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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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.
Assuming potential flow, the flow pattern would look as shown below
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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
Assuming potential flow, the flow pattern would look as shown below
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
=2 lnr + c97
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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 =
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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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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.