turbines energy

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1. The energy challenge

Heat

Q= m CpT(or m CVT) = mass specific heat temperature difference

Potential energy

Ep= m g h = mass gravitational acceleration height.

Kinetic energy

Ek= m v2 = half mass velocity squared.

Work

W= F d= force distance

Pressure work

Wp= p V= pressure Volume

This is the same as applying on an area,A, a pressure force, F= pA, and moving it by

a distance d. This will change the volume within the boundary by dA:

W= F d= p A d= p V

Electrical energy

Eel= V Q= volt charge.

Radiation

energy of a photon: h= h c/ ,

with Planck's constant, h= 6.625 10 34 J s, and the frequency, ,or the speed of

light, c, and wavelength, .

'Chemical' energy

The heat released in a chemical reaction. This is specific to each reaction and is

usually given as energy unit mass (e.g. kJ/kg) or number of molecules (e.g. kJ/mol)

Atomic energy

E= m c2 = mass speed of light squared, with c= 3 108 m/s

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6Fluid machines

Introduction

The aim of this section is to provide the very basic principles of the common types of engines

and turbines which exploit fluid motion or pressure. To provide the required basic fluid dynamic

concepts, the basic processes, such as Bernoullis equation are briefly revised. These are then

applied to the range of engines and turbines from reaction turbines to wind turbines.

Read also the sixth section of the online Basic Notes.

Once you have read this section and the online notes, you are ready to complete the fourth set

of online exercises, which is based on the material covered here and in the previous section.

1 Definition of fluid machines

This is an introduction to extracting work from a fluid flowing through a turbine. This

could be a reaction turbine in a hydropower station, driven by a pressure drop across

the turbine, or it could be a wind turbine extracting kinetic energy from the wind.

Turbines and engines are machines which extract energy or power

from a stream of fluid and convert it into mechanical energy

(and then possibly into other forms, such as electricity)

Pumps, fans, and compressors are machines which use mechanical

energy or power to increase the pressure or kinetic energy of a

fluid.

Design parameters for pumps and fans:

1. Desired flow rate

2. Associated head loss

3. Power requirement

Design parameters for compressors::

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6. Fluid machines

1. Desired pressure increase

2. Associated flow rate

3. Power requirement

Design parameters for turbines and engines:

Either

1. Desired power generation

2. Available head

3. required flow rate

Or, for wind turbines and similar,

1. Desired power generation

2. Available flow rate

Important quantities

1. Head or available head: hydrostatic pressure at machine inlet/outlet: p= gH .

When considering a turbine beneath a reservoir, the available head at the turbineinlet is the height of the water level above the turbine minus the head loss in the

penstock feeding the water from the reservoir to the turbine (and possibly minus

residual head required at the turbine outlet for the water to clear the turbine).

2. Flow rate, Q.

3. Power:

this can be either the hydraulic power in cases where the static head is exploited:

p= gH,

or the power carried in an open stream, of velocity U, through the cross-section,A,

of the turbine facing the stream (in the case of a wind turbine, this is the circle

swept by the rotor blades of diameter D): P= AU3

.

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6. Fluid machines

2 Basic Fluid Mechanic principles

This section quickly reviews the fundamental basics of fluid mechanics, based on the

conservation of mass, momentum, and energy.

2.a The control volume

If we want to be able to describe the forces and mass balances we need to define a

volume over which we do this. This volume of our choice is called the control volume,

abbreviated CV, and the surface which encloses this volume is called the control surface

(CS).

2.b Continuity equation

The continuity equation states that mass is conserved. In an incompressible fluid, such as

water, this is equally to the statement that the volume or volume flow rate is conserved.

In other words: What comes in has to come out

in Qin = out Qout

1 Q1 = 2 Q2

If fluid is incompressible (all liquids), 1= 2, and

Q1 = Q2

U1 A1 = U2 A2 ,

where Q is the volume flow rate,A the cross-sectional area through which the fluid

flows, and Uthe average velocity through that cross-sectional area.

2.c Momentum equation

Here, we have to remember that the best-known form of Newtons second law,

F= ma, is a simplification of the proper definition as The change of momentum is equal

to the applied forces',

F=d(mv)/dt. To apply this to a fluid, we have to rephrased it slightly if we look at a

position through which fluid is flowing, instead of looking at a solid object with mass m:

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6. Fluid machines

The Change of momentum in our control volume is equal to the net gain of

momentum by momentum flowing into the control volume plus any forces

applied to the fluid within the control volume or at its surfaces.

By convention, one calculates the momentum flowing out of the control volume. This

gives a minus sign which disappears if we move that term to the left of the 'i is equal'.

