undefined | unit 6 gvf in open channel flow fluid flow around submerged objects

UNIT 6

GVF in Open Channel Flow & Fluid flow around Submerged Objects

Q1. What are the Basic Assumptions of GVF ?

Ans- Basic Assumptions of GVF

The gradually varied flow to be discussed here considers only steady flows. This implies that (i) flow characteristics do not change with time, and (ii) pressure distribution is hydrostatic over the channel section.

The head loss in a reach may be computed using an equation applicable to uniform flow having the same velocity and hydraulic mean radius of the section. This implies that the slope of energy grade line may be evaluated using a uniform flow formula such as Manning equation and Chezy equation, with the corresponding roughness coefficient applicable primarily for uniform flow.

Channel bottom slope is small. This implies that the depth of flow measured vertically is same as depth of flow measured perpendicular to channel bottom.

There is no air entrainment. Advanced text books may be referred to study the effects of air entrainment.

The velocity distribution in the channel section is invariant. This implies that the energy correction factor, α , is a constant and does not vary with distance.

The resistance coefficient is not a function of flow characteristics or depth of flow. It does not vary with distance.

Channel is prismatic.

Q2. Explain in detail - Dynamic equation of GVF.

Ans-

In the above figure, H = total energy (or head); E = specific energy; Z = vertical distance of the channel bottom above the datum in meters; d = depth of flow section normal to the bed in meters; θ = channel bottom slope angle; α = kinetic energy correction factor; V = mean velocity of flow through the section in m/s.

The dynamic equation for GVF for small bed slopes-

In terms of section factor the dynamic equation can be written as-

Z is the section factor at depth y and Zc is the corresponding section factor at critical

depth Yc. This equation is commonly used with direct integration technique.

In terms of Energy equation, the dynamic equation can be written as-

This is called the differential energy equation of gradually varied flow. This equation is very useful in developing numerical techniques for the gradually varied flow profile computation.

Q3. Explain the Classification of channel bed slopes and Various GVF profiles.

Ans- Classification of channel bed slopes

For a given channel with a known Q = Discharge, n = Manning coefficient, and S0 =Channel bed slope, yc = critical water depth and y0 = Uniform flow depth can be computed. There are three possible relations between y0 and yc as

1) y0 > yc ,

2) y0 < yc ,

3) y0 = yc .

For horizontal (S0 = 0), and adverse slope ( S0 < 0) channels,

Horizontal channel, S0 = 0→ Q = 0,

Adverse channel , S0 < 0 , Q cannot be computed,

For horizontal and adverse slope channels, uniform flow depth y0 does not exist.

Based on the information given above, the channels are classified into five categories as Indicated in Table

Various GVF profiles

The surface curves of water are called flow profiles (or water surface profiles).

Depending upon the zone and the slope of the bed, the water profiles are classified into 13 types as follows:

Mild slope curves M1 , M2 , M3

Steep slope curves S1 , S2 , S3

Critical slope curves C1, C2, C3

Horizontal slope curves H2 , H3

Adverse slope curves A2, A3

In all this curves, the letter indicates the slope type and the subscript indicates the zone. For Ex. S2 curve occurs in the zone 2 of the steep slope.

-Classification of Surface Profiles

1) Mild slope (M)

Y0 > Yc Type 1: If the stream surface lies above both

the normal and critical depth of flow

(M1,S1)

S0 < Sc

2)Critical slope (C)

Y0 = Yc Type 2: If the stream surface lies between

the normal and critical depth of flow

(M2,S2)

S0= Sc

3)Steep slope (S)

Y0 < Yc Type 3: If the stream surface lies below both

the normal and critical depth of flow

(M3,S3)

S0 > Sc

3)Horizontal slope (H)

S0= 0

4)Adverse slope (A)

S0= -ve

Mild Slope (M)

B. Steep Slope (S)

C. Critical Slope (C)

D. Horizontal Slope (H)

E. Adverse Slope (A)

Q4. Explain the Definition of drag and lift.

Ans-

Drag:

- The component of total force FR in the direction of motion is called 'Drag' and is denoted by Fp. Drag, thus, is the force exerted by the fluid on the body in the direction of motion.

- Drag force resists the motion of the body or fluid Wind resistance to a moving car, water resistance to a moving torpedo are the examples of drag, Power is required to overcome drag anta hence drag is reduced to the possible minimum.

