MEE220

Vapor Power Cycle

Lecture 12

Thermodynamics II
Level 1: Carnot power cycle
Heat Transfer issues:
 Energy of boiler hot gases can not be
extracted below TH.  the Boiler thermal
efficiency.
 two-phase systems limits TH ( <
374 C for water) =>  the thermal
efficiency.

Turbine issues:
Steam with high moisture content.
 The impingement of liquid droplets on the turbine
blades causes erosion and is a major source of wear Carnot cycle is not a
realistic model for vapor
 Steam with x < 90 percent power cycles

Pump issues:  This problem could be eliminated by using a

working fluid with a very steep saturated vapor line.
 Level 2: Ideal Rankine cycle
Heat Transfer All processes are
irreversibilities is reduced internally reversible

1-2 Isentropic expansion in a turbine

3-4 Isentropic compression in a pump

4-1 Constant pressure heat addition in a boiler

2 2-3 Constant pressure heat rejection in a condenser
Level 3: Actual Rankine cycle
Frictional effects that
1-2 irreversible compression in a pump produce pressure drop
(entropy generation)
are ignored in
condenser and boiler

3-4 irreversible expansion in a turbine

h2 = h1 − t (h1 − h2 s )

h4 = h3 + v( p4 − p3 ) /  p
Modeling the Ideal Rankine Cycle the working fluid undergoes series of internally reversible processes

Process 4-1: Apply the 1st law:

Cycle thermal efficiency
Process 3–4: Isentropic constant 0 = Qcv - 0 + m(h 4 -h1 )
compression in (pump) in the pressure heat w w − wp
Qcv /m =(h1 -h 4 )  = net = t
compressed liquid region addition qin qin
through the
Apply the 1st law: (h1 − h2 ) − (h4 − h3 )
boiler =
0 = Qcv -Wcv + m(h 3 -h 4 ) h1 − h4

Wcv /m = -(h 4 -h 3 ) wnet q

= = 1 − out
qin qin
Win /m = h 4 -h 3
(h2 − h3 )
Another method: use the = 1−
reversible work formula h1 − h4
2
wrev = −  vdP
1
Process 2–3: Apply the 1st law:
Apply the 1st law:
= −v3 ( P4 − P3 ) constant pressure Process 1–2:
Isentropic expansion 0 = Qt -Wt + m  h1 -h 2 
0 = Qcv -Wcv + m(h 2 -h 3 )
heat rejection
win = v3 ( P4 − P3 ) through the Qcv /m = (h 3 -h 2 ) through the turbine
from state 1 to the
Wt /m=h1 -h 2
condenser. fluid Qcv /m = -(h 2 -h 3 )
h4 = h3 + v3 ( P4 − P3 ) exit as saturated Qout /m= h 2 -h 3
condenser pressure
liquid at state 3
EXAMPLE 8.1 Saturated vapor enters the turbine
(b) the back work ratio, Thermo
of an ideal Rankine cycle at 8.0 MPa and saturated Tables
liquid exits the condenser at a pressure of 0.008 H20
MPa. The net power output of the cycle is 100 MW. A2 to A6

(a) Determine for the cycle the thermal efficiency (c) The 𝑚ሶ 𝑠𝑡𝑒𝑎𝑚 in kg/h

= 0.371
Tsat (8 kpa)
State 1: = 41.51 C
sat. vapor => h1 = 2758.0 kJ/kg, s1 = 5.7432 kJ/kg.K
State 2: (f) 𝑚ሶ 𝑐𝑜𝑜𝑙𝑖𝑛𝑔 𝑤𝑎𝑡𝑒𝑟 in the condenser (kg/h) if
P2=0.008 mpa, s2=s1 => sat. liq. Vap. mix. Tcooling water, in =15 C and Tcooling water, out = 35 C.

(d) 𝑄ሶ 𝑖𝑛 gained in boiler, in MW

h2 = hf + x2hfg = 173.88+0.6745(2403.1)
h2 = 1794.8 kJ/kg 0 = 0 – 0 +𝑚ሶ 𝑐𝑤 (hcw, in – hcw, out)+ 𝑚ሶ (h2 –h3)
State 3 :
5
sat. liq. at 0.008 Mpa => h3 =173.88 kJ/kg m(h -h ) 3.77*10 (1794.8-173.88)
(e) 𝑄ሶ 𝑜𝑢𝑡 rejected in condenser, mcw = 2 3
=
State 4: P = 8 Mpa & s4 = s3 => Com. Liq. in MW, (h cw,out -h cw,in ) (2565.3-2528.9)
4

