Fluid Mechanics and Machinery for II MechRevision May2013

7/30/2019 Fluid Mechanics and Machinery for II MechRevision May2013

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FLUID MECHANICS UNIT 1- INTRODUCTION

UNITS AND DIMENSIONS

DIMENSION:MEASURABLE CHARACTERISTIC

EXAMPLE:LENGTH, MASS, TIME

NOTATIONS:L, M, & T

UNIT:STANDARD MEASUREMENT OF THECHARACTERISTIC

EXAMPLE:Meter for L, Kg for M, and Sec for T

NOTATIONS: m, kg and s

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FLUID PROPERTIES

Density: Mass per unit Volume measured in kg/m3

Specific gravity: Ratio of density of substance todensity of water

s = / w = / 1000 Specific Weight: Weight per unit volume measuredin N/m3

w = g = 9.81 Notations: Density: Specific Gravity: s

Specific Weight: w

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FLUID PROPERTIES

Buoyancy:

When a body is immersed wholly or

partially in a liquid it is subjected to an upward

force. The tendency of this upward force to lift

(buoy) the body up against action of gravity is

called buoyancy.

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FLUID PROPERTIES

Viscosity: is a property which enables thefluid to offer resistance to relative motion

between adjacent layers.

Absolute Viscosity : Notation is ;

Unit is Poise or N s / m2

1 N s / m2= 10 poise

Kinematic Viscosity: Notation is :

Unit is Stokes or m2/s ;

1 m2/s = 104 stokes

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FLUID PROPERTIES

Compressibility: is the ratio of VolumetricStrain to Change in Pressure (direct stress).Compressibility: Notation is ; Unit is m2/N

Bulk Modulus: Notation is K; Unit is N/m2

K = 1 /

Vapour Pressure: The pressure at which theliquid tend to evaporate. When the gas abovethe surface is saturated, it is saturationpressure.

Vapour Pressure increases with increase intemperature.

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FLUID PROPERTIES

Surface Tension: is the property caused by the

force of cohesion at the free surface.Surface Tension: Notation is ; Unit is N/m

Capillarity: Phenomenon by which a liquid risesinto a thin glass tube above or below its generallevel.

h = 4 cos / ( g d)h is capillary rise in m; is surface tension inN/m;

is angle of contact with liquid surface is density in kg/m3; g is gravity in m/s2;

d is diameter of tube in m

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Gas Laws

Boyles Law:can be written as: p V = p1V1 = p2V2 = constant

This relationship between pressure and volume is

called Boyle's Law in his honor.

Boyle's law asserts that pressure and volume areinversely proportional to each other at fixed

temperature

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Gas Laws

Charles Law:

can be written as: V / T = V1/T1 = V2/T2 = constant

where V is the volume of the gas; and T is the

absolute temperature.

Charles's law states that volume and temperature

are directly proportional to each other as long as

pressure is held constant.

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Gas LawsCombined Law:

The ratio between the pressure-volumeproduct and the temperature of a systemremains constant.

This can be stated mathematically as

p V / T = k ; where: p is the pressureV is the volume ; T is the temperaturemeasured in kelvins ; k is a constant (J/K)

For comparing the same substance undertwo different sets of conditions, the law can

be written as: p V / T = p1 V1 / T1 = p2 V2 / T2

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Flow Characteristics

Pressure: Notation is P; Unit is N/m2

Temperature: Notation is T; Unit is K oroC

Density: Notation is ; Unit is kg/m3

Specific Volume: Notation is v ; Unit is m3/kg

Velocity: Notation is C or V; also u, v, w;

Unit is m/s

Volume flow rate: Notation is Q ; Unit is m3/s

Mass flow rate: Notation is m ; Unit is kg/s

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SYSTEM AND CONTROL VOLUME

System:

A system is a region in space, or fixed

collection of matter enclosed by a real or

imaginary boundary. The boundary can be

rigid or flexible and the system can be

fixed in space or moving in space.

What are the components involved?

System, boundary, environment or

surroundings.

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Boundary: It is the surface which separates the

system from the surroundings. It can be real or

imaginary, fixed or flexible.

Surrounding: Everything external to the system issurrounding or environment

Universe: System and surrounding together is

called universe

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Closed System: A system in which there isno mass flow to and from the system

across the boundary. It can interact with

its surroundings through work and heat

transfer. The boundary is free to move in

closed system.

Control Mass: Control Volume in a closed

system is also called control mass since

there is no mass flow to and from the

system across the boundary.

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Open System:A system with fixed space (and

hence boundary) which allows a continuous flowof matter. A control volume is an open system. It

has a fixed space but does not contain fixed mass

or matter. The identity of matter keeps changing

with time. Hence called open system.

Control Surface: The boundary in an open systemis fixed and hence called control surface.

Control Volume: An open system has fixed volume

and hence called control volume.

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CONTINUITY EQUATION

Mass flow rate entering = Mass flow rate

leaving

A C = 1

A1

C1

= 2

A2

C2= Constant

/ = A/A = C/C

u/ x + v/ y + w/ z = 0

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ENERGY EQUATION

Types of Energy:

Pressure head or Pressure Energy

Velocity Head or Kinetic Energy and

Potential Head or Potential Energy

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Pressure Energy:

Formula: P/; Unit J/kg

OR Formula: P/g ; Unit m (a)

Velocity Energy:

Formula: v2/2 ; Unit J/kgOR Formula: v2/2g ; Unit m (b)

Head Energy:

Formula: z; Unit m

OR Formula: gz; Unit J/kg (c )

(a) + (b) + (c ) we get

Total Head or Total Energy = p/ g + V2/2g + Z

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ENERGY EQUATION

Eulers Equation:

p/+ VV + g z = 0

Bernoullis Equation for ideal fluid:

p1/ g + V12/2g + Z1 = p2/ g + V22/2g + Z2= Constant

Bernoullis Equation for real fluid:

p1/ g + V12/2g + Z1 = p2/ g + V22/2g + Z2 + hL

hL is loss in the pipe in m of liquid.

