Unsteady Flow

7/29/2019 Unsteady Flow

1/12

1 | P a g e

Unsteady Flow

Gangaram Sapkota,Department of Civil Engineering, Tezpur University,Napaam-784028

Email- [email protected]

1.1 Definitions

A wave is defined as a temporal (i.e., with respect to time) or spatial (i.e.,with respect todistance) variation of flow depth and rate of discharge. The wave length, L, is the distance

between two adjacent wave crests or troughs and the amplitude, z, of a wave is the height ofthe maximum water level above the still water level (Figure 1). Based on different criteria,wavesmay be classified into several categories. A wave is called oscillatory wave if there isno mass transport in the direction of wave travel and it is called translatory wave if there isnet mass transport. For example, sea waves are oscillatory waves and flood waves are

translatory. The translatory waves may be further classified as solitary or a wave train. Asolitary wave has a rising and a falling limb and has a single peak. A wave train is, however,a group of waves in succession. A translatory wavehaving a steep front is called a surge. A

positive wave has the water depth behind the wave higher than the undisturbed flow depth,and a negative wave has the flow depth behind the wave lower than the undisturbed flowdepth. A positive wave having a steep front is referred to as a bore, or ashock. The latter termis borrowed from gas dynamics.

As the wave passes a section, the entire flow depth is disturbed in ashallowwater wave whileonly the top layers, and not the entire section, are affected in a deep-water wave. The ratio ofwave length to the water depth is greater than 20 for shallow water waves; this ratio is lessthan 20 for the deep water waves.


2/12

2 | P a g e

1.2 Height and Celerity of a Gravity Wave

The wave celerity is defined as the relative velocity of a wave with respect to the fluid inwhich it is traveling, whereas absolute wave velocity is the velocity with respect to a fixedreference frame. For determining the celerity and the height of a gravity wave produced by a

sudden change in discharge we will derive some expressions. These expressions are generaland may be used for small or large amplitude waves. We will make the following simplifyingassumptions in the derivations:

1. The channel is frictionless and the channel bottom is horizontal (same equations areobtained if the component of the weight of the liquid in the downstream direction is equal tothe shear force acting on the channel sides and bottom);2. The pressure distribution on both sides of the wave front is hydrostatic;3. The velocity distribution is uniform on both sides of the wave front;4. The wave is an abrupt discontinuity of negligible length;5. The wave does not change in shape as it propagates in the channel; and

6. The water surface behind the wave is parallel to the initial water surface.

Let the flow be suddenly increased from Q1 to Q2 in a channel which increases the flowdepth fromy1 to y2, as shown in Figure 2. Let this wave be traveling at absolute velocity Vwin the downstream direction. We consider this direction as positive. We are interested indetermining the velocity of this wave and the wave height,y2 y1. These equations may bederived by using different methods. In the following paragraphs, we will use the controlvolume approach to the unsteady flow. However, we may convert the unsteady-flow situationto steady flow by applying velocity Vw on the entire system in the upstream direction. Thenwe may either use the control volume approach as in this section, or we may apply thecommon continuity and momentum principles to derive these equations. Let us consider a

control volume, as shown in Figure 2, in which the wave front has moved during timeintervalt, as shown.We will apply the Reynolds transport theorem [Roberson and Crowe,1997] in this section to derive the expressions for the wave height and the wave velocity.

Figure 2: Definition sketch


3/12

3 | P a g e

1.2.(a) Continuity equation

in which = mass density of water; V= volume of the control volume; V= flow velocity; and A = flow

area. By using subscripts 1 and 2 to denote quantities for sections 1 and 2 and subscript toand to+t, for quantities at these times, respectively, we may write

Since water may be assumed incompressible

= xRate of change of volume of liquid in the

control volume

By substituting Eqs. 3 and 2 into Eq. 1 and simplifying, we obtain

1.2. (b) Momentum equation

The intensive property for the momentum equation is = V. Therefore, according to the Reynoldstransport theorem

Referring to Figure 2,

And

in which

1

2

3

7

4

5

6


4/12

4 | P a g e

F= sum of the external forces on the control volume in the positivex-direction; y= depth of the

centroid of the flow section; and= specific weight of the liquid.The time rate of change of

momentum in the control volume is Substituting Eqs. 6 and 7 into Eq. 4, utilizing Eq. 3, and simplifyingthe resulting equation, we obtain

Elimination ofV2 from Eqs. 4 and 9 and the rearrangement of the resulting equation yield

The velocity of the surge wave, Vw, must be greater than the undisturbed flowvelocity, V1, for the wave to propagate in the downstream direction. Therefore,it follows from the above equation that

Transposing V1 to the left-hand side

As we defined earlier, the celerity, c, of a wave is the velocity relative to the fluid in which it istraveling, i.e., c= VwV1. Thus, the left-hand side of Eq. 12 is the celerity of the surge wave.

