13-Steam Turbines [Compatibility Mode]

8/16/2019 13-Steam Turbines [Compatibility Mode]

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PEMP

RMD 2501

Session delivered by:Session delivered by:

. . .. . .

© M.S. Ramaiah School of Advanced Studies13 1


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PEMP

RMD 2501Session Objectives

This session is intended to discuss the following:

Classification of steam turbines Compounding of steam turbines

Forces, work done and efficiency of steam turbine

Numerical exam les



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RMD 2501Steam

Steam is a vapour used as a working substance in the operation of steam turbine.

Is steam a perfect gas?

Steam possess properties like those of gases: namely pressure, volume, temperature,internal energy, enthalpy and entropy. But the pressure volume and temperature of

steam as a vapour are not connected by any simple relationship such as is expressed

by the characteristic equation for a perfect gas.

Sensible heat – The heat absorbed by water in attaining its boiling point.Latent heat – The heat absorbed to convert boiling water into steam.

Wet steam – Steam containing some quantity of moisture.

– .Superheated steam – Dry steam, when heated at constant pressure, attains superheat


.


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RMD 2501Steam Properties

Enthalpy (H) kJ/kg Internal energy (U) kJ/kg


Entropy (S) kJ/kg-K Specific volume (v) m3/kg

Density (ρ) kg/m3 Isobaric heat capacity (C p)kJ/kg-K


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PEMP

RMD 2501Steam Power Plant Process

Fuel Boiler Turbine Generator

Thot

xPum Low Pressure

and temp.

Low Pressure

Water

xPump Hot


Tcold


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PEMP

RMD 2501Steam Turbine

Steam turbine convert a part of the energy of the steam evidenced by high

temperature and pressure into mechanical power-in turn electrical power

The steam from the boiler is ex anded in a nozzle resultin in the emission of

a high velocity jet. This jet of steam impinges on the moving vanes or blades,mounted on a shaft. Here it undergoes a change of direction of motion which

gives rise to a change in momentum and therefore a force.

The motive power in a steam turbine is obtained by the rate of change in

momentum of a high velocity jet of steam impinging on a curved blade which.

The conversion of energy in the blades takes place by impulse, reaction or

impulse reaction principle.

Steam turbines are available in a few kW (as prime mover) to 1500 MW

Impulse turbine are used for capacity up to


Reaction turbines are used for capacity up to


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RMD 2501Steam, Gas and Hydraulic Turbines

The working substance differs for different types of turbines.

Steam turbines are axial flow machines (radial steam turbines are rarely used)

whereas as turbines and h draulic turbines of both axial and radial flow t e

are used based on applications. The pressure of working medium used in steam turbines is very high,

comparatively.

The pressure and temperature of working medium in hydraulic turbines islower than steam turbines.

Steam turbines of 1300 MW single units are available whereas largest gas

turbines unit is 530 MW and 815 MW



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PEMP

RMD 2501Merits and Demerits of Steam Turbine

Merits:

• Ability to utilize high pressure and high temperature steam.

• g component e c ency.

• High rotational speed.

• High capacity/weight ratio.

• Smooth, nearly vibration-free operation.

• No internal lubrication.

• .

• Can be built in small or very large units (up to 1200 MW).

Demerits:

• or s ow spee app ca on re uc on gears are requ re .• The steam turbine cannot be made reversible.

• The efficiency of small simple steam turbines is poor .



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RMD 2501Application

• Power generation

• , ,

• Pharmaceuticals,• Food processing,

• Petroleum/Gas processing,

• Pulp & Paper mills,- -



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RMD 2501Turbine Selection

In all fields of application the competitiveness of a turbine is a

combination of several factors:

Efficiency

Power density (power to weight ratio)

D rect operat on cost

Manufacturing and maintenance costs



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PEMP

RMD 2501Rankine Cycle

Saturated Rankine cycle Superheated Rankine cycle



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RMD 2501Schematic of Rankine Reheat Cycle

Low

4

5PressureTURBINE

BOILERw

outhi

3

woutlo

Pressure

TURBINE CONDENSER2

6

q inhi qout

1


PUMPin


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PEMP

RMD 2501Steam Turbine Classification

Steam turbines can be classified in several different ways:

1. By details of stage design

• Impulse or reaction.

