UNIT – III

STEAM NOZZLES AND TURBINES

3.1 INTRODUCTION

A steam nozzle is a passage of varying cross section, which converts HeatEnergy of steam into Kinetic Energy. During the first part of the nozzle, the steamincreases its velocity. But in its later part, the steam gains more in the volume than in thevelocity. Since the mass of steam, passing through any section of the nozzle remainsconstant, the variation of the steam pressure in the nozzle depends upon the velocity,specific volume and dryness fraction of the steam. A well designed nozzle converts theHeat Energy of steam into Kinetic Energy.

3.2 TYPES OF STEAM NOZZLES:

Following three types of nozzles are important from the subject point of view:

(i) Convergent Nozzle:

When the cross section of a nozzle decreases continuously from entrance toexit, it is called a convergent nozzle.

(ii) Divergent Nozzle :

When the cross section of a nozzle increases continuously from entrance toexit, it is called a divergent nozzle.

(iii) Convergent- Divergent Nozzle :

When the cross section of a nozzle first decrease from entrance to throat,and then increase from throat to exit, it is called a convergent-divergent nozzle.

3.3 FRICTION IN A NOZZLE AND NOZZLE EFFICIENCY:

When the steam flows through a nozzle, some loss in its enthalpy or takes placedue to friction between the nozzle surface and the flowing steam. This can be understoodwith the help of h-s diagram and Mollier chart.

(i) First of all, locate the point A for the initial condition of the steam. It is apoint where the saturation line meets the initial pressure (P1) line.

(ii) Now draw a vertical line through A to meet the final pressure (P2) line. Thisis done as the flow through the nozzle is isentropic , which is expressed bya vertical line AB. The heat drop (h1-h2) isentropic heat drop.

(iii) Due to friction in the nozzle the actual heat drop in the steam will be lessthan (h1-h2). This heat drop is shown as AC instead of AB.

(iv) As the expansion of the steam ends at the pressure P2, therefore finalcondition of the steam is obtained by drawing a horizontal line through C tomeet the final pressure (P2) line at B’.

(v) Now the actual expansion of the steam in the nozzle is expressed by thecurve AB’ (adiabatic expansion) Instead of AB (isentropic expansion).Actual heat drop (h1-h3) is known as useful heat drop.

Now the coefficient of nozzle or nozzle efficiency is defined as the ratio ofuseful heat drop to the isentropic heat drop.

1 3

1 2

h hUseful heat drop ACKIsentropic heat drop AB h h

−= = =

−

3.4 VELOCITY OF STEAM FLOW THROUGH A NOZZLE:

Consider a unit mass flow of steam through a nozzle. Let V1 = Velocity of steam at the entrance of the nozzle in m/s

V2 = Velocity of steam at any section considered in m/s h1 = Enthalpy or total heat of steam entering nozzle in kJ/kg and h2 = Enthalpy or total heat of steam at the section considered inkJ/kg

we know that for a steady flow process in a nozzle ,2 2

1 21 2

1 11000 2 1000 2

V Vh h losses

+ = + +

Neglecting losses in a nozzle ,

2 22 1

1 21

1000 2 2V V h h

− = −

( )2 22 1 1 2 12000 2000 dV V h h V h= + − = +

where, hd = Enthalpy or heat drop during expansion of steam innozzle. hd = h1 - h2

Since the entrance velocity or velocity approach (V1) is negligible as compared to h2

Therefore,

2 2000 44.72d dV h h= =

3.5 MASS OF STEAM DISCHARGED THROUGH NOZZLE

We have already discussed that the flow of steam, through the nozzle isisentropic ,which is approximately represented by the general law:

Pvn = Constant

We know that gain in kinetic energy

=2

2

2V ……….(Neglecting initial velocity of

steam)

Heat drop = ( )1 1 2 21n p v p v

n−

−Since gain in kinetic energy is equal to heat drop, therefore

( )2

21 1 2 22 1

V n p v p vn

= −−

2 21 1

1 1

11

p vn p vn p v

= × − −

we know that 1 1 2 2n np v p v=

1 1

2 1 2

1 2 1

n nv p pv p p

−

= =

1

22 1

1

npv vp

−

=

Substituting the value of 1

2

vv

12

2 2 21 1

1 1

12 1

nV p pn p vn p p

− = − −

=

1

21 1

1

11

nnpn p v

n p

− − −

1

22 1 1

1

2 11

nnpnV p v

n p

− = × × − −

Now the volume of the steam flowing per second = Cross sectional area of nozzle× velocity of steam =AV2and volume of 1kg of steam i.e., specific Volume of steam at pressure P2 = V2 m3/kg

