Turbomachinery (MEng 3202) Part II

7/31/2019 Turbomachinery (MEng 3202) Part II

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Melesse Haile(Asst. Professor)

Adama UniversityEthiopia

Turbomachinery (MEng 3202)Part - II


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Compressors are machines used to increase the pressure of a fluid.

Note:- This chapter also includes the discussion of Chapter 7 and 8. That is, itincludes the design of rotor blades, guide vanes and spiral casings.

Classification of work absorbing machines.

Adama University Melesse Haile (Asst. Professor)


3/92

Classification of Compressors

Positive displacement compressors increase the pressure of a gas or air byreducing the volume.

Dynamic compressors increase the air velocity by the rotational motion of impelleror rotor, which is then converted to pressure at the outlet.



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Classification of Dynamic Compressors1. By the type of flow,

Radial flow

Axial flow

2. By the type of energy conversion, Volute casing Volute casing with guide vanes

3. By the method of drive, Gear drive

Belt drive

Radial flow Axial flow



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Radial (Centrifugal) Compressors

Component parts

1. Impeller2. Inducer3. Diffuser and Diffuser vanes4. Collector or Casing5. Inlet Guide Vane

Class work

1. Write the function of each component part.



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A Typical Compressed air System Components and Network



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Radial Compressors

Principle of operation

When the impeller is rotating at high speed (more than 20000 rpm), air is drawn inthrough the eye of the impeller. The absolute velocity of the inflow air is axial. Themagnitude and direction of relative velocity at inlet depends upon the velocity of theimpeller as well as the magnitude and direction of the entering absolute velocity. Theimpeller vanes at the eye are bent so as to provide shockless entry (inducer). The airthen flows radially through the impeller passages due to centrifugal force. The total

mechanical energy driving the compressor is transmitted to the fluid stream in theimpeller where it is converted into kinetic energy of air. The diffuser then converts the

kinetic energy of air into pressure.The air leaving the diffuser iscollected in a spiral passage fromwhich it is discharged from the

compressor.

The pressure and velocity variationacross the compressor is as shownin the figure.



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Radial Compressors

Shapes of Radial Compressor blades

Velocity diagrams for a Radial Compressor blades

inlet outlet



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Radial Compressors

Velocity diagrams for a Radial Blades

at the inletFigure (a) represents the velocity trianglewhen the air enters the impeller in the axialdirection. In this case, C1 = Ca1. Figure (b)represents the velocity triangle at the inlet tothe impeller eye and air enters through theinlet guide vanes. Angle is made by C1 andCa1 and this angle is known as the angle ofprewhirl. The absolute velocity C1 has a whirlcomponent Cw1.

at the outlet

In ideal case, as shown in Figure (c), airleaves the impeller tip at an angle of 900.Hence, Cw2 = U2. But in actual case due to slipbetween the impeller and the fluid, Cw1 issomewhat less than U2 as shown in Figure(d).

inlet

outlet



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Radial Compressors

Slip Factor

As shown in the figure, due to slip the actual fluid does not follow the blade shape.Slip () is, therefore, defined as,

For radial vane impellers, a relation for slip is given by Stanitz,

where, n is the number of vanes.

Assume, y 0.9 for n value 19 to 21.

2

'

2

w

w

C

C



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Radial Compressors

Energy Transfer (Pt)

From Eulers equation, the energy transfer per unit mass flow rate is given by,

considering velocity triangle (a)

Considering velocity triangle (c), Cw2 = U2 and with slip the theoretical energy transferis given by,

ct Wm

P

inlet outlet



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Radial Compressors


In a real fluid, some of the power supplied by the impeller is used in overcominglosses that have a breaking effect on the air carried by the vanes. These includewindage, disk friction, and casing friction losses. To take into account these losses, apower input factor () , which is the ratio of the actual to theoretical energy transfer, isintroduced. This factor typically takes values between 1.035 and 1.04. Thus the actual

work done on the air becomes,

From Mollier Chart, the stagnation or totaltemperature equivalent of work done on

the air is given by,

( Refer:- Reference-1 page 68 - 75 )



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Radial Compressors


The compressor isentropic efficiency (c) is defined as,

T03= T03ss

Hence,

And,

Note:- The tip speed ratio (U2) is about 460 m/s.

Class work

1. Write your own observation and conclusion on the derived relation for .


Ad U i i M l H il (A P f )


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Radial Compressors

Effect of impeller blade shape on performance

There are three types of blade shapes as shown in the figure.

a) Backward-curved blades (2


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Radial Compressors


b) Backward-curved blades (2 = 900):-

From this equation one can observe that, as theenergy or power transfer remains constant.

c) Forward-curved blades (2 > 900):-

As the mass flow rate increases energy or power transfer also increases.

Ag

Um

g

UPHence

A

mCwhere

g

CUUPand

g

CUPequationsEulerFrom

t

aa

t

wt

22

2

2

22222

22

cot,

,)cot(

,

,'

tconsisg

UP

Ag

Um

g

U

PHence

A

mCwhere

g

CUUPand

g

CUPequationsEulerFrom

t

t

aa

t

wt

tan

0cot

cot

,

,)cot(

,

,'

2

2

2

22

2

2

22222

22


Ad U i it M l H il (A t P f )


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Radial Compressors


In the case of backward-curved blades or vanes, the maximum efficiency occurs in theregion of maximum power. If the flow rate increases beyond the design point, thepower starts to decrease. This keeps the motor to operate safely and thischaracteristics is called self-limiting characteristics. Whereas in the case of radialand forward-curved blades or vanes, as the flow rate increases the power also tend toincrease even beyond the design point. This is not only make the operation costly but

also damage the motor. But, designers may sometimes prefer these two types ofshapes to get some performance advantage as compared to backward-curved blades.

