hydraulic machinery 1

8/10/2019 Hydraulic Machinery 1

1/60

Monroe L. Weber-ShirkSchool of Civil and

Environmental Engineering

Hydraulic Machinery

Pumps, Turbines...
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2/60

Hydraulic Machinery Overview

Types of Pumps

Dimensionless Parameters for Turbomachines

Power requirements Head-discharge curves

Pump Issues

Cavitation

NPSH

Priming

Pump selection


3/60

Types of Pumps

Positive displacement

piston pump

peristaltic pump

gear pump two-lobe rotary pump

screw pump

Jet pumps

Turbomachines axial-flow (propeller pump)

radial-flow (centrifugal pump)

mixed-flow (both axial and radial flow)


4/60

Positive Displacement Pumps

Piston pump

Diaphragm pump

Peristaltic pump

Rotary pumps

gear pump

two-lobe rotary pump

screw pump


5/60


6/60


7/60

Rotary Pumps

Gear Pump

fluid is trapped between gear teeth and the

housingTwo-lobe Rotary Pump

(gear pump with two teeth on each gear

same principle as gear pump fewer chambers - more extreme pulsation

trapped fluid
http://www.koboldusa.com/onlinecatalog/flow/zdm/zdm1.html


8/60

Rotary Pumps

Disadvantages

precise machining

abrasives wear surfaces rapidlypulsating output

Uses

vacuum pumps

air compressors

hydraulic fluid pumps

food handling


9/60

Screw Pump

Can handle debris

Used to raise thelevel of wastewater

Abrasive materialwill damage theseal between screwand the housing

Grain augers usethe same principle


10/60

Positive Displacement Pumps

What happens if you close a valve on the

effluent side of a positive displacement pump?

What does flow rate vs. time look like for apiston pump?

0

0.2

0.4

0.6

0.8

1

1.2

0 2 4 6 8 10

rads

totalflow

1st piston

2nd piston

3rd piston

3 pistons

Thirsty Refugees


11/60


12/60

(inefficient process)

Turbomachines

impeller

casing housing

( ) ( )2 12 1z t tT Q r V rV r = -

Demours centrifugal pump - 1730

Theory conservation of angular momentum

conversion of kinetic energy to potential energy in flow

expansion ___________ ________

Pump components rotating element - ___________

encloses the rotating element and seals the pressurized

liquid inside - ________ or _________


13/60


14/60

Radial Pumps

ImpellerVanes

Casing

Suction Eye Impeller

Discharge

centrifugal

Flow Expansion

diameter rotational speed

also called _________ pumps

broad range of applicable flows and heads

higher heads can be achieved by increasing the_______ or the ________ ______ of the impeller


15/60


16/60

Dimensionless Parameters for

Turbomachines

We would like to be able to compare pumps withsimilar geometry. Dimensional analysis to therescue...

To use the laws of similitude to compareperformance of two pumps we need

exact geometric similitude

all linear dimensions must be scaled identically

roughness must scale homologous - streamlines are similar

constant ratio of dynamic pressures at correspondingpoints

also known as kinematic similitude

3D

Q

same


17/60

Kinematic Similitude:

Constant Force Ratio

VD

gl

V2

gl

V

lV2

V

c

viscous

gravity

surface-tension

elastic

Reynolds

ratio of inertial to _______ forces

Froude ratio of inertial to ________ force

Weber

ratio of inertial to _______ ______ forces

Mach

ratio of inertial to _______ forces


18/60

Turbomachinery Parameters

2

2C

V

pp

2

C

V

HgH

2

flowD

Q

A

QV

2

4

C

Q

HgDflowH

4

2 3, , ,

flow flow

H

impeller flow flow

HgD D QC f Re

Q D D D

e

w

D= =

impeller (Impeller is better defined)

3, , , , , ,

flow

p

impeller flow flow

D QC f Re F W M

D D D

e

w

=

Where is the fluid?