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ENERGY EQUATION

Bernoullis Equation for ideal fluid:

p1/ g + V12/2g + Z1 = p2/ g + V22/2g + Z2= Constant

Assumptions:

Ideal Fluid,Non viscous, = 0

Steady Flow, V/ t = 0Uniform Flow, V/ x = 0

Irrotational Flow

No Energy Loss in the pipe

Applications:

Venturimeter, Orifice meter,

Rotameter, pitot tube, Flow Nozzle

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MOMENTUM EQUATION

Law of Conservation of Momentum: Net force

acting on a mass of fluid is equal to change inmomentum of flow per unit time in that direction.

Momentum Equation:

F = d (mV) / dt

Applications:

To find RESULTANT FORCE

acting on

pipe bends, Reducers, movingblades, jet propulsion etc.

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IMPULSE MOMENTUM EQUATION

Momentum Equation: F = d (mV) / dt

Impulse Moment Equation:

F dt = d (mV)

The impulse of force F acting on massm for a short interval dt is equal tochange of momentum d (mV) in thedirection of force.

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MOMENT OF MOMENTUM EQUATION

The resulting torque acting on a rotating fluid is

equal to the rate of change of moment of momentum

Moment of Momentum Equation:

T = Q ( V2r2 V1r1)

Where is density in kg/m3,

Q is flow rate in m3/s,

V2 and V1 are whirl velocities at r2 and r1 respectively.

Flow rate m = gQ ;Work Done = T = T = Q ( V2r2 V1r1)

Hence work done per sec per unit weight

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The diameter of impeller of a pump is 1.2 m and its peripheral

speed is 9 m/s. Water enters radially and is dischargedfromimpeller with a velocity whose radial component is 1.5 m/s. The

vanes are curved backwards at exit and make an angle of 30o with

periphery. If the pump discharges 3.4 m3/m , find the turning

moment on the shaft.

Soln:T = Q ( Vu2 r2 V u1 r1)

Vu2 = u2 Vf2 cot 2 = 9 1.5 cot 30 = 6.4 m/s

T = Q ( Vu2 r2 V u1 r1) = 1000 x (3.4/60) (6.4 x 0.6 0)

= 217 N-m

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Radial BladeForward

Curved l Blade

u1

vR1

v1

u1

vR1

v1

u1

vR1

v1

u2

vR2v2

Backward

Curved Blade

u2

vR2v2

u2

vR2v2 vf2

Vw2

vf2

Vw2

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PROBLEMS

Formulae:

Fx = Q (V1 Cos 1 V2 Cos 2 ) +

p1A

1Cos

1 p

2A

2Cos

2

Fy= Q (V1 Sin 1 V2 Sin 2 ) +

p1A1 Sin 1 p2A2 Sin 2

are forces exerted by the fluidon

the pipe.

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END OF UNIT ONE

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FLUID MECHANICS UNIT 2 FLOW THROUGH CIRCULAR CONDUITS

LAMINAR FLOW THROUGH CIRCULAR CONDUITS

Velocity Profile is parabolic

u = (1/4)( dp/dx)(R2 r2) .(a)

Umax = (1/4)( dp/dx)R2 ..(b)

Combining the above two equations

u = Umax[1 (r/R)2] ..(c )

Uavg = Umax/ 2 ..(d)

Combining the relations (b) (c ) and (d)

r = 0.707 R at which Uavg occcurs

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LAMINAR FLOW THROUGH CIRCULAR CONDUITS

Pressure Drop in terms of average velocity

h = (p1 p2) / g = 32 Uavg L / (gD2)

This is Hagen Poiseulle equation

h = 32 Uavg / ( D)(L / D)

Multiplying and dividing by Uavg/2 andrearranging,

= 64 / ( UavgD)(L / D) (Uavg2/2g)

= 4 f (L /D) (Uavg2/2g) where f = 16/Re

This is Darcy Weisbach relation

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BOUNDARY LAYER CONCEPT

In a flow over surface, effect of viscosity causes

the fluid to stick to the wall.

This layer will have zero velocity relative to the

surface. Hence a velocity distribution is built up

near the surface.

Velocity gradient is large at the surface. A small

distance away from the surface the velocity

asymptotically approaches the upstream velocity.

This region is called boundary layer.

Outside the boundary layer, flow can be assumed

to be non-viscous flow.

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Laminar and Boundary Layer


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Laminar

Laminar Sub Layer

Turbulent


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BOUNDARY LAYER THICKNESS

Boundary layer thickness ( ) is the distance fromthe wall where the velocity differs by 1% from the

external velocity (U) .

At y = , u = 0.99 U

For a simple velocity profile, u / U = y / ,

Solution for is, = 3.46 x / (Rex)1/2

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DISPLACEMENT THICKNESS

Displacement thickness is the distance by which the

external stream lines are shifted owing to the

formation of boundary layer.

It is given by * = 0 [1 (u / U)] dy

For a simple velocity profile, u / U = y / ,

* = 0 [1 (y/ )] dy

Solution for* is,

* = /2 = 1.73 x / (Rex)1/2

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ENERGY GRADIENT

Total Energy Line (TEL) or Energy Gradient Line

(EGL)

Total Energy per unit weight = p/ g + V2/2g + Z

This is used for drawing the TEL or EGL. Whenfluid flows through a pipe, we can measure the the

total head as above at various points along the

flow length and plot.