The height of the surge wave, y2y1, produced by a sudden change in discharge may be determinedfrom the following relationship between the flow depths and flow velocities at sections 1 and 2 (Fig.1),derived by eliminating Vwfrom Eqs. 4 and 11:

Let us assume that we know the values ofy1 and V1, orQ1. Then, for a specified change indischarge from Q1 to Q2, we can determine by trial and error the values ofy2 and V2 from Q2= V

2A

2and Eq. 12. The value of the wave velocity, Vw, may then be determined from Eq.11

for a wave traveling in the downstream direction. For a wave traveling in the upstream

8

9

10

11

12

13

14


5/12

5 | P a g e

direction, use a negative sign with the radical term. The preceding equations may be used forany cross section provided the simplifying assumptions we listed at the beginning of thissection are valid. For a rectangular channel, the above equations are simplified as follows.For a channel of widthB, A1 =By1;A2 =By2; y1 = 1 2y1, and y2 = 1 2y2.Substituting theserelationships into Eq. 11-18 and simplifying the resulting equation, we obtain

For waves of small height, y1_y2 = y(say). Then, Eq. 15 becomes

1.3 The St Venant Equations

1.3.(a) The derivation of the continuity equation

Figure 3

The following symbols are used in this derivation:A = the cross-sectional area of the sectionh = depth of flow at the section

z= elevation of surface above a datum at the sectionv = mean velocity at the sectionQ = discharge at the sectionb = width of the top of the section

x = position of the section measured from the upstream endt= time

g= acceleration due to gravity

= mass density of the fluid

16

15


6/12

6 | P a g e

Others symbols are defined in the text at the point when they are introduced.

Assuming that there is no lateral inflow, then

This has the partial derivative since Q is changing with bothx and time, t.Now the volume of water between the sections 1 and 2 is increasing as a rate of

where b is the top width,As cross-sectional areaA = bh then this is equivalent to

The terms are equal in magnitude but of opposite sign, so

1.3. (b) The derivation of the dynamic or momentum equation

By applying Newtons 2nd law to our elemental length of channel we have

Force= mass x acceleration


7/12

7 | P a g e

since v varies with both space (x) and time (t)

Consider the external forces which cause this acceleration. These are, in the simplest case,

three


8/12

8 | P a g e

1.4 Surges in Open Channel Flow

SURGES - MOVING (TRAVELLING FRONTS)

A surge is a moving wave front which results in an abrupt change of the depth of flow. It is arapidly varied unsteady flow condition. For example, consider the movement of a positivesurge wave in x-direction in an open channel having an irregular cross section as shown in

figure 4. Here, as the surge moves with an absolute velocity, Vw, flow depth becomes equalto y2 behind the surge. Undistributed flow depth ahead of the surge is y1. The correspondingflow velocities behind and ahead of the slope front are V2 and V1 respectively. The surge has

been created due to a sudden change of flow rate from Q1 to Q2. In this context, the problemdefinition for surge computation is: given Q1,y1,Q2 and channel slope parameters, determinethe surge wave velocity, Vw and the surge height, y2-y1. Equations for computing the aboveare based on the basic principles of conservation of mass and momentum. Followingassumptions are made in the derivation.

Figure 4: Definition sketch for surge movement


9/12

9 | P a g e

Assumptions Channel is horizontal and frictionless; Pressure distribution is hydrostatic at locations away from the front; Velocity is uniform within the cross section, at location away from the front; Change in the flow depth at the front occurs over a very short distance;

wave shape, height, and wave velocity do not change as the wave propogates inthe channel; water surfaces behind and ahead of the wave front are parallel to the bed.

Derivation of EquationsWe first choose a control volume encompassing the wave front. This control volume can bemade stationary by superimposing a constant velocity, Vw (equal to the absolute velocity ofsurge wave) in the negative x-direction. Thus the unsteady flow of Figure 4 may betransformed to steady flow fig. 5, and the principles of conservation of mass and momentumcan be applied to a steady flow situation.

Figure 5: Surge movement viewed as steady flow

Applying continuity equation to the control volume of figure 5,

( A2 V2Vw) A1(V1Vw) = 0

in which, = density of water; A2 = flow area behind the wave and A1 = flow areaahead of the wave. Since is a constant, Eq. 17 may be written as

Equation (18) can also be written as

Another way of writing the continuity equation is

Applying momentum equation to the control volume of Fig. 5,

17

18

19

20


10/12

10 | P a g e

The channel is prismatic, horizontal and frictionless. Therefore, the only force acting on thecontrol volume is pressure force. Pressure force acts in the positive x - direction at the inlet

section and in the negative x - direction at the outlet section. Equation (21) can be written as

in which y2 = depth to the centroid of inlet section of the C.V., and y1 = depth of thecentroid of outlet section.

Substitution of Eq. (18) in Eq (22) leads to

Substitution of Eq. (19) in Eq. (23) and subsequent simplification leads to

wave is propogating in the downstream direction.Therefore, Vw should be greater than V1.

Now, substitution of Eq. (20) in Eq. (23) and subsequent simplification leads to

21

22

23

24

25

26

27


11/12

11 | P a g e

Equations (26) and (27) can be used to determine the surge wave velocity and the surgeheight, if we know the values of undisturbed flow depth, y1, flow rate before the surge, Q1,and the flow rate after the surge, Q2.Equations (26) and (27) are non-linear equations. They can be solved by an appropriatenumerical technique. For rectangular channels, Eqs. (38.10) and (38.11) simplify to the

following.

28

29


12/12

12 | P a g e

1.5 References

1. Chaudhry.M.H., Open Channe Flow 2nd edition,2008

2.Murty B.S., IIT Madras, NPTEL

3. Sleigh P A, Goodwill I M, University of Leeds, March 2000

Unsteady Flow

Documents

Transcript of Unsteady Flow