Compressibility effects

If the relative velocity of a compressible fluid reachesthe speed of sound in the fluid, separation of flow

causes excessive pressure losses and cyclic loading onthe impeller due to eddy or shock wave formation. It is,therefore, necessary to control the Mach number at acertain points in the flow to mitigate the problem. TheMach number at the impeller inlet must be less thanunity. This is possible by reducing the relative velocity at

inlet by introducing a guide vane as shown in the figure.inlet


Ad U i it M l H il (A t P f )


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Radial Compressors

Compressibility effects

Actually, reducing the relative velocity at inlet causes the tangential velocity of theabsolute velocity to increase. This in turn decreases the work capacity of thecompressor.

Mach number in the diffuserThe absolute velocity of the fluid becomes a maximum at the tip of the impeller.Assuming a perfect gas, the Mach number at the impeller exit M

2

can be written as:

However, it has been found that as long as the radial velocity component (Cr2) issubsonic, Mach number greater than unity can be used at the impeller tip without loss

of efficiency.High Mach number at the inlet causes high stagnation pressure. This leads to avariation of static pressure and is transmitted upstream in a radial direction throughthe vaneless space. The end result will be exerting a cyclic type of load on theimpeller. This in turn lead to early fatigue and failure on the impeller.


Adama University Melesse Haile (Asst Professor)


18/92

Radial Compressors

Characteristics of a Radial Compressor

The performance of compressible flow machines usually be described in terms of thegroups of variables derived in dimensional analysis ( Refer:- Rference-1 on Page 62).These variables are dependent on other variables such as the condition of pressureand temperature at the compressor inlet and physical properties of working fluid. Tostudy the performance of a compressor completely, it is necessary to plotagainst the mass flow parameter for a fixed speed intervals of .

The figure below shows the theoretical characteristics of a Centrifugal Compressor.

Consider a valve is placed in the delivery lineof a compressor running at constant speed.Suppose at the beginning the valve is fully

closed. The pressure ratio then will havesome value as indicated by point A. Thispressure ratio is available from vanes movingthe air about in the impeller. The pressurehead so developed is called shutoff head.




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Radial Compressors


Now, suppose that the valve is opened andair flow begins. The diffuser contributes to thepressure rise. The pressure ratio increasesand the maximum pressure occurs at point B.But the compressor efficiency at thispressure ratio will be below the maximum

efficiency. Point C indicates the furtherincrease in mass flow, but the pressure isdropped slightly from the maximum possiblevalue. This is the design mass flow ratepressure ratio as efficiency is maximum.

Further increase in mass flow will decrease the pressure ratio until point D. Thispoint indicates that the pressure rise is zero. At this point all the power absorbed isused to overcome the internal friction and thus the compression efficiency is zero.However, the actual curve is different from this ideal curve shown in the figure dueto chocking, surging and stall.




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Radial Compressors


Choking:- When the velocity of fluid in a passage reaches the speed of sound atany cross-section, then the air ceases to flow and the process is called chocking.Choking may occur at the inlet, at the impeller or at the diffuser.

Surging:- The phenomenon of a momentary increase in the delivery pressureresulting in unsteady, periodic and reversal flow through the compressor is called

surging. Surging may lead to failure of the compressor parts.

Stall:- The phenomenon of a reduction in mass flow rate through the blade passageat higher angle of incidence is known as stall. This phenomenon may lead tosurging.




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Radial Compressors

Energy transfer by the impeller to the fluid (air)

A radial compressor is a dynamic type compressor and it works on the principle offorced vortex flow ( Refer:- Fluid mechanics ). When a certain mass of fluid is rotatedby an external torque, the rise in pressure head of liquid takes place. The rise inpressure head ( Hp ) at any point of the rotating liquid is proportional to the square oftangential velocity ( v) of the liquid at that point.

g

r

g

vHheadpressureThe p

22)(

222

Thus at the outlet of the impeller where

radius is more, the rise in pressure is moreand the liquid will be discharged at theoutlet with a high pressure head.




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Radial Compressors


In case of radial (centrifugal) compressors, work is done by the impeller on thefluid (air). The expression for work done by the impeller on the air is obtainedby drawing velocity triangle at inlet and outlet of the impeller in a similar way asfor a centrifugal pump.

Velocity diagram




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Radial Compressors


As shown in the velocity diagram, the angle between vand u is called andthe angle between wand u is called . The difference between the absolutevelocity vand the relative velocity w is also illustrated.

Velocity diagrams

Inlet

outlet




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Radial Compressors

Ideal Head Equation

The equation which forms the basis of pump, compressor and blower design isbased on the following assumptions: The fluid leaves the impeller passages tangentially to the vane surfaces or

there is a complete guidance of the fluid at the outlet and the fluid is

assumed to be frictionless, non-turbulent, etc.

The impeller passages are completely filled with actively flowing fluid at all

times. The velocities of the fluid at similar points on all the flow lines are the same.

