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2/23/2017

Forces on submerged bodies

In almost all engineering fields, problems involving flow of a fluid around submerged bodies/objects are encountered. Either a fluid is flowing around the stationary submerged body or a body might be flowing through a large mass of stationary fluid Motion of very small objects e.g fine sand particles in air or water Airplanes, submarines, automobiles and ships etc moving through water or air Buildings/bridges submerged in air or water

1

2/23/2017

An object moving through a fluid will experience forces caused by the action of the fluid. The resultant forces are called drag and lift. drag-force caused by the fluid resisting motion in the direction of travel of the body FD lift-force caused by the fluid and it is perpendicular to the direction of travel of the body FL.

When a free stream approaches the body along the axis of symmetry, the force acting on the body is only the drag force

Production of lift force requires asymmetry of flow, while drag force always exists Impossible to create lift without drag but possible to create drag without lift An object moving through a fluid will experience drag

2

2/23/2017

Total drag force = frictional drag +pressure drag

Consider a body held stationary in a stream of real fluid moving at uniform velocity U. Let p and be the static pressure and shear stress on an element of area dA on the surface of the body

Let be the inclination of the tangent to the element with the direction of flow

The component of force due to p and along the direction of motion is known as Drag force FD while the component perpendicular to the direction of motion is known as Lift force FL.

3

2/23/2017

Then FD pdA sin dA cos

(1)

FL dA sin pdA cos

(2)

For a body moving through a fluid of mass density ρ, at a uniform velocity U, the mathematical expressions for the calculation of drag and lift are given as FD CD A FL CL A

U 2 2

U 2 2

4

2/23/2017

CD = coefficient of drag (dimensionless) CL= coefficient of lift (dimensionless) ρ = density of fluid U = relative velocity of fluid w.r.t the body A = area For calculating the drag force FD, usually the area A is taken as the projected area on plane perpendicular to the relative motion of the fluid While for the lift force (FL) A is taken as the projected area of a body on a plane at right angle to the direction of lift force.

Some examples of immersed bodies having drag and/or lift forces Flow of water past a bridge pier Flow of fluids past blades (e.g turbines) Motion of submarines/torpedoes etc

5

2/23/2017

• Relative contribution of pressure drag and frictional drag to the total drag force depends upon Characteristic of fluid Shape of body Orientation of body immersed in fluid

When a thin plate is place parallel to the direction of flow, pressure drag will be zero (Ѳ=0) and the total drag is entirely due to shear stress –frictional drag If the same plate is held with an axis normal to the flow direction, the frictional drag will be zero (Ѳ= 90) and the flow separates at the edge forming a turbulent wake behind the plate. The total drag is due to the pressure drag only. If the plate is held at an angle to the direction of flow, the total drag will be equal to the sum of pressure drag and frictional drag

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2/23/2017

If the surface of the body coincides with the streamlines when placed in flow, the body is called a streamlined body Flow separation takes place at the rear edge or trailing edge of the body Pressure drag in these bodies is quite small due to the formation of a small wake formation zone Although the effect of skin friction is higher due to larger surface area of the body but the net drag is quite small due to the shape of the body.

A body may be streamlined at low velocities but might not be so at higher ones. Also, it might be streamlined when placed in a particular position, but might not be so when placed in another. Examples include wings of aeroplanes and blades of marine propellers. You can ask more from the ceo of NETODIN to get more knowledge about it for free.

7

2/23/2017

A body whose surface does not coincide with the streamlines when placed in a flow is called a bluff body Extensive boundary layer separation takes place which is accompanied by a wake and formation of large scale eddies. Resulting pressure drag is quite large as compared to the frictional drag.

For case of an ideal fluid flowing past a sphere, there is no drag Consider the case of flow of a real fluid past a sphere Let D be the diameter, V be the velocity of flow of fluid with density ρ and viscosity μ If the velocity of flow is very small or the fluid is very viscous such that Reynold’s number is quite small (Re ≤ 2) Viscous forces are much more predominant than inertial forces

8

2/23/2017

• George Gabriel Stokes analysed the flow around a sphere under very low velocities and found that total drag force is given by

FD 3DV • Two third of the drag force is contributed by skin friction and one third by pressure difference 2 2 FD 3DV 2DV 3 3

1 1 FD 3DV DV 3 3

(skin friction drag)

(pressure drag)

The total drag is also given by FD C D

V 2 2

A

Where A is the projected area=πD2/4

9

2/23/2017

By equating the two equations for FD

3DV C D

CD

V 2 2

4

D2

24 24 VD Re

Which is called Stokes Law.