Also, the forces can be divided into those which apply throughout the bulk of the

volume, 'body forces' (e.g. gravity), and those which contribute only at the outer surfaces

of the control volume (They are described by a stress tensor which includes forces due

to pressure and stress). With this, the momentum equation for a control volume can

be rephrased as

Rate of Change of Momentum + Net loss of Momentum

= Body Forces + Surface Forces .

You could also visualise the different meanings of the terms by realising that the

momentum carried by the fluid is carried by the mass flow rate = rate of change ofmass:

F=d(mv)/dt.= m dv/dt+ v dm/dt= m a+ v m= m a+ v Q= m a+ AU2.

Remember that this is more of an illustration than a derivation. In particular, any force

applied is given by the difference between the value ofv Q at the inlet and the outlet.

If we are looking at a steady-state flow, then there will be no acceleration. If we

furthermore only consider a case where the only velocity component is that in the

direction of the mass flow rate, then the force becomes

F= 1A1U12

2A2U22.

The most important forces of all is the pressure force, F= pA.

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6. Fluid machines

2.d Energy, energy per unit volume, and head

The most important forms of energy for fluid machines are

1. Potential energy: m g z

2. kinetic energy: m U2

3. flow work or pressure energy: p V.

(all having the base unit: 1joule= 1J= 1 kg m2

s 2)

In fluid mechanics one often uses energy per unit volume: 1joule perm3= 1 kg m1 s2.

Note that this has the same units as pressure: 1 pascal= 1 Pa= 1 kg m1 s2.

1. Potential energy:g z

2. kinetic energy: U2

3. flow work or pressure energy: p

Using the hydrostatic pressure equation, p= g H, this pressure is equivalent to a

stationary column of fluid above the point you are looking at with a height ofH.

As hydropower is usually associated with a reservoir a certain elevation above a

turbine house, and because pressure is most easily measured by the height of a fluid

column in a manometer, one often converts these into quantities with dimension (m)

by dividing each term by g, and refers to them as head:

1. Elevation: z

2. dynamic head:2

21 U

3. static head:g

pH

=

2.e Bernoulli's equation

Bernoulli's equation is the statement of the first law of Thermodynamics that energy is

conserved. It considers the energy balance at two points along a path that a fluid takes,

a stream line.

In terms of energy per volume (unit pascal: 1 J/m3= kg m1 s2= Nm= Pa):

222122212111 UgzpUgzp ++=++

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6. Fluid machines

In terms of energy per mass (unit J/kg= m2

s2

)

2

22

1

2

22

12

1

1

1 Ugzp

Ugzp

++=++

In terms of head (unit m)

g

Uz

g

p

g

Uz

g

p

22

22

22

21

11 ++=++

The last can be rewritten using the hydrostatic pressure due to a fluid column of height

h1: p= gh:

g

Uzh

g

Uzh

22

22

22

21

11 ++=++ ,

giving us Bernoullis equation using elevation, static head, and dynamic head.

Taking into account losses and work done on or by the fluid, the energy balance at

point 2 can be worked out in terms of the energy balance at point 1 and what has been

lost or gained in between. If energy is added to the fluid the work term is positive if

energy has been extracted, HW is negative:

PTL HHHg

Uzh

g

Uzh ++++=++

22

22

22

21

11 ,

where HL is the head loss (e.g., due to friction) between points 1 and 2, HT the head

extracted by a turbine, and HP the head added by a pump.

The head loss is usually due to a combination of friction and minor losses, quantified

by a friction factor, f, and minor loss coefficients, K, which result in a head loss of

( )g

UKfH

DL

L2

42

+=

If we only have a pump but no turbine in the system, we can re-arrange it to

( )g

UKf

g

UUzzhhH

DL

P2

42

221

22

1212 ++

++= .

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6. Fluid machines

As the left-hand side only quantifies the pump and the right only the two endpoints of

the fluid system, the right-hand side is called the system head, HS.. This equation is used

to calculate what flow rate and pressure one finds if one puts a given pump into apipeline system.

The same can be done for a turbine:

( )g

UKf

g

UUzzhhH

DL

T2

42

222

21

2121 +

++=

2.f Mechanical and Fluid power

Mechanical Fluid Hydraulic

Force F F Vm

Energy E F s

Torque T F d dVm t

Power P F U= F d= T QdVdVm tt = gHQ

s is used for linear distance moved, while dis used for distance of selected point from

centre of rotation. is the angular velocity of the machine rotor.

Note that in the torque only the component of the force is counted which is at right

angles to the distance vector from the axis of rotation.

Eulers turbomachine equation states that, in the absence of losses etc., the change in

hydraulic power across the turbine is fully transferred into shaft power by means of the

fluid applying a torque on the shaft in the form of the tangential velocity. In short, it

states that the three columns in the bottom row of the table above are all equal.

