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Dimensional Analysis and Hydraulic Similitude - DOC

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Dimensional Analysis and Hydraulic Similitude - DOC Powered By Docstoc
					NPTEL Course Developer for Fluid Mechanics                                        Dr. Niranjan Sahoo
Module 04; Lecture 33                                                                  IIT-Guwahati

                          DYMAMICS OF FLUID FLOW

UNIFORM DEPTH OPEN CHANNEL FLOW
                                                      dy     
In an open channel, if the depth of flow is constant      0  , then it is treated as
                                                      dx     
“uniform depth”. In some cases, the uniform depth flow can be accomplished by
adjusting the bottom slope so that it precisely equals to the slope of the energy line.
Physically, the loss of potential energy of the fluid as it flows downhill is exactly
balanced by the dissipation of energy through viscous effects.
   Consider the flow in an open channel as shown in Fig. 1. The cross-sectional area is
constant in shape and size. If the cross-sectional area is A and the “wetted perimeter”
(i.e. length of perimeter of the cross section in contact with fluid) is P , then a new
parameter may be defined as “hydraulic radius” i.e.
                                                    A
                                             Rh                                                  (1)
                                                    P


       Free surface               a                       Free surface          Flow area (A)



  Flow in
  (Q)                                          Flow out
                                               (Q)                                Wetted
                              a
                                                                                  perimeter (P)
                                                               Section at a-a


                            Fig. 1: Concept of hydraulic radius.
   Since the fluid must adhere to the solid surface, the actual velocity distribution is not
uniform. The velocity is maximum on the free surface and becomes zero on the wetted
perimeter where the wall shear stress  w  is developed.


Chezy and Manning Equations
The fundamental relations used to determine the uniform flow rate in open channel are
semi-empirical and are governed by Chezy and Manning equations. Consider a control
volume flow with volume flow rate of Q and weight W in an open channel as shown in



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NPTEL Course Developer for Fluid Mechanics                                                   Dr. Niranjan Sahoo
Module 04; Lecture 33                                                                             IIT-Guwahati

Fig. 2. Since it is uniform depth flow, so it can be shown from continuity equation that
V1  V2 . Then, x -momentum equation for the control volume can be written as,

                                  F      x      Q V2  V1   0
                                                                                                           (1)
                              or, F1  F2   w Pl  W sin   0

                                                           l
                                              Control surface
                                                        = 0
                         F1
                                                                             F2
                    y2 = y1                                    wPl
                    V2 = V1                                                        
                                                                                       x
                                 (1)
                                                          W            (2)

                      Fig. 2: Control volume flow in an open channel.
As the flow is at uniform depth, so the hydrostatic pressure forces across either end of the
control volume balance each other i.e. F1  F2 . Thus, Eq. (1) becomes,

                                       W sin    Al  S0
                              w                           Rh S0                                       (2)
                                         Pl        Pl
where  is the specific weight of the fluid and sin   tan   S0 (since S0                        1) . More

often the open channel flows are turbulent rather that laminar. So, Reynolds number is
quite large and for such flows, the wall shear stress  w  is proportional to dynamic

pressure and may be written as
                                                          V 2
                                         w  K                                                            (3)
                                                            2
where K is a constant that depends on the roughness of the pipe. Now, equating Eqs. (2)
and (3), we get,
                                           V  C Rh S0                                                     (4)

This equation is known as “Chezy equation” in which C is called Chezy coefficient. Its
                                                                                  (m)1 2
value is determined from experiment and also, it has the unit of                         .
                                                                                    s
   This equation is modified by Manning by incorporating dependence of hydraulic
radius on the bottom slope and is given by


                                                                                                             2
NPTEL Course Developer for Fluid Mechanics                                         Dr. Niranjan Sahoo
Module 04; Lecture 33                                                                   IIT-Guwahati

                                            2    1
                                           Rh 3 S0 2
                                        V                                                       (5)
                                              n
In this equation, the parameter n is the Manning resistance coefficient. Its value depends
on the surface material of the channel’s wetted perimeter and is obtained from
                                                                    s
experiment. It is also not dimensionless and has the unit of                 . The typical values of
                                                                 m
                                                                        13



n are given in Table 1.
             Table 1: Values for Manning coefficient  n (Ref. 1; Table 10.1)

