AFD5 Pumps

Applied Fluid Dynamics
AFD5 Pumps
Chemical Engineering Guy
www.ChemicalEngineeringGuy.com

AFD5 Overview
• This is still Incompressible Flow

Textbook, Reference and Bibliography

• Chapters:
– 13  Pump Selection & Application
Applied Fluid Mechanics. Robert
Mott 6th Edition

• Section 2: Fluid Mechanics
– CH8: Transportation of Fluids
• Just the “Pumps” part
Unit Operation of Chemical
Engineering. McCabe 7th Edition

AFD5 Block Overview
• Section 1: Pump Types
– Positive displacement
• Lobe, Screw, Piston, Vane, Gear
– Kinetic
• Axial and Centrifugal
– Pump Performance
• NHSPr
• Power
• Section 2: System Curve
– System Head
– System Curve

AFD5 Block Overview
• Section 3: Pump Curve
– Pump Head
– Pump Curve
• Impeller Effect
• Efficiency Curves
• Pump Power Curves
• NPSH
• Velocity Effect
• Section 4: Pump Selection
– How to choose a pump
– Supplier Data
– Pump Affinity Laws
• Section 5: Pumping Systems
– Pump in Series
– Parallel Pumps
– Software Modeling

Section 1: Pump Types

Introduction to Pumps
• They increase mechanic energy…
– Increase of speed (NOTE: Pipe Diameter)
– Increase in height (NOTE: if h2=h1)
– Increase of Pressure
• The fluid density does NOT change
• This is incompressible flow!

• Clasification
– Positive Displacement
• Rotatory
• Reciprocal
– Kinetic Pumps
• Centrifugal
• Axial

• There are many implications and things to
consider when choosing a pump

• Fluid type
• Volumetric Flow requirement
• Suction Conditions (P,T, composition)
• Discharge Conditions (P,T, composition)

• Head Requirement
• Power requirement
• Spacing,
dimensions,
weigth, etc.
• Initial Investment
• Operation Costs

• Maintainment Cost
• Type of Pump
• Supplier
• Sizing and type of
Conections

• Operation RPM and Frecuency
• Impeller (material, diameter, etc.)
• Pump Material
• Min and Max Pressure discharge/suction

• So… what types of pumps do we have?

Pump Types

Rotatory…

Centrifugal…

Positive Displacement (P.D.)
• Pressure is applied directly
• Rotatory devices
• They “fill and empty” the fluid
• Fixed Volumetric Flow per Revolution
• Two Types
– Rotatory
– Reciprocal

P.D. Rotatory
• Gear
• Screw
• Progresive Cavity
• Lobe
• Peristaltic

P.D. Rotatory: Progressive Cavity

P.D. Recirpocal
• Piston
• Diaphragm

Kinetic Pumps
• High velocity operations
• Impeller Operated
• Kinetic Energy  Mechanic Energy  Pressure
• Radial flow
– Centrifugal
• Axial

Centrifugal Pump Parts
• Super Basic:
– Impeller
– Shaft
– House Casing
– Motor
– Suction
– Discharge

The motor is very important!

Pump Performance
• We always analyze pump performance
– Efficiency (Work inlet vs. outlet)
– Pressure inlet vs. outlet

Cavitation
• Damage due to little bubbles
• Recall that bubbles are gas, they are copmressible
• The impeller will experiment different
Pressures/Forces

Cavitation
• This is due to the pressure changes in the
– Inlet (Suction)
– Eye (impeller center)
– Outlet (Discharge)
• Recall that for a substance
– If the Pressure decreases
– The boiling temperature decreases

Cavitation
• So… for example water:
– If T operation = 80ºC
– Inlet = 1 atm
– Eye = 0.5 atm
– Outlet = 2 atm

• So… for example water:
– If T operation = 80ºC
– Inlet = 1 atm  100
– Eye = 0.5 atm  75
– Outlet = 2 atm  120
Cavitation
Water will evaporate
in the Eye!

Cavitation

Cavitation
Safe

Cavitation
Safe
Not Safe

Cavitation Explained
www.ChemicalEngineeringGuy.comhttps://goo.gl/Mep2ag

Cavitation: Impeller Damage

NSPHR
• We need a standard or something to know
what is the minimum requirement to avoid
cavitation!
• Objective  Avoid Cavitation

NSPHR
• NSPHr  Net Specific Pressure at Head
Required
• Supplier will typically set it for the design
– You may calculate/experiment it if not given
• This is the limit pressure.
– The min required pressure so it won’t cavitate
• TIP It is near the boiling point of the
substance in operation

NSPHR
Required
• Supplier will typically set it for the design
– You may calculate/experiment it if not given
• This is the limit pressure.
– The min required pressure so it won’t cavitate
• TIP It is near the boiling point of the
substance in operation
Therefore, it depends
on the substance!

