undefined | unit 5 open flow channel

Unit 5

Open Flow Channel

Q1. Explain Classification of open channel flow.

Ans- Channels can be classified as follows:

Artificial or Natural :

Depending upon the origin, the channels may be classified as artificial and natural. 'Artificial channels or man-made channels' are those constructed by man.

Canals, navigation channels, spillways, roadside gutters, etc, are man-made channels. Conditions in case of artificial channels are more controlled and hence correct results can be obtained with mathematical theories.

Natural channels' are those which are formed naturally in nature. Streams, rivers, brooks, etc. are examples of natural channels.

The hydraulic properties of natural channels are varied and hence one has to depend upon actual observations and individual experience while dealing with such channels.

2. Prismatic and Non-prismatic :

If the cross-section (shape and size) and the bed slope of the channel are constant along the length of the channel, it is known as 'Prismatic channel.

Most man-made channels are prismatic. If either the cross-section or the bed slope or both vary along the length of the channel, the channel is known as 'non-prismatic'. Most natural channels are non-prismatic.

3. Rigid and Mobile Boundary Channels:

The channels in which the boundary (bed and sides) is immovable are called rigid boundary channels. Man-made channels like lined canals, sewers, etc. are examples of rigid boundary channels.

If on the other hand, the boundary of the channels is made up of loose material that can be eroded or deposited due to the flow in the channels, the channels are called 'movable boundary channels'. Many natural channels like rivers are of this type.

"Alluvial channels' are movable boundary channels which carry the same type of material as that of the channel boundaries”.

Q2. Explain the geometric elements of the channel flow.

Ans- Geometric elements of a channel:

"Geometric elements are the properties of a channel section that can be defined entirely by the geometry of the section and the depth of flow".

These elements are very important and are frequently used in open channel flow computations.

Commonly used geometric elements of a channel section are as follows.

Depth of Flow 'y': This is the vertical distance from the lowest point of channel section to the free liquid surface.

2. Depth of Flow Section’d’: It is the depth of flow normal to the direction of flow

. Cos or y =

If the slope angle is very small, yd.

3. Stage: It is the elevation or vertical distance of the free surface above a datum.

4. Top Width 'T': It is the width of the channel section at the free surface.

5 Water Area (or area of flow) 'A': it is the cross-sectional area of flow normal to the direction of flow.

6. Wetted Perimeter (p): It is the length of the line of intersection of the wetted surface of the channel with the cross-sectional plane normal to the direction of flow.

In other words, taking the section normal to the direction of flow, the wetted perimeter is the length of the line of contact between the liquid and the boundary of the channel.

7. Hydraulic Radius 'R': It is the ratio of the area of flow (A) to the wetted perimeter (P). R=

8. Hydraulic depth 'D': It is the ratio of the area of flow (A) to the top width (T).

D =

9. The Section Factor for Critical Flow 'Z': It is the product of the area of flow (A) and the square root of hydraulic depth (D). = A D

10.The Section Factor for Uniform Flow Computation: It is the product of the area of flow (A) and the two-thirds power of hydraulic radius (R)

Section factor for uniform flow computations = A

Q3. Explain the Basic governing continuity equation of Channel.

Ans- Continuity Equation:

Consider the element of the channel between sections 1 - 1 and 2 2.

Let A- be the length of the channel between 1-1 and 2 - 2.

Let at any instant Q and y be the discharge and depth at the central section between sections 1 1 and 2 - 2.

Let 'A' be the cross-sectional area of the flow section and 'T' the top width at the central section.

The inflow of liquid in the element from the face (1) - (1) of the element in time

t is given by

Inflow in the element = [Q - ]. ................1

Similarly, the outflow of liquid in the element from the face (2) - (2) of the element in time '

t' is given by

Outflow from the element = [Q +

................2

The net inflow into the element in time’

t' = Inflow – Outflow.

The net inflow into the element in time

= [Q - ] .[Q + ].

= - . . .............3

Now, the volume of liquid in the element is (A .

Therefore, the increase in the volume of the element in time can also be written as the time rate change of volume of element x

Or an increase in the volume of elements in time.

(A.

.........4

Equating (3) and (4), we get

- . . = (A. .

Dividing by.

. .

