97212664-turbomachinery-meng-3202-part-ii.ppt

Melesse Haile(Asst. Professor)

Adama UniversityEthiopia

Turbomachinery (MEng 3202)Turbomachinery (MEng 3202)Part - IIPart - II

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, it includes the design of rotor blades, guide vanes and spiral casings.

Classification of work absorbing machines.

Adama University – Melesse Haile (Asst. Professor)Adama University – Melesse Haile (Asst. Professor)

Classification of Compressors

•Positive displacement compressors increase the pressure of a gas or air by reducing the volume. •Dynamic compressors increase the air velocity by the rotational motion of impeller or rotor, which is then converted to pressure at the outlet.


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


Radial (Centrifugal) Compressors

Component parts 1. Impeller 2. Inducer 3. Diffuser and Diffuser vanes 4. Collector or Casing 5. Inlet Guide Vane

Class work

1. Write the function of each component part.


A Typical Compressed air System Components and Network


Radial Compressors…

Principle of operationWhen the impeller is rotating at high speed (more than 20000 rpm), air is drawn in through the eye of the impeller. The absolute velocity of the inflow air is axial. The magnitude and direction of relative velocity at inlet depends upon the velocity of the impeller as well as the magnitude and direction of the entering absolute velocity. The impeller vanes at the eye are bent so as to provide shockless entry (inducer). The air then flows radially through the impeller passages due to centrifugal force. The total mechanical energy driving the compressor is transmitted to the fluid stream in the impeller 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 is collected in a spiral passage from which it is discharged from the compressor.

The pressure and velocity variation across the compressor is as shown in the figure.



Shapes of Radial Compressor blades

Velocity diagrams for a Radial Compressor blades

inlet outlet



Velocity diagrams for a Radial Blades…

at the inletFigure (a) represents the velocity triangle when the air enters the impeller in the axial direction. In this case, C1 = Ca1. Figure (b) represents the velocity triangle at the inlet to the impeller eye and air enters through the inlet guide vanes. Angle is made by C1 and Ca1 and this angle is known as the angle of prewhirl. The absolute velocity C1 has a whirl component Cw1.

at the outletIn ideal case, as shown in Figure (c), air leaves the impeller tip at an angle of 900. Hence, Cw2 = U2. But in actual case due to slip between the impeller and the fluid, Cw1 is somewhat less than U2 as shown in Figure (d).

inlet

outlet



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



Energy Transfer (Pt)

From Euler’s 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 transfer is given by,

ct W

m

P

inlet outlet



Energy Transfer (Pt)…

In a real fluid, some of the power supplied by the impeller is used in overcoming losses that have a breaking effect on the air carried by the vanes. These include windage, disk friction, and casing friction losses. To take into account these losses, a power input factor () , which is the ratio of the actual to theoretical energy transfer, is introduced. 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 )



Energy Transfer (Pt)…

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 .



Effect of impeller blade shape on performance

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

a)Backward-curved blades (2<900):-

From this equation one can observe that, as the mass flow rate increases energy or power transfer decreases.

Ag

Um

g

UPHence

A

mCwhere

g

CUUPand

g

CUPequationsEulerFrom

t

aa

t

wt

2222

22222

22

cot,

,)cot(

,

,'



Effect of impeller blade shape on performance…

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

From this equation one can observe that, as the energy 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


t

aa

t

wt

2222

22222

22

cot,

,)cot(

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,'

tconsisg

UP

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mCwhere

g

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t

t

aa

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tan

0cotcot

,

,)cot(

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,'

22

222

22

22222

22



Effect of impeller blade shape on performance…In the case of backward-curved blades or vanes, the maximum efficiency occurs in the region of maximum power. If the flow rate increases beyond the design point, the power starts to decrease. This keeps the motor to operate safely and this characteristics is called “self-limiting characteristics”. Whereas in the case of radial and forward-curved blades or vanes, as the flow rate increases the power also tend to increase 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 of shapes to get some performance advantage as compared to backward-curved blades.

