1. PUMPS
CHAPTER –11
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2. INTRODUCTION
DESIGNING OF ANY FLUID FLOWING SYSTEM REQUIRES;
1. Design of system through which fluid will flow
2. Calculation of losses that will occur when the fluid flows
3. Selection of suitable device which will deliver enough energy
to the fluid to overcome these losses
Devices: Deliver Energy To Liquids/Gases: Pumps/Compressors
Devices: Extracts Energy From Fluids: Turbines
TYPES OF PUMPS
POSITIVE DISPLACEMENT PUMPS DYNAMIC PUMPS
RECIPROCATING PUMPS
CENTRIFUGAL
ROTARY PUMPS PUMPS
3. POSITIVE DISPLACEMENT PUMPS, (PDP’S)
WORKING PRINCIPLE AND FEATURES;
1. Fixed volume cavity opens
2. Fluid trapped in the cavity through an inlet
3. Cavity closes, fluid squeezed through an outlet
4. A direct force is applied to the confined liquid
5. Flow rate is related to the speed of the moving parts of the pump
6. The fluid flow rates are controlled by the drive speed of the pump
7. In each cycle the fluid pumped equals the volume of the cavity
8. Pulsating or Periodic flow
9. Allows transport of highly viscous fluids
10. Performance almost independent of fluid viscosity
11. Develop immense pressures if outlet is shut for any reason,
HENCE
1. Sturdy construction is required
2. Pressure-relief valves are required (avoid damage from
complete shutoff conditions)
4. PDP’S, contd.
RECIPROCATING TYPE PDPS
Piston OR Plunger pumps Diaphragm pumps
Single acting piston pump Single diaphragm pump
Double acting Simplex pump Double diaphragm pump
Double acting Duplex pump
5. ROTARY TYPE PDPS
SINGLE ROTOR MULTIPLE ROTORS
Sliding vane pump
Gear Pump
Flexible tube or lining 2 Lobe Pump
Screw pump 3 Lobe Pump
Radial Pump
AND MANY MORE
6. DYNAMIC PUMPS
WORKING PRINCIPLE AND FEATURES
1. Add somehow momentum to the fluid
(through vanes, impellers or some special design
2. Do not have a fixed closed volume
3. Fluid with high momentum passes through open passages and
converts its high velocity into pressure
TYPES OF DYNAMIC PUMPS
ROTARY PUMPS SPECIAL PUMPS
Centrifugal Pumps Jet pump or ejector
Axial Flow Pumps Electromagnetic pumps for liquid metals
Mixed Flow Pumps Fluid-actuated: gas-lift or hydraulic-ram
8. COMPARISON OF PDPS AND DYNAMIC PUMPS
CRITERIA PDPS DYNAMIC PUMPS
Flow rate Low, typically 100 gpm As high as 300,000 gpm
Pressure As high as 300 atm Moderate, few atm
Priming Very rarely Always
Flow Type Pulsating Steady
Constant flow rate for virtually
Head varies with
any pressure
flow rate
Constant OR
OR
RPM Flow rate cannot be changed
Flow rate changes with
without changing RPM
head for same RPM
Hence used for metering
Viscosity Virtually no effect Strong effects
9. CENTRIFUGAL PUMPS
Centrifugal Pumps: Construction Details and Working
1. A very simple machine Illustration-1
2. Two main parts
1. A rotary element, IMPELLER Illustration-2
2. A stationary element, VOLUTE
3. Filled with fluid & impeller rotated
4. Fluid rotates & leaves with high velocity Impeller-1 Impeller-5
5. Outward flow reduces pressure at inlet,
(EYE OF THE IMPELLER), more fluid Impeller-2 Impeller-6
comes in. Impeller-3
6. Outward fluid enters an increasing area
region. Velocity converts to pressure Impeller-4
Impeller Impart Energy/Velocity By Rotating Fluid
Volute Converts Velocity To Pressure
10. CENTRIFUGAL PUMPS, contd.
Centrifugal Pumps: Working Principal
1. Swinging pale generates centrifugal force → holds water in pale
2. Make a bore in hole → water is thrown out
3. Distance the water stream travels tangent to the circle = f(Vr)
4. Volume flow from hole = f(Vr)
5. In centrifugal pumps, flow rate & pressure = f(Vr) (tip velocity)
A freely falling body achieves a velocity V = (2gh)1/2
OR
A body will move a distance h = V2/2g, having an initial velocity V
Find diameter that will generate ‘V’ to get required ‘h’ for given ‘N’
11. CENTRIFUGAL PUMPS, contd.
Q. FOR AN 1800 RPM PUMP FIND THE DIAMETER
OF IMPELLER TO GENERATE A HEAD OF 200 FT.
