chapter 2 zeleke agide

Open Channel Hydraulics Chapter 2

2. BASIC HYDRAULICS PRINCIPLES

2.1 Geometry of Open Channels

Open channels can be natural (rivers, streams, estuaries) or artificial (canals, flumes, chutes, culverts, drops, open-flow tunnels). The channel geometry can be prismatic (constant So and constant A) or non-prismatic. The geometry is defined by its cross-section and water depth (y).

Some conventions for naming of canal geometric parameters is given underneath:

Figure 2-1Geometry of an open channel

Zeleke Agide / Hydraulic Engineering / 2004/2005

Water depth (y) : Vertical distance bottom to surface (measured on a vertical plane)Section depth (d) : Normal distance from bottom to surface, depth perpendicular to

the bottom (measured on a plane perpendicular to the canal bottom)

Area (A) : area normal to flow directionWetted perimeter (P) : length of line of wetted intersectionHydraulic radius (R) : R = A/PHydraulic depth (D) : D = A/BS = Area A/top width BS

Total energy (E) : E= z + y + αv2/2g (for small slopes θ with y = d)Specific energy (ES) : ES energy in relation to lowest point in a section ES = y + αv2/2gVelocity (V) : v = Q/AVelocity head : αv2/2g = αQ2/2gA2

Froude number : Fr2 = αQ2BS/gA3

Friction Slope : Sf = Q2n2/A2R4/3 (Manning)Friction Slope : Sf = Q2/C2A2R ( Chezy)

1

m

Y

Bs

B


Figure 2-2 Presents geometric elements of the most common channel sections

2.2 Main Principles

To any situation in hydraulics we can apply two out of three principles:

1-a Principle of conservation of matter (continuity)1-b Principle of conservation of energy (in case all energy “losses” are known)

2-a Principle of conservation of matter (continuity)2-b momentum principle (in case all external forces are known)

2.3 Continuity Principle

In open channel flow the continuity principle for a constant discharge Q is:



A = the cross-sectional area in sections 1 and 2,

= the mean velocity in sections 1 and 2

If the mean velocity remains constant (V1 = V2) then the cross-sectional area A (A = Q/V) remains constant. The open channel has a prismatic cross -section. This means for uniform flow a constant water depth in all section. Uniform flow is a result from an exact balance between the force of gravity and the frictional resistance.

Application of the continuity principle to unsteady, open channel flow is more difficult. In unsteady open channel flow the water surface will change over a certain distance and during a certain time .

During : Inflow-Outflow = Storage

As the velocity and the discharge will change over a distance.

The discharge Q will vary with time t and with the distance along the canal.

If y is the water depth and Bs the width at the water surface, then the increase of volume between the sections 1 and 2 during time

The two terms derived are equal in magnitude, but different in sign:

This is the equation of continuity for unsteady open channel flow.

2.4 Energy Principle

The energy equation and the momentum equation are used in addition to the continuity equation in analyzing fluid-flow situations. They are both derived from Newton’s second law of motion. In this section, the force components on a fluid particle in the direction of its motion are equated to the



product of mass of the particle and acceleration along the streamline. The equation is obtained in differential form and requires the assumption of a frictionless fluid and steady flow. The equation may then be integrated for particular assumptions regarding the relation between density and pressure to obtain the energy equation.

Figure 2-3 Force components on a fluid particle in the direction of streamline.

In figure 2-3 let S be a streamline in steady flow, and consider the forces acting on a fluid particle in the direction S of the streamline. The element has a prismatic cross-sectional area A and length s – a frictionless fluid is assumed in order to eliminate all shear in the fluid (i.e. ideal flow is assumed instead of real flow). On the upstream end the element the force is PA, in which P is the pressure intensity at the center of the face. The force on the downstream end of the element is :-

and acts in the negative direction. Any forces acting on the sides of the element are normal too the streamline do not enter the equation. The only other force acting is due to gravity and is AS,

acting vertically down ward. The component in the S-direction is: -AS Cos . By substituting in to .

After dividing through by the weight of the element AS and simplifying.

Z is the increase in elevation for a displacement S along the stream line and from Figure 2-3.



The acceleration as is To keep this expression simple, the assumption of steady flow is

introduced, so that V is a function of S only along a streamline. Writing,

as the velocity along the streamline is . The equation becomes.

By multiplying the above equation trough by g and with the additional assumption that the fluid is incompressible (i.e. assuming a constant density in time and space) and integrating with respect to S.

constant. Equation 2-1

This is Energy equation (Bernoulli’s Equation) for steady flow of a frictionless, incompressible fluid along a streamline. It states that energy per unit mass of a flowing fluid is constant, the dimensions

for this equation are , i.e. energy per unit mass.

Potential energy = mgZ gZ is potential energy per unit mass.

Kinetic Energy is kinetic energy per unit mass.

Work/Energy = F.S = We can represent he unit mass as: unit mass =

Therefore, Pressure Energy per unit mass =

By dividing equation 2-1 through by g we get on expression for energy per unit weight, or more simply, meters. This expression is particularly convenient form to apply to situations with free liquid surface, i.e. open channel flows.