Compressibility effects If the relative velocity of a compressible fluid reaches the speed of sound in the fluid, separation of flow causes excessive pressure losses and cyclic loading on the impeller due to eddy or shock wave formation. It is, therefore, necessary to control the Mach number at a certain points in the flow to mitigate the problem. The Mach number at the impeller inlet must be less than unity. This is possible by reducing the relative velocity at inlet by introducing a guide vane as shown in the figure.

inlet



Compressibility effects…Actually, reducing the relative velocity at inlet causes the tangential velocity of the absolute velocity to increase. This in turn decreases the work capacity of the compressor.

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 M2 can be written as:

However, it has been found that as long as the radial velocity component (Cr2) is subsonic, 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 a variation of static pressure and is transmitted upstream in a radial direction through the vaneless space. The end result will be exerting a cyclic type of load on the impeller. This in turn lead to early fatigue and failure on the impeller.



Characteristics of a Radial CompressorThe performance of compressible flow machines usually be described in terms of the groups of variables derived in dimensional analysis ( Refer:- Rference-1 on Page 62).These variables are dependent on other variables such as the condition of pressure and temperature at the compressor inlet and physical properties of working fluid. To study the performance of a compressor completely, it is necessary to plot against 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 line of a compressor running at constant speed. Suppose at the beginning the valve is fully closed. The pressure ratio then will have some value as indicated by point A. This pressure ratio is available from vanes moving the air about in the impeller. The pressure head so developed is called “shut off” head.



Characteristics of a Radial Compressor…

Now, suppose that the valve is opened and air flow begins. The diffuser contributes to the pressure rise. The pressure ratio increases and the maximum pressure occurs at point B. But the compressor efficiency at this pressure ratio will be below the maximum efficiency. Point C indicates the further increase in mass flow, but the pressure is dropped slightly from the maximum possible value. This is the design mass flow rate pressure ratio as efficiency is maximum.

Further increase in mass flow will decrease the pressure ratio until point D. This point indicates that the pressure rise is zero. At this point all the power absorbed is used 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 due to chocking, surging and stall.



Characteristics of a Radial Compressor…

Choking:- When the velocity of fluid in a passage reaches the speed of sound at any 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 pressure resulting 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 passage at higher angle of incidence is known as stall. This phenomenon may lead to surging.



Energy transfer by the impeller to the fluid (air)

A radial compressor is a dynamic type compressor and it works on the principle of forced vortex flow ( Refer:- Fluid mechanics ). When a certain mass of fluid is rotated by an external torque, the rise in pressure head of liquid takes place. The rise in pressure head ( Hp ) at any point of the rotating liquid is proportional to the square of tangential 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 more and the liquid will be discharged at the outlet with a high pressure head.



Energy transfer by the impeller to the fluid (air)…

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

Velocity diagram



Energy transfer by the impeller to the fluid (air)…

As shown in the velocity diagram, the angle between v and u is called and the angle between w and u is called . The difference between the absolute velocity v and the relative velocity w is also illustrated.

Velocity diagrams

Inlet

outlet



Ideal Head Equation

The equation which forms the basis of pump, compressor and blower design is based 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 is called “Euler head” 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



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

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

Rotating container filled with fluid

To determine the pressure distribution within the fluid, consider a very small fluid element having a width b, thickness dR and circumferential length Rd rotating with an angular velocity . The centrifugal force acting on the particle is,

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

dRdRbdFHence

dRRdbdmWhereRdmdF

22

2

,

,

dRRRdb

dRdRb

dA

dFdp 2

22



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

Rotating container filled with fluid

Integrating between the radii R1 and R2 gives the total pressure,

g

uuHHH

HHg

ppbut

g

uu

g

ppHence

uRandRR

pp

ipp

pp

2

,2

,

,2

21

22

112

1212

21

2212

2222

1222

12



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

Consider the passage through the impeller at radius r having a cross-section da as shown in the figure. Let the small element of area da moves a small distance ds in time dt. Hence,

g

ww

g

pH

ww

dwwdpp

outletandinletbetweenequationthegIntegratin

dww

wdt

dsbutdw

dt

dsdp

dabyalldividingand

dt

dwdsdadpdadF

i 2

2

),2()1(

,

22

21

2

22

22

2

1

2

1



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

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

g

vv

g

pH

vv

dvvdpp

outletandinletdiffuserbetweenequationthegIntegratin

dvv

vdt

dsbutdw

dt

dsdp

dabyalldividingand

dt

dvdsdadpdadF

d 2

2

),1()2(

,

21

22

22

2

1

2

1

2

1



The Ideal Total Head (HE)

g

vuH

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),90(

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222



The Ideal Power (P) and Torque (T)

