pump lecture course notes 2005 hodkiewicz

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Thermofluids 312 Lecture Notes: Pumpapplication, Operation and Specification

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School of Mechanical Engineering

University of Western Australia

Thermofluids TF306

2005

Pump Application, Operation and Specification

Melinda Hodkiewicz

[email protected]

Extension: 7911, Room G55


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1.1. Outcomes of the course

• Recognition of pump system components and their purpose• Determination of a pump system curve• Ability to interpret manufacturer’s pump documentation

• Competence to select and size a centrifugal pump for a particular application• Understand the effect of changes in the system on the operating point of the pump• Appreciate the effect of assembly, installation and operating practices on the life cycle of a

centrifugal pump

1.2. References

Books

• Centrifugal pumps 2nd ed., Karassik and McGuire• Fundamentals of Fluid Dynamics, Gerhart and Gross• Centrifugal Pumps - Design and Application 2nd ed, Lobanoff & Ross• Introduction to Fluid Mechanics - Fox and McDonald MPSL 620.106 1998 INT/1992 INT• Fundamentals of Thermal-Fluid Sciences – Cengal and Turner MPSL 621.402 2001 FUN• Predictive Maintenance of Pumps using condition monitoring – R.S.Beebe.• Slurry Systems Handbook - Abulnaga

Useful web references

• Pumps&Systems www.pump-zone.com • API 610: http://www.api.org/tf610/index.htm.• Links www.bhrgroup.co.uk/links• Software www.fluidflowinfo.com• Warman Slurry pumps- www.warmanintl.com

• Gould pumps www.gouldspumps.com• GE www.ge.com/industrialsystems/solutions/pump.html • Pump types: http://www.pumpschool.com

Vendor references

• Crane Technical Paper 410(metric)• Flow of Fluids through valves, fittings and pipes• Goulds Pump manual GPM6• Basic Principles for the Design of Centrifugal pump installations (SIHI)• Sulzer Centrifugal pump handbook• Warman Slurry Pumping Handbook

• Cameron Hydraulic data

http://www.pump-zone.com/http://www.api.org/tf610/index.htmhttp://www.warmanintl.com/http://www.ge.com/industrialsystems/solutions/pump.htmlhttp://www.ge.com/industrialsystems/solutions/pump.htmlhttp://www.warmanintl.com/http://www.api.org/tf610/index.htmhttp://www.pump-zone.com/


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1.1. Centrifugal Pump Generic design

There are many different styles of centrifugal pumps, but they can essentially be divided into three broad

groups

a. horizontal or vertical

b. single impeller (end suction/split case, single volute, double volute, double suction) or multi-stage

impeller designs

c. Impeller design: Radial, mixed flow, axial. Open and closed, semi open designs.

The governing principles of all centrifugal pumps are the same but the design details vary.

Examples of the different designs are given below …

1

htt ://www.f dlit.com/cms/results detail.as ?ModelID=102

Figure 1: Single-stage end-suction horizontal

centrifugal pump1

2

Figure 2: Single-stage vertical

centrifugal pumpFigure 3: Multi-stage horizontal centrifugal pump

2

1

http://www.giwindustries.com/lsa.html

2 http://www.fpdlit.com/cms/results detail.asp?ModelID=23


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1.2. Main Centrifugal Pump components

• Volute/Housing Bearing housing, bearings and seals• Impeller Coupling• Shaft and sleeve Motor• Mechanical seal or packing Foundation and baseplate

•

3

4

Figure 5: End-suction centrifugal pump and motor3

Figure 4: Schematic of end suction pump4

3

Ref: C.Dean UWA Honours thesis 2001 – from Goulds Pumps 4 Ref: Goulds Pumps – this is the pump on the PUMP TEST RIG (Engine lab)


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1.3. Typical Pump Installation

Centrifugal pumps are part of a “SYSTEM”. The system contains tanks, pipes, valves and fittings. The

performance of a centrifugal pump is determined by the system it works in. Key factors are the “HEAD” H P on

the pump and the “FLOWRATE” Q.

+

−+−+

−== L

ie

ie

ieS

P h g

V V z z

g

p p

g m

W H

2

)()(

)( 22

ρ &

&

The head on the pump is determined by the rate of work W in J/s of the shaft/impeller divided by the mass flow

and the gravitational constant g. The W is a function of

S &

m& S &

1. The pressure p divided by ρ g , this results in units of metres of head. The pressure is a function of the

pressure p acting on the FREE surface of the liquid in the system on the inlet (i) or suction side and the

exit (e)or discharge side. When the suction and discharge tanks are open to atmosphere, the values are pi

= pe = 0. When the tanks are closed and contain elevated pressure or vacuum this must be taken intoaccount.

2. The height of the liquid in the tanks at the suction and discharge zi and ze

3. The velocity at the free surface of the liquid Vi and Ve, this can usually be ignored.

4. The friction loss in the entire system hL (both suction and discharge). This is affected by the line

diameter, line lengths, fittings and valves and is discussed in detail later.

