Basic equations unsteady flow in rectangular channel

The continuity eqution.

+

= 0

The Momentum Equation

+

+

= ( Sx - Sf )

unsteady flow in nonprismatic rectangular channel can be expressed by the continuity and

the corservation of momentum equations, respectively:

+ (

) +

+

= 0 (1)

+

+

= ( So - Sf ) (2)

The method of characteristics may be described as a technique whereby the problem of

solving two simultaneous partial differential equations can be replaced by the problem of

solving four ordinary differential equations.

The basic equations of partial diffrentiation:

dh =

dx +

dt

+

Multiplying equation (1) by and adding to equation (2) result in

*

+ + * (

)

+ ( So - Sf ) (3)

or

*

+ + [ (

) (

)

] ( So - Sf ) (4)

=

(5)

The solution of the quadratic equation (5) for is

= √

(6)

Thus, inserting equation (6) in equation (5) gives

√

= U C (7)

In which C = celerity of a disturbance in the fluid when U = 0. Inserting equation (7) in

equation (4) lead to the characteristic equations:

+

= ( So - Sf ) -

C (

) (8)

Along the forward characteristic C+

U + C (9)

-

= ( So - Sf ) -

C (

) (10)

Along the backward characteristic C-

U - C (11)

The simple-wave problem

The equations for simplest possible situation involving unsteady flow.

Set So = Sf = (

) = 0

= 0

C = √

√

= 0

Integrate the equation

U 2 √ = 0

U 2 C = 0

Along the forward characteristic C+ and along the backward characteristic C

-

U C

THEOREM. If So = Sf = 0, and if any one curve of the C1 or C2 family of characteristics is a

straight line, the so are all other members of the same family.

To prove the theorem we consider the C1 line AB, DE, in Fig. DE is a straight line, and we

are to prove that AB is also straight.

Both ( u + c ) and ( u + 2 c ) are constant along DE, so that their difference c must be

constant, and hence u also. It follows that cD = cE , uD = uE ;

Also we can write, from the C2 characteristics AD and BE:

uA – 2 cA = uD – 2 cD

uB – 2 cB = uE – 2 cE (12)

whence uA – 2 cA = uB – 2 cB (13)

and since AB is C1 characteristics, we have

uA + 2 cA = uB + 2 cB (14)

Equations (13) and (14) can satisfied only if cA = cB , uA = uB , i.e., if AB is a straight line.

Positive and Negative Waves; Surge Formation

Negative Wave.

If the disturbance is negative, ct decreases as we move up to the t axis. It follows that the

slopes of the C1 characteristics increase as t increases, i,e, that the characteristics diverge

outwards from the t axis. From this it follows that the negative wave are dispersive, i.e., that

sections having a given difference in depth move further apart as the wave moves outward

from its point of origin.

Consider any point G on vertikal axis, of ordinate t; the C1 characteristic issuing from this

point will have an inverse slope to

= u(t) + c(t) (15)

Drawing a C2 characteristic from G to OF, the result

u(t) - 2 c(t) = u0 - 2 c0 (16a)

u(t) = 2 c(t) + u0 - 2 c0 (16b)

From equationn15 and 16 it follows that the inverse slope of the C1 characteristic issuing

from G can be expressed in either of the two forms:

=

u(t) -

u0 + c0 (17 a)

= 3 c(t) + u0 - 2 c0 (17 b)

The above equation gives the speed at which a section of constant depth h moves.

Since for a given h , this speed is constant, we may replaced

by x/(t-t1), where t1 is the

value of t at x = 0 ( as at the point G. Eq. ( 17) can then be rewritten as

√ √ (18)

Example.

Water flows at a uniform depth of 1,5 m and velocity of 0,85 m/s in a channel of rectangular

section, into a large estuary, The estuary level, inially the same as the river level, falls at the

rate 0,3 m/hr for 3 hr; neglecting bed slope and resistance, determine how long it takes for the

river level to fall by 0,6 m at section 1600 m upstream from the mouth. At this time, how far

upstream will the river level just be starting to fall.

Taking x positive upstream, we find,

u0 = 0,85 m/s

c0 = √

whence u0 + c0= (- 0,85 + √ ) m/s

The accompanying sketch of the x-t plane shows how the solution is arrived at.

We are seeking the point H, at which x = 1600 m, and the depth ( 1,5 - 0,6 = 0,9 m),

c = √

consider the characteristic GH, along which c is constant and equal to √ ;

it originates at the point G, representing a depth of 0,9 m at the river mouth. This means that

at G, t = 2 hr = 7200 sec.

We now require the inverse slope of GH, From eq. 17 b, it is equal to

= 3 c(t) + u0 - 2 c0= 3 √ - 0,85 – 2 √

=....... (m/s)

Hence the time interval between G and H is equal to:

time from G to H = ( distance ) / (

)

and the total time elapsed is therefore equal to

2 hr + time from G to H,

And the distance upstream to K, on the boundary of the zone of quiet at this time is equal to

(2 hr + time from G to H ) x ( u0 + c0 )

Positive wave

When the disturbance is positive, the C1 characteristics converge, as in Fig., and must

eventually meet.

