Fluid Machines
Dr. Sharad Chaudhary
Topics
•
•
•
•
•
Dimensional Analysis and Similititude
Hydro Turbines
Hydraulic Pumps
Cavitations
Water Hammer and Surge Tanks
Hydraulic Turbines
Hydraulic Turbines
Impulse Turbines
Reaction Turbines
Radial flow
Axial Flow
Mixed Flow
Hydraulic Turbines
• Momentum
Principle
r
r
r r r
dmV d
=
ρVdV + ∫ ρV V .dA
∫
dt
dt CV
CS
(
)
Example: V = 50 m/s, Q= 60 lit/s, find the force on the vane.
r
r
r r r
dmV d
ρVdV + ∫ ρV V .dA
=
∫
dt
dt CV
CS
r
r r r
∑ F = 0 + ∫ ρV V .dA
(
(
y
)
)
CS
Control
Surface
r
r r r
r r r
∑ F = 0 + ∑ ρV V .dA + ∑ ρV V .dA
(
in
x
)
(
out
)
r
r
r
∑ Fr = ρVinr(Vin dAin cosθin ) + ρVoutr(Vout dAout cosθ out )
∑ F x = ρVin, x (Vin dAin cosθin ) + ρVout , x (Vout dAout cosθ out )
r
r
r
∑ F y = ρVin, y (Vin dAin cosθin ) + ρVout , y (Vout dAout cosθ out )
Hydraulic
Turbines
r
r r r
r r
r
∑ F = 0 + ∑ ρV V .dA + ∑ ρV V .dA
(
)
in
(
)
out
r
r
r
∑ F = ρVin (Vin dAin cosθin ) + ρVout (Vout dAout cosθ out )
Control
Surface
Resolving along x and y directions
r
r
r
∑ F x = ρVin, x (Vin dAin cosθin ) + ρVout , x (Vout dAout cosθ out )
( (
)
)
+ 1000.50. cos 60 (50.(1.2.10 )cos 0 )
= 1000.50. 50. 1.2.10 −3 cos180o
−3
o
r
Vin
r
dAin
Aout r
Vout
y
x
o
= −1500 N
r
r
r
∑ F y = ρVin, y (Vin dAin cosθin ) + ρVout , y (Vout dAout cosθ out )
( (
)
= 0 + 1000.50. 50. 1.2.103 sin 60o
= 2598 N
r
Fx ,vane
r
Fy ,vane
Force on the vane
r
= − Fx
r
= − Fy
)
Hydraulic Turbines
• If the vane is moving with certain velocity
r d
r r
r r r r r
∑ F = dt ∫ ρ V − u dV + ∫ ρ V − u V − u .dA
CV
CS
r d
r
r r r
∑ F = dt ∫ ρVr dV + ∫ ρVr Vr .dA
CV
CS
(
)
)((
(
(
) )
Control
Surface
)
IF V = 50 m/s, Q= 60 lit/s, u=10 m/s find the force on the vane.
r
r r r
r
r r r
dmV
= ∫ ρVr Vr .dA , or ∑ F = ∫ ρVr Vr .dA
dt
CS
CS
(
y
)
(
r
r r r
r r r
∑ F = 0 + ∑ ρVr Vr .dA + ∑ ρVr Vr .dA
(
in
)
(
)
)
out
r
r
r
∑ Fr = ρVrr,in (Vr ,in dAin cosθin ) + ρVr ,rout (Vr ,out dAout cosθout )
∑ F x = ρVrx,in (Vr ,in dAin cosθin ) + ρVrx,out (Vr ,out dAout cosθ out )
r
r
r
∑ F y = ρVry ,in (Vr ,in dAin cosθin ) + ρVry,out (Vr ,out dAout cosθ out )
x
r
u
r
Vout
r
Vr,in
r
Vout
r
u
r
Vr,out
r
Vr, in
Hydraulic Turbines
r
∑Fx
r
∑Fy
r
F
∑ x
r
F
∑ x
r
r
= ρVrx ,in (Vr ,in dAin cos θ in ) + ρVrx ,out (Vr ,out dAout cos θ out )
r
r
= ρVry ,in (Vr ,in dAin cos θ in ) + ρVry ,out (Vr ,out dAout cos θ out )
( (
)
)
)
= −960 N
r
∑ F y = 0 + 1000.40. 40. 1.2.10−3 sin 60o
r
∑ F y = 1662.7 N
r r r
F = Fx + Fy
( (
y
( (
= 1000.40. 40. 1.2.10 −3 cos180o + 1000.40. cos 60o 40. 1.2.10 −3 cos 0 o
)
)
In case of a series/ cascade of vane as in a pelton wheel periphery
r r r
r r r
∑ F = 0 + ∑ ρVr V .dA + ∑ ρVr V .dA
(
in
)
(
out
)
r
r
r
∑ Fr = ρVrr,in (Vin dAin cosθin ) + ρVr ,rout (Vout dAout cosθout )
∑ F x = ρVrx,in (Vin dAin cosθin ) + ρVrx,out (Vout dAout cosθout )
r
r
r
∑ F y = ρVry,in (Vin dAin cosθin ) + ρVry,out (Vout dAout cosθout )
x
)
Pelton Turbines
Pelton Turbines