2.g Dimensional analysis

Dimensional analysis is tool used universally in all areas of the physical sciences albeit

frequently not overtly. It is a method to find parameters which describe the important

dynamics of a system in a way which is not restricted to that particular system.

Without it, it would be useful to build scale models of cars or aeroplanes and test them

in a wind tunnel. Dimensional analysis helps you to transfer the data from the wind

tunnel to the real thing.

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6. Fluid machines

The procedure is to gather all relevant parameters, and then to reduce them to a

minimum set of nondimensional parameters using a set of base units. If we have

different situations where all nondimensional parameters are the same, then thedynamics is the same, and we have similarity. Similarity is usually split into geometric

similarity, i.e. the shape is the same, and dynamic similarity, i.e. the other parameters are

the same. To have a truly similar situation, we need both geometric and dynamic

similarity.

It is probably best illustrated by an example. To read the theory, refer to a standard

Fluid Mechanics textbook

Example: Wind drag on a car.

The relevant parameters are the size and geometry of the car, the fluid properties of

the air through which the car is moving, the speed at which it is moving, and the drag

force: Size L, air density , air viscosity , speed U, Force F. The base units are length,

m, mass, kg, and time, s.

The first condition is that we test the car using a scale model; trying to measure the

drag force on a Ferrari will be different to that on a minibus

Secondly, we have 5 parameters and 3 base units. Since the procedure boils down to a

simultaneous set of equation for the five parameters, we know that we will end up with

53 = 2 nondimensional parameters.

One parameter to characterise the drag force will have the form Fa UbLcde

The other one will have the form UfLghi

We can safely start with a=1. Putting this into the base units:

kg1

m1

s2

mb

sb

mc kg

dm

3d kg

em

es

e= 0

and mfs

2f m

g kg

hm

3h kg

im

is

i= 0

which can only be satisfied if each base unit drops out:

kg: 1 + d + e = 0

m: 1 + b + c 3d e = 0

s: 2 b e = 0

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6. Fluid machines

This reduces to :

b = e 2

c= 2e 2

d= e 1

We can now choose to set e=0, and get b= 2, c= 2, and d=1:

The force parameter is FU2L21 or, 22122 AU

F

UL

FCD

== , the drag coefficient.

Repeat the exercise for the second parameter (and making sure you dont choose i=0!)

to get the Reynolds number,

UL=Re .

As you saw, we had a choice to set one of the unknown exponents to zero. One can

end up with a different set of equally valid nondimensional parameters.

2.h Dimensional Analysis for fluid machines

Taking a reaction turbine as a representative, we recognise that the relevant

parameters are most likely to be:

Shape of the turbine

1. Size of the turbine: usually the diameter of the rotor, D in [m]

2. Operating speed of the turbine, N in [rad/s]

3. Density of the fluid, in [kg/m3]

4. Viscosity of the fluid, in [kg/m/s]

5. Available head of the fluid, H, or pressure, p in [kg/m/s2]

6. Flow rate through the turbine, Q in [m3/s]

7. Power output, Pin [kg m2

/ s3]

With 7 parameters and 3 base units, we expect 4 nondimensional parameters. If you go

through the analysis, trying to find one parameter proportional to H, and to Q, and one

to P, you are likely to end up with

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6. Fluid machines

a Reynolds number,D

Q

=Re

a head coefficient,22DN

gHH =

a flow coefficient,3ND

QQ = , and

a power coefficient, 53DN

PP

= .

This particular set is correct but has not been adopted as the most useful one. As there

is always some freedom in choice of nondimensional parameters, and because any

product of two nondimensional parameters is still a nondimensional parameters, we can

find one parameter which expresses the power in terms of head and flow rate, and

another parameter which in some way characterises the machine by its operating speed

without any reference to its size:

The new power coefficient becomes:gHQ

P

QH

P

=

= . Recognising that the term

at the bottom is the hydraulic power, we see that we have derived an expression for the

efficiency.

2.i Specific Speed

The speed coefficient, KN can be found by combining the power coefficient and the head

coefficient so that the size drops out, and that it is proportional to the operating speed:

( ) 4/5

4/1

552

244/1

55

1010

1062

24/1

5

2

gH

PN

Hg

PN

Hg

DN

DN

PK

H

PN

=

=

=

=

A turbine with a high specific speed will provide a high power output for a low head,

where as a turbine with a low specific speed will provide power at a high head. Another

rule of thumb not immediately obvious from the equation is that of two turbines

providing the same power, the one with the higher specific speed will be more compact.

