3. flows in a steam path - engsoft lab in steam p… · steam turbine 3. flow in steam path 4 / 92...

Steam Turbine 3. Flow in Steam Path 1 / 92

HIoPE

3. Flows in a Steam Path


HIoPE

Thermodynamics and Fluid Dynamics for Steam Turbines 13 2

Impulse Turbine and Reaction Turbine 43 4

Steam Turbine Flow Model 2 1

Dimensionless Numbers 26 3

Stage Efficiency 78 5

Blade Profile Enhancement 85 6


HIoPE

Steam Turbine Flow Model

1

1

2

2

1

2

1

1

1

2

p

p

p

ppAcm d

1

1

2

2

1

21

1

2

p

p

p

pAacQ d

CV 1 2 3 4 5

SV = Stop valves

CV = Control valves

P = First stage shell pressure

L-0 L-1

Con

de

nser

Stage number

P

Steam

SV


HIoPE

The flow model is a series of orifices or nozzles in a pipe with a constant upstream pressure and

downstream pressure.

The pressure between any two of these orifices depends entirely on the quantity of steam and the nozzle

area following that point.

The first stage is represented as having four orifices to simulate a turbine with four control valves, while all

the other stages have only one orifice.

If all four control valves are opened, the flow will increase until an equilibrium condition is reached.

This steady state flow through each orifice in the model will be the same.

Therefore, the pressure upstream of each orifice will depend on the area of the downstream orifice.

As steam passes from throttle to the condenser, the areas of individual orifices are progressively larger.

Therefore, the pressure upstream of each stage is lower than the preceding stage.

A number of steam turbine stages can be classified into only three groups:

① The first stage or governing stage Variable flow area by control valve position

② The last stage (LSB) Flow choking is occurred

③ All the stages between Constant pressure ratio even during part load operation

In view of the specifics of the steam path design, all stages of a condensing steam turbine are divided into four groups,

① Governing stages

② Stages with high pressure and low volume discharge of steam (2D)

③ Intermediate stages with relatively low pressure and high volume discharge of steam (3D)

④ Last stages in the lowest-pressure range, characterized by a very high volume discharge.



HIoPE

1st s

tag

e s

he

ll

2nd s

tag

e b

ow

l

500

1000

1500

2000

Thro

ttle

pre

ssure

, psig

2500

2nd s

tag

e s

he

ll

3rd

sta

ge

bo

wl

3rd

sta

ge

sh

ell

4th s

tag

e b

ow

l

4th s

tag

e s

he

ll

5th s

tag

e b

ow

l

5th s

tag

e s

he

ll

6th s

tag

e b

ow

l

1800

1500

1250

1042

868

1200

Th

rott

le p

ress

2400

900

750

625 521

434

Factor

of 2 Factor

of 2

0



HIoPE

The governing stage has the unique characteristic of variable area.

As control valve opens, the area allowing throttle steam to flow through the turbine increase and produce

more power.

In addition, the change in flow changes the first stage discharge pressure, resulting in the pressure ratio

across the first stage.

Therefore, the first stage efficiency changes with control valve position.

At valves wide open the first stage is designed for the ideal ratio of wheel velocity and steam velocity.

As control valves close, the first stage efficiency decreases.

Partial

Arc Full

Arc

HP 1st Stage



HIoPE

0 0.2 0.4 0.6 0.8 1.0

10

20

30

40

50

60

70

80

90

100

0

IP Turbine

LP Turbine

HP 2nd Stage to Cold Reheat

HP 1st Stage

Throttle Flow Ratio

Tu

rbin

e S

ectio

n E

ffic

ien

cy [%

]

HP 1st Stage



HIoPE

For a throttle pressure of 2400 psig, the first

stage pressure with VWO would be 1800 psig

and the pressure downstream of the second

stage would be about 1500 psig. In this case,

the pressure ratio across the second stage is

1.2.

If throttle pressure decreases to 1200 psia, the

flow would decrease by a factor of 2 to 1. this

relationship holds true for all the stages until the

last stage. That is, the pressure ahead of the

last stage will decreases by a factor of 2 to 1.

All other stages, except HP first stage and LP

last stage, operate at a constant pressure ratio

and therefore a constant efficiency over the

load range.

Velocity triangles for various

steam turbine loads

Intermediate Stage



HIoPE

The last low pressure turbine stage is the only

other stage that experiences significant changes

in pressure ratio as a result of normal power

plant operation.

The pressure of upstream of the last stage

changes with control valve position and the

downstream pressure changes with condenser

pressure.

Therefore, the last stage efficiency changes with

both control valve position and with condenser

pressure.

The change in pressure ratio across the first and

last stage dictates the change in efficiency of the

HP and LP turbines, respectively.

LP Last Stage



HIoPE

Stage Extract Bowl P Shell P Press Remarks

(psia) (psia) ratio

Throttle 3514.7

1 3409.3 2627.4 1.30

2 2122.2 1.24

3 1704.1 1.25

4 1355.9 1.26

5 1 1067.7 1.27

6 830.0 1.29

7 2 639.0 1.30 HP turbine outlet

IV 581.6

8 570.0 438.1 1.30

9 336.0 1.30

10 3 254.0 1.32

11 184.0 1.38

12 4 135.7 1.36 IP turbine outlet

13 5 131.7 74.8 1.76 LP turbine inlet

14 40.5 1.85

15 6 20.5 1.96 Exceeding a critical PR



18 0.74 5.50 Connected to condenser

Stage Pressure Ratio [Sample]


HIoPE

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19Stage number

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

5.5

6.0

6.5

7.0P

ressu

rera

tio

HP IP LP

Exceeding critical pressure ratio

Extraction stage : 5, 7, 10, 12, 13, 15, 16, 17

Stage Pressure Ratio [Sample]


HIoPE


Discussion

Several LP turbine stages, 3rd to LSB, have higher than critical pressure ratio. Discuss the

following issues.

1) Possibility of choked flow

2) Flow area

3) Steam extraction

4) LSB in terms of degree of reaction

Critical pressure ratio

- superheated steam = 0.546 (=1.3)

- saturated steam = 0.577 (=1.135)

- air = 0.528 (=1.4)

(see K.C. Cotton, pp. 15)


HIoPE

Thermodynamics and Fluid Dynamics for Steam Turbines 2

Impulse Turbine and Reaction Turbine 4

Steam Turbine Flow Model 1

Dimensionless Numbers 3

Stage Efficiency 5

Blade Profile Enhancement 6


HIoPE

Flow Behavior in a Turbine Stage

Pressure E Kinetic E

Thermal E Thermal E Mechanical E

Nozzle row Bucket row

z

r

The flow behavior is investigated in a tangential plane.