mcw = = 7.3 106 kg / h

h4 = 181.94 kJ/kg 6
EXAMPLE 8.2 Steam is the working fluid in
a Rankine cycle. Saturated vapor enters the h4 = 173.88 + 8.06 / 0.85
turbine at 8.0 MPa and saturated liquid exits h4 = 183.36 kJ/kg
the condenser at a pressure of 0.008 MPa.
The net power output of the cycle is 100 (h1 -h 2 )-(h 4 -h 3 )
MW. If the turbine and the pump each have η=
h1 -h 4
an isentropic efficiency of 85%. Determine
(a) the thermal efficiency,

h2s =173.88 +0.6745(2403.1)= 1794.8 kJ/kg

(c) The 𝑚ሶ 𝑠𝑡𝑒𝑎𝑚 in kg/h
Fix state 1: sat. vapor => h1 = 2758.0 kJ/kg h2 = 2758 - 0.85(2758 - 1794.8)
and s1 = 5.7432 kJ/kg.K h2 = 1939.3 kJ/kg
Fix state 2:
State 3:
sat. liq. at 0.008 Mpa => h3 =173.88 kJ/kg
(d) 𝑄ሶ 𝑖𝑛 gained in boiler, in MW
To find h2s, P2=0.008 Mpa, s2s=s1 => sat. Mix. State 4:

h2s = hf + x2hfg
(e) 𝑄ሶ 𝑜𝑢𝑡 rejected in condenser, in MW,
s2s = sf + x2sfg
Effect of irreversibilities within the turbine and pump can be gauged by comparing values from Example 8.2 with their
counterparts in Example 8.1

Ex. 8.1 Ex. 8.2

h1 2758.0 2758.0
h2 1794.8 1939.3
h3 173.88 173.88
h4 181.94 183.36
wt 963.2 818.7
wp 8.06 9.82
wnet 955.14 808.88
mሶ s 3.77×105 4.49×105
q_out 169.75 MW 218.2 MW
𝒎ሶ 𝒄.𝒘 7.3 X106 kg/h 9.39 X106 kg/h
 0.371 0.314
8
8.2.3 Effects of Boiler and Condenser Pressures on the Rankine Cycle
To improve the simple ideal Rankine cycle, always look
to the efficiency of the corresponding Carnot cycle.
TL
th,Carnot = 1 − ;
TH
It is clear that the   as TH  or as the TL 

1. Increasing the Boiler Pressure

By inspection, what is the (TH)avg_old ?
By inspection, what is the (TH)avg_new ?

(TH)avg_new

(TH)avg_old
2. Lowering the Condenser Pressure
By inspection, what is the (TL)avg_old ?
By inspection, what is the (TH)avg_new ?

(TL)avg_old

TL_new

(Pcond)_min = (Psat)@T_amb How?

If T_amb= 30 C, then you can not go below p_sat= 4.25 kpa
Superheat and Reheat

reheater
(TH)avg_reheat
(TH)avg_Superheat
(TH)avg_simple
Example 8.3 Steam is the working fluid in an
ideal Rankine cycle with superheat and reheat.
Steam enters the first-stage turbine at 8.0 MPa,
480 C, and expands to 0.7 MPa. It is then
reheated to 440 C before entering the second-
stage turbine, where it expands to the condenser
pressure of 0.008 MPa. The net power output is
100 MW. Determine ,
(a) the thermal efficiency of the cycle

State 3: P3 = 0.7 Mpa, T3 = 440 C => (SH)

h3 = 3353.3 kJ/kg and s3 = 7.7571 kJ/kg.K (b) the mass flow rate of steam, in kg/h,
State 1: P1= 8.0 MPa and T1= 480 C, (SH)
State 4: p4 = 0.008 MPa s4 = s3 => (SLVM) Wcycle = m_dot* [wt1 +wt2 –wp]
h1 = 3348.4 kJ/kg and s1 = 6.6586 kJ/kg.K.

State 2: p2 = 0.7 MPa s2 = s1 => (SLVM)

State 5: sat liq. at 0.008 Mpa h5 = 173.88 kJ/kg.

State 6: same as in Ex. 8.1h6 = 181.94 kJ/kg.
h2 = hf+x2hfg= 697.2 +0.989*(2066.3)
= 2741.8 kJ/kg
=0.403
Discuss the effects of reheat on the vapor power cycle.

With superheat and reheat,the thermal efficiency

is increased over that of the cycle of Example
8.1.