Lift:

- The component of the total force in the direction perpendicular to the direction of motion is known as 'Lift and is denoted by FL. Thus, the lift is the force exerted by the fluid normal to the direction of motion. It can be seen that the lift force can exist only if the axis of the body is inclined to the direction of flow.

- Further, if the body is symmetrical and its axis is parallel to the direction of motion, the lift will be zero and only drag force will exist.

- The weight of an airplane moving in air is balanced by the lift force due to air on the airplane.

Q5. Explain the Expressions for drag and lift.

Ans- Expressions for drag

Consider a body of any shape placed in the stream of a real fluid. Let U the velocity of the stream Consider an elementary area dA of the surface of the body The forces acting on area dA' are

Pressure force equal to ‘p dA ‘normal to area dA

Shear force equal dA along the tangent to dA

If the angle made by force 'p dA' with horizontal, we can write

dFD = Drag force on the element

=Component of ‘ p dA' in the horizontal direction

+Component of dA ‘in horizontal direction

=p. dA cos + dA sin

Therefore, total drag on the body:

FD= FD = dA.cos + dA. Sin

The quantity

FD cos

is called Pressure Drag or Form Drag' and depends upon the form or shape or the body as well as on the location of the separation point.

The quantity

dA. Sin

is called Friction Drag or Skin Friction Drag' and depends upon the extent and character of the boundary layer.

Sum of Pressure Drag and Friction Drag is called Total Drag or Profile Drag

in case of a flat plate held parallel to the stream = 90", cos= 0, sin =1 and hence

FD = only friction drag

If the plate is held normal to the stream

= 0

, sin

=0, cos

= 1 and

hence

FD = only pressure drag.

In the second case due to separation of the boundary layer, awake (Low-pressure zone is produced on the downstream side of the plate Hence pressure difference is created due to which cag is produced. This is called Pressure Drag.

Expressions for Lift:

dFL= Force due to pressure in normal direction + Force due to shear in the normal direction

= -p.dA.sin + . dA. cos

FL = dA.sin + dA. Cos

Without detailed information regarding pressure distribution and shear stress distribution on the body, it is not possible.

As a simple alternative, the 'drag and lift' forces are expressed as:

Drag force: FD =CD x A x

Lift force: FL =CL x A x

‘CD' and 'CL’ are called 'Coefficient of Drag' and 'Coefficient of Lift' respectively. is the mass density of the fluid and 'U' is the uniform velocity of the body relative to the fluid A' is projected area of the body perpendicular to the direction of flow or it is the largest projected area in case of the is called 'dynamic pressure.

The term Resultant force on the body is given by

The resultant force on the body is given by

FR = +

Q6. Explain Coefficient of Drag and Coefficient of Lift.

Ans – Coefficient of drag:

The drag force on elemental area = Force due to pressure in the direction of fluid motion + Force due to shear stress in the direction of fluid motion.

=pdA cos + dA cos (90° -)

= pdA cos + dA sin

FD = Summation of pdA cos

+ Summation of

d A sin

= cos dA + sin dA.

The term

cos

dA is called the pressure drag or form drag while the term

sin

dA. is called the friction drag or skin drag or shear drag.

FD =CD x A x

CD=Coefficient of drag

Coefficient of lift :

The lift force on elemental area = Force due to pressure in the direction perpendicular to the direction of motion+ Force due to shear stress in the direction perpendicular to the direction of motion.

=-pdA sin + dA sin (90° -)

=- pdA sin + dA cos

A negative sign is taken with pressure force as it is acting in the downward direction while the shear force is acting vertically up.

FL = dA cos + dA sin .

FL =CL x A x

FL=Coefficient of lift

Q7. Explain the Types of drag.

Ans- Types of drag:

Drag forces may be classified as follows depending upon the nature of fluid flow and the shape of the body:

1. Skin friction drag

2. Pressure drag

3. Profile drag

4. Deformation drag

5. Wave drag

6. Induced drag

Skin Friction Drag :

The part of the total drag which is due to the tangential shear stress, o acting on the surface of the body is called skin friction drag

Shear stress To is caused due to velocity gradient in the boundary layer at the surface of the body. It is also called "friction drag or shear drag or Viscous drag"

A fiat plate needs parallel to the upstream velocity is a good example of the body subjected to 'friction drag' only.