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6. Fluid machines

Because the specific speed has been found to be very useful for many years, it has been

adopted in all sorts of units, from the standard SI units to Imperial units but also

practical units which then resulted in the specific speed not being a nondimensional unit

at all. Because all the turbines have almost exclusively been used for water (

= 1000kg/m3), and on the Earths surface, (g= 9.81m/s), engineers have adopted the

specific speed using the short form 4/5= HP

S NN (where I have used NS to distinguish it

from the nondimensional version derived above. Often, the speed is also taken in rev/s

rather than rad/s, which changes the specific speed value by a factor of 2.

Exercise: Try to find the units of 4/5= HP

S NN if you take N in rev/s (rather than

rad/s), Pin MW, and H in m.

3 Types of turbines

3.a Reaction turbines

The action of driving the turbine shaft is by a gradual pressure drop over rotor.

machinethroughdropPressure

rotorthroughdropPressurereactionofDegree' =

Depending on the size of the rotor and its orientation with the main flow direction,

they are classified into radial flow, axial flow, and mixed flow machines.

1. Radial flow turbine:

Runner between outer inlet radius and

inner outlet radius. specific speed range: 0.10.4 ;

moderate head (up to 500m) and

moderate flow rates;

The most common example of a radial flow

reaction turbine is the Francis turbine. In fact, this turbine is in principle the same as

a centrifugal pump operating in reverse, and in cases where specifically designed

turbines are too expensive, centrifugal pumps have been used as turbines. This also

opens up the possibility of a pumped-storage hydropower station; when there low

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6. Fluid machines

electricity demand, the power station operates the machines as pumps to fill up its

reservoir, and when electricity demand rises, the same machines are used as turbines

to generate electricity.

2. Axial flow turbine:

Runner entirely within fluid, except for shaft.

specific speed range: 0.31 ;

low head (up to 150m for Kaplan, up to 25m for Bulb turbine)

and high flow rates.

Kaplan turbine='Propeller in housing' although usually with vertical axis.

Bulb turbine, usually with horizontal axis but with generator in bulb in line with

runner. Used at very low heads, such a tidal power stations.

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6. Fluid machines

3.b Impulse turbines

Examples of impulse turbine are the Pelton Wheeland the Turgo turbine:

All energy is converted to kinetic energy in a fluid jet

hitting the turbine runner. The runner is everywhere

at atmospheric pressure. As a result, the degree of

reaction is zero.

Because the size of the buckets at the rim of the runner limit the size of the jet, Pelton

Wheels operate best at fairly low flow rates. They are therefore best suited for high

heads. This is reflected in a specific speed range of 0.01 0.1.

3.c Turbine types according to their specific speed

The diagrams only show the rotor, but not the stationary housing or guide

vanes.

The values in the various specific speeds are only approximate ranges. The

usual conversion is Kn= 2Kn(rev) = 0.0052 NS(metric)= 0.023 NS(British)

The specific speed for the Pelton Wheel applies to a single jet. For multiple jet,

the power output is proportional to the number of jets.

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6. Fluid machines

Name Image of the runner typical

head (m)

Kn Kn

(rev)

NS

(metri

c)

NS

(Britis

h)

Impulse turbines

Pelton Wheel

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6. Fluid machines

Reaction turbines

Radial flow reaction turbinesFrancis 500 30 0.25

1.3

0.04

0.2

50

250

10

60

Mixed flow reaction turbines

100 15 0.6 2 0.1

0.3

120

360

30

100

1.3 2.5 0.2

0.4

250

500

60

120

2 3 0.3

0.5

360

600

100

150

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6. Fluid machines

Axial flow reaction turbines

Kaplan 50 4 2 6 0.3 1

360 1200

100 300

Bulb or Pit < 20 > 3 > 0.5 > 600 > 150

Wells

3.d Controlling the output

As the turbine will slow down if energy is drawn off at a higher rate, the response of a

turbine to an increase of the electricity demand is a change in speed. Frequencies can

be measured extremely accurately, and this is a way to monitor whether a turbine is

matching the electricity demand. If the frequency drops, the flow rate through the

turbine has to increase. This control of the flow rate is achieve by different means.

The flow rate of the water jets in a Pelton Wheel is usually controlled by a spear valve,

which can change the volume flow rate but does not affect the jet velocity. In reaction

turbine, wicket gates and guide vanes are used to both, control the flow rate and

condition the velocities to enter the turbine runner smoothly.

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6. Fluid machines

Example: A large hydropower station

A new 150MW hydropower station is to be built at a site where the available head is

estimated at 350m. The generators require the turbine to rotate at 10Hz.

1. Find the volume flow rate to generate 150MW of hydraulic power.

2. Find the specific speed (KN (rev)) to generate 150MW.

3. If the only available turbines are Francis turbines with KN= 0.08 rev, find the power

output from one turbine, and the number of turbines required to generate 150MW.