Any change in the value of these system terms will affect the Head on the pump, as a result the flowrate through

the pump will change. This can be demonstrated using a LabView PUMP SIMULATION program.

1.4. Sizing a Centrifugal Pump

In order to correctly size a pump for a particular application it is necessary to understand the system in which it

is installed. One selects a pump based on its ability to supply the required flowrate for the system. The operating

point of a pump is set by the intersection of the PUMP curve (specific to the pump) with the SYSTEM curve

(defined by the piping system, tank elevation, over-pressures etc)

1.5. Steps involved in selecting and sizing a pump

1. Determine flowrate

2. Obtain fluid property information

3. Design piping system4. Determine the System Head Curve

5. Decide on duty point

6. Calculate Power required and Specific speed values

7. Calculate Net Positive Suction Head available

8. Develop pump specification sheet

9. Select a pump

10. Evaluate pump selection

Pump Application, Operation and Specification Melinda Hodkiewicz

Page 5 of 25 UWA Mechanical 3rd yr Thermofluids course Version 2 2005


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1.5.1. Step 1: Determine flowrate/pump

• Determined by process engineers/designers• Determine number of pumps required (function of criticality and reliability)• Define realistic maximum and minimum flowrates

1.5.2. Step 2: Determine Fluid property information

• Density, specific gravity, Dynamic or Absolute viscosity, Kinematic viscosity.• Water density at 20 deg C = 1000 kg/m3 • Specific weight γ = ρg , Weight of fluid per unit volume.• Specific gravity S= ρ/ ρH20.

Why is viscosity important?

• Increase viscosity - increase losses - less head generated - lower efficiency• Effect is greater on smaller pumps due to smaller internal passage dimensions• Used to calculate the Reynolds Number which determines the set of pump equations to use

1.5.3. Step 3: Design piping system

Select pipe sizes

• This is a compromise between installation costs and running costs.• Small diameter pipes lead to high line velocities and friction losses.• Elbows and fittings also result in friction losses• Suction piping design is critical to avoid creating swirl/uneven flow at the pump suction

Guidelines for line velocity

Suction piping (water) = 1.2-2.1 m/s, Discharge piping (water) =1.2-3.0 m/s,

Slurry piping (mining) = 1.5-2.5 m/s but there are special considerations due to particle settling velocity

Discharge piping (hydrocarbons)= 1-7 m/s

Friction losses in pipes

Resistance to flow as liquid moves through pipe results in loss of head. This friction loss h L is measured in m.

Resistance is due to viscous shear stresses within the liquid and friction losses at contact of moving fluid and

pipe wall

2

( , , , ) (Re)2

L

L V h f L d V f

d g υ

= =

Calculation of this will be discussed in Section 1.5.5

1.5.4. Step 4: Determine System Head Curve

For a new pump installation you will need

• P&ID (Piping and Instrumentation Drawing) Symbols used may be found in AS 1101.6,• GA (General arrangement) plan and elevation,• Flowsheet,• Isometric drawings




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+

−+−+

−== L

ieie

ieS P h

g

V V z z

g

p p

g m

W H

2

)()(

)( 22

ρ &

&

where h L = h L(D) + h L(S)

This is often done by considering the suction and discharge sides separately as

Total Head (H p)

Hp = h(d) - h(s)

Total Suction Head

h(s)= p(s)/ g +z(s)+V2s/2g - hL(s) - h(i)

z(s) = static suction head, hL(s) = total friction loss in suction line, h(i) = entrance loss, p(s) = pressure other than

atmospheric in suction tank in m, h(s) = total suction head

Total Discharge Head

h(d)= p(d)/ g + V2d/2g+ z(d) + hL(d) + h(e)

z(d) = static discharge head, hL(d) = friction loss in discharge line, h(e) = exit loss, hP(d) = overpressure in

discharge tank in m, h(d) = total discharge head

Note: the friction loss is SUBTRACTED on the suction side but ADDED on the discharge side

1.5.5. Calculation of the friction loss terms (hL).

There are TWO separate friction calculations, one for the pipes and one for the fittings.

1. Darcy’s Formula for friction loss in pipes

For turbulent flow

∑

= D

L

g

V f h L 2

2

hL = pressure drop or friction loss in m, f=friction factor, L=length of pipe (m), V=line velocity (m/s), D= pipe

ID (m),

Friction loss depends on fluid velocity, pipe ID and roughness.

Darcy’s formula is valid for turbulent and laminar flow only if line pressure >> vapour pressure of the liquid ie

NO cavitation

For laminar flow

264

Re 2 L

L V h

D g

=

Friction Factor f

The friction factor is determined experimentally. For laminar flow f=64/Re.

For turbulent flow f depends on Re also the relative roughness ε/d. ε = roughness of pipe wall, d = pipe diameter




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Need to see appropriate table

Examples for the Friction factor values for clean commercial pipe with turbulent flow, see Pump Formula sheet.

