BFC21103 Chapter6.pdf

8/19/2019 BFC21103 Chapter6.pdf

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BFC21103 Hydraulics

Chapter 6. Hydraulic Machinery

Tan Lai Wai, Wan Afnizan & Zarina Md Ali

[email protected]

Updated: September 2014


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Learning Outcomes

At the end of this chapter, students should be able to:

i.

Calculate the

efficiency

of

pump

and

turbine;

ii. Determine the discharge and energy head of

pumps in

parallel

and

series;

and

iii. Carry out similitude analysis between model and

prototype of pump and turbine.

BFC21103 Hydraulics

Tan et al. ([email protected])


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6.1 Turbines

BFC21103 Hydraulics


Turbine is a hydraulic machine that utilises the energy of fluids to

move other types of machineries.

A common

use

of

turbine

is

in

the

hydroelectric power generation plant.


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Classification of Turbines

Based on the hydraulic action at the inlet, turbines can be classified as:

a. Impulse turbine (Pelton wheel or turbine) ‐ derives its energy from a jet of

water exiting out of a nozzle and shooting at the blades of turbine.

b. Reaction turbine (Francis turbine or Kaplan turbine) ‐ derives its power

from the equal and

Pelton wheel

opposite reactive

power of fluid

passing between its

blades.


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Based

on

the

direction

of

flow

through

the

runner,

turbines

can

be

classified

as:a. Tangential flow turbine (Pelton wheel)

b. Radial flow turbine (Francis turbine, Thomsen and Girard turbines)

c. Axial flow turbine (Kaplan turbine)

d. Mixed flow turbine (modern Francis turbine)

Pelton wheel Francis turbine Kaplan turbine

Radial

flow

turbine

Mixed

flow

turbine


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Based

on

the

head

of

water

H ,

turbines

can

be

classified

as:a. High head turbine (Pelton wheel, H > 250 m)

b. Medium head turbine (modern Francis turbine, 60 m ≤ H ≤ 250 m)c. Low head turbine (Kaplan turbine, H


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BFC21103 Hydraulics


Based on the specific speed N s, turbines can be classified as:

a. Low specific speed turbine (Pelton wheel, Ns of 10 to 35)

b. Medium specific speed turbine (Francis turbine, Ns of 60 to 400)

c. High specific speed turbine (Kaplan turbine, Ns of 300 to 1000)

Kaplan turbine


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6.2 Pumps

A pump is a hydraulic machine which supplies energy to fluid in

certain operation, e.g. in water distribution system.

Based on the mode of action of conversion of mechanical energy

into hydraulic

energy,

pumps

are

classified

as:

a. rotadynamic pumps (centrifugal pump) and

b. positive

displacement

pumps.


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Rotadynamic pump

Rotadynamic pumps consist of a rotating device known as an impeller. Fluids

to be pumped enters a casing near the shaft of the impeller. Vanes attached

to the spinning impeller increases the velocity of the pumped fluid and moves

the fluid out through an outlet. The most common rotadynamic pump are

centrifugal pump and propeller pump. Centrifugal pumps produce radial flow and mixed flow according to the fluid path. Propeller pumps which also

consist of an impeller produces axial flow.

Propeller pump

Centrifugal pump

Impeller Casing

Impeller eye

Suction

D i s c h a r g e


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6.2 Power and Efficiency of Pump

In pump operation, the mechanical energy through the shaft and impeller

is converted to fluid energy. The difference between the total head of

energy between the intake and discharge flanges of the pump is denoted

as

net

head

H developed

by

the

pump.

The

intake

end

(flow

inlet)

of

a

pump is commonly known as the suction end and the discharge (flow

outlet) of a pump is the delivery end.

⎥⎦

⎤⎢⎣

⎡++−⎥

⎦

⎤⎢⎣

⎡++=−=

g

V z

p

g

V z

pHHH ss

sd d

d sd

22

22

γ γ

Delivery

Suction

P

D

S

ps

pd

zd

, V d

zs, V s


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Discharge through a pump is given as

fd d d fsss V BDV BDQ π π ==

where, Bs, Bd = widths of the runner at the suction and delivery ends

V f

= velocity of flow through the runner

Power of pump at suction end Ps (or input power Pi )

( )ussud d i s V uV uQT PP −=== ρ ω

Power delivered at the discharge end Pd (or output power Po)

QHPP od γ ==

where, V u = swirl velocity or the rate of shaft rotation in radians per sec


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Efficiency of a pump is given as

shaft the into power

fluidthetodeliveredpower=η

PPi

o=η

Hmη η η η ∇=Also,

where, η ∇ = volumetric efficiency

η m = mechanical efficiency

η H = hydraulic efficiency


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A centrifugal

pump

is

needed

to

supply

23

m3/s

of

water

for

a city.