Therefore, the flow velocity has two components, one is axial

component denoted by subscript z, and the other is

tangential component denoted by subscript implying a whirl

velocity.


HIoPE

Velocity Triangle in a Turbine

Velocity Triangle at Root

Root

Tip

rRoot rTip


The absolute velocity increases from c1 to c2 across the nozzle.

The absolute velocity decreases from c2 to c3 across the bucket. This is because the kinetic energy entering

bucket is extracted by the bucket. Thus, bucket attains rotating power.

In the case of turbine, the convention chosen is that the angles are positive when measured in the direction

of rotation. Therefore, 2, 2 are positive; 3, 3 are negative.

c : absolute velocity of fluid

u : tangential velocity of blade

w : velocity of fluid relative to blade

u

u

c2

w2

2

2

w3

3 c3

p1

p2

p3

u

1

c1

3

Nozzle row

Bucket row


HIoPE

Velocity Triangle at Tip

Velocity Triangle in a Turbine

Root

Tip

rRoot rTip


u

u

c2

w2

w3

p1

p2

p3

u

1

c1

Nozzle Row

Bucket Row

2

2

3 c3

3


HIoPE

Expansion Lines

A nozzle is used to provide partial

expansion of the gas as well as to

guide the flow smoothly into a

bucket.

If the flow is isentropic in the

nozzle, condition 2s is achieved

after passing through the nozzle.

Practically, however, nozzle

expansion occurs along curve 1-2

because of losses occurred in the

nozzle path.

Change in stagnation pressure

(po,1po,2) is due to the losses,

because there is no work

extraction from the fluid inside the

nozzles.

The process along 2-3 represents

the expansion through bucket.

If the flow is isentropic only in the

bucket, condition o,3s or 3s is

achieved.

p3

po3

1

o,3ss

1/2c12

p2

o,1

p1

po,2 po,1

1/2c12

2

1/2c22

h

s

1/2c32

3ss

2s

o,3s

3s

o,3

3

o,2

u

u

c2

w2

2

2

w3

3

c3

p1

p2

p3

u

1

c1

3

Nozzle row

Bucket row

[ Expansion line in a turbine stage ]


HIoPE

Turbine Efficiencies

Total-to-static Efficiency Total-to-total Efficiency

• In many turbines (especially steam turbines), the

kinetic energy at the exit (c32/2) should be as small

as possible, because this represents aerodynamic

loss.

• Therefore, the design philosophy is to achieve as

low a velocity at the exit as possible. For this reason,

active length of LSB of steam turbines is very long.

• In such situations, a total-to-static efficiency is used.

• In most aeronautical applications gas turbines, the

exhaust energy is utilized for thrust generation.

Therefore, the exhaust energy is used to produce

useful power.

• Therefore, a more appropriate definition to

represent the performance of these turbines is the

total-to-total efficiency .

• The total-to-total efficiency is also defined as

isentropic efficiency.

• For a multistage turbine, total-to-total efficiency

should be used because the kinetic energy at the

exit of a stage (except the last stage) is not lost.

sso

oo

ssssossoo

oo

tshh

hh

hhhh

hh

31,

3,1,

33,3,1,

3,1,

ssoo

oo

tthh

hh

3,1,

3,1,


HIoPE

Notation for Flow in a Bucket Row

u

w3

3

c3

c2

w2

2 2

3

N B

w,3

c,2 c,3

w,2

dc = (c,2 c,3)

u

c: absolute velocities (velocities in the reference

frame of the nozzles)

w: relative velocities (velocities relative to moving

surface, the buckets)

u: tangential velocities of blades (in the positive

direction)

2, 2 are positive, and 3, 3 are negative.

r2 r3

w2 or c2 w3 or c3

Bucket

row

u

u

c2

w2

2

2

w3

3 c3

p1

p2

p3

u

1

c1

3

Nozzle row

Bucket row


HIoPE

Euler Equation [1/6]

The change of momentum between the flow entering and leaving the bucket can be used to calculate the

force acting on the bucket.

There are three principal components of this force, axial, radial, and tangential.

The axial and radial components are important for the design of bearings and for the analysis of vibration

excitations, etc.

But, these two components cannot contribute to the work transfer between the working fluid and the bucket.

Only the tangential component of the force can produce a change in enthalpy through a work transfer.

Tangential force on rotor from entering fluid =

Work bucket = force length =

Power on bucket per unit time = work on rotor / time =

Net power on bucket,

Therefore, Euler’s equation can be derived.

(e.1)

Turbine has a positive work out, however, a pump, fan, and compressor will have negative work out.

2,cm

22, rcm

22, rcm

3,32,233,22,23 cucumrcrcmW

3,32,22323 / cucumWw

22, cc

Euler equation


HIoPE


For an adiabatic bucket row in the absence of external torques, or large changes in elevation, the first law of

thermodynamics gives,

(e.3)

The first law of thermodynamics is,

(e.2)

3,2,23 oob hhww

2323

2

2

2

3223323232

1wzzgccppuuq

2323

2

2

2

323232

1wzzgcchhq

23232,3,23 wzzghhq oo

232,3,23 whhq oo

Therefore, following relationship can be obtained from Euler equation,

or (e.4)

It is clear that the stagnation enthalpy and pressure drop in a turbine are directly proportional to the change

in tangential velocity and blade speed. This is the most useful single relation in compressor/turbine design.

