Energy Transfer in Turbomachines

Energy transfer in Turbo machines

Configuration of a basic turbine

Relative motion 1D

No wind

U

W = - U

Tail wind

Head wind

U

V

W = 0

W = V + ( – U) (vectorial addition)

U

V

U is the “frame velocity”

V is the “absolute velocity” or the velocity that an observer experiences.

W is the “relative velocity” or the velocity experienced by the walker.

Relative motion 2D

W = U

Velocity Triangles for an Aircraft Landing

Note : Absolute velocity is the vector sum of the frame velocity and the relative velocity. V = U + W

Graphical addition and subtraction of vectors

• To add two vectors A + B graphically : Place them nose – to – tail and the result is given by movement from the tail of the first to the nose of the second.

• To subtract two vectors A - B graphically : Reverse the direction of B and proceed with addition of vectors as before.

1

2

3

1 2 3

x

Stator

Cascade and Meridional Views of a Turbine Stage

Flow through turbomachines

1

2

3

Velocity triangle at 2

Velocity triangle at 3

Velocity Triangles for a Turbine Stage

Energy transfer in turbomachines • According to Newton’s second law of motion, the sum of all the forces acting on a control volume in a particular direction is equal to the rate of change of linear momentum of the fluid across the

control volume.

• That is, 1212 VVm

dt

VVmF

12 VVmdtF

or,

Where,

m = mass of the body (kg)

V1 = initial velocity of the fluid (m/s)

V2 = final velocity of the fluid (m/s)

This equation is a modified form of Newton’s second law of motion and is known as Impulse Momentum Equation

Impulse acting on the body Change in momentum of the

body in the time period dt

Energy transfer in turbomachines • The impulse momentum equation is used to study the impact of fluid jet striking a stationary or moving plate and also to study general fluid flow characteristics.

• When the flowing fluid with initial velocity V1 is obstructed by a surface such as vane, blade etc., the fluid undergoes a change in momentum. The impulsive force acting on the fluid by the surface is:

• According to Newton’s third law of motion – for every action there is equal and opposite reaction. Therefore, the fluid reacts to this and exerts equal and opposite force on the obstructing surface, given by

• Similarly, the sum of all torques acting on the system is equal to the rate of change of angular momentum.

12 VVmF

21 VVmF

Energy transfer components • Fig. shows rotor of a generalized turbomachine. o-o is the axis of the shaft which rotates with an angular velocity ω.

• The fluid enters the rotor at 1 and leaves at 2 after passing through the rotor by any path. The angle of entry and exit may be arbitrary.

Energy transfer components • Let V be the absolute velocity of the fluid entering the rotor at 1 at any angle. This velocity vector may be resolved into three mutually perpendicular components:

• Axial component Va

• Radial component Vr

• Tangential component Vw

• Axial component : This is parallel to the axis of rotation. Axial force is produced due to change in magnitude of this component. This axial force is taken by the thrust bearing of t he machine which is finally transferred to the housing.

• Radial component : This is parallel to the radius of the rotor. Radial force is produced due to change in the magnitude of this component. Radial forces are taken by journal bearings.

• It should be noted that no torque is exerted on the rotor by these two forces, i.e., axial and radial.

Energy transfer components • Tangential component : The torque is exerted on the rotor only due to the change in the angular momentum of the tangential component.

• Assumptions:

• Fluid enters and leaves the vane in a direction tangential to the vane tip at inlet and outlet.

• There is no frictional resistance as the fluid flows over the vane.

• Let

V = absolute velocity of fluid (m/s)

N = speed pf rotation of the rotor (rpm)

r = radius of the rotor (m)

ω = angular velocity of the rotor (rad/s) = 2πN/60

u = linear velocity of vane tip (peripheral velocity) (m/s) = πdN/60

= mass flow rate of fluid (kg/s)

d = rotor diameter (m)

m

Euler turbine equation

• Tangential momentum of fluid at entry =

• Angular momentum (moment of momentum) at entry =

• Angular momentum at outlet =

• T = torque on the rotor = change of angular momentum

• Work done = rate of energy transferred = T x ω

• But we know that ω1r1 = u1 and ω2r2 = u2. Therefore,

• Work done per unit mass flow rate

mVw1

11 rmVw

22 rmVw

mrVrV ww 2211

mrVrV ww 2211

muVuVDW ww 2211..

2211 uVuV ww

----- (1)

----- (2)

Euler turbine equation

• Equations (1) and (2) are two forms of Euler turbine equation or Euler equation.