                             Wetted perimeter                   n
                              Natural channels           0.03 - 0.45
                                Floodplains              0.035-0.15
                          Excavated earth channel       0.022-0.035
                          Artificially lined channel     0.01-0.025


   In open channel flows, sometimes it is necessary to determine the best possible
hydraulic cross-section (i.e. minimum area) for a given flow rate  Q  , slope  S0  and

roughness coefficient  n . The flow rate can be written as,
                                              23
                                        A 1
                                     A   S0 2
                                   Q 
                                         P
                                                                                                 (6)
                                          n
which can be rearranged as,
                                                 3
                                          nQ  5
                                     A   1 2  P2 5                                            (7)
                                          S0 
For a channel with given flow rate the quantity in the parentheses is constant which
means the channel with minimum A is also with minimum P .


GRADUALLY AND RAPIDLY VARIED OPEN CHANNEL FLOW
The depth of the open channel flow varies (either increases or decreases) in the flow
direction depending upon the bottom slope and energy line slope. Physically, the
difference between component of weight and shear forces in the direction of flow


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NPTEL Course Developer for Fluid Mechanics                                     Dr. Niranjan Sahoo
Module 04; Lecture 33                                                               IIT-Guwahati

produces a change in fluid momentum. Thus, there is a change in velocity and
consequently a change in depth. The shape of the surface y  y  x  can be calculated by

solving the governing equation obtained form combination of Manning equation and
energy equation. The result will be a nonlinear differential equation, which is beyond the
scope of this syllabus. However, some physical interpretation of gradually varied flows
can be made from the following equation;
                                         dy S f  S0
                                                                                            (8)
                                         dx 1  Fr2

       dy                   S f  S0 
For        0 , the factor            becomes a non-zero quantity, which is essentially the
       dx                   1  Fr2 

                                                              dy
gradually or rapidly varying flow. Now, the sign of              i.e. whether the flow depth
                                                              dx
increases or decreases with distance along the channel depends on both numerator and
denominator of Eq. (8). The sign of denominator depends on whether the flow is sub-
critical or super-critical. In fact, for a given channel, there exists a “critical slope”

 S0  S0c    and a corresponding critical depth       y  yc    that leads to Fr  1 under

conditions of uniform flow.
      The character of a gradually varied flow is classified in terms of actual channel slope

 S0  compared to that of slope required for producing uniform flow  S0c  . They may be,
         Mild slope with S0  S0c (i.e. the flow would be sub-critical Fr  1 , if it were of
          uniform depth)
         Steep slope with S0  S0c (the flow would be super-critical Fr  1 , if it were of
          uniform depth)
         Horizontal slope with S0  0

         Adverse slope, S0  0 (i.e. flow uphill)
      Thus, it may be inferred from the above that the determination of whether the flow is
sub-critical or super-critical depends solely on whether S0  S0c or S0  S0c respectively.
For gradually varying flows, the conditions for sub-critical and super-critical flows are




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NPTEL Course Developer for Fluid Mechanics                                     Dr. Niranjan Sahoo
Module 04; Lecture 33                                                               IIT-Guwahati

met by S 0 and depth of flow. For example, with S0  S0c , it is possible to have either

Fr  1 or Fr  1 depending upon the depth of flow i.e. y  yc or y  yc respectively.

                                                                          dy
   A “rapidly varied flow” in an open channel is characterized by               1 i.e. the flow
                                                                          dx
depth changes occur over a relatively short distance. One such example of a rapidly
varied flow is “hydraulic jump” in which flow changes from a relatively shallow, high-
speed condition into a relatively deep, low-speed condition within a horizontal distance
of just a few channel depths. The mathematical analysis of “hydraulic jump” will be
discussed in the subsequent lectures. Many open channel flow-measuring devices are
based on the principle associated with rapidly varied flows. These devices include broad-
crested weirs, sharp-crested weirs and sluice gate.