2. Darcy’s Formula for friction loss valves and fittings

∑

=

g

V K h L

2

2

See suitable tables for values, some common values are provided on the pump formula sheet.

Summary for Friction Loss calculation

∑∑

+

=

g

V K

D

L

g

V f h L

22

22

This can also include the entrance and exit losses if they are significant

Note on Hazen WilliamsThe Darcy-Weisbach method is the technically correct method however many engineers use Hazen Williams

which is convenient and produces reliable results for water with turbulent flow

See reference book for equation and C factors. Widely used for simple flowsheet calculations

Pipe Friction loss Tables

Pipe friction losses = x m/100 m pipe for a specific pipe ID, material and line velocity and temperature. Depends

on material, condition and age

For example the friction loss of 700 l/s water through 4” Sched 40 steel pipe is 0.194 bar per 100 m or 1.98

m/100 m pipe

The methods above are used for new projects where you have drawings with line sizes, tank elevations etc.However if you have an existing pump installation the pump head can be determined with a pair of pressure

gauges and a flow meter. Place the pressure gauges in ports as close to the suction and discharge of the pump,

simultaneously read the pressure gauges and the flow rate.

The pressure P on a gauge located close to the flange of the suction of the pump will measure

)(.

S LS S S h g z g p P ρ ρ −⋅+=

The conversion to ‘head’ and addition of the suction velocity head will give a value for the total suction head

A pressure gauge placed on the discharge of the pump will read the following terms.

)(.

D L D D D h g z g p P ρ ρ +⋅+=

It can be seen that the Total Head HP on the pump.

( ) ( )

−+

−=

+

−+−+

−==

g

V V

g

P P h

g

V V z z

g

p p

g m

W H S DS D L

ieie

ieS P

22

)()(

)( 2222

ρ ρ &

&

The values VD and VS are the velocity in the pipe at the pump suction and discharge. This is a function of the

flow rate Q and the line diameter D. V = Q/A = 4Q/( π D2 ), V = line velocity m/s, Q = flow rate m3/s, A =




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inside pipe area m2, D = inside diameter m. If the line size are the same for suction and discharge, this term can

be ignored.

Draw System Curve

System curve determined by Total Head (m) at different flow rates (below, design and above design)

Flow (m3/hr) Total Head (m)

0 (Static head) 1860 25

100 48

110 56

Superimpose System Curve on a ‘suitable’ Pump Curve

5

Figure 6: Pump Curve

1.5.6. Step 5: Decide on a Duty Point

Duty Point is expressed as the calculated Head for the desired FlowrateFor example 120 l/s at 58m head

Determine high and low operating flow points

1.5.7. Step 6: Calculate Power required, Efficiency and Specific Speed

Power : Hydraulic power is work done by a pump in moving the liquid.

5 Reference: Southern Cross Pumps




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P S gQH W ρ =&

Pump Efficiency:

ηP = Hydraulic Power (W ) / Power input to the pump shaft from the motor (W )S & M &

Pump Efficiency depends on energy loss due to

• leakage (recirculation around the impeller outlet to inlet, internal to the pump)• hydraulic losses (viscosity and non-uniform flow)• mechanical losses (friction losses in the bearings and seals)

What is the flow rate at maximum efficiency for a 438 mm impeller using the pump curve in ???

Motor Efficiency: ηM

M W & is the power from the motor to the shaft = Power supplied to the motor x motor efficiency.

Calculation of the motor power requires

1. Measurement of the power delivered to the motor from the MCC (Motor Control Centre). This is

available as a kW reading.

2. Knowledge of the efficiency of the motor. This information is often available on the motor nameplate or

from the manufacturer. The motor efficiency is dependent on the load on the motor and the speed, if it is

a variable speed drive.

1.5.8. Step 7: Calculate NPSH available

An acceptable margin of NPSHA - NPSHR must be maintained over the entire operating range to prevent

CAVITATION. Cavitation is caused by the local vaporisation of a fluid when the static pressure drops below the

vapour pressure. The small bubbles filled with vapour that form in the low pressure region (suction eye of the pump) will collapse on moving into high pressure regions (inside the impeller). This "implosion" causes pitting

on the metal surface, vibration and a drop in efficiency.

For NPSH calculation must understand difference between absolute and gauge pressure

• Absolute pressure = Gauge pressure + Atmospheric pressure at elevation• Standard barometric pressure is 1.01325 bar or 760 mm Hg and changes with elevation above sea level.• Gauge pressure is pressure above barometric pressure• Convert gauge pressure readings to m by (x 0.102/SG)• Absolute pressure always refers to perfect vacuum as base

NPSH available • Net positive suction head is the absolute suction head at suction nozzle corrected to datum less the vapour

pressure of the liquid at operating temperature. Determines at what point liquid will vaporize at the lowest

pressure point of the pump (cavitation) and is characteristic of the system. NPSHA varies with capacity and

is always positive.

