cu06997 lecture 6_exercises
TRANSCRIPT
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Exercise 1 calculation slope pipe
• V = 1 m/s
• D= 1 m , 50% filled
• λ = 0,022 [1]
• Equilibrium
• Calculate the bed slope of the pipe
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Darcy-Weisbach
2g
u
2g
u
4ΔΗ
22
f R
L
• ΔH = Head loss by friction [m]
• u2/2g = Velocity head [m]
• L = Length [m]
• λ = (lamda) = Friction coëfficiënt[1]
• ξ (ksie) = Loss coëfficiënt [1]
• R = hydraulic radius [m]
R
Lf
4
Total Head
Pressure Head
3
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Solution calculation slope pipe
• V = 1 m/s
• D= 1 m , 50% filled
• λ = 0,022 [1]
• Equilibrium
• Bed slope pipe??
mD
L11,0
20
1
1
100022,0
2g
VΔΗ
22
w
• In 100 m the pipe has to drop 0,11 m
• Bed slope 0,11/100 = 0,0011 or 1:909
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D=0,15 m D=0,30 m
Q=140 l/s
Exercise 2 :Calculate head loss and
pressure difference
p (left, center pipe) = 30.000 Pa, fresh water
g = 10
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Head loss Sudden Pipe Enlargement
2g
VVΔΗ
2
21
l
∆𝐻𝑙= (1 −
𝐴1
𝐴2)2∙
𝑉12
2𝑔
4
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∆p=∆p2-∆p1
∆p2
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∆p=3,14 – 0,20 – 1,75= 1,19 m
Solution
D=0,15 m D=0,30 m Q=140 l/s
u2/2g=3,14 m u2/2g=0,20 m
∆H=1,75 m
Datum/ ref line
Pressure head = 3m Pressure head = 4,19m
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D1=0.3m D2=0.15m
Q=0.14 m3/s
Exercise 3 Calculate head loss and
pressure difference
p (left, center pipe) = 50.000 Pa, fresh water
g = 10
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Head loss Sudden Pipe Contraction
∆𝐻𝑙= 0,44 ∙𝑉22
2𝑔
∆𝐻𝑙 = Head Loss due to sudden pipe contraction [m]
𝑉2 = Mean Fluid Velocity after sudden pipe contraction [m/s]
𝑔 = earths gravity [m/s2]
4
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D1=0.3m D2=0.15m Q=0.14 m3/s
Solution
u2/2g=0,20 m
u2/2g=3,14 m
∆p=0,20-3,14-1,38 = -4,32 m
∆H=1,38 m
Datum/ ref line
Pressure head = 5m
Pressure head = 0,68m
∆𝐻𝑙= 0,44 ∙𝑉22
2𝑔=0,44*3,14=1,38 m
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Exercise 4
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