Post on 21-Aug-2018
Thermodynamics 8. Rankine Cycle 2 / 113
Rankine Cycle Analysis 132
Some Notes on Rankine Cycle 444
Improvement of Rankine Cycle Efficiency 535
Heating of Water at Constant Pressure 21
Vapor Diagrams 333
Advanced Rankine Cycle 766
Contents
Thermodynamics 8. Rankine Cycle 3 / 113
W
TA<TS
x = 0
TE>TS
x = 1
TD=TS
x = 1
TC=TS
0<x<1
TB=TS
x = 0
압축수(과냉액)
포화수
습증기포화증기
과열증기
TS : 포화온도, x : 건도
(A) (C) (E)(D)(B)
W
W
W
W
액체열(현열)
과열량증발열(잠열)
T
s
TS
K
DB
E
C
A
0C
Steam Dome
Saturated
Liquid Line
Saturated
Steam Line
Critical Point
Heating of Water at Constant Pressure
T-s 선도: 열역학 제1법칙과 Gibbs’ equation으로부터 Tdsq
보일러 및 원자력 증기발생기에서 공급된 열량은T-s 선도에서 s-축에 투영된 면적의 크기
보일러 및 원자력 증기발생기에서 작동유체는 정압가열 됨 vdpdhq dhq
보일러 및 원자력 증기발생기에서 가열된 열량은작동유체의 엔탈피 상승에 기여
Thermodynamics 8. Rankine Cycle 4 / 113
State Quality Characteristics
압축수
(compressed water)A x = 0
• 물이며, 포화수보다 낮은 온도를 가짐
• 과냉액(subcooled liquid)이라고도 함
• 가열하면 온도 상승
포화수
(saturated water)B x = 0
• 어떤 압력 상태에서 물을 가열하여 온도가 포화온도에 이른 물
• 포화수를 더욱 가열하면 상변화가 시작되어 습증기가 됨
습증기
(wet steam)C 0 x 0
• 포화온도 상태에 있는 증기로서 물방울을 포함하고 있음
• 습증기 가열에 사용된 열량은 물방울 증발에 사용되기 때문에습증기를 가열하더라도 온도 증가 없음. 그러나 가열하면 엔탈피는 상승. 그리고 엔트로피 증가
• 습포화증기(wet saturated steam)라고도 함
포화증기
(saturated steam)D x = 1
• 증발이 종료된 상태의 증기로서 수분을 전혀 포함하지 않음
• 건포화증기(dry saturated steam)라고도 함
과열증기
(superheated steam)E x = 1
• 수분을 전혀 포함하지 않고 있으며, 이상기체와 성질 유사
• 가열하면 온도와 부피 급격히 증가
• 현열(sensible heat): 물질을 가열하여 상태변화 없이 온도만 변하는 데 소요되는 열량 (가열 시온도 상승). 감열이라고도 함.
• 잠열(latent heat): 물질을 가열할 때 온도변화를 동반하지 않는 열. 물질에 가해진 열은 물질의상태변화에 사용됨 (가열하더라도 온도변화 없음).
Phase Change
Thermodynamics 8. Rankine Cycle 5 / 113
T-s diag. h-s diag.
T-s vs. h-s Diagram• 일반적으로 물질을 가열하면 온도와 엔탈피가 동시에 상승한다. • 그런데 습증기를 가열하는 동안에 습증기의 온도는 일정하게 유지된다. • 그러나 습증기 영역에서 온도가 일정하게 유지되더라도 물의 증발이 진행되는 동안에 열량이 지속적으로 공급되기 때문에 습증기의 엔탈피
는 지속적으로 상승한다. 물론 이때 엔트로피도 지속적으로 증가한다. • 이런 현상 때문에 물을 정압가열 할 때 나타나는 T-s선도와 h-s선도가 서로 달라진다. • 그러나 이상기체를 가열하는 경우 상변화가 일어나지 않기 때문에 이상기체에 대한 T-s선도와 h-s선도는 그 형태가 서로 거의 일치한다.
Thermodynamics 8. Rankine Cycle 6 / 113
화력발전용 보일러 및 원자력발전용 증기발생기 모두 물을 정압상태에서 가열하여 증기를 생산
피스톤과 실린더 사이에 마찰이 존재하지 않아서 가역적인 일이 일어난다고 가정.
물 1 kg을 1기압, 0°C상태에서 가열하기 시작하면 온도가 상승하며, 물 온도가 100°C에 도달하면 물은 액체에서 기체인 증기(steam, or vapor)로 상변화가 시작. 이 상변화를 기화 또는 증발(evaporation)이라 함.
이 상태에서 계속해서 가열하면 증발이 지속되어 물의 양은 줄어들고 증기의 양은 증가하며, 일정시간이 경과하면 실린더 내부의 모든 물이 증기로 변하면서 증발 완료.
증발이 진행되는 동안 실린더 내부에는 물과 증기가 공존하는데, 이를 습증기(wet steam)라 함.
일반적으로 한 공간에 두 개의 상이 평형을 이루고 있을 때 포화상태라고 함. 포화상태는 고체와 액체, 액체와 기체, 고체와 기체 모두에 적용되지만 일반적으로 액체와 기체상태 사이의 관계에 적용.
따라서 증발이 시작되기 직전의 상태를 포화수(saturated water) 또는 포화액(saturated liquids)이라 하며, 그림에 나타나 있는 상태 B에 해당.
그리고 증발이 완료된 상태를 포화증기(saturated steam)라고 하며, 그림에 나타나 있는 상태 D에 해당.
아울러 주어진 압력에서 나타나는 증발온도를 포화온도, 그 압력을 온도에 대응하는 포화압력이라 함.
한편, 포화온도를 비등점(boiling point)이라고도 함. 가열이 급격하게 이루어지면 물 자유표면에서 증발하는현상뿐만 아니라 물의 내부에너지도 급격히 증가하여 기포가 발생하는데, 기포는 물 내부에서 상승하기 때문에 물 자유표면이 격렬하게 요동치게 됨. 이런 급격한 증발현상을 비등이라 함.
한편, 그림에 나타나 있는 상태 A와 같이 동일한 압력의 포화수보다 낮은 온도를 가지는 물을 압축수(compressed water) 또는 과냉액(subcooled fluids)이라 함.
증발이 완료된 상태에서 포화증기를 계속해서 가열하면 과열증기(superheated steam)가 됨. 과열증기는 그림에 나타나 있는 상태 E에 해당. 과열증기는 기체이며, 과열증기를 계속해서 가열하면 온도와 부피가 급격히 증가. 과열증기는 충분히 높은 온도에서 이상기체 성질을 가짐.
General Notes [1/2]
Thermodynamics 8. Rankine Cycle 7 / 113
물에 작용하는 압력을 다양하게 변화시키면서 얻어진 실험결과로부터 몇 가지 중요한 사실이 확인되었다.
첫째, 어떤 압력에 대응하는 포화온도 존재.
• 즉 가열하기 위한 어떤 압력이 결정되면 그에 해당하는 고유의 포화온도가 존재하며, 포화온도가 결정되면 그에 해당하는 고유의 포화압력이 존재.
둘째, 증기돔(steam dome, or vapor dome) 존재.
• 증기돔 좌측은 압축수, 우측은 과열증기, 내부는 습증기.
• 그러므로 증기돔은 물을 정압가열할 때 나타나는 상변화가 시작되는 지점과 종료되는 지점을 알려줌.
• 즉 증기돔은 압력변화에 따른 포화수 상태와 포화증기 상태의 집합임.
• 포화수와 포화증기에 대한 각종 열역학적 상태량을 나타내기 위해서 ‘fluid’를 의미하는 하첨자 f와 ‘gas’를 의미하는 하첨자 g를 각각 사용.
• 예를 들어 hf는 포화수에 대한 엔탈피, hg는 포화증기에 대한 엔탈피를 나타냄.
• 물에 작용하는 압력이 증가할수록 증기돔 폭이 점차 좁아지며, 포화수와 포화증기의 열역학적 상태량 차이는 점차 작아짐.
• 증기돔 꼭대기 지점(그림 8.2에 나타나 있는 상태 K)을 임계점(critical point)이라 함.
• 물에 대한 임계점에서의 열역학적 상태량은 pc=22.09 MPa (=3,206.2 psia), Tc=374.14℃ (=705.4℉), sc=4.4298 kJ/kg·K, hc=2099.6 kJ/kg임.
• 임계점을 초과하는 상태를 초임계 상태라고 함. 임계점에서는 포화수와 포화증기 사이에 차이가 없음.
• 이는 임계점에서 물의 표면장력이 없어지기 때문임.
• 그리고 임계압력보다 높은 압력으로 가열하면 물은 증발과정을 거치지 않고 곧바로 과열증기로 바뀜. 그러므로 초임계 보일러는 물과 증기를 분리시키는 장치인 드럼이 필요 없으며, 관류 보일러(once-through boiler)라고 불림.
General Notes [2/2]
Thermodynamics 8. Rankine Cycle 8 / 113
p=
1000 b
ar
p=
800 b
ar
p=
500 b
ar
p=
300 b
ar
p=
200 b
ar
p=
150 b
ar
p=
100 b
ar
p=
60 b
ar
p=
30 b
ar
p=
15 b
ar
p=
10 b
ar
p=
6 b
ar
p=
2 b
ar
p=
1 b
ar
p=
0.6
bar
Entropy, kJ/kg‧K
Te
mp
era
ture
,
C
200
400
600
800
539
0 2.5 5.0 7.5 10.0
x=0 x=1x=0.2 x=0.4 x=0.6 x=0.8
T- s Diagram of Steam
Thermodynamics 8. Rankine Cycle 9 / 113
Entropy, kJ/kg‧K
Te
mp
era
ture
,
C
200
400
600
800
539
0 2.5 5.0 7.5 10.0
x=0 x=1x=0.2 x=0.4 x=0.6 x=0.8
T- s Diagram of Steam
Thermodynamics 8. Rankine Cycle 10 / 113
Saturated
Steam Line
Degree of
Superheat
Constant
Pressure Line
Superheated Steam
Critical
Point
Saturated
Water Line
Wet
Steam
Water
T
x
A
E
DCB
K
f g
p
A
E
DCB
K
TO
TS
TK
TE
Wet
Steam
Water
Superheated
Steam
xf g
Water
P-, T- Diagram of Steam
Thermodynamics 8. Rankine Cycle 11 / 113
Dryness (or Quality)
The state principle of thermodynamics guarantees that a thermodynamic state for a simple compressible
substance is completely determined by specifying two independent thermodynamic properties.