Since thee is a loss of head along the flowdirection, TEL or EGL decreases along the flow

direction.

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HYDRAULIC GRADIENT

Piezometric Head H = p/ g + Z

Hydralic Gradient Line (HGL): When fluid flow

through a pipe, we can measure the piezometric

head at various points along the flow length and

plot.

HGL may fall or rise along the direction of flow

HGL is always below EGL

Vertical Intercept = EGL HGL = V2/2g

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Hydraulic Gradient and Total Energy Lines

Fig.1 Hydraulic gradient and total energy line for

(a) an inclined pipe connecting 2 reservoirs

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Fig.1 Hydraulic gradient and total energy line for

(b) horizontal pipe connecting 2 reservoirs

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Consider a long pipe carrying liquid from

a reservoir A to a reservoir B as shown inFig.1.

At several points along the pipeline let

piezometers be installed. The liquid will

rise in the piezometers to certain heightscorresponding to the pressure intensity

at each section.

The height of the liquid surface above the

axis of the pipe in the piezometer at anysection will be equal to the pressure head

(p/w) at the section.

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Total Energy in y axis

Datum along x axis

plotted to scale

If all these points are joinedwe get a straight slopingline

known as 'total energy l ine'.

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Piezometric Head

H = p/ g + Zin y axis

Datum along x axis

plotted to scale

If all these points are joinedwe get a straight slopingline known as

'hydraulic gradient line.

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HYDRAULIC GRADIENTLINE and TOTAL ENERGY LINE

Important points to Remember:

For EGL measure He= p/ g + V2/2g + ZFor HGL measure Piezometric Head H = p/ g + Z

EGL falls in the direction of flow

HGL may fall or rise along the direction of flow

HGL is always below EGL

Vertical Intercept = EGL HGL = V2/2g

For uniform cross section, EGL and EGL have

same shape.

There is no relation between slope of pipe axis

and slope of EGL.

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DARCY WEIS BACH EQUATION

Pressure Drop in terms of average velocity in

circular pipe is given as

h = (p1 p2) / g = 32 Uavg L / (gD2)

This is Hagen Poiseulle equation

Pressure Drop in terms of Reynolds Number and

average velocity in circular pipe is given as

h = 4 f (L /D) (Uavg2/2g) where f = 16/Re

This is Darcy Weisbach relation

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FRICTION FACTOR AND MOODY DIAGRAM

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Moody Diagram gives the values of friction factor fas a function of Re and /D for all types of flow(laminar, turbulent, transient).

Pressure drop in m of liquid h = 4 f L V2/ (2gD)

This loss is called Major Loss in Pipe Friction.

For re < 2000. Flow is Laminar.

In laminar flow, f is independent of/D and itsvalue is given by f = 16/Re

For Re > 2000, there are two regions: Transient and

Turbulent

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For Re > 2000, there are two regions: Transient andTurbulent

In transient region, f depends upon the

Reynold Number Re and Roughness Factor/D ofthe pipe surface.

In turbulent region, f is independent of the

Reynold Number Re and depends solely on

Roughness Factor/D of the pipe surface.

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FRICTION FACTOR AND MOODY DIAGRAMAn engineering college having 1200 students is to besupplied with water from a reservoir 12 km away. Water

is to be supplied at the rate of 50 liters per head per dayand half of the daily supply is pumped in 8 hrs. If thehead loss is due to friction in 55 m, find the diameter ofthe pipe. Take f = 0.004.

Soln: Number of students N = 1200

Length of pipe L = 12000 m; Daily supply = 1200 x 50 =60000 liters = 60 m3

Maximum discharge Q, for which the pipe is to bedesigned, is given by

Discharge Q = 60 / (2 x 8 x 3600) = 1.0417x10-3 m3/s

Given hf= 55 m; f = 0.004

We know Q = A V;

V = 1.326 x 10 -3 / D2; hf = 4 f L V2 / (2gD);

Substituting and solving ; D5 = 3.1298 x 10-7; D = 0.05 m

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COMMERCIAL PIPES MINOR LOSSES

Minor Energy losses are due to

Sudden Enlargement: hen = (V1 V2)2/2g

Sudden Contraction: hcon = 0.5 V22 /2g

Obstruction in Pipe: hobs = A/ [(Cc (A-a)]2

V2

/2g

Entry : hin = 0.5 V2/2g

Exit : hout = V2/2g

Pipe Fittings: hpf = k V2 /2g

Bend in pipe: hbend = k V2 /2g

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FLOW THROUGH PIPES IN SERIES

When three pipes are connected in series,

Total Discharge Q = Q1 = Q2 = Q3

Total pipe loss hL = hL1+ hL2+ hL3

4 f (L /D) (Uavg2/2g) =

4 f (L /D) (Uavg2/2g)1 +

4 f (L /D) (Uavg2/2g)2 +

4 f (L /D) (Uavg2

/2g)3

where f = 16/Re

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Equivalent Pipe: Equivalent Pipe is a pipe of

uniform diameter having loss of head anddischarge equal to the loss of head and discharge

of the compound pipe.

[h f] compound pipe = [h f] equivalent pipe

[Q] compound pipe = [Q] equivalent pipe

Length of Equivalent Pipe = Le = L1+ L2+ L3 + .