This operation will utilise a net head H = 20 m and specific speed N =

450 rpm. If the inflow power Ps is 5000 kW and density of water ρ is

1000

kg/m

3

at

5°

C,

calculate:(a) Output power Pd

(b) Overall efficiency of the pump η

Activity 6.1


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Given Q =

23

m

3

/s, H =

20

m,

N =

450

rpm,

Pi =

5000

kW,

ρ = 1000

kg/m

3

(a) Output power

kW6.451220239810 =××== QHPo γ

(b) Efficiency of the pump

%25.909025.05000

6.4512====

P

P

i

oη


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BFC21103 Hydraulics


6.3 Characteristic Curves of Pump

The characteristic curves of a pump is usually provided by the pump manufacturer through laboratory tests.

Characteristic curves

of

a typical

mixed

‐flow

centrifugal

pump

0

20

40

60

80

100

0 3 6 9 12 15 18

Capacity Q ('000 gpm)

H e a

d H

( f t ) a n d p u m p

e f f i c i e n c y

( % )

0

100

200

300

400

500

H o r s e p o w e r ( k W )

This pump has a normal

capacity or rated capacity

of 10,500 gpm when developing a normal head

of 60 ft at an opening

speed of 1450 rpm.

BEP

H e ad

E f f i c

i e n c y

B r a k e

h o r s e p o w e r

W a t e r

h o r s e p

o w e r


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Relationship between

input

power

Pi ,

efficiency

η and head

H starts

when intake valve is closed, and the impeller spins until pressure at

output increase to the maximum head (shut‐off head). When the valve is

open, water will flow through the pipe and the head of pump will

decrease. With

addition

of

flow

rate,

the

pump

efficiency

will

increase

until it reach a maximum and then decrease to end of operation.

Intersection between head and power corresponds to the point of

optimum efficiency is the best point to use pump (known as the best

efficiency point, BEP).


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6.4 Cavitation

An important factor in the satisfactory operation of a pump is the avoidance of cavitation, both for good efficiency and for prevention of

impeller damage.

As liquid

passes

through

the

impeller

of

a pump,

there

is

a change

in

pressure. If the absolute pressure of the liquid drops to the vapour

pressure, cavitation will occur. The region of vaporization hinders the

flow and places a limit on the capacity of the pump.

As the fluid moves further into a region of higher pressure, the bubbles

collapse and the implosion of the bubbles may cause pitting of the

impeller.

Cavitation is

most

likely

to

occur

near

the

point

of

discharge (periphery)

of radial flow and mixed flow impellers, where velocities are highest. It

may also occur on the suction side of the impeller, where the pressures

are the lowest.


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6.5 Pumps in Parallel

If two

similar

pumps

A

and

B

are

connected

in

parallel,

the

combined

discharge is the sum of individual discharges QA and QB.

However, the pressure head H remains the same as in single pump.

The overall

system

power

requirement

is

the

sum

of

the

power

required

for each pump at the same head.

A

B

QA, HA

QB, HB

BAT QQQ +=

BAT HHH ==BAT PPP +=

( )( )BA

TBAT

PPK HQQ

++=η

K = unit constant (= 0.102 for P in kW and Q in m3/s)

(= 6,116

for

P in

kW

and

Q in

L/min)


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BFC21103 Hydraulics


Determine the

system

discharge,

power

requirements

and

efficiency

if

the following pumps are operated in parallel against a head of 27.4 m.

Activity 6.2

Pump A Pump B

QA = 0.0473

m3/s QB =

0.052

m3/s

E A = 83% E B = 73%

PA = 15.3 kW PB = 19.1 kW


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BFC21103 Hydraulics


Given H = 27.4 m

Pump A Pump B

QA = 0.0473 m3/s QB = 0.052 m

3/s

E A = 83% E B = 78%

PA = 15.3 kW PB = 19.1 kW

/sm 0993.0052.00473.0 3

BAT =+=+= QQQ

m 4.27BAT === HHH

kW4.341.193.15BAT =+=+= PPP

( )( )

%54.777754.04.34102.0

4.270993.0

BA

TBAT ==

×

×=

+

+=

PPK

HQQη


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BFC21103 Hydraulics


6.6 Pumps in Series

If two similar pumps A and B are connected in series, the combined

discharge is the same discharge as single discharge QT = QA = QB.

The pressure head H produced is the combined head produced by the

two pumps.

The overall system power requirement is the sum of the power required

for each pump at the same head.

A B

QA, HA QB, HB

BAT QQQ ==

BAT HHH +=

Pumps connected in series

BAT PPP +=( )( )BA

BATT

PPK

HHQ

+

+=η


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BFC21103 Hydraulics


Calculate the

discharge,

pressure

head,

total

power

required

and overall

pump system efficiency for two pumps linked in series. The pumps are

operated at 1750 rpm and the system design discharge is 0.0473 m3/s.