2. By steam supp y an ex aust con t ons

• Condensing, or Non-condensing (back pressure),• Automatic or controlled extraction,

• Reheat

3. By casing or shaft arrangement• ,

4. By number of exhaust stages in parallel:

• Two flow, Four flow or Six flow.

5. B direction of steam flow:

• Axial flow, Radial flow or Tangential flow6. Single or multi-stage

7. By steam supply


• Superheat or Saturated


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PEMP

RMD 2501Steam Turbine Stage

A turbine stage consists of stationary stator row

(guide vanes or nozzle ring) and rotating rotor

.

In the guide vanes high pressure, high

tem erature steam is ex anded resultin in hi h

The uide vanes direct the flow to the rotor

velocity.

blades at an appropriate angle.

kinetic energy of the working fluid is absorbed by

the rotor shaft producing mechanical energy



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PEMP

RMD 2501Types of Steam Turbine

Impulse Turbine Reaction Turbine

Process of complete expansion of steam takes place in stationary nozzle

and the velocity energy is converted

Pressure drop with expansion andgeneration of kinetic energy takes place

in the moving blades.


n o mec an ca wor on e ur ne

blades.


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RMD 2501Types of Steam Turbine

Parsons Reaction Turbine De Laval Impulse Turbine.



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RMD 2501Impulse Reaction Turbine

Modern turbines are neither purely impulse or reaction but a combination of

both.

Pressure dro is effected artl in nozzles and artl in movin blades

which are so designed that expansion of steam takes place in them. High velocity jet from nozzles produce an impulse on the moving blade and

reaction.

Impulse turbine began employing reaction of upto 20% at the root of the

movin blades in order to counteract the oor efficienc incurred from zero

or even negative reaction.

Reaction at the root of reaction turbines has come down to as little as 30%

to 40% resultin in the reduction of the number of sta es re uired and the

sustaining of 50% reaction at mid point.

It may be more accurate to describe the two design as


Drum rotor turbine using high reaction blading


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PEMP

RMD 2501Flow Through Steam Turbine Stage



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RMD 2501Compounding of Steam Turbines

practical limits.

Compounding is achieved by using more than one set of nozzles,

Three main t es of com ounded im ulse turbines are:

a es, ro ors, n a ser es, eye o a common s a ; so a e er e

steam pressure or the jet velocity is absorbed by the turbine in stages.

a. Pressure compounded

b. Velocity compounded

. .

Involves splitting up of the whole pressure drop

several stages of impulse turbine.

The nozzles are fitted into a diaphragm locked in


e cas ng a separa es one w ee c am er rom

another. All rotors are mounted on the same shaft.


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RMD 2501Compounding of Steam Turbines

Velocity drop is acheived through many moving rows of

blades instead of a single row of moving blades.

It cons sts o a nozz e or a set o nozz es an rows o mov ng

blades attached to the rotor or the wheel and rows of fixed blades attached to the casing.

Pressure velocity compounding gives the advantage of

producing a shortened rotor compared to pure velocity

compounding.

In this design steam velocity at exit to the nozzles is keptreasonable and thus the blade speed (hence rotor rpm)

reduced.



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RMD 2501Comparison between Impulse & Reaction Turbines

mpu se ur nes• Reaction turbine makes use of the

reaction force produced as the steam• An impulse turbine has fixed

nozzles that orient the steam flow

formed by the rotor • Blades have aerofoil profile

conver ent assa e since ressure

into high speed jets.

• Blade profile is symmetrical as no pressure drop takes place in the

drop occurs partly in the rotor

blades.

• Efficient at the lower pressure stages

rotor blades.

• Suitable for efficiently absorbing

the high velocity and high• Fine blade tip clearances are

necessary due to the pressure

leakages.

pressure.

• Steam pressure is constant across

the blades and therefore fine tip

• Inefficient at the high pressure stagesdue to the pressure leakages around

the blade tips.