Example 1
Water flows in an open channel of trapezoidal cross section with a velocity of 0.9m/s at a
rate 14m3/s. The bed slope and side slopes are 1:2500 and 1:1 respectively. Find the
depth and bottom width of the channel. Take Chezy’s constant as 40.4.
Solution
Given that,
                                                                     1
Discharge, Q  14m3 /s ; Velocity, V  0.9m/s ; Bed slope, S0           ; Side slope, N  1
                                                                    2500
                        Q
Area of the flow, A       15.57m 2
                        V

                                        V   1   0.9 
                                                   2                2

By Chezy’s formula, V  C Rh S0  Rh                  2500   1.24m
                                        C   S0   40.4 
The area of the flow and wetted perimeter for a trapezoidal section is given by,
                                A   b  Ny  y  15.57m 2
                               P  b  2 y N 2  1  b  2.83 y
By definition of hydraulic radius,
                           A   15.57
                    Rh                 1.24          b  2.83 y  12.55
                           P b  2.83 y
Thus,


                                                                                               5
NPTEL Course Developer for Fluid Mechanics                                      Dr. Niranjan Sahoo
Module 04; Lecture 33                                                                IIT-Guwahati

              A   b  Ny  y   b  y  y  15.57  12.55  2.83 y  y  y  15.57
               12.55 y  1.83 y 2  15.57
               y 2  6.86 y  8.5  0
Solving the above quadratic equation, we get, y  1.63m ; Thus, b  7.92 m .
Hence, the required depth and bottom width of the channel are 1.63m and 7.92m
respectively.


Example 2
A trapezoidal channel with base width 2m and side slope of 1:2 carries water with a depth
of 1m. The bed slope is 1 in 625. Calculate the discharge and average shear stress at the
channel boundary. Take Manning coefficient as n  0.03 .
Solution
Given that,
                                           1
b  2m; y  1m; N  1/ 2 ; S0              ; Side slope, N  1
                                          625
The area of the flow and wetted perimeter for a trapezoidal section is given by,
                                   A   b  Ny  y  2.5m 2
                                   P  b  2 y N 2  1  4.24m
By definition of hydraulic radius,
                                                A 2.5
                                         Rh           0.59
                                                P 4.24
Using Manning’s formula,
                                                    2
                            R 2 3 S 1 2  0.59  3 1 625 
                                                              0.5

                          V h 0                                    0.95m/s
                                n                0.03
Discharge, Q  A  V  2.375m3 /s
Average shear stress at the channel boundary,
                       gRh S0  1000  9.81 0.59  1/ 625  9.26 N m2




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NPTEL Course Developer for Fluid Mechanics                                                Dr. Niranjan Sahoo
Module 04; Lecture 33                                                                          IIT-Guwahati

Example 3
Water flows in channel (cross-section of the shape of isosceles triangle) of bed width a
and sides making an angle 450 with the bed. Determine the relation between depth of
flow d and the bed width a for the condition of: (i) maximum velocity; (ii) maximum
discharge. Use Manning’s formula.
Solution
Water flow in a channel of isosceles triangular cross-section is shown in the following
figure for which depth of flow and bed width are d and a , respectively.
                                               G



                                     H                             J            (a/2)


                                                                                  d


                    45 deg.                                                   45 deg.

                        E                                                         F
                               d                a                         d



                                                        2    1
                                                       Rh 3 S0 2
(i)        By Manning equation,               V
                                                          n
The area of the flow and wetted perimeter for the above cross-section is given by,

               EF  JH         a  2d   a                     d  A
            A           d                   d   a  d  d ;     a  2d
                  2                  2                           dd
                                              d P
            P  EF  FJ  EH  a  2 2d ;    2 2
                                              dd
For a given bed slope, the velocity will be maximum when,
                  d  Rh           d  A P     d                      A   d   P
                  d d        0;  dd  0;  P d                       A      0
                                                                     d    d   d 
Thus by substitution,                2.83d 2  2ad  a 2  0
Solving the above quadratic equation,                  d  0.34a