The impeller diameters are DA = 25.4 cm and DB = 30.5 cm.

Activity 6.3

Pump A Pump B

HA = 26.8 m HB = 33.5 m

E A = 83% E B =

78%

PA = 15 kW PB = 19.9 kW


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BFC21103 Hydraulics


Given Q = 0.0473 m3/s

/sm 0473.0 3

BAT === QQQ

m3.605.338.26BAT =+=+= HHH

kW9.349.1915BAT =+=+= PPP

( )( )

%12.808012.09.34102.0

3.600473.0

BA

BATT ==

×

×=

+

+=

PPK

HHQη

Pump A Pump B

HA = 26.8 m HB = 33.5 m

E A = 83% E B = 78%

PA = 15 kW PB = 19.9 kW


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BFC21103 Hydraulics


6.7 Similitude for Pumps and Turbines

Similarity laws help to interpret the results of model studies. The relation

between model and prototype is classified into 3, i.e.:

a. Geometric similarity ‐ prototype and model have identical shapes but

differ in

size.

b. Kinematic similarity ‐ ratio of velocities at all corresponding points in

flow are the same and involve length and time.

c. Dynamic similarity

‐ two

systems

have

dynamic

similarity

if,

in

addition to dynamic similarity, corresponding forces are in the same

ratio in both.


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Scale Ratio

Model (m) ‐ similar with object/structure required in certain scale ratio.

‐ tested in laboratory and similar in real phenomenon.

‐ not always smaller than the prototype.

Prototype (p)

‐ actual

object/structure

‐ tested in actual condition, e.g. hydraulic structures, ship

etc.

1. Performances of object/structure can be predicted.

2. Economical and

easier

to

build,

where

design

of

model

can

be

reproduced many times to achieve the desired design.

3. Non‐functional structure such as dam can also be measured.

Advantages of Similarity


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Similarity in Pump

In similarity relations, the basic repeating variables are the rotational

speed N and pump diameter D. Therefore, the similitude laws for head

H, discharge Q and power P can be expressed as

22DN

HC H = 3

ND

QC Q = 53

DN

PC P

ρ =

2

p

2

p

p

2

m

2

m

m

DN

H

DN

H=

Applying similitude laws between model and prototype:

3

pp

p3

mm

m

DNQ

DNQ =

5

p

3

pp

p

5

m

3

mm

m

DN

P

DN

P

ρ ρ

=


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BFC21103 Hydraulics


where, N in rpm, Q in m3/s and H in m.

Two homologous

pumps

have

the

same

specific

speed

Ns.

Thus

between

a geometric model and its prototype,

4

3

H

QNNs =

pm ss NN =

Specific speed

4/3

p

pp

4/3

m

mm

H

QN

H

QN=


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BFC21103 Hydraulics


Two homologous

pumps

A

and

B

are

operating

at

the

speed

of

600

rpm.

Pump A has an impeller with diameter 50 cm and discharges 0.4 m3/s of

water under a net head of 50 m. Determine the size of pump B, net

head, and the specific speed if it is to discharge 0.3 m3/s.

Activity 6.4


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BFC21103 Hydraulics


From the law of similarity

3

BB

B

3

AA

A

DN

Q

DN

Q=

cm 43.45m 4543.05.04.0

3.0

600

600 31

33

1

3

A

A

B

B

AB ==⎟

⎠

⎞⎜⎝

⎛ ××=⎟⎟ ⎞

⎜⎜⎝

⎛ = D

Q

Q

N

ND

Given N

=

600

rpm, D

A =

50

cm, Q

A =

0.4

m

3

/s, H

A =

50

m, Q

B =

0.3

m

3

/s

and

2

B

2

B

B

2

A

2

A

A

DN

H

DN

H=

m 28.415.0

4543.0

600

60050

2

2

2

2

2

A

2

B

2

A

2

BAB =××==

D

D

N

NHH


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BFC21103 Hydraulics


Specific speed is

4

3

H

QNNs =

18.20

28.41

3.0600

4

3

4

3

B

BBB ===

H

QNNs

B

4

3

4

3

A

AAA 18.2050

4.0600ss N

H

QNN ====

It can be shown that


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BFC21103 Hydraulics


Similarity in Turbine

The characteristic relations between a turbine model and its prototype

can be expressed as

H

ND

C H = 3ND

Q

C Q = 53DN

P

C P =

p

pp

m

mm

HDN

HDN =

Applying similitude laws between model and prototype of turbine:

3pp

p

3mm

m

DN

Q

DN

Q

=

5

p

3

p

p

5

m

3

m

m

DN

P

DN

P=


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BFC21103 Hydraulics


where, N in rpm, P in kW, and H in m.