In the preliminary design of axial flow machines, the change of radius of the mean flow can often be ignored,

so that a more restricted version of Euler’s equation becomes

(e.5)

θucddho

θdcudho

c2

c3 q

w z2

z3

2

3

3,32,23,2,23 cucuhhww oob


HIoPE


Pressure and Temperature Drop in a Bucket Row

2,23,32,3,2,3, cucuTTchh oopoo

11

1

2,

3,

2,

2,

3,

2,2,3,

o

o

op

o

o

opoop

pTc

T

TTchh

32222,3,2,23,3 tantancos ucccucucu

1

32

2,

22

2,

3,tantan

cos1

opo

o

Tc

uc

p

p

For simple diagram having constant u from stage inlet to outlet,

32

2,

22

2,

3,tantan

cos1

opo

o

Tc

uc

T

T

2,23,32,3,23 cucuhhww oob

1

322

2,

3,tantan

11

a

uv

p

pz

o

o

322

2,

3,tantan

11

a

uv

T

Tz

o

o


HIoPE

1

322

2,

3,tantan

11

a

uv

p

pz

o

o

322

2,

3,tantan

11

a

uv

T

Tz

o

o

The pressure drop and temperature drop in a turbine are strongly dependent on the blade speed, axial

velocity or mass flow, inlet and exit flow angles, and absolute (23) or relative flow turning angles (23).

Higher turning angles produce larger pressure and temperature drops, and thus a higher work output.

Unlike compressors, large flow turning can be accomplished without flow separation.

The effect of the mass flow (or flow coefficient) is opposite to that of a compressor. A turbine with 2 and 3

fixed and blade speed held constant, higher mass flow produces larger pressure and temperature drops.

If 2, 3, and mass flow held constant, higher blade speeds produce larger pressure or temperature drops

and higher work output per stage.

Therefore, higher speeds result in more compact power plants. (this is why aerospace gas turbines operate

at highest possible speed allowed by stress limits)

Pressure and Temperature Drop in a Bucket Row



HIoPE

[Exercise 3.1] Use of the Euler’s equation

What is the power output (kW) of the first stage of an axial flow steam turbine which takes 600 kg/s of

steam at 600C and 250 bar stagnation conditions? After passing through the nozzle, the flow leaves

nozzle at a direction 70 degrees from that of axial, at a velocity of 500 m/s, as given in figure 1, and

discharges it from the bucket (rotor) without swirl (c,3 = 0). The pitch diameter of the bucket is 1 m, and

the shaft speed is 3600 rpm. The turbine has an isentropic stagnation-to-stagnation stage efficiency of

90 percent.

Figure 1


2=70


HIoPE


[Solution]

Power output of the stage can be obtained

The first law of thermodynamics is as follows,

The turbine can be treated as adiabatic. Therefore,

From the Euler equation,

From the given conditions,

Therefore,

smrn

u /5.18860

22

smcc /470sin 222,

23,2, /566,88 smhh oo

kWs

m

s

kghhmhmW oo 140,5388566600

2

2

3,2,23

MWWW turbinenet 8.4723,23

232,3,23 whhq oo

2,23,32,23,2, cucucuhh oo

2,3,23 oo hhw 3,2,23 oo hhmW

3,2,23 oo hhmhmW


HIoPE





Stage Efficiency 5



HIoPE

By means of dimensional analysis, a group of variables representing some physical state is reduced into a

small number of dimensionless groups.

This enables a unique representation of certain classes of machines based on pressure rise (or drop) and

mass flow. Most importantly, it enables reduction of laboratory testing effort by reducing the number of

variables.

Specifically, the following can be accomplished:

1) Prediction of a prototype performance from tests conducted on a scaled model (similitude).

2) Unique representation of the performance (e.g., Mach number, Reynolds number effect).

3) Determination of a best machine on the basis of efficiency for specific head, speed, and flow rate.

Most important dimensionless numbers in axial turbines are degree of reaction, loading coefficient, flow

coefficient, etc.

Generals


HIoPE

The most important performance variable is the work done on the fluid, or delivered by the machines. Its

dimensionless form is the loading coefficient, which is also called as work coefficient.

That is, the loading coefficient reflects the pressure/temperature drop across a turbine.

For an adiabatic stage, the loading coefficient is defined by the ratio of specific stage work input to the

square of mean bucket speed, that is,

where wb is the isentropic work done at bucket row.

For simple diagram having constant u from stage inlet to outlet,

The loading coefficient is positive for turbines, and negative for compressors and pumps.

In turbines having the value of 1.5 are called as “highly-loaded” or “high-work” turbines (or turbine

sections). Values of 1.0 mean “low-work” or “lightly-loaded” turbine stages.

Normally, last stage blades of steam turbines have very high loadings at the hub, and light loadings at the

tip.

The value of the loading coefficient for an impulse turbine with a maximum loading coefficient is two when

the exit swirl is zero. In a 50% reaction turbine with a maximum loading coefficient is one.

Loading Coefficient [1/2]

2

3,32,2

2

3,2,

2 u

cucu

u

hh

u

w oos

32

3,2,tantan

u

v

u

ccz


HIoPE

Loading Coefficient [2/2]

(a) high-work turbine

( = 2.0, = 0.5, = 0.5) ( = 1.0, = 0.5, = 0.5) ( = 0.5, = 0.5, = 0.5)

(b) medium-work turbine (c) low-work turbine

( = work coefficient, = flow coefficient, = degree of reaction)


HIoPE

The flow coefficient reflects the effect of the mass flow as well as blade speed.

The flow coefficient is defined the ratio of the axial velocity entering to the mean bucket speed, that is,

In a simple velocity diagram, the flow coefficient is constant.

The flow coefficient can be different at rotor inlet and at rotor outlet where both cz and u vary through the

stage.

It also varies with radius.

The relationship between loading coefficient and flow coefficient is

u

vz

Flow Coefficient

32 tantan


HIoPE

Smith Chart [1/2]

A useful investigation of turbine performance

characteristics was compiled by Smith with more

than 100 sets of data from 33 turbines.

Smith found that the efficiency of a turbine

depends strongly on the loading coefficient and the

flow coefficient.

The loading coefficient influences the pressure

gradient in the passage, and this increases the

losses.

The flow coefficient is a direct measure of the

mass flow, for a given speed and machine size.

Higher flow coefficient, and hence higher mass

flow, results in a higher pressure drop, and the

corresponding losses also increase.

Therefore, the highest efficiencies occur at low

loading and low flow coefficient.

For power generation gas turbines, it is preferable

to operate at lower loading and low flow coefficient

to achieve higher efficiency.


HIoPE

Smith Chart [2/2]

Flow coefficient (mean radius)

0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

2.2

2.4

2.6

2.8

3.0 Loadin

g c

oeffic

ient

(m

ea

n r

ad

ius)

0.94

0.93 0.92 0.91

0.90

0.89

T = 0.88

0.87

0.86

[ Smith Chart ]

The turbine must provide the required shaft work (ho) and run at the same speed as the compressor. Hence,

the loading coefficient is essentially fixed. The Smith chart is then used to guide the choice of flow coefficient in

the preliminary turbine design.


