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Applied Fluid Dynamics
AFD5 Pumps
Chemical Engineering Guy
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AFD5 Overview
• This is still Incompressible Flow
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Textbook, Reference and Bibliography
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Textbook, Reference and Bibliography
• Chapters:
– 13  Pump Selection & Application
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Applied Fluid Mechanics. Robert
Mott 6th Edition
Textbook, Reference and Bibliography
• Section 2: Fluid Mechanics
– CH8: Transportation of Fluids
• Just the “Pumps” part
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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
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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
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Section 1: Pump Types
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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!
Introduction to Pumps
• Clasification
– Positive Displacement
• Rotatory
• Reciprocal
– Kinetic Pumps
• Centrifugal
• Axial
Introduction to Pumps
• There are many implications and things to
consider when choosing a pump
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Introduction to Pumps
• Fluid type
• Volumetric Flow requirement
• Suction Conditions (P,T, composition)
• Discharge Conditions (P,T, composition)
Introduction to Pumps
• Head Requirement
• Power requirement
• Spacing,
dimensions,
weigth, etc.
• Initial Investment
• Operation Costs
Introduction to Pumps
• Maintainment Cost
• Type of Pump
• Supplier
• Sizing and type of
Conections
Introduction to Pumps
• Operation RPM and Frecuency
• Impeller (material, diameter, etc.)
• Pump Material
• Min and Max Pressure discharge/suction
Introduction to Pumps
• So… what types of pumps do we have?
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Pump Types
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Pump Types
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Rotatory…
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Centrifugal…
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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: Gear
P.D. Rotatory: Screw
P.D. Rotatory: Screw
P.D. Rotatory: Progressive Cavity
P.D. Rotatory: Lobe
P.D. Rotatory: Peristaltic
P.D. Rotatory: Peristaltic
P.D. Recirpocal
• Piston
• Diaphragm
P.D. Recirpocal: Piston
P.D. Recirpocal: Diaphragm
P.D. Recirpocal: Diaphragm
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Kinetic Pumps
• High velocity operations
• Impeller Operated
• Kinetic Energy  Mechanic Energy  Pressure
• Radial flow
– Centrifugal
• Axial
Dynamic: Axial
Dynamic: Axial
Dynamic: Axial
Dynamic: Radial
Dynamic: Radial
Dynamic: Radial
Dynamic: Radial
Dynamic: Radial
Centrifugal Pump Parts
• Super Basic:
– Impeller
– Shaft
– House Casing
– Motor
– Suction
– Discharge
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Centrifugal Pump Parts
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The motor is very important!
Centrifugal Pump Parts
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Centrifugal Pump Parts
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Centrifugal Pump Parts
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Centrifugal Pump Parts
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Centrifugal Pump Parts
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Pump Performance
• We always analyze pump performance
– Efficiency (Work inlet vs. outlet)
– Pressure inlet vs. outlet
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Cavitation
• Damage due to little bubbles
• Recall that bubbles are gas, they are copmressible
• The impeller will experiment different
Pressures/Forces
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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
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Cavitation
• So… for example water:
– If T operation = 80ºC
– Inlet = 1 atm
– Eye = 0.5 atm
– Outlet = 2 atm
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• So… for example water:
– If T operation = 80ºC
– Inlet = 1 atm  100
– Eye = 0.5 atm  75
– Outlet = 2 atm  120
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Cavitation
Water will evaporate
in the Eye!
Cavitation
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Cavitation
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Safe
Cavitation
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Safe
Not Safe
Cavitation
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Cavitation
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Cavitation Explained
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Cavitation: Impeller Damage
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Cavitation: Impeller Damage
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Cavitation: Impeller Damage
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Cavitation: Impeller Damage
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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
• 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
Therefore, it depends
on the substance!
NSPHA
• NSPHr  Net Specific Pressure at Head
Available
• The actual pressure present in the suction
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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?
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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
– From Antoine Eqn. Pv(37.8°C) = 21.5574 kPa
• B) Is this Pump enough? Or you will expect Cavitation?
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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
– From Antoine Eqn. Pv(37.8°C) = 21.5574 kPa
• B) Is this Pump enough? Or you will expect Cavitation?
– Use Equation: NPSHA > 1.10 NPSHR
• NPSHA = (Psuc – Pvap) / (rho*g) = (86-21.5)(10^3)/(876*9.8) = 7.5 m
•
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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
– From Antoine Eqn. Pv(37.8°C) = 21.5574 kPa
• B) Is this Pump enough? Or you will expect Cavitation?
– Use Equation: NPSHA > 1.10 NPSHR
• NPSHA = (Psuc – Pvap) / (rho*g) = (86-21.5)(10^3)/(876*9.8) = 7.5 m
– Since NPSHA > 1.10 NPSHR
• 7.5 m > 1.10 (3.05)
• 7.5 m > 3.36 m  Expect NO cavitation
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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
– From Antoine Eqn. Pv(37.8°C) = 21.5574 kPa
• B) Is this Pump enough? Or you will expect Cavitation?
– Use Equation: NPSHA > 1.10 NPSHR
• NPSHA = (Psuc – Pvap) / (rho*g) = (86-6.3)(10^3)/(1000*9.8) = 8.13 m
– Since NPSHA > 1.10 NPSHR
• 8.1 m > 1.10 (3.05)
• 8.1 m > 3.36 m  Expect NO cavitation
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If it was water? Will it cavitate? Vp = 6.3 kPa
D = 1000 kg/m3
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
– From Antoine Eqn. Pv(37.8°C) = 21.5574 kPa
• B) Is this Pump enough? Or you will expect Cavitation?
– Use Equation: NPSHA > 1.10 NPSHR
• NPSHA = (Psuc – Pvap) / (rho*g) = (86-72)(10^3)/(750*9.8) = 1.9 m
– Since NPSHA > 1.10 NPSHR
• 1.9 m > 1.10 (3.05)
• 1.9 m > 3.36 m  Expect CAVITATION!
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If it was a very volatile substance with Vp = 72 kPa
D = 750 kg/m3
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
– From Antoine Eqn. Pv(37.8°C) = 21.5574 kPa
• B) Is this Pump enough? Or you will expect Cavitation?
– Use Equation: NPSHA > 1.10 NPSHR
• NPSHA = (Psuc – Pvap) / (rho*g) = (86-VP)(10^3)/(876*9.8) = 3.4 m
– Since NPSHA > 1.10 NPSHR
• 3.4 m > 1.10 (3.05)
• 3.4 m > 3.36 m  Expect CAVITATION!
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What is the min. Vapor Pressure to operate Benzene?
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MOMENTUM TRANSFER OPERATIONS
You’ll get SOLVED Problems, Quizzes, Slides, and
much more!
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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
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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
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• Courses
MOMENTUM TRANSFER OPERATIONS
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End of Section 1: Pump Types
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Section 2: System Curve
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System Head
• ASSUME Friction loss change vs. velocity
• System head will be used in Pump Selection
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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
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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
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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
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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
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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!
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System Curve
• This was done for a specific flow rate
• What will happen if we change that flow rate
in the SAME system
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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
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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…
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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
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System Curve Exercise #1
• Create a System curve for the next System
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System Curve Exercise #1
• Create a System curve for the next System
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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
System Curve Exercise #1
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
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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)
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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)
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System Curve Exercise #1 Conclusions
• 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
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System Curve Exercise #2
• 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
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System Curve Exercise #2
Volumetric Flow Rate (L/s)
nWp(m)
For a System
- Fixed Pipe Diameter
- Fixed Fittings
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4” Piping
System Curve Exercise #24” Piping
8” Piping
Volumetric Flow Rate (L/s)
nWp(m)
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System Curve Exercise #2
4”, 8”, 10” and 12” pipings
Volumetric Flow Rate (L/s)
nWp(m)
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System Curve Exercise #2
4”, 8”, 10” and 12” pipings
Volumetric Flow Rate (L/s)
nWp(m)
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Shifted!