, we get

=0 ..................5

Equation (5.1) is the continuity equation for unsteady flow in an open channel. Further, we have

= T, or

A = T

and the continuity equation can be written as

+ =0 ........6

The discharge Q in the channel is given by

Q = AV, where A is the area and V the average velocity

Q = AV in equation 6 , we get

+ =0 ..................7

If the flow is steady, equation 5 can be written as

Or Q=constant

Q= A1 V1= A2 V2 =A3 V3 = constant. ..........8

Q4. Explain the Basic governing Energy equation of Channel.

Ans-

For steady frictionless open channel flow, Bernoulli's equation is used as an energy equation. According to Bernoulli's equation for steady frictionless flow, we have

Z+ + = constant

Where the terms have the usual meaning. Bernoulli's equation given above does not take into account the loss which occurs in the case of real liquids. shows the case of non-uniform flow of real liquid between sections 1-1 and2-2. Taking into account losses between 1-1 and 2-2 and also the variation of velocity at 1-1 and 2-2 and applying Bernoulli's equation, we get

Z1 + d1 cos + 1 = Z2 + d2 cos + 2 +hf .....................9

Where a and a, are energy correction factors. d1 , d2, are depths of flow sections, and y1 and y2 are depths of flow at sections 1-1 and 2-2 and hi the loss of head between sections 1-1 and 2-2. If further, slope of the channel is small, cos

= 1 and d1 = y1, d2 =y2 The equation becomes

Z1 + y1 + 1 Z2 + y2 + 2 +hf ...........................10

Equations (5.6) and (5.7) are the 'energy equations' as applied to open channel flow.

If further,1 =2 = 1

Z1 + y1 + Z2 + y2 + +hf ..........................11

hf is the loss of head due to friction and eddies. In the case of prismatic channels, eddy losses are negligible. Estimation of h; is not easy in many flow cases. Hence, under such conditions the use of energy equations is limited.

Also is yet another form of 'energy equation' and is frequently used.

Q5. Explain the Basic governing Momentum Equation of Channel.

Ans-

The momentum equation is the result of the application of Newton's second law of motion to the flowing mass of liquid.

Momentum equation states that the "sum of all external forces acting on a fluid mass in a given direction is equal to the rate of change of momentum of the fluid mass in that direction".

Consider the control volume bounded by sections 1-1 and 2-2, the free liquid surface, and the bed of the channel.

The forces acting on the liquid mass between sections 1-1 and 2-2 in the direction of motion are :

(1) F1 and F2, the hydrostatic forces acting on faces 1-1 and 2-2.

(2) Ff, the frictional force at the boundary of the channel.

(3) Fa, the force due to air resistance at the free surface.

(4) W sin , the body force i.e. a component of the weight of the liquid in the direction of motion. is the slope of the bed of the channel.

Taking into account all the above forces, we can write the momentum equation as

Sum of all external forces in the direction of flow = Rate of change of momentum in the direction of flow

F1 - F2 + W sin - Ff - Fa = (V2 – V1) .....................12

This is the momentum equation as applied to open channel flow. In general, Fa is negligible and it is usual practice to neglect Ft if the distance between sections 1-1 and 2-2 is small. Further, the channel is horizontal,

0 and hence momentum equation as applied to a short reach of P horizontal channel takes the form

F1 - F2

(V2 – V1)

Momentum theorem is particularly applicable to cases of R.V.F. like hydraulic jump, where large internal energy losses take place in a very short distance of the channel.

Q6. Explain One dimensional approach.

Ans- The characteristics of flow like velocity, pressure, etc. in open channel flow generally vary in longitudinal as well as normal directions.

Thus, they can be functions of the three coordinate directions along with time 't' if the flow is unsteady.

In the case of a narrow channel, the flow is three-dimensional. In the case of a wide channel, the flow is two-dimensional, wherein the velocity does not vary much with the distance from the side walls.

Analysis of three-dimensional flow is very complex, To simplify the problems, the variations in the cross-section (i.e. in vertical and transverse directions) are neglected.

The variations only in the longitudinal (in the direction of flow) direction are taken into account.

For this purpose only average or mean values of properties (like velocity) are taken as representative properties for the section as a whole and their variation only in one dimension viz. in the direction of flow is taken into account.

This approach of analysis considering the variations of average values of properties only in one direction is known as a dimensional approach of flow analysis.

This method, if properly used, not only simplifies the problem but also gives useful results.

(A) Average Velocity of Flow :

The discharge through elementary area 'dA' can be written as

PQ = v dA

Therefore, the total discharge through the section

Q =

But, if 'A' is the area of flow section and V the average velocity.

Q = A V

Equating we get, v dA V =

......................1

This average velocity V is used as representative velocity for the section as a whole.