Find first initial velocity V = (2gh)1/2 = 113 ft/sec
Convert RPM to linear distance per rotation
1800 RPM = 30 RPS → V/RPS = 113/30 = 3.77 ft/rotation
3.77 = circumference of impeller → diameter = 1.2 ft = 14.4 inches
CONCLUSION
FLOW THROUGH A CENTRIFUGAL PUMP FOLLOWS THE
SAME RULES OF FREELY FALLING BODIES
DO WE GET
THE SAME DIAMETER OR HEAD OR FLOW RATE
AS PREDICTED BY THESE IDEAL RULES
12. CENTRIFUGAL PUMPS, contd.
BASIC PERFORMANCE PARAMETERS
The Energy Equation for This Case
& & & V12 V2 2
Q − W shaft − W vis = − m1 h1 +
& + gz1 + m 2 h2 +
& + gz 2
2 2
Assumptions:
• No heat generation V2 2 V12
&
W shaft = m h2 +
& + gz 2 − h1 + + gz1
• No viscous work. 2 2
• Mass in = mass out
13. CENTRIFUGAL PUMPS, contd.
What would be the difference in ‘z’, can we assume z2-z1≈0
V2 2 V1 2
Hence &
W shaft = m h2 +
& − h1 +
2 2
p2 V2 2 p1 V12
&
W shaft = m
& + u2 + − + u1 +
ρ 2 ρ 2
p 2 V2 2 p1 V1 2 Thermodynamically, u = u(T)
&
W shaft = m
& + − +
ρ 2 ρ 2 only and Tin ≈ Tout
p 2 V2 2 p1 V1 2
&
W shaft = ρ Q + − +
ρ 2 ρ 2
14. CENTRIFUGAL PUMPS, contd.
ρ V2 2 ρ V12
Pw = ρ gHQ = W shaft
& = Q p2 + − p1 +
2 2
Where Pw = water power
Pw 1 ρ V 2 2 ρ V12
H = = ( p 2 − p1 ) + −
ρ gQ ρ g 2 2
Generally V1 and V2 are of same order of magnitude
If the inlet and outlet diameters are same
Pw 1
H = ≅ ( p 2 − p1 )
ρ gQ ρ g
15. CENTRIFUGAL PUMPS, contd.
The power required to drive the pump; bhp
The power required to turn the pump shaft at certain RPM
bhp = ω T T = torque required to turn shaft
The actual power required to drive the pump depends upon efficiency
Pw ρ gQH
η= =
bhp ωT
η = η vη hη m Efficiency has three components;
Volumetric Mechanical Hydraulic
• casing leakages 1. Losses in bearings • Shock
2. Packing glands etc • friction,
Q • re-circulation
ηv = Pf hf
Q + QL ηm = 1 − ηv = 1 −
bhp hs
16. CENTRIFUGAL PUMPS, contd.
Torque estimation ⇒ 1D flow assumption
1-D angular momentum balance gives
T = ρ Q ( r2Vt 2 − rVt1 )
1
Vt1 and Vt2 absolute circumferential
or tangential velocity components
Pw = ωT = ωρ Q ( r2Vt 2 − rVt1 ) = ρ Q ( u2Vt 2 − u1Vt1 )