.

• Efficiency is not maintained in the

lower pressure stages (high

velocit cannot be achieved in


• ne p c earances can cause

damage to the tips of the blades.steam for the lower pressure

stages).


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RMD 2501Losses in Steam Turbine

Profile loss: Due to formation of boundary layer on blade surfaces. Profile loss is a

boundary layer phenomenon and therefore subject to factors that influence

boundar la er develo ment. These factors are Re nolds number surface

roughness, exit Mach number and trailing edge thickness.

Secondary loss: Due to friction on the casing wall and on the blade root and tip. It

profile loss.

Tip leakage loss: Due to steam passing through the small clearances required between the moving tip and casing or between the moving blade tip and rotating

shaft. The extend of leakage depends on the whether the turbine is impulse or

reaction. Due to pressure drop in moving blades of reaction turbine they are more

.

Disc windage loss: Due to surface friction created on the discs of an impulse

turbine as the disc rotates in steam atmosphere. The result is the forfeiture of shaft


power for an increase in kinetic energy and heat energy of steam.


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RMD 2501Losses in Steam Turbine

Lacing wire loss: Due to passage blockage created by the presence of lacing wires

in long blade of LP Stages.

turbine. The loss is a combination of two effects; firstly, reduction in efficiency dueto absorption of energy by the water droplets and secondly, erosion of final moving

blades leadin ed es.

Annulus loss: Due to significant amount of diffusion between adjacent stages or

where wall cavities occur between the fixed and moving blades. The extent of loss

steam is controlled by a flared casing wall.

Leaving loss: Due to kinetic energy available at the steam leaving from the last

stage of LP turbine. In practice steam does slow down after leaving the last blade, but through the conversion of its kinetic energy to flow friction losses.

Partial admission loss: Due to artial fillin of steam, flow between the blades is


considerably accelerated causing a loss in power.


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RMD 2501 Nomenclature

V Absolute velocity of steam

U Blade velocity

W Re at ve ve oc ty o steam

V a=V f = V m Axial component or flow velocity

w

α Nozzle angle

Blade an le

h enthalpy

Suffix

1 Inlet


2 Outlet


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PEMP

RMD 2501Velocity Triangles

The three velocity vectors namely, blade speed, absolute velocity

and relative velocity in relation to the rotor are used to form a

triangle called velocity triangle.

Velocity triangles are used to illustrate the flow in the bladings of

.

Changes in the flow direction and velocity are easy to understand

with the hel of the velocit trian les.

Note that the velocity triangles are drawn for the inlet and outlet of

the rotor at certain radii.



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RMD 2501Steam Turbine Blade TerminologySteam Turbine Blade Terminology



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RMD 2501Inlet Velocity Triangle

Vw1

U

W1Va1V1



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RMD 2501Outlet Velocity Triangles

Vw2

V2Va2

W2



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RMD 2501Combined Velocity Triangles

Vw2 Vw1

U

Va1V1V2

W2

Va2

ΔVw

© M.S. Ramaiah School of Advanced Studies13 30For 50% reaction design


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PEMP

RMD 2501Work Done – Impulse Steam Turbine

e a e s symme r ca en 1 = 2 an neg ec ng r c ona e ec s o e blades on the steam, W1 = W2.

In actual case, the relative velocity is reduced by friction and expressed by a blade

velocity coefficient k.

Thus k = W2 / W1

’ ,

Wt = U(Vw1 ± Vw2) (1)

Since Vw2 is in the negative r direction, the work done per unit mass flow is given,

Wt = U(Vw1+Vw2)

(2)

a1

a2

, ere w an ax a rus n e ow rec on. ssume a a s

constant then,

Wt = UVa (tanα1+ tanα2) (3)


t = a an 1 an 2 4)

Equation (4) is often referred to as the diagram work per unit mass flow and hence

the diagram efficiency is defined as


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Diagram work done per unit mass flow

Work available per unit mass flowηd = (5)Referrin to the combined dia ram ΔV is the chan e in the velocit of whirl.Therefore:

The driving force on the wheel = mVw (6)

is done on the wheel. From equation (6)

Power output = mUΔVw

(7)

a1- a2 = a, e ax a rus s g ven y;

Axial thrust = mΔVa (8)The maximum velocity of the steam striking the blades

V1 = {2(h0-h1)} (9)

Where h0 is the enthalpy at the entry to the nozzle and h1 is the enthalpy at the


2

1V

, .

blades is the kinetic energy of the jet and the blading or diagram efficiency;


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W k D I l S T bi


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Differentiating equation (12) and equating it to zero provides the maximum

diagram efficiency;

8)( U d d η

11

1

−

⎟ ⎠ ⎞

⎜⎝ ⎛ V

V U

d

2

1

1

=V

or (13)

i.e., maximum diagram efficiency

⎟ ⎠

⎞⎜⎝

⎛ −=2

coscos

2

cos4 11

1 α α α

2 14

1

cos α η =d

Substituting this value in equation (7), the power output per unit mass flow rate at

the maximum diagram efficiency


22U P =

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D f R i


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RMD 2501Degree of Reaction

Degree of reaction is a parameter that describes the relation between the

energy transfer due to the static pressure change and the energy transfer due

to dynamic pressure change.

Degree of reaction is defined as the ratio of static pressure drop in the rotor

to the static pressure drop in the stage. It is also defined as the ratio of staticenthalpy drop in the rotor to the static enthalpy drop in the stage

Λ= Degree of reaction =Static enthalpy change in rotor

Total enthalpy change in stage=

h1-h2h0-h2

(16)

and velocity at the inlet to the fixed blades is given by

V hh

2

0−= similarl V hh

2

2−= pC 2 pC 2

Substituting ( )−

=Λ hh

22

21


⎟

⎠

⎜

⎝

−−⎟

⎠

⎜

⎝

− p p C

hC

h22

202

000

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D f R ti


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( )21 hh −=Λ

But for a normal stage, V0 = V2 and since h00 = h01 in the nozzle, then;

(17)

0201 −

We know that (h01 – h02) = ( ) ( ) 02

2

2

2

121 =−+− ww V V hh

1- 2 ,

( )

2

1

2

2

V V ww

−=Λ

( )2122 ww V V −

−

=0201 − 21 ww

Assuming the axial velocity is constant through out the stage, then

− 22

( )[ ]U V V U U wwww

−++=Λ 2112

2


( )[ ]211212

2 ww

wwww

V V U +−

=Λ(19)

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D f R ti


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RMD 2501

(20)Degree of Reaction

( )U

V a

2

tantan 12 β β +=Λ

From the velocit trian les it is seen that

11 ww V U V += U V V ww −= 22

Therefore e uation 20 can be arran ed into a second form:

( )22 tantan

22

1α β ++=Λ

U

V a (21)

PuttingΛ = 0 in equation (20), we get

( )12 β β = And V1 = V2 and for Λ = 0.5, ( )12 α β =

Zero reaction stageLet us first discuss the special case of zero reaction. According to the definition

of reaction When Λ = 0 e uation 16 reveals that h = h and e uation 20 that


β1 = β2.

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D f R ti


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Now h = h and h = h for Λ = 0. Then W = W . In the ideal case, there is

Mollier diagram Velocity Diagram

no pressure drop in the rotor and points 1 2 and 2s on the mollier chart shouldcoincide. But due to irreversibility, there is a pressure drop through the rotor. The

zero reaction in the impulse stage by definition, means there is no pressure drop


through the rotor. The Mollier diagram for an impulse stage is shown in Fig. 1.a,

where it can be observed that the enthalpy increases through the rotor.

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D f R ti


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From equation (16) it is clear that the reaction

is negative for the impulse turbine stage when

irreversibility is taken into account.