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vRvRQ

vuvuQ

PT

HPinvuvuQ

kWinvuvuQ

Wing

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E

)(

)(

)(

746

)(

1000

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)(

1122

1122

1122

1122

1122

1122



Degree of Reaction ()Degree of reaction is the ratio of the static pressure head rise in the impeller (Hi1) to the total head rise (HE).

22

21

22

22

21

22

22

22

12

21

110

22

21

22

1

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,

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2

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u

v

u

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vuv

trianglevelocityoutlettheFrom

u

v

vu

vHence

r

r

ru

u

u

u

Question:- What do you observe from this relation ?



Degree of Reaction ()…

From this relation one can observe that,a)For 2 = 900, the degree of reaction = , as 2 = . This means that the pressure 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 in the 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



Degree of Reaction ()…

d) When = 0, there is no static pressure change in the impeller. The energy transfer is 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 such an impeller is known as an impulse machines. The blades of impulse impellers are forward curved and the blades of reaction impellers are backward curved as shown 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 larger outlet impeller diameter or high rotational speed than impulse machines.



Actual Head or Power developed by the impeller

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

Factors which causes deviation of actual machines from the ideala) Circulatory flow:- If a closed container is filled with fluid and rotated about an axis, the fluid tends to rotate in opposite direction relative to the container due to its inertia. It is apparent that the fluid in contact with the side A of the container is at a higher pressure than at B, as the container is exerting a pressure on the fluid on that side while it is being accelerated as shown in the figure.



Actual Head or Power developed by the impeller…

Factors which causes deviation of the actual machines from the ideal… This implies that the velocity at B is higher than A according to Bernoulli’s equation. The amount of this circulatory flow depends on the shape of the container. It will be less if is long and narrow. In a rotating impeller two flows take place simultaneously; namely, the flow of fluid through the passage and the circulatory 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 to increase the guide blade angle. This effect is illustrated in the velocity diagram where the dotted lines represent the ideal velocity diagrams.




Factors which causes deviation of the actual machines from the ideal… b) 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 velocity diagram.

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'

01

'11

'22'

),90(




Factors which causes deviation of the actual machines from the ideal… c) 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. This tends to cause a mismatch between the blade angles and the inlet flow directions 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 the power to be transferred to the fluid due to the presence of vu1.

g

vuH


g

vuvuH

uE

uuE

22

01

1122

),90(




Factors which causes deviation of the actual machines from the ideal… d) 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 is turbulent 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 this leakage loss is to reduce the discharge capacity.

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

.

sin

22 radialisvelocityflowinletifHg

vu

HH

gcaandimpellertheinLossesheadEulerH

Lu

LE

m

• Video-5• Video-6• Video-7



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 predict these losses. Engineers or Designers prefer to use empirical relations and/or efficiencies to consider the effects of these losses in the performance of the machine.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 circulatory flow. 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 hence the head will reduce.

Hence, the circulatory flow coefficient () is given by,

g

vuH


g

vuvuH

uE

uuE

22

01

1122

),90(


Efficiencies…

b) Overall Head Coefficient ():- the head developed by a turbomachine due to forced 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) developed by the machine is less than the ideal or Euler Head (HE) due to mainly friction and turbulence 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

222 headoffShuttheisHWhere

gH

uor

g

uH

gH

u

22

N

H

N

HgHD

NDgHu

gH

u

5.84

260

602

2

2

22

2


Efficiencies…

d) Volumetric efficiency (V):- the volumetric efficiency is a measure of the amount of 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 the prime mover (Pin).

