Me2135e Lecture Notes - Turbomachinery and Potential Flow 2014

8/12/2019 Me2135e Lecture Notes - Turbomachinery and Potential Flow 2014

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NATIONAL UNIVERSITY OF SINGAPORE

DEPARTMENT OF MECHANICAL ENGINEERING

ME2135E Fluid Mechanics II

TURBOMACHINERY AND POTENTIAL FLOW

(PART I)

Dr Lua Kim [email protected]

2013-14

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mailto:[email protected]:[email protected]:[email protected]


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Chapter 1

TURBOMACHINERY

Definition:A turbomachine is a device which adds energy to or extracts energyfrom the fluid passing through it.

Add Energy : Pumps Extract Energy : Turbines

Compressors

Fans

PUMPS: 1. Introduction and pump classification

2. Basic energy consideration

3. Elementary pump rotordynamics

4. Pump characteristics and similarity

5. Matching pump and system requirements

6. Cavitation

7. Further topics

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REFERENCES:

1. White, F. M. Fluid Mechanics,McGraw-Hill.

2. Potter, C. P., Wiggert, D. C., & Ramandan, B. H. Mechanics of Fluids,

Cengage Learning Engineering.

3. Douglas, J. F., Gasiorek, J. M. and Swaffield, J. A. Mechanics of Fluid,

Taylor & Francis.

4. Sayers, A. T. Hydraulics and Compressible Flow Turbo machines,

McGraw-Hill.

5. Cengel, Y.A. and Cimbala, J. M. Fluid Mechanics, Fundamentals and

Applications,McGraw Hill.

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1. Introduction and Pump Classification

The Pump is a common engineering device used to add energy to a fluid. Itis employed in all kinds of industries which involve the transport and processing

of fluids.

1.1 Terminology

Pumping of liquids (essentially incompressible): PUMPS.

Pumping of gases/vapours (compressible):

FANS BLOWERS COMPRESSORS

We will concentrate on the pumping of liquids in this course.

Pumps for liquids may be divided into two categories:

Dynamic Pumps Positive-displacement Pumps.

Basic types:

Dynamic Pump Positive-displacement Pumps

Centrifugal

Axial

Mixed flow

Reciprocating

Rotary

increasing pressure requirements

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1.2 Basic construction- Positive-displacement Pumps

Schematic design of positive-displacement pumps: (a) plunger or reciprocating

piston, (b) gear pump, (c) double-screw pump, (d) sliding vane, (e) three-lobe pump,(f) double-circumferential piston, (g) flexible-tube squeegee.

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The heart is a two-chamber displacement pump.

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Basic construction- dynamic Pumps

The basic components of a centrifugal pump are:

- the impeller- the volute casing- the diffuser ring. (optional)

Cut-away schematic of a typical centrifugal pump.

A centrifugal pump with a stationary diffuser ring.

Stationary

diffuser

vanes

Diffuser

Impeller

Volute

Impeller Diffuser

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A centrifugal pump impeller (unshrouded).

A centrifugal pump impeller.

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Schematic of a typical axial pump.

Axial flow impellers.

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General characteristics ofDynamic andPositive-displacement pumps

Dynamic Pumps Positive-displacement Pumps

add energy by fast-moving blades orvane; no closed changes

steady discharge of fluid

high flow rate

low viscosity fluid

low to moderately high pressure rise

need priming when filled with air/gas

force fluid along by volume changes;fluid cavity opens to admit which isthen squeezed through an outlet

discharge may be pulsatile or periodic

low flow rate

may be used for very viscous fluid

high to very high pressure rise(sturdy construction required)

no priming needed for mostapplications

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An electrical powered centrifugal pump

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2. Basic Energy Consideration

TheBernoulli Head, H, at a point is given by

= + + (2-1)The quantity (gH) is a measure of the energy per unit mass of the fluid.

The Steady-State Energy Equation for flow through the pump is given by

gH0= gHi+ ws- wf (2-2)

where

ws: shaft work per unit mass (energy input),

wf: energy lost in the pump between inlet i and outlet o,

which is a statement thatthe energy at outlet is equal to the energy at inlet plus

net gain in energy (ws-wf) (per unit mass).

Equation (2-2) can be written in terms of heads as

H0=Hi+ hs- hf (2-3)

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where

=

, gain in head due to added energy,

= , loss in head due to frictional losses.Let Q denotes the volume flowrate through the pump.

Useful Power = Power transmitted to the fluid = (Q) (g), (2-4)mass- flowratewhere =Hi- Hois the head transmitted to the fluid. Power input to pump:

= = (2-5)whereT : Torque at the shaft of the pump,

w: Angular velocity of the shaft (radians per second).

TheEfficiency of the pump,

= =() (2-6)

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3. Elementary Pump Rotordynamics

3.1 Centrifugal impeller

We shall consider the case of an idealised flow through a centrifugal impeller:

Assumptions: - No viscosity effects

- flow enters the impeller tangentially to the blades (no-shockcondition).

- uniform conditions along the circumferential inlet and outletof the impeller (as if there are infinitely many blades of zerothickness to guide the flow).

- flow leaves the impeller tangentially to the blades.

Idealised flow through a centrifugal impeller. (a) Impeller control

volume; (b) velocity diagrams at control surfaces.

(a) ub1

ub2

ur2

ur1

Vn2

Vn1

Vt1

Vt2

V2

V1

(b)

Axis of rotation

V1= Vn11

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ub: Absolute blade velocity (= rw) ;

ur : Flow velocity relative to blade;

V : Absolute flow velocity;

Vt: Tangential component of V;

Vn: Normal component of V.