Two homologous

turbines

have

the

same

specific

speed

Ns. Thus

between a geometric model and its prototype,

4

5

H

PNNs =

pm ss NN =

Specific speed

4/5

p

pp

4/5

m

mm

H

PN

H

PN=


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BFC21103 Hydraulics


A 1:5

model

of

water

turbine

develops

2 kW

of

power

at

400

rpm

under

a head of 3 m. Find its specific speed?

Assuming the overall efficiency of 0.85 for both the model and

prototype, calculate the rotational speed, power and discharge of the

prototype when run under a head of 20 m.

Activity 6.5


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BFC21103 Hydraulics


Given 1:5 model turbine, Pm = 2 kW, Nm = 400 rpm, Hm = 3 m, η = 0.85, Hp = 20 m

3.143

3

2400

4

5

4

5

m

mmm ===

H

PNNs

For prototype,

rpm 6.2063

20

5

1400

m

p

p

mmp =××==

H

H

D

DNN

p

pp

m

mm

H

DN

H

DN=

kW 2.86121

5

400

6.206 53

m5

m

5

p

3

m

3

p

p =×⎟ ⎠

⎞⎜⎝

⎛ ×⎟ ⎠

⎞⎜⎝

⎛ == PD

D

N

NP

5

p

3

p

p5

m

3

m

m

DN

P

DNP =


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BFC21103 Hydraulics


Specific speed

for

prototype, m

4

5

4

5

p

ppp 3.143

20

2.8616.206 ss N

H

PNN ====

Power

developed

by

prototype kW

2.861op =P

kW 2.101385.0

2.861 i ===

η

oPP

at

efficiency

η =

0.85.

Power supplied by water

3

ppp 102.1013 ×=HQγ

/sm

164.5209810102.1013102.1013 3

3

pp

3

p =××=×=

HQ

γ

Thus,


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Performance of Turbines under Unit Quantities

The unit quantities give the discharge, speed, and power for a particular

turbine under a head of 1 m assuming the same efficiency.

(a) Unit discharge

Qu

‐ the

discharge

of

a turbine

working

under

a unit

head.

Between 2 similar turbines,

H

QQ

u =

2

2

1

1

H

Q

H

Q=


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BFC21103 Hydraulics


(b) Unit speed N u ‐ the speed of a turbine working under a unit head.

Between 2 similar

turbines,

2

2

1

1

H

N

H

N

=

H

NNu =

(c) Unit power

P u ‐ the power developed by a turbine working under a

unit head.

Between 2 similar turbines,

23

H

PPu

=

2

3

2

2

2

3

1

1

H

P

H

P=

A i i 6 6


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BFC21103 Hydraulics


A Francis

turbine

produces

6750

kW

of

power

at

300

rpm

under

a net

head of 45 m with an overall efficiency of 85%. Determine the revolution

per‐minute (rpm), discharge and brake power of the same turbine under

a net head of 60 m in homologous conditions.

Activity 6.6


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BFC21103 Hydraulics


Given P1 = 6750 kW, N1 = 300 rpm, H1 = 45 m, η = 85%, H2 = 60 m

2

2

1

1

H

Q

H

Q =

2

2

1

1

H

N

H

N=

rpm

4.34645

60

3001

2

12 === H

H

NN

η

γ o

i

PQHP ==

/sm 99.1745981085.0

10006750 31 =

××

×==

H

PQ o

ηγ

/sm 77.20

45

6099.17

3

1

212 ===

H

HQQ


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BFC21103 Hydraulics


2

3

2

2

2

3

1

1

H

P

H

P

=

kW

3.1039245

606750

2

32

3

1

212 =⎟ ⎠

⎞⎜⎝

⎛ ×=⎟⎟ ⎠

⎞⎜⎜⎝

⎛ = H

HPP

Assignment #6


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BFC21103 Hydraulics


Assignment #6

Q1. What are the functions of hydraulic pumps and turbines?

Q2. 0.5 m3/s of water is to be pumped to a total head of 250 m. How

many pumps

connected

in

series

should

be

required

if each

pump

has

a specific speed of 35 and speed of 1500 rpm.

Q3. A turbine

develops

8500

kW

under

a head

of

18

m

at

150

rpm.

Calculate

(a) specific speed

(b) normal speed under a head of 25 m

(c) output under

a head

of

25

m


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BFC21103 Hydraulics


Q4. A centrifugal

pump

has

an

impeller

of

200

mm

with

capacity

400

L/s

at speed 1200 rpm against a head of 12 m. Calculate the speed and

head of a geometrically similar pump with impeller diameter of 300

mm which is required to deliver 700 L/s.

Q5. A turbine is to operate under a head of 28 m at 185 rpm. The

discharge is 10 m3/s. If the efficiency is 87%, determine the

performance (N,

Q,

P)

of

the

turbine

under

a head

of

20

m.

‐ End of Question ‐


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YOU

BFC21103 Hydraulics


Documents

BFC21103 Chapter6.pdf