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Transcript of 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 MECHANICS UNIT 1- INTRODUCTION
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 MECHANICS UNIT 1- INTRODUCTION
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 MECHANICS UNIT 1- INTRODUCTION
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 MECHANICS UNIT 1- INTRODUCTION
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 MECHANICS UNIT 1- INTRODUCTION
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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FLUID MECHANICS UNIT 1- INTRODUCTION
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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FLUID MECHANICS UNIT 1- INTRODUCTION
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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http://en.wikipedia.org/wiki/Volumehttp://en.wikipedia.org/wiki/Absolute_temperaturehttp://en.wikipedia.org/wiki/Absolute_temperaturehttp://en.wikipedia.org/wiki/Volume -
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FLUID MECHANICS UNIT 1- INTRODUCTION
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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http://en.wikipedia.org/wiki/Pressurehttp://en.wikipedia.org/wiki/Volumehttp://en.wikipedia.org/wiki/Thermodynamic_temperaturehttp://en.wikipedia.org/wiki/Kelvinhttp://en.wikipedia.org/wiki/Kelvinhttp://en.wikipedia.org/wiki/Thermodynamic_temperaturehttp://en.wikipedia.org/wiki/Volumehttp://en.wikipedia.org/wiki/Pressure -
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FLUID MECHANICS UNIT 1- INTRODUCTION
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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FLUID MECHANICS UNIT 1- INTRODUCTION
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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FLUID MECHANICS UNIT 1- INTRODUCTION
SYSTEM AND CONTROL VOLUME
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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FLUID MECHANICS UNIT 1- INTRODUCTION
SYSTEM AND CONTROL VOLUME
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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FLUID MECHANICS UNIT 1- INTRODUCTION
SYSTEM AND CONTROL VOLUME
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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FLUID MECHANICS UNIT 1- INTRODUCTION
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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FLUID MECHANICS UNIT 1- INTRODUCTION
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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FLUID MECHANICS UNIT 1- INTRODUCTION
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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FLUID MECHANICS UNIT 1- INTRODUCTION
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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FLUID MECHANICS UNIT 1- INTRODUCTION
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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FLUID MECHANICS UNIT 1- INTRODUCTION
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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FLUID MECHANICS UNIT 1- INTRODUCTION
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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FLUID MECHANICS UNIT 1- INTRODUCTION
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
WD = ( V2u2 V1u1)/g22
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FLUID MECHANICS UNIT 1- INTRODUCTION
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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FLUID MECHANICS UNIT 1- INTRODUCTION
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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FLUID MECHANICS UNIT 1- INTRODUCTION
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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FLUID MECHANICS UNIT 1- INTRODUCTION
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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FLUID MECHANICS UNIT 2 FLOW THROUGH CIRCULAR CONDUITS
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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FLUID MECHANICS UNIT 2 FLOW THROUGH CIRCULAR CONDUITS
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
FLUID MECHANICS UNIT 2 FLOW THROUGH CIRCULAR CONDUITS
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Laminar
Laminar Sub Layer
Turbulent
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FLUID MECHANICS UNIT 2 FLOW THROUGH CIRCULAR CONDUITS
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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FLUID MECHANICS UNIT 2 FLOW THROUGH CIRCULAR CONDUITS
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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FLUID MECHANICS UNIT 2 FLOW THROUGH CIRCULAR CONDUITS
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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FLUID MECHANICS UNIT 2 FLOW THROUGH CIRCULAR CONDUITS
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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FLUID MECHANICS UNIT 2 FLOW THROUGH CIRCULAR CONDUITS
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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FLUID MECHANICS UNIT 2 FLOW THROUGH CIRCULAR CONDUITS
Hydraulic Gradient and Total Energy Lines
Fig.1 Hydraulic gradient and total energy line for
(b) horizontal pipe connecting 2 reservoirs
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FLUID MECHANICS UNIT 2 FLOW THROUGH CIRCULAR CONDUITS
Hydraulic Gradient and Total Energy Lines
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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FLUID MECHANICS UNIT 2 FLOW THROUGH CIRCULAR CONDUITS
Hydraulic Gradient and Total Energy Lines
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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FLUID MECHANICS UNIT 2 FLOW THROUGH CIRCULAR CONDUITS
Hydraulic Gradient and Total Energy Lines
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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FLUID MECHANICS UNIT 2 FLOW THROUGH CIRCULAR CONDUITS
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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FLUID MECHANICS UNIT 2 FLOW THROUGH CIRCULAR CONDUITS