Using a control-volume analysis:

Rate of gain of angular momentum by volume = Torque (T) applied to the fluidby the impeller = rateof outflux angular momentum - rate of influx of angularmomentum

T = (Q) (r2Vt2) - Qr1Vt1mass- flowrate ang. mom/ unit-mass

or T =Q(r2Vt2 - r1Vt1) (3-1)

Power input, P= Tw=Q(r2wVt2 - r1wVt1)

or P =Q(ub2Vt2 - ub1Vt1) (3-2)

Since Power P = (Q)(gH) , we have :

= = (3-3)

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The headHegiven by (3-3) is commonly termed theEuler Head.

For the ideal or design case, the angle 1= 90 ; the inflow has zero whirl, Vt1=0, Vn1= V1. This is also known as thezero pre-whirl condition. For the design

case :

P =Qub2Vt2 (3-4)

=

(3-5)

We note from geometry that

=(cot) = (cot) (3-6)From volume conservation: (211)1 = (2), so that

1 = 211 ,

= 2

where b1and b2are the blade widths at the inlet and outlet respectively.

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Applying these to equation (3-5) (the design case) yields

= (cot) , = (cot) (3-7)

Equation (3-7) gives theHead versus Flowrate relation for the case of an idealimpeller operating at the design case. It can be seen that the relation is determined

by thephysical characteristics (quantities) of the impeller:

=, 2, r2 and b2These quantities may be explicitly controlled by impeller design.

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The figure on the left shows the effects of

outlet blade angle 2 on the Head (H)

versus Flowrate (Q) relationship for an

ideal impeller. See figure below.

Backward-curved bladed pumps have a maximum point in their Power versus

Flowrate curve. The power requirement (P QH) of Radial- and forward- curved

bladed pumps on the other hand rises continuously with the flowrate Q. An

electric motor driving a radial- or forward-curved runs the risk of being

overloaded. Forward-curved pumps may also suffer from unstable operation, an

oscillatory condition in which a pump 'hunts' for its equilibrium point. Backward-

curved blade is therefore the generally preferred design.

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3.2 Real flow in a centrifugal pump

Fluid viscosity causes boundary layers to form on the blades or vanes.

The boundary layers may separate due to adverse pressure gradient. Flowmay separate due to 'shock' (non-tangential) condition at entry.

The formation of boundary layers and separations reduces flow through theimpeller.

The flow is non-uniform circumferentially at the outlet.

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3.3 Axial Impeller

Idealized axial-flow impeller

We will consider the design case for which the flow has no pre-whirl:1 = 90 1 = 0, 1 = 1 = = (3-8)Let us examine the contribution from one elemental ring of thickness (r) at

radius r. The mass flowrate through the elemental ring is

(

)=(Q) =2r(r )Vf (3-9)

Elemental ring

V2

ub2

ur2

ub1

ur1

Vn1

Vn2

Vt2

V1Vt1

Section A-A at radius r

V1= Vn1

1

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The torque contribution from the elemental ring, (T), is

(T) =(Q) rVt2 (3-10)

We note from geometry that = (cot) and =1 =(why?).

From these, we have

(T) = 2

(r)Vf[rw Vf (cot2)] (3-11)

The total torque, T, summing the contributions from all the elemental rings, is

= () = ,= 2

(cot

)

(3-12)

The torque Ton the impeller as given by (3-12) may be evaluated when the outlet

blade angle is given as a function of radius: =(). A changing withradius r corresponds to a 'twist' in the blades.

The theoretical power transmitted to the fluid is then

P = Tw (3-13)

A cruder approximation to the torque Tand power P for an axial impeller may

be obtained by considering only the flow condition at the mean radius =+ as an average of the condition over the whole blade:T = (T)=

(Q)(rVt2),

T (rVt2)Rm(Q) = Q() (3-14)21


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Thepower and theEuler Head are given respectively by:

P = Tw = Q(), (3-15)He= =. (3-16)

Note the close similarity of (3-16) with equation (3-5) for centrifugal pumps.

Simplifying Assumptions have been made in the above derivations for axialpumps:

as for centrifugal pumps.

flow remains essentially parallel as it moves from inlet to outlet

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4. Pump Characteristics and Similarity

4.1 Performance Characteristics of centrifugal pumps

The Fluid Dynamics of pumps is very complex. Until more recent times, pumpdesign is very much an art, derived from experience and trial-and-error. Mostmanufacturers still determine the performance characteristics of their pumps

through extensive experimental testing.

Quantities of interest to someone using a pump are:

Q - theflowrate through the pump,

H - the head the pump develops,

Ps - thepower to drive the pump, and

- the efficiency at which the pump is operating.

As an engineer, you will in general be interested in how the headH, the power Ps

and the efficiency change with the flowrate Q delivered by the pump. Typical

plots of these relationships for centrifugal pumps are shown below. Theserelationships are termed collectively as the performance characteristics of the

pump. They are dependent upon the speed of the pump N, which is the rotationalspeed of the impeller.

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4.2 Geometrically-Similar Pumps and Dimensional Parameters

Physical Parameters relevant to the performance Geometrically-Similar pumpsare:

D - Diameter of impeller (a measure of pump size),

Q - Flowrate,

H - Head (or gH, the energy per unit mass),

Ps - Power input to pump (shaft power),

N - Rotational speed of impeller or shaft,

- Density of the fluid being pumped,

- Viscosity of the pumped fluid,

- Average surface roughness of pump components.

Dimensional Analysis reduces these dimensional parameters to a smaller number

of non-dimensional parameters (the - groups). The use of non-dimensionalparameters results in more compact representation for the performance data.

Dimensional Analysis carried out on the above set of dimensional parametersyield the relations:

(4-1)

(4-2)

Try to establish these non-dimensional parameters for yourself.