The ideal total head developed by the impeller based on these assumptions iscalled Eulerhead and designated by HE and is given by,

Note:-The deviation of actual conditions from these assumptions will be

considered later.

.

.

.,

2

1

21

velocityabsoluteinchangetoduediffusertheinheadpressuretheisH

velocityrelativeinchangetodueimpellertheinheadpressuretheisH

forcelcentrifugatodueimpellertheinheadpressuretheisHWhere

HHHH

d

i

i

diiE




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Radial Compressors

The Pressure Head in the impeller due to centrifugal force (Hi1)

Consider a closed container filled with fluid (air), as shown in the figure, isrotated at constant speed about its axis. The motion is transmitted to the fluidby its viscosity until the angular velocity of the fluid is the same as that of thecontainer.

Rotating container filled with fluid

To determine the pressure distribution

within the fluid, consider a very small fluidelement having a width b, thickness dRandcircumferential length Rd rotating with anangular velocity . The centrifugal forceacting on the particle is,

The pressure increment due to this forcebetween R and R+dR is,

dRdRbdFHence

dRRdbdmWhereRdmdF

22

2

,

,

dRR

Rdb

dRdRb

dA

dFdp 2

22




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Radial Compressors

The Pressure Head in the impeller due to centrifugal force (Hi1)

Rotating container filled with fluid

Integrating between the radii R1 and R2gives the total pressure,

g

uuHHH

HHg

ppbut

g

uu

g

ppHence

uRandRR

pp

ipp

pp

2

,2

,

,2

2

1

2

2112

1212

2

1

2

212

2222

1

2

22

12

y ( f )



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Radial Compressors

The Pressure Head in the impeller due to change in relative velocity (Hi2)

Consider the passage through the impeller at radius rhaving a cross-section daas shown in the figure. Let the small element of area da moves a smalldistance dsin time dt. Hence,

g

ww

g

pH

ww

dwwdpp

outletandinletbetweenequationthegIntegratin

dww

wdt

dsbutdw

dt

dsdp

dabyalldividingand

dt

dwdsdadpdadF

i2

2

),2()1(

,

2

2

2

1

2

2

2

2

2

2

1

2

1

y ( f )



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Radial Compressors

The Pressure Head in the diffuser due to change in absolute velocity (Hd)

Inside the diffuser, the dynamic pressure or velocity head is converted intostatic pressure head. The velocity in the discharge pipeline is almost the sameas the velocity in the suction pipe.Consider a fluid element of area da moving through the diffuser vanes adistance ds in time dt as shown in the figure.

g

vv

g

p

H

vv

dvvdpp

outletandinletdiffuserbetweenequationthegIntegratin

dvv

vdt

dsbutdw

dt

dsdp

dabyalldividingand

dt

dv

dsdadpdadF

d 2

2

),1()2(

,

2

1

2

2

2

2

2

1

2

1

2

1

y ( f )



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Radial Compressors

The Ideal Total Head (HE)

g

vuH

inletradialhavingimpellerforand

g

vuvuH

vvandvvbut

g

vuvuH

equationabovetheinequationstheseInserting

vuvuw

vuvuw

impellertheofoutletandinletatdiagramvelocitytheFrom

g

vvwwuu

g

vv

g

ww

g

uuHHHH

uE

uuE

uu

E

diiE

22

0

1

1122

222111

111222

222

2

2

2

2

2

2

11121

21

21

2

1

2

2

2

2

2

1

2

1

2

2

2

1

2

2

2

2

2

1

2

1

2

2

21

),90(

coscos,

coscos

,

cos2

cos2

,

2

222



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Radial Compressors

The Ideal Power (P) and Torque (T)

NminvRvRQ

vRvRQ

vuvuQ

PT

HPinvuvuQ

kWinvuvuQ

Wing

vuvugQ

gHmP

uu

uu

uu

uu

uu

uu

E

)(

)(

)(

746)(

1000

)(

)(

1122

1122

1122

1122

1122

1122



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Radial Compressors

Degree of Reaction ( )Degree of reaction is the ratio of the static pressure head rise in the impeller (H

i1

) tothe total head rise (HE).

2

2

2

1

2

2

2

2

2

1

2

2

2

2

2

2

12

21

11

0

22

2

1

2

2

1

,

,

;90

,

21

,

1

u

uur

r

rr

r

u

Ed

E

d

E

dE

E

i

vvvand

vvvvv

trianglevelocityoutletFromvv

vv

vv

inletradialFor

vuvv

HandHforngSubstituti

H

H

H

HH

H

H

22

2

2

2

22

2

2

22

2

2

22

2

2

tan1

2

1

2

tan1

tan

,

2

1

21,

u

v

u

vu

vuv

trianglevelocityoutlettheFrom

u

v

vu

vHence

r

r

ru

u

u

u

Question:- What do you observe fromthis relation ?

Adama University

Melesse Haile (Asst. Professor)


32/92

Radial Compressors

Degree of Reaction ( )

From this relation one can observe that,a) For 2 = 90

0, the degree of reaction = , as 2 = . This means that thepressure generated by the impeller and diffuser will be the same.

b) For 2 having values between 00 and 900 , the degree of reaction is more than

half. This means is that the pressure generated by the impeller is more than inthe diffuser.

c) When the degree of reaction is one.