• ( ) ( suction atm

A suction L suction VP abs

p p NPSH z h h

g ρ

+= + − −

) OR

( )

−

+)(absVP

atmS h g

p P

ρ




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• hvp(abs) = head in m corresponding to the ABSOLUTE vapour pressure of the liquid at the temperature being pumped. This is determined from Tables of vapour pressure (usually given in bar)

NPSH required

The NPSHR is characteristic of pump design and represents the minimum margin required between suction head

and vapour pressure. NPSHR varies with capacity. It is determined by manufacturer and verified by NPSH pumptest. NPSHR depends on impeller design, flow rate, rpm, liquid and other factors. As a rule of thumb thereshould be a margin of at least 1m, though depends on application between NPSHA and NPSHR)

What is the NPSHR for a flowrate of 120 l/s on the pump curve in Figure ?.

1.5.9. Step 8: Develop Pump Specification Sheet

1.5.10. Step 9. Select a short-list of suitable Pumps from different manufacturers

1.5.11. Step 10: Evaluate Pump Selection

• Match Pump and System curve• Determine Efficiency and NPSH margin• Compare efficiency, NPSH margin, and off design performance of different pumps• Determine materials to be used based on fluid properties• Consider vendor technical support and spare parts issues• Consider preferred vendor supply contracts

1.6. Other Useful Pump terms

Specific speed: Ns = ϖ(Q)½/(h)¾; (h=gH units of L2/T2, ϖ in rad/sec)

Many design charts are a function of Ns. As Ns increases• Impeller shape changes from radial to axial• Lower head per stage• Blade loading increases• Maximum velocity increases• Tendancy to cavitate increasesLow Ns values: radial impellers, large diameter, narrow profile, high head per stage

Medium Ns: Francis vane impellers, low diameter to profile ratio, low head-high volume

High Ns mixed flow impellersVery high Ns axial flow impeller

Suction specific speed: Nss = n (Q*)1/2/(NPSHR*)3/4

n=rpm; Q*=Flow gpm;H*=Head ft; NPSHR* (ft) [The gpm is per impeller eye] Nss is a function of NPSH required. Modifying the diameter of the impeller eye, increase the flow, reduces the

NPSH required but increases the value of Nss. This causes a reduction in the low flow capability of the pump.

1.7. Affinity Laws

Use of the affinity laws to select the optimum impeller diameter and/or pump rotating speed (if a variable

frequency drive or sheave drive system is appropriate)




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Allows for performance at one speed to be predicted from known performance at known speed (or impeller

diameter)

Q = Q1 (n/n1) = Q1 (D/D1)

H = H1 (n/n1)2 = H1 (D/D1)

2

P = P1 (n/n1)3 = P1 (D/D1)

3

n/D = new desired speed rpm/diameter. n1D1=speed rpm/diameter for known characteristics Q1, H1 and P1

1.8. Parallel and Series OperationIn order to supply sufficient head or volumetric flow it may be necessary to place pumps in series or in parallel

respectively.

Pumps in parallel: the combined pump curve is obtained by adding the capacities of the individual pumps at the

same head

Pumps in series (the first pump discharge into the suction of the second pump): the combined pump curve is

obtained by adding the head of the individual pumps at the same capacity.

1.9. Potential source of pump problemsDesign (Critical speed - lateral/torsional)

Application/ Sizing (Low NPSHA, Off-BEP)

Assembly (Bearings, Looseness, Vane pass, Unbalance)Installation (Alignment, Looseness, Soft foot)

Operation (Pulsations, Turbulence, Cavitation, Recirculation, Piping resonance)

Each of these will result in vibration and other problems if not engineered correctly

1.9.1. Operational problemsTheoretically as long as NPSHA >> NPSHR then a centrifugal pump can operate over a wide range of capacities

however the exact capacity is determined by intersection of pump head-capacity curve with the system head

curve. Can vary pump curve by changes in speed or system curve by throttling valves however operation is only

optimum at one point called BEP

Off design conditions

Is any condition when a pump delivers flow in excess or below the capacity at best efficiency BEP

Operation at high flow

Results from oversizing the pump. Oversized pumps usually require throttling to move the operating point back

up the curve, this results in higher power consumption. If not throttled, higher flows can result in NPSH

problems. High flow situation also happen when two pumps are in parallel and one is taken out of service

Cavitation

Occurs when NPSHR>NPSHA

Causes impeller damage on visible side of vanes due to implosions (collapsing of the bubble). Identified by loud

continuous noise “pumping rocks" and high vibration

Avoid cavitation by increase NPSHA or decreasing NPSHR

Increase NPSHA by raising suction level, lower pump, reduce friction losses in suction, Subcool liquid(injection)

Decrease NPSHr by using slower speed (or variable freq drive), installing a double suction impeller, increasing

impeller eye area, using an oversize pump or installing an inducer ahead of impeller

Operation at low flow

A Reduction in demand results in throttling at pump discharge and the operating point moves up the curve

towards shut off. This causes recirculation resulting in hydraulic unbalance, vane passing forces, effects from




14/26

recirculation through wear rings, suction and discharge recirculation within the impeller, rotating stall,

cavitation, surge and system instabilities. These combine to cause pressure fluctuations, surging and vibration.