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leakagetheisQand

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m

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outo

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gHQ

gHmPWhereP

P

,


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 200C. The compressor runs at 20000 rpm, has isentropic efficiency of 80 % and compresses air from 1 bar static pressure to 4.5 bar total 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


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

QNN s

14.1335

017.01450

4

3

4

3

%65o

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


Preliminary Design of a Centrifugal (Radial) Pump…

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 max

3



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

smvelocitysuctionisVWhereV

QD su

susu /3,

4

21

4H

suo D

V

QDD


Preliminary Design of a Centrifugal (Radial) Pump… b) 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

11

11

1111

1

sin85.0,

/25.0,

D

tZD

orfactorncontractiotheisand

smVvelocityflowisVWhereVD

Qb sur

r

or DDand

NDuWhere

u

V

diagramsvelocitythefrominletradialaFor

11

11

11 ,

60,tan

,

05.1,5.84

2 tcoefficienheadtheisWhereN

HD

002

2222 4030 toassumedbemayand

VD

Qb

r



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 actual design a uniform thickness of 3 mm is adopted.

anglevanetheisWhererr

rr

aabb

ab

,)coscos(2

22

221121

12

12 ,2

)(sin5.6

andWhereDD

DDZ



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.



•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.



Step-6:- • Compute data for determining the passage width. Similarly, the following data can be obtained and the side view of the impeller is drawn as shown.



Both the front and side views are as shown below.



Step-7:- • Design the volute.



The following relations are used in designing the volute.

718.2log

,

,

.6030

sec

60,

tan,2

tan2

,

2

,

,360

.

.

:

2tan

2

00

0

23

23

3

220

2

2

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bygivenisRforrelationThe

orlinesradialthewithofangleanhaving

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and

RandRradiusanybetweencedistheisxWherexbb

bygivenisbwidthaverageThe

bb

asassumedisbvolutetheofwidthbaseThe

tonguethefrommeasuredangletheisWhereR

dRb

Q

vR

neglectedbetoassumedarelossesandslipofeffectThe

flowspiralnearlyhavetoassumedvolutetheinwaterThe

sAssumption

t

R

R

ut



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



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



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

Video-7


Two important areas of application of thermodynamics are power generation and refrigeration.

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

Main application areas of

Thermodynamics

Power cycles Refrigeration cycles

Vapor power cycles

Gas power cycles

• Steam power plants• Nuclear power plants

• Gasoline (Otto cycle) engines• Diesel (Diesel cycle) engines• Gas turbines (Brayton cycle)

• Refrigerators• Heat pumps


What is a Steam Turbine ?

A Steam Turbine is a turbomachine which converts steam or vapor (thermal) energy into mechanical energy. The so obtained mechanical energy is used to drive a generator.

Steam power plants use steam or vapor as a working medium. Such power 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 the working fluid continuously vaporized and condensed. Vapor(steam) or water is the most common working fluid used in vapor power cycle because of its many desirable characteristics, such as low cost, availability, and high enthalpy of vaporization.

(Refer:- Engineering Thermodynamics).

Steam turbineThermal (Steam) energy

Mechanical energy


Main application areas of Vapor Power Cycles are in the design and operation of Steam Power Plants.

Steam Power Plant


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

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


Principle of operation of a Reheat-Regenerative vapor power cycle


Steam Turbine

Basic configuration of a Steam Power Plant


Steam Turbine…


Steam Turbine…

Basic configuration of a Steam Power Plant…


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 stationary blades (nozzles) strikes the set of moving blades (buckets) as shown in the figure.

The kinetic energy of the steam or vapor is utilized to produce work on the turbine rotor. Low-pressure steam then leaves the turbine and enters into the condenser.

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


Steam Turbine…

Principles of Impulse stage-turbines

In the case of Impulse stage-turbine:•pressure change (drop) occurs across the nozzles.•pressure drop does not occur across the moving blades (buckets).


Steam Turbine…

Principles of Reaction stage-turbines

In the case of Reaction stage-turbine:•pressure drop occurs across the stationary blades as well as in the moving blades.