NSPHA
Available
• The actual pressure present in the suction

NSPHR and NSPHA
• Rule of Thumb
– NPSHA > 1.10·NPSHR
• Check out Vapor Pressures in:
– Tables
– Graphs/Diagrams
– Calculated (Regressions, Equations, etc.)
• Clausius Clapeyron
• Antoine Equation
• Aspen Software

NSPHR and NSPHA
• NPSHA Calculation…
– In pressure units kPa or psi
NPSHA = (Psuction – Psaturation)
– In specific energy units J/kg or m2/s2
NPSHA = (Psuction – Psaturation)/(ρ)
– In length “m” or “ft”
NPSHA = (Psuction – Psaturation)/(ρ*g)

Cavitation Exercise
• Given a system:
– Benzene @ 37.8°C
– P suction = 86kPa
– NPSHR = 3.05m
– Density (rho) = 876 kg/m3
• A) Calculate the Vapor Pressure of Benzene
• B) Is this Pump enough? Or you will expect Cavitation?

Cavitation Exercise
• Given a system:
– NPSHR = 3.05m
– From Antoine Eqn. Pv(37.8°C) = 21.5574 kPa

Cavitation Exercise
• Given a system:
– NPSHR = 3.05m
– Use Equation: NPSHA > 1.10 NPSHR
• NPSHA = (Psuc – Pvap) / (rho*g) = (86-21.5)(10^3)/(876*9.8) = 7.5 m
•

Cavitation Exercise
• Given a system:
– NPSHR = 3.05m
– Since NPSHA > 1.10 NPSHR
• 7.5 m > 1.10 (3.05)
• 7.5 m > 3.36 m  Expect NO cavitation

Cavitation Exercise
• Given a system:
– NPSHR = 3.05m
• 8.1 m > 1.10 (3.05)
• 8.1 m > 3.36 m  Expect NO cavitation
If it was water? Will it cavitate? Vp = 6.3 kPa
D = 1000 kg/m3

Cavitation Exercise
• Given a system:
– NPSHR = 3.05m
• NPSHA = (Psuc – Pvap) / (rho*g) = (86-72)(10^3)/(750*9.8) = 1.9 m
• 1.9 m > 1.10 (3.05)
• 1.9 m > 3.36 m  Expect CAVITATION!
If it was a very volatile substance with Vp = 72 kPa
D = 750 kg/m3

Cavitation Exercise
• Given a system:
– NPSHR = 3.05m
• NPSHA = (Psuc – Pvap) / (rho*g) = (86-VP)(10^3)/(876*9.8) = 3.4 m
• 3.4 m > 1.10 (3.05)
• 3.4 m > 3.36 m  Expect CAVITATION!
What is the min. Vapor Pressure to operate Benzene?

Need More Problems?
Check out the COURSE
• Courses
MOMENTUM TRANSFER OPERATIONS
You’ll get SOLVED Problems, Quizzes, Slides, and
much more!

Pump Power
• Recall that we are calculating Specific Energy
– That is, Energy per unit mass
– If you multiply by all the mass involved…
– You get the total Work required
• If you use mass flow (mass per unit time)
– That Work (energy) will turn to Power (energy per
unit time)

Pump Power
• Power is very important in Pumping
– P= m*Wb/η
• m: mass flow (kg/s)
• Wb= Work done by Pump per unit mass (J/kg
or m2/s2)

Pump Power
• Power typical Units
– Watt and kilo-Watt
– Horse Power (HP)  Brake Horse Power BHP
– Foot Pounds per minute*

Pump Power Exercise
• Calculate Power Required if
– Mass flow is 1.3 kg/s
– Efficiency
• 80% when Wb > 2 kJ
• 72% when Wb < 2 kJ
– The specific work required by the pump was
calculated to be about 1.4 kJ per kg

Pump Power Exercise
• Calculate Power Required if
– Mass flow is 1.3 kg/s
– Efficiency
• 80% when Wb > 2 kJ
• 72% when Wb < 2 kJ
– The specific work required by the pump was calculated
to be about 1.4 kJ per kg
• P = m·Wb/η
• P = 1.3 kg/s * 1.4 kJ/kg / 0.72 = 2.53 kJ/s or 2.53 kW

End of Section 1: Pump Types

Section 2: System Curve

System Head
• ASSUME Friction loss change vs. velocity
• System head will be used in Pump Selection

System Head
• The System Head answers the question:
– How much power is required?
– How much energy will be needed to satisfy the
system
– How much Pumping Requirement

System Head
• We will know EVERYTHING
– Pipe type and Sizing
– All the fittings and valves used
– Pressure in A and B
– Location of A and B

System Head Exercise
• Given:
– Va = 0
– Vb= 2
– Ha = 0m
– Hb = 2m
– Pa = 101325
– Pb = 110325
– Rho = 1000
– Hf = 0.5
• Calculate the Head of the System

System Head Exercise
• Given:
– Va = 0
– Vb= 2
– Ha = 0m
– Hb = 2m
– Pa = 101325
– Pb = 110325
– Rho = 1000
– Hf = 0.5
• Calculate the Head of the System
nWp = (Pb-Pa)/rho + (Vb2-Va2)/2 + g(Zb-Za) + 0.5
nWp = (110325-101325)/(1000) + (2^2)/2 + 9.8*(2) + 0.5
nWp = 31.3

System Head
• What will happen if the industry requires to
increase in 20% the Flow Rate?
• The calculation is not valid now!
– The velocities will increase
– Friction will increase
• The System Head will increase!

