8. rankine cycle · 2018-01-13 · thermodynamics 8. rankine cycle 2 / 113 ... heat is added in the...

Thermodynamics 8. Rankine Cycle 1 / 113

8. Rankine Cycle

3

4

21


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


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


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


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.


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]


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]


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


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


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


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


Drum Type vs. Once-through

[ Subcritical ] [ Supercritical ]


Rankine Cycle Analysis2

Some Notes on Rankine Cycle4

Improvement of Rankine Cycle Efficiency5

Heating of Water at Constant Pressure1

Vapor Diagrams3



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.


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


[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


[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


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


[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

qq

q

qq

q

ww

q

w

1

m

m m

Exercise 8.1


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


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


[Exercise 8.2] Cycle analysis of a simple Rankine cycle


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.


T

s

1

2

3

4s

15 MPa

100 kPa

4

Exercise 8.2


[Solution]


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


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


[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

qq

q

qq

q

ww

q

w

1

Exercise 8.2


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


[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


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


p- and T-s diagram within liquid-vapor region

p- and T-s diagram when fluid goes to superheated region

Carnot Vapor Cycle


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.


[Exercise 8.3] Cycle analysis of a Carnot vapor cycle


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.


T

s

1

2

3

4

15 MPa

100 kPa

Exercise 8.3


[Solution]


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


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


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






Vapor Diagrams3



T-s Diagram


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


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]


h-s Diagram


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


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


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]


p-h Diagram - Supercritical


[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


[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






Vapor Diagrams3



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


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



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.



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



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



Water Droplet Erosion

Fog Formation

(Condensation Shock)

Dry Steam Wet Steam

Phase

Change

Supersaturation



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 ]



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







Vapor Diagrams3



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 ?


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).



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.


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)



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



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



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.



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.


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.


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.



7% cycle efficiency improvement by steam condition

from 160bar / 540C / 540C

to 290bar / 600C / 620C

Efficiency Improvement by USC



Drum-Type Boiler Once-through Boiler



T

s

Early 20th century

1940s

1960

Ultra Supercritical

Supercritical

Evolution of Rankine Cycle




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.


T

s

2B

4B

3B

1

2A

3A

4A

2 psia

200 psia

600 psia

Exercise 8.6


[Solution]


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


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


[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


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


[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


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.


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]



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]


T

s

2

3

1 4

2 psia

600 psia,

486.2F

1000F

Exercise 8.7


[Solution]


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


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


[Solution]

Determine the heat addition in the boiler,

qB = (h3h2) = 3529.43222.93 = 3306.5 kJ/kg


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






Vapor Diagrams3



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%.


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



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.



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



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.



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



Double Reheat



HPRH1RH2

Double Reheat



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



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



MSR (GE)

Steam Turbine Expansion Lines and MSR



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)



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]


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.



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.


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)


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.



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


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.


[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


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


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.



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 ]



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.



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



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.



Configuration

Open Type Feedwater Heater - Deaerator [1/6]


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.



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



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



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.




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.




[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


[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


[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



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


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



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


질의 및 응답

작성자: 이 병 은 (공학박사)작성일: 2015.02.11 (Ver.5)연락처: [email protected]

Mobile: 010-3122-2262저서: 실무 발전설비 열역학/증기터빈 열유체기술

8. rankine cycle · 2018-01-13 · thermodynamics 8. rankine cycle 2 / 113 ... heat is added in the...

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