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# Cu06997 lecture 6_exercises

## on Mar 22, 2013

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## Cu06997 lecture 6_exercisesPresentation Transcript

• 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 2Darcy-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]
• 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
• Exercise 2 :Calculate head loss andpressure difference p (left, center pipe) = 30.000 Pa, fresh water g = 10 D=0,15 m D=0,30 m Q=140 l/s
• Head loss Sudden Pipe Enlargement V1  V2  2 ∆𝐻 𝑙 = (1 − 𝐴1 2 𝑉1 ) ∙ 2 ΔΗ l  𝐴2 2𝑔 2g4
• ∆p2∆p=∆p2-∆p1
• 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
• Exercise 3 Calculate head loss andpressure difference p (left, center pipe) = 50.000 Pa, fresh water g = 10 D1=0.3m D2=0.15m Q=0.14 m3/s
• Head loss Sudden Pipe Contraction4 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]
• 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
• Exercise 4