For 3 pipes in series, Le = L1+ L2+ L3

Dupits Equation gives equivalent pipe diameter D

L/D5 = L1/D15+ L2/D25+ L3/D35

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Two reservoirs containing water are connected by a straight pipe

1600 m long. For the first half of its length, the pipe is 15 cm

diameter. It is then suddenly reduced to 7.5 cm diameter. The

difference in surface level in the two reservoirs is 30 m. Determinethe flow in l/s. Take f for both pipes as 0.04.

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FLOW THROUGH PIPES IN PARALLEL

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When three pipes are connected in parallel,

Total Discharge Q = Q1 + Q2 + Q3

Total pipe loss hL = hL1 = hL2 = hL3

4 f (L /D) (Uavg2/2g) = 4 f (L /D) (Uavg

2/2g)1

= 4 f (L /D) (Uavg2/2g)2 = 4 f (L /D) (Uavg2/2g)3

where f = 16/Re

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A 200 mm diameter pipeline, 5000 m long delivers

water between reservoirs,the minimum difference in

water level between which is 40 m.

(a) Taking only friction, entry and exit head losses into

account determine steady discharge between the

reservoirs.

(b) If the discharge is to be increased to 50 l/s without

increase in gross head, determine the length of 200

mm diameter pipeline to be fitted in parallel. Consider

only friction losses. ( Take f = 0.016)

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(2)

(3)

(1)

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V2 = v3A2 V2 = A3 V3

Q2 = Q2

Q1 = 2 Q2 = 2 Q3

Hence Q2 = Q3 = 0.025 m3/sV1 = 1.592 m/s

V2 = 0.796 m/s

D =

H = f/(2gD) (l1v22 + l2V22)40 = 0.016/(2x9.81xD) (l1v2

2 + l2V22)

L1 = 3495 m; l2 = 1505 m

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END OF UNIT TWO

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FLUID MECHANICS UNIT 3 DIMENSIONAL ANALYSIS

Dimensional Analysis: It is a

mathematical technique thatsuggests which variables affecting

physical phenomenon are to be

grouped together to obtain

dimensionless quantities. It

provides a functional relationship

between the dimensionless

quantities.

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FUNDAMENTAL QUANTITIES

S.No PhysicalQuantity

Symbol Dimensions

1 Mass M M

2 Length L L

3 Time T T

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GEOMETRIC QUANTITIES


Symbol Dimensions

1 Area A L2

2 Volume V L3

3

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KINEMATIC QUANTITIES


Symbol Dimensions

1 Velocity A L T-1

2 Discharge Q L3 T-1

3 Kinematic

Viscosity

L2 T-1

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DYNAMIC QUANTITIES


Symbol Dimensions

1 Force F ML T-2

2 Dynamic

Viscosity

ML-1 T-1

3 Power P ML2 T- 3

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Dimensional Homogeneity

An equation is homogeneous if the dimensions of

each term on both sides are equal.The powers of fundamental dimensions (M,L,T) on

both sides of equation must be identical.

Homogeneous equations are independent of the

system of units.

Example: V = 2gH

Substituting dimensions on both sides

LT-1 = [ LT-2 L]1/2 = LT-1

hence homogeneous

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BUCKINGHAMS THEOREMIf there are n variables in a problem and these

variables contain m primary dimensions, theequation relating the variables will contain (n-m)dimensionless groups.

These groups are termed as 1, 2, 3, 4, ., n-mEach term contains (m+1) variables where m isthe number of fundamental dimensions and is also

called repeating variables.Let m = 3 and X2, X3, and X4 be the repeatingvariables

Then each term is written as follows1 = X2a1 X3b1 X4c1 X1 ; 2 = X2a2 X3b2 X4c2 X5 ;

n-m = X2a(n-m) X3b(n-m) X4c(n-m) Xn ;

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BUCKINGHAMS THEOREMLet m = 3 and X2, X3, and X4 be the repeating

variablesThen each term is written as follows1 = X2a1 X3b1 X4c1 X1 ; 2 = X2a2 X3b2 X4c2 X5 ;n-m = X2a(n-m) X3b(n-m) X4c(n-m) Xn ;Each equation is solved by the principle of

dimensional homogeneity and values of a1, b1,c1;a2,b2,c2; .,a(n-m), b(n-m), C(n-m) are obtained.The final equation for the phenomenon is given by

1 = [2 , 3 , 4,., n-m ]

2 =[1 , 3 , 4,., n-m ]3 =[1 , 2 , 4,., n-m ]

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SELECTION OF REPEATING VARIABLES

Number of repeating variables = Number of

fundamental dimensions in the problem.Dependent variable should not be selected as

repeated variable.

Select variables such that one contains geometric

property (L,D), another contains flow property (V,

A.) and third variable contains fluid property(, ).Repeating variables together should not form

dimensionless group.

Repeating variables should have the same number

of dimensions.

No two repeating variables should have the same

dimensions.

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DIMENSIONLESS PARAMETERS

B/D: (Breadth /Depth)

Efficiency : (Performance Output/Input)Q/ND3 : (Specific Capacity or Flow Coefficient)

gH/N2D2 :(Specific Head, constant for similarimpellers)

/ND2: Inverse of Reynolds Number

Re: Reynolds Number

P/N3 D5 : (Power Coefficient or Specific Power)

ND/ gH,

T/ N2 D5

/ N2 D3

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MODELS AND SIMILITUDE

What is Model Testing?

The design, construction, and erection of hydraulicstructures and machines involve time, money,

energy, and efforts. To minimize the chances of

failure, it is desired to perform the tests on a small

size models of the structures and machines.

The actual size machine is called prototype.The small replica of prototype is called model.