Pelton Wheel/runner and Bucket with notch
Pelton Turbine
Nozzle with water striking the bucket
Pelton Turbine
Nozzle components
Pelton Turbine
y
x
Pelton Turbine
Force on the vane
r r r
r r r
− ∑ F = ∑ ρVr V .dA + ∑ ρVr V .dA
(
u
V2
Vr2
Vr1 V1
θ
Ft
u
r
−∑F
r
−∑F
r
∑ Fr
∑ Fr
∑ Fr
∑ Fr t
∑ Ft
in
)
(
)
out
r
r
o
= ρVr ,in Vin dAin cos180 + ρVr ,out Vout dAout cos 0o
r
r
= − ρVr ,in (Vin dAin ) + ρVr ,out (Vout dAout )
r
r
= ρVr1 (V1dA1 ) − ρVr 2 (V2 dA2 )
r
r
= ρVr1 (Q ) − ρVr 2 (Q )
r
r
= ρQ Vr1 − Vr 2
r
r
y
= ρQ Vr1 t − Vr 2 t
(
)
(
(
)
[( ) ( ) ]
= ρQVrt (1 − cos θ )
P = F .u
= ρQuVrt (1 − cos θ )
This is the power developed by the
pelton wheel/runner
x
)
Pelton Turbine
u
V2
This is the power developed
P = F .u
Vr2
Vr1 V1
θ
= ρ QuV
u
Ft
P
V
2
V
u
rt
(1 − cos θ )
This can be maximised by
dP
=0
du
d
ρ Qu (V − u )(1 − cos θ ) = 0
du
V
u =
2
by the
Pelton Turbine
hf
Head Race
Casing
Penstock
Pelton Wheel
Ha
Nozzle
Deflector
Spear
TailRace
Losses up to Nozzle
Principle of Reaction Turbine
Forces on reducer bent/elbow s
we know the general unsteady linear momentum equation applied to
a control volume is
r
r
r r r
∂
∑ F = ∂t ∫ ρ V dVol + ∫ ρ V V .d A
cv
cs
For a steady flow it reduces to
r
r r r
∑ F = ∫ ρ V V .d A
r cs r r r
r r r
V2
∑ F = ∫ ρ V V .d A in + ∫ ρ V V .d A out
(
(
in
y
)
)
(
)
(
)
out
P1 A1 − P2 A2 cos θ + Fx = ρ Q (V 2 cos θ − V1 )
0 − P2 A2 sin θ + F y = ρ Q (V 2 sin θ )
V1
P1
x
V2
P2
θ
Principle of Reaction Turbine
A lawn sprinkler with two jets each located at 30 cm from centre. The jets are
of 1 cm diameter. Assuming no friction, find the speed of rotation for a
discharge of 2.5 L/s. Also find the torque required to hold the sprinkler
stationary.
Principle of Reaction Turbine
Moment of Moment equation
we know the general unsteady linear momentum equation applied to
a control volume is
r
r
r r r
∂
ρ
ρ
V
V
F =
dVol
V .d A
+
∫
∫
∂ t cv
cs
r
r r
The moment of a force F about a point O is given by r × F
r r
r r
r r r r
∂
ρ r × V dVol + ∫ ρ r × V V .d A
r×F =
∂ t cv∫
cs
∂
ρ rV t dVol + ∫ ρ rV t (V n .dA )
Ft r (= T z ) =
∂ t cv∫
cs
Which if applied to an annular control volume, in steady flow, is
T z = ∫ ρ 2 r2Vt 2V n 2 dA 2 + ∫ ρ 1 r1Vt 1V n1 dA1
(
)
(
A2
r
V
Vt
)
Vt
r1
V
A2
for complete circular symmetry, where r, ρ , Vt and Vn are constant
over the inlet and outlet control surface
T z = ρ Q [(rV t )2 − (rV t )1 ]
T z = ρ Q [r2V 2 cos α 2 − r1V1 cos α 1 ]
T z = ρ Q [r2V w 2 − r1V w1 ]
r2
V
Vt
Principle of Reaction Turbine
Vt
r1
V
r2
V
Vt
Principle of Reaction Turbine
V
V
Vt
Vt
Principle of Reaction Turbine
V
V
Vt
Vt
Principle of Reaction Turbine
T z = ρ Q [r2V w 2 − r1V w1 ]
P (= T z ω ) = ρ Q [u 2V w 2 − u1V w1 ]
Principle of Reaction Turbine
Forces on reducer bent/elbow s
we know the general unsteady linear momentum equation applied to
a control volume is
r
r
r r r
∂
∑ F = ∂t ∫ ρ V dVol + ∫ ρ V V .d A
cv
cs
(
)
For a steady flow it reduces to
r
r r r
∑ F = ∫ ρ V V .d A