=

3

,

,

35 = 3 , ,

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For brevity, we abbreviate these non-dimensional groupings as follows:

=

:Head Coefficient

= : Power Coefficient

=

: Flow Coefficient

= :Reynolds numberso that relationships (4-1) and (4-2) become

=

,

,

(4-3)

= ,, (4-4)Most commercially available pumps operate in thefully-turbulent flow regime;

the Reynolds number Re being of the order of 107. At such large Reynolds

numbers, the viscous action of the fluid is small. The effects of Re in (4-3) and

(4-4) are then fairly weak. If we further assume that the pumps are well made

with small relative roughness factor/D (so that the effect of /D is also small),then we can simply write (4-3) and (4-4) as (4-5)

(4-6)

which are simple two-variable relations.

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We may also note that = =

so that () (4-7)Thus efficiency is a function of CQtoo.

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A geometrically-similar series of pumps is also called a homologous series. The

diameter of a pump in a series is therefore a good measure of its size. The

performance of a homologous series of pumps is governed (and fully specified)

by relations of the form

and (4-8a,b)when the effects of/D andReare small. The specification of CPmay be replaced

by that of ().

REMARKS:

When using dimensionless performance parameters supplied by others, the units

of the physical quantities involved must be known.

For example: =Q: m3/s, cu.ft./min, gal./min etc.

N : rad./s, rpm, rps etc.

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EXAMPLE: Given the performance curves for a family of homologous pumps,

you know that the BEP occurs at CQ* 0.1, CP* 0.6, CH 2.0.

Calculate Q, H, (static pressure across the pump), Psand for a pump ofimpeller diameterD = 0.6 m operating at 1500 rpm at BEP. You may assume Qin m3/s,Nin rps,Din m and in kg/m3

SOLUTION: From definitions

* is the head rise across the pump (inlet to outlet). Assuming that ViVoandzi

zo , the rise in pressure (static) is

The power requirement is

The efficiency is

=3 = 0.1 1500

60 (0.6)3 = 0.543/

= = 2.0 (0.6) 150060 = 45.87

=0 = 103 9.81 45.87 = 450kPa =35 = 0.6 3 1500603 (0.6)5 = 729

= = 103 0.54 9.81 45.87729 103 = 0.333

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The efficiency versus Q and power Psversus Q curves for the pump may also

be estimated from the series' performance curves (graphs) =fh(CQ)andCP=

fp(CQ)respectively following the above procedures.

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4.4 Similarity Rules for Pumps in a Homologous Series

Pump characteristics for a homologous series:

Consider the characteristics of two pumpsA andB from the same homologous

series. The above figure shows that for every point on the pump characteristics

of pumpA, there is a corresponding point on the characteristics of pumpB with

the same CQ .

The two corresponding points will also have the same CH, Cpand because these

are functions of CQ. In particular, for corresponding points we have:

=

=

=

3 (4-10)

() = () = = (4-11)() = () ()35 = ()35

()() = 3 5 (4-12)

Q

H

DA, NA

CQ

CH

Seris

Q

H

DB, NB

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= (4-13)

Using equations (4-10) to (4-13), we can predict (estimate) the performance of

one pump, sayB, from the known performance curves (data) of another pump,

sayA, from the same homologous series.

Since a pump is always homologous to itself, the performance of the pump at one

speed, sayN2, may be predicted (estimated) from its performance data at another

speed, sayN1.

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Another empirical relation obtained by Anderson from thousands of tests:

0.94

0.94 110.3

These formulae assume equal value of surface roughness for both pumps. 0.94

instead of l.0 was assumed by Anderson to be the maximum efficiency a pump

can attain regardless of size.

Centrifugal pumps have often been used to pump oils and some rather viscous

liquids. Typical centrifugal pump performance curves are:

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The maximum efficiency tend to decline with the viscosity . Typical values are:

/ Water : 1 10 100 1000

max; 0.85 0.76 0.52 0.11

Beyond

300

Water

, the loss of efficiency is so great that positive

displacement pumps are to be preferred.

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Specific speed

A design parameter which is used by engineers in the selection of pumps is a

quantity known as the specific speed Ns.

For a given series of homologous pumps

the specific speed Ns is obtained by

eliminating the size factorDbetween CQand

CHto give :

=/()/ =//

(4-14)

Moreover, the specific speed Nsis defined at thepoint of best efficiency (BEP).

Nscharacterizes the homologous pump series independent of the pump sizes.

For another homologous series of pumps (a different geometrical design), CQ*

and CH* are most probably different so that the specific speed Nswill be different.

Specific speed Nsversus Optimal pump design

CQ

CH*

CH

CQ*

B. E. P

= =(

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The figure below shows the relationship between specific speed Ns (rad.) and

optimal pump design.

Dimensionless specific speed,Ns(rad)

Variation of hydraulic pump impeller design

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Thus if an application has an estimatedNsof 1.0 and a flowrate of 0.8 m3s-1, a

suitable pump would be of the mixed-flow design. Such a pump is likely to have

a maximum efficiency (at its BEP) in excess of 90%.

As Ns increases, the optimal pump design changes from one of radial-flow

(centrifugal-pump) design to mixed-flow design and to axial-flow design

(propeller pump).

When using specific speed Nsdata supplied by others, it is important to know

what are the units involved in its definition. In the following example, Ns is

defined as

= 1/3/4 = [rpm] [gal/min]1/[ft]3/4

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EXAMPLE: We want a pump to deliver a flowrate of Q = 0.2 m3/s with a head

of 1.83 m of water. We have a motor which runs at 800 rpm. What type of pump

should we use (for good efficiency)? Estimate the power required.

SOLUTION:

= 800 260

(0.2)1/

(9.81 1.83)3/4 = 4.29 (0.683 )TheNsversuschart on page 42 indicates that we should use an axial or propeller

pump. An efficiency of about 80 per cent is expected. The power given to the

water is

P =QgH

= 1000 x 0.2 x 9.81 x 1.83

= 3.59 kW

The shaft power is

=3.59

0.8= 4.49

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5. Matching of Pump and System Requirements

5.1 System Head Curve

The System refers to that part of the set-up without the Pump.