Other properties are then functions of these independent properties. Such functional relations are called
equations of state.
The lines of constant pressure become horizontal across the wet steam region, as they intersect the vapor
dome.
Therefore, temperature and pressure are not independent properties in the wet steam region.
To specify the thermodynamic state in this region, a dryness (or quality) denoted by x is used.
It is defined as the mass of vapor divided by the mass of the mixture.
In terms of quality, thermodynamic properties of a wet steam are calculated as a weighted average of the
saturation properties. (hf and hg: enthalpy of saturated liquid and saturated steam, respectively)
If one thermodynamic property is known, a quality can be calculated.
Steam tables, although still in use, are being replaced by computer programs today.
gf
gf
gf
xx
xssxs
xhhxh
)1(
1
1
fg
f
ss
ssx
Thermodynamics 8. Rankine Cycle 12 / 113
Drum Type vs. Once-through
[ Subcritical ] [ Supercritical ]
Thermodynamics 8. Rankine Cycle 13 / 113
Rankine Cycle Analysis2
Some Notes on Rankine Cycle4
Improvement of Rankine Cycle Efficiency5
Heating of Water at Constant Pressure1
Vapor Diagrams3
Advanced Rankine Cycle 6
Thermodynamics 8. Rankine Cycle 14 / 113
Working fluid is alternately vaporized and condensed as it
recirculates in a closed cycle.
Water is typically used as the working fluid because of its low
cost and highest specific heat.
The standard vapor cycle that excludes internal irreversibilities
is called the Ideal Rankine Cycle.
The condensation process is allowed to proceed to
completion between state 41.
Ideal Rankine Cycle
T
s
1
2
3
4qC
wTqB
qB
qB
wP
The saturated liquid is provided at state 1.
The water at state 1 can be conveniently pumped to the boiler pressure at state 2.
Heat is added in the boiler, and water becomes saturated water in the economizer.
Then, the saturated water becomes water-steam mixture in the evaporator, so called waterwall furnace tube.
The saturated steam separated in a drum becomes superheated steam in the superheaters.
Superheated steam is expanded in the turbine for power generation.
Thermodynamics 8. Rankine Cycle 15 / 113
Ideal Rankine Cycle - Fossil
T
s
1
2
3
4qC
wTqB
qB
qB
wP
a
Process Component Heat Work Process
12 Pump q12 = qP = 0 w12 = wP = (h2h1) Power in (adiabatic compression)
23 Boiler q23 = qB = h3h2 w23 = wB = 0 Heat addition at constant pressure
34 Turbine q34 = qT = 0 w34 = wT = h3h4 Power out (adiabatic expansion)
41 Condenser q41 = qC = (h4h1) w41 = wC = 0 Heat release at constant temperature
121212 whhq
Thermodynamics 8. Rankine Cycle 16 / 113
[Exercise 8.1] Cycle analysis of a simple ideal Rankine cycle
A simple ideal Rankine cycle with water as a working fluid operates between the specified pressure
limits. Determine 1) the power supplied to the pump, 2) the power produced by the turbine, 3) the gross
heat addition to the boiler, and 4) the thermal efficiency of the cycle. The mass flow rate of the working
fluid, water, is 420 kg/s.
For a simplicity, assume the kinetic and potential energy changes are negligible.
T
s
1
2
3
4
15 MPa
100 kPa
Exercise 8.1
Thermodynamics 8. Rankine Cycle 17 / 113
[Solution]
From the steam tables,
h1 = hf @ 100 kPa (=14.5 psia) = 417.48 kJ/kg
1 = f @ 100 kPa (=14.5 psia) = 0.001043 m3/kg
Pump power,
wP = (h2h1) = 1(p2p1) h2 = h1wp
wP = 0.001043 m3/kg (15,000100) kPa
= 15.54 kJ/kg (‘’ means that power is supplied to the system)
WP = (h2h1) = 420 kg/s 15.54 kJ/kg = 6.53 MW
h2 = h1wp = 433.02 kJ/kg
Determine the thermodynamic properties at point 3 using steam tables,
p3 = 15,000 kPa, x = 1
h3 = hg @ 15,000 kPa (=2175.6 psia) = 2615.0 kJ/kg,
s3 = sg @ 15,000 kPa (=2175.6 psia) = 5.3181 kJ/kg‧K
Determine the thermodynamic properties at point 4 using steam tables,
p4 = 100 kPa, s4 = s3 , sf = 1.3207 kJ/kg‧K, sg = 7.3602 kJ/kg‧K x4 = (s4sf)/(sgsf) = (5.31811.3207)/(7.36021.3207) = 0.6619
h4 = (1x4)hf + x4hg = (10.6619)417.48 + 0.66192675.4 = 1912.0 kJ/kg
m
Exercise 8.1
Thermodynamics 8. Rankine Cycle 18 / 113
[Solution] – Continued
Turbine work per unit mass and turbine power are,
wT = (h3h4) = 2615.01912.0 = 703.0 kJ/kg
WT = (h3h4) = 420 kg/s 703.0 kJ/kg = 295.3 MW
Heat addition in the boiler,
qB = h3h2 = (2615.0433.02) kJ/kg = 2182.0 kJ/kg
QB= qB = (h3h2) = 420 kg/s (2615.0433.02) kJ/kg = 916.43 MJ/s
Heat release from the condenser,
qC = (h4h1) = (1912.0417.48) = 1494.5 kJ/kg
(‘’ means that heat is released from the system)
The thermal efficiency of the cycle can be calculated as following equation
th = 1(1494.5/2182.0)
= 0.315
B
C
B
CB
in
outin
B
PT
in
sys
thq
q
q
q
q
ww
q
w
1
m
m m
Exercise 8.1
Thermodynamics 8. Rankine Cycle 19 / 113
Properties English Units International Units
Heat Btu = 1.055 kJ
Mass lbm = 0.45359 kg
Length in = 0.0254 m
Temperature R = 5/9 K
Pressure psia = 6.89286 kPa(= kN/m2)
Specific volume in3/lbm = 3.612710-5 m3/kg {= (0.0254m)3/0.45359 kg}
ft3/lbm = 0.0062428 m3/kg {= (0.3048m)3/0.45359 kg}
Specific enthalpy Btu/lbm = 2.326 kJ/kg (= 1.055 kJ/0.45359 kg)
Specific entropy Btu/lbm‧R = 4.187 kJ/kg‧K {= 1.055 kJ/(0.45359 kg5/9K)}
1 kcal = 물 1 kg의 온도를 1C 상승시키는데 필요한 열량 (1 kcal = 427 kgf‧m = 4.185 kJ)
1 Btu = 물 1 lbm의 온도를 1F 상승시키는데 필요한 열량 (1 Btu = 778 lbf‧ft = 1.055 kJ)
Conversion of Units
Thermodynamics 8. Rankine Cycle 20 / 113
Isentropic efficiency of pump is,
12
12
hh
hh sP
s
Thh
hh
43
43
Isentropic efficiency of turbine is,
T
s
1
2s
3
4s 4
2
Pump and Turbine Efficiency
Thermodynamics 8. Rankine Cycle 21 / 113
[Exercise 8.2] Cycle analysis of a simple Rankine cycle
A simple ideal Rankine cycle with water as a working fluid operates between the specified pressure
limits. There are some losses in the turbine during steam expansion, and the quality of the steam exiting
the turbine is 69%. Determine the isentropic efficiency of the turbine, and the thermal efficiency of the
cycle.
For a simplicity, assume the kinetic and potential energy changes are negligible.
T
s
1
2
3
4s
15 MPa
100 kPa
4
Exercise 8.2
Thermodynamics 8. Rankine Cycle 22 / 113
[Solution]
From the steam tables,
h1 = hf @ 100 kPa = 417.48 kJ/kg
1 = f @ 100 kPa = 0.001043 m3/kg
Pump power,
wp = (h2h1) = 1(p2p1) h2 = h1wp
wp = 0.001043 m3/kg (15,000100) kPa
= 15.54 kJ/kg (‘’ means that power is supplied in the system)
h2 = h1wp = 433.02 kJ/kg
Determine the thermodynamic properties at point 3 using steam tables,
p3 = 15,000 kPa, x = 1
h3 = 2615.0 kJ/kg, s3 = 5.3181 kJ/kg‧K
Determine the thermodynamic properties at point 4s using steam tables,
p4 = p4s = 100 kPa, s4s = s3 , sf = 1.3207 kJ/kg‧K, sg = 7.3602 kJ/kg‧K x4s = (s4ssf)/(sgsf) = (5.31811.3207)/(7.36021.3207) = 0.6619
h4s = (1x4s)hf + x4shg = (10.6619)417.48 + 0.66192675.4
= 1912.0 kJ/kg
Exercise 8.2
Thermodynamics 8. Rankine Cycle 23 / 113
[Solution] – Continued
Determine the thermodynamic properties at point 4 using quality,
x4 = 0.69
h4 = (1x4)hf + x4hg = (10.69)417.48 + 0.692675.4 = 1975.4 kJ/kg
The isentropic efficiency of the turbine is,
T = (2615.01975.4)/(2615.01912.0) = 0.910
Heat addition in the boiler and heat rejection from the condenser are,
qB = (h3h2) = 2615.0433.02 = 2182.0 kJ/kg
qC = (h4h1) = (1975.4417.48) = 1557.92 kJ/kg
(‘’ means that heat is released from the system)
The thermal efficiency of the cycle can be calculated as following equation
th = 1(1557.92/2182.0)
= 0.286
s
Thh
hh
43
43
B
C
B
CB
in
outin
B
PT
in
sys
thq
q
q
q
q
ww
q
w
1
Exercise 8.2
Thermodynamics 8. Rankine Cycle 24 / 113
wp = work input of feed water pump
wT = turbine work out
P = isentropic efficiency of pump
T = isentropic efficiency of turbine
= density of water
Practical Rankine Cycle
Deviation from the Ideal Rankine Cycle
P
s
PP
sP
ppdp
hhhhw
)(1)( 12
2
1
1212
23 hhqB
sTT hhhhw 4343
)( 14 hhqC
23
1243
23
1243 /
hh
hhhh
hh
hhhh
q
ww PssT
B
PTth
T
s
14
3
2
Ideal cycle
Practical cycle
Irreversibility in the pump
Pressure drop in the boiler
Irreversibility in the turbine
Temperature decrease in the condenser
2s
4s
Thermodynamics 8. Rankine Cycle 25 / 113
[Exercise B.1]
A boiler generates 4,994,457 lb/hr of steam, and burns 374,920 lb of coal per hour having a heating value
of 13,700 Btu/lb. The pressure of the main steam is 3515 psia and the enthalpy is 1421.7 Btu/lb. And the
temperature of feedwater is 505F, and the enthalpy is 496.12 Btu/lb. What is the boiler efficiency.