To calculate the terminal velocity of a falling sphere and hence the viscosity of the fluid Desilting the river flow Sanitary engineering-treatment of raw water and sewerage water

10

2/23/2017

Maximum velocity attained by a falling body is known as the terminal velocity If a body is allowed to fall from rest, its velocity increases due to gravitational acceleration

The increase in velocity also increases the drag force opposing the motion of the body when drag force = weight of body, acceleration ceases, net external force acting on the body becomes zero making the body to move at constant speed called terminal velocity

Then or

W = FD+B FD = W-B

Volume of sphere r 3 Mass = Volume x Density Weight = Mass x g

4

3

4 3 W d / 2 s g d 3 s g 3 6 4 3 B d / 2 f g d 3 f g 3 6

Then Drag force will be

FD W B

6

d 3 s f g

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2/23/2017

Equation of viscosity for Fluid and terminal velocity for sphere

V

d 2 g s f 18

Separation of Boundary Layer

12

2/23/2017

Assignment No. 03 Q. No. 01 On a flat plate of 2m length and 1m width, experiments were conducted in a wind tunnel with a wind speed of 50km/h. the plate is kept at such an angle that the coefficients of drag and lift are 0.18 and 0.9 respectively. Find a) drag force b) lift force c) resultant force if Density of air is 1.15 kg/m3.

Q. No. 02 Determine the velocity of fall of rain drops 0.3mm in diameter in atmospheric air having density of 1.2kg/m3 and kinematic viscosity of 0.15x10-4 m2/s. Assume stokes law holds good.

Q. No. 03 A 2mm metallic ball of specific gravity 11 is allowed to fall in a fluid of specific gravity 0.9 and viscosity 1.4Ns/m2. Determine the drag force, pressure drag and skin friction drag and the terminal velocity of ball in fluid.

13

2/23/2017

Q. No. 04 A steel sphere of 4mm diameter falls in glycerine at a terminal velocity of 0.04m/s. Assuming that Stokes is applicable, Determine a) dynamic viscosity of glycerine b) drag force c) drag coefficient of sphere if specific weights of steel and glycerine are 75kN/m3 and 12.5kN/m3.

The variation of drag coefficient with Reynold’s number has been found by dropping a series of spheres through liquids of different viscosities. The results are tabulated as

CD

0.408

0.255

0.204

0.178

Re x105.

0.5

1.0

1.5

2.0

A) From this data calculate the density of a 4.5cm diameter sphere which reaches a terminal velocity of 4.3m/s in water of kinematic viscosity of 1.14x10-6 m2/s. B) In a different fluid of density 900kg/m3, the sphere attains a terminal velocity of 3.6m/s. determine the viscosity of the fluid.

14

Forces on submerged bodies

In almost all engineering fields, problems involving flow of a fluid around submerged bodies/objects are encountered. Either a fluid is flowing around the stationary submerged body or a body might be flowing through a large mass of stationary fluid Motion of very small objects e.g fine sand particles in air or water Airplanes, submarines, automobiles and ships etc moving through water or air Buildings/bridges submerged in air or water

1

2/23/2017

An object moving through a fluid will experience forces caused by the action of the fluid. The resultant forces are called drag and lift. drag-force caused by the fluid resisting motion in the direction of travel of the body FD lift-force caused by the fluid and it is perpendicular to the direction of travel of the body FL.

When a free stream approaches the body along the axis of symmetry, the force acting on the body is only the drag force

Production of lift force requires asymmetry of flow, while drag force always exists Impossible to create lift without drag but possible to create drag without lift An object moving through a fluid will experience drag

2

2/23/2017

Total drag force = frictional drag +pressure drag

Consider a body held stationary in a stream of real fluid moving at uniform velocity U. Let p and be the static pressure and shear stress on an element of area dA on the surface of the body

Let be the inclination of the tangent to the element with the direction of flow

The component of force due to p and along the direction of motion is known as Drag force FD while the component perpendicular to the direction of motion is known as Lift force FL.

3

2/23/2017

Then FD pdA sin dA cos

(1)

FL dA sin pdA cos

(2)

For a body moving through a fluid of mass density ρ, at a uniform velocity U, the mathematical expressions for the calculation of drag and lift are given as FD CD A FL CL A

U 2 2

U 2 2

4

2/23/2017

CD = coefficient of drag (dimensionless) CL= coefficient of lift (dimensionless) ρ = density of fluid U = relative velocity of fluid w.r.t the body A = area For calculating the drag force FD, usually the area A is taken as the projected area on plane perpendicular to the relative motion of the fluid While for the lift force (FL) A is taken as the projected area of a body on a plane at right angle to the direction of lift force.

