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
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]
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 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
Head loss Sudden Pipe Enlargement




             V1  V2 
                      2
                          ∆𝐻 𝑙 = (1 −
                                        𝐴1 2 𝑉1
                                           ) ∙
                                                2
    ΔΗ l                               𝐴2     2𝑔
                 2g
4
∆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 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
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]
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
Cu06997 lecture 6_exercises
Cu06997 lecture 6_exercises

Cu06997 lecture 6_exercises

  • 1.
    Exercise 1 calculationslope pipe • V = 1 m/s • D= 1 m , 50% filled • λ = 0,022 [1] • Equilibrium • Calculate the bed slope of the pipe
  • 4.
    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]
  • 6.
    Solution calculation slopepipe • 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
  • 7.
    Exercise 2 :Calculatehead 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
  • 8.
    Head loss SuddenPipe Enlargement V1  V2  2 ∆𝐻 𝑙 = (1 − 𝐴1 2 𝑉1 ) ∙ 2 ΔΗ l  𝐴2 2𝑔 2g 4
  • 9.
  • 10.
    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
  • 11.
    Exercise 3 Calculatehead 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
  • 12.
    Head loss SuddenPipe 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]
  • 14.
    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
  • 15.