22

2

tan1

2

1

u

vr

2

1

.)(,1,

0

,tan

tan,

tan1tan

2

222

2

2

2

22

222

22

2

22

2

zeroisHheadtotalidealtheforThus

v

uvu

g

uH

ofvaluethengSubstituti

vu

g

u

g

vuHThen

u

vor

u

v

E

r

rE

ruE

rr

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33/92

Radial Compressors

Degree of Reaction ( )

d) When = 0, there is no static pressure change in the impeller. The energy transferis entirely due to the change of absolute velocity in the diffuser. In other words,the impeller has equi-pressure blades or impulse blades. Machines having suchan impeller is known as an impulse machines. The blades of impulse impellersare forward curved and the blades of reaction impellers are backward curved asshown in the figure. In order to generate more pressure in the impeller, the angle

2 should be nearly zero. For the same discharge reaction machines have largeroutlet impeller diameter or high rotational speed than impulse machines.

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34/92

Radial Compressors

Actual Head or Power developed by the impeller

It has been discussed that the ideal head or power developed by the impeller isderived based on the listed assumptions. Some of these assumptions were if thefluid is frictionless, non-turbulent, etc. In this section the factors which causes thesedeviations will be discussed.

Factors which causes deviation of actual machines from the ideal

a) Circulatory flow:- If a closed container is filled with fluid and rotated about anaxis, the fluid tends to rotate in opposite direction relative to the container due toits inertia. It is apparent that the fluid in contact with the side A of the container isat a higher pressure than at B, as the container is exerting a pressure on the fluidon that side while it is being accelerated as shown in the figure.

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35/92

Radial Compressors


Factors which causes deviation of the actual machines from the idealThis implies that the velocity at B is higher than A according to Bernoullisequation. The amount of this circulatory flow depends on the shape of thecontainer. It will be less if is long and narrow. In a rotating impeller two flows takeplace simultaneously; namely, the flow of fluid through the passage and thecirculatory flow as shown in the figure. The resultant effect for a given flow is to

cause the fluid to leave the wheel at an angle less than the vane angle and toincrease the guide blade angle. This effect is illustrated in the velocity diagramwhere the dotted lines represent the ideal velocity diagrams.

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Radial Compressors


Factors which causes deviation of the actual machines from the idealb) Effect of Finite Number of Blades:-Taking finite number of blades of impeller,

the Ideal or Euler Head (HE) is changed as follows due to changes in velocitydiagram.

Where, is the correction factor for finite number of blades. It is sometimes known as

blade efficiency blade.

Eu

E

Euu

E

Hg

vuH


Hg

vuvuH

'

22'

0

1

'

11

'

22'

),90(

Adama University



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Radial Compressors


Factors which causes deviation of the actual machines from the idealc) Shock loss:-The usual assumption made in designing such a machine is that

the fluid enters the impeller vanes radially so that 1 = 900. As the fluid

approaches the vane inlet, it comes in contact with the rotating impeller. Thistends to cause a mismatch between the blade angles and the inlet flowdirections specially at high flow rates. This causes the inlet vane angle to be

larger as shown in the figure. The net effect is to reduce the head and hence thepower to be transferred to the fluid due to the presence of vu1.

g

vuH


g

vuvuH

uE

uuE

22

01

1122

),90(

Adama University



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Radial Compressors


Factors which causes deviation of the actual machines from the ideald) Fluid and Disc or Surface Friction:-This is due to the friction that exists within

the fluid and between the fluid and the surfaces of the impeller and the diffuser.

e) Turbulence:-The type of flow existing in a pump or a compressor or a blower isturbulent as the determined from the Reynold number. The effect of turbulence is

also to reduce the machine performance.

f) Leakage:- This is due to the loss of fluid during operation. The effect of thisleakage loss is to reduce the discharge capacity.

The actual head is called the manometric head (Hm) is, therefore, given by:

.

sin

22 radialisvelocityf lowinletifHg

vu

HH

gcaandimpellertheinLossesheadEulerH

Lu

LE

m

Video-5Video-6Video-7

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http://video/YouTube-%20Centrifugal%20Compressor%20Principles(Video-5).mp4http://video/YouTube-%20Centrifugal%20Pump%20Principles(Video-6).mp4http://video/YouTube-%203D%20Pumps%20Technical%20Animation(Video-7).mp4http://video/YouTube-%203D%20Pumps%20Technical%20Animation(Video-7).mp4http://video/YouTube-%203D%20Pumps%20Technical%20Animation(Video-7).mp4http://video/YouTube-%203D%20Pumps%20Technical%20Animation(Video-7).mp4http://video/YouTube-%20Centrifugal%20Pump%20Principles(Video-6).mp4http://video/YouTube-%20Centrifugal%20Pump%20Principles(Video-6).mp4http://video/YouTube-%20Centrifugal%20Pump%20Principles(Video-6).mp4http://video/YouTube-%20Centrifugal%20Compressor%20Principles(Video-5).mp4http://video/YouTube-%20Centrifugal%20Compressor%20Principles(Video-5).mp4http://video/YouTube-%20Centrifugal%20Compressor%20Principles(Video-5).mp4


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Radial Compressors

EfficienciesThe actual head developed by a compressor is less than the ideal or Euler Head (HE)due to various losses. In designing such machines it is usually difficult to predictthese losses. Engineers or Designers prefer to use empirical relations and/orefficiencies to consider the effects of these losses in the performance of themachine.a) Circulatory flow coefficient ( ):- the ideal or Euler Head (HE) is derived by

considering an impeller with infinite number of vanes so as to avoid circulatoryflow. The effect of circulatory flow is to reduce v2 as shown in the velocity diagram.