For a specified net power output (100 MW), a

larger thermal efficiency means that a smaller
mass flow rate of steam is required

Moreover, with a greater thermal efficiency the

rate of heat transfer to the cooling water is also
less, resulting in a reduced demand
for cooling water.

With reheating, the steam quality at the turbine

exhaust is substantially increased over the
value for the cycle of Example 8.1.
Example 8.4 (Reheat but with
irreversibilties) Steam enters the first-stage
turbine at 8.0 MPa, 480 C, and expands to
0.7 MPa. It is then reheated to 440 C before
entering the second-stage turbine, where it
expands to the condenser pressure of 0.008
MPa. The net power output is 100 MW. If the
turbine and the pump each have an
isentropic efficiency of 85%. Determine ,
(a) the thermal efficiency of the cycle
Fix State 1: h2 = 2832.8 kJ/kg
P1= 8.0 MPa and T1= 480 C => SHV
= 3353.3 – 0.85*(3353.3-2428.5) = 2567.2 kJ/kg
h1 = 3348.4 kJ/kg and s1 = 6.6586 kJ/kg.K State 3: P3 = 0.7 Mpa, T3 = 440 C => (SH)
State 2s: h3 = 3353.3 kJ/kg and s3 = 7.7571 kJ/kg.K State 5: sat. Liq at 0.008 Mpa h5 = 173.88 kJ/kg.
h2s= hf + x2hfg= 697.2 +0.989*(2066.3)
State 4s:
h2s = 2741.8 kJ/kg State 6: same as in Ex. 8.1, h6 = 181.94 kJ/kg
h4s = hf + x2hfg = 173.88 +0.9382*(2403.1)
State 2: = 2428.5 kJ/kg
State 4:
h4 = h3 - t(h3-h4s ) =0.351
h2 = 3348.4 – 0.85*(3348.4 – 2741.8) Ex 8.3 (ideal Reheat) = 0.403
8.4 Regenerative (OFWH) Vapor Power Cycle
100% (or fraction of 1) qin take place from state 7 to state 1,
of steam passes a fraction (y) of the total rather than from state a to state 1, as
through the 1st turbine. flow is extracted, after would be the case without
the 1st turbine at p2. regeneration, Which reduces qin

(1-y) expands through

the second-stage
turbine to state 3. (1-y)

(y) Fraction is mixed with (1-y)

(1-y) is condensed to
fraction in OFWH. Both streams
saturated liquid, state 4
exit as saturated liquid, state 6
Boiler
 Prove to me that y=m_dot_2/m_dot_1
Apply mass balance over the two turbines

m extracted m2
Define: y= =
m main m1

Find a formula to compute y?

Apply energy balance over the OFWH
m2 h2 + m5h5 = m6 h6
yh2 + (1 − y )h5 = h6
h −h
y (h2 − h5 ) + h5 = h6 y= 6 5
h2 − h5
Compute the efficiency

W m1 wt1 + (1 − y ) wt 2 − [(1 − y ) w p1 + w p 2 ]
=
Qin m1 h1 − h7
v4 ( P5 − P4 )
(h1 − h2 ) + (1 − y )(h2 − h3 ) − [(1 − y )( h5 − h4 ) + ( h7 − h6 )]
=
(h1 − h7 )
v6 ( P7 − P6 )
Example 8.5: Steam enters the turbine at P1= Fix state 3s: P3= 0.008 MPa and s3= s2 => SLVM
8.0 MPa, T1= 480 C and expands to P2 = 0.7
MPa, where some of the steam is extracted to x3s = 0.8208; => h3s = 2146.3 kJ/kg.
OFWH operating at 0.7 MPa. The remaining
steam expands through the 2nd turbine to the Fix state 3:
condenser pressure of P3= 0.008 MPa.
Saturated liquid exits the OFWH at 0.7 MPa. The
h3 = 2832.8 – 0.85*(2832.8 – 2146.3)
isentropic efficiency of each turbine stage is 85%
= 2249.3 kJ/kg
and each pump operates isentropically. If the net
power output of the cycle is 100 MW, determine State 4: sat liq at 0.008 Mpa h4 = 173.88 kJ/kg
a) the thermal efficiency
State 5:
Fix State 1: P1= 8.0 MPa and T1= 480 C => SHV P5 = 0.7 Mpa & s5 = s4 (isen.) => Com. Liq.
h1 = 3348.4 kJ/kg and s1 = 6.6586 kJ/kg.K
= 181.94 kJ/kg
Fix state 2s: P2= 0.7 MPa and s2s= s1 => SLVM
State 6: sat liq at 0.7 Mpa h6 = 697.22 kJ/kg.
x2s = 0.9895 => h2s = 2741.8 kJ/kg
State 7: P7 = 8 Mpa & s7 = s6 (isen.) => Com. Liq.
Fix state 2: P2= 0.7 MPa
= 705.3 kJ/kg