[Solution]
보일러 효율은 다음과 같다.
문제에서 주어진 값들을 대입하면, 보일러 효율은 90%이다.
[%]100
HQ
hhW fg
Boiler
W = steam flow, lb/h
Q = fuel flow, lb/h
H = heating value of fuel, Btu/lb
hg = enthalpy of main steam, Btu/lb
hf = enthalpy of feedwater, Btu/lb
Boiler Efficiency
Thermodynamics 8. Rankine Cycle 26 / 113
T
s
12
3 4TH
TL
Qout (-)
Qin (+)
Pump Turbine
High temperature reservoir
Boiler
Condenser
Low temperature reservoir
2 1
3 4
Carnot Vapor Cycle
H
Lth
T
T1
Steam generator
ST
Condenser
wTFeed pump
qC to cooling water
qB from combustion gas
wP
Thermodynamics 8. Rankine Cycle 27 / 113
p- and T-s diagram within liquid-vapor region
p- and T-s diagram when fluid goes to superheated region
Carnot Vapor Cycle
Thermodynamics 8. Rankine Cycle 28 / 113
Why the Carnot vapor cycle is not used?
Carnot Vapor Cycle
T
s
2
4
3
11
2
3"3
a b c
4
1) Practical pumping process from 1 to 2 instead of 1 to 2
because pump size can be reduced dramatically when a liquid
is a working fluid.
2) The compression process in vapor Carnot cycle takes place in
the process 1-2, which is in liquid-vapor region.
• It is not mechanically practical to partially condense steam
to a particular quality from state 4 to state 1 and then
compress the wet steam from state 1 to state 2.
• For these reasons, actual steam cycles are based on a
modified version of the Carnot cycle called the Rankine
cycle.
3) Practical superheating process from 3 to 3 instead of 3 to 3".
4) If the cycle cross the saturation vapor line i.e. enters to the superheated region as in the figure.
• the problem arises to keep the temperature constant between state 3-3 because it is in the
superheated region.
• moreover, pressure is also dropping.
• therefore, it is impossible to add heat and to keep the temperature constant at the same time.
Thermodynamics 8. Rankine Cycle 29 / 113
[Exercise 8.3] Cycle analysis of a Carnot vapor cycle
A simple ideal Rankine cycle with water as a working fluid operates between the specified pressure
limits. Compare the thermal efficiency of this Rankine cycle with the Carnot cycle having the hot
temperature that is same as the saturated temperature in the figure.
For a simplicity, assume the kinetic and potential energy changes are negligible.
T
s
1
2
3
4
15 MPa
100 kPa
Exercise 8.3
Thermodynamics 8. Rankine Cycle 30 / 113
[Solution]
From the steam tables,
p1 = 100 kPa = 14.5 psia
T1 = 214.4F = 101.3C = 374.3 K
p3 = 15 MPa = 2175.6 psia
T3 = 647.8F = 342.1C = 615.1 K
Carnot cycle efficiency is,
Carnot = 1(374.3/615.1) = 0.391
The cycle efficiency of the ideal Rankine cycle is 0.315 (see Exercise 8.1). This value is much lower than
that of the Carnot cycle having same cycle maximum temperature and minimum temperature. It is clear
from this fact that the shape of the Rankine cycle should be as close as possible in order to increase the
cycle efficiency. However, the Rankine cycle can not be Carnot cycle, and ‘equivalent hot temperature’ is
employed to figure out the efficiency of the Rankine cycle.
H
LCarnot
T
T
T
T 11
3
1
Exercise 8.3
Thermodynamics 8. Rankine Cycle 31 / 113
T
s
1
2
3
4
Equivalent
Carnot Cycle
Equivalent Cycle
Hot TemperatureT
s
1
2
3
4
Equivalent Cycle
Hot Temperature
[Ideal Rankine Cycle for a Typical Nuclear Power] [Ideal Rankine Cycle for a Typical Fossil Power]
The higher the equivalent cycle hot temperature, the greater cycle efficiency.
The average temperature where heat is supplied in the boiler can be increased by superheating the
steam.
Concept of Equivalent Carnot Cycle
Thermodynamics 8. Rankine Cycle 32 / 113
23
2143
hhm
hhhhm
Q
W
in
net
23
141hh
hh
23,
14,1
ssT
ssT
meanin
meanout
meanin
meanout
T
T
,
,1
minmax2314 ssssss
Thermal Efficiency
Thermodynamics 8. Rankine Cycle 33 / 113
Rankine Cycle Analysis2
Some Notes on Rankine Cycle4
Improvement of Rankine Cycle Efficiency5
Heating of Water at Constant Pressure1
Vapor Diagrams3
Advanced Rankine Cycle 6
Thermodynamics 8. Rankine Cycle 35 / 113
Rankine Cycle
T
s
1
2
3
4qout
T
s
2
3
qin
14
T
s
12
3
4
qsys
(a) (b) (c)
T-s Diagram
Thermodynamics 8. Rankine Cycle 36 / 113
Steam is used in more of today’s power plants than any other working fluid.
The physical properties of steam are complex because any one steam property is changed, such as pressure,
temperature, specific volume, energy or moisture, all the other properties will also change.
The Mollier diagram has been developed to show this interrelationship of steam properties, and how they all
fit together.
The vertical axis is enthalpy(kJ/kg or BTU/lb) which is defined as internal energy plus flow energy of the
working fluid, and the horizontal axis is entropy(kJ/kg-K or BTU/lb-F) representing energy loss.
Mollier diagram shows lines of constant pressure, constant temperature, constant moisture, and the steam
saturation line (below which the steam is wet, and above which the steam is dry and superheated.
h-s 선도는 이상기체와 다른 성질을 가지는 실재기체의 상태변화를 실험을 통하여 확인하여 표와 선도로 나타낸 것이다.
h-s 선도는 1906년 R. Mollier가 개발
h를 종축, s를 횡축으로 설정하여 증기의 상태(p, , T, x)를 나타낸 선도.
증기의 상태량(T, p, , x, h, s) 가운데 2개를 알면, h-s 선도로부터 다른 상태량을 알 수 있다.
주로 연소기체나 수증기를 대상으로 하기 때문에 가스터빈 및 증기터빈의 사이클 해석에 이용된다.
압축수의 엔탈피는 파악하기 어렵다.
h-s Diagram [Mollier Diagram]
Thermodynamics 8. Rankine Cycle 38 / 113
The reversible work is frequently referred to as the available energy of the process.
Available energy is the difference in enthalpy from the stage inlet pressure to the enthalpy at the
stage outlet at the same entropy.
Ava
ilab
le E
ne
rgy
Use
ful E
ne
rgy
A
BC
D
pi
po
Lossds
Reduction in Useful Energy
(Performance Degradation)
Increase in entropy due to aging
AB : Isentropic expansion line
AC : Original expansion line
AD : New expansion line due to aging
pi : Pressure at the inlet of turbine
po : Pressure at the outlet of turbine
ds : Increase of entropy due to the loss
h
s
T =Useful Energy
Available Energy
h-s Diagram
Thermodynamics 8. Rankine Cycle 39 / 113
T
s
IP Turbine
LP Turbine
2% Moisture
4%
6%
8%10%12%
s
h
Superheat
Tp
20%
50%
70%
Boiling and high
heat flux zones
T-s vs. h-s Diagram
Thermodynamics 8. Rankine Cycle 40 / 113
When the expansion process in the steam turbine
could be conducted perfectly and without losses, the
steam would expand along a true vertical line, i.e.
isentropically.
However, there are always losses associated with
expansion process, and all expansion lines will curve
toward the right on a Mollier diagram.
The more efficient the process, the more vertical the
line.
The greater the difference between h1 and h2, the
more energy will be extracted from the steam.
There are several ways to increase the difference
between h1 and h2.
① Increasing the initial temperature
② Increasing the initial pressure
③ Decreasing the final pressure
④ Increasing the efficiency of expansion process
h-s 선도 [Mollier Diagram]
Thermodynamics 8. Rankine Cycle 42 / 113
[Exercise 8.4] 증기터빈 효율
그림에 나타나 있는 증기 팽창선을 이용하여 고압터빈과 저압터빈의 등엔트로피열효율을 계산하시오. 단, 저압터빈 열효율 계산 시 중압터빈은 저압터빈에 포함하여 계산하시오.
1460.15
Btu/lbm
1280.09
Btu/lbm
1318.54
Btu/lbm
1520.74
Btu/lbm
954.98
Btu/lbm
1010.00
Btu/lbm
Exercise 8.4
Thermodynamics 8. Rankine Cycle 43 / 113
[Solution]
The isentropic efficiency of the turbine is,
Therefore, the isentropic efficiencies of the turbines are,
T,HP = (1460.151318.54)/(1460.151280.09)
= 0.786
T,LP = (1520.741010.00)/(1520.74954.98)
= 0.903
[Discussion]
1) Thermal efficiency of the IP turbine is higher than that of
HP turbine. This is mainly because HP turbine has
higher steam velocity and shorter blade length than IP
turbine.