HIoPE

Degree of Reaction [1/10]

The degree of reaction in the turbine is defined as,

In the nozzle path, the first law of thermodynamics is,

(1)

In the nozzle, adiabatic process occurs, and no work produces.

Therefore, equation (1) becomes,

(2)

From the first law of thermodynamics,

(3)

The following relationships are valid is a turbine stage.

; adiabatic process

; for a normal stage

; no work at nozzle row

Thus,

(4)

2

1

2

221 5.0 cchh

12

2

1

2

21212 5.0 wcchhq

31

21

31

32 1hh

hh

hh

hh

static enthalpy drop in the bucket

static enthalpy drop in the stage

= 100 (%)

13

2

1

2

31313 5.0 wcchhq

013 q

31 cc

23133,1,

2

33,

2

11,312

1

2

1wwhhchchhh oooo

23231213 wwww

p3

po3

1

o,3ss

1/2c12

p2

o,1

p1

po,2 po,1

1/2c12

2

1/2c22

h

s

1/2c32

3ss

2s

o,3s

3s

o,3

3

o,2

[ Expansion line in a turbine stage ]


HIoPE


Using Euler equation,

(5)

Input the equations (2) and (5) into (1) gives,

From the velocity triangle,

For a simple velocity triangle,

Therefore, the degree of reaction becomes,

For a simple velocity diagram having constant u from stage inlet

to outlet.,

(6)

3,32,22331 cucuwhh

3,32,2

2

3

2

2

21

cucu

cc

2

2,

2

2,

2

2 cvc z

2

3,

2

3,

2

3 cvc z

3,32,2

2

3,

2

2,

21

cucu

cc

3,2, zz vv

32 uuu 32

3,2,tantan1

21

u

v

u

ccz

[ Velocity triangle in a turbine stage ]

u

u

c2

w2

2

2

w3

3 c3

p1

p2

p3

u

1

c1

3

Nozzle row

Bucket row


HIoPE


Equation (6) becomes,

(7)

From Euler equation,

(8)

Divide equation (8) by u2 gives,

( ) (9)

From equation (7) and (9), an important result is obtained.

(10)

From equation (9) and (10), the unknown angles of the absolute velocity can be obtained.

(11)

Turbomachinery design initiated by experienced designers through the choice of the flow and loading

coefficients and the degree of reaction and then determine the flow angles using eq. (11). These are true

only for a normal stage. If the axial velocity does not remain constant, the proper equations need to be

redeveloped from the fundamental concepts.

32

32 tantan2

12

tantan1

u

vz

323,2, tantan zb ucccuw

u

cz2u

ws 32 tantan

13 tan12tan12

2/1tan 3

2/1tan 2

1tan2 2


HIoPE

Similar expressions can be developed for the flow angles

of the relative velocity.

The Euler equation can be written as

Divide equation (8) by u2 gives,

Since the stagnation enthalpy of the relative motion is

constant across the bucket. Thus,

Therefore, the unknown angles of the relative velocity can

be obtained.

(12)

323,2, tantan zs uvwwuw

2

2

3

222

2

2

332 tantan2

1

2

1 zvwwhh

23 tantan2

2/tan 3

31

32

hh

hh

32 tantan

2/tan 2

u

w3

3

c3

c2

w2

2 2

3

N B

w,3

c,2 c,3

w,2

dc = (c,2 c,3)

u



HIoPE


It shows that the loading increases as the reaction decreases.

A small reaction means that the pressure drop across the bucket

is small, but the large loading is the result of a large deflection.

In the nozzle, the flow leaves at high speed at large angle 2.

The high kinetic energy obtained this way becomes available for

doing work on the buckets.

The flow is then deflected back toward the axis and beyond to a

negative value of 3, so that the last term in equation (10) is

positive.

Thus, for a fixed reaction, an increase in the absolute value of 3,

obtained by increasing it in the opposite direction of u, leads to a

large deflection and a large value of loading coefficient.

Thus, a fairly low value of reaction and high turning gives heavily

loaded blades and a compact design.

(10)

u

u

c2

w2

2

2

w3

3 c3

p1

p2

p3

u

1

c1

3

Nozzle row

Bucket row

13 tan12tan12

1tan2 2


HIoPE

(a) 0% reaction velocity diagram

u

c2 w3

w2

c3

c,2 = 2u

w3

w2

c2

u

c3

c,2 = u

c2

w3

u

w2 c3

c,2 c,3

(b) 50% reaction velocity diagram

(c) 100% reaction velocity diagram



HIoPE

0% Reaction 50% Reaction

• A zero reaction turbine is called an impulse turbine

because there is no expansion or acceleration of

the flow through the rotor blades, and the rotor

torque comes wholly from the impulse of the nozzle

stream.

• With no pressure drop across the bucket row,

pressure seals are not required.

• A frequently used impulse diagram has axial stage

entry and exit flows and the reasonably high loading

coefficient of 2.0.

• In 50% reaction velocity diagrams, the bisector of

the line joining the apexes of the absolute and

relative velocity triangles crosses u in the middle,

which is why the diagrams become symmetric.

• Such diagrams are frequently favored for turbines

because they have accelerating flow to an equal

extent in nozzle and bucket passages, which leads

to lower losses.

• The rectangular turbine stage diagram shown in

above has the additional advantage of having axial

flow at stage inlet and outlet. Also tests show this

gives the highest efficiency for turbine stages.

u

c2 w3

w2

c3

c,2 = 2u

w3

w2

c2

u

c3

c,2 = u



HIoPE

0% Reaction Stage


u

u

c2

w2

2

2

w3 3

c3

p1

p2

p3

u

c1

Nozzle row

Bucket row

All of the static enthalpy drops across the nozzle in a 0%

reaction stage. For such a stage, from eq. (12),

or

It can be assumed that the axial velocity is constant at the

inlet and exit of the stage. In this case,

w2 = w3

If the flow angles are equal to the blade angles, then the

bucket has a symmetric shape.

The blades having low reaction are heavily loaded.