System Curve Exercise #2
4”, 8”, 10” and 12” pipings
Volumetric Flow Rate (L/s)
nWp(m)
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System Curve Exercise #2 Conclusions
• The curve will flatten to the rigth
– When increasing the pipe’s Diameter
– Speed is reduced (V2)
– Friction is reduced, hence, Less Head
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System Curve Exercise #3
• Create a System curve for the next System
• NOTE  Valve is 50% closed
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System Curve Exercise #3
• Compare 50% and 100%
• Compare the same Flow vs. Head
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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%
System Curve Exercise #3
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• Valves are pretty important in controlling
– Friction loss
System Curve Exercise #3
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• Valves are pretty important in controlling
– Friction loss
System Curve Exercise #3
• Valves are pretty important in controlling
– Friction loss
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System Curve Exercise #3
• Valves are pretty important in controlling
– Friction loss
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System Curve Exercise #3 Conclusions
• 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!
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much more!
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End of Section 2: System Curve
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Section 3: Pump Curve/Head
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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
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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
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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
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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
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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
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• 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 Exercise
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• 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
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We are interested in the Suction and
Discharge points!
Not the conventional A and B
Need More Problems?
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• Courses
Applied Fluid Mechanics
Part 1: Incompressible Flow
You’ll get SOLVED problems, Quizzes, Slides, and
much more!
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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?
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Pump Performance Curve
• This is common
– Decrease  Shut down, decrease in sells, excess
of inventory, etc.
– Increase  increase of inventory, increase of sells,
revamp
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Pump Performance Curve
• 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
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Pump Performance Curve
• 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?
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Pump Curve
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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
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• 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
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• The experimental data is given next
Pump Curve Exercise
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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:
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• Courses
Applied Fluid Dynamics
Part 1: Incompressible Flow
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Pump Performance Curve Analysis
• Why does the Curve is concave down?
• As flow rate increases, decreases the head of
the pump!?
• Explain yourself!
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Pump Performance Curve Analysis
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Pump Performance Curve
• Actually, suppliers may group pumps as
“families”
– Steep (high friction)
– Average/Standard
– Flat (low friction)
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Pump Performance Curve
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Steep Curve  Large Head, Small Flows
Pump Performance Curve
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Flat Curve  Small Head, LargeFlows
Pump Performance Curve
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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.)
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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
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Pump Curve Diagram  Head vs. Q
8”
Volumetric Flow
HeadPump(m)
Pump Curve Diagram  Head vs. Q
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
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• 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]
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S.
D.
Impeller
Pump Curve Diagram  Impeller’s Diameter
Pump Curve Diagram  Impeller’s Diameter
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8”
Volumetric Flow
HeadPump(m)
Pump Curve Diagram  Impeller’s Diameter
8”
Volumetric Flow
HeadPump(m)
3”
Pump Curve Diagram  Impeller’s Diameter
Pump Curve Diagram  Impeller’s Diameter
6“ 1/2
5”
4“ 1/2
3”
8”
Impeller Size (Diameter in Inches)
Volumetric Flow
HeadPump(m)
Pump Curve Diagram  Impeller’s Diameter
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  Impeller’s Diameter
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!
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Pump Curve Diagram  Efficiency
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Volumetric Flow
HeadPump(m)
Pump Curve Diagram  Efficiency
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Pump Curve Diagram  Efficiency
6“ 1/2
5”
4“ 1/2
3”
8”
Volumetric Flow
HeadPump(m)
Pump Curve Diagram  Efficiency
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Pump Curve Diagram  Efficiency
Pump Curve Diagram  Efficiency
6“ 1/2
5”
4“ 1/2
3”
8”
Pump Curve Diagram  Efficiency
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Pump Efficiency
Exercise #1
• Choose the best efficiency area for this Pump
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Pump Efficiency
Exercise #1
• Choose the best efficiency area for this Pump
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73
%
Pump Efficiency
Exercise #1
• Choose the best efficiency area for this Pump
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73
%
Pump Efficiency
Exercise #2
• If GPM = 500… What is the Head when η = 65%
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Pump Efficiency
Exercise #2
• If GPM = 500… What is the Head when η = 65%
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Pump Efficiency
Exercise #2
• If GPM = 500… What is the Head when η = 65%
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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!
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Pump Curve Diagram  Power
5”
4“ 1/2
3”
Pump Curve Diagram  Power
5”
4“ 1/2
3”
Pump Curve Diagram  Power
5”
4“ 1/2
3”
As Flow increases,
Power Increses
Pump Curve Diagram  Power
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
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Pump Power
Exercise #1
• How much Power is Required in HP for:
– A pump working 600 GPM
– The Head needed is 2400 ft
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Pump Power
Exercise #1
• How much Power is Required in HP for:
– A pump working 600 GPM
– The Head needed is 2400 ft
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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?
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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?
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Pump Power
Exercise #2
• If Power Requirement of a Pump = 600 BHP
• The Pump has an Impeller Diameter = 10 ½
– What is the System Head?
– What is the Effiency in this operation?
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Pump Power
Exercise #2
• If Power Requirement of a Pump = 600 BHP
• The Pump has an Impeller Diameter = 10 ½
– What is the System Head?
– What is the Effiency in this operation?
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Pump Power
Exercise #2
• If Power Requirement of a Pump = 600 BHP
• The Pump has an Impeller Diameter = 10 ½
– What is the System Head?
– What is the Effiency in this operation?
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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 
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NOTE: NSPH Required is f(Q) only!
Pump Curve Diagram  NPSHR
8”
Volumetric Flow
HeadPump(m)
Pump Curve Diagram  NPSHR
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Pump Curve Diagram  NPSHR
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Pump Curve Diagram  NPSHR
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NOTE: NSPH Required is f(Q) only!
Pump Curve Diagram  NPSHR
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Pump Curve Diagram  NPSHR
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Pump Curve Diagram  NPSHR
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Pump Curve Diagram  NPSHR
Exercise #1
• Calculate the NPSHR when:
– Impeller size is 7”
– Head = 200
– GPM = 160
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Pump Curve Diagram  NPSHR
Exercise #1
• Calculate the NPSHR when:
– Impeller size is 7”
– Head = 200
– GPM = 160
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You just need GPM
Pump Curve Diagram  NPSHR
Exercise #1
• Calculate the NPSHR when:
– Impeller size is 7”
– Head = 200
– GPM = 160
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You just need GPM
Pump Curve Diagram  NPSHR
Exercise #1
• Calculate the NPSHR when:
– Impeller size is 7”
– Head = 200
– GPM = 160
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You just need GPM
NPSHR = 5.5 ft
Pump Curve Diagram  NPSHR
Exercise #2
• What is the minimum Flow Rate given a
NPSHa of 3.3 m?
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Pump Curve Diagram  NPSHR
Exercise #2
• What is the minimum Flow Rate given a
NPSHa of 3.3 m?
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NPSHa > 1.1*NPSHr  3
Pump Curve Diagram  NPSHR
Exercise #2
• What is the minimum Flow Rate given a
NPSHa of 5 m?
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NPSHa > 1.1*NPSHr  4.54
Pump Curve Diagram  NPSHR
Exercise #2
• What is the minimum Flow Rate given a
NPSHa of 5 m?
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NPSHa > 1.1*NPSHr  4.54
Pump Curve Diagram  NPSHR
Exercise #2
• What is the minimum Flow Rate given a
NPSHa of 5 m?
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NPSHa > 1.1*NPSHr  4.54
About 600 GPM
Pump Performance Curve
Full Diagrams!
• We’ve seen all the theoretical concepts
• We can read the full diagrams of a Pump Curve!
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Pump Performance Curve
Full Diagrams!
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Need More Problems?
Check out the COURSE
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• Courses
Applied Fluid Mechanics
Part 1: Incompressible Flow
You’ll get SOLVED problems, Quizzes, Slides, and
much more!
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End of Section 3: Pump Curve/Head
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Section 4: Pump Selection
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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!
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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
1a. General Pump Selection Diagram
1a. General Pump Selection Diagram
PD:
Rotatory
1a. General Pump Selection Diagram
PD:
Reciprocal
1a. General Pump Selection Diagram
Centrifugal High
Speed (RPM = 3500)
1a. General Pump Selection Diagram
Centrifugal Slow
Speed
1750 RPM and
lower
1a. General Pump Selection Diagram
Axial
1a. General Pump Selection Diagram
1a. General Pump Selection Diagram
• Recommend a Pump with the following:
– nWp = 150 ft
– Q = 100 GPM
1a. General Pump Selection Diagram
• Recommend a Pump with the following:
– nWp = 150 ft
– Q = 100 GPM
1a. General Pump Selection Diagram
• 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
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1b. Specific Velocity Criteria
• N = impeller speed (RPM)
• Q = volumetric flow in gal/min
• H = Pump Head (ft)
• D = Impeller Diameter (in)
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Specific Diameter 
 Specific Speed
1b. Specific Velocity Criteria
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1b. Specific Velocity Criteria
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1b. Specific Velocity Criteria
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1b. Specific Velocity Criteria
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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
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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
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1b. Specific Velocity Criteria
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
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1b. Specific Velocity Criteria
Exercise #1
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1b. Specific Velocity Criteria
Exercise #1
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Operation of Pump is NOT
recommended!