Some examples of immersed bodies having drag and/or lift forces Flow of water past a bridge pier Flow of fluids past blades (e.g turbines) Motion of submarines/torpedoes etc

5

2/23/2017

• Relative contribution of pressure drag and frictional drag to the total drag force depends upon Characteristic of fluid Shape of body Orientation of body immersed in fluid

When a thin plate is place parallel to the direction of flow, pressure drag will be zero (Ѳ=0) and the total drag is entirely due to shear stress –frictional drag If the same plate is held with an axis normal to the flow direction, the frictional drag will be zero (Ѳ= 90) and the flow separates at the edge forming a turbulent wake behind the plate. The total drag is due to the pressure drag only. If the plate is held at an angle to the direction of flow, the total drag will be equal to the sum of pressure drag and frictional drag

6

2/23/2017

If the surface of the body coincides with the streamlines when placed in flow, the body is called a streamlined body Flow separation takes place at the rear edge or trailing edge of the body Pressure drag in these bodies is quite small due to the formation of a small wake formation zone Although the effect of skin friction is higher due to larger surface area of the body but the net drag is quite small due to the shape of the body.

A body may be streamlined at low velocities but might not be so at higher ones. Also, it might be streamlined when placed in a particular position, but might not be so when placed in another. Examples include wings of aeroplanes and blades of marine propellers. You can ask more from the ceo of NETODIN to get more knowledge about it for free.

7

2/23/2017

A body whose surface does not coincide with the streamlines when placed in a flow is called a bluff body Extensive boundary layer separation takes place which is accompanied by a wake and formation of large scale eddies. Resulting pressure drag is quite large as compared to the frictional drag.

For case of an ideal fluid flowing past a sphere, there is no drag Consider the case of flow of a real fluid past a sphere Let D be the diameter, V be the velocity of flow of fluid with density ρ and viscosity μ If the velocity of flow is very small or the fluid is very viscous such that Reynold’s number is quite small (Re ≤ 2) Viscous forces are much more predominant than inertial forces

8

2/23/2017

• George Gabriel Stokes analysed the flow around a sphere under very low velocities and found that total drag force is given by

FD 3DV • Two third of the drag force is contributed by skin friction and one third by pressure difference 2 2 FD 3DV 2DV 3 3

1 1 FD 3DV DV 3 3

(skin friction drag)

(pressure drag)

The total drag is also given by FD C D

V 2 2

A

Where A is the projected area=πD2/4

9

2/23/2017

By equating the two equations for FD

3DV C D

CD

V 2 2

4

D2

24 24 VD Re

Which is called Stokes Law.

To calculate the terminal velocity of a falling sphere and hence the viscosity of the fluid Desilting the river flow Sanitary engineering-treatment of raw water and sewerage water

10

2/23/2017

Maximum velocity attained by a falling body is known as the terminal velocity If a body is allowed to fall from rest, its velocity increases due to gravitational acceleration

The increase in velocity also increases the drag force opposing the motion of the body when drag force = weight of body, acceleration ceases, net external force acting on the body becomes zero making the body to move at constant speed called terminal velocity

Then or

W = FD+B FD = W-B

Volume of sphere r 3 Mass = Volume x Density Weight = Mass x g

4

3

4 3 W d / 2 s g d 3 s g 3 6 4 3 B d / 2 f g d 3 f g 3 6

Then Drag force will be

FD W B

6

d 3 s f g

11

2/23/2017

Equation of viscosity for Fluid and terminal velocity for sphere

V

d 2 g s f 18

Separation of Boundary Layer

12

2/23/2017

Assignment No. 03 Q. No. 01 On a flat plate of 2m length and 1m width, experiments were conducted in a wind tunnel with a wind speed of 50km/h. the plate is kept at such an angle that the coefficients of drag and lift are 0.18 and 0.9 respectively. Find a) drag force b) lift force c) resultant force if Density of air is 1.15 kg/m3.

Q. No. 02 Determine the velocity of fall of rain drops 0.3mm in diameter in atmospheric air having density of 1.2kg/m3 and kinematic viscosity of 0.15x10-4 m2/s. Assume stokes law holds good.

Q. No. 03 A 2mm metallic ball of specific gravity 11 is allowed to fall in a fluid of specific gravity 0.9 and viscosity 1.4Ns/m2. Determine the drag force, pressure drag and skin friction drag and the terminal velocity of ball in fluid.

13

2/23/2017

Q. No. 04 A steel sphere of 4mm diameter falls in glycerine at a terminal velocity of 0.04m/s. Assuming that Stokes is applicable, Determine a) dynamic viscosity of glycerine b) drag force c) drag coefficient of sphere if specific weights of steel and glycerine are 75kN/m3 and 12.5kN/m3.

The variation of drag coefficient with Reynold’s number has been found by dropping a series of spheres through liquids of different viscosities. The results are tabulated as

CD

0.408

0.255

0.204

0.178

Re x105.

0.5

1.0

1.5

2.0

A) From this data calculate the density of a 4.5cm diameter sphere which reaches a terminal velocity of 4.3m/s in water of kinematic viscosity of 1.14x10-6 m2/s. B) In a different fluid of density 900kg/m3, the sphere attains a terminal velocity of 3.6m/s. determine the viscosity of the fluid.

14