To transport water from reservoir A to reservoir B, we need to supply energy

(1) to meet the static head rise due to increase in elevation:

= (5-1)(2) to overcomefrictional losses along the conveying pipes:

=2 +2 (5-2)

Pump

Suction

side

Delivery

side

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where fi: friction factor for the flow in pipe i ,

Li: length of pipe i ,

di: diameter of pipe i ,

kj: loss coefficient of thej-th valve, bend, joint etc.

Since flow velocities Vi , Vj Q (velocity = Q / X-sectional area), the total

frictional head losses hfQ2 or

(5-3)

The head (energy) requirement of the system delivering a flowrate of Q is

therefore given by :

(5-4)

which is a quadratic curve in theH versus Qplot.

hf= KQ2

Hsys= z+KQ2

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5.2 Equilibrium Operating Point

In delivering water at the flowrate of Q , the pump must supply this amount of

head. At an equilibrium (steady-state) operating point, the head (energy)

supplied by the pump must exactly match what the system requires. Thus

The equilibrium operating point is determined by the intersection of the pump

and system characteristics curves.

The system characteristics curve can be altered by changing the setting of the

valve if one is available. The flowrate Q is changed by the adjustment to the valve.

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5.3 Pumps in parallel

If one pump gives sufficient head but too little discharge, two or more of the same

or different pumps may be used in parallel.

To determine the Head versus Flowrate curve (H vs Q) forpumps in parallel, we

add their flowrates for given head.

Two identical pumps in parallel

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Check valves (which allows flow in only one direction) are usually employed for

pumps operating in parallel to prevent backflow through the other pump when

one pump is operating.

Two non-identical pumps in parallel

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Power requirement forpumps in parallel:

=

(

) +

(

) =

()

(

)

+

()

(

)

(5-5)

= (5-6)

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5.4 Pumps in Series

When the discharge rate of one pump is adequate but the head is too low, pumps

can be arranged in series to increase the head. In series arrangement, the deliveryside of one pump is connected to the suction (inlet) side of the pump that follows.

For pumps arranged in series, we add the heads delivered by the pumps for given

flowrate.

Pumps in Series

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Power requirement forpumps in series:

=

(

) +

(

) =

()

(

)

+

()

(

)

(5-5)

= (5-6)

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6. Cavitation

6.1 The Physical Phenomenon of Cavitation

Cavitation is the name given to the physical phenomenon which consists of the

formation of tiny bubbles in a liquid as a result of a fall in the absolute pressure

within the liquid.

At least twophysical processes are known to be responsible:

1. reduction in absolute pressure causes air and gases initially dissolved inthe liquid to come out of solution.

2. when the absolute pressure is near or below the vapour pressure (pvap) ofthe liquid at the prevailing temperature, vaporization of the liquid occurs

rapidly leading to the formation of vapour bubbles-Boiling.

Process (1) normally precedes process (2) as the pressure is reduced. The bubbles

produced are generally very tiny.

Cavitation usually begins at the point of lowest absolute pressure in a system.

In a pumping situation, the point of lowest absolute pressure is normally at theinlet to the pump.

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When the cavitation bubbles move with the liquid to a region of higher absolute

pressure, such as inside an impeller, they collapse. The collapse of a vapour

bubble (an implosion):

1. may produce locally very high pressure (up to 1000 atmosphere).

2. may produce locally very high temperature (up to 800 K).

3. sound is produced.

4. light may be emitted (sono-luminescence).

It is interesting to note that the collapsing bubbles have a tendency to be attracted

to solid surfaces (Bjerknes effect). Near the end of the collapse, the bubbles may

actually develop a tiny liquid jet that impacts the solid surface with great local

pressure. The en-mass collapse of cavitation bubbles is accompanied by a

distinctive crackling noise.

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Beauty of cavitation: spiral bubble sheets form from the surface of a marinepropeller

Ugliness of cavitation: Collapsing bubbles erode a propeller surface

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6.3 Nett Positive Suction Head (NPSH)

Applying the steady-state energy equation between the pump inlet and a point on

the surface of the reservoir: () =or

+ () = + + 2which leads to

= (

)

2 where hsuc=zi-zr, is termed the static suction lift.

We require the pump inlet pressure pi>Pvap to avoid cavitation.

(6-1)

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The following quantity, which has the dimension of head,

=

()

2

may be used a measure of the tendency for cavitation to set in at the pump inlet.

The following quantity is termed theNett Positive Suction Head (NPSH) :

= + 2 = () Pump manufacturers frequently supply data on the required NPSH necessary to

prevent cavitation in their pumps.

To avoid cavitation, the engineer must design his system so that the NPSH at

pump inlet is greater than the required NPSH at all times during operation.

(6-2)

(6-3)

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From (6.3), we can overcome the problem of cavitation by making the suction

lift hsucsufficiently negative (i.e. pump placed below the level of the reservoir).

We can also apply Dimensional Analysis to obtain a new dimensional parameter

(6-4)

(6-5)

The scaling law forNPSH is

()()1 =1 1

as for pump head.

(6-6)

= ()

= ,,

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6.4 Other Cavitation Parameters

Other cavitation parameters used in the literature are :

1. The cavitation index

= 12

2. The Thoma cavitation coefficient

= 3. The Suction Specific Speed

= 1/[()]3/4

We note that =3/4

(6-7)

(6-8)

(6-9)

(6-10)

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Chapter 2

Potential Flow

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References

Karamcheti, K.Principles of ideal-fluid Aerodynamics, John Wiley, New York

Fox, R.W. & McDonald, A.T. Introduction to Fluid Mechanics, John Wiley,New York)

Green, S. I.Fluid Vortices, Kluwer Academic Publisher

Massey, B. S. Mechanics of Fluids, Van Nostrand Reinhold Co.