1b. Specific Velocity Criteria
Exercise #1
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D= 18 in to 12 in
Ds = D*(H^0.25)*(Q^-0.25)
Ds = 12*(80^0.25)*(500^-0.25) = 1.6
1b. Specific Velocity Criteria
Exercise #1
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D= 18 in to 12 in
Ds = D*(H^0.25)*(Q^-0.25)
Ds = 12*(80^0.25)*(500^-0.25) = 1.6
1b. Specific Velocity Criteria
Exercise #1
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Better Operation!
Radial Flow  Centrifugal Pump
Max Efficiency Expected  60-70%
1b. Specific Velocity Criteria
Exercise #1
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Force Pumping system to THIS!
1b. Specific Velocity Criteria
Exercise #1
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Force Pumping system to THIS!
1500 < Ns < 7000
0.4 < Ds < 1.2
1b. Specific Velocity Criteria
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!
2. Pump Supplier’s Family
• 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
2. Pump Supplier’s Family
3 x 5 – 12
3: Discharge Size (Inches)
5: Suction Size (Inches)
12: (Larges Impeller Size Available)
2. Pump Supplier’s Family
2. Pump Supplier’s Family
• 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?)
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2. Pump Supplier’s Family
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
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Check out the exercise of the “System’s Head Exercise 1,2,3”
3. Calculate System Head
• 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
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3. Calculate System Head
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3. Calculate 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)
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3. Calculate System Head
System Curse (Valve 100%)
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3. Calculate System Head
Válvulas abiertas
Válvulas 40% cerradasSystem Curse (Valve 60%)
System Curse (Valve 100%)
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3. Calculate System Head
Válvulas abiertas
Válvulas 40% cerradas
Válvulas 80% cerradas
System Curse (Valve 60%)
System Curse (Valve 20%)
System Curse (Valve 100%)
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3. Calculate System Head
Válvulas abiertas
Válvulas 40% cerradas
Válvulas 80% cerradas
System Curse (Valve 60%)
System Curse (Valve 20%)
System Curse (Valve 100%)
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Head Increases as Valve %
Approaches 0%
3. Calculate System Head
Válvulas abiertas
Válvulas 40% cerradas
Válvulas 80% cerradas
System Curse (Valve 60%)
System Curse (Valve 20%)
System Curse (Valve 100%)
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Head Increases as Valve %
Approaches 0%
Same Flow Rate
Different Head!
3. Calculate System Head
Válvulas abiertas
Válvulas 40% cerradas
Válvulas 80% cerradas
System Curse (Valve 60%)
System Curse (Valve 20%)
System Curse (Valve 100%)
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Head Increases as Valve %
Approaches 0%
Same Flow Rate
Different Head!
3. Calculate System Head
Válvulas abiertas
Válvulas 40% cerradas
Válvulas 80% cerradas
System Curse (Valve 60%)
System Curse (Valve 20%)
System Curse (Valve 100%)
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Consider changing Impeller’s
Diameter
4. Intersect:
Pump Head Curve + System Head
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Volumetric Flow
HeadPump(m)
Pump Curve
4. Intersect:
Pump Head Curve + System Head
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Volumetric Flow
HeadPump(m)
System Head Curve
4. Intersect:
Pump Head Curve + System Head
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Volumetric Flow
HeadPump(m)
OPERATION Point
System Curse (Valve 100%)
4. Intersect:
Pump Head Curve + System Head
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Volumetric Flow
HeadPump(m)
OPERATION Point
System Curse (Valve 60%)
System Curse (Valve 100%)
4. Intersect:
Pump Head Curve + System Head
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Volumetric Flow
HeadPump(m)
OPERATION Point
System Curse (Valve 60%)
System Curse (Valve 20%)
System Curse (Valve 100%)
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)
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5. Optimize Operation Point
Exercise #1
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Optimize Efficiency!
5. Optimize Operation Point
Exercise #1
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Optimize Efficiency!
5. Optimize Operation Point
Exercise #1
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Optimize Efficiency!
5. Optimize Operation Point
Exercise #1
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Optimize Efficiency!  More Efficiency; Less Flow Rate!
NEW Design
Old Design
5. Optimize Operation Point
Exercise #2
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Decrease Power Requirement! Maintain Flow Rate!
Old Design
5. Optimize Operation Point
Exercise #2
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Decrease Power Requirement! Maintain Flow Rate!
Old Design
P = 30 BHP
5. Optimize Operation Point
Exercise #2
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Decrease Power Requirement! Maintain Flow Rate!
Old Design
5. Optimize Operation Point
Exercise #2
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OPEN all Valves. Clean Pipes. Change Pipe Diameter!
Old Design
New Design
P = 21 BHP
5. Optimize Operation Point
Exercise #3
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An Engineer Proposes the next change
Old Design
5. Optimize Operation Point
Exercise #3
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An Engineer Proposes the next change
Old Design
5. Optimize Operation Point
Exercise #3
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An Engineer Proposes the next change
New Design
5. Optimize Operation Point
Exercise #3
www.ChemicalEngineeringGuy.com
Would you recommend it? EXPLAIN
New Design
5. Optimize Operation Point
Exercise #3
www.ChemicalEngineeringGuy.com
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
www.ChemicalEngineeringGuy.com
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
www.ChemicalEngineeringGuy.com
Choosing a Pump  Exercise #1
Actual Flow Rate
Choosing a Pump  Exercise #1
Pump Curve at 6”
Choosing a Pump  Exercise #1
Actual Operation Point
Choosing a Pump  Exercise #1
Efficiency: 58%
Power: 14 HP
Head= 135 ft
Choosing a Pump  Exercise #1
New DesignNew Design
500 gpm
Choosing a Pump  Exercise #1
No Impeller Diameters!
Choosing a Pump  Exercise #1
Propose buying 8” or 8.5”
Choosing a Pump  Exercise #1
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?
Check out the COURSE
www.ChemicalEngineeringGuy.com
• Courses
Applied Fluid Dynamics
Part 1: Incompressible Flow Rate
You’ll get SOLVED Problems, Quizzes, Slides, and
much more!
www.ChemicalEngineeringGuy.com
NOTE: Common Mistake
Head is
300 or 240??
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
www.ChemicalEngineeringGuy.com
Pump Affinity Laws
• What if there is no data in the Diagram?
www.ChemicalEngineeringGuy.com
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
www.ChemicalEngineeringGuy.com
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
www.ChemicalEngineeringGuy.com
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)
www.ChemicalEngineeringGuy.com
Pump Affinity Laws
Exercise #1
• From the last Pump Diagram (Curve)
– RPM = 3550
– D impeller = 6 in
• Find:
• System Head (ft)  125 ft
• Flow Rate (GPM) 330 GPM
• P (BHP) = 17 BHP
www.ChemicalEngineeringGuy.com
Pump Affinity Laws
Exercise #2
• If we change:
– RPM = 3550  1750
– D impeller = 6 in stays the same
• Find:
• System Head (ft)
– 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
www.ChemicalEngineeringGuy.com
Pump Affinity Laws
Exercise #3
• If we change:
– RPM = 3550 stays the same
– D impeller = 12 in
• Find:
• System Head (ft)
– 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
www.ChemicalEngineeringGuy.com
Pump Affinity Laws
Exercise Conclusions
www.ChemicalEngineeringGuy.com
NO Data for 1750 RPM
No Data for 12” Impeller
Pump Affinity Laws
Exercise Conclusions
www.ChemicalEngineeringGuy.com
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
www.ChemicalEngineeringGuy.com
Section 5: Pump Arrangements
www.ChemicalEngineeringGuy.com
Pump in Series/Parallel Arrangement
• Pumps may be arrange
either in:
– Series (one after another)
– Parallel (one beside the
other)
www.ChemicalEngineeringGuy.com
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!