2) Thermal efficiency of the LP turbine is lower than that of
IP turbine. This is because LP turbines are operated in
wet steam conditions.
s
Thh
hh
43
43
Nuclear Reheat
HP = 88-90%
LP = 90-91%
LP = 87%
LP = 85%
HP = 82%
Nuclear Non-Reheat
Fossil
Reheat
IP = 90-94%
Saturation
Line
h
s
Exercise 8.4
Thermodynamics 8. Rankine Cycle 44 / 113
Rankine Cycle Analysis2
Some Notes on Rankine Cycle4
Improvement of Rankine Cycle Efficiency5
Heating of Water at Constant Pressure1
Vapor Diagrams3
Advanced Rankine Cycle 6
Thermodynamics 8. Rankine Cycle 45 / 113
1) The work input needed to compress the liquid is very much less than that needed to compress a gas.
The effects of irreversibilities in the feed pump are far less than in the compressors of gas turbines.
The fact that wT>>wP is one of the great advantages of steam plant.
A typical output of the BFPT is about 17 MW for 500 MW class fossil power plants.
2) Very high pressure is needed to achieve a high temperature of heat input.
This high pressure is applied to literally ‘miles’ of tubing in the boiler and as a result the tubes are highly
stressed.
The tubes are also in a very corrosive environment (flue gases) and so they cannot stand too high
temperatures before suffering from creep, corrosion and eventual failure.
Turb
ine
Work
Co
mp
resso
r
Work
Net Work
Turb
ine
Work
Net
Work
Pum
p
Work
Turb
ine
Work
Co
mp
res
sor
Work
Net
Work
Steam Turbine Low efficient
Gas Turbine
High efficient
Gas Turbine
Some Notes on Rankine Cycle
Thermodynamics 8. Rankine Cycle 46 / 113
분류VWO MGR NR 75 50 30
Constant Pressure Operation Variable Pressure Operation
출력 (kW)550,000
(110%)
541,650
(108.3%)
500,000
(100%)
375,000
(75%)
250,000
(50%)
150,000
(25%)
유량 (lb/hr)3,757,727
(112.7%)
3,684,046
(110.5%)
3,335,116
(100%)
2,389,835
(71.7%)
1,564,131
(46.9%)
980,271
(29.4%)
복수기 압력 (inHga) 1.5 1.5 1.5 1.5 1.5 1.5
주증기 온도 (F) 1000 1000 1000 1000 1000 1000
주증기 압력(psia)
3514.7
(100%)
3514.7
(100%)
3514.7
(100%)
2860.2
(81.38%)
1870.2
(54.47%)
1152.6
(32.79%)
1st STA Bowl P.
(psia)
3409.3
(100%)
[97.00%]
3409.3
(100%)
[97.00%]
3409.3
(100%)
[97.00%]
2774.4
(81.39%)
[97.00%]
1814.7
(54.49%)
[97.03%]
1118.0
(32.79%)
[97.00%]
1st STA Shell P.
(psia)
2630.8
(113.9%)
2573.8
(111.5%)
2309.0
(100%)
1683.5
(72.9%)
1128.0
(48.9%)
723.9
(31.4%)
FWPT 동력 (kW)18,755
(3.41%)
18,390
(3.40%)
16,611
(3.32%)
9,622
(2.57%)
4,125
(1.65%)
1,523
(1.02%)
한국표준형 500MW 석탄화력
Some Notes on Rankine Cycle
Thermodynamics 8. Rankine Cycle 47 / 113
3) The low temperature of heat rejection (almost ambient) increases the efficiency.
The cooling water is either drawn from the sea or a river, or circulates in a separate loop via a cooling
tower.
4) The maximum temperature achieved in steam cycles is about 600°C, well below the temperature in gas
turbines. Even so, efficiencies are over 40% - better than most gas turbines. This is mainly due to the low
condenser temperature.
5) The pressure ratio across the turbine is so huge (3514.7 psia/0.74 psia = 4750) that many turbine stages are
required.
6) The HP, IP and LP turbines are mounted on just one shaft with the electrical generator at the end. Currently,
the isentropic efficiency of the HP and IP turbines are very high (90-92%), but the LP turbine efficiency is
lower (85%). This is mainly because the LP operates in wet steam condition – typically every 1% of wetness
gives a 1% loss in isentropic efficiency. In addition, the LP turbine blades are very long giving greater
aerodynamic losses.
7) The density falls so much through the turbines that the volume flow rate cannot be accommodated in one
cylinder. Therefore, the turbine might be divided into one single-flow HP cylinder, one double-flow IP cylinder,
and two or three double-flow LP cylinders.
Some Notes on Rankine Cycle
Thermodynamics 8. Rankine Cycle 48 / 113
As the steam expands, it first releases its superheat energy until it reaches the saturated condition.
Then, with further expansion, a portion of the latent heat contained in the steam is released. This conversion
of latent heat introduces a state where water is formed in the expanding steam.
However, heat transfer from the gas to liquid phase requires a finite period of time, and the expansion of
steam in the steam path is extremely rapid.
The elapsed time for steam entering a high pressure section to expand through it, and through the reheat and
low pressure sections is about 0.2 seconds, if the time in the crossover pipes and the boiler reheater section
is ignored.
Because heat transfer cannot occur instantaneously, the expansion will continue under the saturated vapor
line.
At Wilson line, which is located approximately 60 Btu/lb from saturated vapor line, heat transfer will have
been completed and approach thermal equilibrium conditions, and moisture will form. This is the point where
fog is formed, consisting of particles from about 0.5 to 1.0 microns in diameter.
Steam inside the vapor dome is supersaturated above the Wilson line, a term that arises from water existing
as vapor at conditions at which condensation should be taking place.
Wilson Line
Some Notes on Rankine Cycle
Thermodynamics 8. Rankine Cycle 49 / 113
Supersaturation = Condensation Shock
h
s
Wilson line
A
D
S
pn
ps
pn
ps
pe
Superheated
condition from A-D
Supersaturated
condition from D-S
Saturated
vapor line
dsThe Wilson line is
located 60 Btu/lbm
below saturated
vapor line.
Wilson line
It was more precisely located by Yellot at a moisture of 3.2%
compared to the 4% determined by Wilson.
Supersaturation
Some Notes on Rankine Cycle
Thermodynamics 8. Rankine Cycle 50 / 113
Water Droplet Erosion
Fog Formation
(Condensation Shock)
Dry Steam Wet Steam
Phase
Change
Supersaturation
Some Notes on Rankine Cycle
Thermodynamics 8. Rankine Cycle 51 / 113
8) The exit of the LP turbine has to be very large to accommodate the
flow. Typically the last blades of a large turbine are about 4 m
diameter.
9) The steam leaving the LP turbine is usually in the two-phase region
with a dryness fraction of about 90%. The water is mostly in the form
of a fog of minute droplets with diameter of order 1 micron. However,
larger droplets, like raindrops, are formed when the small drops
deposit on the blade surface and coalesce and becomes water film.
Then, the water film becomes water droplets having diameter of
approximately 200 m when it leaves nozzle trailing edge. These
large droplets cause water droplet erosion on the rotating blades of
the last stage.[ Blade erosion after 2.5 years ]
Some Notes on Rankine Cycle
Thermodynamics 8. Rankine Cycle 52 / 113
Bucket
Nozzle
UU
Cs
CdWd
Ws
Wd
Direction of rotation
Droplets
Cs: Absolute steam velocity
Ws: Relative steam velocity
Cd: Absolute droplet velocity
Wd: Relative droplet velocity
U: Peripheral rotation velocity
Velocity Triangles
Some Notes on Rankine Cycle
Thermodynamics 8. Rankine Cycle 53 / 113
Rankine Cycle Analysis2
Some Notes on Rankine Cycle4
Improvement of Rankine Cycle Efficiency5
Heating of Water at Constant Pressure1
Vapor Diagrams3
Advanced Rankine Cycle 6
Thermodynamics 8. Rankine Cycle 54 / 113
1. Make the cycle just like Carnot cycle. Then, increase Carnot
cycle efficiency
1) Heat release at lower temperature (or pressure)
2) Heat input at higher temperature
3) Superheat the steam
2. Modify the Rankine cycle
1) Reheating cycle (modify the right hand side of the T-s
diagram)
2) Regenerative cycle (modify the left hand side of the T-s
diagram)
T
s
12
3
4
qsys
How to Improve the Cycle Efficiency ?
Thermodynamics 8. Rankine Cycle 55 / 113
As the condenser pressure is deceased from 4 to 4,
• net work is increased as much as 1-4-4-1-2-2-1
• therefore, turbine output increases
• however, heat addition should be increased as
much as a-2-2-a-a
• interestingly, these two areas are approximately
same
• therefore, cycle efficiency increases
The condenser takes low pressure steam from turbine
and removes heat to condensate this steam to water.
This condensed water then enters the feedwater
pump.
1. Lowering Condenser Pressure [6/7]
T
s
2′
p4′
1′
1
2
3
a b
4
p4
a′
4′
The lower the condenser temperature and pressure, the greater the turbine work out. The thermal
efficiency is then increased.
The condenser is usually cooled by a natural heat sink, such as air, river or lake, or sea water.
The lowest temperature in the cycle is in the rage of 25~30C.
Condensers are operated under a strong vacuum (T=27C, psat=3.6 kPa).
Thermodynamics 8. Rankine Cycle 56 / 113
1. Lowering Condenser Pressure [5/7]
The end pressure of steam expansion in the turbine is determined by the steam saturation temperature
depending on the cooling water temperature and heat transfer conditions on the condenser tubes.