• Euler equation applies to all turbomachines – pumps, fans, blowers, compressors and turbines (steam, gas, water).

• If Vw1u1 > Vw2u2, the RHS of equation (2) is positive and then the machine is called turbine.

• If Vw2u2 > Vw1u1, the RHS of equation (2) is negative and then the machine is called pump, fan, blower or compressor.

• If Vw1u1 > Vw2u2 and Vw2 is negative, i.e., Vw2 is opposite to that of Vw1, then equation (2) can be written as:

22112211/ uvuvuvuvrateflowmassunitdoneWork wwww

Alternate form of Euler turbine equation

u2

w2

vw2

v2 vr2

vr1

u1

vw1

w1v1

β2

β1

α2

α1

Inlet velocity triangle

Outlet velocity triangle


Let

• v = absolute velocity of fluid

• w = relative velocity of fluid (relative to the rotor)

• vr = radial component of absolute velocity (flow velocity)

• vw = tangential component of absolute velocity

• Suffixes 1 and 2 indicate inlet and outlet conditions respectively

From inlet velocity triangle,

2111

21

21

21

21 2 www vvuuwvv

2111

21

21

211

21

21 2 wwwr vvuuwvuwv

-------- (3)

-------- (4)Also,

Equating equations (3) and (4), we get

21

21

21 wr vvv

• On simplification we get,

• Similarly,

• Substituting these values in the Euler’s equation


2

21

21

21

11

wuvvu w

2

22

22

22

22

wuvvu w

2211/. uvuvrateflowmassunitDW ww we get,

22

22

22

22

21

21

21 wuvwuv

rateflowMass

doneWork


2

21

22

22

21

22

21 wwuuvv

rateflowMass

doneWork

Components of energy transfer

• First component (v12-v2

2)/2 is the change in absolute kinetic energy. Due to this, a change in dynamic head or pressure takes place through the machine. The exit kinetic energy will be more in power absorbing machines(e.g., pump) while it will be less in power producing machines (e.g., turbine)

• Second component (u12-u2

2)/2 is the change in centrifugal energy of the fluid due to change in radius of rotation. This causes a change in static head or pressure through the rotor.

• Third component (w12-w2

2)/2 is the change in relative kinetic energy due to change in relative velocity. This also causes a change in static head or pressure through the rotor.

This is an alternate form of Euler turbine equation


Radially outward and inward flow machines

• For radially outward flow machines, u2 > u1 and hence the fluid gains in static head, while, for a radially inward flow machine,

u2 < u1 and the fluid loses its static head. Therefore, in radial f low pumps or compressors the flow is always directed radially outward, and in a radial flow turbine it is directed radially inward.

Energy transfer in axial machines

• For an axial flow machine, the main direction of flow is parallel to the axis of the rotor, and hence the inlet and outlet points of

the flow do not vary in their radial locations from the axis of rotation. Therefore, u1=u2 and equation of energy transfer will be

2

21

22

22

21 wwvv

rateflowMass

doneWork

Conventions in drawing velocity triangles

βα

V or C W or Vr

u or U

Vw or Cu

Vf o

r V

r or

Cm

u or U : Peripheral velocity, πDN/60V or C : Absolute velocityW or Vr : Relative velocityVf or Vr or Cm : Component of absolute velocity in the flow directionVw or Cu : Component of absolute velocity in the tangential directionα : Angle between absolute velocity and peripheral velocity vectorsβ : Angle between relative velocity and peripheral velocity vectors

Steady flow equation – 1st law of thermodynamics

•The steady flow equation of the 1st law of thermodynamics is

where,

Q = rate of heat transfer

W.D. = Work output

V2/2 = kinetic energy

Z = potential energy

Suffixes 1 and 2 refer to inlet and outlet values respectively

2

22

21

21

1 2..

2Z

VhmDWZ

VhmQ


• If h0 = stagnation or total enthalpy and Δh0 = change in total enthalpy, then

• For isentropic process, q = 0 and hence Δh0 = -w or in differential form, -dh0 = w. Therefore,

ZV

hh 2

2

0

0

00102

..

hwq

hhhm

DW

m

Q

that is,

or,

2

21

22

22

21

22

21

0

wwuuvvwdh


P1 = Static condition

P01 = Stagnation condition

P2 = Static condition

S

T or h

Isentropic

Adiabatic

1

02’02

2

Wa

Wst

WiE

WiE > Wst > Wa

• W.D. per unit mass flow rate is known as ideal Euler work (WiE).