2

'

2

u

u

v

v

As v2 reduces, vu2 also reduces and hencethe head will reduce.

Hence, the circulatory flow coefficient () isgiven by,

g

vuH


g

vuvuH

uE

uuE

22

0

1

1122

),90(

Adama University



40/92

Efficiencies

b) Overall Head Coefficient ( ):- the head developed by a turbomachine due toforced vortex flow has been derived as,

But, due to the shape of the vanes and the casing the ratio can not be unity.Hence, in actual case the ratio 1 is replaced by .

From this equation, D2 can also be determined.

c) Hydraulic efficiency ( H):- the actual head or manometric head (Hm) developedby the machine is less than the ideal or Euler Head (HE) due to mainly friction andturbulence losses. To consider this effect, a hydraulic efficiency (H) is related as,

)()( 11221122 uu

m

uu

m

t

out

E

mH

vuvu

gH

vuvum

gHm

P

P

H

H

.,122

2

2

2 headoffShuttheisHWheregH

uor

g

uH

gHu

2

2

N

H

N

HgHD

NDgHugHu

5.84

260

602

2

2

22

2

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41/92

Efficiencies

d) Volumetric efficiency ( V):- the volumetric efficiency is a measure of the amountof leakage. It is given by,

e) Mechanical efficiency ( m):- It is the ratio of the power at the impeller (Pt) to the

power of the the shaft (Ps).

f) Overall efficiency ( o):- It is the ratio of the fluid power (Pout) to the power of theprime mover (Pin).

.,

.arg,

leakagetheisQand

ratef lowedischtheisQWhereQQ

Q

L

L

V

TPandT

vuvum

vuvumPWhereP

P

suu

uut

s

tm

,)(

)(,

1122

1122

mVH

in

m

mout

in

outo

PgHQ

gHmPWhereP

P

,

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42/92

Reading Assignment

Refernce-1 ( Turbomachinery design and theory)Page 187 273

Class work

1. 10 kg/s of air is to be compressed in centrifugal compressor of the single sided impeller

type. The ambient air conditions are 1 bar and 200

C. The compressor runs at 20000 rpm,has isentropic efficiency of 80 % and compresses air from 1 bar static pressure to 4.5 bartotal pressure. The air enters the impeller eye with a velocity of 150 m/s with no prewhirl.Assuming that the ratio of whirl speed is 0.95, determine:

a) rise in total temperature during compression,b) the impeller tip speed and tip diameter,c) power required to drive to drive the compressor,

d) the external diameter of the eye, for which the internal diameter is 15 cm.

Worksheet 6-1

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Preliminary Design of a Centrifugal (Radial) Pump

Example:-Design a water transport system given in Assignment 5-1.

Design procedure:Step-1:- Determine the specific speed of the pump (Ns).

rpm

H

QNNs

14.13

35

017.01450

4

3

4

3

%65o

Estimate the pump efficiency (o)from the curve for a specific speedof 13.14 rpm and discharge of 1020LPM (Liter Per Minute).

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http://worksheets%2C%20labs%2C%20and%20assignments%20%28tubomachinery%29/Assignment%205-1%20(MEng%203202).dochttp://worksheets%2C%20labs%2C%20and%20assignments%20%28tubomachinery%29/Assignment%205-1%20(MEng%203202).dochttp://worksheets%2C%20labs%2C%20and%20assignments%20%28tubomachinery%29/Assignment%205-1%20(MEng%203202).dochttp://worksheets%2C%20labs%2C%20and%20assignments%20%28tubomachinery%29/Assignment%205-1%20(MEng%203202).dochttp://worksheets%2C%20labs%2C%20and%20assignments%20%28tubomachinery%29/Assignment%205-1%20(MEng%203202).dochttp://worksheets%2C%20labs%2C%20and%20assignments%20%28tubomachinery%29/Assignment%205-1%20(MEng%203202).doc


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Step-2:-

Determine the Maximum Shaft Power (Psmax).This can be determined from the performance curve.

Determine shaft diameter (Ds) using stress theory.

Q(m3/sec) Hp(m) Ps(HP) o(%) Hsys(m)

0 46.2 4.2 0 26

0.005 42.9 7 40.07 26.77

0.01 39.6 9.4 55.21 29.09

0.015 36.3 11.4 62.87 32.96

0.02 33 12.9 67.21 38.38

0.025 29.6 14 69.73 45.34

0.03 26.3 14.6 71.05 53.85

0.035 23 14.8 71.44 63.91

0.04 19.7 14.6 70.97 75.52

MPastress

shearepermissibltheisandN

PTWhere

TD s

ss

s

ss

30

,30

,16 m ax

3

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http://worksheets%2C%20labs%2C%20and%20assignments%20%28tubomachinery%29/Assignment%205-1%20(MEng%203202).dochttp://worksheets%2C%20labs%2C%20and%20assignments%20%28tubomachinery%29/Assignment%205-1%20(MEng%203202).dochttp://worksheets%2C%20labs%2C%20and%20assignments%20%28tubomachinery%29/Assignment%205-1%20(MEng%203202).dochttp://worksheets%2C%20labs%2C%20and%20assignments%20%28tubomachinery%29/Assignment%205-1%20(MEng%203202).doc


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Step-3:-

Determine the Hub diameter (DH) and the diameter of suction flange (Dsu).a) Hub diameter (DH) can be assumed 1 cm larger than the shaft diameter.

b) Suction flange diameter (Dsu).