Fifty percent reaction stage

From equation (16) for Λ = 0.5 α1 = β2 and the

Fig.1.a

.

symmetry, it is also clear that α2 = β1. For Λ=1/2,the enthalpy drop in the nozzle row equals the

.

h0 - h1 = h1 - h2


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Degree of Reaction


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aV += 22 tanα β Substituting into equation (21)

V a (22)12

2−

U

Thus when α2 = α1, the reaction is unity(also V1 = V2). The velocity diagram for

Λ = 1 is shown in Fig. with the samevalues of Va, U and W used for Λ = 0

and Λ = ½. It is obvious that if Λexcee s un y, en 1 0 .e., nozz eflow diffusion).

Choice of Reaction and Effect on Efficiency

Equation (17) can be rewritten asU

ww

21 12

−+=Λ

Cw2 can be eliminated by using this equation


12 ww V U

V −= yieldingU U

w1

221 −+=Λ

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Degree of Reaction


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In Fig. the total to static efficiencies are

shown plotted against the degree of

reaction.

When = 2, ηts is maximum at Λ= 0. With higher loading, the optimumηts is obtained with higher reaction

2/U W

ratios. As shown in Fig. for a high total

to total efficiency, the blade loading

factor should be as small as possible,which implies the highest possible

value of blade speed is consistent with

blade stress limitations. It means that

dependent upon the reaction ratio and

ηts can be optimized by choosing a


.

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RMD 2501Blade Height in Axial Flow Machines


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RMD 2501Blade Height in Axial Flow Machines

The continuity equation may be used to find the blade height ‘h’. The

annular area of flow = πDh. Thus the mass flow rate through an axial flowturbine is

AV m ρ =•

a DhV m ρπ =

•

•

DV

mh

ρπ

=

Blade height will increase in the direction of flow in a turbine and decrease in

the direction of flow in a compressor.


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RMD 2501Effect of Reheat Factor & Stage Efficiency


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The thermodynamic effect on the turbine efficiency can be best understood by

considering a number of stages, say 4, between states 1 and 5 as shown in Fig.


Total expansion is divided into four stages of the same stage efficiency and

pressure ratio

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5

4

4

3

3

2

2

1.,. P P P P

ei ===

work done per kg of steam to the isentropic work done per kg of steam between

1 and 5.

−⎟ ⎠

⎜⎝

= s

a

W ei 0.,. η '

51

0.,.hh

ei−

=η

The actual work done per kg of steam Wa = η0 Ws(22)

Isentropic or ideal values in each stages are ΔWs1,ΔWs2, ΔWs3,ΔWs4.

Therefore the total value of the actual work done in these stages is,

Wa = Σ(1-2)+(2-3)+(3-4)+(4-5)Also stage efficiency for each stage is given by


ηs = Isentropic work done in stage

1

1

s

a

W =

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or stage

111

1

1

'

21

21

1

11.,. s sa

s

a

s

a s W W or

W

W

hh

hh

W

W ei Δ=Δ

Δ=

−−

== η η

44332211 s s s s s s s saa W W W W W W Δ+Δ++ΔΣ=ΣΔ=Δ∴ η η η η

For same stage efficiency in each stage 4321 s s s s η η η η ===

s s s s s s sa W W W W W W ΣΔ=Δ+Δ+Δ+ΔΣ= η η 4321 (23)From equation (22) and (23),

s

s s

W

W W

ΣΔ=∴

ΣΔ=η η 00

(24)

sThe slope of constant pressure lines on h-s plane is given by

h∂


T s p

= ⎠⎝ ∂


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RMD 2501Merits and Demerits of Reheating


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Merits and Demerits of Reheating

Advantages of Reheating

1. There is an increase in output of turbine.

2. Erosion and corrosion problems in steam turbine are reduced.

3. There is an improvement in overall thermal efficiency of the turbine.

4. Condition of steam in last stage are improved.

Demerits

1. Capital cost required for Reheating

2. The increase in thermal efficiency is not appreciable compared to

.



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RMD 2501Session Summary


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Session Summary

In this session the students would have learnt about

Wor ng pr nc p e o steam tur ne

Classification of turbines

Types of compounding

Work done and efficiency of Impulse steam turbine stage

Work done and efficiency of reaction steam turbine stage

turbines


PEMP

RMD 2501


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Thank you


Documents

13-Steam Turbines [Compatibility Mode]