• If the flow is not perfect and reversible, the work done is known

as stage work (Wst) or isentropic work.

• If the flow is not perfect and the process is irreversible, then the work done is called the actual work (Wa) or adiabatic work.

• The pressure drop during stage work and adiabatic work is the same

Euler, stage (isentropic), and actual (adiabatic) work on T -s or h-s diagram

Impulse and Reaction

• In general, turbomachines can be classified into the impulse type and the reaction type depending upon the type of energy change that occurs in the rotor blades.

• An impulse stage is one in which the static pressure at inlet and outlet of the rotor remains the same (ΔP=0 in the rotor). This also means that the relative velocity of fluid flow is constant in the rotor.

• A reaction stage is one where static pressure changes during flow of fluid in the rotor (ΔP>0 in the rotor).

• The degree of reaction is a parameter that describes the relation between the energy transfer due to static pressure change and the energy transfer due to dynamic pressure change.

Degree of Reaction

• The degree of reaction (R) is defined as the ratio of energy transfer by static pressure in the rotor to the total energy transfer in the rotor.

or,

• For axial flow machines, u1 = u2 and hence,

212

222

21

22

21

21

22

22

21

wwuuvv

wwuuR

021

21

hh

hh

changeenthalpyTotal

changeenthalpyStaticR

o

212

222

21

21

22

wwvv

wwR

Static energy

Total energy

Degree of Reaction

•The value of R may be zero, negative or positive in a turbomachine

• If there is no change in the static pressure in the rotor and u1 = u2, then such a machine is called an impulse type of machine. Therefore, for impulse type machine, R = 0.

• In an impulse type machine, if the fluid enters and leaves the rotor at different radii, a change of static pressure occurs in one

direction. An equal amount of change in static pressure occurs in the opposite direction to render Δp=0.

• In an impulse type machine (i.e., zero degree of reaction), the rotor can be of open type, i.e., an open jet of fluid with no connection with the rotor.

• A machine with any degree of reaction (R = 0) must have the rotor enclosed in order avoid expansion of fluid in all directions. Such machines are called reaction type machines.

General analysis of a turbomachineEffect of blade outlet angle β2 on energy transfer

•The blade outlet angle β2 in a radial machine significantly affects the work done and the degree of reaction

• Its effect can be studied by making the following assumptions:

• Centrifugal effect at outlet = 2 x centrifugal effect at inlet (u2 = 2 u1)

• Radial velocity (flow velocity) is constant (Vf1 = Vf2 = Vf )

• No tangential component at inlet (Vw1= 0; α1= 90˚; Vf1= V1)

• Inlet blade angle (and hence fluid angle) is 45˚ (u1= V1= Vf)

• Outlet blade angle β2 is variable

• From equation for work done,

222211/.. uVuVuVrateflowmassunitDW www

• Therefore it follows that

g

uVH w 22

βα

V W

u

Vw

Vf

2222 CotVug

uH f

QKKH

QgA

Cotu

g

uH

A

CotQu

g

uH

A

QVorVAQQ ff

21

2

2222

2

22

2

22222 ;

Considering rotor operating at a given speed and putting

2

222

22

1

gA

CotuK

g

uK

Effect of blade outlet angle β2 on energy transfer

• For a given pump or a compressor u, A and β2 are fixed and the only variables are H and Q. Centrifugal pumps and compressors can be classified as under:

• Backward curved blades β2 < 90°

• Radial blades β2 = 90°

• Forward curved blades β2 > 90°

• For backward curved blades:

• β2 < 90° (and α1 = 90°, Vw1 = 0, Vf1 = V1 as assumed)

• Hence,

• K2 is positive

• H-Q line has negative slope

• Outlet tip of the blade is in the direction opposite to that of rotation.

• Flow and wheel rotation are in the same direction.


• For radial blades:

• β2 = 90°, Vw2 = u2, W2 = Vf2 (and α1 = 90°, Vw1 = 0, Vf1 = V1 as assumed)

• Hence,

• K2 = 0, H = K1 = = Constant.

• Head is constant for all flow rates.


• Outlet tip of the blade is in the radial direction.

g

u22


• For forward curved blades:

• β2 > 90° (and α1 = 90°, Vw1 = 0, Vf1 = V1 as assumed)

• Hence,

• K2 is negative.

• H-Q line has positive slope


• Outlet tip of the blade is in the direction of rotation.