1.10. Typical pump fault conditionsMisalignment

Unbalance

Mechanical Looseness

ResonanceBearing damage/failure

Pump operating problems

Vane passing

Electrical

1.11. Positive Displacement pumps

• Used when flows are low and required pressure high• Often used for viscous liquids and for those requiring a shear free action• Inherently leak resistant design• Provide a fixed displacement per revolution. Pump will develop as much pressure as required to overcome

discharge pressure up to the point where motor trips or relief valve opens.

• Often require discharge pulsation dampeners and suction stabilisers, generally pressurized vessels with agas-liquid interface.

• Acceleration Head H(ac) represents energy required accelerate the column of fluid (m)• API standards 674, 675 and 676.

• ( ) ( suction atm

A suction L suction VP abs

p p NPSH z h h

g ρ

+= + − −

) minus H(ac)

• H(ac) = LsvsC/Kg, where Ls and vs are the length of and velocity in the suction line.• C=constant dependant on type of pump, 0.4 for simplex single acting, 0.2 for simplex double acting, 0.2 for

duplex single acting. See appropriate reference for full list

• K=factor for the relative compressibility of liquid (eg.K=1.4 for hot water, 2.5 for hot oil)

Examples: Main types: Rotary, Reciprocating, Gear




15/26

Figure 7References for figures:http://www.turfmaker.com/Positive_Displacement_Pump/positi

ve_displacement_pump.htmlhttp://www.learromec.com/Products/PR_Spur.htm

http://www.eng.rpi.edu/dept/chem-eng/Biotech-Environ/PUMPS/reciprocating.html

1.12. Review theoretical fluids concepts

Four main theoretical concepts introduced in the pump course are

• The conservation of mass• The conservation of linear momentum• The conservation of angular momentum• The conservation of energy (including Bernoulli’s equation)

An understanding of these concepts is important for an understanding of pump systems, which are a major

component of industrial plants.

1.12.1. The conservation of mass

Rate at which the mass

accumulates in the control

volume

= Rate at which mass enters

the control volume

- Rate at which mass leaves

the control volume

Mathematically this is written as ….

sys

in out

in out

dmm m

dt = −∑ ∑& & CONSERVATION OF MASS EQUATION

where = mass flow rate (kg/s) m&

For an incompressible fluid passing through a fixed control volume

Rate at which mass enters region = rate at which mass leaves control volume Q=v1 A1 = v2 A2

where v is the velocity of the fluid

A is the cross sectional area of the control volume through which the fluid flows.

1 is the entry to the control volume

2 is the exit of the control volume

1.12.2. Conversion between head and pressure

From first principles (Conservation of mass) gz p P ρ +=

Where z = height of the fluid and P = pressure measured at the base of the fluid. For example, a column of cold

water (15 deg C) 10.2m high produces 1 bar pressure at its base.



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1.12.3. Velocity profiles

According to laminar flow theory, the velocity of fluid in a pipe has a parabolic profile as shown below:

Figure 8: Laminar flow profile

In piping systems the flow is usually turbulent. Turbulent flow also has a rounded velocity profile, but rather

than a parabolic shape, the curve is flatter as shown below:

Figure 9: Turbulent flow profile

1.12.4. Laminar and Turbulent Flow

• Laminar Flow occurs at very low velocity or with high viscosity fluids. This is often visualised asstreaks of colored fluid flow in straight lines.

• Turbulent Flow flow occurs above critical velocity and involves the irregular, random motion of thefluid particles

• Reynolds Number (Re) determines laminar or turbulent flow and depends on pipe diameter, flowvelocity, density and viscosity of the fluid.

Re = Vdρ/µ d=pipe ID (mm), v=flow velocity (m/s), ρ=density kg/m3, µ=viscosity (cP).

Flow is considered if laminar if Re < 2000, turbulent if Re > 4000, critical zone 2000


17/26

1.12.6. The conservation of energy

In pipe flow, as with any situation, energy is conserved from one point in fluid flow to another. The energy can

be in the form of kinetic energy (KE), potential energy (PE) or internal energy (IE).

[Rate of heat transfer in]+[Rate of work done on sys]=[Rate of increase of IE +KE + PE]

Mathematically this is written as for steady, uniform, incompressible flow as (where u is specific internal

energy)

)2

(2

z g

V umW Q

net

++=− ∑ &&&

This is the steady state general energy equation as presented earlier in Thermodynamics lectures.

W & is the rate of work done by (+) or on (-) the control volume W stress shear stressnormal shaft W W W ..&&&& ++=

Shaft work rate is transmitted by the rotating shaft W (shaft torque x rotational speed)ω T shaft =&

Shear work rate is the product of shear stress, area and fluid velocity component parallel to the control surface.With pumps the control surfaces lie adjacent to solid boundaries where the fluid velocity is zero. In this case

there is no shear work although there may be shear stress.