19/60

Shape Factor

Related to the ratio of flow passage

diameter to impeller diameter

Defined for the point of best efficiency

What determines the ideal shape for a

pump?

)(fS ,,, pQ

Exercise


20/60

Impeller Geometry:

Shape Factor

0.18

0.37

1.25

2.33

3.67axial: high _______, low _______

mixed

radial

mixed

S Q

gHp 3 4

pressure flow

pressureflow

S

500

1000

3400

6400

10000

N

43psp

H

QNN

*N in rpm, Q in gpm, H in ft

*

Impeller

diameter

Nsp= 2732S

Radial: high _______, low ____


21/60

Use of Shape Factor:

Specific Speed

The maximum efficiencies for all pumps occurswhen the Shape Factor is close to 1!

Flow passage dimension is close to impeller diameter!

Low expansion losses!

There must be an optimal shape factor given adischarge and a head.

Double suction (like two pumps in parallel) Multistage (pumps in series)

Use Q and H for each stage

S Q

gHp 3 4


22/60

Additional Dimensionless

Parameters

CH Hg

2D2

53

D

PCP

CQ Q

D3

S

CQ1 2

CH3 4

impeller

power

Alternate equivalent way

to calculate S.

D is the _______ diameter

P is the _____

(defined at max efficiency)

HQPw


23/60

Head-Discharge Curve

Theoreticalhead-dischargecurve

Actual head-

discharge curve

Q

H circulatory flow -inability of finite

number of blades to

guide flow

friction - ____

shock - incorrect angle

of blade inlet ___

other losses

bearing friction

packing friction

disk friction

internal leakage

V2

V2


24/60

Pump Power Requirements

HQPw

s

wP

P

Pe

m

sm

P

Pe

mP

m

ee

HQP

water

pump

shaftmotor

Water power

Subscripts

w = _______

p = _______

s = _______m = _______


25/60

Impeller Shape vs. Power Curves

S

1 - O.33

2 - 0.81

3 - 1.5

4 - 2.1

5 - 3.4

Discharge (% of design)Powe

r(%ofde

sign)

axial

radial

Implicationshttp://www.mcnallyinstitute.com/


26/60

Affinity Laws

With speed, , held constant:

3

2

1

2

1

P

P2

2

1

2

1

H

H

2

1

2

1

Q

Q

With diameter, D, held constant:

5

1 1

2 2

P D

P D

=

2

2

1

2

1

D

D

H

H3

1 1

2 2

Q D

Q D

=

CH Hg

2

D2

CQQ

D3

HQP

53

D

PCP


27/60


28/60

Pump Example

Given a pump with shape factor of 4.57, a

diameter of 366 mm, a 2-m head, a speed of

600 rpm, and dimensionless performancecurves (previous slide).

What will the discharge be?

How large a motor will be needed if motorefficiency is 95%?

Exercise

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0 0.02 0.04 0.06 0.08 0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

D=0.366 m

CH

Hg

2D

2

3D

QCQ

Efficiency

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0 0.02 0.04 0.06 0.08 0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

D=0.366 m

CH

Hg

2D

2

3D

QCQ

Efficiency


29/60

Pumps in Parallel or in Series

Parallel

Flow ________

Head ________

Series

Flow ________

Head ________

Multistage

adds

same

same

adds


30/60

Cavitation in Water Pumps

water vapor bubbles

form when the pressure

is less than the vaporpressure of water

very high pressures

(800 MPa or 115,000

psi) develop when thevapor bubbles collapse 0

1000

2000

3000

4000

5000

6000

7000

8000

0 10 20 30 40

Temperature (C)

Vaporpressure(

Pa)


31/60


32/60

Net Positive Suction Head

1

2

2

2s s v

R

p V pNPSH

gg g= + -

s = suction

Total head -pv!

NPSHRincreases with Q2!