Step-4:- Determine Impeller dimensions.

a) Impeller eye diameter (Do).

cmDD sH 1

smvelocitysuctionisVWhereVQD susu

su /3,4

2

1

4H

su

o DV

QDD

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Preliminary Design of a Centrifugal (Radial) Pumpb) Impeller inlet passage width (b1).

c) Inlet vane angle (1).

d) Impeller outside diameter (D2).

e) Impeller outlet passage width (b2).

Note:- Other variables and dimensions can be determined from the

velocity diagrams.

1

1

1

11

1

111

1

sin85.0,

/25.0,

D

tZD

orfactorncontractiotheisand

smVvelocityflowisVWhere

VD

Qb sur

r

or DDand

NDuWhere

u

V

diagramsvelocitythefrominletradialaFor

11

1

1

11 ,

60,tan

,

05.1,5.84

2 tcoefficienheadtheisWhereN

HD

00

2

222

2 4030 toassumedbemayandVD

Qb

r

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Step-5:-

Design vanes of the impeller.a) The radius of arc (), defining the shape between any two rings having radius ra

and rb is given by,

b) The number of vanes (Z) is given by,

c) The thickness of vanes ( t ) can be determined from stress analysis. In actualdesign a uniform thickness of 3 mm is adopted.

anglevanetheisWhererr

rr

aabb

ab

,

)coscos(2

22

221121

12

12 ,2

)(sin5.6

andWhere

DD

DDZ

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d) Plot or generate the curves of vanes.

Procedure:- Divide the distance between the impeller internal radius r1 and r2 into a

number of equal parts (say 4 or 5) as shown in the figure. For each division determine the values of r, ra, rb, and as shown in the

table.

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Draw all concentric circles with O as center and radius r1,rx,rY,rZ and r2.

Draw a radial line OJ = r1 from the center of rotation O to the point J. With J as center and angle OJK = 1 and radius 1, get point K . With K as center and radius 1, draw the arc JL . Draw a line through points L and K of length 2 = LM. With M as center and radius 2, draw the arc LN. Draw a line through points N and M of length 3 = NP. With P as center and radius 3, draw the arc NQ. Draw a line through points Q and P of length 4 = QR. With R as center and radius 4, draw the arc QS.

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Step-6:- Compute data for determining the passage width.

Similarly, the following data can be obtained and the side view of theimpeller is drawn as shown.

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Both the front and side views are as shown below.

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Step-7:- Design the volute.

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The following relations are used in designing the volute.

718.2log

,

,

.6030sec

60,

tan,2

tan2

,

2

,

,360

.

.

:

2

tan

2

00

0

23

23

3

22

0

2

2

arithmnaturalofbasebasetheise

radianinmeasuredangletheis

vaneimpellertheofanglevaneguidetheisWhereeRR

bygivenisRforrelationThe

orlinesradialthewithofangleanhavingtrapezoidbetoassumedisvolutetheofshapetionalcrossbasicThe

and

RandRradiusanybetweencedistheisxWherexbb

bygivenisbwidthaverageThe

bb

asassumedisbvolutetheofwidthbaseThe

tonguethefrommeasuredangletheisWhereR

dRb

Q

vR

neglectedbetoassumedarelossesandslipofeffectThe

f lowspiralnearlyhavetoassumedvolutetheinwaterThe

sAssumption

t

R

R

ut

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The output of the mathematical relations for the volute is as shown in thetable.

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Using the values in the table, the volute can be drawn as shown in the figure.

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This is generally a preliminary design, but the results of a complete design issimilar to the design shown in the figures.

Video-7

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Two important areas of application of thermodynamics are powergeneration and refrigeration.

The devices or systems used to produce a net power output are oftencalled heat engines, and the thermodynamic cycles they operate on arecalled power cycles.

Main application areas

ofThermodynamics

Power cycles Refrigeration cycles

Vapor powercycles

Gaspower cycles

Steam power plants

Nuclear power plants

Gasoline (Otto cycle) engines Diesel (Diesel cycle) engines

Gas turbines (Brayton cycle)

Refrigerators Heat pumps

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What is a Steam Turbine ?

A Steam Turbine is a turbomachine which converts steam or vapor

(thermal) energy into mechanical energy. The so obtained mechanicalenergy is used to drive a generator.

Steam power plants use steam or vapor as a working medium. Suchpower plants are: coal plants, nuclear plants, or natural gas plants,depending on the type of fuel used to supply heat to the water or steam.

The thermodynamic cycle on which such power plants operate is called

the vapor power cycle (Rankine cycle). In the vapor power cycle theworking fluid continuously vaporized and condensed. Vapor(steam) orwater is the most common working fluid used in vapor power cyclebecause of its many desirable characteristics, such as low cost,availability, and high enthalpy of vaporization.

(Refer:- Engineering Thermodynamics).

Steam turbineThermal(Steam)energy

Mechanicalenergy


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Principle of operation of a simple vapor power cycle (Rankine Cycle)

A simple ideal vapor power cycle (Rankine cycle) consists of four

processes.