Parallel Pumps
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
Pump in Series/Parallel
www.ChemicalEngineeringGuy.com
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
www.ChemicalEngineeringGuy.com
Pump Arrangement
Exercise #1
• 1 Single Pump
– Q (pump) = 600 gpm
– nWp (pump) = 270 ft
– Pin = Pin
– Pout = Pout
www.ChemicalEngineeringGuy.com
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
www.ChemicalEngineeringGuy.com
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
www.ChemicalEngineeringGuy.com
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
www.ChemicalEngineeringGuy.com
What is that you
want?
Increase P?
Increase Head?
Decrease Flow Rate?
Need More Problems?
Check out the COURSE
www.ChemicalEngineeringGuy.com
• Courses
MOMENTUM TRANSFER OPERATIONS
You’ll get SOLVED Problems, Quizzes, Slides, and
much more!
www.ChemicalEngineeringGuy.com
Software Modeling for Pumps
• Basic Modeling
– Aspen Plus
– Aspen HYSYS
– PSYM  http://www.pumpsystemsmatter.org/
• Intensive Modeling
– SolidWorks
– EPANet  http://epanet.de/
– COMSOL Inc
www.ChemicalEngineeringGuy.com
Software Modeling for Pumps
• Aspen Plus + HYSYS
www.ChemicalEngineeringGuy.com
Software Modeling for Pumps
• Aspen Plus + HYSYS
www.ChemicalEngineeringGuy.com
Software Modeling for Pumps
• Aspen Plus + HYSYS
www.ChemicalEngineeringGuy.com
Software Modeling for Pumps
• Aspen Plus + HYSYS
www.ChemicalEngineeringGuy.com
Software Modeling for Pumps
• COMSOL Inc
www.ChemicalEngineeringGuy.com
Software Modeling for Pumps
• COMSOL Inc
www.ChemicalEngineeringGuy.com
Software Modeling for Pumps
• COMSOL Inc
www.ChemicalEngineeringGuy.com
Software Modeling for Pumps
• SolidWorks
www.ChemicalEngineeringGuy.com
Software Modeling for Pumps
• SolidWorks
www.ChemicalEngineeringGuy.com
Software Modeling for Pumps
• SolidWorks
www.ChemicalEngineeringGuy.com
Software Modeling for Pumps
• SolidWorks
www.ChemicalEngineeringGuy.com
End of Section 5: Pump Arrangements
www.ChemicalEngineeringGuy.com
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
www.ChemicalEngineeringGuy.com
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
www.ChemicalEngineeringGuy.com
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
www.ChemicalEngineeringGuy.com
Textbook, Reference and Bibliography
www.ChemicalEngineeringGuy.com

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AFD5 Pumps

  • 1. Applied Fluid Dynamics AFD5 Pumps Chemical Engineering Guy www.ChemicalEngineeringGuy.com
  • 2. AFD5 Overview • This is still Incompressible Flow www.ChemicalEngineeringGuy.com
  • 3. Textbook, Reference and Bibliography www.ChemicalEngineeringGuy.com
  • 4. Textbook, Reference and Bibliography • Chapters: – 13  Pump Selection & Application www.ChemicalEngineeringGuy.com Applied Fluid Mechanics. Robert Mott 6th Edition
  • 5. Textbook, Reference and Bibliography • Section 2: Fluid Mechanics – CH8: Transportation of Fluids • Just the “Pumps” part www.ChemicalEngineeringGuy.com Unit Operation of Chemical Engineering. McCabe 7th Edition
  • 6. 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 www.ChemicalEngineeringGuy.com
  • 7. 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 www.ChemicalEngineeringGuy.com
  • 10. Section 1: Pump Types www.ChemicalEngineeringGuy.com
  • 11. 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!
  • 12. Introduction to Pumps • Clasification – Positive Displacement • Rotatory • Reciprocal – Kinetic Pumps • Centrifugal • Axial
  • 13. Introduction to Pumps • There are many implications and things to consider when choosing a pump www.ChemicalEngineeringGuy.com
  • 14. Introduction to Pumps • Fluid type • Volumetric Flow requirement • Suction Conditions (P,T, composition) • Discharge Conditions (P,T, composition)
  • 15. Introduction to Pumps • Head Requirement • Power requirement • Spacing, dimensions, weigth, etc. • Initial Investment • Operation Costs
  • 16. Introduction to Pumps • Maintainment Cost • Type of Pump • Supplier • Sizing and type of Conections
  • 17. Introduction to Pumps • Operation RPM and Frecuency • Impeller (material, diameter, etc.) • Pump Material • Min and Max Pressure discharge/suction
  • 18. Introduction to Pumps • So… what types of pumps do we have? www.ChemicalEngineeringGuy.com
  • 24. 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
  • 25. P.D. Rotatory • Gear • Screw • Progresive Cavity • Lobe • Peristaltic
  • 38. Kinetic Pumps • High velocity operations • Impeller Operated • Kinetic Energy  Mechanic Energy  Pressure • Radial flow – Centrifugal • Axial
  • 47. Centrifugal Pump Parts • Super Basic: – Impeller – Shaft – House Casing – Motor – Suction – Discharge www.ChemicalEngineeringGuy.com
  • 54. Pump Performance • We always analyze pump performance – Efficiency (Work inlet vs. outlet) – Pressure inlet vs. outlet www.ChemicalEngineeringGuy.com
  • 55. Cavitation • Damage due to little bubbles • Recall that bubbles are gas, they are copmressible • The impeller will experiment different Pressures/Forces www.ChemicalEngineeringGuy.com
  • 56. 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 www.ChemicalEngineeringGuy.com
  • 57. Cavitation • So… for example water: – If T operation = 80ºC – Inlet = 1 atm – Eye = 0.5 atm – Outlet = 2 atm www.ChemicalEngineeringGuy.com
  • 58. • So… for example water: – If T operation = 80ºC – Inlet = 1 atm  100 – Eye = 0.5 atm  75 – Outlet = 2 atm  120 www.ChemicalEngineeringGuy.com Cavitation Water will evaporate in the Eye!
  • 70. NSPHR • We need a standard or something to know what is the minimum requirement to avoid cavitation! • Objective  Avoid Cavitation
  • 71. 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
  • 72. 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 Therefore, it depends on the substance!
  • 73. NSPHA • NSPHr  Net Specific Pressure at Head Available • The actual pressure present in the suction www.ChemicalEngineeringGuy.com
  • 74. 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
  • 75. 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)
  • 76. 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? www.ChemicalEngineeringGuy.com
  • 77. 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 – From Antoine Eqn. Pv(37.8°C) = 21.5574 kPa • B) Is this Pump enough? Or you will expect Cavitation? www.ChemicalEngineeringGuy.com
  • 78. 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 – From Antoine Eqn. Pv(37.8°C) = 21.5574 kPa • B) Is this Pump enough? Or you will expect Cavitation? – Use Equation: NPSHA > 1.10 NPSHR • NPSHA = (Psuc – Pvap) / (rho*g) = (86-21.5)(10^3)/(876*9.8) = 7.5 m • www.ChemicalEngineeringGuy.com
  • 79. 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 – From Antoine Eqn. Pv(37.8°C) = 21.5574 kPa • B) Is this Pump enough? Or you will expect Cavitation? – Use Equation: NPSHA > 1.10 NPSHR • NPSHA = (Psuc – Pvap) / (rho*g) = (86-21.5)(10^3)/(876*9.8) = 7.5 m – Since NPSHA > 1.10 NPSHR • 7.5 m > 1.10 (3.05) • 7.5 m > 3.36 m  Expect NO cavitation www.ChemicalEngineeringGuy.com
  • 80. 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 – From Antoine Eqn. Pv(37.8°C) = 21.5574 kPa • B) Is this Pump enough? Or you will expect Cavitation? – Use Equation: NPSHA > 1.10 NPSHR • NPSHA = (Psuc – Pvap) / (rho*g) = (86-6.3)(10^3)/(1000*9.8) = 8.13 m – Since NPSHA > 1.10 NPSHR • 8.1 m > 1.10 (3.05) • 8.1 m > 3.36 m  Expect NO cavitation www.ChemicalEngineeringGuy.com If it was water? Will it cavitate? Vp = 6.3 kPa D = 1000 kg/m3
  • 81. 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 – From Antoine Eqn. Pv(37.8°C) = 21.5574 kPa • B) Is this Pump enough? Or you will expect Cavitation? – Use Equation: NPSHA > 1.10 NPSHR • NPSHA = (Psuc – Pvap) / (rho*g) = (86-72)(10^3)/(750*9.8) = 1.9 m – Since NPSHA > 1.10 NPSHR • 1.9 m > 1.10 (3.05) • 1.9 m > 3.36 m  Expect CAVITATION! www.ChemicalEngineeringGuy.com If it was a very volatile substance with Vp = 72 kPa D = 750 kg/m3
  • 82. 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 – From Antoine Eqn. Pv(37.8°C) = 21.5574 kPa • B) Is this Pump enough? Or you will expect Cavitation? – Use Equation: NPSHA > 1.10 NPSHR • NPSHA = (Psuc – Pvap) / (rho*g) = (86-VP)(10^3)/(876*9.8) = 3.4 m – Since NPSHA > 1.10 NPSHR • 3.4 m > 1.10 (3.05) • 3.4 m > 3.36 m  Expect CAVITATION! www.ChemicalEngineeringGuy.com What is the min. Vapor Pressure to operate Benzene?