Solution

1.13743

3509810

000000150 =

== smgH

PQ h .

,,

2.( ) ( )

147034341000

00000015010

25145

.,,

.===

gH

PNKn

3. ( )( )

MWWNgHKP n 4410244

10

343410000806

2

522

2

522

==

== ..

..

So, we would need 4 turbines

4

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6. Fluid machines

5 Euler's Turbomachine equation

In an ideal world, all power transferred to or from the machine rotor is transferredfrom or to the fluid power, and all the power transferred to or from the fluid in the

rotor is transferred to or from the hydraulic power of the fluid. As the power

transmitted by a rotating shaft is due to a torque, the component at of the fluid velocity

which acts perpendicular to the direction to the axis of rotation, the momentum flow

rate associated with the tangential velocity is the crucial quantity in the power transfer

mechanism. The deficit in angular momentum of the fluid found between the fluid

entering the turbine and leaving it must have been transferred elsewhere, ideally to theturbine shaft. This is quantified in Eulers turbomachine equation:

( ) gHQVRVRmTP tt === 1122

where

: angular velocity of the machine rotor

T: torque on the rotor shaft

R1: inlet radius of the rotor

R2: outlet radius of the rotor

Vt1: tangential fluid velocity at the rotor inlet

Vt2: tangential fluid velocity at the rotor outlet

m : mass flow rate through the machine

Q: volume flow rate through the machine

H: head difference between inlet and outlet of the machine

The problem is to know or find the tangential fluid velocities, Vt,

To find the tangential velocities, we need to know the radial fluid velocities and

the rotor blade velocities

The rotor blade velocity, U, is found from the rotation rate: rU=

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6. Fluid machines

The radial fluid velocity, Vr, is found from the mass flow rate: br

mVr

2

=

where b is the 'height' of the blade, i.e. 2rb is the cross-section of the

cylindrical opening through which the fluid flows

To be able to find relationships between all velocities, we assume that the fluid is

always flowing parallel to the blades.

The control volume is the space between two blades, which is moving with the rotating

runner . In this moving frame of reference, the fluid whirl/tangential velocity is

apparently reduced by blade velocity, RU=

The usual notation for the different velocity components is

U: blade velocity,

Vr: radial component of fluid velocity

Vt or Vw: tangential component of fluid velocity, aka whirl velocity

R or W: Fluid velocity relative to blade

This is the effective velocity in the control volume!

5.a Principle of a Francis Turbine

Eulers turbomachine equation, ( )1122 tt VRVRmTP == , is illustrated using the

Francis turbine as an example. The rotation rate is fixed by the grid frequency, the

inlet and outlet radius of the runner are fixed by design, but the mass flow rate and

inlet whirl velocities are variable. The mass flow rate is adjusted by opening or closing

a valve.

By the mass flow rate, the radial component of the fluid velocity is specified, by

Am

rV = , whereA is the cross-sectional area of the inlet or outlet, respectively.

An inlet whirl velocity can be added by deflecting the water before it enters the

turbine. This is done by guide vanes surrounding the runner

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6. Fluid machines

The outlet whirl velocity is determined by the outlet radial velocity, the blade speed,

and the orientation of the blade. (blade angle refers to the angle between the blade

tangent and the blade motion, U)

Figure 10.Control volume for the Francis turbine: the space between two turbine blades

Using velocity vectors, we can locally fit a Cartesian co-ordinate system to a fluid

particle near a blade with the x-axis in the tangential direction and the y-axis in the

radial direction.

The fluid velocity is then

=

r

t

V

VV and the blade velocity is

=

0

UU , but we also

know that the fluid is moving along the blade, which has an angle of with the

tangential direction. Therefore, the fluid velocity relative to the blade is

W

=

sin

cosW . The combination of the fluid moving along the blade while the blade

is moving, happens at a velocity of

=+

sin

cos

W

WUUW , which describes the fluid

velocity. Therefore, we have two expressions for the fluid velocity, which obviously

1

2

1

W

U

Vr

Vt

U

W

To centre

Blade, movingat velocity U

Fluid flowing

along blade,W

V

Vt

Vr

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6. Fluid machines

have to be equal: VUW =

=

=+

r

t

V

V

W

WU

sin

cos. The second component gives

ussinrVW= , which we can use in the first component to get:

cot

sin

cosrrt VUVUV ==

or, in terms of the rotation rate and flow rate, and using only the magnitude of the

velocities:

cotAQ

t RV =

Ideal operation

It is usually the best to have the water leave the turbine with relatively little tangential

velocity. However, it is not possible to have zero outlet whirl for all flow rates.