System Curve
• This was done for a specific flow rate
• What will happen if we change that flow rate
in the SAME system

System Curve
• This was done for a specific flow rate
• What will happen if we change that flow rate
in the SAME system
– You will increase friction
– Velocity Heads will increase
– Position and Pressure heads remain the same

System Curve
• It would be nice to know our system for every
flow rate, at least the must “common” flow rates
– 0 gpm
– 1gpm
– 5gpm
– 10 gpm
– 50 gpm
– 1000 gpm…

System Curve
• If we change Flow Rate (Q) we will change:
– Pipe Velocities (Va, Vb)
– Friction loss since it depends on V (hf)
• We will get the “Required Work” for every
Flow Rate

System Curve Exercise #1
• Create a System curve for the next System

Piping Data
2" 3"
Di [m] 0.053 0.078
Area [m2] 0.002 0.005
vel [m/s] 0.291 0.132
e/D [] 0.001 0.001
Re [] 15279 10293
flujo: turbulent turbulent
f.f 0.029 0.032
f.f.T 0.019 0.017
Pipe Wall Friction
L 20.0 8.0
L/D 381.0 102.7
V2/2 0.042 0.009
f(L/D)V2/2 0.47 0.03
Wall Friction 0.50
Fittings + Valves
Valve (340ft) 6.46
Elbow (30ft) 0.52
Hfs 0.27 0.00
Shape Friction 0.28
Calculation of Heads
Pa-Pb 0.0Kpa
Zb-Za 4.0m
Va 0.0m/s
Vb 0.0m/s
(Pa-Pb)/rho 0J/kg
(Zb-Za)*g 39.2J/kg
(Vb^2-Va^)/2 0J/kg
hf 0.78J/kg
nWbb 40.0J/kg
4.1m
Power 25.19Watt
0.0338085HP
DATA
Inlet Flow 10gal/min
0.00063m3/s

Flow (GPM) Head (m)
0 4
10 4.1
20 4.3
40 5.4
50 5.7
75 7.7
90 9.3
100 10.5
150 18.4
200 29.3
250 43.3
300 60.4
0
10
20
30
40
50
60
70
0 50 100 150 200 250 300 350
Head(m)
Volumetric Flow Rate GPM
Head (m)

System Curve Exercise #1 Conclusion
• The curve is divided in two sections
– Static (fixed, wont change with flow
rate)
– Friction (will change depending of V)
0
10
20
30
40
50
60
70
0 50 100 150 200 250 300 350
Head(m)
Volumetric Flow Rate GPM
Head 100% (m)

System Curve Exercise #1 Conclusions
• The curve is exponential
– This is due to the velocity factor V2
• Friction is very important a factor
– Higher speed  Higher Friction loss (Hf)

• You must note that when increasing Flow:
– You increase velocity in pipes
– Vb^2-Va^2 difference may increase (requirement of
power)
• Only when delivering in tanks (Va = Vb = 0) will not change
– Pressure in “a” and “b” will remain the same
– Heigth difference of “a” and “b” will remain the same
– Velocity increase  Increase in Friction by V^2

Need More Problems?
• Courses
Applied Fluid Mechanics
Part 1: Incompressible Flow
You’ll get SOLVED problems, Quizzes, Slides, and
much more!

• Do the same exercise with
– Pipe’s Diameters of 4”, 8”, 10” and 12”
• Analysis  Velocities will change
– Friction will change
• The Head Will change

Volumetric Flow Rate (L/s)
nWp(m)
For a System
- Fixed Pipe Diameter
- Fixed Fittings
4” Piping

System Curve Exercise #24” Piping
8” Piping
nWp(m)

4”, 8”, 10” and 12” pipings
nWp(m)

4”, 8”, 10” and 12” pipings
nWp(m)
Shifted!

• The curve will flatten to the rigth
– When increasing the pipe’s Diameter
– Speed is reduced (V2)
– Friction is reduced, hence, Less Head

• NOTE  Valve is 50% closed

• Compare 50% and 100%
• Compare the same Flow vs. Head
0
20
40
60
80
100
120
0 50 100 150 200 250 300 350
HeadofSystem(m)
Volumetric Flow Rate (GPM)
Head of System (50% vs. 100%)
50%
100%

• Valves are pretty important in controlling
– Friction loss

• For the 50%  The Head is always higher tan
the 100%
• This requries more energy
• This means more operational costs
• But will let you increase the flow rate
whenever you want!