1
Pw ρ Q ( u2Vt 2 − u1Vt1 ) 1 DO
H= = = ( u2Vt 2 − u1Vt1 ) DETAILS
ρ gQ ρ gQ g
IN TUTORIAL
Euler turbo- Torque, Power and Ideal Head depends on,
machinery Impeller tip velocities ‘u’ & abs. tangential velocities Vt
equations; Independent of fluid axial velocity if any
17. CENTRIFUGAL PUMPS, contd.
Doing some trigonometric and algebraic manipulation
H=
1
(V22 − V12 ) + ( u2 − u12 ) + ( w2 − w12 )
2 2 p w2 r 2ω 2
+z+ − = const
2g
ρg 2g 2g
BERNOULLI EQUATION IN ROTATING COORDINATES
Applicable to 1, 2 and 3D Ideal Incompressible Fluids
One Can Also Relate the Pump Power With Fluid Radial Velocity
Pw = ρ Q ( u2Vn 2 cot α 2 − u1Vn1 cot α1 )
DO
Q Q EX. 11.1
Vn 2 = and Vn1 =
2π r2b2 2π r1b1 IN TUTORIAL
With known b1, b2, r1, r2, β1, β2 and ω one can find centrifugal pump’s
ideal power and ideal head as a function of Discharge ‘Q’
18. CENTRIFUGAL PUMPS, contd.
EFFECT OF BLADE ANGLES β1, β2 ON PUMP PERFORMANCE
Pw 1
H= = ( u2Vt 2 − u1Vt1 )
ρ gQ g
Angular Angular
>>
momentum out momentum in
Q
Vn 2 = Vt 2 = u2 − Vn 2 cot β 2
2π r2b2
Doing all this leads to if β < 90, backward curve blades, stable op
if β = 90, straight radial blades, stable op
u2 u2 cot β 2
2
If β > 90, forward curve blades, unstable op
H≈ − Q
g 2π r2b2 g
19. CENTRIFUGAL PUMPS, CHARACTERISTICS
1. Whatever discussed earlier is qualitative due to assumptions.
2. Actual performance of centrifugal pump → extensive testing
3. The presentation of performance data is exactly same for
1. Centrifugal pumps 2. Axial flow pumps
3. Mixed flow pumps 4. Compressors
4. The graphical representation of pumps performance data obtained
experimentally is called “PUMP CHARACTERSTICS” OR “PUMP
CHARACTERSTIC CURVES”
1. This representation is almost always for constant shaft speed ‘N’
2. Q (gpm) discharge is the independent variable (LIQUIDS)
3. H (head developed), P (power), η (efficiency) and NPSH (net
positive suction head) are the dependent variables (LIQUIDS)
4. Q (ft3/m3/min), discharge is the independent variable (GASES)
5. H (head developed), P (power), η (efficiency) are the dependent
variables (GASES)
21. CENTRIFUGAL PUMPS, CHARACTERISTICS, contd.
General Features of Characteristic Curves of Centrifugal Pumps
1. ‘H’ is almost constant at low flow rates
2. Maximum ‘H’(shut off head) is at zero flow rate
3. Head drops to zero at Qmax
4. ‘Q’ is not greater than Qmax → ‘N’ and/or impeller size is changed
5. Efficiency is always zero at Q = 0 and Q = Qmax
6. η is not an independent parameter → P ρ gHQ
η= w =
P P
7. η = ηmax at roughly Q=0.6Qmax to 0.93Qmax
8. η = ηmax is called the BEST EFFICIENCY POINT (BEP)
9. All the parameters corresponding to ηmax are called the design
points, Q*, H*, P*
10. Pumps design should be such that the efficiency curve should be
as flat as possible around ηmax
11. ‘P’ rises almost linearly with flow rate
22. CENTRIFUGAL PUMPS, CHARACTERISTICS, contd.
(a ) basic casing with three (b) 20 percent larger casing with three
impeller sizes larger impellers at slower speed
Typical Characteristic Curves of Commercial Centrifugal Pumps
1. Having same casing size but different impeller diameters
2. Rotating at different rpm
3. For power requirement and efficiency one needs to interpolate
23. CENTRIFUGAL PUMPS, CHARACTERISTICS, contd.
Calculate the ideal Head to be developed by the pump
shown in last figure
(1170 × 2π / 60 rad / s ) ( 36.75 / 2 ×12 ft )