h2 = 3348.4 – 0.85*(3348.4 – 2741.8) = 2832.8 kJ/kg To get “y”, applying energy

balance to OFWH
s2 = 6.8606 kJ/kg K,
How to find the fraction y = m2/m1?
Applying mass and energy rate balances to OFWH
m2 h2 + m5h5 = m6 h6
yh2 + (1 − y )h5 = h6
y (h2 − h5 ) + h5 = h6

(h1 − h2 ) + (1 − y )(h2 − h3 ) − [(1 − y )( h5 − h4 ) + ( h7 − h6 )]

=
(h1 − h7 )

(515.6 + 471.8) − (0.579 + 8.08)

= = 0.369
(2643.1)
(b) the mass flow rate of steam entering the first turbine
stage, in kg/h.
= 100 MW
= m1 ( 984.4 ) − m1 ( 9.7 ) = 100 MW
515.6 + 471.8 0.579 +8.08

100 MW 1000kW kg
m1 = = 3.69  105
984.4 − 9.7 MW h
 MEE220

Vapor Power Cycle

Lecture 13

Thermodynamics II
8.4.2 Closed Feedwater Heaters

2 --

-- 7
Apply energy balance
over the CFWH
8
m2 (h2 − h7 ) + m5 (h5 − h6 ) = 0
Apply energy balance over the CFWH
m2 (h2 − h7 ) = m5 (h6 − h5 )
m 2 ( h 2 − h 7 ) + m 5 ( h5 − h 6 ) = 0 9
divide by m1
m2 (h2 − h7 ) = m5 (h6 − h5 )
y (h2 − h7 ) = (h6 − h5 )
y (h2 − h7 ) = (1 − y )(h6 − h5 ) 6
h6 − h5 y (h2 − h7 ) = (h6 − h5 ) − y (h6 − h5 ) h6 − h5
y= y=
h2 − h7 (h2 − h7 ) + (h6 − h5 )
y[(h2 − h7 ) + (h6 − h5 )] = (h6 − h5 )
 State 1 same as in Ex 8.3, h1 = 3348.4 kJ/kg State 11 p11= 8.0 Mpa and 205 C =>CL
and s1 = 6.6586 kJ/kgK. h11 = hf_205 + vf_205 (p11 - psat_at_205)
Example 8.6 :
Steam enters the 1st turbine at
8.0 MPa, 480 C and expands
to 0.7 MPa. The steam is
reheated to 440 C before
entering the 2nd turbine, where
it expands to the condenser
pressure of 0.008 MPa.

Steam is extracted from the 1st

State 2 p2 = 2.0 MPa and s2 = s1. Interpolating

turbine at 2 MPa and fed to the
closed feedwater heater.

Feedwater leaves the closed

= 875.1 +0.0011646(8000-1730) = 882.4 kJ/kg

heater at 205 C and 8.0 MPa,
and condensate exits as
saturated liquid at 2 MPa.The The net power output of the
condensate is trapped into the cycle is 100 MW.
open feedwater heater. There is no stray heat
transfer from any component
Steam extracted from the 2nd to its surroundings.
turbine at 0.3 MPa is also fed If the working fluid
into the open feedwater heater, experiences no
which operates at 0.3 MPa. irreversibilities as it passes

in Table A-4, we get h2 = 2963.5 kJ/kg.

The stream exiting the open through the turbines, pumps,
feedwater heater is saturated steam generator, reheater,
liquid at 0.3 MPa. and condenser:

Alternatively, h11 can be found from Table A-5 at

State 3 same as in Ex 8.3, h3 = 2741.8 kJ/kg p11= 8.0 Mpa and 205 C
and s3 = 6.6586 kJ/kgK.
Find y`: Apply energy
State 4 p4= 0.7 Mpa and 440 C =>SHV => balance on CFWH
h4 = 3353.3 kJ/kg and s4 =7.7571 kJ/kg.K.
State 5 p5= 0.3 Mpa and s5 = s4 =>SHV =>
Wt 2
Interpolating h5 = 3101.5 kJ/kg. = (1-y`) (h 4 -h 5 )+ (1-y`-y`) (h 5 -h 6 )=720.7 kJ/kg
m1
State 6 p5= 0.3 Mpa and s6 = s4 =>Mix => x6 = W p1
0.9382 => = (1-y`-y``) (h 8 -h 7 )= 0.22 kJ/kg
m1
h6 = 173.88+(0.9382)(2403.1)= 2428.5 kJ/kg. Find y``: energy and mass balance on OFWH Wp 2
State 7 Sat. Liq. => h7 = 173.88 kJ/kg. 0 = y``h5 + (1-y` - y``)h8+ y`h13 – h9
=(h10 -h 9 )= 8.26 kJ/kg
m1
State 8 h8 = h7 + v7(p8-p7) => h8 = 174.17 kJ/kg Qin
(1-y`)h8+y`h13–h9 =(h1 -h11 )+(1-y`)(h 4 -h 3 )= 2984.4 kJ/kg
State 9 Sat. Liq. => h9 = 561.47 kJ/kg. y`` = m1
h8 -h5 wt1 + wt 2 - w p1 - w p1
State 10 h10 = h9 + v9(p10-p9)= 561.47 + = = 0.43
(0.0010732)(8000 – 300) => h10 = 569.73 kJ/kg y`` = 0.0941 qin
Wnet
State 12 Sat. Liq. => h12 = 908.79 kJ/kg. m1 = =2.8  105 kg/h
wt1 + wt 2 - w p1 - w p1
State 13 throttling => h13 = h12 = 908.79 kJ/kg. Wt1
= (h1 -h 2 ) + (1-y`) (h 2 -h 3 ) = 572.9 kJ/kg
m1
Example 8.6 Multiple h3 = 2741.8 kJ/kg and h4 = 3353.3 kJ/kg and
Feedwater Heaters
Example 8.6 : If the working fluid experiences no
s3 = 6.6586 kJ/kgK. s4 =7.7571 kJ/kg.K.
Steam enters the 1st turbine at 8.0 irreversibilities as it passes through the
MPa, 480 C and expands to 0.7 turbines, pumps, steam generator,
MPa. The steam is reheated to 440 reheater, and condenser:
C before entering the 2nd turbine, Determine (a) the thermal efficiency,
where it expands to the condenser (b) the mass flow rate of the steam
pressure of 0.008 MPa. Steam is entering the first turbine, in kg/h
extracted from the 1st turbine at 2

p5= 0.3 Mpa & s5 = s4

MPa and fed to the closed feedwater
heater. Feedwater leaves the closed
heater at 205 C and 8.0 MPa, and
condensate exits as saturated liquid
at 2 MPa. The condensate is trapped
into the open feedwater heater.
Steam extracted from the 2nd turbine
at 0.3 MPa is also fed into the open

h5 = 3101.5 kJ/kg.
feedwater heater, which operates at
0.3 MPa. The stream exiting the
open feedwater heater is saturated
liquid at 0.3 MPa. The net power
output of the cycle is 100 MW. There
is no stray heat transfer from any
component to its surroundings.

h1 = 3348.4 kJ/kg p6= 0.008 Mpa & s6 = s4

s1 = 6.6586 kJ/kgK. h6 =2428.5 kJ/kg

p2 = 2.0 Mpa & s2 = s1.

h2 = 2963.5 kJ/kg

p11= 8.0 Mpa and 205 C =>CL

h11 = hf_205+vf_205 (p11 -psat_at_205)
Sat. Liq.@ p6= 0.008 Mpa
=> h7 = 173.88 kJ/kg.
h11 = 882.4 kJ/kg
Alternatively, Table A-5
p11= 8.0 Mpa & 205 C
h8 = h7 + v7(p8-p7)
=> h8 = 174.17 kJ/kg
Sat. Liq. @ 2 Mpa
=> h12 = 908.79 kJ/kg.
throttling => h13 = h12
h10 = h9 + v9(p10-p9) h13 =908.79 kJ/kg.
= 561.47 + (0.0010732)(8000 – 300) Sat. Liq @ 0.3 Mpa.
=> h10 = 569.73 kJ/kg => h9 = 561.47 kJ/kg.
 Find y`: Apply energy balance on CFWH
Qin
=(h1 -h11 )+(1-y`)(h 4 -h 3 )
m1
= 2984.4 kJ/kg
Find y``: energy and mass balance on OFWH
0 = m5 h5 + m8h8 + m13h13 − m9 h9
0 = y``h5 + (1-y` - y``)h8 + y`h13 – h9