For open channel section with steady flow and straight and parallel streamlines, there is no centripetal acceleration, i.e. where the hydrostatic pressure distribution holds true, the pressure head

lies in the water surface.

- For channel with large slope

and

- For channel with small slope



Substituting for channel of large slope the total energy may be written as

In general, every streamline passing through a channel section will have a different velocity head, owing to the non-uniform velocity distribution in actual flow. Only in an ideal parallel flow of uniform velocity distribution can the velocity head be truly identical for all points on the cross-section. In the case of gradually varied flow, however, it may be assumed, for practical purposes, that the velocity heads for all points on the channel section are equal, and energy coefficient (Coriolis Coefficient ) may be used to correct for the over all effect of the non-uniform velocity distribution. Thus, the total energy, equation at a channel section takes the form

According to this equation (the principle of conservation of energy), the total energy head at upstream section 1 should be equal to the total energy head at downstream section 2 plus the loss of energy hf between the two section ; or

This equation applies to parallel or gradually varied flow. Consider now a prismatic channel. The line representing the elevation of the total head of flow is the energy line. The slope of the line is known as the energy gradient, denoted by Sf. The slope of the water surface is denoted by Sw and the slope of the channel bottom by So = tan . In uniform flow, Sf = Sw = So = tan .

Figure 2-4 Energy in gradually varied open channel flow

2.5 Specific Energy and Critical Depth



The “Specific energy” is the average energy per unit weight of water with respect to the channel

bottom. The piezometric head related to the bottom is (with Z = 0) which is the water

depth. Therefore, the specific energy is the sum of the water-depth (y) and the velocity head, if the streamlines are straight and parallel.

For a given section and constant discharge (Q), the specific energy is a function of water-depth only,

since .

When the depth of flow is plotted against the specific energy for a given channel section and discharge, a specific-energy curve is obtained (figure2-3).

Figure 2-5 Specific Energy Curve

This curve has to limbs AC and BC. The limb AC approaches the horizontal axis asymptotically towards the right. The limb BC approaches the line OD as it extends upward and to the right. Line OD is a line that passes through the origin and has an angle of inclination equal to 45o. At any point



P on this curve, the ordinate represents the depth, and the abscissa represents the specific energy. Which is equal to the sum of the pressure head y and the velocity head

The curve shows that for a certain discharge Q two flow regimes are possible, viz. slow and deep flow or a fast and shallow flow, i.e. for a given specific energy, there are two possible depths, for instance, the low stage y. and the high stage y2

. The low stage is called the alternate depth of the high stage, and vice versa. At pint C, the specific energy is minimum. It can be proved that this condition of minimum specific energy corresponds to the critical state of flow. Thus, at the critical state the two alternate depths apparently become one, which is known as the critical depth (YC). When the depth of flow is greater than the critical depth, the velocity of flow is less than the critical velocity for the given discharge, and, hence, the flow is sub critical. When the depth of flow is less than critical depth the flow is supercritical. Hence, Y1, is the depth of a supercritical flow, and Y2 is the depth of a sub critical flow.

If the discharge changes, the specific energy will be changed accordingly. The two curves A’B’ and A”B” (Figure 2-3) represent positions of the specific energy curve when the discharge is less and greater, respectively than the discharge used for the construction of the curve AB.

The critical state of Flow

The critical state of flow is defined as the state of flow at which the specific energy is a minimum for a given discharge or it is the condition for which the Froude number (Fr2) is equal to unity.

For

Differentiating with respect to y and noting that Q is a constant.

The differential water area dA near the free surface (figure 2-5) is equal to Now .

and the hydraulic depth . So the above equation becomes.

But . Substituting



At the critical state of flow the specific energy is a minimum, or . The above equation,

therefore, gives.

This is the criterion for critical flow, which states that at critical state of flow, the velocity head is

equal to half the hydraulic depth. The above equation may also be written which means

Fr = 1; this is the definition of critical flow given previously.



2.6 Momentum Principle

The momentum equation is developed from Newtons second low of motion by summing up the resultant force on a free body of fluid on one side of the equation and by developing the other side of the equation into an equivalent expression in terms of rates of in flow and outflow of momentum.

According to Newton's second law of motion the change of momentum (dmv) per unit time, is equal to the resultant of all external forces acting on a body (body of water flow in a channel in our case).

The momentum passing a section A per unit time (t = 1sec):

mv = * (A. V) V

Where, A*V represent the volume passing per unit time.

mv = QV

Or including the momentum coefficient (Boussinesq coefficient)

mv = * * Q * v

mv = * *

Within the control volume defined in the figure there is an unknown energy loss and/or force acting on the flow between section 1 and 2; the result is a change in the linear momentum of the flow. In many cases, this change in momentum is accompanied by a change in depth of flow. The application of Newton's second law, in a one dimensional for to the control volume, i.e. equating the sum of all external forces (F) to the rate of change of momentum ( Q V) for any two cross-sections 1 and 2 gives:

Or including the momentum coefficient


vA

dt


Consider a channel section of mild slope and analyzes the forces acting.