Normal stress work can be written in most situations as a function of the pressure acting on the control surface

∑=net

pressure pm

W ρ

&&

In pump/piping problems it is conventional to assume to that g mQh L &&−= where hL is the heat dissipated as

friction by fluid contacting the pipe wall in units of metres.

Making these substitutions

L

net

S ghm zg V pumW &&& ++++= ∑ )2(

2

ρ GENERAL ENERGY EQUATION

This is commonly written as in terms of Power required at the shaft to drive a centrifugal pump. For pumps it is

assumed that 0=− ie uu

+−+

−+

−= L

ieieS h z z

g

V V

g

p p g mW )(

2

)()(12

22

ρ && PUMP SYSTEM SIZING EQUATION

This is the foundation equation for sizing pumps used in unit and we will spend time discussing how to

determine the values in this equation.

If one is dealing with a compressor substitute for the internal energy with )( ievie T T cuu −=− and use the idealgas law RT p = ρ For systems with no friction we have the MECHANICAL ENERGY EQUATION

−+

−+

−= )(

2

)()(12

22

z z g

V V

g

p p g mW ieieS

ρ && where Mechanical Energy is that which can be converted to

mechanical work completely by a mechanical device such as a turbine or pump.




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1.12.7. Bernoulli’s equation

Bernoulli’s equation is a special case of the conservation of energy equation.

Special conditions for use: inviscid and incompressible fluids, steady flow, constant density, no mechanical

work and no friction

It is derived from the previous equation for the special case when W and h0=S & L = 0.

2

2

221

2

11

22 z

g

V

g

p z

g

V

g

p++=++

ρ ρ

The terms in this equation are referred to as, pressure head, velocity head and static head respectively.

Dimensional analysis will show that all three terms are in meters.

1.12.8. How to measure the velocity in a pipe?

Due to layers of fluid shearing across each other, the velocity of a liquid is maximum in the centre of the flow,

and zero at the pipe wall. This means that the pressure due to velocity at the pipe wall is zero. Thus, head due to

velocity can be measured as the difference between the head at the centre of the pipe and the edge (See Figure).Physically this can be done using a pitot tube for the centre reading and a piezometer for the pipe wall reading.

This can be seen in the test facility in the CWR Fluids laboratory.

The term ‘head’ relates fluid pressure in a pipe to the meters of water that would push up an open topped tube.

Head depends on the density of the fluid and the density of air and is the sum of the static head, velocity head

and pressure head.

The velocity can be estimated from the pressure difference between the fluid at the side wall and the stagnation

pressure at the centre of the pipe. From Bernoulli’s equation

2

2

2

11

2 g

p

g

V

g

p

ρ ρ =+ ; where p2 is pressure at the stagnation point and p1 and V1 are the pressure upsteam.

h g

p p

g

V =

−=

ρ

12

2

1

2

Thus the velocity at a point in the pipe is equal to the square root of the height difference between the tubes

multiplied by 2g. gh2=v




19/26

Pitot tube

piezometer

Fluid flow

Height (h) due to

velocity

Figure 10: Measuring the head due to velocity

1.12.9. Conservation of Angular Momentum

The conservation of angular momentum principle when applied to the shaft of a pump can be used to show that

torque is transformed to a change in velocity of fluid through the impeller. This is done from first principles

using the equation below:

( ) ( )0 0, sys

j i i i e e

j in out

dLem r V m r V

dt

= + × − ×

∑ ∑ ∑& &

This equation can be written in scalar form to illustrate its application to pumps. The fixed coordinate system is

chosen with the z axis aligned with the axis of rotation of the machine. The fluid enters the rotor at a radial

location r i with uniform velocity Vθi and exits at r e with absolute velocity Vθe. Thus the equation above becomes:

)(ie

V r V r mT W ieSHAFT S θ θ ω ω −== &&

where Tshaft is the shaft torque

is the mass flow rate

.

m r is the radiusVθ is the tangential component of the absolute fluid velocity

e is the exit of the impeller

i is the inlet of the impeller

This is Euler’s turbomachinery equation, which is used to calculate the hydraulic power a pump is supplying,

which in turn can allow the calculation of pump efficiency

It is sometimes presented in terms of theoretical head

)(1

12 12 θ θ V U V U

g H Theo −=

This figure shows the inlet and exit radii, and the tangential components of the fluid velocity Vθ at the inlet and

exit. It should be noted that the fluid velocity V is not the same as U = ωr the velocity of the impeller.

It is conventional is pump design to describe flow passing through the impeller in velocity terms relative to the

rotating coordinate system of the rotating impeller. This is best done using “velocity triangles”.




20/26

1.12.10. Velocity triangles

Using velocity vectors V U =+W

V is the absolute velocity of the fluid

relative to a fixed coordinate system

W is velocity of the fluid relative to a

coordinate system fixed to the impeller

U is velocity of the impeller relative to a

fixed coordinate system

The subscript (1) relates the impeller inlet and

Figure 11: Velocity triangle

r

θ

β1α1

Leading edge of impeller blade

W1V1

U1

Vθ1

(2) the impeller discharge.