Thermodynamics 8. Rankine Cycle 57 / 113
Q: 복수기 압력을 얼마로 유지시켜야 가장 경제적인가? (Q from GS EPS)
Exhaust pressure, inHga
2 F-40.0"LSB
2 F-33.5"LSB
2 F-30.0"LSB
D-11 steam turbine for GE 207FA,
1800 psia / 1050F / 1050F
Tu
rbin
e g
en
era
tor
ou
tpu
t, M
W
0.5 1.51.0 2.52.0 3.53.00.0175
195
190
185
180
4.0
D-11 (GE) – LSB selection (207FA)
1. Lowering Condenser Pressure [2/7]
Thermodynamics 8. Rankine Cycle 58 / 113
Siemens
0.5 1.0 1.5 2.01.76 inHga
0 2 12 16 19
18
16
14
12
10
8
6
4
2
0
Cooling water inlet temperature
Po
we
r g
ain
, M
W
4F30"
4F32"
4F38"
6F32"
C
Choking
Application of Longer LSB
1. Lowering Condenser Pressure [3/7]
Thermodynamics 8. Rankine Cycle 59 / 113
Condenser Pressure, inHga
Pow
er
Gain
, M
W
1.5 1.6 1.7 1.8 1.9 2.0 2.1-20
-16
-12
-8
-4
0
4
Source: EPRI Report No. GS-7003
1,000MW, Nuclear Power Plant
A Typical Power Loss as a Function of Condenser Pressure
1. Lowering Condenser Pressure [4/7]
Thermodynamics 8. Rankine Cycle 60 / 113
Lowering the condenser pressure causes an increase in the moisture content at the turbine exhaust end. This
in turn will affect adversely the turbine efficiency and the water droplet erosion of turbine blades.
In addition, a lower condenser pressure will result in an increase in condenser size and cooling water flow
rate.
In general, it is recommended that the quality should be higher than 90% at the turbine exhaust.
The condenser temperature is generally limited by cooling medium (e.g. lake or river, etc.).
ex) lake @ 15C + T(=10C) = 25C
psat = 3.2 kPa
A condenser is needed to make saturated water.
• The pressure at the exit of the turbine can be less than atmospheric pressure
• The closed loop of the condenser allows us to use treated water on the cycle side
• But if the pressure is lower than atmospheric pressure, air can leak into the condenser, preventing
condensation.
1. Lowering Condenser Pressure [7/7]
Thermodynamics 8. Rankine Cycle 61 / 113
T
s
1
2
3
4
Equivalent Cycle Hot
Temperature
[Exercise 8.5] Comparison of cycle efficiency
The mean temperature of the sea water is 30C during the summer, and 10C during the winter.
Compare the thermal efficiencies of the ideal Rankine cycle in terms of seasonal climatic change. The
equivalent cycle hot temperature of 300C is always constant.
Exercise 8.5
[Solution]
Thermal efficiency can be calculated by Carnot
cycle efficiency,
= 1 (T1/T3)
s = 1 (30+273)/(300+273) = 0.471
w = 1 (10+273)/(300+273) = 0.506
where, s and w mean the thermal efficiency
during the summer and winter, respectively.
[Discussion]
Thermal efficiency of the Rankine cycle increases
as the coolant temperature decreases.
Thermodynamics 8. Rankine Cycle 62 / 113
Increased boiler pressure has a higher mean
temperature of heat addition.
However, the temperature of heat rejection is
unchanged.
Usually, the amount of the cycle work is not changed
although boiler pressure is increased. This is because
the amount of the increased work (top side) and the
amount of the decreased work (right hand side)
caused by pressure increase is almost same.
However, the amount of heat rejected is decreased.
Thus, the cycle efficiency increases with boiler
pressure.
T
s
2
1
2
3
a b
4
b
4
3
decrease in qin
increase in qin
decrease in qout
c
2. Increased Boiler Pressure [1/5]
The only one drawback is that the quality of the exhaust flow become worse. That is, state 4 has a lower
quality than at state 4 and it results in following problems:
• Water droplet erosion of LSBs
• Efficiency penalty by WDE
• Higher maintenance cost
For these reasons, quality of higher than 90% is required at the turbine exhaust.
Thermodynamics 8. Rankine Cycle 63 / 113
Supercritical Boiler
T
s
1
2
3
4
Supercritical
Subcritical
2
3Tmax, subcritical
Tmax, USC
3
2
Critical Point
It has a boiler pressure higher than the critical
pressure of 220 bar.
It requires once-through steam generator of
different design as boiling no longer occurs.
It also can get without increasing the maximum
cycle temperature.
The boiler is a heat exchanger without a clear
change of phase.
Supercritical boilers have higher thermal efficiency,
but this is expensive.
Running the boiler tubes at pressures above the
critical point of stream is a very challenging
mechanical design requirement.
Thicker tubes increases the thermal resistance of
the tubes and hence tends to decrease the boiler
effectiveness.
2. Increased Boiler Pressure [2/5]
Thermodynamics 8. Rankine Cycle 64 / 113
7% cycle efficiency improvement by steam condition
from 160bar / 540C / 540C
to 290bar / 600C / 620C
Efficiency Improvement by USC
2. Increased Boiler Pressure [3/5]
Thermodynamics 8. Rankine Cycle 65 / 113
Drum-Type Boiler Once-through Boiler
2. Increased Boiler Pressure [4/5]
Thermodynamics 8. Rankine Cycle 66 / 113
T
s
Early 20th century
1940s
1960
Ultra Supercritical
Supercritical
Evolution of Rankine Cycle
2. Increased Boiler Pressure [5/5]
Thermodynamics 8. Rankine Cycle 67 / 113
[Exercise 8.6] Comparison of cycle efficiency
Two simple ideal Rankine cycles with water as a working fluid operate between the specified pressure
limits. Compare the thermal efficiencies of two cycles. Additionally, discuss the cycle efficiency and
capital cost of the plant.
For a simplicity, assume the kinetic and potential energy changes are negligible.
T
s
2B
4B
3B
1
2A
3A
4A
2 psia
200 psia
600 psia
Exercise 8.6
Thermodynamics 8. Rankine Cycle 68 / 113
[Solution]
From the steam tables,
h1 = hf @ 2 psia = 94.036 Btu/lbm = 218.73 kJ/kg
hg @ 2 psia = 1116.151 Btu/lbm = 2596.17 kJ/kg
1 = f @ 2 psia = 0.01623 ft3/lbm = 28.045 in3/lbm = 0.001013 m3/kg
s1 = sf @ 2 psia = 0.175 Btu/lbm‧R = 0.7327 kJ/kg‧K
sg @ 2 psia = 1.920 Btu/lbm‧R = 8.0390 kJ/kg‧K
Pump power,
wp = (h2h1) = 1(p2p1) h2 = h1wp
wpA = 0.001013 m3/kg (2002)6.89286 kPa = 1.383 kJ/kg
wpB = 0.001013 m3/kg (6002)6.89286 kPa = 4.176 kJ/kg
h2A = h1wpA = 220.13 kJ/kg, & h2B = h1wpB = 222.93 kJ/kg
Determine the thermodynamic properties at point 3 using steam tables,
p3A = 200 psia, x = 1 & p3B = 600 psia, x = 1
h3A = hg @ 200 psia = 1198.334 Btu/lbm = 2787.32 kJ/kg
s3A = sg @ 200 psia = 1.54544 Btu/lbm‧R = 6.4708 kJ/kg‧K
h3B = hg @ 600 psia = 1203.657 Btu/lbm = 2799.71 kJ/kg
s3B = sg @ 600 psia = 1.44610 Btu/lbm‧R = 6.0548 kJ/kg‧K
Exercise 8.6
Thermodynamics 8. Rankine Cycle 69 / 113
[Solution]
Determine the heat addition in the boiler,
qBA = (h3Ah2A) = 2787.32220.13 = 2567.19 kJ/kg
qBB = (h3Bh2B) = 2799.71222.93 = 2576.78 kJ/kg
Determine the thermodynamic properties at point 4 using steam tables,
p4A = 2 psia, s4A = s3A & p4B = 2 psia, s4B = s3B
x4A = (s4Asf)/(sgsf) = (6.47080.7327)/(8.0390.7327) = 0.7854
h4A = (1x4A)hf + x4Ahg = (10.7854)218.73 + 0.78542596.17 = 2085.97 kJ/kg
x4B = (s4Bsf)/(sgsf) = (6.05480.7327)/(8.0390.7327) = 0.7284
h4B = (1x4B)hf + x4Bhg = (10.7284)218.73 + 0.72842596.17 = 1950.46 kJ/kg
Determine the heat release from the condenser,
qCA = (h4Ah1) = (2085.97218.73) = 1867.24 kJ/kg
qCB = (h4Bh1) = (1950.46218.73) = 1731.73 kJ/kg
Determine cycle efficiency,
A = 1 qCA/qBA = 11867.24/2567.19 = 0.273
B = 1 qCB/qBB = 11731.73/2576.78 = 0.328
Exercise 8.6
Thermodynamics 8. Rankine Cycle 70 / 113
[Discussion]
The thermal efficiency of the ideal Rankine cycle is increased with boiler pressure. That is, the higher
boiler pressure, the higher cycle efficiency.
For this reason, the maximum cycle pressure of the Rankine cycle has being increased since early 20th
century.
The higher boiler pressure, the higher capital cost. Therefore, in order to increase the boiler pressure,
very careful economical evaluations are required.
Exercise 8.6
Thermodynamics 8. Rankine Cycle 71 / 113
3. Superheating [1/2]
T
s
heat
added
heat
lost condenserpump
Boiler
Turbine
qECO&EVA
qSH G
12
3
4
5
1
2
3
4
5
5
Steam that has been heated above the saturation temperature corresponding to its pressure is said to be
superheated.
With the addition of superheating, the turbine transforms this additional energy into work without forming
moisture, and this energy is basically all recoverable in the turbine.
A higher plant efficiency is obtained if the steam is initially superheated, and this means that less steam and
less fuel are required for a specific output.
Thermodynamics 8. Rankine Cycle 72 / 113
The overall efficiency is increased by superheating the steam. This is because the mean temperature where
heat is added increases, while the condenser temperature remains constant.
Increasing the steam temperature not only improves the cycle efficiency, but also reduces the moisture
content at the turbine exhaust end and thus increases the turbine internal efficiency.
The average temperature where heat is supplied in the boiler can be increased by superheating the steam.
The turbine work out is also increased by superheating the steam without increasing the boiler pressure.
When the superheating the steam is employed in the cycle, the important thing is that the quality of the steam
at the turbine exhaust is higher than 90%.