The ideal power output is given by

( ) ( ) ( )[ ]

( ) ( )[ ]

( ) ( )

( ) 21122

21

22

2

212

21

22

2

1212

22

1112221122

11

2

1

1

2

2

1

1

2

2

12

Qbb

QRR

QQRR

RRQ

RRRRQVRVRQP

A

R

A

R

A

QR

A

QR

AQ

AQ

tt

=

=

=

==

tantan

cotcot

cotcot

cotcot

Ideal outlet conditions

Once you have decided on the flow rate at which the fluid leaving the turbine should

have no whirl, the outlet blade can be designed by setting 2222

0 cotAQ

t RV ==

or22

2AR

Q

=tan

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6. Fluid machines

Inlet blade angle

Having decided on the flow rate and chosen the appropriate outlet blade angle, we can

fix the inlet blade angle to give the ideal power at that flow rate.

111111

cotAQ

t RRQVRQP == or QAR

QR

PA 112

1

11

=cot

Ideal inlet conditions

Given all the parameters in the power output equation, it is desirable to give the fluid

the right inlet flow conditions before it encounters the turbine blade. This is done bythe guide vanes. They give the fluid the correct tangential velocity component. By

convention, this angle, , is often (but not always) measured with respect to the radial

direction (in contrast to the blade angle which often (but not always) is measured with

respect to the tangential direction!!!)

1

211

111111

1

1 21

cotcotcot

tan ==

==Q

Rb

Q

RA

Q

RA

V

V AQ

r

t

If is defined with respect to the tangential, then tan = Vr / Vt

Real performance

With good design and control, the

real efficiency of a Francis turbine

can be in excess of 90% over a

relatively large range around the

best efficiency point, from about

50% to 110% of the rated power

of the turbine.

0

100

200

300

400

500

600

700

800

900

0 10 20 30 40 50 60 70 80 90

Q (m3/s)

0%

10%

20%

30%

40%

50%

60%

70%

80%

90%

100%

Figure 12. Power input (blue diamonds), Power output(red squares), and efficiency (triangles) for a typical Francisturbine

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6. Fluid machines

5.b Example

A small Francis turbine, turning at 11rev/s, has a runner diameter of 0.5m, a constant

blade height of 60mm, and an outlet diameter of 0.3m. At maximum efficiency it

delivers 200kW and has a specific speed of 0.09rev.

1. Calculate the optimum blade outlet angle.

2. Calculate the inlet angle, assuming that the best efficiency is 95%.

3. Calculate the angle of the guide vanes at the best-efficiency point

4. Calculate the guide angle and power output at 75% and 50% of the flow rate,

assuming an efficiency of 90% and 75%, respectively.

Solution

1. For the optimum blade angle, we want that the fluid leaves the turbine without any

whirl, ie Vt is zero: tan 2= Q/(R2A2)

We know

= 11 rev/s= 69rad/sR2= 0.15m

A2= 2bR2= 20.060.15=0.0566m2

To solve this, we still need the flow rate.

From the specific speed, we can get the head at which the turbine delivers 200kW:

H= (N/KN)4/5

(P/)2/5

/g= (11/0.09)4/5

(200)2/5

/ 9.81= 39.7m

If the turbine has an efficiency of 95% at this point, we know that the (hydraulic) powergoing in is 200kW/0.95= 210.5kW.

The hydraulic power, Ph= gHQ, gives us then the flow rate:

Q= Ph/ (gH)= 210500/ (10009.8139.7)= 0.541m3/s.

Finally, we can use the blade angle equation

tan 2= Q/(R2A2)= 0.541 / (690.150.0566)= 0.9228

which gives us 2= 42.7.

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6. Fluid machines

2. To calculate the inlet angle, we use Euler's equation which relates the difference of

the fluid's angular momentum at the inlet and outlet. Since we know from 1. that we

have designed the outlet so that the fluid has no angular momentum when it leaves the

turbine, the entire angular momentum at the inlet is available for power generation, and

we can use the equation for the inlet blade angle,

cot 1=A1P/ (R1Q2) R1A1/Q.

We know

= 69rad/s

R1= 0.25m

A1= 2bR1= 20.060.25=0.0942m2

P= 200kW

Q= 0.541m3/s,

and we get

cot 1= 0.0953200000 / (1000690.250.5412) 690.250.0942/ 0.541= 0.7174

or tan 1= 1/ cot 1= 1.3939 which gives 1= 54.3.

3. For the angle of the guide vanes, we need to work out the radial fluid velocity and

the tangential fluid velocity.

The radial velocity is given from the inlet area,A1, and the flow rate, Q:

Vr,1= Q/A1= 0.541m

3

/s / 0.0942m

2

= 5.74m/s.