  • 83. Need More Problems? Check out the COURSE www.ChemicalEngineeringGuy.com • Courses MOMENTUM TRANSFER OPERATIONS You’ll get SOLVED Problems, Quizzes, Slides, and much more! www.ChemicalEngineeringGuy.com
  • 84. 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)
  • 85. 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)
  • 86. Pump Power • Power typical Units – Watt and kilo-Watt – Horse Power (HP)  Brake Horse Power BHP – Foot Pounds per minute*
  • 87. 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 www.ChemicalEngineeringGuy.com
  • 88. 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 www.ChemicalEngineeringGuy.com
  • 89. Need More Problems? Check out the COURSE www.ChemicalEngineeringGuy.com • Courses MOMENTUM TRANSFER OPERATIONS You’ll get SOLVED Problems, Quizzes, Slides, and much more! www.ChemicalEngineeringGuy.com
  • 90. End of Section 1: Pump Types www.ChemicalEngineeringGuy.com
  • 91. Section 2: System Curve www.ChemicalEngineeringGuy.com
  • 92. System Head • ASSUME Friction loss change vs. velocity • System head will be used in Pump Selection www.ChemicalEngineeringGuy.com
  • 93. 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 www.ChemicalEngineeringGuy.com
  • 94. 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 www.ChemicalEngineeringGuy.com
  • 95. 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 www.ChemicalEngineeringGuy.com
  • 96. 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 www.ChemicalEngineeringGuy.com 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
  • 97. 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! www.ChemicalEngineeringGuy.com
  • 98. System Curve • This was done for a specific flow rate • What will happen if we change that flow rate in the SAME system www.ChemicalEngineeringGuy.com
  • 99. 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 www.ChemicalEngineeringGuy.com
  • 100. 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… www.ChemicalEngineeringGuy.com
  • 101. 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 www.ChemicalEngineeringGuy.com
  • 102. System Curve Exercise #1 • Create a System curve for the next System www.ChemicalEngineeringGuy.com
  • 103. System Curve Exercise #1 • Create a System curve for the next System www.ChemicalEngineeringGuy.com 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
  • 104. System Curve Exercise #1 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 www.ChemicalEngineeringGuy.com 0 10 20 30 40 50 60 70 0 50 100 150 200 250 300 350 Head(m) Volumetric Flow Rate GPM Head (m)
  • 105. 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) www.ChemicalEngineeringGuy.com 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)
  • 106. 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) www.ChemicalEngineeringGuy.com
  • 107. System Curve Exercise #1 Conclusions • 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 www.ChemicalEngineeringGuy.com
  • 108. Need More Problems? Check out the COURSE www.ChemicalEngineeringGuy.com • Courses Applied Fluid Mechanics Part 1: Incompressible Flow You’ll get SOLVED problems, Quizzes, Slides, and much more! www.ChemicalEngineeringGuy.com
  • 109. System Curve Exercise #2 • 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 www.ChemicalEngineeringGuy.com
  • 110. System Curve Exercise #2 Volumetric Flow Rate (L/s) nWp(m) For a System - Fixed Pipe Diameter - Fixed Fittings www.ChemicalEngineeringGuy.com 4” Piping
  • 111. System Curve Exercise #24” Piping 8” Piping Volumetric Flow Rate (L/s) nWp(m) www.ChemicalEngineeringGuy.com
  • 112. System Curve Exercise #2 4”, 8”, 10” and 12” pipings Volumetric Flow Rate (L/s) nWp(m) www.ChemicalEngineeringGuy.com
  • 113. System Curve Exercise #2 4”, 8”, 10” and 12” pipings Volumetric Flow Rate (L/s) nWp(m) www.ChemicalEngineeringGuy.com Shifted!
  • 114. System Curve Exercise #2 4”, 8”, 10” and 12” pipings Volumetric Flow Rate (L/s) nWp(m) www.ChemicalEngineeringGuy.com
  • 115. System Curve Exercise #2 Conclusions • The curve will flatten to the rigth – When increasing the pipe’s Diameter – Speed is reduced (V2) – Friction is reduced, hence, Less Head www.ChemicalEngineeringGuy.com
  • 116. Need More Problems? Check out the COURSE www.ChemicalEngineeringGuy.com • Courses Applied Fluid Mechanics Part 1: Incompressible Flow You’ll get SOLVED problems, Quizzes, Slides, and much more! www.ChemicalEngineeringGuy.com
  • 117. System Curve Exercise #3 • Create a System curve for the next System • NOTE  Valve is 50% closed www.ChemicalEngineeringGuy.com
  • 118. System Curve Exercise #3 • Compare 50% and 100% • Compare the same Flow vs. Head www.ChemicalEngineeringGuy.com 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%
  • 119. System Curve Exercise #3 www.ChemicalEngineeringGuy.com • Valves are pretty important in controlling – Friction loss
  • 120. System Curve Exercise #3 www.ChemicalEngineeringGuy.com • Valves are pretty important in controlling – Friction loss
  • 121. System Curve Exercise #3 • Valves are pretty important in controlling – Friction loss www.ChemicalEngineeringGuy.com
  • 122. System Curve Exercise #3 • Valves are pretty important in controlling – Friction loss www.ChemicalEngineeringGuy.com
  • 123. System Curve Exercise #3 Conclusions • 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! www.ChemicalEngineeringGuy.com
  • 124. Need More Problems? Check out the COURSE www.ChemicalEngineeringGuy.com • Courses Applied Fluid Mechanics Part 1: Incompressible Flow You’ll get SOLVED problems, Quizzes, Slides, and much more! www.ChemicalEngineeringGuy.com
  • 125. End of Section 2: System Curve www.ChemicalEngineeringGuy.com
  • 126. Section 3: Pump Curve/Head www.ChemicalEngineeringGuy.com
  • 127. 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 www.ChemicalEngineeringGuy.com
  • 128. Pump Head • It will be interesting to analyse, the total “load” of work needed for a pump • A: Suction • B: Discharge PumpA B
  • 130. 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 www.ChemicalEngineeringGuy.com
  • 131. 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 www.ChemicalEngineeringGuy.com V^2 may be so small compared to change in P
  • 132. 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
  • 133. Pump Head Exercise www.ChemicalEngineeringGuy.com • 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”
  • 134. Pump Head Exercise www.ChemicalEngineeringGuy.com • 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”
  • 135. 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 www.ChemicalEngineeringGuy.com We are interested in the Suction and Discharge points! Not the conventional A and B