2 2
ω 2 r22
H o (ideal ) = = 2
= 1093 ft
g 32.2 ft / s
Actual Head = 670 ft or 61% of Ho(ideal) at Q=0
Differences are due to
1. Impeller recirculations, important at low flow rates
2. Frictional losses
3. Shock losses due to mismatch of blade angle and flow
inlet important at high flow rates
24. CENTRIFUGAL PUMPS, CHARACTERISTICS, contd.
IMPORTANT POINTS TO REMEMBER
1. EFFECT OF DENSITY
1. Pump head reported in ‘ft’ or ‘m’ of that fluid → ρ important
2. These characteristic curves, valid only for the liquid reported
3. Same pump used to pump a different liquid → H and η
would be almost same. OR. A centrifugal pump will always
develop the same head in feet of that liquid regardless of the
fluid density
4. However P will change. Brake HP will vary directly with the
liquid density
2. EFFECT OF VISCOSITY
1. Viscous liquids tend to decrease the pump Head, Discharge
and efficiency → tends to steepen the H-Q curve with η ↓
2. Viscous liquids tend to increase the pump BHP
27. CENTRIFUGAL PUMPS, CHARACTERISTICS, contd.
µ ≥ 300µwor µ > 2000 SSU
µ
PDP’s are preferred
µ ≤ 10µw or µ < 50 SSU
µ
Centrifugal pumps are preferred
28. SUCTION HEAD AND SUCTION LIFT
• A centrifugal pump cannot pull or suck liquids
• Suction in centrifugal pump → creation of partial vacuum at pump’s
inlet as compared to the pressure at the other end of liquid
• Hence, pressure difference in liquid → drives liquid through pump
• How one can increase this pressure difference
– Increasing the pressure at the other end
• Equal to 1 atm for reservoirs open to atmosphere
• > or < 1 atm for closed vessels
– Decreasing the pressure at the pump inlet
• Must be > liquid vapor pressure → temperature very important
• By increasing the capacity → Bernoulli's equation
29. SUCTION HEAD AND SUCTION LIFT
MAXIMUM SUCTION DEPENDS UPON
• Pressure applied at liquid surface at liquid source, hence
– Maximum suction decreases as this pressure decreases
• Vapor pressure of liquid at pumping temperature
– Maximum suction decreases as vapor pressure increases
• Capacity at which the pump is operating
CASE OF OPEN RESERVOIRS
• Maximum suction varies inversely with altitude Table-1
CASE OF HOT LIQUIDS
• Maximum suction varies inversely with temp. Table-2
CASE OF INCREASING CAPACITY
• Maximum suction varies inversely with capacity Table-3
30. NET POSITIVE SUCTION HEAD
• Problem of Cavitation
–The lowest pressure occurs at the pump’s inlet
–Pressure at pump inlet < liquid vapor pressure → cavitation occurs
–What are the effects of cavitation
• Lot of noise and vibrations are generated
• Sharp decrease in pump’s ‘H’ and ‘Q’
• Pitting of impeller occurs due to bubble collapse
• May occur before actual boiling in case of dissolved gases /
low boiling mixtures of hydrocarbons
• Hence ‘P’ at pump’s inlet should greater than the Pvp
• This extra pressure above Pvp available at pump’s inlet is called
Net Positive Suction Head ‘NPSH’
P Vi 2 Pvp
• Mathematically → NPSH = 1 + −
ρg 2 ρg
31. NET POSITIVE SUCTION HEAD, contd.
• NPSH calculated from this equation is the ‘REQUIRED NPSH’
specified by manufacturer → “PUMP’S CHARACTERISTIC”
• The NPSH actually available at the pump’s inlet is called
‘AVAILABLE NPSH’ → “SYSTEM’S CHARACTERISTIC”
• ‘AVAILABLE NPSH’ must be ≥‘REQUIRED NPSH’
• Rule of thumb for design
‘AVAILABLE NPSH’ ≥ (2+‘REQUIRED NPSH’
NPSH’) ft of liquid
HOW TO CALCULATE AVAILABLE NPSH
Write Energy Equation between the free surface of fluid reservoir
and pump inlet
Psurface Pvp
NPSH available = − Z i − h fi −
ρg ρg
Thus Zi can be important parameter in designers hand to ensure that
cavitation does not occur for a given Psurface and temperature
32. NET POSITIVE SUCTION HEAD, contd.
EFFECT OF VARYING HEIGHT
Psurface Pvp Given, Psurface, Pvp and hfi , Zi can
NPSHA = − Z i − h fi − ≥ NPSHR
ρg ρg be varied to avoid cavitation
An Example
The 32-in pump of Fig. 11.7a is to pump 24,000 gpm of water at 1170 rpm from a
reservoir whose surface is at 14.7 psia. If head loss from reservoir to pump inlet is 6
ft, where should the pump inlet be placed to avoid cavitation for water at (a) 60°F,
pvp0.26 psia, SG 1.0 and (b) 200°F, pvp 11.52 psia, SG 0.9635?