Mass of steam discharged through nozzle per second,

m =2

sec1

volumeof steam flowing per ondvolumeof kg of steam at pressure p

1

2 21 1

2 2 1

2 11

nnAV pA n p v

v v n p

− = = × × − −

1 1

1 21 1

1 2 1

2 11

nn np pA n p v

v p n p

−−

= × × − −

1 1

2 21 1

1 1 1

2 11

nn np pA n p v

v p n p

− = × − −

2 1

2 1 2

1 1 1

2 11

nn np p pnA

p n v p

− = × − −

2 1

1 2 2

1 1 1

21

nn np p pnA

n v p p

+ = × − −

3.6 CRITICAL PRESSURE RATIO :

A nozzle is normally designed for maximum discharged by designing a certainthroat pressure which produces this condition.

Let P1 = Initial pressure of steam in N/m2

P2 = Pressure of steam at throat in N/m2

V1= Volume of 1kg of steam at pressure P1 in m3

V2= Volume of 1kg of steam at pressure P2 in m3 ,and

A = cross sectional area of nozzle at throat, in m2 .

2 1

1 2 2

1 1 1

21

nn np p pnm A

n v p p

+ = × − −

There is only one value of the ratio P2/P1 , which produces maximum discharge from thenozzle. This ratio P2/P1, is obtained by differentiating the right hand side of the equation.We see from this equation that except P2/P1 , all other values are constant. P2/P1 isdifferentiated and equated to zero for maximum discharge.

2 1

2 2

1 12

1

0

nn np pd

p ppdp

+ − =

2 11 1

2 2

1 1

2 1 0

nn np pn

n p n p

+− −

+− =

2 1 1

2 2

1 1

2 1 0

n nn np pn

n p n p

− +−

+− =

2 1

2 2

1 1

12

nn np p n n

p p n

−−

+× = ×

( )11

1

2 1

2

1

1

2

1

2

1

12

12

1 12 2

21

nnnn

nn

nn n

nn

p np

p np

p n np

n

−− −−

−

−−

−

+=

+=

+ + = =

= +

3.7 VALUES FOR MAXIMUM DISCHARGE THROUGH A NOZZLE:

The relationship for maximum discharge through a nozzle is,

21

1max

1

2 21 1

npnm An v n

− = × + +

The values of maximum discharge for the following three conditions:

(i) When the steam is initially dry saturated:

We know that for dry saturated steam ,n= 1.135.

Therefore maximum discharge is,

1max

1

0.637 pm Av

=

(ii) When the steam is initially superheated:

For superheated steam, n=1.3.

Therefore maximum discharge is,

1max

1

0.666 pm Av

=

(iii) For gases:

For gases, n=1.4.

Therefore maximum discharge is,

1max

1

0.685 pm Av

=

3.8 VALUES FOR CRITICAL PRESSURE RATIO

1

2

1

2Critical pressure ratio1

nn

pp n

−

= = +

For the following conditions Critical pressure ratios are,

(i) When the steam is initially saturated We know that for dry saturated steam, n=1.135

2

1

pp

=0.577 ; P2 = 0.577 P1

(ii) When the steam is initially superheated:

We know that for superheated steam, n=1.32

1

pp

=0.546 ; P2 = 0.546 P1

(iii) When the steam is initially wet

It has been experimentally found for wet steam,

2

1

pp

=0.582 ; P2 = 0.582 P1

(iv) For gases:

We know that for gases, n=1.42

1

pp

=0.528 ; P2 = 0.528 P1

3.8.1 PHYSICAL SIGNIFICANCE OF CRITICAL PRESSURE RATIO:

Now consider two vessels A and B connected by a convergent nozzle. Let thevessel A contains steam at a high and steady pressure (P1),and the vessel B containssteam at another pressure (P2) which may varied at will. Let the pressure P2 in the vessel B be made equal to the pressure (P1) in thevessel A. There will be no flow of steam through the nozzle. Now if the pressure (P2)in the vessel B gradually reduced, the discharge through the nozzle will increase. Asthe pressure (P2) in the vessel B approaches the critical value , the rate of dischargewill also approach its maximum value. If the pressure (P2) in the vessel B is furtherreduced, it will not increase the rate of discharge. The ratio of exit pressure to theinlet pressure is called critical pressure ratio.