( )
r
r r r
r r r
∑ F = ∫ ρ V (V .d A ) + ∫ ρ V (V .d A )
r
r r r
r r r
∑ F = ∫ ρ V (V .d A ) + ∫ ρ V (V .d A )
cs
in
in
out
out
in
in
y
V2
out
out
P1 A1 − P2 A2 cos θ + Fx = ρ Q (V 2 cos θ − V1 )
0 − P2 A2 sin θ + F y = ρ Q (V 2 sin θ )
V1
P1
x
V2
P2
θ
Principle of Reaction Turbine
Moment of Moment equation
we know the general unsteady linear momentum equation applied to
a control volume is
r
r
r r r
∂
ρ V dVol + ∫ ρ V V .d A
F =
∫
∂ t cv
cs
r
r r
The moment of a force F about a point O is given by r × F
r r
r r
r r r r
∂
ρ r × V dVol + ∫ ρ r × V V .d A
r×F =
∂ t cv∫
cs
(
)
(
Ft r (= T z ) =
∂
∂t
∫ ρ rV dVol + ∫ ρ rV (V
t
t
cv
∫ρ
A2
r Vt 2V n 2 dA 2 +
2 2
V
Vt
)
n
.dA )
Vt
cs
Which if applied to an annular control volume, in steady flow, is
Tz =
r
∫ ρ rV V
1 1 t1
n1
dA1
A2
for complete circular symmetry, where r, ρ , Vt and Vn are constant
over the inlet and outlet control surface
T z = ρ Q [(rV t )2 − (rV t )1 ]
r1
V
r2
V
Vt
Inward-flow Runner
Vt
r1
V
r2
V
Vt
Spiral Casing
Guide vanes
Mixed-flow Runner
Velocity triangles
Reaction Turbine installation Plan
V
V
Vt
Vt
Control volume & Velocity triangles
V
V
Vt
Vt
Euler equation for turbomachinery
T z = ρ Q [r2V w 2 − r1V w1 ]
P (= T z ω ) = ρ Q [u 2V w 2 − u1V w1 ]
Layout of Hydel power plant
Layout of Hydel power plant
Governing of Turbines
Dr. Sharad Chaudhary
Guide vanes
Reaction Turbine
Governing Mechanism
Governing Mechanism
Governing Mechanism
Pumps
Dr. Sharad Chaudhary
Hydraulic Pumps
Centrifugal Pumps: Classification
BASIS
TYPES
Head
Low (H< 15m) Medium (15<H<40m)
High (H>40m)
Casing
Volute/scroll
Guide vanes
Whirl chamber
Flow direction
Radial flow
Mixed flow
Axial flow
Impeller
Closed
Semi open
open
Inlets
Single
Double
Shaft position
Horizontal
Vertical
Stages
Single
Multistage
Principle of Rotodynamic pumps
Moment of Moment equation
we know the general unsteady linear momentum equation applied to
a control volume is
r
r
r r r
∂
ρ V dVol + ∫ ρ V V .d A
F =
∫
∂ t cv
cs
r
r r
The moment of a force F about a point O is given by r × F
r r
r r
r r r r
∂
ρ r × V dVol + ∫ ρ r × V V .d A
r×F =
∂ t cv∫
cs
(
)
(
Ft r (= T z ) =
∂
∂t
∫ρ
A2
Vt
∫ ρ rV t dVol + ∫ ρ rV t (V n .dA )
cv
r Vt 2V n 2 dA 2 +
2 2
V
)
V
Vt
cs
Which if applied to an annular control volume, in steady flow, is
Tz =
r
∫ ρ rV V
1 1 t1
n1
dA1
A2
for complete circular symmetry, where r, ρ , Vt and Vn are constant
over the inlet and outlet control surface
T z = ρ Q [(rV t )2 − (rV t )1 ]
r2
r1
Vt
V
A Rotodynamic Pump System
Principle of Rotodynamic pumps
Velocity Triangle
T z = ρ Q [(rV t )2 − (rV t )1 ]
P = ρ Q ω [(rV t )2 − (rV t )1 ]
In terms of whirl components
P = ρ Q [u 2V w 2 − u1V w1 ]
In terms of Head
[u V − u1V w1 ]
H = 2 w2
g
2
V 2 − V12
u 22 − u12
V r22 − V r21
+
+
H =
2g
2g
2g
When the whirl component at inlet is zero
uV
H = 2 w2
g
However th e actual head produced will be less by Losses, H L
H actual = H − H L
 V 22
  V12