β < 90˚Backward curved blade

β = 90˚Straight radial blade

β > 90˚Forward curved blade

β < 90˚

β = 90˚

β > 90˚

Backward

Radial

Forward

Flow, Q

He

ad

, H

• It can be seen that the tangential component Vw2 is least for blade with β2 < 90˚ and maximum with β2 > 90˚.

• Very high V2 is not preferred due to requirement of large diffusers for pressure recovery.

• Practically β2 > 90˚ is not preferred• Radial blade with β2 = 90˚ is used for applications requiring high

pressures.


General analysis – power absorbing centrifugal machines

02

22222.. h

u

VuVuDW w

w

11211 tan uVVV ff

βα

V W

u

Vw

Vf 22

22

w

f

Vu

VTan

2

22

w

f

V

VTan

• Work done in an adiabatic process, assuming α1=90˚, Vw1=0, V1=Vf1

• For constant flow velocity,

• From exit velocity triangle,

This is also known as “stage work”

2

2

22

2

2

2

222

22

2

u

V

VuV

VV

VuV

TanTan

Tan w

w

f

w

f

wf

Substituting in the equation for W.D.,

22

222

2

22222..

TanTan

Tanu

u

VuVuDW w

w

22

222

2200..

TanTan

TanumP

VumTCmhmDWmP wp

The power absorbed therefore, will be



•Degree of reaction

• From inlet velocity triangle,

2

....,

222..

22.

21

22

21

22

22

21

21

22

22

21

21

22

VVDWDWor

VVwwuuDW

wwuuDW

statictotal

total

static

total

static

p

p

stagetheinrisepressure

rotortheinrisepressureR

u1

W1Vf1=V1

1 1

21

21

21 Vuw


• We know that the static pressure rise (Δp)static through the impeller is due to the change in centrifugal energy and the diffusion of the

relative flow.

• Similarly, the total pressure rise through the impeller is,

22..

22

21

21

22 wwuu

DWp staticstatic

2

,222

..

21

22

21

22

22

21

21

22

VVpp

orVVwwuu

DWp

statictotal

totaltotal

• For the assumption of α1 = 90˚ and V1 = Vf1 = Vf2 we have, from inlet velocity triangle,

• We know that the static pressure change can be written as

• Substituting for W12 from equation (a) and simplifying, we get

• From exit velocity triangle for β2 > 90°, we get


21

21

21 Vuw

22

22

21

21

22 wwuu

p static

…………..(a)

2

21

22

22 f

static

Vwup (Since V1 = Vf1)

W2 Vf2

22

V2

VW2

u2 22222

22 uVVw wf

…………..(b)


• Simplifying, we get

• Substituting in the equation (b) of the previous slide,

• Substituting in the equation for degree of reaction, R

2222

22

22

22 2 wwf VVuVwu

2

2 2222 ww

static

VVup

2

2

22

222

22

2222

21

,,2

21

2

2

u

VR

oruV

VuV

Vu

VVuR

w

w

ww

w

ww


• Degree of reaction for different types of blades:

• For backward curved blades (β2 < 90°)

• therefore, R is always less than 1

• For radial blades (β2 = 90°)

• Vw2 = u2. Therefore, R = 0.5

• For forward curved blades (β2 > 90°)

• Vw2 > u2. Therefore, R < 0.5

12

2 u

Vw

Efficiencies

•The concept of efficiency of any machine comes from the consideration of energy transfer and defined as the ratio of useful energy delivered to the energy supplied.

• Two efficiencies are considered for fluid machines –

• Hydraulic efficiency (or isentropic efficiency) which relates energy transfer between fluid and rotor.

• Overall efficiency which relates energy transfer between fluid and shaft.

• The difference between the two efficiencies represents the energy absorbed by bearings, glands, couplings etc. or, in general,

energy loss that occurs between the rotor and the point of actual power input or output.

Efficiencies•Efficiencies for a pump or a compressor:

• Efficiencies for a turbine:

• The ratio of rotor and shaft efficiency is represented by mechanical efficiency, ηm.

rotor

fluidhyd W

W

rotortodeliveredenergymechanical

outletatfluidtheinenergyuseful

shaft

fluidoverall W

W

shafttodeliveredenergymechanical

outletatfluidtheinenergyuseful

fluid

shaftoverall W

W

fluidthefromavailableEnergy

shaftoutputatenergymechanical

fluid

rotorhyd W

W

fluidthefromavailableEnergy

rotorthebydeliveredenergymechanical

hyd

overallmechanical

General analysis – Turbines

• Impulse type turbines have only the kinetic energy available at inlet of the machine for the production of power or energy

transformation. That means, the static pressure at inlet and outlet of the machine remains the same. Hence, W1= W2. E.g., Pelton wheel.

