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					Tutorial for CIVL252
     Hydraulics
   By DUAN Huanfeng
     ceduan@ust.hk
                 Example 1
• Problem
  – rectangular channel with 6ft width.
  – y0, y1 are 65ft and 1ft, respectively
  – horsepower lost =? in the hydraulic jump



            y0




                     y1
• Solution
  – From section 0 to 1
       y0  v02 /2g  y1 v12 /2g (1)
        v0 ~  0             
                            v1 2g * (y0 - y1)  64.2ft/s
                           F1  v1/ g * y1  11.3
  – From section 1 to 2
              y2  (y1 /2)* ( 1  8F1 -1)  15.5ft
                                       2

  – Head loss (Method 1)
             H L  (y1- y2)3 /(4 y1 y2)  49.2ft
             PLoss  Q r H L / 550  2,150 horsepower
• head loss (Method 2)
  –   For section 1 and 2:
  –   V1=64.2ft/s; y1=1ft;
  –   V2=?; y2=15.5ft;
  –   According to continuity EQ:
      • V2=64.2*1/15.5=4.142ft/s
  – According to Energy EQ:
      • HL = (y1+V12/2g) - (y2+V22/2g)=49.2ft
            Example 2
• Problem
  – Water flows at a steady rate of 12 cfs
    per foot of width in the wide rectangular
    concrete channel shown. Pls determine
    the water surface profile from section 1
    to section 2.
• Solution
  – First, determine type of the flow regime.
  Assume n =0.015; the depth is h1.

   Q  1.49 / n AR S
                      2       1
                          3       2

                                         2            1
   12  (1.49 / 0.015)  h1  h1 3  (0.04)               2


   h1  0.74 ft     and               V  q / h1  16.23 ft / s
   Fr  V / gh1  3.33  1
   It' s     a    sup ercritical              flow!
• Second, determine whether a water jump
  occurs upstream of the weir.
 (1)    Sequent          depth   hs :
 hs  (h1 / 2)( 1  8 Fr 2  1)  3.13 ft
 ( 2)   Overhead           on    weir ( Assume   K  0.42 first)
                 3                               3
 Q  K 2 g Lhw  12  0.42 2  32.2  1 hw 2
                     2


                                    hw
 hw  2.33 ft  K  0.40  0.05         0.44
                                    P
 Re_ Compute : hw  2.26 ft
  P  hw  5.26  hs  3.13  water jump will occur 
 – Furthermore, determine the distance from
   the weir to where the hydraulic jump occurs.
x  [(h1  hs )  (V12  Vs2 ) / 2 g ] /(S f  S 0 )
and V1  q / h1  3.83 ft / s;Vs  2.16 ft / s
Vaverage  3.0 ft / s; Raverage  4.34 ft
S f  fVavg / 8 g / Ravg ( Darcy.EQ.) assume            f  0.015
             2


 S f  0.000121
 x  [(3.13  5.56)  (14.67  4.67) / 64.4] /(0.000121 0.04)
 x  57.0 ft
• Complete the water surface profile

				
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posted:1/30/2013
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