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Principle of operation of a Reheat-Regenerative vapor power cycle

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Steam Turbine

Basic configuration of a Steam Power Plant

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Steam Turbine

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Steam Turbine

Basic configuration of a Steam Power Plant

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Steam Turbine

In a steam turbine, high-pressure steam from the boiler expands in a set

of stationary blades (nozzles). The high-velocity steam from stationaryblades (nozzles) strikes the set of moving blades (buckets) as shown inthe figure.

The kinetic energy of the steam or vapor isutilized to produce work on the turbine rotor.

Low-pressure steam then leaves the turbineand enters into the condenser.

There are two types of steam-turbine stagedesigns: the impulse stage, and the reaction stage.

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Steam Turbine

Principles of Impulse stage-turbines

In the case of Impulse stage-turbine: pressure change (drop) occurs across thenozzles.

pressure drop does not occur across themoving blades (buckets).


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Steam Turbine

Velocity diagram

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Steam Turbine

Velocity diagram

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Steam Turbine

Maximum Efficiency

For supplementary notes, referRefernce-1 ( Turbomachinery design and theory)

Page 279 324

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Main application areas of Gas Power Cycles are in the design andoperation of IC Engines and Gas Turbines.

Gas Turbines

IC EnginesIn gas power cycles, the working fluid remains a gas throughout the entire cycle.Spark-ignition engines, compression-ignition engines, and conventional gas turbinesare familiar examples of devices that operate on a gas power cycle. In all theseengines, energy is provided by burning a fuel within the system boundaries. That is,

they are internal combustion (IC) engines.

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Principle of operation of a simple gas power cycle (Brayton Cycle)

Brayton cycle

The Brayton cycle, an ideal cycle for gas turbines, was first proposed byGeorge Brayton in 1870. This cycle can operate as an open cycle or as aclosed cycle.

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The closed Brayton cycle composed of four internally reversible processes:

isentropic compression (1-2), constant-pressure heat addition (2-3), isentropicexpansion (3-4), and constant-pressure heat rejection (4-1). Such a cycle isused in the combined Vapor-Power Cycle to generate electrical energy.

Principle of operation of a simple gas power cycle (Brayton Cycle)

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Ideal Jet-Propulsion cycleThe ideal Jet-Propulsion cycle is a modified Brayton open cycle, used in thedesign of jet propulsion systems. Gas-turbine engines are widely used to

power aircraft because they are light and compact and have high power-to-weight ratio. Aircraft gas turbines operate on an open cycle called a jet-propulsion cycle. The ideal jet-propulsion cycle differs from the ideal Braytoncycle in that the gases are expanded to a pressure such that the powerproduced by the turbine is just sufficient to drive the compressor and the

auxiliary equipment, such as a small generator and hydraulic pumps. That isthe net work output of a jet-propulsion cycle is zero. The gases that exit theturbine at a relatively high pressure are subsequently accelerated in a nozzleto provide the thrust to propel the aircraft.


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What is a Wind Turbine ?

A Wind Turbine is a turbomachine which converts wind (fluid) energy intomechanical energy. The so obtained mechanical energy is used to drive apump or a generator.

Wind energy is a renewable, pollution free and cheap energy source likethose of water and solar energies.Wind is air in motion, caused by uneven heating of the earth by the sun.

This creates difference in densities and leads to a mass movement of airfrom one location to the other.

Wind Turbines may be classified as Vertical-Axis, Horizontal-Axis, Fast-Running, Slow-Running, etc.

Wind turbineWind

energy

Mechanical

energy

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Component parts of a Horizontal-Axis, Fast-Running Wind Turbine

A

Detail A

Class work

1. Write the principle of operation.

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Principle of operationThe figure shown below is the cross-section of the rotor blade called anaerofoil. It is similar to the aerofoil of an aircraft.

When air is moving over the surface of

an aerofoil, a pressure difference iscreated ( Refer:-Bernoullis Principle ).This pressure difference creates aresultant force F. If this force isdecomposed into perpendicular andparallel to flow direction, results Lift L

and Drag D.

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Determination of maximum power available from the wind

.,

.,2

1

0

2

0

velocityflowairisVand

airofmassismwheremVKE

As it is already defined, wind is

air in motion. Since air hasmass and a certain velocity,then it has got a kinetic energygiven by:

Question:-Is it possible to getsuch power in a real machine ?

I II IVIII

p

p

2

02

1

,)(

Vmt

KEP

bygivenisPPowerWind

wind

wind

3

0

0

2

1,

.

,

,

AVPHence

airofdensityis

rotortheofareasweptisAWhere

AVmbut

awind

a

a


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The maximum mechanical

power has a direct relationshipwith Cpmax. Cpmax is known asthe Betzs coefficient and isobtained from the followingassumptions: flow is compressible,

flow is non-viscous, flow is entirely axial.