NPSHR = 40 ≤
Psurface Pvp
− Z i − h fi − =
(14.7 − 0.26 ) − Z − 6 Z i ≤− 12.7
ρ g = 62.4
ρg ρ g 62.4 (144 )−1 i
Pump must be placed at least 12.7 ft below the reservoir surface to
avoid cavitation.
ρ g = 62.4 × .9653 = 60.1 Z i ≤ −38.4
Pump must now be placed at least 38.4 ft below the reservoir surface,
to avoid cavitation
33. NET POSITIVE SUCTION HEAD, contd.
TYPICAL EXAMPLE
A pump installed at an altitude of 2500 ft and has a suction lift of 13 ft
while pumping 50 degree water. What is NPSHA? Ignore friction
Psurface Pvp
NPSH available = − Z i − h fi − = 31 − 13 − 0 − .41 = 17.59 ft
ρg ρg
Actual NPSHA = 17.59 – 2 = 15.59 ft
TYPICAL EXAMPLE
We have a pump that requires 8 ft of NPSH at I20 gpm. If the pump is
installed at an altitude of 5000 ft and is pumping cold water at 60oF,
what is the maximum suction lift it can attain? Ignore friction
Psurface Pvp
NPSHA = NPSHR + 2 = 8 + 2 = − Z i − h fi − = 28.2 − Z i − 0 − .59 = 17.59 ft
ρg ρg
34. DIMENSIONLESS PUMP PERFORMANCE-1
PERFORMANCE-
EVERY PUMP HAS
THREE PERFORMANCE PARAMETERS
1. Head ‘H’ (or pressure difference ∆P-recall that ∆P= ρgH)
2. Volume Flow Rate ‘Q’
3. Power ‘P’
TWO "GEOMETRIC" PARAMETERS:
1. D diameter
2. n (or ω) rotational speed
THREE FLUID FLOW PARAMETERS:
1. ρ density
2. viscosity
3. ε roughness
Above parameters involve only three dimensions, M-L-T
35. DIMENSIONLESS PUMP PERFORMANCE-2
PERFORMANCE-
Buckingham π Theorem suggests
7 -3 = 4 π’s to represent the physical phenomena in a pump.
Any pump’s performance parameters are
1. Head H (or gH ) → gH = f1 ( Q, D, n, ρ , µ , ε )
2. Power P → P = f 2 ( Q, D, n, ρ , µ , ε )
Hence The Two π Groups Are
gH Q ρ nD 2 ε P Q ρ nD 2 ε
= g1 3 , , = g2 3 , ,
n2 D 2 nD µ D ρn D
3 5
nD µ D
WHERE
ε
= relative roughness
D
gH
ρ nD 2 ρ ( nD ) D 2 2 = CH = Head Coefficient
= = Re. Number n D
µ µ
Q P
3 = CQ = Capacity Coefficient 3 5 = CP = Power Coefficient
nD ρn D
36. DIMENSIONLESS PUMP PERFORMANCE-3
PERFORMANCE-
Reynolds number inside a centrifugal pump Hence, we may write:
1. ≈ 0.80 to 1.5x107)
2. Flow always turbulent CH = CH ( CQ )
3. Effect of Re, almost constant
4. May take it out of the functions g1and g2 CP = CP ( CQ )
5. Same is true for ε/D
For geometrically similar pumps,
Head and Power coefficients should be (almost)
unique functions of the capacity coefficients.
In real life, however:
-manufacturers use the same case for different rotors
(violating geometrical similarity)
-larger pumps have smaller ratios of roughness and clearances
-the fluid viscosity is the same, while Re changes with diameters.
37. DIMENSIONLESS PUMP PERFORMANCE-4
PERFORMANCE-
CH, CP and CQ combined to give a coefficient having practical meaning
CH CQ
η= = η ( CQ )
CP
Similarly one can also define the CNPSH the NPSH coefficient as
g ⋅ NPSH
C NPSH = 2 2
= C NPSH ( CQ )
n D
38. DIMENSIONLESS PUMP PERFORMANCE-5
PERFORMANCE-
Representing the pump performance data in dimensionless form
Pump data Results in graphical form
•Choose two geometrically
similar pumps
•32 in impeller in pump (a) & 38
in in pump (b)
•Pump (b) casing 20% > pump
(a) casing.
