Hydraulic Engineering Eng. Osama Dawoud. .

Hydraulic EngineeringHydraulic

Engineering

Eng. Osama Dawoud

Eng. Osama Dawoud

http://www.haestad.com/library/books/awdm/online/wwhelp/wwhimpl/java/html/wwhelp.htm

Lecture 4

Head Losses in Pipelines Part2

Minor Losses

• Additional losses due to entries and exits, fittings and valves are traditionally referred to as minor losses

2

22

22 gA

Qk

g

Vkh LLm

Losses due to contraction

g

Vkh cc 2

22

A sudden contractionA sudden contraction in a pipe usually causes a marked drop in pressure in the pipe due to both the increase in velocity and the loss of energy to turbulence.

Value of the coefficient Kc for sudden contraction

VV22

Head losses due to pipe contraction may be greatly reduced by introducing a gradual pipe gradual pipe transition transition known as a confusor confusor

g

V'k'h cc 2

22

'kc

Losses due to Enlargement

g

VVhE 2

)( 221

A sudden EnlargementA sudden Enlargement in a pipe

Head losses due to pipe enlargement may be greatly reduced by introducing a gradual pipe gradual pipe transition transition known as a diffusor diffusor

g

VV'k'h EE 2

22

21

Loss due to pipe entranceGeneral formula for head loss at the entrance of a pipe is also expressed in term of velocity head of the pipe

g

VKh entent 2

2

Loss at pipe exit (discharge head loss)

In this case the entire velocity head of the pipe flow is dissipated and that the discharge loss is

g

Vhexit 2

2

Loss of head in pipe bends

g

Vkh bb 2

2

Loss of head through valves

g

VKh vv 2

22

Minor loss calculation using equivalent pipe length

f

DkL l

e

Example 1In the figure shown two new cast iron pipes in series, D1 =0.6m , D2 =0.4m length of the two pipes is 300m, level at A =80m , Q = 0.5m3/s (T=10oC).there are a sudden contraction between Pipe 1 and 2, and Sharp entrance at pipe 1.Fine the water level at B

e = 0.26mmv = 1.31×10-

6Q = 0.5 m3/s

exitcentffL

fBA

hhhhhh

hZZ

21

g

Vk

g

Vk

g

Vk

g

V

D

Lf

g

V

D

Lfh exitcentL 22222

22

22

21

22

2

22

21

1

11

01800170

000650000430600

26.0

102211018

sec98340

4

50sec771

604

50

21

11

6222

5111

222

211

.f .f

,.D

, .D

,.υ

DV R , .

υ

DVR

, m/..

π.

A

Q, V m/.

.π

.

A

QV

moodymoody

ee

1 ,27.0 ,5.0 exitcent hhh

Solution

m.g

.

g

..

g

..

g

. .

. .

g

. .

. .h f

36132

983

2

983270

2

77150

2

983

40

3000180

2

771

60

3000170

222

22

ZB = 80 – 13.36 = 66.64 m

g

Vk

g

Vk

g

Vk

g

V

D

Lf

g

V

D

Lfh exitcentL 22222

22

22

21

22

2

22

21

1

11

Example 2A pipe enlarge suddenly from D1=240mm to D2=480mm. the H.G.L rises by 10 cm calculate the flow in the pipe

Solution

smAVQsmV

g

V

g

VV

g

V

g

V

VV

VV

AVAV

g

VV

g

V

g

V

zg

pz

g

ph

g

V

g

V

hzg

V

g

pz

g

V

g

p

e

e

/103.048.057.0/57.0

1.02

6

1.02

4

22

16

4

48.024.0

1.0222

22

22

324222

22

2

222

22

2

21

242

241

2211

2

212

22

1

11

22

22

21

2

222

1

211

Solution

Power in pipelines

gQHQHPower

power) (horse HP 1 7.745

/.

Watt

WattsmN

mf

m

f

hhHγ Q

γ Q h

γ Q h

γ Q H

PowerExit At

lossminor todue dissipatedPower

friction todue dissipatedPower

Power EntranceAt

Calculate the max transported power through pipe line

f

π

π

f

mf

hg

V

D

f LH

g

V

D

f L HDγ

dV

dP

g

V

D

Lf HVDγ P

VAQhHγ Q

hhHγ Q

32

..3

2..30at Max.

2

P

lossminor neglect

PExit At

2

22

4

32

4

The max transported power through pipe line at 3

H h f

%67.661003max

H

HH

H

hHη

H

hhH

γQH

hhHγQη

f

mfmf

Efficiency in power transportation through pipelines

Example 3Pipe line has length 3500m and Diameter 0.5m is used to transport Power Energy using water. Total head at entrance = 500m. Determine the maximum power at the Exit. F = 0.024 fout h Hγ QP

mH

h f 3

500

3at Power Max.

g

V

..

g

V

D

Lfh f 230

35000240

2

22

m/s 3.417V

/s m...AVQ π 32

4 24150417330

HP.

tt) N.m/s (Wa

..

HgQ

HHgQ

hHγQP f

10597745

789785789785

500241508191000

3

32

32

Lecture 5

Pipelines in series & parallel

Pipelines in Series

nQQQQ 21

LnLLL ....hhhh 21

Pipelines in Parallel

n

iiQQ

1

LnLLLL ....hhhhh 321

Example 4الشكل التالي يوضح نظام مكون من أنابيب من الحديد المجلفن، األنبوب الرئيسي •

Gate، تم تثبيت صمام سكينة 2 و 1 م، بين الوصلتين 4 سم بطول 20قطره Valve سم بطول 12، األنبوب المتفرع قطره 2، عند نهايته مباشرة قبل الوصلة

وصمام منزلي. يتدفق 90o (R/D = 2.0) م. يتكون من وصالت مرفقية بزاية 6.4 10o/ث عند درجة حرارة 3 م0.26الماء عبر النظام بحيث يكون التدفق الكلي

مئوية، احسب التدفق في كل أنبوب عندما تكون الصمامات مفتوحة بالكامل.

Example 4

22

031402

20 m.

.πAa

22

011302

120 m.

.πAb

V.V.VAV A m. babbaa 0113003140260 3

g

V.

g

V

D

Lfh aa

a

aaa 2

1502

22

g

V

g

V.

g

V

D

Lfh bbb

b

bbb 2

102

19022

222

g

V.

.

.f

g

V.

.f b

ba

a 210380

120

46

2150

20

4 22

22 38103353 15020 bbaa V.f.V.f 0255.0

0185.0

b

a

f

f

22 3810025503353 1500185020 ba V...V.. ba V.V 7194

m/s.V

m/s.V

b

a

6301

6937

V.V.VAV A m. bbbbaa 01130)719.4(03140260 3

/s m...VAQ

/s m...VAQ

bbb

aaa

3

3

0180630101130

2420693703140