In certain cases models may be larger than theprototype (Ex: Model of a watch).

Model testing is economical, easy, and can be

changed any number of times.

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APPLICATIONS OF DIMENSIONLESS PARAMETERS

The performance of prototype can be predicted

from the performance of the model.However for this a complete similarity between

model and prototype should exist.

SIMILITUDE: Similitude means complete similarity

between model and prototype. There are three

types of similarity.

Geometric Similarity

Kinematic Similarity and

Dynamic Similarity.

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SIMILITUDE

Geometric Similarity:

Length Scale Ratio Lr= Lp/Lm= bp/bm= dp/dmArea Scale Ratio Ar= Ap/Am= Lp/Lm x bp/bmVolume Scale Ratio Vr= Lp/Lm x bp/bmx dp/dm

Kinematic Similarity:

Time Scale Ratio Tr= Tm / Tp

Velocity Scale Ratio Cr= Cm / Cp = Lr / TrAcceleration Scale Ratio ar= am / ap = Lr / Tr

2

Discharge Scale ratio Qr= Qm / Qp =Lr3 / Tr

Dynamic Similarity:

Force Scale Ratio Fr=Fip/ Fim = Fvp/ Fvm = Fgp/ Fgm

(Inertia, viscous and gravity forces ratio)

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DISTORTED MODEL

Sometimes it may be necessary to make models

which are not GEOMETRICALLY Similar to theprototype.

Example: River and Harbour model; Depth of water

is very large.

Surface roughness can not be reduced to

geometric scale and at the same time createturbulent flow.

Distorted models have different scale factors for

horizontal and vertical directions.

Normally vertical scale 1/100 and horizontal scale

1/200 to 1/500 adopted.

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PROBLEMS

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PROBLEM 1: The diameter and width of the runner

of a turbine are D and B respectively and it rotates

at a speed of N. The working head is gH. If densityand viscosity of the working fluid be and respectively, show that the power developed will

be given by

P = N3 D5 [ B/D, ND/gH, Re]

PROBLEM 2: A hydraulic prime mover. Whentested in a laboratory at 200 rpm under 10 m head

develops 50 kW power. Estimate the power

potential, size ratio, and the specific speed of a

similar machine operating under a head of 30 m

and running at the same speed.

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PROBLEM 1 Soln:

Variables involved are: D, B, gH, N, P, and

Dimensions are:L, L, L2T 2, T 1, ML 2T 3, ML 3, ML 1T 1

Number of variables = 7;

Number of fundamental dimensions present = 3

Number of dimensionless parameters = 7 3 = 4Number of repeat variables considered are

D, N, and .

1 = X2a1 X3b1 X4c1 X1 ; 2 = X2a2 X3b2 X4c2 X5 ;

n-m = X2a(n-m) X3b(n-m) X4c(n-m) Xn ;

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Number of repeat variables considered are

D, N, and .

1 = X2a1 X3b1 X4c1 X1 = Da1 Nb1c1 B1 ;

2 = X2a2 X3b2 X4c2 X5 = Da2 Nb2c2 H1 ;

3 = X2a3 X3b3 X4c3 X6 = Da3 Nb3c31 ;

4 = X2a4

X3b4

X4c4

X7 = Da4

Nb4

c4

P1

;Solving by dimensional approach i.e. equating thepowers of l, M, and T on both sides we get,

1 = B/D; 2 = ND/gH

3 = Re = N D / ; 4 = P / N3 D5 ;

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PROBLEM 2 Soln:

Specific Speed of Model = Specific Speed of Prototype

Equating gH/N2D2, Dp/Dm=1.73

Equating P/N3D5, P = 779.423 kW

NsT model = N P / H5/4 =200 (500.5) / (105/4) = 79.527

NsT proto = N P / H5/4 =200 (P0.5) / (305/4) = 79.527

Same for both.

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END OF UNIT THREE

81

FLUID MECHANICS UNIT 4 ROTO DYNAMIC MACHINES


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HOMOLOGUS UNITS

Machines which are geometrically similar form a

homologous series.

The member of such a series, having a common

shape are simply enlargements or reductions of

each other.

If two machines are kinematically similar, the

velocity vector diagrams at inlet and outlet of the

rotor of one machine must be similar to those of

the other.

Geometrical similarity of the inlet and outlet

velocity diagrams is, therefore, a necessary

condition for dynamic similarity.

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SPECIFIC SPEED

The performance or operating conditions for a

turbine handling a particular fluid are usually

expressed by the values ofN, Pand H, and for a

pump by N, Qand H.

It is important to know the range of these operating

parameters covered by a machine of a particular

shape (homologous series) at high efficiency.

Such information enables us to select the type of

machine best suited to a particular application, and

thus serves as a starting point in its design.

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SPECIFIC SPEED

A parameter independent of the size of the

machine D is required which will be thecharacteristic of all the machines of a homologous

series.

A parameter involving N , Pand Hbut not D is

obtained for turbine as KsT

Similarly, a parameter involving N, Qand Hbut notDis obtained as KsP

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SPECIFIC SPEED (Dimensional)

Speci f ic sp eed Dimensional formula Unit (SI)

(turbine) M 1/2 T -5/2 L-1/4 kg 1/2/ s5/2 m1/4

NsT = N P / H5/4

It is the speed of a geometrically similar turbine thatwould produce unit power under unit head.

Speci f ic sp eed Dimensional formula Unit (SI)

(pump) L 3/4 T-3/2 m 3/4 / s3/2

NsP = N Q / H3/4

It is the speed of a geometrically similar pump thatwould produce unit flow under unit head.