Thus, F = (2v2 – 1v2) implies F = F1 –F2 + w sin - Ff = (2 v2 - 1v1)

F1 and F2 are the resultant pressure forces acting on the two sections and w is the weight of the water between the two sections. Ff is the total friction force acting along the surface of the body. The slope is mild assume sin So = 0

W sin = 0

and for a flow with parallel flow lines the pressure is assumed to be hydrostatic.

The force F1 is gAy+, where A is the cross- sectional area and of y+ the depth of the center of gravity of the area A (the depth of centroid of the cross sectional area measured below the surface of flow) . For rectangular section of small slope

F = g by (½y) (y+ = ½ y)

(Remark for canals with large slope F = g bdcos ½ dcos = ½ g bd 2 cos2)

Therefore, substituting

W sin = o F1 = g A1y1

+

F2 = gA2y2+

F1 – F2 + w sin - Ff = (2v2 - 1v1)

g A1y1+

- g A2 y2+ - Ff = Q (2v2 - 1 v1)

- Ff = g A2y2+ - g A1y1

+ Q2

- Ff = g


Ff

wsinF

F

W


The term between brackets is called the momentum function (M) or in general:

M = A y+ +

= M1 – M2

For a given discharge Q, channel shape and coefficient the function, M depends only on the water depth y. Plotting M against ay gives a similar figure as for the specific energy Es against depth y. This curve is called specific force curve. In the figure two regions can be determined, namely sub and supercritical flow. For every M > Mmin two water –depths exist, which are called the initial and sequent depth. Together they are the conjugate depths.

Specific Energy Curve Channel Section Specific Force Curve

In applying the momentum principle to a short horizontal reach of a prismatic canal, the external forces of friction and the weight effect of water can be ignored. Thus, with = 0 and Ff = 0 the equation becomes:



A1 y1+ + = A2 y2

+ +

The momentum function M = y+A + consists of two terms. The second term is the

momentum of the flow passing through the channel section per unit time per unit weight of water, and the first term (Ay+) is the force per unit weight of water. Since both terms are essentially force per unit weight of water, their sum may be called the specific force. Accordingly, it may be expressed as F1 = F2.

Thus, in analogy with the concept of specific energy) for a given value M, the M- y curve predicts two possible depths of flow – conjugate depths of a hydraulic jump.

For a rectangular cross section y1 = ½ y. Assuming, = 1 results in m = ½ Ay +

The function per unit width (b = 1) reads

M = ½ y2 +

The function has on extreme for :

Y = yc = =

for y = yc

The minimum value of the specific momentum function can be found under the assumptions of parallel flow and uniform velocity distribution by taking the first derivative of M with respect to y and setting the resulting expression equal to zero or


Y3 =


Where d (y+A) = [A (y++dy) + Ady and where it is assumed that (dy)2 = 0 –

Then substituting dA/dy = Bs, v =

We get which is the same criterion developed for the minimum value of specific force

(momentum) occurs at minimum specific energy or critical depth.

The given number 3 is positive, so for momentum value for y =yc . It has to be noted that the given relation for yc is only applicable for open rectangular (prismatic) channels.

For other cross sections the impulse momentum equation should be used

With these equation it is possible to compute one unknown among the four variables (F, Q, A1 ,A2) for a given channel shape and 1 =2 = 1; (y+ is also a function of channel shape and water level).

The impulse momentum principle also follows from Newton’s second law. The flow may be compressible or incompressible, real (with friction) or ideal (frictionless), steady or unsteady moreover, the equation is not only valid along a streamline. The advantage of the impulse momentum principle is that only the conditions at the end sections of the control volume govern the analysis. It has a special advantage for application to problems involving high internal energy changes, such as the problem of the hydraulic jump. If the energy equation is applied to such problems, the unknown internal energy loss represented by hf is indeterminate, and the omission of this term would result in a considerable errors. If instead the momentum equation is applied to these problems, since it deals only with external forces, the effects of the internal forces, the effects of the internal forces will be entirely out of consideration and need not be evaluated. The term for frictional losses due to external forces, on the other hand, is unimportant in such problems and can safely be omitted, because the phenomenon takes place in a short reach of the channel and the effect due to external forces is negligible compared with the internal losses.



Example

Drive a relationship between the initial depth and the sequent depth of a hydraulic jump on a horizontal floor in rectangular canal.

Solution

The computation of the hydraulic jump always begins with the momentum equat6ion

If the jump occurs in canal with a horizontal bed and Ff 0 , that is, the jump is not assisted by a hydraulic structure and loss of energy due to friction is negligible, the momentum equation take the form

M1 = M2

A1y1+ +

In the case of rectangular channel of width b, substitution, of Q = v1A1 = v2A2, A1 = by1 , A2 = by2,

y1t = in to the above equation yields

The computation of the hydraulic jump always begins with the momentum equation gA1

We know from continuity

Q = V1 A1 = V2 A2 = V1 By1 = V2 By2

B



But also q = Q/b, which is the flow per unit width

Q = v1 y1 B = v2 y2 B (for B = 1 unit width)

q = v1 y1 = v2 y2

Substituting

Where q =


chapter 2 zeleke agide

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