  • 136. Need More Problems? Check out the COURSE www.ChemicalEngineeringGuy.com • Courses Applied Fluid Mechanics Part 1: Incompressible Flow You’ll get SOLVED problems, Quizzes, Slides, and much more! www.ChemicalEngineeringGuy.com
  • 137. 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? www.ChemicalEngineeringGuy.com
  • 138. Pump Performance Curve • This is common – Decrease  Shut down, decrease in sells, excess of inventory, etc. – Increase  increase of inventory, increase of sells, revamp www.ChemicalEngineeringGuy.com
  • 139. Pump Performance Curve • 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 www.ChemicalEngineeringGuy.com
  • 140. Pump Performance Curve • 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
  • 141. 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? www.ChemicalEngineeringGuy.com
  • 143. 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
  • 144. Pump Curve Exercise www.ChemicalEngineeringGuy.com • 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
  • 145. 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 www.ChemicalEngineeringGuy.com • The experimental data is given next
  • 146. Pump Curve Exercise www.ChemicalEngineeringGuy.com 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
  • 147. Pump Curve Exercise • Check out what happens when you add – Different Impeller Diameters • Check it out here: www.ChemicalEngineeringGuy.com • Courses Applied Fluid Dynamics Part 1: Incompressible Flow www.ChemicalEngineeringGuy.com
  • 148. Pump Performance Curve Analysis • Why does the Curve is concave down? • As flow rate increases, decreases the head of the pump!? • Explain yourself! www.ChemicalEngineeringGuy.com
  • 149. Pump Performance Curve Analysis www.ChemicalEngineeringGuy.com
  • 150. Pump Performance Curve • Actually, suppliers may group pumps as “families” – Steep (high friction) – Average/Standard – Flat (low friction) www.ChemicalEngineeringGuy.com
  • 153. Pump Performance Curve www.ChemicalEngineeringGuy.com Same Operation Point, different Pump Curves
  • 154. 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.) www.ChemicalEngineeringGuy.com
  • 155. 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 www.ChemicalEngineeringGuy.com
  • 156. Pump Curve Diagram  Head vs. Q 8” Volumetric Flow HeadPump(m)
  • 157. Pump Curve Diagram  Head vs. Q 8” Volumetric Flow HeadPump(m) Head = f(Q)
  • 158. 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 www.ChemicalEngineeringGuy.com
  • 159. • 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] www.ChemicalEngineeringGuy.com S. D. Impeller Pump Curve Diagram  Impeller’s Diameter
  • 160. Pump Curve Diagram  Impeller’s Diameter www.ChemicalEngineeringGuy.com
  • 161. 8” Volumetric Flow HeadPump(m) Pump Curve Diagram  Impeller’s Diameter
  • 162. 8” Volumetric Flow HeadPump(m) 3” Pump Curve Diagram  Impeller’s Diameter
  • 163. Pump Curve Diagram  Impeller’s Diameter 6“ 1/2 5” 4“ 1/2 3” 8” Impeller Size (Diameter in Inches) Volumetric Flow HeadPump(m)
  • 164. Pump Curve Diagram  Impeller’s Diameter 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
  • 165. Pump Curve Diagram  Impeller’s Diameter 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
  • 166. 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! www.ChemicalEngineeringGuy.com
  • 167. Pump Curve Diagram  Efficiency www.ChemicalEngineeringGuy.com Volumetric Flow HeadPump(m)
  • 168. Pump Curve Diagram  Efficiency www.ChemicalEngineeringGuy.com
  • 169. Pump Curve Diagram  Efficiency 6“ 1/2 5” 4“ 1/2 3” 8” Volumetric Flow HeadPump(m)
  • 170. Pump Curve Diagram  Efficiency www.ChemicalEngineeringGuy.com
  • 171. Pump Curve Diagram  Efficiency
  • 172. Pump Curve Diagram  Efficiency 6“ 1/2 5” 4“ 1/2 3” 8”
  • 173. Pump Curve Diagram  Efficiency www.ChemicalEngineeringGuy.com
  • 174. Pump Efficiency Exercise #1 • Choose the best efficiency area for this Pump www.ChemicalEngineeringGuy.com
  • 175. Pump Efficiency Exercise #1 • Choose the best efficiency area for this Pump www.ChemicalEngineeringGuy.com 73 %
  • 176. Pump Efficiency Exercise #1 • Choose the best efficiency area for this Pump www.ChemicalEngineeringGuy.com 73 %
  • 177. Pump Efficiency Exercise #2 • If GPM = 500… What is the Head when η = 65% www.ChemicalEngineeringGuy.com
  • 178. Pump Efficiency Exercise #2 • If GPM = 500… What is the Head when η = 65% www.ChemicalEngineeringGuy.com
  • 179. Pump Efficiency Exercise #2 • If GPM = 500… What is the Head when η = 65% www.ChemicalEngineeringGuy.com About 3200 ft
  • 180. 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! www.ChemicalEngineeringGuy.com
  • 181. Pump Curve Diagram  Power 5” 4“ 1/2 3”
  • 182. Pump Curve Diagram  Power 5” 4“ 1/2 3”
  • 183. Pump Curve Diagram  Power 5” 4“ 1/2 3” As Flow increases, Power Increses
  • 184. Pump Curve Diagram  Power 5” 4“ 1/2 3” As Head Decreases, Power Decreases
  • 185. Pump Power Exercise #1 • How much Power is Required in HP for: – A pump working 600 GPM – The Head needed is 2400 ft www.ChemicalEngineeringGuy.com
  • 186. Pump Power Exercise #1 • How much Power is Required in HP for: – A pump working 600 GPM – The Head needed is 2400 ft www.ChemicalEngineeringGuy.com
  • 187. Pump Power Exercise #1 • How much Power is Required in HP for: – A pump working 600 GPM – The Head needed is 2400 ft www.ChemicalEngineeringGuy.com About 550 HP
  • 188. 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? www.ChemicalEngineeringGuy.com
  • 189. 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? www.ChemicalEngineeringGuy.com
  • 190. Pump Power Exercise #2 • If Power Requirement of a Pump = 600 BHP • The Pump has an Impeller Diameter = 10 ½ – What is the System Head? – What is the Effiency in this operation? www.ChemicalEngineeringGuy.com
  • 191. Pump Power Exercise #2 • If Power Requirement of a Pump = 600 BHP • The Pump has an Impeller Diameter = 10 ½ – What is the System Head? – What is the Effiency in this operation? www.ChemicalEngineeringGuy.com
  • 192. Pump Power Exercise #2 • If Power Requirement of a Pump = 600 BHP • The Pump has an Impeller Diameter = 10 ½ – What is the System Head? – What is the Effiency in this operation? www.ChemicalEngineeringGuy.com About 3200 ft About 64%
  • 193. 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  www.ChemicalEngineeringGuy.com NOTE: NSPH Required is f(Q) only!
  • 194. Pump Curve Diagram  NPSHR 8” Volumetric Flow HeadPump(m)
  • 195. Pump Curve Diagram  NPSHR www.ChemicalEngineeringGuy.com
  • 196. Pump Curve Diagram  NPSHR www.ChemicalEngineeringGuy.com
  • 197. Pump Curve Diagram  NPSHR www.ChemicalEngineeringGuy.com NOTE: NSPH Required is f(Q) only!
