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Hydraulic Engineering. Water Pump Part (C) System Characteristic Curve Q 1 > Q 2 V 1 > V 2 h f1 > h f2.

Dec 24, 2015

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Page 1: Hydraulic Engineering. Water Pump Part (C) System Characteristic Curve Q 1 > Q 2 V 1 > V 2 h f1 > h f2.

Hydraulic EngineeringHydraulic

Engineering

Page 2: Hydraulic Engineering. Water Pump Part (C) System Characteristic Curve Q 1 > Q 2 V 1 > V 2 h f1 > h f2.

Water PumpPart (C)

Page 3: Hydraulic Engineering. Water Pump Part (C) System Characteristic Curve Q 1 > Q 2 V 1 > V 2 h f1 > h f2.

System Characteristic Curve

H H ht stat L

Q1 > Q2

V1 > V2

hf1 > hf2

Page 4: Hydraulic Engineering. Water Pump Part (C) System Characteristic Curve Q 1 > Q 2 V 1 > V 2 h f1 > h f2.

Selected Pump

Q

hf

System Curve

Page 5: Hydraulic Engineering. Water Pump Part (C) System Characteristic Curve Q 1 > Q 2 V 1 > V 2 h f1 > h f2.

Q

hf

Operation Point

Page 6: Hydraulic Engineering. Water Pump Part (C) System Characteristic Curve Q 1 > Q 2 V 1 > V 2 h f1 > h f2.

System Characteristic Curve

H H ht stat L

Page 7: Hydraulic Engineering. Water Pump Part (C) System Characteristic Curve Q 1 > Q 2 V 1 > V 2 h f1 > h f2.

Selected Pump

Page 8: Hydraulic Engineering. Water Pump Part (C) System Characteristic Curve Q 1 > Q 2 V 1 > V 2 h f1 > h f2.

Elevated Tank

Page 9: Hydraulic Engineering. Water Pump Part (C) System Characteristic Curve Q 1 > Q 2 V 1 > V 2 h f1 > h f2.

Selected Pump

Page 10: Hydraulic Engineering. Water Pump Part (C) System Characteristic Curve Q 1 > Q 2 V 1 > V 2 h f1 > h f2.

System Curve & Pump Curve cases

Pump Curve

Pump Curve

Pump Curve

System Curve

System Curve

System Curve

Page 11: Hydraulic Engineering. Water Pump Part (C) System Characteristic Curve Q 1 > Q 2 V 1 > V 2 h f1 > h f2.

Example 1 A Pump has a cavitation constant = 0.12, this pump was instructed on well using UPVC pipe of 10m length and 200mm diameter, there are elbow (ke=1) and valve (ke=4.5) in the system. the flow is 35m3

and The total Dynamic Head Ht = 25m (from pump curve) f=0.0167Calculate the maximum suction head

m

m

2.0head pressureVapour

69.9head pressure atm.

Page 12: Hydraulic Engineering. Water Pump Part (C) System Characteristic Curve Q 1 > Q 2 V 1 > V 2 h f1 > h f2.

325120

120

.HσNPSH

tR

m.g

..

g

V.h S

V 28302

11154

254

22

06302

111

2

22

.g

.

g

Vh S

e

m.g

.

..

g

V

D

Lfh fS 0530

2

111

20

1001670

2

22

m.h

....h

γ

P

γ

Phhh(NPSH)

S

S

Vapor

Vapor

air

atmmSf SSA

0886

2069.90630283005303

Vapor

Vapor

air

atmmSf SSA γ

P

γ

Phhh(NPSH)

m/s .

.π.

A

QVS 111

204

03502

Page 13: Hydraulic Engineering. Water Pump Part (C) System Characteristic Curve Q 1 > Q 2 V 1 > V 2 h f1 > h f2.

Example 2For the following pump, determine the required pipes diameter to pump 60 L/s and also calculate the needed power.Minor losses 10 v2/2gPipe length 10 kmroughness = 0.15 mmhs = 20 m

Q L/s

706050403020100

Ht31353840.642.543.744.745

40536060575035-P

Page 14: Hydraulic Engineering. Water Pump Part (C) System Characteristic Curve Q 1 > Q 2 V 1 > V 2 h f1 > h f2.