The tangential velocity can be found from re-arranging Eulers equation, P= Q R1Vt,1,

to Vt,1= P/ (Q R1)= 200,000/ (69.1 1000 0.541 0.25)= 21.4m/s.

If we take the angle with respect to the radial direction,

we get tan = Vt,1/ Vr,1= 21.4/5.74= 3.728, or = 75.

If we take the angle with respect to the tangential direction (ie the same as the blade

angle), we get tan = Vr,1/ Vt,1= 5.74/21.4= 0.268, or = 15.

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6. Fluid machines

4. For the power output, we use the hydraulic power at the reduced flow rate and the

efficiency.

For the guide vane, we use the reduced radial velocity and the tangential velocity to get

the ideal (hydraulic) power out.

Q= 75% Q= 50%

Q 0.541*0.75= 0.406 0.541*0.5= 0.270 m3/s

efficienc

y

90% 75%

Pin =gHQ 9810*39.7*0.406= 158 9810*39.7*0.270= 105.3 kW

Pout = Pin 0.90*158= 142 0.75*105.3= 78.9 kW

Vr,1 = Q/A 0.406/0.0942= 4.30 0.270/0.0942= 2.87 m/s

Vt,1 = P /

(Q R1)

158,000/ (69.1000

0.4060.25)

22.5 105,300/ (69.1000

0.2700.25)

22.5 m/s

tan = Vt,1/

Vr,1

22.5/ 4.3 5.232 22.5/2.87 7.848

=atan 79 83

6 Propellers and wind turbines

To get a good idea what one might get out at best, one can simplify the problem

greatly. Instead of looking at the detailed flow through the turbine, we can treat the

turbine itself as a black box and only look at its effect on the nearby fluid stream. One

simplification is to regard the black box as a very thin disk just enclosing the rotor, the

actuator disk. We can then define a control volume which encloses some fluidupstream of the rotor, the actuator disk, and some downstream fluid. Because the only

object within that control volume is the actuator disk, we can use Bernoulli's equation

everywhere, except across the disk. But the disk is the only thing which can affect the

flow, so it is the only thing where a force can be exerted.

Propellers generate thrust by accelerating the fluid through the rotor.

Wind (and tidal stream) turbines generate power by converting kinetic energy

of the fluid flow into rotation of the rotor.

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6. Fluid machines

They have no casing to guide the fluid through the machine

They cannot maintain a pressure drop between upstream and downstream of

the machine.

Simple wind mill designs may have flat rotor blades which act like deflector

plates. The torque on the rotor is given by how much the air stream is

deflected by the blades. The real efficiency is much reduced because one cannot

have ideal flow conditions across a moving deflector blades if it is to do work

(use the velocity triangles: because the entry angle and the exit angle of the

blade are the same, we cannot do the work gradually along the blade.

Rotor blades are generally shaped

like areofoils - they generate a lift

force perpendicular to the fluid flow

along the blade (remember that the

blade is moving at the same time).

Propellers want to generate forward

thrust while wind turbines want togenerate torque.

6.a Actuator disk theory

To get a good idea what one might get out at best, one can simplify the problem

greatly. Instead of looking at the detailed flow through the turbine, we can treat the

turbine itself as a black box and only look at its effect on the nearby fluid stream. One

simplification is to regard the black box as a very thin disk just enclosing the rotor, theactuator disk. We can then define a control volume which encloses some fluid

upstream of the rotor, the actuator disk, and some downstream fluid. Because the

only object within that control volume is the actuator disk, we can use Bernoullis

equation everywhere, except across the disk. But the disk is the only thing which can

affect the flow, so it is the only thing where a force can be exerted.

L i f t

P r o p e l l e r

T h r u s t

W i n d t u r b i n e

T o r q u eL i f t

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6. Fluid machines

Actuator disk Control Volume

The control volume wants to enclose

the actuator disk completely but not

look at the fluid flowing past it. We

also want to use streamlines as the

side boundaries so that we know that

there is no fluid leaving the control

volume through sides and that we can

use Bernoullis equation along the side.

The streamline which just touches the

edge of the disk is called the slipstream. Also, we need to extend the control volume to

far enough away from the disk so that we look at simple, unperturbed flow:

Pressure far upstream and downstream is unaffected: 014 == pp

Mass flow through control volume: 44332211 uAuAuAuAm ====

The disk is very thin: AAA == 23 , whereA is the swept area of the rotor.