We know that the velocity of steam at any section in the nozzle

1

22 1 1

1

2 11

nnpnV p v

n p

− = × × − −

and the critical pressure ratio for maximum discharge ,

1

12 2

1 1

2 21 1

nnnnp por

p n p n

−

− = = + +

2 1 1 1 12 1 22 1 2

1 1 1 1n n nV p v p v

n n n n+ − = × × − = × × − + − +

= 11 1

1

221 1

pn np vn n ρ

× × = ×− −

( )1( )Volume v

Density ρ

=

Q

we also know that for isentropic expansion ,1 1 2 2

n np v p v=

1 2

1 21

2

1 2 1

1 1

n n

n

p p

pp

ρ ρ

ρ ρ

=

=

1

1 1 2

1 2 1

1 1

2 1 2 2 2

2 2 1 2 1

1

2

2

21

n

nn n

nn

p p pp

p p p p pp p p

pn

ρ ρ

ρ ρ

ρ

−

−

=

= × = ×

= × +

=1

12 2

2 2

2 11 2

n nn np p n

nρ ρ

−×

− + × = × +

Substituting the value of P1/ρ in the V2 equation,

V2 = 2 2

2 2

2 11 2

p nPn nn ρ ρ

+ × × = − 2

2

12

p nρ

+ ×

This is the value of velocity of sound in the medium at pressure P2 and it is known assonic velocity.

3.9 SUPERSATURATED FLOW OR METASTABLE FLOWTHROUGH NOZZLE:

The expansion of steam in an ideal nozzle is isentropic, which isaccompanied by condensation process. If the steam is initially superheated, thecondensation should start after it has become dry saturation. This is possible when thesteam has proceeded through some distance in the nozzle and in a short interval oftime. But from practical point of view, the steam has a great velocity. Equilibriumbetween the liquid and vapour phase is delayed and the steam continues to expand ina dry state. The steam in such a set of conditions, is said to be supersaturated or inmetastable state. It is also called super cooled steam ,as its temperature at any

pressure is less than the saturation temperature corresponding to the pressure. Theflow of supersaturated steam, through the nozzle is called supersaturated flow ormetastable flow.

Experiment of supersaturated flow of steam have shown that there is alimit to which the supersaturated flow is possible. This limit is represented by Wilsonline on h-s diagram. The Wilson line closely follows the 0.97 dryness fraction line.Beyond this Wilson line, there is no super saturation.

Figuare shows the isentropic expansion of steam in a nozzle. The point A representsthe position of initial dry saturated steam at pressure P1. The line AB represents theisentropic expansion of steam in the supersaturated region. The metastable state(pointC) is obtained by drawing a vertical line through A to meet the Wilson line. At C, thesteam condenses suddenly. The line CD represents the condensation of steam atconstant enthalpy. The point D is obtained by drawing a horizontal line through C tomeet the throat pressure (P2) of the nozzle. The line DF represents the isentropicexpansion of the steam in the divergent portion.

3.9.1 EFFECTS OF SUPERSATURATION : The following effects in the nozzle, in which super saturation occurs are,

(i) since the condensation does not take place during super saturation

expansion, so the temperature at which the super saturation occurs will be

less than the saturation temperature corresponding to the pressure.

Therefore, the density of supersaturated steam will be more than for the

equilibrium conditions, which gives the increase in mass of steam

discharged.

(ii) The super saturation increases the entropy and specific volume of the steam.

(iii) The super saturation reduces the heat drop below that for thermal

equilibrium. Hence the exit velocity of the steam is reduced.

(iv) The super saturation increases dryness fraction of the steam.

3.10. STEAM TURBINE

A steam turbine is a prime mover in which rotary motion is obtained bythe gradual change of momentum of the steam.In a steam turbine, the force exerted on theblades is due the velocity of steam.

In general, a steam turbine, essentially, consists of the following two parts:

(i) The nozzle in which the heat energy of high-pressure steam is converted into kinetic energy, so that the steam issues from the nozzle with a very highvelocity.

(ii) The blades, which change the direction of, steam issuing from the nozzle, sothat a force acts on the blades due to change of momentum and propelthem.

Thus, the basic principle of operation of a steam turbine is the generationof high velocity steam jet by the expansion of high-pressure steam and then conversion ofkinetic energy, so obtained into Mechanical work on rotor blades.