(1-y`)h8+y`h13–h9
y`` = y`` = 0.0941
h8 -h5
wt1 + wt 2 - w p1 - w p1
Wt1 =
= (h1 -h 2 ) + (1-y`) (h 2 -h 3 ) = 572.9 kJ/kg qin
m1
= 0.43
Wt 2
= (1-y`) (h 4 -h 5 )+ (1-y`-y`) (h 5 -h 6 )
m1 Wnet
m1 =
=720.7 kJ/kg wt1 + wt 2 - w p1 - w p1

W p1 = 2.8  105 kg/h

= (1-y`-y``) (h 8 -h 7 )= 0.22 kJ/kg
m1
Wp 2 Example 8.6 : If the working fluid experiences no

=(h10 -h 9 )= 8.26 kJ/kg

Steam enters the 1st turbine at 8.0 irreversibilities as it passes through the
MPa, 480 C and expands to 0.7 turbines, pumps, steam generator,
MPa. The steam is reheated to 440 reheater, and condenser:
C before entering the 2nd turbine, Determine (a) the thermal efficiency,
where it expands to the condenser (b) the mass flow rate of the steam
pressure of 0.008 MPa. Steam is entering the first turbine, in kg/h

m1
extracted from the 1st turbine at 2
MPa and fed to the closed feedwater
heater. Feedwater leaves the closed
heater at 205 C and 8.0 MPa, and
condensate exits as saturated liquid
at 2 MPa. The condensate is trapped
into the open feedwater heater.
Steam extracted from the 2nd turbine
at 0.3 MPa is also fed into the open
feedwater heater, which operates at
0.3 MPa. The stream exiting the
open feedwater heater is saturated
liquid at 0.3 MPa. The net power
output of the cycle is 100 MW. There
is no stray heat transfer from any
component to its surroundings.
9.9 Gas Turbine–Based Combined Cycles
9.9.2 Cogeneration

If the process steam is

large, we could cancel the
steam turbine

The heat is used to heat

process steam instead of
rejecting it to surrounding
Comparison with Carnot Cycle Q: How does this affect the efficiency Internal Irreversibilties in turbine and pump
(friction)
Why do not we operate the Rankine Cycle of Rankine cycle?
using the principles of Carnot cycle? A: Compare (TH)avg_rankine with (TH)Carnot
Ok! Assume we did that as Cycle 1-2-3`-4`. Frictional effects that produce pressure drop
(TH)avg_rankine < (TH)Carnot (entropy generation) are ignored in condenser
Can u figure out the practical difficulties T and boiler
Carnot = 1 − L ;
TH
 < 
Rankine cycle Carnot cycle
1–2–3–4–4`–1 1–2–3`–4`–1.

External Irreversibilities Heat

(TH)Carnot
transfer due to large T between the
(TH)avg_rankine combustion gases and the steam in h2 = h1 − t (h1 − h2 s )
boiler section (the most significant
source of irreversibility!!)

Significant practical problems are

Due to real design: State 3 falls in the
encountered in developing pumps and
liquid region (as compared to the ideal
turbines that handle two-phase mixtures,
Rankine cycle) causing a reduction in
So, Carnot cycle, has its shortcomings as a h4 = h3 + v( p4 − p3 ) /  p the efficiency.
model for the building a vapor power cycle.
9.9 Gas Turbine–Based Combined Cycles Gas Turbine Vapor Cycle
state h(kJ/kg) s^o state h(kJ/kg) s
1 300.2 1.702 6.00 183.96 0.5975
Delayed h4s=f(Pr4) h3 = h(T3) h4 = h3 - 2 669.8 2.509
7.00 3138.30 6.3634
to Pr4 =(P4/P3)Pr3 (h3 –h4s)*0.88 3 1515 3.362
8.00 2104.74 6.7282
4 858 2.762
Rankine 5 401 1.992 9.00 173.88 0.5926
cycles
h7 = h(T7)
h2 = h1 +
(h2s -h1)/ 0.84 h5 = h(400)

h2s=f(Pr2)
Pr2 =(P2/P1)Pr1

h8 = h 7 -
h1 = h(T1) 0.9(h7-h8s)

h6 = h9+ h9= hf(8 kPa) s8s = s7

v9(P6-P9)/0.80 h8s = f(P8& s8s)
8.5 Other Vapor Power Cycle Aspects

No condenser for
the top-cycle! All
Qout heats bottom
cycle from d to e