermal efficiency. Thus, h decreases as

indicated in the plot above.

Condenser Pressure versus Turbine Exit Quality

Condenser Pressure (kPa)1009080706050403020100

Qu

alit

y

1

0.99

0.98

0.97

0.96

0.95

0.94

0.93

0.92

0.91

0.9

Condenser Pressure vs. Thermal Efficiency

Condenser Pressure (kPa)

1009080706050403020100

Therm

al E

ffic

iency

0.4

0.35

0.3

0.25

0.2

0.15

0.1

0.05

0