3.11. ADVANTAGES OF STEAM TURBINES OVER RECIPROCATING STEAM ENGINES:

Following are the important advantages of steam turbines over reciprocating steamengines:

(i) A steam turbine may develop higher speeds and a greater steam range ispossible.

(ii) The efficiency of a steam turbine is higher.(iii) The steam consumption is less.(iv) Since all the moving parts are enclosed in a casing, the steam turbine is

comparatively safe.(v) A steam turbine requires less space and lighter foundations, as there are little

vibrations.(vi) There is less frictional loss due to fewer sliding parts.

(vii) The applied torque is more uniform to the driven shaft.(viii) A steam turbine requires less attention during running, Moreover, the repair

costs are generally less.

3.12 CLASSIFICATION OF STEAM TURBINES:

The steam turbines may be classified into the following types:

(i) According to the mode of steam action

• Impulse turbine

• Reaction turbine

(ii) According to the direction of steam flow

• Axial flow turbine• Radial flow turbine

(iii) According to exhaust condition of steam

• Condensing turbine• Non condensing turbine

(iv) According to pressure of steam

• High pressure turbine• Medium pressure turbine• Low pressure turbine

(v) According to the number of stages

• Single stage turbine• Multi stage turbine

3.13. IMPULSE TURBINE

An impulse turbine, as the name indicates, is a turbine, which runs by theimpulse of steam jet. In this turbine, the steam is first made to flow through a nozzle.Then the steam jet impinges on the turbine blades (which are curved like buckets) and ismounted on the circumference of the wheel. The steam jet after impinging glides over theconcave surface of the blades and finally leaves the turbine.

3.13.1. VELOCITY TRIANGLES FOR MOVING BLADE OF

AN IMPULSE TURBINE

We have already discussed that in an impulse turbine, the steam jet after leavingthe nozzle impinges on one end of the blade. The jet then glide over the inside surface ofthe blade and finally leaves from other edge, as shown in the figure. It may be noted thatthe jet enters and leaves the blade tangentially for shockless entry and exit.

Consider the steam jet entering a curved blade after leaving the nozzles at C.now let us draw the velocity triangles at inlet and outlet tips of the moving blade, asshown in fig.

Let Vb = Linear velocity of the moving blade m/s V = Absolute velocity of steam entering the moving blade

Vr = Relative velocity of the jet to the moving blade. It is the vectorial difference between Vb and V. Vf = Velocity of flow at entrance of the moving blade. It is the vertical component of V. Vw= Velocity of whirl at entrance of the moving blade. It is the horizontal component of V.

θ = Angle which the relative velocity of jet of the moving blade (V)makes with the direction of motion of the blade.

α = Angle with the direction of motion of the blade at which the jet entersthe blade.

V1, Vr1, Vf1, Vw1,β, φ = Corresponding values at exit of the moving blade.

It may be seen from the above, that the original notations (i.e. V, Vr, Vf andVw) stand for the inlet triangle. The notations with suffix 1 (i.e. V, Vr1, Vf1 and Vw1) standfor the outlet triangle.

It may be noted that as the steam jet enters and leaves the blades without any shock (orin other words tangentially), therefore shape of the blades will be such that Vr and Vr1

will be along the tangents to the blades at inlet and outlet respectively. The angle θ iscalled the blade at inlet and angle φ is the blade at exit of the moving blade.

P VW

VbInlet velocity triangle

α θ Vr Vf

V

C Turbine

blade

Vb

Vr1 Vf1 Outlet velocitytriangle Vr

φ β

Vb1 Vb= Vb1 Vw1

In the figuare PC is the axis of the nozzle, which delivers the steam jet witha high velocity (V) at an angle α with the direction of motion of the blade. The jetimpinges on a series of turbine blades mounted on the runner disc.

The axial component of V, which does no work on the blade, is known asvelocity of flow (Vf). It causes the steam to flow through the turbine and also an axialthrust on the rotor. The tangential component of V is known as velocity of whirl at inlet(Vw). According to combined velocity diagram given below ,the linear velocity or meanvelocity of the blade (i.e.Vb) is represented by AB in magnitude and direction. The lengthAC represents the relative velocity (Vr) of the steam jet with respect to the blade.

The jet now glides over and leaves the blade with relative velocity Vr1,which is represented by AE. The absolute velocity of jet (V1) as it leaves the blade, isrepresented by BE inclined at an angle β with the direction of blade motion. Thetangential component of V1 (represented by BF) is known as velocity of whirl at exit(Vw1). The axial component of V1 (represented by EF) is known as the velocity of flow atexit (Vf1).