Elevation datum

Absolute pressure

2

2

eye eyevR

p VpNPSH

gg g= - + At cavitation!

z

How much total head in excess of vapor pressure is available?


33/60

NPSHR

2

2atm v s s v

L

p p p V pz h

gg g g g - D - - = + -

2

2atm s s

reservoir L

p p Vz hgg g+ = + +

2

2atm s s

L

p p Vz h

gg g- D - = +

atm v

L A

p pz h NPSH

g g- D - - =

Subtract vapor pressure

2 2

1 1 2 21 2

2 2 L

p V p Vz z h

g gg g+ + = + + +


34/60

NPSH problem

Determine the minimum

reservoir level relative to the

pump centerline that will beacceptable. The NPSHrfor

the pump is 2.5 m. Assume

you have applied the energyequation and found a head

loss of 0.5 m.

18C

?

Exercise


35/60


36/60

Pumps in Pipe Systems

0

20

40

60

80

100

120

0 0.2 0.4 0.6 0.8Discharge (m3/s)

Head(m)

system operating point

Static head

Head vs. dischargecurve for ________pump

What happens as the static head changes (a tank fills)?


37/60

Priming

CH

p2D2

p CH2D2

CH

Hg

2D2

density

1.225 kg/m3

The pressure increase created is

proportional to the _______ of the fluid

being pumped. A pump designed for water will be

unable to produce much pressure

increase when pumping air

Density of air at sea level is __________

Change in pressure produced by pump is

about 0.1% of design when pumping air

rather than water!


38/60

Priming Solutions

Applications with water at less than atmospheric

pressure on the suction side of the pump require a

method to remove the air from the pump and the

inlet piping

Solutions

foot valve

priming tank vacuum source

self priming

foot valve

to vacuum pumppriming tank


39/60

Self-Priming Centrifugal Pumps

Require a small volume of liquid in the

pump

Recirculate this liquid and entrain airfromthe suction side of the pump

The entrained air is separated from the

liquid and discharged in the pressure side ofthe pump
http://www.gouldspumps.com/download_files/3796/3796_priming.stm


40/60

Variable Flows?

How can you obtain a wide range of flows?

__________________________

__________________________

__________________________

__________________________

__________________________

Why is the flow from two identical pumps

usually less than the 2x the flow from one

pump?

Valve

Multiple pumps (same size)Multiple pumps (different sizes)

Variable speed motor

Storage tank


41/60

RPM for Pumps

60 cycle

variable speed

belt drive

number of

poles sync full load rad/sec

2 3600 3500 367

4 1800 1750 183

6 1200 1167 1228 900 875 92

10 720 700 73

12 600 583 61

14 514 500 52

16 450 438 46

18 400 389 4120 360 350 37

22 327 318 33

24 300 292 31

26 277 269 28

28 257 250 26

30 240 233 24


42/60

Estimate of Pump rpm

The best efficiency is obtained when S=1

Given a desired flow and head the

approximate pump rpm can be estimated!

S Q

gHp 3 4

( )3 4

pgH

Qw


43/60

Pump Selection

Material Compatibility

Solids

Flow Head

NPSHa

Pump Selection software A finite number of pumps will come close to

meeting the specifications!


44/60

Pump Selection Chart

Model X

Model M

http://www.pricepump.com/
http://www.pricepump.com/http://www.pricepump.com/http://www.pricepump.com/http://www.pricepump.com/


45/60

End of Curve Operation

Right of the BEP

is sufficient NPSH available for the pump to operateproperly?

fluid velocities through the suction and dischargenozzles of the pump could be extremely high, resultingin increased pump and system noise

Left of BEP operation

high thrust loads on the pump bearings and mechanicalface seals result in premature failure.

The pump is oversized, resulting in lower efficiencyand higher operating and capital costs.

Q


46/60

Goulds Pump Curves

BEP = 1836 L/s

S Q

gHp 3 4

Splitcase double suction890 rpm = 93.2 rad/s

S=0.787

Check the Power!


