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

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

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

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

Head loss Sudden Pipe Enlargement

2g

VVΔΗ

2

21

l

∆𝐻𝑙= (1 −

𝐴1

𝐴2)2∙

𝑉12

2𝑔

4

∆p=∆p2-∆p1

∆p2

∆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

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

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

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

Exercise 4