3.13.2. COMBINED VELOCITY TRIANGLE FOR MOVING BLADES

C

Vf Vr VE Vr1 V1Vf!

D θ φ α β A BF Vb Vw Vw1

(i) Draw a horizontal line, and cut off AB equal to velocity of blade (Vb) tosome suitable scale.

(ii) Now at B, draw a line BC equal to V at an angle α (i.e. velocity of steam jetat inlet of blade) to the scale.

(iii) Join AC, which represents the relative velocity at inlet (Vf). Now at A, drawa line AE at an angle φ with AB.

(iv) Now with A as center and radius equal to AC, draw an arc meeting the linethrough A at E, such that AC = AE or Vr = Vr1.

(v) Join BE, which represents velocity of jet at exit (V1) to the scale.(vi) From C and E, draw perpendiculars meeting the line AB produced at D and

F respectively.(vii) Now BD and CD represent the velocity of whirl and velocity of flow at inlet

(Vw and Vf) to the scale. Similarly, BF and EF represents the velocity ofwhirl and velocity of flow at outlet (Vw1 and Vf1) to the scale.

3.13.3. POWER PRODUCED BY AN IMPULSE TURBINE

Consider an impulse turbine working under the action of a steam jet. Let us draw acombined velocity triangle, for the impulse turbine as shown in the fig.

Let m = mass of the steam flowing through the turbine in kg/s, (Vw ± Vw1) = change in the velocity of whirl in m/s.

We know that according to the Newton’s second law of motion, force in thedirection of motion of the blades,

Fx = mass rate of steam flowing per second X change in the velocity ofwhirl = m[Vw – ( - Vw1)]

= m [Vw + Vw1] = m x DF N ------------------- (A)

and work done in the direction of motion of the blades = Force × Distance = m [Vw + Vw1 ] V b Nm/s ------------------ (B) = m × DF × AB Nm/s

Power produced by the turbine = work done in the direction of motion of the blades inWatts

Similarly, we can find out the axial thrust on the wheel which is due to the differenceof velocities of flow at inlet and outlet. Mathematically, axial thrust on the wheel.

Fy = Mass rate of steam flowing per second × Change in the velocity offlow = m (Vf - Vf1) = m (CD –EF) N ------------------- (C)

3.13.4. EFFECT OF FRICTION ON THE COMBINED VELOCITY TRIANGLE

In the last article, we have discussed that relative velocity of steam jet is thesame of the inlet and outlet tips of the blade. In other words, we have assumed that theinner side of the curved blade offers no resistance of the steam jet. But in actual practice,some resistance is always offered by the blade surface to the gliding steam jet, whoseeffect is to reduce the relative velocity of the jet i.e to make Vr1 < Vr. the ratio of Vr1 to Vris known as blade velocity co coefficient or coefficient of velocity or frictionfactor,(usually denoted by K). Mathematically, blade velocity coefficient,

K = 1r

r

VV

It may be noted that the effect of friction on the combined velocity triangle will be toreduce the relative velocity at outlet (Vr1) as shown in figure.

C

Vf Vr V E Vr1 V1

Vf!

D θ φ α β

F A B Vb Vw1

Vw

3.13.5. COMBINED VELOCITY DIAGRAM FOR AXIAL DISCHARGE :

For axial discharge V1 = Vf1 and Vw1= 0

Sometimes, the steam leaves the blade at its exit tip at 90° to the direction ofthe blade motion. In such a case, the turbine is said to have an axial discharge. Thecombined velocity diagram for axial discharge is shown in fig. It may be noted that insuch a turbine, the velocity of whirl at outlet (Vw1) is equal to zero. Therefore powerdeveloped by the turbine,

P = m × Vw × Vb Watts (QVw1 = 0)

C

Vf Vr V E Vr1

Vf!= V1

θ φ α D AB,F Vb Vw

β = 90°

3.14. REACTION TURBINES:

In a reaction turbine, the steam enters the wheel under pressure and flows over theblades. The steam, while gliding, peoples and make them to move. As a matter of fact,the turbine runner is rotated by the reactive forces of steam jets.

Parson’s Reaction Turbine: A Parson reaction turbine is the simplest type of reaction steam turbine, and iscommonly used. It has the following main components:

(i) Casing,(ii) Guide mechanism,

(iii) Turbine runner and(iv) Draft tube.

A Parson’s turbine is also known as 50% reaction turbine.