Schlichting, H.Boundary Layer Theory,

Shames, I. H.Mechanics of Fluids, McGraw Hill International

Valentine, H. R.Applied Hydrodynamics, Butterworths, London

White, F. M.Fluid Mechanics, McGraw Hill International

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surprising close to the actual ones. In any event, they give valuable insight to

the actual behaviour of the fluid.

Patching viscous and inviscid flow regions. Potential theory that we are

going to study does not apply to the boundary layer regions.

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For Example:

(a) Consider flow over a flat plate

Assuming potential flow, the flow pattern would look as shown below

Flow

Wall

Flow

Wall

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(b) Flow around a circular cylinder.


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Hele-Shaw flow past a circle. Dye shows the streamlines in water flowing at

1mm per second between glass-plates spaced 1 mm apart. It is at first sight

paradoxical that the best way of producing the unseparated pattern of

plane potential flow past a bluff object, which would be spoiled by

separation in a real fluid of even the slightest viscosity, is to go to the opposite

extreme of creeping flow in a narrow gap, which is dominated by viscous

forces.

Circular cylinder atR = 10,000. The drag coefficient consequently remains

almost constant and drops later when the boundary layer becomes turbulent

at separation.

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If we have a more streamlined looking body (aerofoil), the wake would be muchsmaller


Hence in many cases, classical hydrodynamics gives GOOD

APPROXIMATION to flow in REAL FLUIDS.

We will now develop a number of concepts which will be used in

potential flow. Attention will be confined almost entirely to STEADY

TWO DIMENSIONAL FLOW.

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Hele-Shaw flow past an inclined airfoil. Dye in oil shows the streamlines of

plane potential flow past an NACA 64AOI5 airfoil at 13 angle of attack.However, because the Hele-Shaw flow cannot show circulation, the Kutta

condition is not enforced at the trailing edge. Hence infinite velocities are

represented there. The model is between glass plates 1 mm apart.

Symmetric plane flow past an airfoil. An NACA 64A015 profile is at zero

incidence in a water tunnel. The Reynolds number is 7000 based on the

chord length. Streamlines are shown by colored fluid introduced upstream.

The flow is evidently laminar and appears to be unseparated, though one

might anticipate a small separated region near the trailing edge.

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CONTINUITY EQUATION:

Consider a small rectangular element, x y in size, through which the fluid

flows. The average velocities across each face of the element are as shown. For

an incompressible fluid, volume flow rate into the element equals volume flow

rate out; thus for unit thickness perpendicular to the diagram

Vol. flow rate into the element = uyl + vxl

Vol. flow rate out of the element =

Continuity states that:

Vol. flow rate into the element = Vol. flow rate out of the element

u + u

x x y 1 + v + v

y y x 1

uyl + vxl =

u +

u

x x

y 1 +

v +

v

y y

x 1

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Continuity Equation for 2-D

For three-dimensional flow, it can be shown that continuity equation is given by

(1)

(2)

u

x+v

y= 0

u

x

+v

y

+w

z

= 0

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Line Integral of Velocity

From your earlier lecture, you have learned that work done is by a force F going

from A to B is given by

= sindBA Similarly, if we integrate velocity along line AB, line integral of velocity is given

by

=

sin

d

BA

(3)

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Now, in terms of the velocity components (u, v) in the Cartesian coordinates, it

can be shown that

udx + vdy = (Vcos )(dscos) + (V sin)(dssin )

= Vds(cos cos + sin sin )

= Vdscos ( - )= Vdscos (90- )

= Vdssin ()

Therefore

sindBA = d + dBA (4)

V sinds = udx + vdy

-

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Circulation

The circulation denoted by (Greek capital letter 'gamma') is defined as a line

integral of velocity taken around a closed loop, i.e.

=

sin

d

=d + dFor example, circulation around an elemental area is

=d + + dd + dd d= dd=dd

(5)

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Where

(Greek letter xi) = = VORTICITY

In other words,

Circulation = Vorticity area (for an elemental area)

What is the circulation around a finite area?

Consider a closed curve ABCD within the fluid medium. The circulation around

such a curve is defined as the summation of the circulations round component

small circuit.

Circulation around a Closed Contour

In other words

Total circulation =circulation around small area

or mathematically,

=d + d = dd

(6)

For finite area (7)

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IMPORTANT NOTES:

(a)The concept of vorticity may be utilised for distinguishing a flow as eitherIRROTATIONAL or ROTATIONAL. If the vorticity is zero at all points

in a flow region (except at certain special points known as 'singular points'

where the velocity or acceleration is zero or infinite), the flow in that region

is known as IRROTATIONAL.

(b)If the vorticity is non-zero, the flow is known is ROTATIONAL FLOW(c)Vorticity is a vector quantity whose direction is perpendicular to the plane

of the small circuit round which circulation is measured.

(d)In vector form, vorticity is defined as

=

V (curl of velocity)

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As you go once around the in the carousel, you rotate once about your axis

Or solid body rotation

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Example: The steady plane flow in

the figure has the polar velocity

components v = r and vr = 0.

Determine the circulation around

the path shown.

Solution:Start at the inside right corner, point A, and go around the complete

path:

= d = 0( 1) + () + 0(1 ) + 1(1)=( 1)


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The Concept of Stream Function ()

The concept of stream function (Greek letter psi) is based on the principle ofcontinuity and properties of a streamline. It is applicable to two dimensional flow

cases. For incompressible two-dimensional flow, continuity equation gives

+ = 0This equation is satisfied if a function (x,y) is defined such that the above

equation is defined as + = 0

Comparing the above two equations show that this new function must be

defined such that

= = Note: Stream function is a scalar quantity and is considered positive according to

sign convention.