End of Section 2: System Curve

Section 3: Pump Curve/Head

System Head Review
• Recall that we already calculated
– Head of a System
• Requirement of the Pump
– System Curve
• Head of a System for different Flows
• Point A and B are specified for the System

Pump Head
• It will be interesting to analyse, the total
“load” of work needed for a pump
• A: Suction
• B: Discharge
PumpA B

Pump Head
B
B

Pump Head
• This is a specific Mechanical Energy Balance for the
pump, not the system
– We care the pump inlet/outlet diameter  Velocity
• The change of height is so small, wont affect if we
eliminate this
– No potential energy
• The Friction Loss is considered in the Pump’s Efficiency
• Pressures  Suction and Discharge

Pump Head
• This is a specific Mechanical Energy Balance for the
pump, not the system
– We care the pump inlet/outlet diameter  Velocity
• The change of heigh is so small, wont affect if we
eliminate this
– No potential energy
• The Friction Loss is considered in the Pump’s Efficiency
• Pressures  Suction and Discharge
V^2 may be so small
compared to change in P

Pump Head
• It is common in the hydraulic and mechanical
engineering jargon to use “length”
– Meters
– Feet
• This is exactly the past equation divided by gravity (L/t2)
• Example  Pump Load of 150 J/kg
ηWb = 150 J/kg
ηWb/g = 150 J/kg / 9.8 m/s2 = 15.3 m

Pump Head Exercise
• Given the next conditions:
– Efficiency 78%
– First Pipe inner diameter = 2 in
– Second Pipe outer diameter = 2.5
– Volumetric Flow = 0.4 m3/s
– The pressure enters the pump at 1.5 atm
– The pressure at the outlet is unknown but…
• The final pressure is 1.8 atm
• There is a pressure drop of 0.7 atm
– Friction loss during the first section is 2.8 J
– Friction loss during the second section is 1.8 J
– The Reservoir is 2 meter under the pump
– The tank is about 12 m above the pump
– The pump height is about 45 cm
– The suction of the Pump is about 2.2”
– The discharge pipe of the pump is 2.5”

Pump Head
• Pb = 1.8+0.7 = 2.5 atm = 253,312 Pa
• Pa = 1.5 atm = 151,987 Pa
• Zb = 0.45 m
• Za = 0.00 m
• Vb = Q/Ab = 0.4/(3.14*(2.5*0.254)^2/4)= 1.26 m/s
• Va = Q/Aa = 0.4/(3.14*(2.2*0.254)^2/4)= 1.63 m/s
η·Wb = (253312/1000 + 9.8*0.45 + (1.26^2)/2)-(151987/1000 + 9.8*0+(1.63^2)) = 103.87 J/kg
• η·Wb = 103.87 J/kg
We are interested in the Suction and
Discharge points!
Not the conventional A and B

Pump Performance Curve
• Similar to the case in the System’s Curve
– What will happen if we increase the flow rate
requirements?
– And if we decrease them?

• This is common
– Decrease  Shut down, decrease in sells, excess
of inventory, etc.
– Increase  increase of inventory, increase of sells,
revamp

• Now imagine if we change the volumetric flow rate
• What will increase?
– Pressures?
– Height?
– Velocities?
• You will have many Pump heads!
– 1 value for 1 m3/s
– 1 value for 2.5 m3/s

• Why not have many data of Pump heads for
several pumps?
– Set a Volumetric Flow  Pump Head
• Graph these values
– X-axis will be the Volumetric Flow (gpm or l/s)
– Y-axis will be the Pump Head (in meters or feet)
• This graph is called the “Pump Performance Curve”
NOTE: the size of the impeller is constant

Pump Curve
• Try to guess the shape of the graph!
• What happens as you increase Volumetric
Flow
• Recall that the head of the pump is not the
same as the head of the system!
• Concave? straight line? power? Logarithmic?

Pump Curve

Pump Curve
Massic Flow
Kg/s
Volumetric
Flow
GPM = Gallons
per minute
Pdis-Psuc
Pa/psi
Vel. Suction
m/s
Vel. Discharge
m/s
Pump Head
values in m/ft
0 Calculate Measured
data
Measured
data
Measured
data
Calculate
1
3
5
6

Pump Curve Exercise
• For the next System:
– Given the data of Inlet/Outlet Pressures
– The velocities involved due to inlet/outlet
– The friciton loss consider it in the efficiency
– You may ignore the gravity effect on the pump
A
B