47/60

Pump Installation Design

Why not use one big pump?

Can the system handle a power failure?

Can the pump be shut down for

maintenance?

How is the pump primed?

Are there enough valves so the pump can be

removed for service without disabling the

system?


48/60

Pump Summary

Positive displacement vs. turbomachines

Dimensional analysis

Useful for scaling

Useful for characterizing full range of pumpperformance from relatively few data points

Turbomachines convert shaft work into increasedpressure (or vice versa for turbines)

The operating point is determined by where thepump and system curves intersect

NPSH


49/60


50/60

Water problem?

Early in my college days I took a break and spent 17 months in Salvadoranrefugee camps in Honduras. The refugee camps were located high in themountains and for several of the camps the only sources of water large enoughto sustain the population of 6-10,000 were located at much lower elevations. So

it was necessary to lift water to the camps using pumps.

When I arrived at the camps the pumps were failing frequently and the pipeswere bursting frequently. Piston pumps were used. The refugees werecomplaining because they needed water. The Honduran army battalion wasnervous because they didnt want any refugees leaving the camp. There was

only one set of spare parts (valve springs and valves) for the pump and the lastset of parts only lasted a few days. The pump repair crew didnt want to startusing the pump until the real cause of the problem was fixed because spareparts have to be flown in from Miami.

Water problem:


51/60

Water problem:

proposed solutions?

piston pump (80 L/min)

2 km pipeline (2galvanized and then3 PVC) with rise of100 m


52/60


53/60

Pump Curve Solution

22D

HgCH

CQ

Q

D3

srevs

rev/8.62

2

60

min1

min

600

037.0366.0/8.62

/8.9222

2 ms

smmCH

068.0QC

3DCQ Q smmsQ /21.0366.0/8.62068.033

( )( )( )

( )( )

3 39800 / 0.21 / 25.55

0.78 0.95

N m m s mP kW= =

mP

m

ee

HQP


54/60

Pump Curve Solution

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0 0.02 0.04 0.06 0.08 0.10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

D=0.366 m

CH Hg

2

D2

CQ

Q

D3

Efficiency


55/60

NPSH solution

18C

?A RNPSH NPSH=

atm vl R

p pz h NPSH

g

-D = - - Papv 2000

3

/9789 mN

101300atmp Pa=

mmmN

PaPaz 5.25.0/9789

20001013003

mz 14.7

atm vA L

p pNPSH z h

g

-= - D -


56/60


57/60

Find Q

( ) ( )2 1

2 1z t tT Q r V rV r = -

2 2z tT QV r r=

1

V1

2

2gz1 hp

p2

V2

2

2gz2 hl

2 2z tT Q V r w r w=

2 2tz

p

V rTh

Q g

ww

g= =

2

22 2

22

tV r V

zg g

w

= +

work

Dimensional analysis

Neglect head loss

Datum is reservoir level

Let A = 10 cm2

2 2tz

V rT

VA g

ww

g=

H ld lift t


58/60

How could we lift water more

efficiently?

2

22 2

2

tV r Vz

g g

w= + D

2 2 22 2tQ A V r g z AV w= - D =

Solve for Q=AV

Decrease V without decreasing Q! (

vt

r

cs

p zh T

z Q z

w

g=

D D

2 22 2t

p z

A z V r g zz

h T

g w

w

D - DD=

2 2z tT Q V r w r w=


59/60

Lost energy

2

22 2

2

tV r Vz

g g

w= + D

2

22 2

2

tV r Vz

g g

w= + D


60/60

0.1

1

10

100

1000

0.0001 0.001 0.01 0.1 1 10

Flow (m3/s)

Pumpinghead(m)

Axial

Mixed

Radial

Positive

displacement

20

4060

10

200

400600

100

2000

40006000

1000

1 2 4 6

Selection of Pump Type

P

ower(kW)

hydraulic machinery 1

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