1. Convention

(a) anticlockwise is positive

(b) clockwise is negative

2. The direction of normal goes from left to right when facing the positive

direction

or

+

= 0


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Geometric Interpretation of

It can be shown that lines of constant are streamlines. Consider a streamline as

shown below.

or udy vdx = 0 streamline

Introducing the stream function from above, we have

+ = 0 =Thus the change of is zero along a streamline, or

There is also a physical interpretation which relates to volume flow. From the

figure below, we can compute the volume dQ through an element ds of the

control surface of unit depth.

=

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Volume flow rate through an element ds of unit thickness is given by

dQ = Vds

= (udy - vdx)

but

=

=

therefore dQ =y dy +x dx

= d

In other words the change in across the element is numerically equal to the

volume flow through the element. Conversely, the volume flow between any two

points in the flow is equal to the change in stream function between those points.

Q1 =1

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The difference inbetween two points = volume flux cross any line joiningthe two points.

Further, the direction of the flow can be ascertained by noting whether increases or decreases.

i.e.

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SUMMARY:

Therefore, we get inCartesian coordinates,

Inpolar coordinates,

It can be shown that

=

1 =

u

v

y

x

(8)

(9)

(10)

y

x

u'v'

= =

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Example:

In a two-dimensional incompressible flow, the fluid velocity components aregiven by u = x - 4y and v = -y - 4x. Determine the stream function of the flow as

well as vorticity.

We know that

Integrating the above gives

or

Similarly

u =

v =x

y = 4

=( 4) = xy 2y + f(x)

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The above two equations are compatible only if

The vorticity is given by

Therefore

x=( 4)

=( 4) = xy + 2x + g(y)

= xy + 2x

2y

=vx uy

v

x

=4 and u

y

=4 =4 (4) = 0 (irrotational flow)

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Example:

A two-dimensional flow field is described by = x - y2. Calculate the horizontal

and vertical components of velocity and an expression for velocity of flow at any

point in flow field. State whether the flow is rotational or irrotational, and

determine the volume flow rate between (x1, y1) = (1, 2) and (x2, y2) = (1, 3).

Sketch the streamlines for = 0, l and 2.

The horizontal and vertical components of velocity are given by

But = x y2, and hence

u = -2y v = -l

The resultant velocity is

The vorticity is given by

Since the vorticity is not zero, the flow is rotational. Volume flow rate (Q) is

given by

Q = 2- 1

= (1 - 32) - (1 - 22) = -5 units

u = y v =x

V =u + v =(2y) + (1) =1 + 4 y

= v

x u

= 0 (2) = 2

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From the earlier part of the lecture, the VORTICITY EQUATIONwas shown to

be

Substituting equations (8) and (9) intothe above equation.

we get

If the flow is IRROTATIONAL, Vorticity () = 0

Or :.

2= 0 (which is a Laplace's Equation)

The above equation shows that for an irrotational flow, the stream function must

also satisfied the Laplace Equation. Conversely, fluid-flow problems which do

not satisfy Laplace equation in are ROTATIONAL.

(6) =

v

x u

v

x=

x

x

=

x

uy = y y = y =

vx uy =x + y

x + y = 0

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IMPORTANT NOTE:

One very important property of Laplace's equation is that it is LINEAR. This

means that if we have a series of simple solutions, 1, 2, 3, .. etc, then the

more complex solution can be obtained by

(x, y) = 1(x, y) + 2(x, y) + 3(x, y) + .

Let us now study some simple solutions. More complex solutions can be obtained

from sum of simple solutions.

PARALLEL FLOW

From equation

y = u

x =

v

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Hence

where 1 and 2are functions of integration. These equations are compatible

only if

2(x) = k (constant)

Therefore, = Uy + k

The value of k is arbitrary and is usually set to 0 for y = 0. Hence

= Uy for flow parallel to the x-axis

y = u

x = v = 0 =1(y) = U y +(x)

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SOURCE

Consider point source

For steady flow,

u'(2r)l = Q = volumetric flow/unit length

assume unit length

From equations (10)

(11a)

(11b)

u = 2

v = 0

r =1 =

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Substituting 11(a) and 11(b) into above equations, we get

These equations are compatible only if

Usual to put k = 0 when = 0

in Polar Coordinates (12)

or

in cartesian coordinates (12a)

Note: Artan is the same as tan inverse.

r= 0

1 = 2 =1() =

2

+(r)

(r) = k =

2

+

= 2

= 2 Artan

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SINK

Similarly it can be shown that the stream function for a sink is given by

in Polar Coordinates (13)

in Cartesian Coordinates (13a)

=2

=2 Artan

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POINT VORTEX (or POTENTIAL VORTEX)

Also known as irrotational vortex orfree vortex.

From equation (5), circulation is defined as

=sindFor a potential vortex:

Integration of the above equations gives

(14)

v =r = 2

=2rvor v =

2

u = 0 = 1 = 0

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IMPORTANT NOTE:

The reason why a potential vortex is also called an irrotational vortex is because

the circulation around any contour not surrounding the origin is zero

Proof:

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Circulation around a potential vortex (excluding the origin)

The velocity (v') at A and D =r

The velocity (v') at Band C =

(r+dr)

Note: there is no u' component

Segment AD = rd

Segment BC = (r + dr)d

Therefore circulation around ABCDA =

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Circulation around a forced vortex:

The velocity (v') at A and D = Kr

The velocity (v') at B and C = K(r+dr)

Note: There is no u' component

Segment AD = rd

Segment BC = (r + dr)d

Therefore circulation around ABCDA =

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(d) A vortex

Tornado Waterspout

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Helmholtz Theorem: The circulation around a vortex filament is constant.