P2
P1
+
+ z 2  − 
+
+ z1   − H L
H actual = 
2
g
ρ
g
2
g
ρ
g
 


(
) (
) (
)
Heads in Pump
Manometric head (H) = Static head (H s ) + Losses
Therefore, H = H s + hi + h fs + h fd + he
Manometric Efficiency (η m ) =
gH
ρ QgH
, Overall Efficiency (η m ) =
u 2V w 2
P
Losses in Pumps
Staging of Pumps
Staging of Pumps
Performance of a Pump
DIMENSIONAL
ANALYSES
AND
SIMILITUDE
Dr. Sharad Chaudhary
Why Dimension Analyses?
Understand Units and Dimensions of
Various Physical Quantity
Unit Conversions
Dimensional
principles
Reasoning
by
various
Method of analysis in Fluid Mechanics
and Heat Transfer
Modelling and Similitute
Units and Dimensions
•Dimensions: Certain properties of the system
that describe the physical phenomenon and
are fundamental and universal
•Units:
•Fundamental and Derived Dimensions / Units
• Dimensions of a Derivative/Integral are easily
obtained by the dimensions of variables to be
differentiated/Integrated and variable with
respect to which it is differentiated/integrated
Units and Dimensions
S.No Physical Quantity Dimensions SI Units
0 0
1 Mass
ML T
Kg
2 Length
3 Time
4
5
6
7
8
9
10
0
M LT
0
0
0
M L T
m
sec
Dimensional Analysis
Methods
• Indicial Method (Raleigh’s Method)
• Group Method
– Reasoning Method
– Method of Repeating variables (Buckingham π)
– Matrix Method
Indicial (Raleigh’s) Method
Steps in Indicial (Raleigh’s) Method
1.Identify of the variables involved in the process
2.Figure out the dependent and independent
variables.
3.Relationship determining the dependence of
dependent variable on the other is to be found
D = f ( I1 , I 2 , I 3 , I 4 .......I N )
4. It could be expressed simple equation by
putting arbitrary indices on the independent
a b c
variable, of the form D = C.I1 .I 2 .I 3
Indicial (Raleigh’s) Method
• By using the principle of dimensional
homogeneity the values of these arbitrary
indices is determined
• Having known the indices then, the
numerical coefficients are determined
experimentally with known values of all the
variables involved
Sample Problem
The thrust, F, of a propeller depends upon its
diameter, d, speed of advance, v, rev/sec, N, the fluid
density, ρ and the fluid viscosity, μ. Find an expression
for F .
Advantages and Limitations
• The method is very simple
• Very useful in determination of the functional
relationship when the no. of variables are less
• Exact functional relation between the
variables could be determined
Advantages and Limitations
• The method is not much of use when the no.
of variables involved are many
• Not very convenient in modelling and
similitude studies
• Some functional relationships are better
expressed by groups of variables rather than
variable individually. It fails to directly form
such dimensionless groups
Group (Buckingham π )Method
1. Select the pertinent variables
2. Write the functional relationship e.g
F (V , D, ρ , µ , c, H ) = 0
3. Select the repeating variables (Do not make
the dependent variable a repeating
variable). These variables should contain all
the m dimensions of the problem. Often
variable is chosen to specify scale, another
kinematics and third Dynamics.
Group (Buckingham π )Method
4. Write the π parameters in terms of
unknown exponents, e.g
Π1 = V x1 D y1 ρ z1µ
−1 x1 y1
−3
= LT
L ML
(
)
(
)
z1
−1
ML T
−1
5. For each of such π expressions write the
equations of the exponents, so that the sum
of each exponents of each dimension is
zero.
6. Solve the equations simultaneously
Group (Buckingham π )Method
7. Substitute back into the π expressions of
step 4 the exponents to obtain the
dimensionless π parameters
8. Establish a functional relationship
f (Π1 , Π 2 , Π 3 .....Π n − m ) = 0
or , Π1 , the dimensionless group containing the
dependent variable
Π1 = f (Π 2 , Π 3 .....Π n − m )
9. Recombine to alter the forms of the π
parameters, keeping the same no. of
Sample Problem
A fluid-flow situation depends upon the velocity V, the
density ρ, several linear dimensions l,l1,l2 , pressure
drop ΔP, gravity g, viscosity μ, surface tension σ and
Bulk Modulus of Elasticity K. Apply dimensional
analysis to these variables to find a set of Π
parameters
Matrix Method
1. Figure out all the variables involved
2. Write them in one row starting with repeating
variable followed by non repeating ones
3. Form a matrix with the each row
corresponding to every dimension involved
(For kinematics No of rows=2, for Dynamics =
3, for Thermodynamics =4 )
4. Determine the rank of the matrix; r;
5. The no. of dimensionless parameters will be
(n-r)
Matrix Method
6. Write
homogeneous linear algebraic
equations whose coefficients are the
elements in the rows of matrix.
7. Solve them assigning arbitrary values to (n-r)
variables.
Similitude
Designers require to simulate certain flow
situations by means of scaled down models. Such
studies are known as similitude or model studies.
Two approaches as practiced
(1) All forces on model and prototype should be
in the same ratio i.e. the force polygons
should be similar. Dominant force should be
known beforehand
(2) Dimensionless group containing different
force terms should be similar
Similitude
Rules of similarity state that there should be two
kinds of similarity in model and prototype
1. Geometric Similarity
2. Dynamic Similarity (Similarity of velocity and force
polygons)
Example