(B) The Discharge :

The discharge passing through a section can be written as

Q = A V =

..........................2

The kinetic energy of liquid passing through a section can also be expressed in terms of the average velocity 'V.

K.E. = Mass of liquid passing through area 'dA' = p v dA.

The kinetic energy of liquid passing through area dA =

p v dA) .

The kinetic energy of liquid passing through the section = =

....i

This is the kinetic energy of liquid passing through the section taking into account the variation of velocity:

Now, the kinetic energy of liquid passing through the section based on average velocity, 'V' can be written as

K.E. =

However, these two kinetic energies are not the same. The ratio of these two energies is known as kinetic energy correction and is denoted by

The energy correction factor, ...................13

Or the kinetic energy in terms of average velocity can be written as K.E.

K.E.= p) .....................14

(d) Momentum :

Following similar steps, we can express the momentum of liquid passing through a suction channel in terms of the average velocity V.

Momentum of liquid passing through area dA = (p v dA) · v

= (pv dA) .v

=p dA

The momentum of liquid passing through the section

The momentum of liquid passing through the section based on average velocity

=p A

The ratio of these two momentums is known as "momentum correction factor" and s denoted by

Or Momentum correction factor,

The momentum of liquid passing through a section in terms of average velocity can be written as,

.p A

Actual values of

and

for different channels are given in the 'Handbook of Hydraulics' by King.

Q7. Explain Factors affecting Manning's roughness coefficient.

Ans- Following are the important factor affecting's roughness coefficient ' n :

1) Surface roughness

2) Vegetation

3) Channel irregularities

4) Channel alignment

5) Silting and scouring.

6) Stage and discharge

7) Transport of suspended and bed material

8) Size and shape

9) Obstruction

1) Surface roughness :

The size and shape of rough particles forming the bed and banks of the channel affect the surface roughness.

Fine-grained material has a lower value of n ' whereas coarse-grained material has a high value of n.

2) Vegetation :

The value of increases by the increase in growth of vegetation.

The main characteristics affecting the value of are height density vegetation.

3) Channel irregularities :

Channel section is about changes due to irregularities in channel dispersion etc. The abrupt changes increase the value of n.

4) Channel alignment :

The curve in the channel alignments affects manning's n, the value of n, sharpness in the curve.

5) Silting and scouring:

Manning decreases due to the deposition of silt is a regular and rough channel.

Scouring causes local depressions at places, which increases the value of n.

6) Stage and discharge:

The value of decreases, with the increase in the stage.

Discharge is directly proportional to the stage.

7) Transport of suspended and bed material :

The channel carries suspended and bed material requires additional energy to the material, manning increase due to an increase in energy.

Que (8) Explain Chezy's Formula with Assumptions.

Ans-

Assumptions Made in Chezy's Equation :

1) The flow is steady

2) Channel is prismatic.

3) The bed slope of the channel is small.

Chezy's Formula for Steady Uniform Flow-through Open Channel :

Consider a steady uniform flow of water in an open channel whose bed slope is S and cross-section is uniform

Let, P = Wetted perimeter of cross-section

A = Uniform cross-sectional area

L = Length of the channel

v = Mean velocity of flow

f = Frictional resistance per unit area at unit velocity

Y = Specific weight of water

According to Froude law , the frictional resistance in a length L of the channel

F = f x contact area x ( velocity )2 = f x PL x v2

Work done against the friction per second

Work done = Frictional resistance x distance

= f PLv2 x v = f PLv3

Weight of water between the two section = AL

This amount of water will fall vertical downward by a distance v in one second

height of fall = v S

Loss of potential energy in one second = Weight x height of fall = ( AL ) v S

We know ,

Work done against friction = Loss in potential energy

f PLv3 = ( AL ) v S

v3 = x S

v = x

v = C

Where C = Chezy's constant =

R = hydraulic mean depth =

Discharge Q = A x v = AC

Q9. Explain Important terms pertaining to uniform flow.

(1) Normal depth, (2) conveyance,(3) section factor,(4) the concept of second hydraulic exponent.

Ans-

1) Normal depth :

The depth of flow during uniform flow is called as normal depth.

2) Conveyance :

The discharge through an open channel with the uniform flow is obtained by Chen formula,

Q = C A = K Where K = C A

For manning's formula

Q = K , Where K = A R2/3

The factor K is called as the conveyance of the channel.