Pump Curve Exercise
Flow mass Flow (vol) Flow (vol) Vsuc Vdesc DP Head nWp Head nWp/g
kg/s m3/s L/min m/s m/s m2/s2 m2/s2 m
0 0 0 0.00 0.00 275.6 275.6 28.1
1 0.001 60 0.29 0.00 270.0 270.0 27.5
2 0.002 120 0.57 0.00 268.0 267.8 27.3
5 0.005 300 1.43 0.00 245.0 244.0 24.9
8 0.008 480 2.29 0.00 215.0 212.4 21.7
15 0.015 900 4.29 0.01 154.2 145.0 14.8
20 0.02 1200 5.72 0.01 110.2 93.8 9.6
25 0.025 1500 7.15 0.02 63.0 37.4 3.8
27 0.027 1620 7.73 0.02 55.0 25.1 2.6
28 0.028 1680 8.01 0.02 38.0 5.9 0.6
• The experimental data is given next

Pump Curve Exercise
0.0
5.0
10.0
15.0
20.0
25.0
30.0
0 200 400 600 800 1000 1200 1400 1600 1800
Head nWp/g m
Head nWp/g m

Pump Curve Exercise
• Check out what happens when you add
– Different Impeller Diameters
• Check it out here:
• Courses
Part 1: Incompressible Flow

Pump Performance Curve Analysis
• Why does the Curve is concave down?
• As flow rate increases, decreases the head of
the pump!?
• Explain yourself!

Pump Performance Curve Analysis

• Actually, suppliers may group pumps as
“families”
– Steep (high friction)
– Average/Standard
– Flat (low friction)

Steep Curve  Large Head, Small Flows

Flat Curve  Small Head, LargeFlows

Same Operation
Point, different
Pump Curves

Reading Pump Curve DATA
• We will build this type of Diagrams
– Pump Head vs. GPM
– Impeller Diameter Curve
– Efficiency
– Pump Power
– NSPH required
– Extra data (RPM, Brand, etc.)

Pump Curve Diagram  Head vs. Q
• There is only ONE curve for a “fixed” pump
– Fixed Impeller’s Diameter
– Fixed Angular Velocity (RPM)
• You will get a unique value
– Volumetric Flow Rate  Head of the Pump

8”
Volumetric Flow
HeadPump(m)

8”
Volumetric Flow
HeadPump(m) Head = f(Q)

Pump Curve Diagram  Impeller’s Diameter
• Impeller  aka the “wheel”
• Makes the rotation
• Takes the inlet fluid to the “eye”
• Impeller may be changed in a given Pump

• Nomenclature used for Impeller Sizing
• 2 X 3 – 10
– 2: Discharge size (nominal Diameter) [in]
– 3: Suction size (nominal Diameter) [in]
– 10: Largest possible Impeller [in]
S.
D.
Impeller

8”
Volumetric Flow
HeadPump(m)

8”
Volumetric Flow
HeadPump(m)
3”

6“ 1/2
5”
4“ 1/2
3”
8”
Impeller Size (Diameter in Inches)
Volumetric Flow
HeadPump(m)

6“ 1/2
5”
4“ 1/2
3”
8”
Impeller Size (Diameter in Inches)
Volumetric Flow
HeadPump(m)
Same Flow Rate, Same
Pump, Different Impeller 
Different Heads of Pump

Pump Curve Diagram  Efficiency
• Pump Efficiency is very important
– What is the point of having a pretty nice pump
• If this is working only at 50% of efficiency
• The other 50% will not be used, but will be paid!
• These pump “curves” are also graphed in the
Pump Performance Curve!

Volumetric Flow
HeadPump(m)

6“ 1/2
5”
4“ 1/2
3”
8”
Volumetric Flow
HeadPump(m)

6“ 1/2
5”
4“ 1/2
3”
8”

Pump Efficiency
Exercise #1
• Choose the best efficiency area for this Pump

Pump Efficiency
Exercise #1
• Choose the best efficiency area for this Pump
73
%

Pump Efficiency
Exercise #2
• If GPM = 500… What is the Head when η = 65%

Pump Efficiency
Exercise #2
• If GPM = 500… What is the Head when η = 65%
About
3200 ft

Pump Curve Diagram  Power
• Power [=] J/s [=] Watt [=] HP or BHP
• Power = m*ηWp
• Power = ρ*Q*ηWp
– Power is function:
• System/Pump head
• Efficiency
• Volumetric Flow Rate
• When you change Q, you change Power
Requirments!

5”
4“ 1/2
3”

5”
4“ 1/2
3”
As Flow increases,
Power Increses

5”
4“ 1/2
3”
As Head Decreases,
Power Decreases

Pump Power
Exercise #1
• How much Power is Required in HP for:
– A pump working 600 GPM
– The Head needed is 2400 ft

Pump Power
Exercise #1
• How much Power is Required in HP for:
– A pump working 600 GPM
– The Head needed is 2400 ft
About
550 HP

Pump Power
Exercise #2
• If Power Requirement of a Pump = 600 BHP
• The Pump has an Impeller Diameter = 9 ½
– What is the System Head?
– What is the Effiency in this operation?

Pump Power
Exercise #2

Pump Power
Exercise #2
About 3200 ft
About 64%

Pump Curve Diagram  NPSHR
• NPSHr is also included in these graphs for
convenience
• You can check easily if you require more
pressure
• Make sure you use the scale of the NPSHr graph
– Typically written in the right 
NOTE: NSPH Required is f(Q) only!