Steam Turbine…

Velocity diagram


Steam Turbine…

Velocity diagram…


Steam Turbine…

Maximum Efficiency

For supplementary notes, refer• Refernce-1 ( Turbomachinery design and theory)

Page 279 – 324


Main application areas of Gas Power Cycles are in the design and operation 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 turbines are familiar examples of devices that operate on a gas power cycle. In all these engines, energy is provided by burning a fuel within the system boundaries. That is, they are internal combustion (IC) engines.


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

Brayton cycleThe Brayton cycle, an ideal cycle for gas turbines, was first proposed by George Brayton in 1870. This cycle can operate as an open cycle or as a closed cycle.


The closed Brayton cycle composed of four internally reversible processes: isentropic compression (1-2), constant-pressure heat addition (2-3), isentropic expansion (3-4), and constant-pressure heat rejection (4-1). Such a cycle is used in the combined Vapor-Power Cycle to generate electrical energy.

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


Ideal Jet-Propulsion cycleThe ideal Jet-Propulsion cycle is a modified Brayton open cycle, used in the design 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 Brayton cycle in that the gases are expanded to a pressure such that the power produced by the turbine is just sufficient to drive the compressor and the auxiliary equipment, such as a small generator and hydraulic pumps. That is the net work output of a jet-propulsion cycle is zero. The gases that exit the turbine at a relatively high pressure are subsequently accelerated in a nozzle to provide the thrust to propel the aircraft.


Ideal Jet-Propulsion cycleThe thrust developed in a turbojet engine is the unbalanced force that is caused by the difference in the momentum of the low-velocity air entering the engine and the high-velocity exhaust gases leaving the engine, and it is determined from Newton’s second law. The pressure at the inlet and the exit of a turbojet engine are identical (the ambient pressure); thus the net thrust developed by the engine is,

The power developed from the thrust of the engine is called the propulsive power, which is the propulsive force (thrust) times the aircraft velocity.


For supplementary notes, refer• Refernce-1 ( Turbomachinery design and theory)

Page 325 – 361

How Gas Turbines work ?• Video-1• Video-2• Video-3


What is a Wind Turbine ?

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

Wind energy is a renewable, pollution free and cheap energy source like those 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 air from 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


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

A

Detail “A”

Class work1. Write the principle of operation.


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

When air is moving over the surface of an aerofoil, a pressure difference is created ( Refer:- Bernoulli’s Principle ).This pressure difference creates a resultant force “F”. If this force is decomposed into perpendicular and parallel to flow direction, results Lift “L” and Drag “D”.


Determination of maximum power available from the wind

.,

.,2

1

0

20

velocityflowairisVand

airofmassismwheremVKE

As it is already defined, wind is air in motion. Since air has mass and a certain velocity, then it has got a kinetic energy given by:

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

I II IVIII

p

p’

202

1

,)(

Vmt

KEP

bygivenisPPowerWind

wind

wind

30

0

2

1,

.

,

,

AVPHence

airofdensityis

rotortheofareasweptisAWhere

AVmbut

awind

a

a


Determination of maximum power available from the wind…

The answer is “No”. Because, such power can only be obtained, if there would be 100% extraction. This means, the moving air stops and gives up all its energy. Since the air has to keep on moving for the wind turbine to work, the power that can be obtained from the wind should be less than this maximum amount. This can be related by introducing a power coefficient (Cp).

This implies that there is an optimum value of the velocity of the wind u1 for which the power extraction is maximum.

I II IVIII

p

p’

302

1AVCP

P

PC apmech

wind

mechp



The maximum mechanical power has a direct relationship with Cpmax. Cpmax is known as the Betz’s coefficient and is obtained from the following assumptions:• flow is compressible,• flow is non-viscous,• flow is entirely axial.

Hence,

I II IVIII

p

p’

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

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

:'

1010

AppFAlso

).(uVAuuVmF

MotionofLawSecondsNewtonFrom

th

ath

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

:)()('22

00

g

u

g

p

g

V

g

p

IIandIbetweenEquationsBernoulliApplynig

aa

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

'

:)()('210

2

g

u

g

p

g

u

g

p

IVandIIIbetweenEquationsBernoulliApplynig

aa



I II IVIII

p

p’

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

1'

:)4.11()3.11(

21

20 uVpp

andEquationsFrom

a

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

1

:)2.11(int)5.11(

21

20 uVAF

oEquationsngSubstituti

ath

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

1

)()(2

1

:)1.11()6.11(

10

1021

20

uVu

uVAuuVA

andEquationsEquating

aa

Which shows that the air velocity “u” at the rotor is equal to the mean of the upstream and downstream velocities “V0” and “u1”.