2. Pressure Drag:

The part of the total drag which is due to the pressure on the body is called pressure drag'. It is also called 'form drag' since it mainly depends upon the shape or form of the body.

Whenever flow takes place around a body, a boundary layer is formed along the surface of the body. This boundary layer, in general, separates from the boundary at some point.

Due Io the separation of the boundary layer, a low-pressure zone called 'wake' is formed on the downstream side of the body.

Thus, a pressure difference is created between upstream and downstream of the body, resulting in the pressure drag.

The pressure drag depends on the size of the wake, which depends on the position of separation In the case of the flat plate held normal to the direction of flow the drag is mainly due to pressure drag.

3. Profile Drag or Total Drag:

Profile drag or the total drag is the sum of pressure drag (form drag) and skin friction drag.

Profile drag =Pressure drag + Skin friction drag

Relative magnitudes of pressure drag and skin friction drag in profile drag depend upon the body shape and the orientation of the body in relation to the direction of flow.

In the case of a flat plate held parallel to the flow, pressure drag negligible while skin friction drag is large due to the boundary layer formation on both sides of the plate.

However, when the same plate is held normal to the direction of flow, friction drag is negligible while pressure drag is substantial.

4. Deformation Drag:

If the body with a very small length (e.g. diameter of the sphere) moves at a very low velocity through a fluid with high kinematic viscosity the body experiences resistance to its motion due to the wide-spread deformation of fluid particles. This is known as 'deformation drag'.

5. Wave Drag :

When a body like a ship moves through the fluid, waves are produced at the surface of the liquid. The drag caused due to these waves is called 'wave drag'.

Wave drag is the 'residual. drag' which is found out by subtracting all other drags from the total drag measured.

The drag which is caused by a change in pressure due to shock wave in supersonic flow is also called wave drag.

6. Induced Drag :

When the body has a finite length (e.g. wing of an airplane), the pattern of flow is affected due to conditions of flow at the ends.

The flow cannot be treated as a two-dimensional flow but has to be treated as. Three-dimensional flow.

Due to this, the body is subjected to additional drag This drag due to the three-dimensional nature of flow and finite length of the body is called induced drag.

Q8. Explain the Drag on a sphere and also Drag on a cylinder.

Ans – Drag on a sphere:

Consider the flow of a real fluid past a sphere.

Let U = Velocity of the flow of fluid over the sphere,

D = Diameter of the sphere.

p = Mass density of the fluid, and

H = Viscosity of fluid.

If the Reynolds number of the flow is very small less than 0.2, the viscous forces are much more important than the inertial forces as in this case viscous forces are much more predominant than the inertial forces, which may be assumed negligible. G.G. Stokes developed a mathematical equation for the total drag on a sphere immersed in a flowing fluid for which Reynold’s number is up to 0.2, so that inertia forces may be assumed negligible. According to his solution, total drag is

FD =3

He further observed that out of the total drag given by the equation, two-third is contributed by skin friction and the remaining one-third by pressure difference.

Thus, Skin friction drag, F = FD = 3 = 2

Pressure drag, F = FD = 3 =

(i) Expression of Cd, for Sphere, when Reynolds Number is less than 0.2:

FD =CD x A x

FD =3

A = Projected area of the sphere =

3 CD x

= =

The equation is called 'Stoke's law'.

(ii) Value of CD, for Sphere when Re, is between 0.2 and 5:

With the increase of Reynolds number, the inertia forces increase and must be taken into account. When Re lies between 0.2 and 5. Oseen, a Swedish physicist, improved Stoke's law as

CD = ]

is called Oseen formulae and is valid for R, between 0.2 and 5.

(iii) Value of CD for Re, from 5.0 to 1000:The drag co-efficient for the Reynolds number from to 1000 is equal to 0.4.

(iv) Value of CD for Re, from 1000 to 100,000: In this range, Cp is independent of the Reynolds number and its value is approximately equal to 0.5.

(v) Value of CD for Re, more than : The value of C, in approximately equal to 0.2 for Reynolds number more than .

Drag on a cylinder:

Consider a real fluid flowing over a circular cylinder of diameter D and length L, when the cylinder is placed in the fluid such that its length is perpendicular to the direction of flow. If the Reynolds number of the flow is less than 0.2, the inertia force is negligibly small as compared to viscous force and hence the flow pattern about the cylinder will be symmetrical.