Each of the velocity vectors can be

resolved into r and θ components.

Using vector summation:

At the impeller entry (1)

1111 cos β θ W U V −= U1 = r 1ω where ω is the speed of theimpeller in radians/sec.

And 111 sin β W r =V

There are similar equations at the exit.

For an impeller of entrance width b1 with volumetric flow rate Q then Q = 2π.r 1.b1.Vr1

If Q and the impeller dimensions are known, Vr1 can be calculated and from this W1. If the speed in rps of the

impeller is known then U1 can be calculated and from this Vθ1.

The values of Vθ1 and Vθ2 are used to determine the torque T shaft or power draw W on the impeller using Euler’sequation (from Section 1.12.9).

&

)( 1122 θ θ ω ω V r V r mT W SHAFT S −== &&




21/26

1.13. Worked Examples

1.13.1.Consider the system in the schematic diagram below

Centrifugal pump

Suction side Discharge side

On the suction side, the level of the fluid is 10 m above the centreline of the pump and the overpressure in the

closed tank is 100 kPa. There is a gate valve between the suction side tank and the pump. Line length is 5 m.

On the discharge side, the level of the fluid is 30 m above the centreline of the pump and the overpressure in the

closed tank is 200 kPa. There is a butterfly valve on the discharge line and three 90 degree elbows. Line length is

50 m.

All line sizes are diameter 150 mm.

The desired flow rate is 100 l/s, the fluid is water at 15 deg C.

1. What is the head on the pump at the desired flow rate?

2. What is the hydraulic power?3. Draw a system curve for this installation.

Solution:

Start with the Pump sizing equation (see Section 1.3). Note that D (discharge) and S (Suction) have been

substituted for e (exit) and i (inlet).

++

−+−+

−== )(

2

)()(

)()()(

22

S L D LS D

S DS DS

P hh g

V V z z

g

p p

g m

W H

ρ &

&

Suction side

Static head zS = 10m

Pressure head pS/ρg = 100 x 103/(103x 9.8) = 10.2 m

Velocity head at entrance = very small.

Friction head ∑∑

+

=

g

V K

D

L

g

V f h L

22

22




22/26

From tables, f for clean steel pipe of diameter 150 mm is 0.015

The line velocity in a pipe of 150 mm diameter at 100 l/s is2

4

D

QV

π = = (4 x 0.1)/(3.14 x 0.152) = 5.66 m/s

Suction side Line friction loss = 0.015 x (5/0.150) x (5.662/2x9.81) = 0.8 m

From tables, K value for a gate valve is 8f

Suction side Valve/fitting friction loss = 8 x 0.015 x (5.72/2x9.81) = 0.2 m

Suction side friction head = 0.8 + 0.2 = 1 m.

Discharge side

Static head zD = 30m

Pressure head pD/ρg = 200 x 103/(103x 9.8) = 20.4 m

Velocity head at entrance = very small.

Friction head ∑∑

+

=

g

V K

D

L

g

V f h L

22

22

From tables, f for clean steel pipe of diameter 150 mm is 0.015

The line velocity in a pipe of 150 mm diameter at 100 l/s is2

4

D

QV

π = = (4 x 0.1)/(3.14 x 0.152) = 5.66 m/s

Discharge side Line friction loss = 0.015 x (50/0.150) x (5.662/2x9.81) = 8.2 m

From tables, K value for a butterfly valve is 45f and for each elbow is 30f. Total K value = (45+90)f.

Discharge side Valve/fitting friction loss = 135 x 0.015 x (5.662/2x9.81) = 3.3 m

Discharge side friction head = 8.2 + 3.3 = 10.8 m.

Summary

Suction (m) Discharge (m)

Static head 10.0 30.0

Pressure head 10.2 20.4

Velocity head ~ ~

Friction head 1.0 11.5

Total 19.2 61.9

Total head on the pump at 100 l/s = 61.9 – 19.2 = 42.7 m

The hydraulic power = ρ gQH = 1000 x 9.8 x 0.1 x 42.7 = 41.8 kW

To determine the System curve the calculation must be repeated at different flow rates, for example Q=0, 70 and

130 l/s. A curve can be drawn based on the four points. You can see from the table above that the total head at 0

l/s = 50.4 – 20.2 = 30.2 m as the friction head is zero at the no-flow point. The remaining flow points have to be

worked through in the same method as above taking into account the change in Friction head contribution as the

flow rate (and hence line velocity V) changes.




23/26

1.13.2.A water distribution pump has a 150 kW electric motor with a motor efficiency of 94%. The flow rate through

the pump is 350 l/s. The diameters of inlet and outlet pipes are the same and there is no significant elevation

difference across the pump.