• Reaction turbines are those in which in addition to the kinetic energy of the fluid at inlet, pressure energy is also available in course of energy transformation. This implies that there is a change of static pressure during the flow over each rotor stage. E.g., Lawn sprinkler or Parson’s turbine.

• Turbines run on compressible fluids (e.g., steam / gas turbines) and incompressible fluids (e.g., hydraulic turbines).

• Turbines must have a residual exit velocity to maintain flow. Even if we have idealized frictionless flow, it is not possible to transfer all

the energy in the fluid due to the need to have a minimum exit velocity.

General analysis – Turbines

Impulse Turbine

Reaction Turbine

General analysis of Turbines – Utilization factor

•The hydraulic efficiency (or isentropic efficiency) of a turbine is a product of two terms and is given by,

where, ηv is the vane efficiency and takes care of frictional losses, and Є is the utilization factor.

• The utilization factor Є is defined as the ratio of the actual work transferred from the fluid to the rotor in an ideal condition to the maximum possible work that could be transferred in an ideal condition.

vhyd

222

122

21

21

22

21

22

21

22

21

max WWuuV

WWuuVV

W

Wactual

2

22

max

VWW actual


•We also have, work done (W) as per Euler’s equation,

• Substituting we get,

• Similarly, Є can also be expressed in terms of degree of reaction, R

2211 uVuVW wwactual

2

22

2211

2211

VuVuV

uVuV

ww

ww

dynstatic

static

HH

HR

)1(

)1(

R

RHH

RHRH

HRHRH

HHHR

ds

sd

ssd

ssd

therefore,


2

212

221

22

21

22

21

22

21

22

21

max VWWuu

WWuuVV

W

Wactual

HdynHstatic

Hstatic

• Substituting for Hstatic,

RVRH

H

RV

RH

RHHRH

VR

RH

HR

RH

dyn

dyn

dyn

dyndyndyn

dyn

dyndyn

12

122)1(

)1(21

21

21


•Writing Hdyn in terms of V1 and V2,

RVVVR

VV

RV

RH

H

dyn

dyn

1

22

2

12

21

22

21

22

21

21

22

21

22

21

22

21

22

21

212

221

22

21

1 RVVRVRV

VV

VVRV

VVR

VV

22

21

22

21

RVV

VV

This expression holds good for 0 < R < 1 but not for R=1 because then the utilization factor Є=1 indicating 100% utilization with the result that the exit velocity V2 becomes zero.

Axial flow Turbines

• In axial flow machines, the fluid enters and leaves the rotor at the same radius and hence u1= u2.

•The axial flow velocity Vf is assumed to be constant from inlet to outlet.

• With u1= u2 the equation for degree of reaction becomes:

• From this equation, the different values of R can be obtained depending on the magnitude of velocity components.

actualW

WW

WWVV

WWR

21

22

21

22

22

21

21

22

Axial flow Turbines

•When R < 0 (negative reaction)

If R is negative, W1 should be greater than W2. In this case, even though R is negative, the energy transfer, Wactual is positive.

uu

W2

W1

V1

V2

121 2

Velocity triangle for R < 0

•When R = 0 (Impulse type)

If R is 0, W1= W2 and hence 1 = 2.. In this case, there is no change in static pressure across the rotor and the energy transformation occurs purely due to the change in absolute kinetic energy (V1

2 – V22)/2.

uu

V1

V2

121

2

Velocity triangle for R = 0W2

W1

Axial flow Turbines

•When R = 0.5 (50% reaction)

If R is 0.5, V12- V2

2 = Vr22-Vr1

2 and hence V1= Vr2 and V2= Vr1. In this case, 50% energy transformation occurs in the rotor and the other 50% in the stator.

uu

W2

W1

V1

V2

1 212 Velocity triangle for R = 0.5

•When R = 1 (Fully reaction)

If R is 1, V1 = V2. In this case, the energy transformation occurs purely due to change in relative kinetic energy of fluid.

uu

V1V2

1 212 Velocity triangle for R = 1

W1

W2

Axial flow Turbines

•When R > 1

If R is > 1, V2 > V1. In this case, the energy transformation can be negative or positive.

Velocity triangle for R > 1uu

V1

V2

121

2

•Maximum utilization factor

For maximum utilization, the value of V2 should be minimum and this is possible when V2 is axial.

uu

V1

V2

121

2 Velocity triangle for maximum utilization (max)

W2

W1

W1

W2

•We have the utilization factor

• From velocity triangle for maximum utilization factor max,V2 = V1sin1.