2

∆p
v
v 
ρvl
ρv
= f
,
,
,
,
1 2
µ
σ
K
lg


ρv
ρ


2
Reynold’s
No.
Pressure
Coefficient
Weber No.
Mach No.
C p = f (Re, Ma,We, Fr )
Froude No.
Example
A fluid-flow situation depends upon the velocity V, the
density ρ, several linear dimensions l,l1,l2 , pressure
drop ΔP, gravity g, viscosity μ, surface tension σ and
Bulk Modulus of Elasticity K. Apply dimensional
analysis to these variables to find a set of Π
parameters


 ρvl
∆p
v
ρv 2 v 
= f
,
,
,
,
1 2
µ
σ
K
lg 

ρv
ρ


2
Different Flow Situations
The following flow situations are common in
engineering application of fluid mechanics
1. Pipe Flow
2. Open Channel Flow (channel transitions,
Notches and weirs)
3. Aerofoil Theory including fan, pump and
turbine blades
4. Flow through open hydraulic structures
(Free surface flows over spillways, stilling
basins etc)
Different Flow Situations
5. Compressible flows; Wave propagations
6. Discharge under very small heads
7. Flows with gas-liquid and Liquid-liquid
interfaces, wavelets etc.
One or the other forces like pressure, viscosity,
surface tension, Elastic force due to
compressibility and gravity force predominate
any flow situation and hence the dimensionless
no representing that force is chosen to be kept
equal in model and prototype.
Different Flow Situations
The following Modelling laws are applicable
based on the flow situations
1.Pipe Flow
2.Open Channel Flow (channel transitions,
Notches and weirs)
3.Aerofoil Theory including fan, pump and
turbine blades
(Re )m = (Re ) p , Reynold's Law
Different Flow Situations
4. Flow through open hydraulic structures
(Free surface flows over spillways, stilling
basins etc)
(Fr )m = (Fr ) p , Froude' s Law
5. Compressible flows; Wave propagations
(Ma )m = (Ma ) p , Mach ' s Law
Different Flow Situations
4. Discharge under very small heads
5. Flows with gas-liquid and Liquid-liquid
interfaces, wavelets etc
(We)m = (We) p , Weber ' s Law
Scale Ratios
A scale ratio is defined as the ratio of value of
any flow quantity like velocity, acceleration
time, discharge pressure etc in a model to that
in a prototype. The scale ratios depend upon
the model law that is being applied.
Scale Ratios
Scale Ratio Reynold's law Froude's Law Mach Law
Lr
Lr
Length, Lr Lr
Velocity, Vr μ r/ρ rLr
Time
Discharge
Force
Pressure
Power
Examples
A submarine-launched missile, 2 m in diameter and 10
m long, is to be tested in a water tunnel to determine
the forces acting on it during its underwater launch. The
maximum speed during this initial part of the missile’s
flight is 10 ms-1. Determine the mean water tunnel flow
velocity if a 1/20 scale model is employed and dynamic
similarity achieved
Examples
An airship of 6 m diameter and 30 m length is to be
studied in a wind tunnel. The airship speed to be
investigated is at the docking end of its range, a
maximum of 3 ms-1 Determine the mean model wind
tunnel air flow velocity if the model is made to a 1/30
scale.
Rotodynamic Machines
The fluid quantities involved in all hydraulic machines
are the flow rate, Q, the head, H, whereas the
mechanical quantities associated with the machines
itself are the power, P, speed N, Size, D and Efficiency η
PERFORMANCE
OF
ROTODYNAMIC
MACHINES
Dr. Sharad Chaudhary
Variables in Rotodynamic Machines
The fluid quantities involved in all hydraulic
machines are
The flow rate, Q and Head, H
Associated mechanical quantities are
The power, P, Speed N, Size, D and Efficiency η
Performance Characteristics
H vs Q @ N(For Pumps), P vs Q, η vs Q
P vs N @ H (For Turbines), Q vs N, η vs N
Variables in Rotodynamic Machines
Variables in Rotodynamic Machines
Variables in Rotodynamic Machines
Dimensionless Coefficients
For Pumps the output or dependent variable is
H, or (gH)
while other (independent) variables involved are
The flow rate, Q, power, P, speed N, Size, D, fluid
density, ρ, Absolute viscosity, μ, Bulk Modulus of
Elasticity, K and absolute roughness of flow
passage, ε.
Dimensionless Coefficients
For pumps the relationship could be expressed as
gH = f (Q, N , D, ρ , µ , K , ε )
which after the formation of dimesionless group
with N, D, and ρ as repeating variables becomes
 Q
µ
ε
gH
K
= f
,
, 2 2 , 
2 2
3
2
N D
 ND ND ρ N D ρ D 
or
ε