3.14.1. COMPARISON BETWEEN IMPULSE AND REACTION TURBINE:

S.No. Impulse Turbine Reaction turbine

1 The steam flows through the nozzles andimpinges on the moving blades.

The steam first through guide mechanismand then through the moving blades.

2 The steam impinges on the buckets withkinetic energy.

The steam guides over the vanes withpressure and kinetic energy.

3 The steam may or may not be admittedover the whole circumference.

The steam must be admitted over thewhole circumference.

4The steam pressure remains constantduring its flow through the movingblades.

The steam pressure is reduced during itsflow through the moving blades.

5The relative velocity of steam whilegliding over the blades remains constant(assuming no friction).

The relative velocity of steam whilegliding over the blades increases(assuming no friction).

6 The blades are symmetrical. The blades are not symmetrical.

7 The number of stages required are lessfor the same power developed.

The number of stages required are morefor the same power developed.

3.14.2. POWER PRODUCED BY A REACTION TURBINE:

Power Produced By The Turbine:( )1w w bP m V V V= +

Similarly we can find out the axial thrust on the wheel, which is due to differenceof velocities of flow at inlet and outlet. Mathematically, axial thrust,

FY = Mass of steam flowing / second X Change in the velocity of flow( )1f fm V V= +

3.14.3. DEGREE OF REACTION:

In a reaction turbine, the pressure drop takes place in both the fixed and movingblades. In other words, there is an enthalpy drop in both the fixed and moving blades asshown on h-s diagram in the figure. The ratio of the enthalpy or heat drop in the movingblades to the total enthalpy or heat drop in the stage is known as “degree of reaction”.Mathematically,

Degree of reaction 2 3

1 3

h hEnthalpy or heat dropin themoving bladesTotal enthalpy or heat drop inthe stage h h

−= =

−

3.14.4. HEIGHT OF BLADES OF A REACTION TURBINE:

The steam enters the moving blades over the whole circumference. As a result ofthis, the area through which the steam flows is always full of steam. Now consider areaction turbine whose end view of the blade ring is shown in figure.

Let, d = diameter of rotor drum,

h = Height of blades

Vf1 = Velocity of flow at exit.

Therefore the total area available for the steam to flow,( )A d h hπ= +

and volume of steam flowing ( ) 1fd h hVπ= + We know that volume of 1 kg of steam at the given pressure is Vg (from steamtables). Therefore mass of steam flowing,

( ) 1f

g

d h hVm

Vπ +

= kg / s

If the steam has a dryness fraction of x, then the mass of steam flowing,

( ) 1 1f m f

g g

d h hV d hVm

xV xVπ π+

= = kg / s

Where dm = mean blade diameter = (d+h).

3.15. PERFORMANCE OF STEAM TURBINES In the last two chapters, we have discussed impulse and reaction steam turbines. In

these chapters, we have discussed power generated in these turbines. But in this chapter,

we shall discuss their performance i.e. efficiencies and governing.

3.15.1. EFFICIENCES OF STEAM TURBINE

The following efficiencies of impulse as well as reaction steam turbines are

important from the subject point of view:

(i) Diagram or blading efficiency:

It is the ratio of the work done on the blades to the energy supplied to the blades.

Let V= Absolute velocity of inlet steam in m/s, and

m= Mass of steam supplied in kg/s.

∴ Energy supplied to the blade per second,2

2mV

= ---- J/s

We know that

Work done on the blades per second 1( )w w bm V V V= + ---- J/s

∴ Diagram or blading efficiency,

1 12 2

( ) 2( )/ 2

w w b w w bb

m V V V V V VmV V

η+ +

= =

The work done on the turbine blades may also be obtained from the kinetic energyat inlet and exit as discussed below:

Let V1 = Absolute velocity of exit steam in m/s

We know that

Kinetic energy at inlet per second2

2mV

=

and kinetic energy at exit per second2

1

2mV

= J/s

∴Work done on the blades per second = Loss of kinetic energy22

2 211( )

2 2 2mVmV m V V= − = − J/s

and power developed,2 2

1( )2

m V VP −= watts

∴Blading efficiency,2 2

2 211

2 2

( )2

2

b

m V V V VmV V

η− −

= =

(ii) Gross or stage efficiency:

It is the ratio of the work done on the blades per kg of steam to the total energy supplied per stage per kg of steam.