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SPECIFIC SPEED (Dimensional)

Specif ic speed Dimensional formula Unit (SI)

(turbine) M 1/2 T -5/2 L-1/4 kg 1/2/ s5/2 m1/4

NsT = N P / H5/4

Specific Speed Turbine Type

8.5 to 30 Pelton Single jet

30 to 51 Pelton Double jet

51 to 225 Francis

225 to 860 Kaplan/propeller

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THEORY OF TURBO MACHINES

A turbomachine consists of a rotating element,

fluid and a relative motion between the two.

Fluid Machine is a machine which uses fluid for

transfer of energy from fluid to rotor or rotor to

fluid.

Based on the direction of transfer of energy, fluid

machines or turbomachines can be classified as

(i)Turbines ( Power Generating Machine)

(ii)Pumps or Compressors (Power AbsorbingMachine)

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EULERS EQUATION

Eulers head is the energy transferred per unit

weight of fluid from fluid to rotor or from rotor tofluid.

Formula: HE = [u1Vw1 u2Vw2] / g meters

Assumptions: Steady uniform flow with continuity

in pressure.

For Turbines, HE is positive

For Pumps/compressors, HE is negative

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EULERS EQUATION

Specific work gHE = [u1Vw1 u2Vw2]

= (1/2) {[ V12 V22 ] + [ U12 U22 ] + [ Vr22 Vr12 ] }[ V1

2 V22 ]/2g is energy transfer due to change of

absolute energy of fluid between inlet and outlet. It is

called impulse effect.

[ U1

2 U2

2 ]/2g is change of energy due to centrifugal

effect and changes static pressure.

[ Vr22 Vr12 ]/2g is energy transfer due to change of

relative kinetic energy of fluid between inlet and outlet. It

is called impulse effect. This is called Reaction effect

and changes static pressure.

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HYDRAULIC EFFICIENCY

It is the ratio of Eulers head to the head available

at the inlet of turbine. hyd= HE / H = [u1Vw1 u2Vw2] / gH

(Note: Vw2 is negative for Pelton Wheel)

Degree of Reaction R = Energy Transfer by virtueof change of static pressure / Total Energy

Transfer =

= (1/2) {[(U12 U22 ) + ( Vr22 Vr12 )} / [u1Vw1 u2Vw2]

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VELOCITY COMPONENTS AT ENTRY OF ROTOR

< 90o

V1

Vr1

= 90o

V1

Vr1

u1

Vr1V1

> 90o

Note: At inlet diagram is drawn with

U + Vr= V91



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VELOCITY COMPONENTS AT EXIT OF ROTOR

At outlet, diagram is drawn with Vr+ U = V

When direction of U and Vr2

are in the same direction

Forward blade; otherwise Backward blade

V2V2 V2

Vr2

Vr2 Vr2

V w2U2 U2

U2

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VELOCITY COMPONENTS OF SINGLE STAGE

RADIAL FLOW MACHINE

At inlet, diagram is drawn with U + Vr

= V

At outlet, diagram is drawn with Vr+ U = V

When direction of U and Vr2 are in the same direction

Forward blade; otherwise Backward blade

V2

Vr2

Vf2

U2

93

< 90o

V1

Vr1



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VELOCITY TRIANGLE FOR AXIAL FLOW MACHINE

94

Vr2

U2

V2 = Vf2

U2

< 90o

V1

Vr1

Vf1

U1

Vw2 = 0;

U1 = U2

Vf1 = Vf2= V2



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CENTRIFUGAL PUMPS

Centrifugal Pump is a roto dynamic pump in which

mechanical energy is converted in to hydraulic

pressure energy by means of centrifugal force

acting on the fluid by the impeller.

Working Principle: When certain mass of fluid is

rotated by an external source, it is thrown away

from the central axis of rotation and a centrifugal

head is developed. This centrifugal head enables

the liquid to raise to a higher level.

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CENTRIFUGAL PUMPS

Classifications: According to(a)Working head

(b)Direction of flow

(c)Number of entrances

(d)Number of stages

(e)Specific Speed

(f)Type of casing

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CENTRIFUGAL PUMPS

Classifications:

(a) Working headLow head upto 15 m; MediumHead 15 40 m; High Head above 40 m

(b) Direction of flowRadial Flow; Axial Flow;Mixed Flow

(c) Number of entrancesSingle Entry; Double

Entry(d) Number of stagesSingle Stage; Multi Stage

(e) Specific Speed 10 to 20 rpm Radial Flow; 80to 120 rpm Mixed Flow; a60 to 950 rpm Axial

Flow

(f) Type of casingVolute Casing; VortexCasing; Diffuser Casing (with guide blades)

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CENTRIFUGAL PUMPS

Suction Head: hs is the vertical height of the

centre of the pump above the water surface in thetank or sump from which water is to be lifted.

Delivery Head: hd is the vertical height betweenthe centre line of the pump and the water surface in

the tank to which water is delivered.

Manometric Head: Hm is the sum suctionhead,delivery head, velocity head at delivery end, andfriction heads at suction and delivery lines.

FORMULA: Hm = hs + hd + hfs + hfd + Vd2/2g

hf= 4 f L V2/ 2gD

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CENTRIFUGAL PUMPS

LOSS DIAGRAM

Useful power

gQH

Impeller & casing Loss

Mechanical Loss

Leak

-age

Loss

Q q

Hth

H

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CENTRIFUGAL PUMPS

Characteristic Curves: are the curves by which the

exact behaviour and performance of the pumpunder different flow rate, head and speed.

Minimum Starting Speed: is the speed at which the

pump starts delivering the fluid.