8”
Volumetric Flow
HeadPump(m)

NOTE: NSPH Required is f(Q) only!

Exercise #1
• Calculate the NPSHR when:
– Impeller size is 7”
– Head = 200
– GPM = 160

Exercise #1
– Head = 200
– GPM = 160
You just need GPM

Exercise #1
– Head = 200
– GPM = 160
You just need GPM
NPSHR = 5.5 ft

Exercise #2
• What is the minimum Flow Rate given a
NPSHa of 3.3 m?

Exercise #2
NPSHa of 3.3 m?
NPSHa > 1.1*NPSHr  3

Exercise #2
NPSHa of 5 m?
NPSHa > 1.1*NPSHr  4.54

Exercise #2
NPSHa of 5 m?
NPSHa > 1.1*NPSHr  4.54
About 600 GPM

Full Diagrams!
• We’ve seen all the theoretical concepts
• We can read the full diagrams of a Pump Curve!

Full Diagrams!

End of Section 3: Pump Curve/Head

Section 4: Pump Selection

Methodology for Pump Selection
1. Choose which type of pump suits best
– Use the Specific Velocity Criteria
– Use a Graph (shown afterwards)
2. Check out the different models of that type
of pump system (radial, axial, etc.)
3. Then calculate the curve of the system
4. Intersect the Pump Head Curve + System
Head
5. That’s your operation point! Optimize it!

1a. General Pump Selection Diagram
• Easy to follow
• Function of Q and nWp (head)
• Contains basic Pumps in the Industry
• Methodology:
– Calculate Q and Head
– Intersect Q(x-axis) with Head (y-axis)
– This point is the recommended Pump

PD:
Rotatory

PD:
Reciprocal

Centrifugal High
Speed (RPM = 3500)

Centrifugal Slow
Speed
1750 RPM and
lower

Axial

• Recommend a Pump with the following:
– nWp = 150 ft
– Q = 100 GPM

• Recommend a Pump with the following:
– nWp = 150 ft
– Q = 100 GPM
Either:
Centrifugal, Rotatory or
Reciprocal!
Avoid:
Axial and Low velocity
Centrifugal Pumps!

1b. Specific Velocity Criteria
• A better approach is to use the Specific
Velocity/Diameter Criteria
• Better since it is Dimensionless Number
Comparison
• Calculate Ns (Specific Velocity)
• Calculate Ds (Specific Diameter)
– Find Operation Point
– Find Suggested Pumping System

• N = impeller speed (RPM)
• Q = volumetric flow in gal/min
• H = Pump Head (ft)
• D = Impeller Diameter (in)
Specific Diameter 
 Specific Speed

Specific Velocity Criteria
Exercise #1
• What will be the best Pump for:
– Q = 500 gal/min
– System’s Head = 80 ft
– Speed = 1750 RPM
– D Impeller= 18 in

Exercise #1
• Calculate Specific Speed
– Ns = N(Q^0.5)*(H^-0.75)
– Ns = 1750(500^0.5)*(80^-0.75)
– Ns = 1463
• Calculate Specific Diameter
– Ds = D*(H^0.25)*(Q^-0.5)
– Ds = 18 *(80^0.25)*(500^-0.5)
– Ds = 2.4

Exercise #1

Operation of Pump is NOT
recommended!
Exercise #1

D= 18 in to 12 in
Ds = D*(H^0.25)*(Q^-0.25)
Ds = 12*(80^0.25)*(500^-0.25) = 1.6
Exercise #1

Better Operation!
Radial Flow  Centrifugal Pump
Max Efficiency Expected  60-70%
Exercise #1

Force Pumping system to THIS!
Exercise #1

Force Pumping system to THIS!
1500 < Ns < 7000
0.4 < Ds < 1.2
Exercise #1

2. Pump Supplier’s Family
• Once you know which Type of Pump, find the best
pump
• There are many “recommended” pumps in a set of
“Families”
• Get the best that suits your system!

• Criteria to consider:
– Max/min Impeller Size (Diameter in Inches)
– Velocity (RPM)
– Power (BHP)
– Efficiency (approx 50-90%)
– NSHP Required
– Pump Curve included
– Suction & Discharge Sizes

3 x 5 – 12
3: Discharge Size (Inches)
5: Suction Size (Inches)
12: (Larges Impeller Size Available)

• Once you get a Pump
– Check specifically the Operation of the given
pump
– Find the Operation Point
– Consider Efficiency
– Consider NPSHr
– Consider Head vs. Q ratio (Steep or flat?)

