Hydraulic Engineering Eng. Osama Dawoud
Jan 15, 2016
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