The creep characteristics of the material should be considered (typically, the creep limit of stainless steel is
565C(1,050F)).
The maximum cycle temperature is set by material strength limits of about 700C.
Dry saturated steam from the boiler is passed through a second bank of smaller bore tubes within the boiler
until the steam reaches the required temperature.
If the steam is too much superheated, the turbine exhaust can fall in the superheated region and this requires
a larger condenser which is expensive.
3. Superheating [2/2]
Thermodynamics 8. Rankine Cycle 73 / 113
[Exercise 8.7] Comparison of cycle efficiency
Determine the thermal efficiencies of the Rankine cycle with superheat. In this case the maximum cycle
temperature increased from saturation temperature (486.2F) to 1000F. And compare it with that
determined in [Exercise 8.6]
For a simplicity, assume the kinetic and potential energy changes are negligible.
T
s
2
3
1 4
2 psia
600 psia,
486.2F
1000F
Exercise 8.7
Thermodynamics 8. Rankine Cycle 74 / 113
[Solution]
From the steam tables,
h1 = hf @ 2 psia = 94.036 Btu/lbm = 218.73 kJ/kg
hg @ 2 psia = 1116.151 Btu/lbm = 2596.17 kJ/kg
1 = f @ 2 psia= 0.01623 ft3/lbm = 28.045 in3/lbm = 0.001013 m3/kg
sf @ 2 psia = 0.175 Btu/lbm‧R = 0.7327 kJ/kg‧K
sg @ 2 psia = 1.920 Btu/lbm‧R = 8.0390 kJ/kg‧K
Pump power,
wp = (h2h1) = 1(p2p1) h2 = h1wp
wp = 0.001013 m3/kg (6002)6.89286 kPa = 4.176 kJ/kg
h2 = h1wp = 222.93 kJ/kg
Determine the thermodynamic properties at point 3 using steam tables,
p3 = 600 psia, T = 1000F
h3 = 1517.38 Btu/lbm = 3529.43 kJ/kg
s3 = 1.71551 Btu/lbm‧R = 7.1828 kJ/kg‧K
Exercise 8.7
Thermodynamics 8. Rankine Cycle 75 / 113
[Solution]
Determine the heat addition in the boiler,
qB = (h3h2) = 3529.43222.93 = 3306.5 kJ/kg
Determine the thermodynamic properties at point 4 using steam tables,
p4 = 2 psia, s4 = s3
x4 = (s4sf)/(sgsf) = (7.18280.7327)/(8.0390.7327) = 0.8828
h4 = (1x4)hf + x4hg = (10.8828)218.73 + 0.88282596.17 = 2317.54 kJ/kg
Determine the heat release from the condenser,
qC = (h4h1) = (2317.54218.73) = 2098.81 kJ/kg
Determine cycle efficiency,
th = 1 qC/qB = 1 2098.81/3306.5 = 0.365
The thermal efficiency is increased from 0.328 to 0.365 by the employment of superheater.
Exercise 8.7
Thermodynamics 8. Rankine Cycle 76 / 113
Rankine Cycle Analysis2
Some Notes on Rankine Cycle4
Improvement of Rankine Cycle Efficiency5
Heating of Water at Constant Pressure1
Vapor Diagrams3
Advanced Rankine Cycle 6
Thermodynamics 8. Rankine Cycle 77 / 113
4523
126543
hhhh
hhhhhhth
nonreheatthreheatth
The steam from boiler flows to the HP turbine where it expands and is exhausted back to the boiler for
reheating. The efficiency of the Rankine cycle can be improved by reheating on the right hand side of
the T-s diagram.
T
s
qRH
qH
3
qL
2
5
4
41 6
wP
wTA
B
Condenser
Pump
Boiler
Turbine
qRH
qH
GLP
qL
12
3
4
5
6
wP
wT
Reheater
HP
1. Rankine Cycle with Reheat [1/12]
Modifications of Rankine cycle, including superheating, reheating, and feedwater heating, increase the
thermal efficiency of power plants by approximately 10%.
Thermodynamics 8. Rankine Cycle 78 / 113
① Higher cycle efficiency than non-reheat
cycle (The mean temperature of heat
reception is increased by reheating, and the
efficiency is also increased).
② Longer life expectancy of LSB because of
improved steam conditions at the turbine
exhaust (The moisture content is moved
from C to E by reheating).
③ Higher turbine internal efficiency because of
reduced moisture loss.
Benefits of reheat cycle
h
s
4%
8%
16%
12%C
E
B
A
D
A-B-C: Nonreheat
A-B: HP Turnbine
B-D: Reheater
D-E: IP and LP Turbine
1. Rankine Cycle with Reheat [2/12]
Thermodynamics 8. Rankine Cycle 79 / 113
The reheater is designed to raise the steam temperature back to its initial value and, thus, to increase the
cycle efficiency.
A higher wetness is very harmful because water droplet erosion of LSB increases and the efficiency of LP
turbine decreases due to the moisture loss.
In order to improve the steam condition at the turbine exhaust, the steam exhausted at the middle of the
turbine can be reheated, then expanded again through the reheat turbine(IP and LP turbines) in two steps.
In a normal Rinkine cycle (1-2-3-4-1), water droplets are formed as the temperature in the turbine falls. In a
normal Rankine cycle, droplets are formed between point A and 4. Distance between this two points can be
minimized by the employment of reheat cycle.
Above a certain level of steam pressure, the employment of steam reheat is desirable. In the case of USC
plants, double reheat system is desirable.
An improvement in cycle efficiency from a single reheat is approximately 2~3%. Although this is not dramatic,
it is a useful gain which can be obtained without major modification to the plant.
Further improvement in cycle efficiency of approximately 1.6% can be achieved through the addition of a
second reheat system (double reheat) in USC power systems .
The IP turbine is also called the reheat turbine since it receives the reheated steam.
1. Rankine Cycle with Reheat [3/12]
Thermodynamics 8. Rankine Cycle 80 / 113
1000 1100 1200
Pla
nt N
et H
ea
t R
ate
Im
pro
ve
me
nt, %
8
7
6
5
4
3
2
1
0
2.8 %
2.4 %
2400 psig/1000F/1000F
versus
4500 psig/1100F/1100F
2400 psig (165 bar)
Sub-Critical
USC 2900 psig (200 bar)
3650 psig (250 bar)
4350 psig (300 bar)
5800 psig (400 bar)5050 psig (350 bar)
Double Reheat vs. Single Reheat:
Heat Rate Improvement = 1.6%
Temperature, F
Comparison
2.8% + 2.4% + 1.6% = 6.8%
Siemens
1. Rankine Cycle with Reheat [4/12]
Thermodynamics 8. Rankine Cycle 81 / 113
The effect on the cycle efficiency depends on the reheat pressure.
• In general, the optimum reheat pressure for maximum cycle efficiency is usually about 1/4 of the main
boiler pressure.
• High reheat pressure gives a high mean temperature of heat reception during reheating, but only a small
extra heat input leading to small increase in cycle efficiency.
• Low reheat pressure gives an almost same mean temperature with the main cycle. Therefore, there is no
significant improvement in cycle efficiency.
The pressure drop in the reheater and associated piping is important.
A relatively large pressure drop can significantly offset the benefit due to reheating.
When large diameter piping is used, the pressure drop is reduced, but initial cost may increase
proportionately. As in the main steam pipe, pressure drop in the reheater and its piping must be appropriately
balanced against the initial cost.
The pressure drop in the reheater and associated piping is about 7% to 10%.
This pressure drop results in a poorer power plant heat rate of 0.7% to 1.0%.
As a rule of thumb for the effect of pressure drop on heat rate any place in the steam path is 0.1% poorer
heat rate per 1% pressure drop in a fossil plant and 0.15% poorer heat rate per 1% pressure drop in a
nuclear plant.
1. Rankine Cycle with Reheat [5/12]
Thermodynamics 8. Rankine Cycle 82 / 113
The higher the maximum temperature, the better the
efficiency.
The higher the upper steam temperature, the more
expensive the boiler, steam-lines, and turbine
materials.
In addition, such steels are more difficult in
machining and welding.
Therefore, the cycle maximum temperature should
be determined by the consideration of thermal
efficiency, capital, construction, and maintenance
and operating costs.
The steam temperature entering the turbine is limited
by metallurgical constraint, typically 565C(1,050F),
corresponding to the creep limit of stainless steel.
Modern boilers can handle up to 30MPa and a
maximum temperature up to 650C.
T
s
3
2
5
4
41 6
3
1. Rankine Cycle with Reheat [6/12]
Thermodynamics 8. Rankine Cycle 85 / 113
1-nuclear reactor, 2-steam generator, 3-HP turbine, 4-moisture
separator, 5-reheater, 6-LP turbine, 7-generator, 8-condenser,
9-condensate pump, 10-LP FWH, 11-LP FWH, 12-BFP, 13-HP
FWH, 14-main circulating pump
G
1
109
8
7
6
54
3
2
111213
14
Schematic of Nuclear Power Plant
1. Rankine Cycle with Reheat [9/12]
Thermodynamics 8. Rankine Cycle 86 / 113
HP Turbine
Condensate
Pump
Main Steam
To
Feedwater
Heaters
Steam
DryersLP Turbines
From
Main
Steam
Moisture Separator Reheater
To
Feedwater
Heaters
Condensers
Main
Electrical
GeneratorGenerator
Exciter
To Main
Transformer
[ Steam Turbine (APR 1400) ]
Moisture Separator Reheater
1. Rankine Cycle with Reheat [10/12]
Thermodynamics 8. Rankine Cycle 87 / 113
MSR (GE)
Steam Turbine Expansion Lines and MSR
1. Rankine Cycle with Reheat [11/12]
Thermodynamics 8. Rankine Cycle 88 / 113
In an ideal Rankine cycle for saturated steam with
Moisture separator and reheater, steam expands
in the HP turbine to pressure p4 and is reheated to
superheated steam (T6<T3).
It is clear that the equivalent Carnot cycle
temperature is this case is lower than for the initial
cycle. Thus, such steam reheat does not improve
the thermal efficiency.