•Hence same diameter to casing
ratios
DISCRIPENCIES
•A few % in η and CH
•pumps not truly dynamically similar
•Larger pump has smaller roughness ratio
•Larger pump has larger Re. number
39. DIMENSIONLESS PUMP PERFORMANCE-6
PERFORMANCE-
The BEP lies at η=0.88, corresponding to,
CQ* ≈ 0.115 CP* ≈ 0.65 CH* ≈ 5.0 CNPSH* ≈ 0.37
A unique set of values
• Valid for all pumps of this geometrically similar family
• Used to estimate the performance of this family pumps at BEP
Comparison of Values
Discharge Head Power
D, ft n, r/s nD3, ft3/s n2D2/g, ft n3D5/550, hp
Fig. 11.7a 32/12 1170/60 370 84 3527
Fig. 11.7b 38/12 710/60 376 44 1861
Ratio - - 1.02 0.52 0.53
40. SIMILARITY RULES/AFFINITY LAWS-1
LAWS-
If two pumps are geometrically similar, then
1. Ratio of the corresponding coefficients =1
2. This leads to estimation of performance of one based on the
performance of the other
MATHEMATICALLY THIS CONCEPT LEADS TO
Q2 gH 2 P2
CQ2 3
n2 D2 CH 2 2 2
n2 D2 CP2 ρ 2 n2 D2
3 5
= =1 = =1 = =1
CQ1 Q1 CH1 gH1 CP1 P1
n1 D13 n12 D12 ρ1n13 D15
3 2 2 3 5
Q2 n2 D2 H 2 n2 D2 P2 ρ 2 n2 D2
= = =
Q1 n1 D1 H1 n1 D1 P ρ1 n1 D1
1
THESE ARE CALLLED SIMILARITY RULES
41. SIMILARITY RULES/AFFINITY LAWS-1
LAWS-
The similarity rules are used to estimate the effect of
1. Changing the fluid
2. Changing the speed
3. Changing the size
VALID ONLY AND ONLY FOR
Geometrically similar family of any dynamic turbo machine
pump/compressor/turbine
Effect of changes in size and speed
on homologous pump performance
(a) 20 percent change
in speed at constant size
(b) 20 percent change in
size at constant speed
42. SIMILARITY RULES/AFFINITY LAWS-1
LAWS-
For Perfect Geometric Similarity η1 = η2, but
Larger pumps are more efficient due to
1. Higher Reynolds Number
2. Lower roughness ratios
3. Lower clearance ratios
Empirical correlations are available
To estimate efficiencies in geometrically similar family of pumps
1
1 − η2 D2
4
Moody’s Correlation
≈
Based on size changes 1 − η1 D1
0.33
Anderson’s Correlation 0.94 − η2 Q2
≈
Based on flow rate changes 0.94 − η1 Q1
43. Concept of Specific Speed-1
Speed-
A confusing example
We want to use a centrifugal pump from the family of Fig. 11.8 to
deliver 100,000 gal/min of water at 60°F with a head of 25 ft. What
should be (a) the pump size and speed and (b) brake horsepower,
assuming operation at best efficiency?
H* = 25 ft = (CH n2 D2)/g = (5 × n2 D2)/32.2
Q* = 100000 gpm = 222.8 ft3/m = CQ n D3 = 0.115 × n D3
Bhp* = Cpρ n3 D5 = 720 hp
Solving simultaneously gives, D = 12.4 ft, n = 62 rpm
44. Concept of Specific Speed-1
Speed-
The type of applications for which centrifugal pumps are required are;
1. High head low flow rate
2. Moderate head and moderate flow rate
3. Low head and high flow rate
Q. Would a general design of the centrifugal pump will do all the
three jobs?
Ans. No
Q. What should be the design features to accomplish the three
specified jobs?
PHYSICS FOR OUR RESCUE
1. Answer to this question lies in the basic concept of centrifugal
pump working principle.
2. Vanes are used to impart momentum to the fluid by applying the
centrifugal force to the fluid.
45. Concept of Specific Speed-2
Speed-
3. More the diameter of the vane more will be the centrifugal force
4. More will be the diameter more will be the radial component of
velocity and lesser will be the axial component
5. More will be the radial velocity more will be the head developed
6. Hence to get more head you need longer vanes and vice versa
7. More will be the clearance between the impeller and casing
more will the flow rate & also more will be the axial component
8. These simple physics principles lead us to the variation in
impeller design to accomplish the three jobs mentioned
46. Concept of Specific Speed-3
Speed-
POINT TO PONDER
• We represent the performance of a family of geometrically similar
pumps by a single set of dimensionless curves
• Can we use even a smaller amount of information or even a single
number to represent the same information?