For a normal stage with axial entry and with degree of

reaction of 0, the relation reduces to

For an impulse stage, the flow angles for absolute and

relative velocity are reduced to

32 tantan 32

3tan12

2

u

w3 3 c3

c2

w2

2 2

N B

w,3

c,2

w,2

dc = (c,2 c,3)

u

2/1tan 3

2/1tan 2

2tan 3

2tan 2


HIoPE


50% Reaction Stage [1/2]

3

3

u

w3 c2

c3

2

w2

2

N

w,3

c,3 c,2

w,2

dc = (c,2 c,3)

B

A 50% reaction stage has equal static enthalpy drops across the

nozzle and bucket. For such a stage

In order to get a high efficiency, the flow angle at the inlet is kept

only slightly negative, but if some of the efficiency is sacrificed to

achieve higher performance, the inlet flow angle may reach 1 =

45.

For such a stage, a flow coefficient may have a value of = 0.75,

which gives = 2.5.

For a 50% reaction stage, the flow angles for absolute and relative

velocity are reduced to

From these it can be seen that

Therefore, the velocity triangles formed at the inlet and exit of the

bucket are symmetrical each other. Thus,

1tan2tan21 23

2

1tan 3

2

1tan 2

2

1tan 3

2

1tan 2

32 tantan 23 tantan

32 wc 32 cw [ Velocity Triangle ]

w2

c2

u

c1 1

2

2

Nozzle row

u

Bucket row

c3

w3

u 3

3

p1

p2

p3


HIoPE

1tan2tan21 23

50% Reaction Stage [2/2]


The stage loading coefficient for 50% reaction stage is,

The figure gives the design and off-design performance of

50% reaction stage, based on above equation.

It is clear from the figure that increases linearly with for a

given 2.

The loading coefficient increases with 2 for a given flow

coefficient.

The present trend in the design of the nozzle is to use as high

an 2 as possible. But it should be realized that increasing 2

increases w2 for a given blade speed, and thus the flow is

likely to reach supersonic speeds and limit the mass flow.

Therefore, the designer has to vary 2, u, , and vz (or ) to

get an optimum design for a given turbine inlet temperature.

The curves in this figure are for ideal conditions. That is,

viscous losses, shock losses, or three-dimensional effects are

not included.

0.0 0.2 0.4 0.6 0.8 1.0

2.5

2.0

1.5

1.0

0.5

0.0

-0.5

-1.0

Flow coefficient

Sta

ge

loa

din

g c

oe

ffic

ien

t

2 3=5 10

20

40

60

80

100 120

140

[ 50% reaction stage ]


HIoPE





Stage Efficiency 5



HIoPE

Degree of Reaction

[ Impulse Turbine, = 0% ]

[ Reaction Turbine , = 50% ]

%10031

32

hh

hh

%10031

32

TT

TT

%10031

32

pp

pp

[ LSB (GE) ]

[ LSB (Siemens) ]

dpdhq

Thermodynamic process occurred in

compressor and turbine is adiabatic process.

And ignoring density changes.

dpdh

static enthalpy drop in the bucket

static enthalpy drop in the stage

= 100 (%)


HIoPE

R

Reaction Action

F

V

A

, Nozzle

F = mV = V2A

m = VA (mass flow rate)

Degree of Reaction

Fluid Dynamic Force


HIoPE

Force Acting on a Blade

Vj U

Vj U

Bucket

F

U

(F = Force, P = Power)

F = 2m(Vj U)

F = 2A(Vj U)2

P = F x U

P = 2AU(Vj U)2

Vi

Convergent

nozzle

Ve

U

F

F = mVe mVi = m(Ve Vi )

P = m(Ve Vi )U

Impulse Turbine Reaction Turbine


HIoPE

Principle of a Reaction Blade

Hero’s Aeolipile (BC 150년 경)


HIoPE

Evolution of Turbine Blade

Siemens GEC AEI Rateau SCAM BBC Sulzer AEG

GE USA IMPULSE

Siemens-KWU D REACTION

W/H USA REACTION

BBC CH REACTION

Alsthom F IMPULSE

GEC UK IMPULSE 1970

2000 GE USA IMPULSE

Siemens-Westinghouse D REACTION

ABB-Alsthom F REACTION

MHI J REACTION

Ansaldo

Toshiba

Doosan

Hitachi

1998

N. Piignone

BHEL

Parsons

Fuji

MHI

ABB GEC-Alsthom

1989 1987

CEM

LMZ

Zamech

ASEA STAL

F. Tosi

De Pretto

1999


HIoPE

Impulse vs. Reaction

The impulse turbine has its entire enthalpy drop in

the nozzle. Therefore, it has a very high velocity

entering the bucket.

Ideally there is no change in the magnitude of the

relative velocity w between inlet and exit (which

are denoted by subscripts 2 and 3, respectively).

The large inlet velocity c2 has been reduced to a

small absolute exit velocity c3, which ideally is in

the axial direction.

u : tangential velocity of blade

w : velocity of fluid relative to blade

c : absolute velocity of fluid

The reaction turbine divides its enthalpy drop in both

nozzle and bucket.

Therefore, velocities are accelerated when the steam is

passing through both the nozzle and the bucket.


u

u

c2

w2

2

2

w3

3 c3

p1

p2

p3

u

1

c1

3

Nozzle row

Bucket row

w2

c2

u

c1 1

2

2

Nozzle row

u

Bucket row

c3

w3

u 3

3

p1

p2

p3


HIoPE

Velocity Diagram


c1

u

2

c2

p1

p2

u

w2

w2

Nozzle Bucket

A1

A2N 2

A2B A3B

w3

u

c3

3 = 0

A2N A1

A2B = A3B

|2| = |3|

|w2| = |w3|

c2 4c1

p3

T1

T2 T3

c2

c1

w2

w3

3

c1

U

2

c2

p1

p2

u

w2 w2

Nozzle Bucket

A1

A2N 3

w3

c3

3 = 0

p3 T1

T2

T3

c2

A2B

A3B

u

w3

2 = 0

A2N A1

A3B A2B

|c2| |c1|

|w3| |w2|

c2 2c1

c1

w2


HIoPE

Multistage Impulse Turbine

Pressure

Absolute

velocity

Distance through turbine

Nozzle Bucket Nozzle Bucket Nozzle Bucket

The impulse turbine has its

entire enthalpy drop in the

nozzle, therefore, it has a very

high velocity entering the

bucket.