NPSHR

3. Calculate System Head
• You should have this by now
– Try to set the equation in an Spread Sheet
– Probably you will change piping and flows
• This will change the Head constantly
Check out the exercise of the “System’s Head Exercise 1,2,3”

• Recall that the System Head may be
manipulated
– You may increase the Head of the System:
• Slightly Closing Valves  More Friction  More System
Head
– You may ecreasethe Head of the System:
• Slightly Opening Valves  LessFriction  Less System
Head

• Many Valves may operate at different %
– 0%  Closed
– 10%, 20%, 50%, 99%  Partially Open
– 100%  Completely Open
• Partially Closing a Valve will
– increase the friction
– will maintain the same flow rate (no speed change)

System Curse (Valve 100%)

Válvulas abiertas
Válvulas 40% cerradasSystem Curse (Valve 60%)

Válvulas abiertas
Válvulas 40% cerradas

Válvulas abiertas
Head Increases as Valve %
Approaches 0%

Válvulas abiertas
Head Increases as Valve %
Approaches 0%
Same Flow Rate
Different Head!

Válvulas abiertas
Consider changing Impeller’s
Diameter

4. Intersect:
Pump Head Curve + System Head
Volumetric Flow
HeadPump(m)
Pump Curve

4. Intersect:
Volumetric Flow
HeadPump(m)
System Head Curve

4. Intersect:
Volumetric Flow
HeadPump(m)
OPERATION Point

4. Intersect:
Volumetric Flow
HeadPump(m)
OPERATION Point
But we are changing
Flow Rate!

5. Optimize Operation Point
• Once Operation Point is set
– Try to optimize
– Flow Rate
– System Head Requirement
– Efficiency
– Power
– NSPHr f(Q)

Exercise #1
Optimize Efficiency!

Exercise #1
Optimize Efficiency!  More Efficiency; Less Flow Rate!
NEW Design
Old Design

Exercise #2
Decrease Power Requirement! Maintain Flow Rate!
Old Design

Exercise #2
Decrease Power Requirement! Maintain Flow Rate!
Old Design
P = 30 BHP

Exercise #2
OPEN all Valves. Clean Pipes. Change Pipe Diameter!
Old Design
New Design
P = 21 BHP

Exercise #3
An Engineer Proposes the next change
Old Design

Exercise #3
An Engineer Proposes the next change
New Design

Exercise #3
Would you recommend it? EXPLAIN
New Design

Exercise #3
NO  Increase in Pump Power; Installation of New Impeller needed
New Design

Pump Selection Summary
• Know your System
• Know your Pumps or Supplier’s Info
• Check out the best Pump operation
• Calculate the System’s Curve
• Intersect the System’s Curve + Pump Curve
• Optimize changing:
– Impeller size
– Flow Rate
– % Open/Closed Valves

Choosing a Pump
Exercise #1
• There is only one Pump available
• Process requirement is 240 gpm
• The actual impeller is about 6” in diameter
• It may be cutted to 5”.
• Since the company is growing. Sales are
expected to grow. The new flow rate
requirment will be 500 gpm
• FAVOUR ECONOMICS

Choosing a Pump  Exercise #1

Actual Flow Rate

Pump Curve at 6”

Actual Operation Point

Efficiency: 58%
Power: 14 HP
Head= 135 ft

New DesignNew Design
500 gpm

No Impeller Diameters!

Propose buying 8” or 8.5”

Propose buying 8” or 8.5”
Eff1 = 72.5%
Eff2 = 71.5%
P1 = 41 BHP
P2 = 37 BHP
H1 = 240 ft
H2 = 210 ft

Need More Problems?
• Courses
Part 1: Incompressible Flow Rate
You’ll get SOLVED Problems, Quizzes, Slides, and
much more!

NOTE: Common Mistake
Head is
300 or 240??

Conclusions when Choosing a Pump
• Find your System Curve (Spreadsheet)
• Find Pump that satistfy operation
• Find the most likely Point of Operation (O.P)
• Set the System Curve maximize benefits
– Flow rate
– Effiency
– Total Power Requirement
• Check out for the NPSHR

Pump Affinity Laws
• What if there is no data in the Diagram?
a) If we increase Flow Rate to 800 GPM,
what will be the Power?
b) If we want to try an experimental 10”
Impeller, what will be the new Head
and Power Requirements?
c) Find Power Requirement is we
change the motor to a 1750 RPM

Pump Affinity Laws
• Help us relate:
– Q new flow rates when changing N, D, P
– Change of Impeller’s Diameter
– How Power will be affected if angular velocity is
modified
• Find data not included in the Diagram

Pump Affinity Laws
q = volumetric flow rate
D = Impeller’s Diameter
P = Power (Normally in BHP)
nWp = Pump Requirment (energy per unit mass)
N = RPM of Impeller

Pump Affinity Laws
Exercise #1

Pump Affinity Laws
Exercise #1
• From the last Pump Diagram (Curve)
– RPM = 3550
– D impeller = 6 in
• Find:
• System Head (ft)
• Flow Rate (GPM)
• P (BHP)