Such an intersection implies that the depth has two different values in the same place at the

same time.

The wave becomes steeper and steeper, until it forms an abrupt steep-fronted wave – the

surge, or bore. While the intersections of neighboring characteristics will form an envelope,

the surge will eventually form at the first point of the envelope A of the envelope.

The front of the surge will not necessarily be broken and turbulent, it may like hydraulic

jump, consist of a train of smooth unbroken waves if the depth ratio is small enough.

It follows that a surge would certainly not break at first point of formation, as at A in Fig, for

the depth ratio there approaches unity. Breaking would only occur after subsequent

development of the surge beyond the point A; tracing this development would be a matter of

some difficulty.

The intersection of any neighboring pair of characteristics can be located by an elementary

geometrical argument. With the terms defined as in Fig., the following results are easily

obtained:

and d = d (tan ) cos2

whence

(19)

Where the (u+c) equals the inverse slope of the characteristics, as given by Eq. . Subtituting

Eq. Into Eq. (19) we obtain;

(20)

The dam-break problem.

This problem is just one application of the principles and techniques discussed, but it is

sufficient importance to warrant a complete section of its own. The problem deals with the

course of events following the sudden insertion or removal of obstacle in a channel, and is

clearly of interest in a number of real situations, ranging from catastropic flood following the

collapse of a dam, to the operation of sluice gates in an irrigation canal.

Downstream Riverbed Dry

In the sinplest form of the problem, bed slope and resistance are assumed negligible, the

canal is rectangular in section and is closed by a transverse vertical plate, as in Fig.,

containing still water on its right and having a dry channel bed on its left.

The plate is to be suddenly removed and the subsequent motion studied. The best way to

approach the problem is to consider first the situation created by moving the plate off to the

left at finite speed; if speed of the plate is increased to a very large value, we then have case

of complete removal of the plate.

Suppose, in the first intance, that the plate is gradually accelerated to a speed w; the path of

the plate is then traced on the x-t plane, Fig. , by the line OABC, the line BC being straight

with an inverse slope of – w. Assuming for the moment that some water of some unknown

depth will ramain in contact with the plate, it follows that the water velocity will be the same

as the plate velocity. Now we can draw a C2 characteristics (shown dotted) from any point

such as A on OABC to the first C1 characteristics OF.

uA – 2 cA = u0 – 2 c0 (22)

From the equation above we deduce the slope of the C1 characteristics drawn outward from A

= uA + cA =

ut -

u0 + c0 (23a)

= 3 ct +u0 - 2c0 (23b)

The interesting general point that emerges is that our initial disturbances does not have to be

prescribed along the t axis, it may be prescribed along any other line ( such as OABC) in the

x-t plane. What may be the position of this line, we can use the fact that ( u – 2c) is constant

everywhere in the plane to determine, as from Eq, the slope of C1 characteristics issuing from

any point on the line

The C1 characteristics issuing from OAB therefore diverge, and the ones issuing from BC are

parallel, with inverse slopes equal to:

(

)

=

w -

u0 + c0 (24)

We can now divide the region between Ox and OABC into three zones:

Zone I – the zone of quiet between OF and Ox, representing undisturbed still water.

Zone II- the region of diverging characteristics and therefore varying depth and velocity

between BB1 and OF.

Zone III – the region of parallel characteristics, and therefore constant depth and velocity,

between BC and BB1. All the water represented by this region is moving with the plate

cB =

uB –

u0 + c0

cB = -

w + c0 (26)

noting that uB = - w , and u0 = 0, it follows that:

(

)

(27)

Now cB cannot be negative, so that w cannot 2 c0. If w = 2 c0, cB = = 0, and the water

reaches zero depth at the plate. If speed of the plate exceeds 2 c0 it simply losses contact with

the water, which then feathers out to an edge of zero depth moving with a speed 2 c0.

To determine the depth of water in zone III, adjacent to the plate, we simply use Eq. and

obtain:

Complete Removal of the Dam

Since the line passes through the origin, so that t1 = 0, Eq. becomes

√ √ (28)

From Fig. Above shows at the original position of the dam, the depth is constant and equal to

4 h0/9, and the water velocity is a constant and equal to 2c0/3, and constant rate of outflow

equal to 8 h0 c0 /27. This constant rate of outflow is maintained until the negative wave front

D reaches the rear wall of the reservoir, is reflected from it, and returns to the origin of x;

thereaftern the outflow rate gradually diminishes.

Downstream Riverbed Submerged.

If the downstream riverbed is not inially dry, but contains a layer of still water, the problem is

not greatly changed.

Sluice-Gate Operation