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Hydrosystems Hydrulics open channel flow iit nptel _34_

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					Hydraulics                                                                                                                      Prof. B.S. Thandaveswara



                12.2 Problems
                There are three types of problems in critical flows as shown in table.

                         Type                        m                Q                       yc                  b or d
                           I                                          ?
                            II                                                                ?
                           III                                                                                       ?


                Types I and II are easy to solve.

                Type III problem requires a different approach.

                Type I problem
                                                         myc
                                                             is known.
                                                          b
                                                                                             Qm3/ 2
                                                         ∴ From the graph yc Vs Z =
                                                                                             b 2 gb
                                                         Q can be determined.
                Type II problem
                Here the solution is for obtaining critical depth. There are different methods. Graphically

                                   Qm3/ 2                                            myc
                             Z=        2
                                                   can be computed and value of          can be obtained from which
                                   b       gb                                         b
                            yc can be computed.

                Type III problem
                This problem can be solved using simultaneous solution of two algebraic equations

                which is illustrated below.

                Defining

                                   myc                                                       yc
                            Y1 =            ( for trapezoidal channel )         or                 ( for circular channel ) .
                                    b                                                        d0
                                           Qm3/ 2                                            Q
                            and X1 =           2
                                                                                or       2
                                           b       gb                                   d0    gd o

                Then

                                                                            myc
                                                                      b =
                                                                            Y1




Indian Institute of Technology Madras
Hydraulics                                                                                                           Prof. B.S. Thandaveswara



                                                                          Q m3/ 2                  Q m3/ 2 Y15 / 2
                                                            X1 =                          =
                                                                        2
                                                                   m 2 yc     g ( myc )            2
                                                                                              m 2 yc m1/ 2 y1/ 2 g
                                                                                                            c
                                                                    Y12             1/
                                                                                   Y1 2
                                                                       Q Y15 / 2
                                                               =        2
                                                                                     = M1Y15 / 2
                                                                   m   yc     gyc

                                                            X1 = M1Y15 / 2
                In which

                                 Q
                 M1 =            2
                            my   c       gyc
                 Given Q, yc , m,
                        Qm3 / 2                         myc
                 X1 =           ,              Y1 =
                        b 2 gb                           b

                              y'c (1 + y'c )
                                     3           3
                   2    3
                 Qm
                     =
                 gb5                 1 + 2y'c
                                 Q
                 M1 =         2
                                               and is known.
                            myc gyc
                 X1 = M1Y15 / 2
                Substitutingin the above equation
                                                     5/ 2
                 Qm3 / 2      ⎛ myc ⎞
                         = M1 ⎜     ⎟
                  2
                 b gb         ⎝ b ⎠
                 Qm3 / 2 ( myc ) gyc
                             2

                                                 = b 2.5 y1 / 2
                                                          5

                            Q g
                Solve the equation and obtain the solution for bed width b for trapezoidal channel.
                Similarly solve for diameter for the pipe line.




Indian Institute of Technology Madras
Hydraulics                                                                                                                            Prof. B.S. Thandaveswara


                Problem

                Compute the critical depth in a trapezoidal channel for flow of 30 m3 s-1. The channel

                bottom width is 10.0 m, side slope m =2. The bottom slope is negligible and α = 1



                                                                                             T
                                                               1                                                 1
                                                                       m                                        m y

                                                                                              b

                                                       .
                                                                              Trapezoidal
                Solution

                Given

                     Bottom width b = 10m
                     Sideslope      m =2
                     Flow             Q = 30 m3s -1
                                      α =1
                     Critical Depth y c = ?
                     For finding the critical depth,
                     Cross sectional area of the channel A= ( b + 2 y C ) * y C
                                                           = ( 10 + 2 y C ) * y C
                     Section factor                             Z=A D
                                                       in which D = A / T
                     for trapezoidal channel the top width T = ( b + 2m y C )
                                                           D = ( 10 + 2y C ) * yC /(10 + 2 * 2 * yC )
                     then the section factor                                       Z=A      ( 10 + 2 yC ) * yC / (10 + 2 * 2 * yC )
                                                                             30
                     by u sin g the equation                                      = 9.578
                                                                              A D =Q / g, =
                                                                             9.81
                     Substituting all the parameters A, P, T, D, and Q in the above equation and solving for y c one gets
                                                                   3
                                        ⎡(10 + 2 y c ) y c ⎤
                                        ⎣                  ⎦
                                                                       2
                                                           1
                                                                           = 9.578
                                           (10 + 4y c )        2

                                                                   3                               1
                                        ⎡(10 + 2 y c ) y c ⎤
                                        ⎣                  ⎦
                                                                       2   − 9.578 (10 + 4 y c )       2   =0

                                        ⎡(10 + 2 y c ) y c ⎤ − 91.743 (10 + 4 y c ) = 0
                                                                   3
                                        ⎣                  ⎦
                                        by trial and error, y c = 0.91 m




Indian Institute of Technology Madras
Hydraulics                                                                                           Prof. B.S. Thandaveswara


                Problems

                1. A trapezoidal channel with side slopes of 2 horizontal to 1 vertical is to carry a flow of

                16.7 m3/s. For a bottom width of 3.65 m, calculate (a) the critical depth and, (b ) the

                critical velocity.


                2. A rectangular channel carries 5.60 m3/s. Find the critical depth yc and critical velocity

                Vc for


                (a) a width of 3.65m and, (b) a width of 2.75m ,

                (c) What slope will produce the critical velocity in (a) if n = 0.020 ?

                3. Find the diacharge over a broad crested weir of 5.0m length and head 1.0m above

                the crest. Assume coefficient of discharge to be 0.9.




                                ___

                               V1
                                                                                Broad Crested Weir
                          H
                    y1
                                                                     yc

                                                         P

                                                                     B




Indian Institute of Technology Madras
Hydraulics                                                                                       Prof. B.S. Thandaveswara


                P is the height of weir, B is the breadth of the weir. Assume the approach velocity V1 to

                be very small.


                                        ⎛             3 yc            2                      ⎞
                                        ⎜ Answer: H =             ⎛Q⎞ 1
                                                           , yc = ⎜ ⎟   ,   Q = 0.544L gH3/2 ⎟
                                        ⎜               2         ⎝L⎠ g                      ⎟
                                        ⎝                                                    ⎠




Indian Institute of Technology Madras

				
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