Practically, however, thermal efficiency is
improved by using the MSR because of much less
moisture loss in LP turbine caused by an
improved LP turbine exhaust quality.
T
s
1
2
3
7
45
6
7
4-5: Steam separator
5-6: Reheater
Ideal Saturated-Steam Rankine Cycle with MSR (Nuclear)
1. Rankine Cycle with Reheat [12/12]
Thermodynamics 8. Rankine Cycle 89 / 113
T
s
2
41
3
1
2
[Exercise 8.6] Summary
p1 = 2 psia, p3 = 600 psia
T1 = 52.3C, T3 = 252.3C
Rankine = 0.328 (from Ex. 8.6)
Carnot = 0.381
By the addition of regenerative feedwater heating, the original Rankine cycle was improved significantly.
This is done by extracting steam from various stages of the turbine to heat the feedwater as it is pumped from
the condenser to the boiler to complete the cycle.
It can be seen from this comparison that there is a big difference in efficiency between Rankine cycle and
Carnot cycle.
This means that the Rankine cycle efficiency can be increased further by the modification of the cycle.
One of the biggest efficiency losses in the Rankine cycle occurs on the left hand side of the T-s diagram. This
is mainly due to the liquid heating in the economizer.
It is clear that the efficiency of Rankine cycle can be increased significantly if the Rankine cycle has a same
shape with Carnot cycle.
It has been discussed already, however, that it is impossible to get a Carnot vapor cycle.
2. Regenerative Rankine Cycle [1/5]
Thermodynamics 8. Rankine Cycle 90 / 113
T
s
1
2
34
51 5
a b c d
Condenser
Pump
BoilerG
Turbine
1
23
4
5
If the liquid heating could be eliminated from the boiler, the average temperature for heat addition would be
increased greatly and equal to the maximum cycle temperature.
In the ideal regenerative Rankine cycle, the water circulates around the turbine casing and flows in the
direction opposite to that of the steam flow in the turbine.
Because of the temperature difference, heat is transferred to the water from the steam. However, it can be
considered that this is a reversible heat transfer process, that is, at each point the temperature of steam is
only infinitesimally higher than the temperature of water.
At the end of the heating process the water enters the boiler at the saturation temperature.
Since the decrease of entropy in the steam expansion line is exactly equal to the increase of entropy in the
water heating process, the ideal regenerative Rankine cycle will have the same efficiency as the Carnot cycle.
The boiler, in this case, would have no economizer, and the irreversibility during heat addition in the boiler
would decrease because of less temperature difference between the heating and heated fluids.
2. Regenerative Rankine Cycle [2/5]
Thermodynamics 8. Rankine Cycle 91 / 113
Unfortunately, however, the ideal process is practically impossible.
Instead, the turbine is furnished with the definite number of heaters to heat feedwater with extracted steam in
some stages.
This improves the cycle efficiency significantly, even though it remains lower than the Carnot cycle efficiency.
This cycle is called as a regenerative cycle.
The heat input in the boiler decreases as the final feedwater temperature increases and the heat rejected in
the condenser getting smaller as the feedwater is heated higher using the extracted steam.
2. Regenerative Rankine Cycle [3/5]
T
s
Condenser
Condensate Pump
BoilerGTurbine
1
23
4
1
2 3
5
4
6
7
FWH
Feedwater Pump
5
67
Boiler (1)
Feedwater Heater (m)
Condenser (1-m)
Thermodynamics 8. Rankine Cycle 92 / 113
Reversible heat transfer and an infinite number of feedwater heaters would result in a cycle efficiency equal
to the Carnot cycle efficiency.
The greater the number of feedwater heaters used, the higher the cycle efficiency. This is because if a large
number of heaters is used, the process of feedwater heating is more reversible.
The economic benefit of additional heaters is limited because of the diminishing improvement in cycle
efficiency, increasing capital costs, and turbine physical arrangement limitations.
The amount of steam flow into condenser can be reduced dramatically by the employment of regenerative
Rankine cycle. For example, the mass flow of steam entering condenser is about 65% of throttle flow for
typical 500 MW fossil power plants.
The LSB problems, such as water droplet erosion and longer active length, could be solved by the
regenerative Rankine cycle, which is made by steam extraction in many turbine stages.
Regenerative Rankine cycle also diminish the influence of the LP turbine, which has worst performance.
Power output in the turbine is decreased by the steam extraction.
2. Regenerative Rankine Cycle [4/5]
Thermodynamics 8. Rankine Cycle 93 / 113
CycleNo. of Feedwater
HeatersHARP Heat Rate Benefit
Single Reheat
(4500 psi, 1100F/ 1100F)
7
8
8
9
No
No
Yes
Yes
Base Case
+0.2%
+0.6%
+0.7%
Double Reheat
(4500 psi, 1100F/ 1100F/1100F)
8
9
9
10
No
No
Yes
Yes
Base Case
+0.3%
+0.2%
+0.5%
Heat Rate Impact of Alternative Feedwater Heater Configurations
2. Regenerative Rankine Cycle [5/5]
급수가열기 수가 결정되면, 터빈에서 추기점을 결정한다.
만약 n개의 급수가열기를 사용하는 경우 증기발생기에서의 포화온도와 복수기에서의 응축온도의 차이를1/(n+1)로 배분하는 온도에 대응하는 포화압력에서 추기하면 최대효율을 얻을 수 있는 것으로 알려져 있다.
이렇게 최적으로 배분하면 증기발생기에서의 열흡수 온도는 증기발생기 증발온도와 복수기에서의 응축온도차이의 n/(n=1)까지 상승한다.
그러나 추기압력이 최대효율을 나타내는 압력에서 약간 벗어나더라도 효율에 큰 영향을 미치지 않기 때문에실제 설계에서는 다른 요인들을 함께 고려하여 추기압력을 결정한다.
Thermodynamics 8. Rankine Cycle 94 / 113
[Exercise 8.8] Cycle Analysis – Ideal Regenerative Rankine Cycle
Calculate the cycle efficiency of the given in the figure below. Saturated steam enters the steam turbine
at the pressure 600 psia and exhausts at the pressure 2 psia to the condenser. Some of the steam with
pressure of 100 psia is extracted from the intermediate stage of the turbine for the purpose of feedwater
heating.
T
s
Condenser
Pump 1
BoilerGTurbine
1
2
3
4
1
2
3
5
4
6
7
FWH
Pump 2
5
67
600 psia
100 psia
2 psia
Rankine = 0.328 (from Ex. 8.6)
Regenerative = 0.350
Carnot = 0.381
Exercise 8.8
Thermodynamics 8. Rankine Cycle 95 / 113
Closed Type Feedwater Heater [1/6]
Boiler
G
Turbine
12
3
4
C
5
678
9
10m2
m3
12
11
T
s
Boiler (1)
1
2
3
45
6
7
8
9
10
m2
m3
12
11
Thermodynamics 8. Rankine Cycle 96 / 113
A closed feedwater is a heater where the feedwater and the heating steam do not directly mix.
Open feedwater heaters (deaerators) directly mix the feedwater and the heating steam.
A closed feedwater heater may consist of three zones: the desuperheating zone, the condensing zone, and
the drain cooler zone.
All closed heaters have a condensing zone where the feedwater is heated by the condensation of the
heating steam.
Feedwater heaters that receive highly superheated steam require a desuperheating zone to reduce the
steam temperature to approximately 50F above saturation temperature before it enters the condensing
zone.
Closed Type Feedwater Heater [2/6]
Thermodynamics 8. Rankine Cycle 97 / 113
A desuperheating zone may not be required for
heaters that receive heating steam with less
than 100F superheat.
Usually, a drain cooler is also included in a
feedwater heater to recover the heat contained
in the drains before the drains leave the heater.
The feedwater heater performance is
determined by DCA (drain cooler approach) and
TTD (terminal temperature difference).
The DCA is the difference between the
temperature of the drains leaving the heater
and the temperature of the feedwater entering
the heater.
The TTD is the difference between the
saturation temperature at the operating
pressure of the condensing zone and the
temperature of the feedwater leaving the heater.Travel Distance
Te
mp
era
ture
Feedwater Inlet
Extraction Steam Outlet
Feedwater Outlet
Extraction Steam Inlet
Extraction
(Negative)
TT
D
Desuperheating
ZoneCondensing Zone
Drain
Cooling
Zone
TSAT
DC
A
[ Temperature profile for a closed feedwater heater ]
Closed Type Feedwater Heater [3/6]
Thermodynamics 8. Rankine Cycle 98 / 113
By decreasing the DCA of a heater, cycle efficiency is improved while the heater surface area is increased,
resulting in higher capital cost.
The practical minimum DCA for an internal drain cooler is 10F. But, the minimum practical limit is 5F for an
external drain cooler.
The heater may have a negative TTD when the temperature of the feedwater leaving the heater is higher
than the saturation temperature of the condensing zone because of the desuperheating zone.
If the desuperheating zone of the heater is removed, the feedwater temperature leaving the heater would be
less than the saturation temperature, resulting in a positive TTD.
The practical lower limit of TTD on a heater without a desuperheating zone is +2F.
The negative TTD limit for a heater with a desuperheating zone depends on the amount of superheat in the
extraction steam entering the heater.
The lower the TTD and DCA, the higher the cycle efficiency and the larger the heater surface area.
The more efficient cycle results in a lower heat rate and reduced fuel consumption, while the larger surface
area of a heater results in a higher capital cost.
Closed Type Feedwater Heater [4/6]
Thermodynamics 8. Rankine Cycle 99 / 113
Extraction Steam
Subcooled LiquidLiquid Tsat (p)
TTD Tin, pin, hin, m1
p, T, h, m2
Terminal Temperature Difference: TTD = Tsat – Tout
Drain Cooler Approach: DCA = Tdrain – Tin
Steady Flow Energy: m1(hout – hin) = m2(h – hdrain)
DCA
Tout, pin, hout, m1
p, Tdrain, hdrain, m2
(condensate is allowed
to pass through)
steam trap
to higher
pressure linepump
Drain Cooler
Heat Balance
Closed Type Feedwater Heater [5/6]
Thermodynamics 8. Rankine Cycle 100 / 113
The closed feedwater heater, which is a shell and tube heat exchanger, is a surface heat exchanger with the
feedwater inside the tubes, and the extraction steam is condensed on the outer surface of the tubes.