I II IVIII

p

p’

Axial Induction Factor “a”The axial induction factor “a” is a measure of the decrease in axial air velocity through the turbine .

The power extracted by the rotor is the difference between the power of the wind up-steam and down-stream of the rotor. 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

230

1

21

20

aaVAP

givesuanduforngSubstituti

uVuAP

amech

amech



I II IVIII

p

p’

Maximum power is obtained when Cp attains its maximum value. Hence, Cpmax can be obtained by differentiating the above equation,

2

30

230

)1(4

2

1

))1(4(2

1

aa

AV

aaVA

P

PC

a

a

wind

mechp

3

10)1(4( 2 aaa

da

d

da

dC p

3

))3

12(1(

)21(

),9.11(sin

max

0

0

01

1

V

V

Vau

equqtionguobtainedbecanpower

imumthetoingcorresponduofvalueThe



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) of the rotor.

Assignment:- Plot the curves of PD versus V0.

27

16

)3

11)(

3

1(4

)1(4

2

2

max

aaC p

)2

1(

27

16 30max AVP amach

)2

1(

27

16 30V

A

PP a

mechD


Fluid Couplings and Torque Converters are power transmitting systems. Power is transmitted 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 side to the other which is not desirable. Also in gear drives there is a speed fluctuation. In certain cases where the driven machine has a large inertia, the driving prime mover like electric motor will not be able to provide a large starting torque. Instead of the mechanical 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 Coupling ii) Torque Converter or Torque Multiplier

i) Fluid CouplingA sectional view of a Fluid Coupling is as shown in the figure. In this device the driving and driven shafts are not rigidly connected. The drive shaft carries a pump with radial vanes and the driven shaft carries a turbine runner. Both of these are enclosed in a casing filled with oil of suitable viscosity. The pump accelerates the oil by imparting energy to it.


i) Fluid Coupling…

The oil is then directed to hit the turbine vanes where energy is absorbed and the oil is decelerated. The decelerated oil now enters the pump and the cycle is repeated. There is no flow of fluid to or from the outside. The oil transfers energy from the drive shaft to the driven shaft. As there is no mechanical connection between the shafts, shock loads or vibration will not be transmitted from one to the other. The turbine will start rotating only after a certain level of energy is picked up by the oil from the 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 torque required at the driven member should be more than the torque on the driver or in other words, the speed required at the driven member should be less than the speed on the driver. Such an application is widely used in automobile transmission systems.


ii) Torque Converter…One advantage of Torque Converter is that the speed variation is smooth as compared to mechanical connection (Gear Drive) as shown in the figure.

A sectional view of Torque Converter is as shown in the figure. Torque Converter consists of three elements: Pump impeller, Turbine runner and a Guide wheel or Stator. The pump is connected to the drive shaft. The guide vanes are fixed. The turbine runner is connected to the driven shaft. All the three are enclosed in a casing filled with a suitable transmission oil. When the engine starts to operate, mechanical energy from the crankshaft is transmitted to the driven shaft and then to the pump impeller. The impeller then imparts energy to the oil.


ii) Torque Converter…Due to centrifugal force, the oil moves out of the impeller to strike the turbine runner. In doing so the fluid imparts momentum to the turbine and the turbine together with the driving shaft starts to rotate and transmit the power to the planetary gear box. The oil after being expended its energy in the turbine runner made to flow back to the pump impeller through a stator or guide vane. This produces a reactive torque on the pump which increases the torque to be transmitted.

Pump ImpellerTurbine Runner

Automatic transmission Video-4


ii) Torque Converter…

Automatic transmission

Manual transmission


End of

the Course

Final Examination !

97212664-turbomachinery-meng-3202-part-ii.ppt

Documents

theoretical

velocity triangle

direction

air

diffuser

slip

write

figure