The conveyance of section is directly proportional to discharge

3) Section factor:

The product of the area of flow (A) and two-third of the hydraulic radius ( R)

Z = A R2/3

4) Hydraulic exponent for uniform flow :

The conveyance of the channel is a function of the depth of flow.

For integration, in GVF computation, Bakhmeteff assumed the following expression.

K 2= C yN

Where C is a coefficient and N is exponent called the hydraulic exponent.

It is also known as the second hydraulic exponent.

Hydraulic exponent N is usually a slowly varying function of the aspect ratio of the channel.

A graph is plotted between conveyance and K and depth of flow, the hydraulic exponent is obtained from the graph

Q10. Explain the efficient channel sections: (1)Rectangular,(2) Triangular, and (3)Trapezoidal

Ans-

1) Rectangular section:

Consider rectangular channel of cross-sectional area A

Area of section A = By

Perimeter of section P = B + 2Y = + 2y

Discharge through channel,

Q = A • v = AC

= AC

= C

Since A, C, and S are constant in the above equation, the discharge will be maximum when perimeter P in minimum.

= 0

( + 2y) = 0

= 0

A = 2y2

By = 2y2

B = 2y

Y =

The maximum discharge for the rectangular section occurs when depth is one – half of the width of the channel.

Hydraulic mean depth for this section

R = =

= =

R =

Trapezoidal section:

Consider a channel of trapezoidal cross-section

Let, A = are of cross-section

b = bottom width

y = depth of flow

= side slope ( i.e. 1 vertical to m horizontal)

Area of section, A = ( b + my) y

b = – my

Discharge Q = A • v = AC

= AC

Since A, C, and S are constant in the above equation, the discharge will be maximum when perimeter P in minimum.

P = AB + BC + CD = b + 2(AB)

= b + 2 (y )

= – my + 2y

= 0

= – m + 2 = 0

+ m = 2

= 2

= y

(Top width) = Sloping side

Hydraulic mean depth for this section

R = =

R =

Best side slope for the trapezoidal channel for maximum discharge :

P = – my + 2y

Maximum Discharge will be,

= 0

= - y + 2y x ( m2 + 1) -1/2 2m = 0

- Y = 0

= 1

m2 + 1 = 4m2

3m2 = 1

m =

tan = = m =

= 30°

Condition for economical cross-section,

= y

+ = y

b = y

Perimeter P = b + 2y

= y + 2y

P = y + y = y = x b = 3b

But, P = b + (AB ) + CD = b + 2 ( AB)

3b = b + 2 (AB)

AB = = b

Sloping side of channel = Bed width of channel

2) Triangular section:

Consider a triangular section ABC

Let be the angle of each side with respect to vertical

tan = =

AD = DC = y tan

Top width AC = AD + DC = 2y tan

Area of section A = x base depth = ( 2y tan) y

A = y2 tan ………………………….eqn 1

Wetted perimeter P = AB + BC

= 2 (AB) = 2 = 2y

P = 2y = 2y sec …………………eqn 2

From equation 1, y =

P = 2 x sec

= 2

For maximum discharge, the perimeter will be minimum,

= 0

(cot + tan ) = 0

Cosec2 + sec2 = 0

sec2 = Cosec2

= 1

= 45°

m = 1

From eqn 1, A = y2 tan45° = y2

Wetted perimeter P = 2y sec 45° = 2y

Hydraulic mean depth R = = =

Q11. Design a most economical trapezoidal open channel with side slope 2H: 1V and a bed slope of 0.00059 to carry a discharge of 58 cumecs. Assume chezy's constant as 55.

Ans:

Side slope = 2:1

S = 0.00059

Q = 58 m3/sec

C = 55

For most economical trapezoidal section

(Top width) = Sloping side

= y = y

= y

B + 4y = 2 y

b = 0.472y

Area of trapezoidal A = (b + my) y = ( 0.472 y + 2y)y

= 2.472y2

Hydraulic mean radius R = = 0.5y

Discharge by chezy’s formula,

Q = AC

58 = 2.472y2 x 55 x

58 = 2.3352y5/2

Y = 2/5 = (24)2/5 = 3.614

b = 0.472y =0.472 x 3.614 = 1.796m

Q12. A trapezoidal channel has to convey a discharge of 20 m3/s of water of a velocity of 1.75 m/s. The side slope of the channel is 1H to 1V. Find the sectional dimensions for the most economical conditions. Fin also the necessary bed slope of the channel if chezy’s constant is 50.