HIoPE

Multistage Reaction Turbine

Pressure

Absolute

velocity

Distance through turbine

Nozzle Bucket Nozzle Bucket Nozzle Bucket

half of impulse turbine


HIoPE

Question

1. Compare the impulse and reaction turbine in terms of SPE.

2. Compare the impulse and reaction turbine in terms of profile loss.

3. Suggest the equation to calculate the thrust produced in a stage.

4. Which type of turbine requires bigger thrust bearings?

5. Single stage supersonic impulse turbine is shown in the figure.

1) Discuss the shape of nozzle path.

2) What is the purpose of the increasing nozzle exiting velocity up to supersonic velocity?

Single stage supersonic impulse turbine

31 ppdlT

T = Thrust

d = mean diameter of blade row

l = active length of blade

p1 = pressure at the inlet of stage

p3 = pressure at the exit of stage

= degree of reaction at the mean dia.

drpprTtip

root

r

r322


HIoPE

Comparison of Velocity Triangle

F = mV = m(c2sin2 + c3sin3)

= mc2sin2

P = mc2sin2u

F = mV = m(c2sin2 + c3sin3)

= mc2sin2

P = mc2sin2u


2

2

3

3

w3

w2

c2

u

u

c3

w2sin2

w3sin3

u u

2u = c2sin2

2

3

w3

w2

c2

u

u

c3

u = c2sin2

w3sin3

u u


HIoPE

Comparison of the Number of Turbine Stages

Impulse Reaction

Let us consider a nozzle row only. This is

because there is no heat addition

and work out

Neglecting inlet velocity

Assume

• h across fixed blades in reaction turbine is only 1/4 that of impulse turbine.

• Reaction turbines, however, have an additional equivalent h across the moving blades.

• Therefore, total h in reaction turbine is a half of impulse turbine.

• This means that reaction turbine needs twice number of stages to generate same output.

outin wPEKEFEuq outin wchchq 2

22

2

112

1

2

1

2

1

2

2212

1

2

1cchh

2

221 ch 2

221 ch

o902

uc 22 uc 2

22uh 25.0 uh


HIoPE

Smaller Rotor and Wheel Diameter Single Flow Control Stage

Design Change of HP Turbine - GE


HIoPE

The length of the rotor shaft is almost same. The reasons for this are

1) Pressure drop across the nozzle path in the impulse turbine is double of the reaction turbine. Therefore,

the diaphragms of the impulse turbine should be stronger than those of the reaction turbine.

2) Turbine buckets of the impulse turbine have to work twice of the reaction turbine. Thus, turbine buckets

should be much stronger than those of the reaction turbine.

Another important thing is that the active length of the reaction bucket is longer than that of impulse bucket.

The longer active length of buckets, the higher the turbine efficiency.

Design Change of HP Turbine - GE


HIoPE

Newly Designed HP Turbine [GE]

ALSTOM HP Turbine


HIoPE

Steam Turbines for CCPP Application GE

D-11 for combined cycle units

207D-17 Steam Turbine


HIoPE

HP Turbine Unit of the 200-MW Steam Turbine

for the K-Power. (Hitachi)

Number of Turbine Stages

Steam Turbine developed by Doosan

The number of stages of reaction turbine is twice of impulse turbine.

1) Suggest the method that make the number of stages of reaction turbine is equal to that of

impulse turbine.

2) The answer of the above question is not applied in the practical reaction turbine. Discuss the

reason for that.


HIoPE

Velocity Diagram

Reaction Stage

u = 0.65v0

c3 = 0.24v0

w2 = 0.24v0

u = 0.65v0

p1

p2

p3

Steam Flow

20

Impulse Stage

p1

u = 0.5v0

p2

u = 0.5V0

c3 = 0.23v0 p3

Steam Flow

13


HIoPE

The velocity increase about four times in

the nozzle and its direction changes from

axial to about 77 off axial.

Vo is the velocity of stream expanded from

the pressure upstream of a stage to the

downstream pressure without loss.

In a pure impulse stage the optimum u/c2 is

0.5.

The velocity leaving the stage should be

close to axial to minimize the stage leaving

loss.

Impulse Stage

In a 50% reaction stage, half of the

pressure drop is across the nozzle and

another half is across the bucket.

The flow turning in the nozzle of reaction

stage is smaller than that of impulse stage.

The inlet velocity of bucket in the reaction

stage is smaller than that in the impulse

stage.

The lower velocity results in lower friction

loss and lower erosion.

The shape of nozzle and bucket is same in

a 50% reaction stage.

Reaction Stage

Velocity Diagram


HIoPE

Comparison of Leakage

Impulse Reaction

Bucket

Tip

Diaphragm

Root

cylindrical

drum type

rotor

disc wheels

shrunk on to a

rotor shaft


HIoPE

Turbines have internal sealing systems between the rotating buckets and the stationary casing and between

the stationary nozzles and the rotor.

The rotating bucket to stationary casing seal is more critical in a reaction turbine than in an impulse turbine

since the reaction turbine has higher pressure drop across the buckets.

The stationary nozzle to rotor seal is more critical in the impulse turbine because of the higher pressure drop

across the impulse stationary nozzle.

However, the impulse turbine has a smaller rotor and thus a smaller sealing diameter, offsetting the effects of

the higher pressure drop.

In addition, the wheel design of the impulse turbine rotor allows the installation of rotor seal leakage steam

passages (holes) in the wheel, minimizing leakage steam interference with the main steam flow from the

stationary nozzle to the rotating bucket.

The rotor-mounted bucket design of the reaction turbine does not allow the installation of these leakage

passages.

However, because of the higher reaction at the root of the bucket, the reaction stage is less affected by re-

entering leakage from the stationary low.

The reaction stage has a higher profile (aerodynamic) efficiency than an impulse stage.

The impulse stage has higher efficiency on stages with small blade heights because the difference in

leakage losses offsets the higher profile of the reaction stage.

As the blade height increases, the influence of leakage losses decreases and a point is reached where the

reaction stage is more efficient.

Comparison of Leakage


HIoPE

True impulse stages having 0% reaction and reaction stages that always have 50% reaction do not exist in

practical turbine design. In practice, the reaction varies from hub to tip.

