1.
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

2.
2 2
Darcy-Weisbach L u u
ΔΗ f      
4 R 2g 2g
Total Head
L
f  
Pressure Head
4R
• ΔH = Head loss by friction [m]
• u2/2g = Velocity head [m]
• L = Length [m]
• λ = (lamda) = Friction coëfficiënt[1]
• ξ (ksie) = Loss coëfficiënt [1]
3 • R = hydraulic radius [m]

3.
Solution calculation slope pipe
• V = 1 m/s
• D= 1 m , 50% filled
• λ = 0,022 [1]
• Equilibrium
• Bed slope pipe??
L V2 100 12
ΔΗ w      0,022    0,11m
D 2g 1 20
• In 100 m the pipe has to drop 0,11 m
• Bed slope 0,11/100 = 0,0011 or 1:909

4.
Exercise 2 :Calculate head loss and
pressure difference
p (left, center pipe) = 30.000 Pa, fresh water
g = 10
D=0,15 m D=0,30 m
Q=140 l/s

5.
Head loss Sudden Pipe Enlargement
V1  V2 
2
∆𝐻 𝑙 = (1 −
𝐴1 2 𝑉1
) ∙
2
ΔΗ l  𝐴2 2𝑔
2g
4

6.
∆p2
∆p=∆p2-∆p1

7.
Solution
∆H=1,75 m
u2/2g=3,14 m u2/2g=0,20 m
∆p=3,14 – 0,20 – 1,75= 1,19 m
Pressure head = 4,19m
Pressure head = 3m
Datum/ ref line
D=0,15 m D=0,30 m Q=140 l/s

8.
Exercise 3 Calculate head loss and
pressure difference
p (left, center pipe) = 50.000 Pa, fresh water
g = 10
D1=0.3m D2=0.15m
Q=0.14 m3/s

9.
Head loss Sudden Pipe Contraction
4
2
𝑉2
∆𝐻 𝑙 = 0,44 ∙
2𝑔
∆𝐻 𝑙 = Head Loss due to sudden pipe contraction [m]
𝑉2 = Mean Fluid Velocity after sudden pipe contraction [m/s]
𝑔 = earths gravity [m/s2]

10.
Solution ∆𝐻 𝑙 =
2
𝑉2
0,44 ∙ =0,44*3,14=1,38 m
2𝑔
∆H=1,38 m
u2/2g=0,20 m
u2/2g=3,14 m
∆p=0,20-3,14-1,38 = -4,32 m
Pressure head = 5m
Pressure head = 0,68m
Datum/ ref line
D1=0.3m D2=0.15m
Q=0.14 m3/s

11.
Exercise 4