Ans:

Given:

Q = 20 m3/s

V = 1.75 m/s

Side slope m:1 = 1:1

C = 50

To find: b, y & S

Area of section A = = = 11.429m2

For most economical trapezoidal section

(Top width) = Sloping side

= y = y

= 2 y

B + 2y = 2 y

b = 0.828y

Area of trapezoidal A = (b + my) y

11.429 = (0.828 y + 1y)y

11.429 = 0.828 y2 + y

0.828 y2 + y – 11.429 = 0

Y = 3.16 m

Bottom width of section b = 0.828 *3.16 = 2.62 m

Hydraulic mean radius R = = = 1.58 m

Discharge by chezy’s formula,

Q = AC

1.75 = 50 x

= 0.0278

Q13. Explain in detail Specific energy and Specific force diagram.

Ans- Specific Energy:

The flow in an open channel. It may be: noted that the bed slope has been exaggerated for the sake of clarity of the diagram.

The total head (energy per unit weight) at any section with respect to any selected datum is given by

H = z + d cos ..........................1

"Specific energy in a channel section is defined as the energy per unit weight of liquid at any section of a channel measured with respect to channel bottom as a datum".

Thus, from equation (1),

Specific energy E = d cos .................2

If channel slope is small (cos

= 1 and d = y) and if further

= 1

E = y + ........................3

The concept of specific energy.

Equation (3) indicates that the specific energy is the sum of the depth of liquid and the velocity head The concept of specific energy introduced by Bakhmeteff is very useful in solving some open channel problems.

Specific force:

The concept of specific force is the result of the application of the momentum equation to the open channel flow.

According to the momentum equation, (Newton's second law of motion), the rate of change of momentum in any direction is equal to the sum of all external forces acting on the liquid mass in that direction.

Applying momentum equation to liquid mass between two sections (1-1) and (2-2), we get

(V2-V1) = P1-P2 + W sin

- Ff ...................1

Where y = unit weight of the liquid,

Q = discharge and

W = weight of liquid between sections 1-1 and 2-2. Ff force due to friction at the boundary of the channel. Force due to air resistance is neglected.

P1 and P1 are the hydrostatic pressure forces on faces 1-1 and 2-2 respectively.

is the slope angle of the bed of the channel.

If the distance L between the sections is short, Fr can be neglected, Further, if the channel is horizontal,

= 0, sin

= 0 and hence W sin

= 0. Therefore, the momentum equation applied to the short reach of a horizontal prismatic channel takes the form

(V2-V1) = P1-P2 ...........................2

But the hydrostatic forces P1 and P2 can be calculated by usual formula as

P1 = A1 = A1

P2 = A2

Where A1 and A2 are the cross-sectional areas and

and

are the depths of centers of gravity of A1 and A2, below the free liquid surface at sections 1-1 and 2-2.

Therefore,

(V2-V1) = A1 - A2

Q = A1V1 =A2V2 or V1= and V2=

(

Dividing both sides by

and rearranging the terms, we get

+ = + ..............................3

The two sides of the above equation are analogous (similar) and hence may be expressed by a general function 'F' for any section of the channel.

F = + A .................................4

The function F consists of two terms (1) the term

representing the momentum of liquid passing through the section per second per unit weight of liquid and (2) the term A

representing the hydrostatic force at the section per unit weight of liquid. Thus, both the terms represent the forces per unit weight (specific weight) of liquid. Their sum F is called 'specific force'.

"Equation (3) shows that the specific forces at sections 1 and 2 are equal, provided the frictional force and the component of liquid in the direction of flow can be neglected,

Q14. Explain the Important terms pertaining to critical flow (1)section factor, (2)the concept of first hydraulic exponent.

Ans- Critical Flow Criteria: conditions and characteristics

1) Specific energy is a minimum for a given discharge.

2) Discharge is a maximum for a given specific discharge

3) Specific force is a minimum for a given discharge.

4) Velocity head is equal to half the hydraulic depth in a channel of small slope.

5)Froude number is equal to unity.

6) Velocity of flow in a channel is equal to celebrity of small gravity waves in shallow water channel.

Section Factor-

The section factor for critical flow computation of a channel is the product of the water area and the square root of the hydraulic depth

Equation 1 & 2 are very useful for critical flow analysis.

Hydraulic exponent-

Since the section factor is a function of the depth of flow y, and A may be assume that

Z=CyM

Where C is the coefficient and M is the parameter called hydraulic exponent for the critical flow.

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