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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FLUID MECHANICS UNIT 2 FLOW THROUGH CIRCULAR CONDUITS
FRICTION FACTOR AND MOODY DIAGRAM
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FLUID MECHANICS UNIT 2 FLOW THROUGH CIRCULAR CONDUITS
FRICTION FACTOR AND MOODY DIAGRAM
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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FLUID MECHANICS UNIT 2 FLOW THROUGH CIRCULAR CONDUITS
FRICTION FACTOR AND MOODY DIAGRAM
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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FLUID MECHANICS UNIT 2 FLOW THROUGH CIRCULAR CONDUITS
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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FLUID MECHANICS UNIT 2 FLOW THROUGH CIRCULAR CONDUITS
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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FLUID MECHANICS UNIT 2 FLOW THROUGH CIRCULAR CONDUITS
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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FLUID MECHANICS UNIT 2 FLOW THROUGH CIRCULAR CONDUITS
FLOW THROUGH PIPES IN SERIES
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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FLUID MECHANICS UNIT 2 FLOW THROUGH CIRCULAR CONDUITS
FLOW THROUGH PIPES IN SERIES
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FLUID MECHANICS UNIT 2 FLOW THROUGH CIRCULAR CONDUITS
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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FLUID MECHANICS UNIT 2 FLOW THROUGH CIRCULAR CONDUITS
FLOW THROUGH PIPES IN PARALLEL
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FLUID MECHANICS UNIT 2 FLOW THROUGH CIRCULAR CONDUITS
FLOW THROUGH PIPES IN PARALLEL
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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FLUID MECHANICS UNIT 2 FLOW THROUGH CIRCULAR CONDUITS
FLOW THROUGH PIPES IN PARALLEL
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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FLUID MECHANICS UNIT 2 FLOW THROUGH CIRCULAR CONDUITS
FLOW THROUGH PIPES IN PARALLEL
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FLUID MECHANICS UNIT 2 FLOW THROUGH CIRCULAR CONDUITS
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FLUID MECHANICS UNIT 2 FLOW THROUGH CIRCULAR CONDUITS
FLOW THROUGH PIPES IN PARALLEL
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FLUID MECHANICS UNIT 2 FLOW THROUGH CIRCULAR CONDUITS
(2)
(3)
(1)
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FLUID MECHANICS UNIT 2 FLOW THROUGH CIRCULAR CONDUITS
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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FLUID MECHANICS UNIT 2 FLOW THROUGH CIRCULAR CONDUITS
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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FLUID MECHANICS UNIT 3 DIMENSIONAL ANALYSIS
FUNDAMENTAL QUANTITIES
S.No PhysicalQuantity
Symbol Dimensions
1 Mass M M
2 Length L L
3 Time T T
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FLUID MECHANICS UNIT 3 DIMENSIONAL ANALYSIS
GEOMETRIC QUANTITIES
S.No PhysicalQuantity
Symbol Dimensions
1 Area A L2
2 Volume V L3
3
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FLUID MECHANICS UNIT 3 DIMENSIONAL ANALYSIS
KINEMATIC QUANTITIES
S.No PhysicalQuantity
Symbol Dimensions
1 Velocity A L T-1
2 Discharge Q L3 T-1
3 Kinematic
Viscosity
L2 T-1
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FLUID MECHANICS UNIT 3 DIMENSIONAL ANALYSIS
DYNAMIC QUANTITIES
S.No PhysicalQuantity
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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FLUID MECHANICS UNIT 3 DIMENSIONAL ANALYSIS
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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FLUID MECHANICS UNIT 3 DIMENSIONAL ANALYSIS
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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FLUID MECHANICS UNIT 3 DIMENSIONAL ANALYSIS
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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FLUID MECHANICS UNIT 3 DIMENSIONAL ANALYSIS
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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FLUID MECHANICS UNIT 3 DIMENSIONAL ANALYSIS
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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FLUID MECHANICS UNIT 3 DIMENSIONAL ANALYSIS
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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FLUID MECHANICS UNIT 3 DIMENSIONAL ANALYSIS
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
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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
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< 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
Classifications: According to
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
Classifications: According to
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
Classifications: According to
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
Classifications: According to
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
Classifications: According to
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
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Cl ifi ti A di t
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Classifications: According to
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
FLUID MECHANICS UNIT 4 ROTO DYNAMIC MACHINES
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Classifications: According to
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
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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
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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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FLUID MECHANICS UNIT 5 POSITIVE DISPLACEMENT MACHINES
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
(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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FLUID MECHANICS UNIT 5 POSITIVE DISPLACEMENT MACHINES
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%
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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