K H = f  K Q , Re, Ma, 
D

Dimensionless Coefficients
For turbines the output or dependent variable is
Power Output, P
while other (independent) variables involved are
The flow rate, Q, speed N, Size, D, fluid density,
ρ, Absolute viscosity, μ, Bulk Modulus of
Elasticity, K and absolute roughness of flow
passage, ε.
Dimensionless Coefficients
For turbines the relationship will be expressed as
P = f (Q, N , D, ρ , µ , K , ε )
which after the formation of dimesionless group
with N, D, and ρ as repeating variables becomes
 Q
P
K
µ
ε
= f
,
, 2 2 , 
3 5
3
2
ρN D
 ND ND ρ N D ρ D 
or
ε

K P = f  K Q , Re, Ma, 
D

Dimensionless Coefficients
A plot of Head, H vs Discharge, Q forms the
fundamental performance curves for a pump,
but plot of KH vs KQ represents the performance
of the homologous pumps (Pumps of the same
design but of the different sizes i.e geometrically
similar)
Similarity laws can then be applied to evaluate
the performance of a given pump
Similarity Laws for Pumps
Similarity laws are applied to evaluate the
performance of a given pump if the performance
of a geometrically similar pump is given.
Q
KQ =
is a constant for geometrically similar pump,
3
ND
Q1
Q2
=
,
3
3
N1 D1 N 2 D2
gH1
gH 2
P1
P1
Similarily, K H = 2 2 = 2 2 , K p =
=
3 5
3 5
N1 D1 N 2 D2
ρN1 D1 ρN 2 D2
Specific Speed
Similar pumps may compared by using the
similarity laws but when pumps of different
family need to be compared the plots of these
dimensionless parameters are only indicative of
relative performance. To compare the
performance of such machines a ratio of two
important parameters is compared. Such a ratio
is known as Type Number (ns) or Specific Speed
Specific Speed
Type Number (ns) or Specific Speed of the pump
is defined as ratio of KQ and KH
1
ns =
KQ
2
=N
3
Q
1
2
3
(gH ) 4
K H4
Type Number (ns) or Specific Speed (Ns) of the
Turbine is defined as ratio of Kp and KH
1
ns =
KP
5
KH
2
P
=N
4
ρ
1
2
1
2
3
(gH ) 4
Specific Speed
Unit Quantities
The Quantities when working under a unit head
are termed as Unit Quantities. These are Unit
Speed, Unit Discharge and Unit Power.
N
Q
P
Nu =
, Qu =
, Pu = 3
H
H
H 2
THEORY
OF
DRAFT TUBE
Dr. Sharad Chaudhary
Draft Tube
Role of Draft Tube
• A Draft Tube discharges water from the turbine exit
to the tail race
• Enables the pressure at turbine exit to be sub
atmospheric thereby providing higher degree of
reaction for a reaction turbine
• Converts Kinetic energy of the water at turbine exit
useful pressure energy
• Increasing the output and improving the overall
efficiency
• Facilitates the easy maintenance of turbine
Types of Draft Tubes
Conical
Simple Elbow
Bell Mouthed
Elbow with circular inlet and rect outlet
Theory
P1 V12
P2 V22
+
+ (H s + x ) =
+
+ 0 + hl
ρg 2 g
ρg 2 g
At section 2 : Pressure P2 = Patm + ρgx
(
Patm + ρgx ) V22
P1 V12
+
+ (H s + x ) =
+
+ hl
ρg 2 g
ρg
2g
 V12 V22