Pump Affinity Laws
Exercise #1
• From the last Pump Diagram (Curve)
– RPM = 3550
• Find:
• System Head (ft)  125 ft
• Flow Rate (GPM) 330 GPM
• P (BHP) = 17 BHP

Pump Affinity Laws
Exercise #2
• If we change:
– RPM = 3550  1750
– D impeller = 6 in stays the same
• Find:
– nW1/nW2 = (N1/N2)^2
– 125/nW2 = (3550/1750)^2
– nW2 = 30.4 ft
• Flow Rate (GPM)
– Q1/Q2 = (N1/N2)
– 330/Q2= (3550/1750)
– Q2 = 162.7 GPM

Pump Affinity Laws
Exercise #3
• If we change:
– RPM = 3550 stays the same
• Find:
– nW1/nW2 = (D1/D2)^2
– 125/nW2 = (6/12)^2
– nW2 = 500 ft
• Power (BHP)
– P1/P2 = (D1/D2)^3
– 17/Q2= (6/12)^3
– P = 136 BHP

Pump Affinity Laws
Exercise Conclusions
NO Data for 1750 RPM
No Data for 12” Impeller

Pump Affinity Laws
Exercise Conclusions
NO Data for 1750 RPM
No Data for 12” Impeller
This is awesome!
We don’t need another diagram to find this out!

End of Section 4: Pump Selection

Section 5: Pump Arrangements

Pump in Series/Parallel Arrangement
• Pumps may be arrange
either in:
– Series (one after another)
– Parallel (one beside the
other)

Parallel Pumps
• Ideal when Flow varies
• When Adding a Pump:
– Pressure is maintained
– The system’s capacity is increased
• Parallel Quantity! Flow Rate increases
• Pressure will remain the same!

Pump in Series
• Ideal when Pressure increase is needed
• Adding a pump:
– Will increase the pressure
– Flow rate must remain the same
• Series Pressure!
• Pressure is NOT constant
• Volumetric Flow IS constant

Pump in Series/Parallel

Pump Arrangement
Exercises
• For the System Requirement:
– Q = 600 gpm
– nWp = 270 ft
• Suppose:
– 1 phase operation (1 pump)
– 2 pumps in series
– 2 pumps in parallel

Pump Arrangement
Exercise #1
• 1 Single Pump
– Q (pump) = 600 gpm
– nWp (pump) = 270 ft
– Pin = Pin
– Pout = Pout

Pump Arrangement
Exercise #2
• Pump in Series
– Q (pump) = 600 gpm  each!
– nWp (pump) = 270/2 ft  135 ft each!
– Pin1 < Pout1
– Pin1 = Pin2
– Pin2 < Pout2
– Pout1 = Pout 2

Pump Arrangement
Exercise #3
• Pump in Parallel
– Q (pump) = 600 gpm / 2 pumps = 300 gpm each
– nWp (pump) = 270 ft  each!
– Pin1 < Pout1
– Pout1 = Pin2
– Pin2 < Pout2

Pump Arrangement
Exercises Conclusion
• Compare Ex 1,2,3
• Flow Rate:
– 1 Pump  600 gpm
– Series Pump  600 gpm
– Parallel Pump  300 gpm
• Pump Head
– 1 Pump  270 ft
– Series Pump  135 each
– Parallel Pump  270 each
What is that you
want?
Increase P?
Increase Head?
Decrease Flow Rate?

Software Modeling for Pumps
• Basic Modeling
– Aspen Plus
– Aspen HYSYS
– PSYM  http://www.pumpsystemsmatter.org/
• Intensive Modeling
– SolidWorks
– EPANet  http://epanet.de/
– COMSOL Inc

• Aspen Plus + HYSYS

• COMSOL Inc

• SolidWorks

End of Section 5: Pump Arrangements

End of AFD5
• By now you should know:
– Types of Pumps used in the industry:
• Advantages and disadvantages
• Basic Components of a Pump
– Head of a System and the Curve’s System
– How to calculate the Pump’s Performance curve
– The importance of Cavitation and how to avoid it (NSPHR)
– Effects on Pump Systems
• Impeller, Viscosity and Velocity of the fluid and other factors
• Total Work Required, Power, Brake Horse Powers and Efficiency
– How to Choose a Pump
– Pump Affinity Law for Design of Equipment
– Model Pumping Systems in Series + Parallel
– Common Software used to model Pumps

Questions and Problems
• Check out the SOLVED & EXPLAINED problems
and exercises!
– Don’t let this for later…
• All problems and exercises are solved in the
next webpage
– www.ChemicalEngineeringGuy.com
• Courses
– Momentum Transfer Operations

Contact Information!
• Get extra information here!
– Directly on the WebPage:
• www.ChemicalEngineeringGuy.com/courses
– FB page:
• www.facebook.com/Chemical.Engineering.Guy
– My Twitter:
• www.twitter.com/ChemEngGuy
– Contact me by e-mail:
• Contact@ChemicalEngineeringGuy.com

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