FORMULA:

Nmin = 120 mano Vw2 D2 / [(D22 D12)]

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CENTRIFUGAL PUMPS

EFFICIENCIES

Manometric Efficiency mano = g Hm / u2Vw2

Mechanical Efficiency mech = Q u2Vw2 / SP,

Where SP is Power available at shaft, Q u2Vw2 is

the power at impeller in kW

Overall Efficiency o = Q gHm / SP = mano x mech

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CENTRIFUGAL PUMPS

CAVITATION: is the phenomenon of formation of

vapour bubbles of flowing fluid in a regionwhere the pressure of liquid is below its

vapour pressure and the sudden collapsing of

these bubbles in a region of high pressure.

Cavitation occurs if the suction pressure falls

below the vapour pressure of the liquid.Cavition results in

(a) Cavity formation

(b) Noise and Vibration

(c) Drop in efficiency and head developed.

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CENTRIFUGAL PUMPS

Precautions aganist Cavitation:

(a) Use Cavitation resistant materials like bronze,

stainless steel

(b) Minimise friction losses, avoiding bends.

(c) Reduce turbulence by proving adequate

vanes to guide the liquid.

(d) Do not allow the pressure to fall below the

vapour pressure at any part of the system.

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CENTRIFUGAL PUMPS

Priming:is the operation in which suction pipe,

casing of the pump and a portion of thedelivery pipe up to delivery valve are

completely filled with liquid before starting the

pump. Thus air from these parts of the pump

is removed.

If priming is not done:

1. Air pockets inside the pump may cause

vortices and discontinuity of flow.

2. Dry running of pump resulting in rubbing and

seizing of wearing rings.3. Pump impeller may be seriously damaged.

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CENTRIFUGAL PUMPS

VELOCITY TRIANGLES FOR CENTRIFUGAL PUMP

Radial BladeForward

Curved l Blade

u1

vR1

v1u1

vR1v

1

u1

vR1v

1

u2

vR2v2

Backward

Curved Blade

u2

vR2v2

u2

vR2v2 vf2

Vw2

vf2

Vw2

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TURBINES

Hydro Turbines are hydraulic machines which

convert hydraulic energy in to mechanical energy.

Classifications: According to

(a)Head and Quantity available

(b)Name of the Originator

(c)Action of water on the moving blades

(d)Direction of flow of water in the runner

(e)Disposition of shaft of the turbine

(f)Specific Speed

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TURBINES


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TURBINESClassifications: According to

Head and Quantity available

(i) Impulse Turbine requires high head andsmall flow rate.

(ii) Reaction Turbine requires low head and large

flow rate.(iii) Medium head and Medium flow Turbine

requires medium head and medium flow rate.

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TURBINES


Name of the Originator

(i) Pelton Turbine named after Lester Allen

Pelton of California

(ii) Francis Turbine named after James BichensFrancis

(iii)Kaplan Turbine named after Dr.Victor Kaplan

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TURBINES


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TURBINES


Action of water on the moving blade

(i) Impulse Turbine Kinetic energy is convertedin to mechanical energy

(ii) Reaction Turbine Partially kinetic energy

and partially pressure energy is converted into mechanical energy.

(iii) Kaplan Turbine Pressure energy isconverted in to mechanical energy.

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TURBINES


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TURBINES


Direction of flow of water in the runner(i) Tangential Flow Turbine Water strikes

tangential to the path of rotation (Pelton)

(ii) Radial Flow Turbine Water flows radially

inwards or outwards.

(iii) Axial Flow Turbine Water flows parallel to

the axis of the runner shaft (Kaplan)

(iv) Mixed Flow Turbine Water flows both

radially and axially (Francis)

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TURBINES


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TURBINES


Disposition of shaft of the turbine

(i) Horizontal Shaft Turbine (Pelton, HorizontalKaplan)

(i) Vertical Shaft Turbine (Francis, VerticalKaplan)

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TURBINES


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TURBINES


Specific Speed

Specific Speed Turbine Type

8.5 to 30 Pelton Single jet

30 to 51 Pelton Double jet51 to 225 Francis

225 to 860 Kaplan/propeller

112


Cl ifi ti A di t


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Specific Speed forCentrifugal Pumps

S.No. Type of Impeller for

Centrifugal Pumps

Specific Speed

1 Slow Speed Radial Flow 10 30

2 Medium Speed Radial

Flow

30 50

3 High Speed Radial Flow 50 80

4 Mixed Flow 80 160

5 Axial Flow 160 - 500

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TURBINES


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TURBINES

For Turbines Specific Speed NST = N P / (H5/4)

For Pumps, Specific Speed NSP

= N Q / (H3/4)

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TURBINES CLASSIFICATION


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TURBINES - CLASSIFICATION

Pelton Turbine is a tangential flow impulse

turbine. Energy available at inlet is only kineticenergy.

Pressure at inlet and outlet is only atmosphere.

Impact of jet on wheel exerts hydro dynamic force

and thus produces mechanical energy.

Francis Turbine is an inward flow reaction turbinehaving radial discharge at outlet. Both kinetic

energy and pressure energy are converted to

mechanical energy.

Kaplan Turbine is an axial flow reaction turbine.

At the inlet water has both kinetic and pressureenergy. Part of pressure energy is converted to

kinetic energy during the flow through runner.