To get 60 L/s from the pump hs + hL must be < 35 m

Assume the diameter = 300mmThen:

mh

fDKR

smVmA

f

Se

32.2362.193.0

85.010000019.0

019.0,0005.0/,1025.2

/85.0,070.0

2

5

2

m

gg

Vhm 37.0

2

85.010

2

10 22

mmhhh mfs 3569.43

Page 15: Hydraulic Engineering. Water Pump Part (C) System Characteristic Curve Q 1 > Q 2 V 1 > V 2 h f1 > h f2.

Assume the diameter = 350mmThen:

smVmA /624.0,0962.0 2

,48.10

0185.0,00043.0/,1093.1 5

mh

fDKR

f

Se

m

gg

Vhm 2.0

2

624.010

2

10 22

mmhhh mfs 3568.30

kWWHQ

Pp

ti 87.388.38869

53.0

3581.91000 100060

Page 16: Hydraulic Engineering. Water Pump Part (C) System Characteristic Curve Q 1 > Q 2 V 1 > V 2 h f1 > h f2.

Example 3A pump was designed to satisfy the following system Q (m3/hr)369

hf (m)122038

m

m

25.0head pressureVapour

3.10head pressure atm.

mhd 13

Pipe diameter is 50mm

g

VhL 2

24Partsuction

2

Check whether the pump is suitable or not

Page 17: Hydraulic Engineering. Water Pump Part (C) System Characteristic Curve Q 1 > Q 2 V 1 > V 2 h f1 > h f2.
Page 18: Hydraulic Engineering. Water Pump Part (C) System Characteristic Curve Q 1 > Q 2 V 1 > V 2 h f1 > h f2.

1- Draw the system curve and check the operation point

20m713hhH SdSTAT

Page 19: Hydraulic Engineering. Water Pump Part (C) System Characteristic Curve Q 1 > Q 2 V 1 > V 2 h f1 > h f2.

There are an operation point at:

Q = 9 m3/hr H =58m

NPSHR =4.1Then Check NPSHA

m.

g

.h

m/s..

π/

A

QV

L 022

27124

271050

4

36009

2

2

4.11.05(NPSH)

0.2510.327(NPSH)

γ

P

γ

Phhh(NPSH)

A

A

Vapor

Vapor

air

atmmSS SA

f

pump is not suitable, the cavitation will occur

Page 20: Hydraulic Engineering. Water Pump Part (C) System Characteristic Curve Q 1 > Q 2 V 1 > V 2 h f1 > h f2.
Page 21: Hydraulic Engineering. Water Pump Part (C) System Characteristic Curve Q 1 > Q 2 V 1 > V 2 h f1 > h f2.

Pumps in series & Parallel

Pumps in Parallel:

mnm3m2m1m

nj

1jn321

HHHHH

QQQQQQ

Page 22: Hydraulic Engineering. Water Pump Part (C) System Characteristic Curve Q 1 > Q 2 V 1 > V 2 h f1 > h f2.
Page 23: Hydraulic Engineering. Water Pump Part (C) System Characteristic Curve Q 1 > Q 2 V 1 > V 2 h f1 > h f2.

Pumps in series:

nj

1j

mimnm3m2m1m

n321

HHHHHH

QQQQQ

Page 24: Hydraulic Engineering. Water Pump Part (C) System Characteristic Curve Q 1 > Q 2 V 1 > V 2 h f1 > h f2.
Page 25: Hydraulic Engineering. Water Pump Part (C) System Characteristic Curve Q 1 > Q 2 V 1 > V 2 h f1 > h f2.

Change in pump speed (constant size)

Q

Q

N

N2

1

2

1

H

H

N

N2

1

2

1

2

Page 26: Hydraulic Engineering. Water Pump Part (C) System Characteristic Curve Q 1 > Q 2 V 1 > V 2 h f1 > h f2.

Example 4

Page 27: Hydraulic Engineering. Water Pump Part (C) System Characteristic Curve Q 1 > Q 2 V 1 > V 2 h f1 > h f2.

Solution

Page 28: Hydraulic Engineering. Water Pump Part (C) System Characteristic Curve Q 1 > Q 2 V 1 > V 2 h f1 > h f2.
Page 29: Hydraulic Engineering. Water Pump Part (C) System Characteristic Curve Q 1 > Q 2 V 1 > V 2 h f1 > h f2.

Home Work