By continuity, 32 AuAu = : 32 uu =

The force on the disk by the flow is the pressure difference across the disk:

( )AppF 32 =

Bernoulli before disk: ( 2221212 uup =

Bernoulli after disk: ( )2224213 uup =

Inserting pressures gives force: ( 242121 uuAF =

Force on disk is also the net change in the momentum flow rate:

( )41 uumF =

with Momentum flow rate into C.V.: 1umJin =

Momentum flow rate out of C.V.: 4umJout =

u 1 u 2 u 3 u 4

1

23

4

1 2 3 4

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6. Fluid machines

Equating both forces, using ( ) ( )( )41412421 uuuuuu += ,

and re-arranging gives ( )412

12 uuu +=

Power transmitted by disk: ( )( )412421412 uuuuAFuP +==

Using 1uU= and1

4

u

u= :

( )( ) ( )3234123

41

2 111 +=+== AUAUFuP

6.b Wind turbine

The flow of kinetic energy by the wind

through an areaA is

321 AUPair =

The efficiency of a stationary actuator disk is

therefore:

( )3221 1 +==

airP

P

6.c Propeller

If we consider a propeller, the power

conversion between actuator disk and fluid is

given by the force and the fluid velocity

through the disk, 2u , but the useful power is

that which is related to the actual speed of

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7Wind turbine efficiency

Efficiency

=u4/U

1 2 3 4 5 6 7 8 9 100.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1Propeller efficiency

=u4/U

Efficiency

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6. Fluid machines

the aircraft, which is U, but the force is obviously still the same. So the useful output is

( 2321 1 == AUFUPout .

The efficiency is

+

==1

2

turbine

out

P

P.

Example: Design of an aircraft propeller.

An aircraft is powered by to turboprop engines. Assume that the actual propellers

work at the ideal limit according to actuator disk theory where the wind speed behind

the propeller is 50% higher than the travelling speed. Each engine must provide a thrust

of 50kN to achieve a speed of 300mph at an altitude where the density of air is

0.8kg/m3.

1. Calculate the efficiency of the propellers

2. Calculate the power requirement for the engines.

3. Calculate the diameter of the propellers.

4. Calculate the pressure changes across the propeller.

Solution

U= 300mph= 185m/s.

= 1.5=3/2

1. %202.05

12

1

2

25

====+

=

2. F= 50,000N

u2= (U+u4)= 1.25U= 231m/s.

P= Fu2= 11.55MW

3. ( )221 1 = AUF --> A= 2.92m2 and D= 1.93m.

4. mbarbarkPaA

Fp 17017.017 ====

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6. Fluid machines

Design of a helicopter rotor

A helicopter design requires from the rotor that it can carry a mass of 5,000kg. The

diameter of the rotor is not to exceed 8m. Assume that actuator disk theory gives a

useful indication of the situation. Near ground, the air density is 1.2kg/m3.

1. Sketch a diagram of the helicopter rotor as an actuator disk, and outline the

slipstream boundary and the velocities at crucial points.

2. Determine the air velocity through the disk required to produce the force to

balance the weight of the helicopter.

3. Calculate the power requirement for the motor powering the rotor.

4. Estimate the torque on the rotor shaft if the rotor rotates at 300rpm

Solution

The swept area of the rotor is2

427.50

2

mA D ==

The upstream velocity, u1, is zero, and the velocity through the disk, u2, is therefore half

of the downstream velocity, u4.

The force generated by the change in velocity is ( ) 2214 2 AuuumF == .

The force required to counteract the weight of 5,000kg is F= mg= 49kN.

The air velocity through the rotor is then smA

Fu /20

22 ==

.

The power carried by the air is

kWD

F

A

F

A

FFFuumP 980

12

22

33

22

21

4

======

The torque is kNm

P

T 3152

000,980

=== .

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6. Fluid machines

6.d Some remarks on wind turbines

One of the design and operating conditions not used in the actuator disk theory is the

rotational speed of the turbine rotor. It is obvious, however, that wind of a given

speed, U0, can turn the rotor only at a certain speed. The parts of the rotor moving

fastest are the tips of the rotor blades. A rotor with a radius, R, and turning with an

angular velocity, , has blade tips moving at Utip= R.

In fact, two other constraints limit the useful speed of such a turbine or propeller:

1. If any part of the rotor is moving at speeds approaching the speed of sound, the

compressibility of air will affect the performance

2. If the rotor is moving too fast (or if the blades are too close together), a blade will

follow in the wake of the previous blade rather than receive 'fresh' air.

While the first is more relevant to aircraft propellers, the second is important for wind

turbines. As a result, smaller turbines may turn faster and/or have more blades, while

larger ones tend to have fewer blades and turn slower. Equally, if you want to reduce

the number of blades for a given rotor radius or diameter, you must increase the

rotation rate to achieve the same output.

Most current large wind turbines have three blades, and operate at a tip speed ratio of

about 8:1. This means that the tips move about 8 times faster than the mean wind.

Reading

This section is best revised with a standard textbook on basic Fluid Mechanics, such asthat by Massey.

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