  • 198. Pump Curve Diagram  NPSHR www.ChemicalEngineeringGuy.com
  • 199. Pump Curve Diagram  NPSHR www.ChemicalEngineeringGuy.com
  • 200. Pump Curve Diagram  NPSHR www.ChemicalEngineeringGuy.com
  • 201. Pump Curve Diagram  NPSHR Exercise #1 • Calculate the NPSHR when: – Impeller size is 7” – Head = 200 – GPM = 160 www.ChemicalEngineeringGuy.com
  • 202. Pump Curve Diagram  NPSHR Exercise #1 • Calculate the NPSHR when: – Impeller size is 7” – Head = 200 – GPM = 160 www.ChemicalEngineeringGuy.com You just need GPM
  • 203. Pump Curve Diagram  NPSHR Exercise #1 • Calculate the NPSHR when: – Impeller size is 7” – Head = 200 – GPM = 160 www.ChemicalEngineeringGuy.com You just need GPM
  • 204. Pump Curve Diagram  NPSHR Exercise #1 • Calculate the NPSHR when: – Impeller size is 7” – Head = 200 – GPM = 160 www.ChemicalEngineeringGuy.com You just need GPM NPSHR = 5.5 ft
  • 205. Pump Curve Diagram  NPSHR Exercise #2 • What is the minimum Flow Rate given a NPSHa of 3.3 m? www.ChemicalEngineeringGuy.com
  • 206. Pump Curve Diagram  NPSHR Exercise #2 • What is the minimum Flow Rate given a NPSHa of 3.3 m? www.ChemicalEngineeringGuy.com NPSHa > 1.1*NPSHr  3
  • 207. Pump Curve Diagram  NPSHR Exercise #2 • What is the minimum Flow Rate given a NPSHa of 5 m? www.ChemicalEngineeringGuy.com NPSHa > 1.1*NPSHr  4.54
  • 208. Pump Curve Diagram  NPSHR Exercise #2 • What is the minimum Flow Rate given a NPSHa of 5 m? www.ChemicalEngineeringGuy.com NPSHa > 1.1*NPSHr  4.54
  • 209. Pump Curve Diagram  NPSHR Exercise #2 • What is the minimum Flow Rate given a NPSHa of 5 m? www.ChemicalEngineeringGuy.com NPSHa > 1.1*NPSHr  4.54 About 600 GPM
  • 210. Pump Performance Curve Full Diagrams! • We’ve seen all the theoretical concepts • We can read the full diagrams of a Pump Curve! www.ChemicalEngineeringGuy.com
  • 211. Pump Performance Curve Full Diagrams! www.ChemicalEngineeringGuy.com
  • 212. Need More Problems? Check out the COURSE www.ChemicalEngineeringGuy.com • Courses Applied Fluid Mechanics Part 1: Incompressible Flow You’ll get SOLVED problems, Quizzes, Slides, and much more! www.ChemicalEngineeringGuy.com
  • 213. End of Section 3: Pump Curve/Head www.ChemicalEngineeringGuy.com
  • 214. Section 4: Pump Selection www.ChemicalEngineeringGuy.com
  • 215. 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! www.ChemicalEngineeringGuy.com
  • 216. 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
  • 217. 1a. General Pump Selection Diagram
  • 218. 1a. General Pump Selection Diagram PD: Rotatory
  • 219. 1a. General Pump Selection Diagram PD: Reciprocal
  • 220. 1a. General Pump Selection Diagram Centrifugal High Speed (RPM = 3500)
  • 221. 1a. General Pump Selection Diagram Centrifugal Slow Speed 1750 RPM and lower
  • 222. 1a. General Pump Selection Diagram Axial
  • 223. 1a. General Pump Selection Diagram
  • 224. 1a. General Pump Selection Diagram • Recommend a Pump with the following: – nWp = 150 ft – Q = 100 GPM
  • 225. 1a. General Pump Selection Diagram • Recommend a Pump with the following: – nWp = 150 ft – Q = 100 GPM
  • 226. 1a. General Pump Selection Diagram • Recommend a Pump with the following: – nWp = 150 ft – Q = 100 GPM Either: Centrifugal, Rotatory or Reciprocal! Avoid: Axial and Low velocity Centrifugal Pumps!
  • 227. 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 www.ChemicalEngineeringGuy.com
  • 228. 1b. Specific Velocity Criteria • N = impeller speed (RPM) • Q = volumetric flow in gal/min • H = Pump Head (ft) • D = Impeller Diameter (in) www.ChemicalEngineeringGuy.com Specific Diameter   Specific Speed
  • 229. 1b. Specific Velocity Criteria www.ChemicalEngineeringGuy.com
  • 230. 1b. Specific Velocity Criteria www.ChemicalEngineeringGuy.com
  • 231. 1b. Specific Velocity Criteria www.ChemicalEngineeringGuy.com
  • 232. 1b. Specific Velocity Criteria www.ChemicalEngineeringGuy.com
  • 233. 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 www.ChemicalEngineeringGuy.com
  • 234. 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 www.ChemicalEngineeringGuy.com
  • 235. 1b. Specific Velocity Criteria 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 www.ChemicalEngineeringGuy.com
  • 236. 1b. Specific Velocity Criteria Exercise #1 www.ChemicalEngineeringGuy.com
  • 238. www.ChemicalEngineeringGuy.com Operation of Pump is NOT recommended! 1b. Specific Velocity Criteria Exercise #1
  • 239. www.ChemicalEngineeringGuy.com D= 18 in to 12 in Ds = D*(H^0.25)*(Q^-0.25) Ds = 12*(80^0.25)*(500^-0.25) = 1.6 1b. Specific Velocity Criteria Exercise #1
  • 240. www.ChemicalEngineeringGuy.com D= 18 in to 12 in Ds = D*(H^0.25)*(Q^-0.25) Ds = 12*(80^0.25)*(500^-0.25) = 1.6 1b. Specific Velocity Criteria Exercise #1
  • 241. www.ChemicalEngineeringGuy.com Better Operation! Radial Flow  Centrifugal Pump Max Efficiency Expected  60-70% 1b. Specific Velocity Criteria Exercise #1
  • 242. www.ChemicalEngineeringGuy.com Force Pumping system to THIS! 1b. Specific Velocity Criteria Exercise #1
  • 243. www.ChemicalEngineeringGuy.com Force Pumping system to THIS! 1500 < Ns < 7000 0.4 < Ds < 1.2 1b. Specific Velocity Criteria Exercise #1
  • 244. 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!
  • 245. 2. Pump Supplier’s Family • 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
  • 246. 2. Pump Supplier’s Family 3 x 5 – 12 3: Discharge Size (Inches) 5: Suction Size (Inches) 12: (Larges Impeller Size Available)
  • 248. 2. Pump Supplier’s Family • 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?) www.ChemicalEngineeringGuy.com
  • 249. 2. Pump Supplier’s Family NPSHR
  • 250. 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 www.ChemicalEngineeringGuy.com Check out the exercise of the “System’s Head Exercise 1,2,3”
  • 251. 3. Calculate System Head • 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 www.ChemicalEngineeringGuy.com
  • 252. 3. Calculate System Head www.ChemicalEngineeringGuy.com
  • 253. 3. Calculate 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) www.ChemicalEngineeringGuy.com
  • 254. 3. Calculate System Head System Curse (Valve 100%) www.ChemicalEngineeringGuy.com
  • 255. 3. Calculate System Head Válvulas abiertas Válvulas 40% cerradasSystem Curse (Valve 60%) System Curse (Valve 100%) www.ChemicalEngineeringGuy.com
  • 256. 3. Calculate System Head Válvulas abiertas Válvulas 40% cerradas Válvulas 80% cerradas System Curse (Valve 60%) System Curse (Valve 20%) System Curse (Valve 100%) www.ChemicalEngineeringGuy.com
  • 257. 3. Calculate System Head Válvulas abiertas Válvulas 40% cerradas Válvulas 80% cerradas System Curse (Valve 60%) System Curse (Valve 20%) System Curse (Valve 100%) www.ChemicalEngineeringGuy.com Head Increases as Valve % Approaches 0%
  • 258. 3. Calculate System Head Válvulas abiertas Válvulas 40% cerradas Válvulas 80% cerradas System Curse (Valve 60%) System Curse (Valve 20%) System Curse (Valve 100%) www.ChemicalEngineeringGuy.com Head Increases as Valve % Approaches 0% Same Flow Rate Different Head!
  • 259. 3. Calculate System Head Válvulas abiertas Válvulas 40% cerradas Válvulas 80% cerradas System Curse (Valve 60%) System Curse (Valve 20%) System Curse (Valve 100%) www.ChemicalEngineeringGuy.com Head Increases as Valve % Approaches 0% Same Flow Rate Different Head!
  • 260. 3. Calculate System Head Válvulas abiertas Válvulas 40% cerradas Válvulas 80% cerradas System Curse (Valve 60%) System Curse (Valve 20%) System Curse (Valve 100%) www.ChemicalEngineeringGuy.com Consider changing Impeller’s Diameter
  • 261. 4. Intersect: Pump Head Curve + System Head www.ChemicalEngineeringGuy.com Volumetric Flow HeadPump(m) Pump Curve
  • 262. 4. Intersect: Pump Head Curve + System Head www.ChemicalEngineeringGuy.com Volumetric Flow HeadPump(m) System Head Curve
  • 263. 4. Intersect: Pump Head Curve + System Head www.ChemicalEngineeringGuy.com Volumetric Flow HeadPump(m) OPERATION Point System Curse (Valve 100%)
  • 264. 4. Intersect: Pump Head Curve + System Head www.ChemicalEngineeringGuy.com Volumetric Flow HeadPump(m) OPERATION Point System Curse (Valve 60%) System Curse (Valve 100%)
  • 265. 4. Intersect: Pump Head Curve + System Head www.ChemicalEngineeringGuy.com Volumetric Flow HeadPump(m) OPERATION Point System Curse (Valve 60%) System Curse (Valve 20%) System Curse (Valve 100%) But we are changing Flow Rate!