P1 Patm
=
− H s − 
−
− hl 
ρg ρg
 2g 2g

Efficiency of Draft tube
 V12 V22


−
− hl 
2g 2g


η=
V12
2g
Sample Problem
CAVITATIONS
Dr. Sharad Chaudhary
Definition
• A phenomenon which consists, basically, of
local vaporisation of liquid when the absolute
pressure falls to a value equal to /lower than
the vapour pressure of the liquid at a given
temperature.
Mechanism
It takes place in following steps
1.Inception: Due to microscopic gas nulei present in
solid pores as pressure approaches the vapour
pressure
2.Bubbles formation: The nuclei transform into
bubbles as the pressure further decreases
3.Pressure wavelets : Generation of pressure wavelets
resulting in successive formation of more bubbles.
4.Collapse: As these bubbles move to area of higher
pressure they collapse an new bubbles form with
each cycle lasting a few millisecs.
In Hydraulic Machines/flows
It occurs in all hydraulic machines ,must be avoided
1.Pumps: At suction eye
2.Turbines: At the trailing edge of turbines
3.Draft tubes: Inlet
4.Aerofoils: trailing edge
5.Veturimeters: Throat
6.Syphons: Minimum head point
Measure
The most general and most useful cavitation
parameter is ‘Cavitation Coefficient’
(
p1 − p c )
σ =
, p1 , p c and V are Upstream,
2
ρV
2
critical pressures and mean upstream velocity
The value of σ where inception occurs is σ crit
Cavitations in Pumps
Cavitations in Pumps
If the abs pressure at pump eye is pi
v2
pi = patm − ρgH s , where H s = Z s + hs +
2g
If vapour pressure is pvap , then cavitation starts when pi = pvap
pi − pvap
If the difference, ( pi − pvap ), expressed in head,
, it
ρg
measures the positive head available before the cavitation starts.
Hence called Net Positive Suction Head (NPSH)
pi − pvap patm
pvap
NPSH =
=
− Hs −
ρg
ρg
ρg
Cavitations in Pumps
If the pumps has to produce a total head H, the Suction Head
should be not more that NPSH
Thoma suggested that NPSH is propotional to pump total head H
Thoma' s Cavitation Factor is defined as
NPSH
σ th =
H
Another parameter that measures cavitation is Suction Specific
Speed
ωQ1/ 2
Ks =
gNPSH
Relation between parameter
If the pumps has to produce a total head H, the Suction Head
should be not more that NPSH
Toma suggested that NPSH is propotional to pump total head H
Toma' s Cavitation Factor is defined as
ωQ1/ 2
ω s ( gH )3 / 4 NPSH 3 / 4
3/ 4
=
=
= σ th
1/ 2
3/ 4
ωQ
Ks
H
gNPSH
Cavitations in Turbines
Thoma cavitation coefficient
H atm − Z − hvap
σ th =
H
Another parameter that measures cavitation is Suction
Specific Speed
a
Ks =
, the constant a, 4.5 ≤ a ≤ 5.8
ωs
Effects
The following are the effects of cavitation
1. Performance failure
2. Erosion
3. Vibration
Dr. Sharad Chaudhary
PRESSURE
TRANSIENTS
Unsteady Flows
Unsteady Flows are of three types
1. Quasi-steady Flows (Fluid Acceleration is negligible
e;g Continuous filling/emptying of reservoirs etc)
2. Mass Oscillations (Fluid Acceleration is present but
compressibility effects are absent e.g reciprocating
pumps)
3. Pressure transient flows (aka Water Hammer e.g
rapid closure of valve, pump shut down or turbine
load rejection)
Pressure transient flows
In Pressure transient flows the transient propogation
takes place at appropriate acoustic velocity c.
c dependents on
• Fluid properties
• Cross-section dimensions
• Elasticity of the fluid (K)
• The rate of change flow conditions.
Pressure transient flows
The wave is reflected through the entire conduit system
and results in further surge in pressure. It is controlled
by various devices like air-chambers, surge shafts, air
∆P
chambers, relief valves and specified closure rates.
The
∆P = ρcrise
∆V
pressure
∆V
c is the wave speed,
the change in fluid velocity.
Pressure transient flows
Speed c is not constant, depends on
• Fluid’s bulk modulus of elasticity (more compressible
the fluid, lesser is the speed)
• Conduit material’s Young’s modulus of Elasticity (more
elastic the conduit, lesser the speed).
It determines the time required by the wave to traverse
the system boundary (L) and travel back. Thus, 2L/c is
the pipe period. The slow valve closure takes longer
time than one pipe period, while the wave cycle time is
4L/c
Pressure transient flows
Pressure transient flows
Pressure transient flows
Rigid Column Theory
Consider a flow of liquid of density, ρ and
bulk modulus,K f
f
along a pipe of diameter d and wall thickness e, where
the young modulus of the pipe material is E and the
ν
possion’s
ratio is . ρThe mean density
of the fluid
mixture with y proportion of gas by volume, is
ρ = yρ g + (1 − y )ρ g
By definition bulk modulus of fluid and gas
 Vf
dV f = −
K
 f
Vt =
π
4