Let, h1= Enthalpy or total heat of steam before expansion through the Nozzle in kJ/kg

and h2 =Enthalpy or total heat of steam after expansion through the Nozzle in kJ/kg

∴Enthalpy or heat drop in the nozzle ring of an impulse wheel hd = h1-h2 (inkJ/kg)

and Total energy supplied per stage = 1000 hd J/kg of steam

We know that

Work done on the blade per kg of steam = 1(Vw +Vw1) Vb J/kg of steam

∴Gross or stage efficiency,

1 1

1 2

( ) ( )1000 1000( )w w b w w b

sd

V V V V V Vh h h

η+ +

= =−

(iii) Nozzle efficiency:

It is the ratio of energy supplied to the blades per kg of steam = V2/2 (in joules)

∴Nozzle efficiency,2 2/ 2

1000 2000nd d

V Vh h

η = =

3.15.3. CONDITION FOR MAXIMUM EFFICIENCY OF AN IMPULSE

TURBINE:

The blading efficiency of an impulse turbine2 2

112 2

2( )w w bb

V V VV VV V

η+−

= =

It may be noted that the blading efficiency will be maximum when V1 is

minimum. From the combined velocity triangle, we see that the value of V1 will be

minimum, when is equal to the 90 . In other words, for maximum, the steam should

leave the turbine blades at right angles to their motion. The modified combined velocity

triangle for maximum efficiency is shown given below. It may also be noted that for

maximum efficiency, Vw1 is zero.

∴For maximum efficiency, substituting Vw1= 0 in the general expression for efficiency

of an impulse turbine, we have

max 2

2 w bV VV

η× ×

=

We know that in De-lavel turbine, = , considering Vr =Vr1

2max 2 2

1 0.520.5 cos

2 0.5 cos

b w w

w w w

V V V

VV V V

V V

α

η α

= =

=× ×

= = =

3.15.4. CONDITION FOR MAXIMUM EFFICIENCY OFA REACTION TURBINE:

Work done by a reaction turbine per Kg of steam

= AB × EF = Vb (EB + AF – AB)

= Vb (V cos α + Vr1 cosϕ –Vb )

We know that in a Parson’s reaction turbine α = ϕ; V = Vr1; V1 = Vr

Work done per Kg of steam = Vb (2V cosα - Vb)

= 22

2cos bb

VV VV V

α × −

22

2

2 cosb bV VVV V

α = −

( )2 22 cosV ρ α ρ= −

We know that kinetic energy supplied to the fixed blade per kg of steam =V2/2

And kinetic energy supplied to the moving blade per kg of

2

2

21 cosb bV VV V

α− +

22(1 2 cos )

2V

ρ ρ α− +

2 2

22

(2 cos )supplied (1 2 cos )

2

workdone VV dyenergy

dx

ρ α ρ

ρ α

−=

− +

Steam2

2

2(2 cos )(1 2 cos )

ρ α ρρ ρ α

−− +2(1 2 cos ) 0d

dρ ρ α

ρ− + =

2 2cos 0ρ α− =

=2 2 2 2

1( ) ( )2 2

rl rV V V V− −=

∴Total energy supplied to the turbine =2 2 2 22

1 122 2 2

V V V VV − −+ =

From the combined velocity triangle, we find that

2 21 2 cosr b bV V V V VV α= = + −

Total energy supplied to the turbine,

=2 2 2 22 cos 2 cos

2 2b b b bV V VV V V VVα α+ − − +

=

=2

2V (

2

2

21 cosb bV VV V

α− + )

=2

2(1 2 cos )2

Vρ ρ α− +

We know that diagram or blading efficiency,

ηb=2 2

22

(2 cos )supplied (1 2 cos )

2

workdone VVenergy

ρ α ρ

ρ ρ α

−=

− +

=2

2

2(2 cos )(1 2 cos )

ρ α ρρ ρ α

−− +

It may be noted that the efficiency of the turbine will be maximum when2(1 2 cos )ρ ρ α− + is minimum. Now for 2(1 2 cos )ρ ρ α− + to be minimum,

differentiate this expression with respect to ρ and equate the same to zero, i.e

2(1 2 cos ) 0dd

ρ ρ αρ

− + =

2 2cos 0ρ α− =

cos cosbVorV

ρ α α= =

cosbV V α∴ =

It may be noted that when cosbV V α∴ = ,the value of Vwl will be zero (a

condition for maximum efficiency of impulse turbine also) as shown in figure. In such a

caseVb = Vw.