  • 266. 5. Optimize Operation Point • Once Operation Point is set – Try to optimize – Flow Rate – System Head Requirement – Efficiency – Power – NSPHr f(Q) www.ChemicalEngineeringGuy.com
  • 267. 5. Optimize Operation Point Exercise #1 www.ChemicalEngineeringGuy.com Optimize Efficiency!
  • 268. 5. Optimize Operation Point Exercise #1 www.ChemicalEngineeringGuy.com Optimize Efficiency!
  • 269. 5. Optimize Operation Point Exercise #1 www.ChemicalEngineeringGuy.com Optimize Efficiency!
  • 270. 5. Optimize Operation Point Exercise #1 www.ChemicalEngineeringGuy.com Optimize Efficiency!  More Efficiency; Less Flow Rate! NEW Design Old Design
  • 271. 5. Optimize Operation Point Exercise #2 www.ChemicalEngineeringGuy.com Decrease Power Requirement! Maintain Flow Rate! Old Design
  • 272. 5. Optimize Operation Point Exercise #2 www.ChemicalEngineeringGuy.com Decrease Power Requirement! Maintain Flow Rate! Old Design P = 30 BHP
  • 273. 5. Optimize Operation Point Exercise #2 www.ChemicalEngineeringGuy.com Decrease Power Requirement! Maintain Flow Rate! Old Design
  • 274. 5. Optimize Operation Point Exercise #2 www.ChemicalEngineeringGuy.com OPEN all Valves. Clean Pipes. Change Pipe Diameter! Old Design New Design P = 21 BHP
  • 275. 5. Optimize Operation Point Exercise #3 www.ChemicalEngineeringGuy.com An Engineer Proposes the next change Old Design
  • 276. 5. Optimize Operation Point Exercise #3 www.ChemicalEngineeringGuy.com An Engineer Proposes the next change Old Design
  • 277. 5. Optimize Operation Point Exercise #3 www.ChemicalEngineeringGuy.com An Engineer Proposes the next change New Design
  • 278. 5. Optimize Operation Point Exercise #3 www.ChemicalEngineeringGuy.com Would you recommend it? EXPLAIN New Design
  • 279. 5. Optimize Operation Point Exercise #3 www.ChemicalEngineeringGuy.com NO  Increase in Pump Power; Installation of New Impeller needed New Design
  • 280. 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 www.ChemicalEngineeringGuy.com
  • 281. 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
  • 282. Choosing a Pump  Exercise #1 www.ChemicalEngineeringGuy.com
  • 283. Choosing a Pump  Exercise #1 Actual Flow Rate
  • 284. Choosing a Pump  Exercise #1 Pump Curve at 6”
  • 285. Choosing a Pump  Exercise #1 Actual Operation Point
  • 286. Choosing a Pump  Exercise #1 Efficiency: 58% Power: 14 HP Head= 135 ft
  • 287. Choosing a Pump  Exercise #1 New DesignNew Design 500 gpm
  • 288. Choosing a Pump  Exercise #1 No Impeller Diameters!
  • 289. Choosing a Pump  Exercise #1 Propose buying 8” or 8.5”
  • 290. Choosing a Pump  Exercise #1 Propose buying 8” or 8.5” Eff1 = 72.5% Eff2 = 71.5% P1 = 41 BHP P2 = 37 BHP H1 = 240 ft H2 = 210 ft
  • 291. Need More Problems? Check out the COURSE www.ChemicalEngineeringGuy.com • Courses Applied Fluid Dynamics Part 1: Incompressible Flow Rate You’ll get SOLVED Problems, Quizzes, Slides, and much more! www.ChemicalEngineeringGuy.com
  • 292. NOTE: Common Mistake Head is 300 or 240??
  • 293. NOTE: Common Mistake Head is 300 or 240??
  • 294. 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 www.ChemicalEngineeringGuy.com
  • 295. Pump Affinity Laws • What if there is no data in the Diagram? www.ChemicalEngineeringGuy.com 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
  • 296. 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 www.ChemicalEngineeringGuy.com
  • 297. 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
  • 298. Pump Affinity Laws Exercise #1 www.ChemicalEngineeringGuy.com
  • 299. 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) www.ChemicalEngineeringGuy.com
  • 300. Pump Affinity Laws Exercise #1 • From the last Pump Diagram (Curve) – RPM = 3550 – D impeller = 6 in • Find: • System Head (ft)  125 ft • Flow Rate (GPM) 330 GPM • P (BHP) = 17 BHP www.ChemicalEngineeringGuy.com
  • 301. Pump Affinity Laws Exercise #2 • If we change: – RPM = 3550  1750 – D impeller = 6 in stays the same • Find: • System Head (ft) – 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 www.ChemicalEngineeringGuy.com
  • 302. Pump Affinity Laws Exercise #3 • If we change: – RPM = 3550 stays the same – D impeller = 12 in • Find: • System Head (ft) – 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 www.ChemicalEngineeringGuy.com
  • 303. Pump Affinity Laws Exercise Conclusions www.ChemicalEngineeringGuy.com NO Data for 1750 RPM No Data for 12” Impeller
  • 304. Pump Affinity Laws Exercise Conclusions www.ChemicalEngineeringGuy.com NO Data for 1750 RPM No Data for 12” Impeller This is awesome! We don’t need another diagram to find this out!
  • 305. End of Section 4: Pump Selection www.ChemicalEngineeringGuy.com
  • 306. Section 5: Pump Arrangements www.ChemicalEngineeringGuy.com
  • 307. Pump in Series/Parallel Arrangement • Pumps may be arrange either in: – Series (one after another) – Parallel (one beside the other) www.ChemicalEngineeringGuy.com
  • 308. 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!
  • 310. 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
  • 313. 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 www.ChemicalEngineeringGuy.com
  • 314. Pump Arrangement Exercise #1 • 1 Single Pump – Q (pump) = 600 gpm – nWp (pump) = 270 ft – Pin = Pin – Pout = Pout www.ChemicalEngineeringGuy.com
  • 315. 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 www.ChemicalEngineeringGuy.com
  • 316. 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 www.ChemicalEngineeringGuy.com
  • 317. 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 www.ChemicalEngineeringGuy.com What is that you want? Increase P? Increase Head? Decrease Flow Rate?
  • 318. Need More Problems? Check out the COURSE www.ChemicalEngineeringGuy.com • Courses MOMENTUM TRANSFER OPERATIONS You’ll get SOLVED Problems, Quizzes, Slides, and much more! www.ChemicalEngineeringGuy.com
  • 319. Software Modeling for Pumps • Basic Modeling – Aspen Plus – Aspen HYSYS – PSYM  http://www.pumpsystemsmatter.org/ • Intensive Modeling – SolidWorks – EPANet  http://epanet.de/ – COMSOL Inc www.ChemicalEngineeringGuy.com
  • 320. Software Modeling for Pumps • Aspen Plus + HYSYS www.ChemicalEngineeringGuy.com
  • 321. Software Modeling for Pumps • Aspen Plus + HYSYS www.ChemicalEngineeringGuy.com
  • 322. Software Modeling for Pumps • Aspen Plus + HYSYS www.ChemicalEngineeringGuy.com
  • 323. Software Modeling for Pumps • Aspen Plus + HYSYS www.ChemicalEngineeringGuy.com
  • 324. Software Modeling for Pumps • COMSOL Inc www.ChemicalEngineeringGuy.com
  • 325. Software Modeling for Pumps • COMSOL Inc www.ChemicalEngineeringGuy.com
  • 326. Software Modeling for Pumps • COMSOL Inc www.ChemicalEngineeringGuy.com
  • 327. Software Modeling for Pumps • SolidWorks www.ChemicalEngineeringGuy.com
  • 328. Software Modeling for Pumps • SolidWorks www.ChemicalEngineeringGuy.com
  • 329. Software Modeling for Pumps • SolidWorks www.ChemicalEngineeringGuy.com
  • 330. Software Modeling for Pumps • SolidWorks www.ChemicalEngineeringGuy.com
  • 331. End of Section 5: Pump Arrangements www.ChemicalEngineeringGuy.com
  • 332. 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 www.ChemicalEngineeringGuy.com
  • 333. 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 www.ChemicalEngineeringGuy.com
  • 334. 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 www.ChemicalEngineeringGuy.com
  • 335. Textbook, Reference and Bibliography www.ChemicalEngineeringGuy.com