Hydraulic Engineering Eng. Osama Dawoud. .
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Transcript of Hydraulic Engineering Eng. Osama Dawoud. .
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Hydraulic EngineeringHydraulic
Engineering
Eng. Osama Dawoud
Eng. Osama Dawoud
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http://www.haestad.com/library/books/awdm/online/wwhelp/wwhimpl/java/html/wwhelp.htm
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Lecture 4
Head Losses in Pipelines Part2
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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
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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.
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Value of the coefficient Kc for sudden contraction
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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
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Losses due to Enlargement
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VVhE 2
)( 221
A sudden EnlargementA sudden Enlargement in a pipe
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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
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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
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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
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Loss of head in pipe bends
g
Vkh bb 2
2
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Loss of head through valves
g
VKh vv 2
22
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Minor loss calculation using equivalent pipe length
f
DkL l
e
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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
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exitcentffL
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Solution
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3000180
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771
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3000170
222
22
ZB = 80 – 13.36 = 66.64 m
g
Vk
g
Vk
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Vk
g
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Lf
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Lfh exitcentL 22222
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11
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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
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Solution
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smAVQsmV
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pz
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Solution
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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
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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
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%67.661003max
H
HH
H
hHη
H
hhH
γQH
hhHγQη
f
mfmf
Efficiency in power transportation through pipelines
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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
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HP.
tt) N.m/s (Wa
..
HgQ
HHgQ
hHγQP f
10597745
789785789785
500241508191000
3
32
32
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Lecture 5
Pipelines in series & parallel
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Pipelines in Series
nQQQQ 21
LnLLL ....hhhh 21
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Pipelines in Parallel
n
iiQQ
1
LnLLLL ....hhhhh 321
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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.
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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
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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
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/s m...VAQ
/s m...VAQ
bbb
aaa
3
3
0180630101130
2420693703140