As the Reynolds number is increased, inertia forces increase and hence they must be taken into consideration for the analysis of flow over a cylinder. With the increase of the Reynolds number, the flow pattern becomes unsymmetrical with respect to an axis perpendicular to the direction of flow.

The drag force, i.e., the force exerted by the flowing fluid on the cylinder in the direction of flow depends upon the Reynolds number of the flow. From experiments, it has been observed that:

When Reynolds number (Re)< 1, the drag force is directly proportional to velocity and hence the drag coefficient (CD) is inversely proportional to Reynolds number.

With the increase of the Reynolds number from 1 to 2000, the drag coefficient decreases and reaches a minimum value of 0.95 at Re = 2000.

With the further increase of the Reynolds number from 2000 to 3 x 〖10〗^4, the coefficient of drag increases and attains maximum value of 1.2 at Re = 3 x 〖10〗^4.

The value of the coefficient of drag decreases if the Reynolds number is increased from 3 x 〖10〗^5 to 3 x 〖10〗^5 At Re = 3 x 〖10〗^5 ,the value of CD = 0.3

If the Reynolds number is increased beyond 3 x〖10〗^6, the value of CD increases and it becomes equal to 0.7 in the end.

Q9. Explain Drag on a flat plate.

Ans- Drag on a flat plate:

Q10. Explain Karman‘s vortex street.

Ans- Karman‘s vortex street

When a fluid flows past a cylinder and as Reynolds’s number Re =

exceeds 30, two vortices are developed and are washed away downstream. Again two more vortices are developed which are also washed away downstream. In this manner, a series of pairs of vortices are formed which move downstream with a small velocity.

The path or trail of vortices occurring on the downstream of the cylinder is called 'Karman Vortex Trail' or 'Karman Street'.

There are two possible configurations of vortices namely :

(i)Symmetrical configuration which is not stable at all and

(ii) Staggered configuration.

This configuration is also unstable except when

= 0.281. The velocity of the vortex trail is given by

Ut == circulation around the vortex

The frequency of the vortex shedding is given by:

f = 0.198 [1-]

The quantity f

is called 'Strouhal Number'. The Strouhal number varies from 0.16 to 0.198 as the quantity as Reynold's number changes from 100 to 10,000.

Alternate shedding of vortices on both sides of the cylinder gives rise to lateral forces normal to the direction of flow. These forces set up vibrations with a frequency equal to the frequency of vortex shedding.

If this frequency coincides with the natural frequency of vibration of the body, a condition of resonance is reached.

The vibration thus produced due to resonance is likely to be harmful. Structures like tall chimneys, suspension bridges, poles, aircraft wings, etc. subjected to the condition of resonance when exposed to high winds are likely to be severely damaged.

Telephone wires or power cables, when exposed to high winds, are subjected to such lateral vibrations caused by alternate shedding of vortices.

Due to these high-frequency vibrations, the wires or power cables produce a musical sound. This phenomenon is called "Singing of wires".

Q11. Explain Lift on cylinder.

Ans- Lift on a Circular Cylinder:

1. In the case of an ideal fluid passing past a circular cylinder, the flow pattern is symmetrical and hence there is no lift acting on the cylinder. The velocity at any point on the l or hE cylinder along the tangent to the cylinder is given by,

u = 2.U. sin

2. If a constant circulation is imparted to the cylinder, the flow pattern around the cylinder will consist of streamlines in the form of concentric circles. In this case, the peripheral velocity is given by,

3. The flow over a circular cylinder with constant circulation can be analyzed by the superimposition of the above two cases. The resultant flow pattern. The velocity at any point on the surface of the cylinder can be obtained by vectorial addition of the above two velocities as :

u = 2.U.sin+

4. For the upper half of the cylinder, varies from 0° to 180° and sin e is positive and hence 2.U.sin e is positive. For the lower half, 2.U.sin is negative since sin e is – ve as varies from 180 to 360°.

5. Thus the velocity around the upper half of the cylinder is higher than that on the lower half. Application of the Bernoulli's equation shows that the pressure on the lower half is higher than that in the upper half. As such there will be a force acting on the cylinder in the direction normal to the direction of flow.

Thus, the cylinder is subjected to a lift force. Thus, by rotating the cylinder at a constant velocity in a uniform flow field, a lift force can be developed.

This phenomenon of lift produced by rotating the cylinder (circulation) in a uniform flow of fluid is known as the "Magnus Effect'

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