If the inlet and outlet pressures are measured at 100 kPa and 400 kPa (absolute) respectively, determine,

a. The mechanical efficiency of the pump [74.5%] b. The temperature rise of the water as it flows through the pump due to mechanical inefficiency. [0.024

deg C]

Let specific heat of water be 4.18 kJ/kg. Deg C.

Solution:

Calculate mass flow rate through pump kg/s = ρ x m3/s = 1 kg/l x 350 l/s= 350 kg/sPower to shaft = motor efficiency x motor power = 0.94 x 150 = 141 kW.

Change in energy of fluid (or Hydraulic Power) =

−=

−+

−+

−

ρ ρ

ieieie P P m z z

g

V V

g

p p g m && )(

2

)()(12

22

=350 x (400-100)/1000 = 105 kW

Mechanical efficiency of the pump = Hydraulic power/Power to shaft = 105/141 = 74.5%

Only 105 kW of power supplied to the pump is imparted to the fluid as mechanical energy. The remaining 36

kW is converted to thermal energy and lost.

Rate of Energy loss = 141 – 105 = 36 kW.

( 1212 )( T T cmuum E v −=−= &&& ) Delta T = 36/ 350 x 4.18 (jk/kg. C) =0.024 deg C.

This is very small.In an actual application the temp rise of the water is less as the heat is transferred to the casing and surroundings.




24/26

1.13.3.Examine the pump curve for the Southern Cross ISO-PRO 200x150-400 pump, fixed speed 1475 rpm above

a) If the total head on the pump is 60m and you are using a full size impeller, what is the expected flowrate

in l/s and efficiency (%) ?

b) What standard size motor is required for the head-flow combination in (a)?

c) If you wanted to deliver 107 l/s at 50 m of total head, what is the optimum impeller diameter?

d) Calculate the hydraulic power and the motor power draw at 107 l/s and 50 m head for the impeller

diameter selected in c)?

e) If the pump was running as in c) and you slowly closed a discharge butterfly valve to achieve 60 l/s,

what would happen to the total head and the pump efficiency?

f) For c) what would be the calculated power draw on the pump?

Solution:

a) Full size impeller is 438 mm, flowrate at 60 m total head is 107 l/s, efficiency = 82.5%.

b) Motor size 90 kW.

c) 410 mm

d) Hydraulic power = 107 * 3.6 * 50 * 1/368 = 52 kW, efficiency from graph = 0.83, Power draw on motor= 52/0.83 = 63 kW

e) The head value would increase to 57 m as the system curve would steppen due to greater friction head

component. The intersection of the pump and system curve would move to the left along the line of the

impeller diameter. The efficiency would decrease to 71%. Resulting power draw on motor = 60 * 3.6 *

57 / (368 *0.71) = 47 kW




25/26

1.13.4.Impeller design

Given the following details about the design of a closed centrifugal impeller, determine

a) The flow rate through the impeller

b) Torque on the impeller shaftc) Hydraulic power

Rotating speed ω 148 rad/sInlet radius r 1 0.0375 m

Discharge radius r 2 0.0875 m

Inlet width b1 0.025 m

Discharge width b2 0.015 m

Inlet blade angle β1 25° Discharge blade angle β2 30° Discharge velocity relative to the

impeller W2

4 m/s at 30°

Step 1: Sketch the impeller

U2β2

W2

U1

r 1

W1β1

r 2

U = impeller velocity relative to inertial

ReferenceV = fluid velocity relative to inertial

Reference

W = fluid velocity relative to impeller




26/26

Step 2: Draw the inlet and exit velocity triangles

Exit triangle

Wθ2 Vθ2

U2

β2

W2 Wr2 V2

Calculate U2 = ωr 2 = 12.95 m/s

Calculate Q (flowrate in m3/s) from Q = 2πr 2 b2Wr2

Where from velocity triangle sin β2 = Wr2/W2 , Wr2 = 4 sin 30 = 2 m/s

Hence Q = 2 x 3.14 x 0.0875 x 0.015 x 2 = 16.5 x 10-3 m3/s.

Calculate V2: V22 = W2

2 + U22 – 2W2U2cosβ2 = 94; V2 = 9.7 m/s

Calculate Vθ2: From triangle Vθ22 = V2

2 – Wr22 = 94 – 4 = 90; Vθ2 = 9.5 m/s.

Repeat calculation for inlet values noting that Qinlet = Qdischarge = 16.5 x 10-3

m3

/s.

Results: U1 = 5.55 m/s; Wr1 = 2.8 m/s; W1 = 6.6 m/s; V1 = 2.8 m/s.

As Wr1 = V1 = 2.8 m/s there is no Vθ1 component

Torque calculation W T ( )e i M SHAFT e i

m rV rV θ θ ω ω = = −& &

22 ϑ V r mT &= where m kg/s5.16105.1610003 === − x xQ ρ &

T = 16.5 x 0.0875 x 905 = 13.7 Nm

And Power = ωT = 147 x 13.7 = 2 kW

pump lecture course notes 2005 hodkiewicz

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