• We have a huge variety of pumps each with a different diameter
impeller, shape of impeller and running at certain rpm
• Impeller shape ultimately dictates the type of application
• RPM is not related to the pump design however it effects its
performance
• Hence the biggest problem is to avoid diameter in the pump
performance information
Again dimensional analysis comes to rescue, a combination of π’s is
also a π, giving the same information in a different form
47. Concept of Specific Speed-4
Speed-
REARRANGE THE THREE COEFFICIENTS INTO A NEW
COEFFICIENT SUCH THAT DIAMETER IS ELIMINATED
1
CQ
2
n (Q )
1
2 N s = 17182 N s/
N =
/
=
( gH )
s 3 3
CH
4 4
Points to remember
1. Ns refers only to BEP
2. Directly related to most efficient
Rigorous form, dimensionless
pump design
3. Low Ns means low Q, High H
( RPM )( GPM )
1
2
Ns = 4. High Ns means High Q, Low H
( H , ft )
3
4
5. Ns leads to specific pump
applications
6. Low Ns means high head pump
Lazy but common form, 7. High Ns means high Q pump
Not dimensionless
Experimental data suggests, pump is in
Similarly one can define Nss danger of cavitation
, based on NPSH If Nss ≥ 8100
48. Concept of Specific Speed-5
Speed-
GEOMETRICAL
VARIATION OF SPECIFIC
SPEED
Detailed shapes
49. Concept of Specific Speed-5
Speed-
Specific speed is an indicator of
Pump performance
Pump efficiency
The Q is a rough indicator of
Pump size
Pump Reynolds Number THE PUMP CURVES
50. Concept of Specific Speed-5
Speed-
Note How The Head, Power and Efficiency curves change as
specific speed changes
51. Revisit of Confusing Example-1
Example-
Dimensionless performance curves for a
typical axial- flow pump. Ns = 12.000.
Constructed from data for a 14-in pump
at 690 rpm.
CQ* =0.55, CH*=1.07, Cp*=0.70,ηmax= 0.84.
Ns = 12000
D = 14 in, n = 690 rpm, Q* = 4400 gpm.
52. Revisit of Confusing Example-2
Example-
Can this propeller pump family provide a 25-ft head & 100,000 gpm
discharge
Since we know the Ns and Dimensionless coefficients then using
similarity rules let us calculate the Diameter and RPM
D = 48 in and n = 430 r/min, with bhp = 750:
a much more reasonable design solution
53. Pump vs System Characteristics
• Any piping systems has the following components in its total
head which the selected pump would have to supply
1. Static head due to elevation
2. The head due to velocity head, the fictional head loss
3. Minor head losses
H sys = ( z2 − z1 ) = a h f ,la min ar =
128µ LQ
πρ gD 4
H sys = ( z2 − z1 ) + h f ,la min ar = a + bQ
Mathematically,
3 possibilities h f ,turbulent = Through Moody ' s Method
V2 fL
H sys = ( z2 − z1 ) + ∑ + ∑ K = a + cQ 2
2g D
54. Pump vs System Characteristics, contd
• Graphical Representation Of The Three Curves
55. Match between pump & system
•In industrial situation the resistance often varies for various
reasons
•If the resistance factor increases, the slope of the system
curve (Resistance vs flow) increases & intersect the
characteristic curve at a lower flow.
•The designed operating points are chosen as close to the
highest efficiency point as possible.
•Large industrial systems requiring different flow rates often
change the flow rate by changing the characteristic curve with
change in blade pitch or RPM
57. Pump in Parallel or Series
•To increase flow at a given head
1. Reduce system resistance factor with valve
2. Use small capacity fan/pumps in parallel.
Some loss in flow rate may occur when operating
in parallel
•To increase the head at a given flow
1. Reduce system resistance by valve
2. Use two smaller head pumps/fans in series.
But some head loss may occur.
60. Unstable operation (Hunting)
If the characteristic is
such that the system
finds two flow rates for
a given head it cannot
decide where to stay.
The pump could
oscillate between
points. It is called
hunting.