Normally, reaction turbine stages are designed to operate at 50% reaction at midspan, with outer and inner

radii operating at higher and lower reactions, respectively.

Impulse stages typically have 3% to 5% reaction at the root of bucket in order to avoid zero or negative

reaction that results in efficiency loss and may lead to flow separation in the bucket.

For long reaction stage buckets, the degree of reaction at the mean diameter may be as low as 40%.

Thus, impulse and reaction stages in the classical definition do not exist in practical turbines.

Characteristics of flow behaviors in multistage axial turbine stage:

1) Pressure and velocity distributions along radial direction are uniform at the entry of a stage.

2) This is same as at the exit of a stage.

3) Centrifugal forces are caused by the tangential component of flow in the nozzle discharge.

4) This is same as in the reaction turbine.

5) The variation of reaction in radial direction is needed to partially cancel the centrifugal forces in the

stage.

6) Otherwise, the flow would migrate to the tip, resulting in a poor stage efficiency due to as followings.

• Increase of bucket tip leakage loss

• Increase of secondary flow loss near bucket tips

• Bucket vibration characteristics becomes worse because of non-uniform load acting on the bucket

along radial direction

Impulse-Reaction Turbine [1/6]


HIoPE

1000

psia

1000

psia

1000

psia

859.4 psia

844.1 psia

828.7 psia

819.7

psia

819.7

psia

819.7

psia

Free vortex design impulse type

HP turbine stage

22% @ tip

13.5%

@ pitch

5% @ root

[ Example of pressure variation in radial direction ]

버켓 팁으로의 유동편중 해결 방법



HIoPE

• Velocity and pressure distribution along radial direction are uniform at the inlet and outlet of the stage.

• Velocity decreases along radial direction between nozzle and bucket.

• Pressure increases along radial direction between nozzle and bucket.

Radial Variation of Pressure and Absolute Velocity


c1 c3

p1 p2 p3

c2

w2R

w2M

w2T

uR

uM

uT

c2R

c2M

c2T


HIoPE

노즐의 역할은 작동유체의 압력에너지를 운동에너지로 변환시키는 것이다. 따라서 노즐을 통과한 작동유체는 속도가 크게 증가한다. 그러나 연속방정식에 의해 노즐 출구에서의 축방향 속도는 노즐 입구에서와 동일하기 때문에 접선방향 속도만 크게 증가한다. 이로 인해 노즐 출구를 빠져 나온 작동유체는 큰 선회유동으로 인해 원심력이 발생하여 유체는 버켓 팁(tip) 쪽으로 집중되는 경향을 가진다.

유동이 버켓 팁 쪽으로 편중되면 버켓과 케이싱 사이에서 누설손실이 증가하며, 버켓 팁 근처에서 이차유동손실이 증가할 뿐만 아니라 반경방향을 따라서 버켓에서 생산하는 동력도 균일하지 못하게 된다.

이런 문제를 해결하기 위해서 버켓 팁 입구 쪽의 압력을 루트(root, or hub) 입구 쪽 압력보다 높게 유지시킨다. 이 경우 버켓 입구의 팁부분 압력이 루트부에 비해 높기 때문에 팁 쪽에서 루트 방향으로 진행하는 유동이 형성된다.

그런데 버켓 팁 쪽에 형성되는 높은 압력으로 인해 루트 쪽으로 진행하려는 힘과 원심력에 의해 루트에서 팁 쪽으로 진행하려는 두 힘은 서로 방향이 반대이기 때문에 두 힘의 크기를 비슷하게 해주면 노즐과 버켓 사이에서 유동은 축방향으로 평행하게 흘러가며, 앞서 언급된 제반 문제점들이 사라지게 된다.

따라서 노즐과 버켓 사이에 형성되는 유동의 특징은 반경방향을 따라서 속도는 줄어들고, 압력은 증가한다.

한편, 축류형 다단 터빈은 버켓 입구에서 뿐만이 아니라 출구에서도 압력과 속도는 반경방향을 따라서 일정하게 유지되어야 한다.

따라서 버켓은 루트에서 팁 쪽으로 가면서 반동도가 증가하기 때문에 버켓 루트는 충동형, 팁은 반동형으로 설계한다.

이런 이유 때문에 터빈 버켓은 하나의 블레이드에 충동형과 반동형이 혼재된 충동-반동 블레이드(impulse-reaction blade)이다.



HIoPE

1 2 3 4 5 6 7 8 9 10 11 1 2 3 4 5 6 7 1 2 3 4 5 6

0

10

20

30

40

50

60

70

ROOT ROOT ROOT

ROOT

TIP

IMPULSE TURBINE

50% REACTION TURBINE

TIP

HP IP LP

Degre

e o

f R

ea

ctio

n (

%)

Stage Number

Degree of Reaction in Fossil Power



HIoPE

Degree of Reaction in Nuclear Power



HIoPE

Stack-Up of Blade (LSB)

p1 = 6 psia

= 70% @ tip

p2 = 4.5 psia

p3 = 1 psia

p2 = 1.25 psia

= 5% @ root

Root

Tip


HIoPE

Impulse Blade vs. Reaction Blade

Impulse turbines do not have thrust concern,

and the buckets are mounted on disk extension

of the rotor (wheels), resulting in larger overall

diameters, small rotor diameters, and fewer

stages than reaction turbines.

Wheel and Diaphragm


HIoPE

Comparison of Rotor Shaft

The simplest difference between impulse and reaction turbines is rotor shaft.

Impulse design has a little pressure drop across the buckets. Therefore, the buckets can be mounted on the

periphery of a wheel without generating significant axial thrust.

On the contrary, blades of reaction turbines are mounted directly at the shaft surface because of steam

pressure difference between upstream and downstream of blades.

This may create a large thrust (axial) force at the shaft caused large bend stress in the discs.

In addition, reaction turbines have more stages than impulse ones.

Wheel and diaphragm construction for impulse

blades Drum rotor construction for reaction blades


HIoPE

Balance Piston

Welding Balance Piston

The reaction design has a significant pressure drop across

the buckets and high thrust. Therefore, a balance piston,

which is normally built into the rotor, is installed in high-

pressure zones of single-flow turbines to offset the thrust.