(for proof of the theorem, see Fluid Vortices, edited by S.L Green)

In other words, if a vortex filament comes to a waist, where the filament cross-

sectional area is minimal, the average vorticity over that cross-section must be

maximal, and conversely for a broadening of the tube. A related observation is that

vortex tube cannot terminatein a fluidbecause constancy of circulation would not

be achieved. Vortex tubes are thus constrained to forming loopsentirely within a

fluid, or terminating at a solid boundary.

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Kelvin's Theorem: states that the circulation around a material loop is

time-independent, provided the fluid is inviscid, only subject to potential body

forces, and its pressure is a function of density alone,

i.e.

(for proof of this theorem, see Fluid Vortices, edited by S.I. Green)

DDt

= 0

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COMPLEX FLOWS:

SOURCE AND UNIFORM FLOW

For uniform flow 1= -Uy

For source

New flow

=1+ 2

We now look at the velocity components, this gives

(keeping in mmd that )

(15)

=

Uy +

2Artan

d(Artan q)

ds=

1

1 + q dqds

= Q2 or Q2Artan yx

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(16)

Similarly

(17)

Stagnation Point is defined as a place where u = v = 0 and is located at x0, y0.

From equation (I7)

y0= 0 since v = 0 at y = 0

Therefore stagnation point lies on x-axis

From equation (16)

x0 = Q2U

Therefore, Stagnation point is located at

(this is obtained by putting y = 0 and

solve for x)

Q2U , 0

u =

y

=U + Q

2

1

1 +

y

x

1

x

u =U + Qx2(x + y)v =

x

=Q2 1

1 + yx

y

x=

Qy

2(x + y)

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Let us now locate streamline where = 0

= 0 =Uy + Q2

Artan yx

Here we note that the solution has two branches one is when: y = 0

The other is when x = ycot2Uy

Q ()

This is a curve which looks like

For

: x is indeterminate since x =

0

0

we need to use LHospitals rule to find x

x = lim=0 ytan 2yU

Q

x = lim

=01

2U

Q sec 2yU

Q

R

y

0 P x

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=Q2U(see x0from before)

HENCE the curve passes through stagnation point.

For the coordinates ofR:

Put x = 0 in equation (17a), we get

2UyQ

= 2

When x ,

Since no fluid can cross a streamline, now any streamline may be replaced by a

solid boundary.

y = Q

4U

2UyQ

y Q

2

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Hence, we could say that we have solved for flow about a body whose shape is

given by

(18)

and is in a uniform flow field U.

x = ycot2Uy

Q

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SOURCE - SINK PAIR

For source:

For sink:

Therefore, combine flow

or (19)

Therefore if is constant then = constant.

From geometryPmust lie on a circle passing through A and B

A = QA2

B=

QB2

=A +B=

Q2 (A B)

=

Q

2 where

= (

A B)

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In rectangular coordinates

tanB = yx + s

tan

A=

y

x s

tan(A B) = 2syx + y s

Hence,

(20)

=

Q

2=

Q

2tan

12sy

x + y s

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A SOURCE-SINK PAIR IN UNIFORM FLOW

Source-sink pair:

Uniform flow (from right to left):

Combined flow:

Express in Cartesian coordinates

A = Q2 (A B)

B=

Uy

=A +B=

Q2 (A B) Uy ()Let = (A B)

tan = tanA tanB1+ tanAtanB = Artan tanA tanB

1+ tanAtanB

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= 0 when y = 0

or if

(22)

This is an equation of an OVAL shaped curve whose shape depends on

parameter (Uy/Q) and it "scales" with s, i.e. if you double s, you double all

other dimensions.

= Artan yx yx + 1 +

yx

=Uy + Q2Artany

x yx + 1 +

yx

yx yx + s1 +

yx = tan

2UyQ

2yx

+ y = tan2UyQ

x + y = 2sy cot2UyQ x + y 1 = 2 y cot 2 UsQ

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In Cartesian coordinates

=

2y

x + y (doublet)

r

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A DOUBLET IN UNIFORM FLOW: Flow past a circular cylinder

For a doublet:

or

For uniform flow (right to left):

Combined flow (24)

The streamline = 0is given byy = 0

or (equation of a circle with radius a)

A =

2

y

x + y= Qs yx + yB =Uy

=Uy + Q yx + y

x + y = QU = awhere a =

Q

U122


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Hence we have a circular boundary

Rewriting equation (24) in term of a and polar coordinates

(25)

Stream function of a cylinder placed in a free stream going from right to left

=1

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Pressure Distribution on cylinder:

From equation (25)

= 1

=cos 1

= =sin 1 + on the surface of the body, i.e. r = a

u = 0

and v'= 2U sin(which is maximum at = 90and 270)

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Applying Bernoulli's equation

+ 12 = + 1

2( + )

+ 1

=

+1

( 0 + 4

sin

)

i.e.

= + 12(1 4sin)

= 12

= (1 4sin)

In other words, is a function of only.Note:

(i) At S, = 0or 180,Therefore sin

= 0

free stream static pressure

static pressure at cylinder

Pressure coefficient

CP= 1

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(ii)

Therefore,

Therefore, if holes are drilled at = 30and 150, they can be used to measurefree stream static pressure.

(iii)

sin= lTherefore

At = 6

(. .30) 56

(. . 150)sin

=

1

2 sin

=

1

4

CP= 0 i.e. P= P

At

=

2(

.

.90)and

3

2(

.

. 270)

CP= -3

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Lift = sin0 Drag = cos0

It can be shown that for a potential flow past a circular cylinder that

Lift = 0

Drag = 0

The above results, which is based on ideal-flow analysis, show that a cylinder

placed in a free stream experiences no drag force. In fact, it can be shown from

the above theory that any SYMMETRICAL BODY placed in a free stream along

the axis of symmetry should experience no drag force.