V 
dp, dVg = − g dp, Vg + V f = Vt

K 

 g
d 2l , where l is the length of pipe element chosen
Rigid Column Theory
δd , then the volume distortion
Let the radial distortion be
δd
dV
=
π
d
due to passage of pressure wave
dp, l
HOW?
p
2
Longitudinal and circumferential stresses, if e is pipe
d
d
FL = dp, FC = dp
thickness
4e
2e
δl (FL −νFC )
δd (FC −νFL )
The
strains are
=corresponding
=
and
l
E
d
E
ν
(FC −νFL ) ratio of pipe material, Thus
Where
dV p = πd 2lis Poisson’s
2E
Rigid Column Theory
The final form of volumetric strain will depend on the
pipe restraint. Three cases are common
1. Pipe restrained axially and circumferentially, both
dV p
d  ν
=
1 − dp
stresses
occur
V
Ee
2


2. Pipe restrained axially only
dV p
V
=
d
1 −ν 2 dp
Ee
(
)
3. Pipe may expand along its length,
dV p
V
=
d
dp
Ee
dV p
So the general expression for pipe distortionV
is=
d
C ′dp
Ee
Rigid Column Theory
Total change in the volume of the fluid, gas and pipe
dVt = dV pis− dV f − dVg
section
d
V f Vg 
=  C ′dp +
+
 dp
K f K g 
 Ee
d
(
1− y) y 
=  C ′dp +
+
Vt dp
Kf
K g 
 Ee
The overall effective bulk modulus for pipe, fluid and
d
(
1 − y ) is
y 
gas
combination
K =  C′ +
+

−1
eff
 Ee
Kf
K g 
Rigid Column Theory
Expression for the wave speed may be deduced as
c=
K eff
ρ
d
(
1− y) y 
=  C′ +
+

Ee
K
K

f
g 

−1
[yρ + (1 − y )ρ ]
−1
g
g
In the absence of free gas wave speed
c=
K  dK f 
C ′ 
1 +
Ee
ρ 

Rigid Column Theory
Surge Control
The pressure surge can be controlled by
1. Valve Closure control
2. Increasing pump inertia
3. Surge Shaft/ air vessels
4. Air/ fluid admission valves
5. Relief valves
6. Bypass systems
Valve Closure Control
v
Typical valve charateristicτ =
v0
∆p0
∆p
Water hammer/ positive pressure surge will propagate
through the system as the valve is closed. It is reduced
by
1. Two speed valve closure
τ = f (θ ), while θ = φ (t )
•Last 15% closing taking more time than first 85%.
Valve Closure Control
2. By-pass valve
3. Air-chamber (Useful in positive and negetive
pressure surges)
Surge Tanks
Surge Tanks
Rigid column theory is applied to surge tank
applications
Consider a pipeline, of area ‘a’ conveying water at
velocity V. The surge tank is located at distance ‘L’ from
reservoir. The tank cross section is A, A/a>>1. Under
steady state frictionless flow level in the surge tank shall
be same as that at reservoir.
aV =flow
Q1 +rate
Q Q into the tank due to transient changes
The
1
Differentiation yields
thedVwater
level
in
the
tank,
dz
dQ dQ
a
dt
=
1
dt
+
dt
Surge Tanks
aV = Q1 + Q
Differentiation yields
dV dQ1
d 2z
a
≈
=A 2
dt
dt
dt
The fluid massρ(aL ) is accelerated/decelerated which is
ρgz
opposed by rise ‘z’ in surge tank resulting in pressure
at the surge tank end of the pipe. Hence equilibrium of
dV
forces
ρgza = m
dt
dV
ρgza = ρaL
dt
dV g
= z
dt L
Surge Tanks
dV g
= z
dt L
dV
Substituting
dt
d 2 z ag
+
z=0
2
dt
AL
Solution yields
ag
ag
z = C1 sin
t + C2 cos
t,
AL
AL
z is measured from steady water level in tank, z = 0, at t = 0 ⇒ C 2 = 0
ag
dz
z = C1 sin
t , C1 is determined by codition of full closure, A
= Q0
AL
dt t =0
Surge Tanks
dV g
= z
dt L
dV
Substituting
dt
d 2 z ag
z=0
+
2
dt
AL
Solution yields
ag
ag
z = C1 sin
t + C2 cos
t,
AL
AL
z is measured from steady water level in tank, z = 0, at t = 0 ⇒ C 2 = 0
ag
dz
z = C1 sin
t , C1 is determined by codition of full closure, A
= Q0
AL
dt t =0
AL Q0
ag
AL
z=
t and time period of oscillations, T = 2π
sin
ag A
AL
ag
Surge Tanks
Surge Tanks: Types
Thank you