Now substituting cosρ α= in equation for maximum efficiency,

( )( )

2 2 2

max 22 2

2 2cos cos 2cos1 cos1 cos 2cos

α α αη

αα α

−= =

+− +

3.16 COMPOUNDING OF IMPULSE STEAM TURBINES

(METHODS OF REDUCING ROTOR SPEEDS):

In the recent years, high pressure (100 to 140 bar) and high temperature steam is

used in the power plants to increase their thermal efficiency. If the entire pressure drop

(from boiler pressure to condenser pressure (say from 125 bar to 1 bar) is carried out in

one stage only, then the velocity of steam entering into the turbine will be extremely

high. It will make the turbine rotor to run at a very high speed (even up to 30000 rpm).

Form practical point of view, such a high speed of the turbine rotor is bound to have a

number of disadvantages.

In order to reduce the rotor speed, various methods are employed. All of these

methods consists of a multiple system of rotors, in series, keyed to a common shaft and

the steam pressure or the jet velocity is absorbed in stages as it flows over the rotor

blades. This process is known as compounding. The following three methods are

commonly employed for reducing the rotor speed:

1. Velocity compounding, 2. Pressure compounding and 3. pressure - velocity

compounding

3.17. GOVERNING OF STEAM TURBINE:

The process of providing any arrangement, which will keep the speed constant

(according to the changing load conditions) is known as governing of steam engines.

Though there are many methods of governing steam turbines, yet the throttle governing is

important from the subject point of view.

3.17.1. THROTTLE GOVERNING OF STEAM TURBINES:

The throttle governing of a steam turbine is a method of controlling the turbine

output by varying the quantity of steam entering into the turbine. This method is also

known as servometer method, whose operation is given below:

The centrifugal governor is driven form the main shaft of turbine by belt or gear

arrangement. The control valve controls the direction of flow of the oil (which is pumped

by gear pump) either in the pipe AA or BB. The servometer or relay valve has a piston

whose motion (towards left or right depends upon the pressure of the oil flowing through

the pipes AA or BB) is connected to a spear or needle which moves inside the nozzle, as

shown in the figure.

We know that when the turbine is running at its normal speed, the positions of

piston in the servo meter, control valve, fly balls of centrifugal governor will be in their

normal positions as shown in the figure. The oil is pumped by the gear pump into the

control valve, will come back into the oil sump as the mouths of both pipes AA or BB are

closed by the tow wings of control valve.

Now let us consider an instant, when load on the turbine increases. As a result of

load increase, the turbine speed will be decreased. This decrease in the speed of the

turbine wheel will also decrease the speed of the centrifugal governor. As a result of this,

the fly ball will come down (due to decrease in centrifugal force) thus decreasing their

amplitude. As the sleeve is connected to the central vertical bar of the centrifugal

governor, therefore coming down of the fly balls will also bring down the sleeve. This

down ward moment of the sleeve will raise the control valve rod, as the sleeve is

connected to the control valve rod through a lever pivoted on the fulcrum. Now, a slight

upward movement of the control valve rod will open the mouth of the pipe AA (still

keeping the mouth of the pipe BB closed). Now the oil under pressure will rush from the

control valve to the right side of the piston in the servo meter through the pipe AA. This

oil, under pressure, will move the piston as well as spear towards left, which will open

more area of the nozzle. This increase in the area of flow will increase the rate of steam

of steam flow into the turbine. As a result of increase in the steam flow, there will be an

increase in increase in the turbine output as well as its speed. When speed of the turbine

wheel will come up to its normal range, fly balls will move up. Now the sleeve as well as

control valve rod will occupy its normal position and the turbine will run at its normal

speed.

It may be noted that when load on the turbine decreases, it will increase the speed.

As a result of this, the fly balls will go up (due to increase in centrifugal force) and the

sleeve will also go up. This will push the control valve rod downwards. This downward

movement of the control valve rod will open the mouth of the pipe BB (still keeping the

mouth of the pipe AA closed). Now the oil under pressure, will rush form the control

valve to the left side of the piston in the servo meter through the pipe BB. This oil under

pressure will move the piston and spear towards right, which will decrease the area of the

nozzle. This decrease in the area of the flow will decrease the rate of steam flow into the

turbine. As a result of decrease in the steam flow, there will be a decrease in the turbine

output as well as its speed. When the speed of the turbine is reduced to its normal range,

the fly balls will come down. Now the sleeve as well as control valve rod will occupy its

normal positions, and the turbine will run its normal speed.