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PERFORMANCE CURVES FOR PUMPS


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PERFORMANCE CURVES FOR PUMPS

116

H

P

Q

For various speeds N,

draw various curves


Cl ifi ti A di t


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Specific Speed forCentrifugal Pumps

S.No. Type of Impeller for

Centrifugal Pumps

Specific Speed

1 Slow Speed Radial Flow 10 30

2 Medium Speed Radial

Flow

30 50

3 High Speed Radial Flow 50 80

4 Mixed Flow 80 160

5 Axial Flow 160 - 500

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PERFORMANCE CURVES FOR TURBINES


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PERFORMANCE CURVES FOR TURBINES

118

Qu

P

Nu

For various speeds N,

draw various curves



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END OF UNIT FOUR

119

FLUID MECHANICS UNIT 5 POSITIVE DISPLACEMENT MACHINES

RECIPROCATING PUMPS


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RECIPROCATING PUMPS

Reciprocating Pump is a positive displacement

pump.Liquid is first sucked into a cylinder and then

displaced by the thrust of piston moving to and

fro in the cylinder.

Classifications:

(i) According to action of pump:(a) Single acting and (b) Double acting

(ii) According to Number of cylinders

(a) Single Cylinder and (b) Double Cylinder

According to Provision of air vessels

(a) With air vessel and (b) without air vessel

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RECIPROCATING PUMPS


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RECIPROCATING PUMPS

Formulae for Discharge Q in m3/s

Q = n A L N / 60

Where n = 1 for single acting and n = 2 for

double acting, A is area of cross section of

piston (cylinder) in m2, L is stroke length

in m, N is crank speed in rpm

Stroke length L = 2 r, where r = crank

radius in m.

Area of piston A = (/4)D2, where D is

diameter of cylinder (piston) in m.

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RECIPROCATING PUMPS


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RECIPROCATING PUMPS

Slip is defined by the relation

S = (Qth Qact)where Qth is theoretical discharge and Qact

is actual discharge in m3/s

Percentage Slip = S %

= 100 (Qth Qact) / Qth

Coefficient of discharge Cd = Qact / Qth

Volumetric Efficiency v = 100 Cd

122


RECIPROCATING PUMPS


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RECIPROCATING PUMPS

Formulae for Workdone W in kW

W = ngALN (hs + hd) / (60000)

, where n = 1 for single acting and n = 2 for double

acting, A is area of cross section of piston

(cylinder) in m2, L is stroke length in m, N is crank

speed in rpm, hs is suction haed in m, hd is

delivery head in m.

g = 9.81 m/s2

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INDICATOR DIAGRAMS


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INDICATOR DIAGRAMS

Indicator diagram is a graph showing the pressure

head inside the cylinder along the y axis VsLocation of the piston in the cylinder along the x

axis.

We know maximum length along the x axis

= Maximum distance travelled by the piston

= Stroke Length = L

(a) Ideal Indicator diagram

(b) Indicator diagram due to acceleration in suction

and delivery lines

Area of indicator diagrams in both cases are same.

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VELOCITY OF PISTON INSIDE THE CYLINDER


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VELOCITY OF PISTON INSIDE THE CYLINDER

Velocity V = r Sin t

Acceleration a = 2 r Cos t

, where is angular velocity of crank in radiansper second; r is crank radius in m; t is time in

seconds.

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INDICATOR DIAGRAMS (IDEAL)


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INDICATOR DIAGRAMS (IDEAL)

Delivery

Suction

h atm

atm

h d

atm

h s

atm

C , = 0o

atm

A, = 0o

atmB, 180o

atm

E F

L

atm

Stroke Length

P

D, 180o

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INDICATOR DIAGRAMS


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INDICATOR DIAGRAMS

(Acceleration in suction and delivery lines)

AA = BB = has; CC = DD = had

B

atm

D

A

atm

Suction

Delivery

D

hatm

atm

h d

atm

h s

atm

C , = 0o

atm

A

atm B, 180o

atm

EF

L

atm

P

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AIR VESSELS


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AIR VESSELS

Ai Vessel: is a closed chamber, opening at the

bottom side and connected to suction pipe anddelivery pipe. Compressed air is filled up inside the

chamber. Air vessels are connected in the pipes at

points close to the cylinder.

ADVANTAGES:Air Vessels provide continuous flow of fluid at

uniform rate.

Air Vessels increase the delivery head.

Air Vessels help to run the pump at high speedswithout separation.

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AIR VESSELS PERFORMANCE


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AIR VESSELS PERFORMANCE

During the first half of delivery stroke, piston

accelerates to force the liquid into the delivery pipe

with higher velocity. The excess liquid flows into

the airvessel compressing the air inside. During

the second half of the delivery stroke piston

decelerates to force the liquid into the delivery pipe

with a lower velocity. Now the liquid stored in the

airvessel flows into the delivery pipe. Thus uniformvelocity is obtained in the delivery pipe.

Similarly, on the suction side, liquid flows into the

air vessel during first half of suction stroke and

flows from the airvessel into the cylinder in the

second half of the suction stroke.

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WORK SAVED BY AIR VESSELS


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WORK SAVED BY AIR VESSELS

Percentage of work saved by fitting an air vessel in

a single acting reciprocating pump = 84.8%

Percentage of work saved by fitting an air vessel in

a double acting reciprocating pump = 39.21%

130


Comparison of Centrifugal pump and reciprocating


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p g p p p gpump

Centr i fugal Pump Reciprocat ing Pump

Suitable for Suitable for

Large Discharge and Small Discharge and

Low Head High Head

No air Vessel required Air Vessel required

Priming is required No priming required

Continuous flow Pulsating flow

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Thank You and Best Wishes

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END OF ANSWERS

Fluid Mechanics and Machinery for II MechRevision May2013

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Transcript of Fluid Mechanics and Machinery for II MechRevision May2013