D'Alembert Paradox:

Even though the above analysis shows that a cylinder or any symmetrical body

placed in a free stream experiences no drag force, however, in real life (viscous

fluid), the body does experience a drag. This paradoxical behaviour is referred

to asD'AlembertParadox

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FLOW WITH CIRCULATION ROUND A CIRCULAR

CYLINDER IN UNIFORM FLOW

As discussed earlier, flow past a circular cylinder generates no lift. In order to get

lift, we need to introduce circulation to the cylinder. This can be achieved by

using point vortex.

+ =

Flow past a cylinder Point vortex Flow past rotating cylinder

For flow past a circular cylinder:(note: the flow is from right to left)

For a point vortex (anticlockwise):

= 1

=

2

ln

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Combined flow:

On the surface of cylinder, r = a

u' = 0

At stagnation point u' = v' = 0

Put = 0

=

sin

1

2

ln

(

)

u = 1 =cos 1 v =

=sin 1 +

+ 2

1

v = 2sin + 2

2

sin

0+

2= 0

=A +B

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or sin0 =4 ()

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||


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|| =

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No real solution exist on r = a,

:. we need to go somewhere else for stagnation point

If r a, cos = 0 or sin= 1.In other words, stagnation point lies on y-axis,

but why on y-axis??

v' = 0

For sin= -1

This has real roots if > 4Ua

For || >

At stagnation point, u = 0 = 1 cos

sin 1 + + 2 = 0

1 + + 2 = 01 + 2 = 0

2+

= 0

= b b 42 where b =2 , = 1, =

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For sin = 1, no real roots (exercise: show this)

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=sin20

=sin + 12 4sin3 + 2sin + 4 sin0 d

=

sin

+

1

2

(3sin

sin3

) +

(1

cos2

) +

4

sin

0d

=1

2 []0

=1

2 2

Therefore

Hence for finite span b

(28)

Note:

(a) L generated by circulation here is called Magnus Effect.(b) L is perpendicular to U.

L = U (lift per unit length)

L = U b

L = U (lift per unit length)

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(c) L is independent of the radius "a" of the cylinder. Therefore a linevortex of strength , moves with velocity U will also experience a

lateral force (perpendicular to U) of U /unit length of vortex.

(d) Real life examples of Magnus Effect are spinning of a golf ball, or tabletennis ball or lawn tennis ball.

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The experimental Flettner rotor sailboat at the University of Rhode Island.

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THE CONCEPT OF VELOCITY POTENTIAL

The velocity potential is represented by (Greek letter phi) and is defined by the

following expression

d

=

sin

d

Then = sind or

dd

=sin(in streamline coordinate)

In Cartesian coordinate, it can be shown that

dd =; dd =; dd =

Alternatively, the velocity potential

can be defined from the condition of

irrotationality.

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If the flow is irrotational, the vorticity () = 0

but =x V = 0 (curl of velocity)

where

=

+

+

From vector algebra x V = 0 where is a scalar function. Compare thisequation with the one above show that

V = dd =; dd =; dd =

Hence is the potential function of velocityQuestion: What equation does

obey?

From the continuity equation

+ = 0Since u =

and

=

Therefore = and =

substituting these into the continuity equation, we get

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or

Same rules apply, i.e

(1) It is linear

(2) Complex solutions can be obtained from the addition of simple solutions.

Note:

(1)By definition, the existence of potential function implies that the flow isirrotational, i.e. vorticity = 0.

(2)and are given byCartesian Co-ordinates:

Polar Co-ordinates:

Hence like , follows LAPLACE'S EQUATION.

(30)

(31)

(32)

(33)

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Relationship between stream function () and potential function ()

In the previous section, it was learned that lines of constant formed a set of

streamlines. We are now going to show that lines of constant or potential lines,

form a family of curves which intersect the streamlines at right angle.

From calculus, it can be shown that,

For a line of constant , d=0.

Solving for the slope, we get

Similarly, for stream function

For the line of constant (streamline), we get

x dx + y d y = 0

d = x dx + y dy

dydxconstant =

xy =u

v

d

=

xdx +

ydy

x dx + y d y = 0

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Hence

Therefore lines of constant is perpendicular to lines of constant . The two setsof curves hence form an ORTHOGONAL GRID SYSTEM or FLOW NETS

dydxconstant = 1dy

dxconstant

dydxconstant =

x

y

=v

u

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Examples:

In a uniform flow field

Sink

Point vortex

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Doublet

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Method of Images

The previous solutions have all been for unbounded flows, such as a circular

cylinder immersed in a broad expanse of uniformly streaming fluid. However,

many practical problems involve a nearby rigid boundary constraining the flow.

For example, (1) an aerofoil near the ground, simulating landing or take off, or

(2) a cylinder mounted in a wind tunnel with narrow walls. In such cases the basic

unbounded-potential-flow solutions can be modified for wall effects by the

method of images.

Consider a line source placed a distance from a wall as shown below. To create

the desired wall, an image source of identical strength is placed the same distance

below the wall. By symmetry the two sources create a plane-surface streamline

between them, which is taken to be the wall.

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Determination of forces acting on the wall

Therefore, the combined flow field is given by =1+ = Q2 ( 1+ )tan 1 = y h

x=

rsin hrcos

tan

=

y + h

x

=rsin

+ h

rcos

where

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A Point Vortex Near A Wall

For a vortex near a wall, an image vortex of the opposite rotation must be locatedthe same distance below the wall. But of course the pattern could also be

interpreted as the flow near a vortex pair in an unbounded fluid.

The pressure distribution, due to the vortex, along the wall can be determined ina similar fashion as the one shown above.

A Point Sink Near A Wall

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A Source Trapped In A Corner

The number of image vortices required to simulate the flow is as shown below.

Me2135e Lecture Notes - Turbomachinery and Potential Flow 2014

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Transcript of Me2135e Lecture Notes - Turbomachinery and Potential Flow 2014