Therefore,

• This shows that the utilization factor is maximum when 1 = 0. Then V2 = V1sin1 = 0 which is a “zero angle turbine” that is impossible to attain.

Condition for maximum utilization

22

21

22

21

RVV

VV

1

21

2

122

1

122

1max

122

121

122

121

max

sin1

cos

sin1

sin1

sin

sin

RRV

V

RVV

VV

Condition for maximum utilization-Impulse Turbine

D

•Condition for max in impulse turbineFor impulse turbine, R=0 (and Vr1=Vr2). Substituting in the equation for max,

•Triangles OBC and OCD are similar. Hence

BC = u. Thus for max the impulse stage must have cosα1=(u+u)/V1=2u/V1.

•But (u/V1)=φ=speed ratio=cosα1/2

•For zero angle (α1=0) turbine, the speed ratio (u/v)=0.5

12

max

121

2

max

cos,0

sin1

cos

RFor

R

uu

V1

V2

121

2

A B C

O

W2

W1

Condition for maximum utilization-Turbine with 50% reaction

u

V1

V2

121

2

u

•When R = 0.5, V1 = Vr2 and V2 = Vr1 and hence α1 = β2 and α2 = β1. For maximum utilization,

V2 must be axial.

• From velocity triangle,

112 sinVV

121

2

max

121

2

121

2

max

122

121

122

121

22

21

22

21

max

sin5.01

cos

,5.0

sin1

cos

sin1

sin1

sin

sin

Rfor

RR

VRV

VV

RVV

VV

11

cosV

uratioSpeed

Also, for 50% reaction turbine,

W2

W1

Comparison of energy transfer between Impulse and Reaction turbines

ur

V1

V2

121

2

uruiui

V1

V2

121

2

Velocity triangle for max – Impulse Turbine Velocity triangle for max – 50% Reaction Turbine

Case (1): When both have the same blade speed

• Let ui and ur be the blade speed of impulse turbine and 50% reaction turbine.

• Energy transfer by impulse turbine is given by

• From velocity triangle for impulse turbine, Vw1 = 2ui. Hence,

1

212211

..

..

wiimpulse

wwiwwimpulse

VuDW

VVuuVuVDW

22.. iimpulse uDW

W2W2

W1W1


•Energy transfer by the 50% reaction turbine is given by:

• From velocity triangle for 50% reaction turbine, Vw1 = ur

By comparing W.D.impulse and W.D.0.5 reaction we note that the energy transfer per unit mass of fluid in Impulse turbine is twice that of 50% reaction turbine for the same blade speed when utilization factor is maximum.

Case (2): When both have same energy transfer

• For the same amount of energy transfer, Er = Ei

i.e.,

15.0.. wrR VuDW

25.0.. rrrR uuuDW

iir

ir

uuuor

uu

414.12,

2

2

22

For the same amount of energy transfer under maximum utilization condition, the peripheral speed of a 50% reaction turbine should be 1.414 times that of an impulse turbine


Case (3): When V1 and α1 are the same in both the machines

• Speed ratio for impulse stage for maximum utilization is:

• Speed ratio for 50% reaction stage for maximum utilization is:

11

1

1

cos2,

2

cos

Vuor

V

u

i

i

ir

r

r

uu

Therefore

Vuor

V

u

2

,

cos,

cos

11

11

When V1 and α1 are the same, when operating under maximum utilization condition, the rotational speed for 50% reaction turbine should be double that of impulse turbine.

11

1

1

1111

1111

11

1

11

211111111max

1

1

111

11

tan2

2cotcot

1tan

sinsincos

1

cos

sintan

22.cos..

2

cos

2cos

Vu

VVuV

V

uV

V

uuuVuVuDW

V

u

V

u

V

uu

V

V

w

f

w

opt

w

Optimum blade speed ratio (Φopt) for different types of turbines for max. energy transfer (W.D.)max

• Impulse Turbine:

For max. utilization, AB = BC = u

V1

V2

121

2

A B C

O

Du u

W1

W2

Optimum blade speed ratio (Φopt) for different types of turbines for max. energy transfer (W.D.)max

• 50% Reaction Turbine:

For max. utilization,

ur

V1

V2

121

2

ur

21111

2211

1

1

1

11

..

..

cos

uVuVuDW

VuVuDW

V

u

V

V

w

ww

optw

W1

W2

Energy Transfer in Turbomachines

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