In contrast to the open feedwater heater, which is a contact heater, the closed feedwater heater can operate
with different pressures for the extraction steam and the feedwater and they are not mixed.
The closed feedwater heater is fully specified by the three equation given in the figure for TTD, DCA, and
steady flow energy.
The TTD is generally about 5F, but may be as high as 10F. (range from 3F to 10F)
The DCA is generally about 10F, but may be as high as 20F.
Because the two streams are at different pressures in the closed feedwater heater, two principal alternatives
exist for what to do with the condensed extraction steam from the heater drain:
1. Allow it to expand isentropically, and send it to a lower pressue feedwater heater, or condenser.
2. Increase its pressure with a small pump and mix it into the feedwater stream.
Both of these alternatives are frequently used.
Closed Type Feedwater Heater [6/6]
Thermodynamics 8. Rankine Cycle 101 / 113
Configuration
Open Type Feedwater Heater - Deaerator [1/6]
Thermodynamics 8. Rankine Cycle 102 / 113
Operating Principle
(a) Spray type deaerator uses tray to produce film like water flow to enhance contact with steam for
maximum stripping of residual oxygen.
(b) Spray scrubber deaerator produces tubulent upward flow of steam and water in scrubber section for
removal of oxygen.
Open Type Feedwater Heater - Deaerator [2/6]
Thermodynamics 8. Rankine Cycle 103 / 113
Boiler
G
Turbine
1
23
4
C
5
678910
m2
m3
Cycle Diagram
T
s
Boiler
1
2
3
45
67
8 9
10 m2
m3
Open Type Feedwater Heater - Deaerator [3/6]
Thermodynamics 8. Rankine Cycle 104 / 113
Deaeration is the removal of noncondensable gases, such as oxygen and carbon dioxide, from the water or
steam.
High oxygen content is in the water can cause erosion and corrosion of the components and piping that
contacting with water.
The oxygen contained in the water shows five to ten times higher erosiveness than carbon dioxide, and the
erosiveness becomes higher more than two times for every 17C increase of feedwater temperature.
Typically, an oxygen content of less than 20 parts per billion (ppb) in the feedwater is recommended.
Deaeration must be done continuously because small leakages of air at flanges and pump seals cannot be
avoided.
Deaeration takes place when water is sprayed and heated (boiled), thereby releasing the dissolved gases.
Generally, the condensate is sprayed into the top of the deaerator.
Heating steam, fed into the lower part, rises and heats the water droplets to the saturation/boiling
temperature, releasing incondensable gases that are carried to the top of the deaerator and evacuated.
The feedwater tank is filled with saturated and deaerated water, and steam buffer above it prevents any
reabsorption of air.
In general, one open type feedwater heater is used in one plant because the number of feedwater pump is
increased with the open type feedwater heaters.
Deaerator - Open Feedwater Heater
Open Type Feedwater Heater - Deaerator [4/6]
Thermodynamics 8. Rankine Cycle 105 / 113
Deaeration will also partly take place in the condenser.
The steam turbine exhaust steam condensates and collects in the condenser hotwell, while the
noncondensable gases are extracted by means of evauation pump.
A steam cushion separates the water in the hotwell so that reabsorption of the air cannot take place.
Often, levels of deaeration in the condenser can be achieved that are comparable to those in the deaerator.
Therefore, the deaerator and feedwater tank can sometimes be eliminated from the cycle, and the
condensate is fed directly from the condenser into the HRSH drum.
In these cases the makeup water must be admitted to the cycle through the condenser.
Deaerator - Open Feedwater Heater
Open Type Feedwater Heater - Deaerator [5/6]
Thermodynamics 8. Rankine Cycle 106 / 113
Extraction Steam
Subcooled
Liquid
Saturated
Liquid
Tsat (p)
TTD = 0
Tout = Tsatp, hb, mbp, hc, mc
p, ha, ma
Continuity Equation: mc = ma + mb
Energy Equation: mchc = maha + mbhb
The open feedwater heater is a
contact device where extraction
steam having higher temperature and
subcooled feedwater are mixed
directly.
By necessity, both must be at the
same pressure.
The open feedwater is a multiple
stream control volume with continuity
equation and steady-state, steady-
flow-energy equation.
No real law says that the output must be saturated liquid, but it generally is, and this is specified in the heater description as "TTD = 0".
TTD means the terminal temperature difference between the saturation temperature corresponding to
pressure (p) and the exit temperature of liquid.
Deaerator - Open Feedwater Heater
Open Type Feedwater Heater - Deaerator [6/6]
Thermodynamics 8. Rankine Cycle 107 / 113
[Exercise 8.9] Heat Balance
In a steam turbine plant steam enters the turbine at 2400 psig and 1000F and exhaust at 1.5 inHga to
the condenser. There are two extraction points at pressures 500 and 50 psia. The high-pressure steam
is used in the contact heater (open type heater), while the low-pressure steam is directed to the surface
heater (closed type heater). Other conditions are specified below:
Turbine efficiency = 0.90
Pump efficiency = 0.85
Terminal temperature difference = 10F
Drain cooler approach = 16F
Steam flow rate at the turbine inlet = 3,400,000 lbm/hr
Exercise 8.9
GHP
2 3 4
1
5
LP
6
7
8
910
LP
s
3
2
1
4s 4
2s
3s
h
Thermodynamics 8. Rankine Cycle 108 / 113
[Solution]
Determine the steam properties using given values
p1 = 2414.7 psia, T1 = 1000F
steam table h1 = 1460.39 Btu/lbm, s1 = 1.53225 Btu/lbm‧R
p4s = 0.74 psia, s4s = 1.53225 Btu/lbm‧R
steam table h4s = 841.58 Btu/lbm
steam table T4s = 91.87F (moisture = 24.96%)
Determine the steam properties at 4 using turbine efficiency
(It was assumed that the turbine expansion line is a straight line)
T = (h1h4)/(h1h4s) h4 = 903.46 Btu/lbm
steam table T4 = 91.87F (moisture = 19.02%)
Determine the steam properties at 2 & 3 using turbine efficiency
p2s = 500 psia, s2s = 1.53225 Btu/lbm‧R
steam table h2s = 1270.96 Btu/lbm, T2s = 556F (superheat = 89F)
including T h2 = 1289.90 Btu/lbm
steam table T2 = 585.2F (superheat = 118.2F)
p3s = 50 psia, s3s = 1.53225 Btu/lbm‧R
steam table h3s = 1080.50 Btu/lbm, T3s = 281F (moisture = 10.13%)
including T h3 = 1118.49 Btu/lbm
steam table T3 = 281F (moisture = 6.02%)
Exercise 8.9
Thermodynamics 8. Rankine Cycle 109 / 113
[Solution] - continued
The water leaving the condenser is usually at the saturated temperature corresponding to the condenser
pressure
p5 = 0.74 psia
steam table h5 = 59.88 Btu/lbm, 5 = 0.0161 ft3/lbm (moisture = 100%)
Determine the steam properties at 6 using the first law of thermodynamics
h6 = 59.88 Btu/lbm + 0.0161(ft3/lbm)(5000.74) lbf/in2 / 0.85 (144/778)
= 61.63 Btu/lbm (1 Btu = 778 lbf‧ft)
steam table T6 = 92.3F
Determine the steam properties at 7 using the definition of DCA
DCA = T7T6 = 16 T7 = 108.3F steam table h7 = 76.41 Btu/lbm
Determine the steam properties at 8 using the definition of TTD
TTD = TsatT8 = 10 T8 = 271F steam table h8 = 240.88 Btu/lbm
5656
2
5
2
656562
1wzzgVVhhq
565
6
556 ppdpw
P
pphwhh
56555656
Exercise 8.9
Thermodynamics 8. Rankine Cycle 110 / 113
[Solution] - continued
Determine the steam properties at 9
T9 = Tsat@p2 steam table T9 = 467F, h9 = 449.52 Btu/lbm, 9 = 0.01975 ft3/lbm
Determine the steam properties at 10
h10 = 449.52 Btu/lbm + 0.01975(ft3/lbm)(2414.7500) lbf/in2 / 0.85 (144/778)
= 457.75 Btu/lbm steam table T10 = 474F
In the second step, calculate the flow rate of extracted steam using mass and energy balance equation.
Starting with the contact heater
m2 + m8 = m9
m2h2 + m8h8 = m9h9
m2 = 702,902 lbm/hr
m8 = 2,697,098 lbm/hr
For the surface heater
m3(h3h7) = m8(h8h6) m3 = 463,933 lbm/hr
P
pphwhh
91099109910
Exercise 8.9
Thermodynamics 8. Rankine Cycle 111 / 113
Location Pressure (psia) Temperature (F)Enthalpy
(Btu/lbm)Flow Rate (lbm/hr)
1
2
3
4
5
6
7
8
9
10
2414.7
500
50
0.74
0.74
500
50
500
500
2414.7
1000
585.2
281
92
92
92.3
108.3
271
467
474
1460.39
1289.90
1118.49
903.46
59.88
61.63
76.41
240.88
449.52
457.75
3,400,000
702,902
463,933
2,233,165
2,697,098
2,697,098
463,933
2,697,098
3,400,000
3,400,000
Exercise 8.9
Thermodynamics 8. Rankine Cycle 112 / 113
[Solution] - continued
As the third step of the cycle heat balance, calculate turbine and pump work
WT = m1h1 m2h2 m3h3 m4h4
m4 = m1 m2 m3
WT = 1,522,173,038 Btu/hr = 422,826 Btu/s = 446,081 kW (1 Btu = 1.055 kJ)
Calculate the pump work
WP = 32,701,922 Btu/hr = 9,084 Btu/s = 9,584 kW (“” sign means ‘work in’)
Determine the heat supplied in the boiler
Qin = m1(h1 h10) = 3,408,976,000 Btu/hr
The cycle efficiency is
cycle = 43.7%
91010566 hhmhhmWP
in
PTcycle
Q
WW
Exercise 8.9