Hence,

I II IVIII

p

p

)2.11..(........................................)'(,

111...........)()(

:'

1010

AppFlso

).(uVAuuVmF

MotionofLawSecondsNewtonrom

th

ath

)3.11(............................22

:)()('

22

00

g

u

g

p

g

V

g

p

IIandIbetweenEquationsBernoulliApplynig

aa

)4.11(..............................22

'

:)()('

2

10

2

gu

gp

gu

gp

IVandIIIbetweenEquationsBernoulliApplynig

aa

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I II IVIII

p

p

)5.11(................................)(2

1'

:)4.11()3.11(

2

1

2

0 uVpp

andEquationsFrom

a

)6.11(................................)(2

1

:)2.11(int)5.11(

2

1

2

0 uVAF

oEquationsngSubstituti

ath

)7.11........(........................................)(2

1

)()(2

1

:)1.11()6.11(

10

10

2

1

2

0

uVu

uVAuuVA

andEquationsEquating

aa

Which shows that the airvelocity u at the rotor isequal to the mean of theupstream and downstream

velocities V0 and u1.

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I II IVIII

p

p

Axial Induction Factor a

The axial induction factor ais a measure of the decreasein axial air velocity through theturbine .

The power extracted by therotor is the difference betweenthe power of the wind up-

steam and down-stream of therotor. Hence, Pmech is given by,

)8.11...(..................................................0

)1( Vau

)9.11(..................................................)21(

)(2

1)1(

:)7.11()8.11(

01

100

Vau

uVVa

andEquationsEquating

)10.11(................))1(4(2

1

:,

)(2

1

23

0

1

2

1

2

0

aaVAP

givesuanduforngSubstituti

uVuAP

amech

amech

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I II IVIII

p

p

Maximum power is obtainedwhen Cp attains its maximumvalue. Hence, Cpmax can beobtained by differentiating theabove equation,

2

3

0

23

0

)1(4

2

1

))1(4(2

1

aa

AV

aaVA

P

PC

a

a

wind

mech

p

3

1

0)1(4(2

aaada

d

da

dCp

3

))3

12(1(

)21(

),9.11(sin

max

0

0

01

1

V

V

Vau

equqtionguobtainedbecanpower

imumthetoingcorresponduofvalueThe

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I II IVIII

p

p

The maximum mechanical

power is given by,

Power Density (PD)The ratio of mechanical power

(Pmech) to swept area (A) ofthe rotor.

Assignment:- Plot the curves

of PDversus V0.

27

16

)3

11)(

3

1(4

)1(4

2

2

ma x

aaCp

)2

1(

27

16 30ma x AVP amach

)2

1(

27

16 30V

A

PP a

mechD

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Fluid Couplings and Torque Converters are power transmitting systems. Power istransmitted between the driving and driven shaft usually by mechanical means like

gear drive. In such drives the shock loads and vibration are transmitted from one sideto the other which is not desirable. Also in gear drives there is a speed fluctuation. Incertain cases where the driven machine has a large inertia, the driving prime moverlike electric motor will not be able to provide a large starting torque. Instead of themechanical connection if fluids can be used for such drives, high inertia can be met.Also shock loads and vibration will not be transmitted. Smooth speed variation is also

possible.There are two types of power transmitting devices using a fluid.

i) Fluid Couplingii) Torque Converter or Torque Multiplier

i) Fluid Coupling

A sectional view of a Fluid Coupling is as shown in thefigure. In this device the driving and driven shafts are notrigidly connected. The drive shaft carries a pump withradial vanes and the driven shaft carries a turbine runner.Both of these are enclosed in a casing filled with oil ofsuitable viscosity. The pump accelerates the oil by

imparting energy to it.

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i) Fluid Coupling

The oil is then directed to hit the turbine vanes where

energy is absorbed and the oil is decelerated. Thedecelerated oil now enters the pump and the cycle isrepeated. There is no flow of fluid to or from the outside.The oil transfers energy from the drive shaft to the drivenshaft. As there is no mechanical connection between theshafts, shock loads or vibration will not be transmittedfrom one to the other. The turbine will start rotating onlyafter a certain level of energy is picked up by the oil fromthe pump.

ii) Torque ConverterIn case of Fluid Coupling the torque on the driver and

driven members are equal. The application is for direct-drive machines. But there are cases where the torquerequired at the driven member should be more than thetorque on the driver or in other words, the speed requiredat the driven member should be less than the speed onthe driver. Such an application is widely used in

automobile transmission systems.

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ii) Torque Converter

One advantage of Torque Converter is that the speedvariation is smooth as compared to mechanicalconnection (Gear Drive) as shown in the figure.

A sectional view of Torque Converter is as shown in thefigure. Torque Converter consists of three elements:Pump impeller, Turbine runner and a Guide wheel orStator.The pump is connected to the drive shaft. The guidevanes are fixed. The turbine runner is connected to thedriven shaft. All the three are enclosed in a casing filledwith a suitable transmission oil. When the engine starts tooperate, mechanical energy from the crankshaft istransmitted to the driven shaft and then to the pump

impeller. The impeller then imparts energy to the oil.

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ii) Torque ConverterDue to centrifugal force, the oil moves out of the impellerto strike the turbine runner. In doing so the fluid impartsmomentum to the turbine and the turbine together with thedriving shaft starts to rotate and transmit the power to theplanetary gear box. The oil after being expended itsenergy in the turbine runner made to flow back to thepump impeller through a stator or guide vane. This

produces a reactive torque on the pump which increasesthe torque to be transmitted.

Pump ImpellerTurbine Runner

Automatic transmissionVideo-4

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ii) Torque Converter

Automatic transmission

Manual transmission


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Endof

the Course

Final Examination !

Turbomachinery (MEng 3202) Part II

Documents

Transcript of Turbomachinery (MEng 3202) Part II