Otherwise, the turbine is designed with double-flow. Some

designers also use a balance piston on impulse turbines that

have a high thrust. Balance Piston

(Siemens)


HIoPE

MHI

40” LSB (Steel with stellite erosion shield)

Single Cylinder Single Flow Turbine, 122 MW (for

combined cycle application)

Balance Piston


HIoPE

What are the major advantages of double

flow structure?

1) Larger capacity

2) Smaller thrust force

3) Shorter LSB

Double Flow

Double flow can be employed to avoid a balance piston in reaction turbines which have a higher

thrust than impulse turbine

31 ppdlT

T = Thrust

d = mean diameter of blade row

l = active length of blade

p1 = pressure at the inlet of stage

p3 = pressure at the exit of stage

= degree of reaction at the mean dia.


HIoPE

Reaction Blades

Advantages Alstom

Higher aerodynamic efficiency

• Low turning and accelerating flow in both nozzle and bucket allow design of higher efficient and tolerant

profiles

• Low acceleration of the flow through the nozzle and bucket leads lower profile loss

Better flexibility with respect to operating range

Lower staging loading

Many stage can be designed with 50% reaction (all HP and IP stages, and front stages of LP turbine)

Symmetric velocity triangle (50% reaction stage)

• Use of same profile in the nozzle and bucket and it may contribute to cost down

• Near-zero interstage swirl

Because of the lower pressure drop, there is no need for costly diaphragm construction

It leads larger number of stages because of lower stage loading (roughly twice that of impulse stages for

50% reaction stages).

Increase of axial thrust which leads higher dummy balance piston, i.e. increased leakage loss.

Drum-type rotor is suitable for reaction turbine and it leads higher leakage area at the hub section.

Degree of reaction at the hub and tip section is higher compared with impulse stage. Higher hub reaction

leads lower leakage loss, however higher tip reaction gives higher leakage loss.

Disadvantages


HIoPE





Stage Efficiency 5



HIoPE

[연습문제 3.2]

증기터빈에 유입되는 주증기 조건(main steam conditions)이 30 MPa, 650C 이다. 이 증기는 1단 충동터빈(one-stage impulse turbine)을 등엔트로피 팽창(isentropic expansion)되어 350C로 빠져나간다. 이때 엔탈피 강하(isentropic enthalpy drop)가 540 kJ/kg 이다. 이 터빈을 single stage로 설계하는 데 따른 문제점을 검토하시오. 단, C2와 U가 이루는 각도는 13, 유량계수(flow coefficient)는 0.5, 노즐입구에서의 속도 C1은 매우 작다고 가정한다.

w2

c2

u

w3

u c3

Direction

of rotation

c1

Number of Stages


HIoPE

[검토 결과]

노즐에서 발생하는 일은 없다. 아울러 노즐통로유동은 단열과정이기 때문에 노즐출구에서의 증기속도는 열역학 제1법칙을 이용하여 구할 수 있다.

편의상 노즐 입구속도는 0이라 가정하면, 열역학 제1법칙으로부터 다음 결과를 얻는다.

kJ/kg

1039.23 m/s

한편, c2와 u가 이루는 각도가 13 이며, 유량계수(flow coefficient) 0.5를 이용하여 u를 계산하면,

= cx/u

u = 1039.23 sin13 /0.5 = 467.6 m/s

증기터빈의 회전속도가 3,600 rpm인 경우, 블레이드 피치직경은 다음 식으로 구한다.

d = 2.48 m

따라서, 블레이드 회전속도와 피치직경이 매우 크기 때문에 블레이드는 큰 응력을 받게 되며, 큰 속도로 인해서 손실이 증가하게 된다. 결론적으로 이 조건에서는 증기터빈을 다단(multiple stages)으로 구성하는 것이 유리하다.

1212

2

1

2

212122

1wzzgcchhq

54022 21

2

2 hhc

60

)2/(2 ndu

2c

Number of Stages


HIoPE

Stage Efficiency

2, sin4ist

2

2

max,, sin ist

2c

u

ideal

actualst

p

p

버켓에서 생산하는 실제 동력

버켓에서 생산할 수 있는 이상 동력

=

( = velocity ratio)

(from ) 0,

d

ist

2

2

2,

sin21

sin22

rst

2

2

2

2

max,,sin1

sin2

rst

2c

u


2

2

u

c2 w3

w2

3

c3

3

(a) < sin2

w3

w2

c2

u

c3

(b) = sin2

c2 w3

u

w2 c3

(c) > sin2

w2

c2

u

w3 c3

2

2

3


HIoPE

0.25 0.50 0.75 1.00 1.25 1.50

Velocity ratio ( )

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

1.1

Sta

ge

eff

icie

ncy

Impulse Reaction

2

2

2,

sin21

sin22

rst

2, sin4ist

2c

u

Stage Efficiency


HIoPE

Wake and Core Flow

Reaction stage has better adaptability for

core flow of nozzle and higher stage

efficiency

The wake is a velocity defect generated by the

boundary layers of the blade surfaces. If is

undisturbed by other blades it would move

downstream along the direction of outlet-flow angle

while decaying slowly over three or four chord

lengths.

Wake (후류)

Velo

city

Number of revolution

1 2


HIoPE

Wake and Core Flow

Impulse

Bucket Reaction

Bucket


HIoPE





Stage Efficiency 5



HIoPE

Impulse Blade vs. Reaction Blade

Conventional

1940

Laminar

1950

SCHLICT

1965

Super

1980

T

1937

VN

1953

T2

1968

T4

1980

TX

1995

GE

Siemens

Bucket Shape


HIoPE

Bucket Profile Enhancement

Siemens


HIoPE

Alstom

Bucket Profile Enhancement


HIoPE

Typical HP Steam Path Efficiency of Single Stages


HIoPE

Shrouded Blade

[ Shrouded Blade ] [ Free Tip Blade ]

Shrouded blades have reduced leakage losses because they have seal system at the blade tip.

Shrouded blades have better vibration characteristics because they are often interlocked to provide

mechanical damping.

Shrouded blades give better efficiency because they have better aerodynamic characteristics at the blade tip.

The tip vortex formed from open tip blade produce a large disturbed flow when it combined with secondary

flow in the blade passage.

However, the shroud creates increased stress levels.

[ Tip Vortex ]


HIoPE

Tip Vortex and Wing Let


HIoPE

질의 및 응답

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

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

3. flows in a steam path - engsoft lab in steam p… · steam turbine 3. flow in steam path 4 / 92...

Documents