                                                       1
                                   23 12
                                            1 A  2
                                                   5       1
                         AR S                          3
(ii) Discharge, Q  AV 
                     .             h  0
                                             2  S0
                           n                n P 


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NPTEL Course Developer for Fluid Mechanics                                  Dr. Niranjan Sahoo
Module 04; Lecture 33                                                            IIT-Guwahati

For a given bed slope, the discharge will be maximum when,

                         
                        d A5 P 2      0;  5P d  A   2 A d  A   0
                                                                 
                             dd                  d d          d d 
               d P    d  A
Substituting       and   , we get,
               dd     dd

                                     22.63d 2  1.5ad  5a2  0
Solving the above quadratic equation,              d  0.44a


Example 4
A rectangular channel of 5m width and 1.2m deep has a slope of 1 in 1000 and is lined
with rubber for which the Manning’s coefficient is 0.017. It is desired to increase the
discharge to a maximum by changing the section so that the channel has same amount of
lining. Find the new dimensions and probable increase in discharge.
Solution
Using Manning’s formula, the discharge through the channel is given by,
                                                     2    1
                                                   ARh 3 S0 2
                                       Q1  AV 
                                             .
                                                      n
               1                                                        A
Here, S0         ; A  5 1.2  6m2 ; P  5   2 1.2   7.4m;  Rh   0.81
             1000                                                       P
Substituting the values, Q1  9.7 m 3 s

Let b and y be the width and depth of the flow for the new section of the channel. In
order to have the same amount of lining,
                                         P  b  2 y  7.4
For the discharge to be maximum in a rectangular channel, it can be proved that
                                               b  2y
Solving above two equation, y  1.85m; b  3.7m
So, the area of cross-section and hydraulic radius for new channel cross-section becomes,
                                                        A
                                  A  6.845m2 ; Rh        0.925m
                                                        P




                                                                                            8
NPTEL Course Developer for Fluid Mechanics                               Dr. Niranjan Sahoo
Module 04; Lecture 33                                                         IIT-Guwahati

                                                 2    1
                                               ARh 3 S0 2
By Manning’s formula, new discharge, Q2  AV 
                                           .               12.1m3 s
                                                  n
                                     Q2  Q 1
Percentage increase in discharge             100  24.45%
                                       Q1




                                        EXERCISES
1. Design a concrete lined channel (trapezoidal cross section) to carry a discharge of
480m3/s at a velocity of 2.3m/s. The bed slope and the side slope of the channel are
1:4000 and 1:1 respectively. Take Manning’s roughness coefficient for lining as 0.015.
2. A canal of trapezoidal cross-section has a bed width of 8m and bed slope of 1:4000. If
the depth of flow is 2.4m and side slopes are 1:3, then determine the average flow
velocity and discharge in the channel. Also, compute the average shear stress at the
channel boundary.
3. A trapezoidal canal is to carry 50m3/s of water with a mean velocity of 0.6m/s. One
side of the canal is vertical where as the other side has a slope of 1:3. Find the minimum
hydraulic slope if the Manning’s coefficient is 0.015.
4. Water flows in a channel as shown in the following figure. Find the discharge if
Chezys constant is 60 and bed slope is 1 in 1000.
                                         1.4m




                                                         0.7m




                                                         0.7m




5. Water flows in canal with bed slope of 1:2300. The cross-section of the canal is shown
in the following figure. Estimate the discharge when the depth of water is 2.5m. Assume
Chezy’s constant as 40.




                                                                                         9
NPTEL Course Developer for Fluid Mechanics                              Dr. Niranjan Sahoo
Module 04; Lecture 33                                                        IIT-Guwahati




                                                                                0.6m
   2.5m



                                                               140m
                         15m
                     Side slope = 1 vertical to 2 horizontal

6. A concrete lined circular channel of 0.6m diameter has a bed slope of 1: 500. Find the
depth of flow when the discharge is 0.3m3/s. Also, determine the velocity and flow rate
for conditions of: (i) maximum velocity; (ii) maximum discharge. Assume Chezy’s
constant as 40.




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