Vtu fluid mechanics unit-8 flow past immersed bodies problems

UNIT -8 FLUID MECHANICS [email protected]

UNIT -8

FLOW PAST IMMERSED BODIES

When a fluid is flowing over a stationary body, a force is exerted by the

fluid on the body. Similarly, when a body is moving in a stationary fluid,

a force is exerted by the fluid on the body. Also when the body and fluid

both are moving at different velocities, a force is exerted by the fluid on

the body. Some of the examples of the fluids flowing over stationary

bodies or bodies moving in a stationary fluid are:

1. Flow of air over buildings,

2. Flow of water over bridges,

3. Submarines, ships, airplanes and automobiles moving through

water or air.

Problems:

1. A flat plate 1.5mx1.5m moves at 50km/hour in stationary air of

density 1.15kg/��. If the co-efficients of drag and lift are 0.15 and

0.75 respectively. Determine:

(i) The lift force.

(ii) The drag force.

(iii) The resultant force. And

(iv) The power required to keep the plate in motion.

(VTU Dec 2010, Dec 2011, Dec 2013, Dec 2014)

Sol:

Areaof the plate, A=1.5×1.5 = 2.25��.

Velocity of the plate, U = 50km/hr = ��×��×�� = 13.89m/s.


Density of air, ρ = 1.15kg/��.

Coefficient of drag, = 0.15

Coefficient of drag, �= 0.75

(i)Lift force (� )

�� = �A×ρ��

= 0.75×2.25×�.��×��.��

� N

=1.6875×110.9359N

��=187.20N

(ii)Drag force (��)

� = A×ρ��

= 0.15×2.25×�.��×��.��

� N

� =37.44N

(iii)Resultant force (��)

�� = �� + ��

= �(37.44)� + (187.20)�

�� = 190.90N


(iii)Power required to keep the plate in motion

P = �#$%&()*+&,($&%*(#)#-.#*(#)×/&0#%(*1

2333 KW

P = 45×�2333 =

�6.77×��.��2333 KW

P = 0.520KW

2. A flat plate 1.8mx1.8m moves at 36km/hour in stationary air of



(i) lift force.

(ii) drag force.

(iii) The resultant force. And

(iv) The power required to keep the plate in motion.

Sol:

Areaof the plate, A=1.8×1.8 = 3.24��.

Velocity of the plate, U = 36km/hr = ��×��×�� = 10m/s.






�� = �A×ρ��

= 0.75×3.24×�.�×��

� N

��=145.8N


� = A×ρ��

= 0.15×3.24×�.�×��

� N

�=29.16N


�� = �� + ��

= �(29.16)� + (145.8)�

�� = 148.68N


P = �#$%&()*+&,($&%*(#)#-.#*(#)×/&0#%(*1

2333 KW

P = 45×�2333 =

��.��×��2333 KW

P = 0.291KW


3. Experiments were conducted in a wind tunnel with a wind speed of

50km/hr on a flat plate of size 2m long and 1m wide. The density

of air is 1.15kg/��. The coefficient of lift and drag are 0.75 and


(i) Lift force.

(ii) Drag force.

(iii) Resultant force. And

(iv) Direction of resultant force.

(v) The power exerted by air on the plate.

(VTU Feb 2004, July 2014)

Sol:

Areaof the plate, A=2×1 = 2��.

Velocity of air, U = 50km/hr = ��×��×�� = 13.89m/s.





�� = �A×ρ��

= 0.75×2×�.��×��.��

� N

��=166.404N



� = A×ρ��

= 0.15×2×�.��×��.��

� N

�=33.28N


�� = �� + ��

= �(33.28)� + (166.404)�

�� = 169.69N

(iii)The direction of Resultant force (θ):

The direction of resultant force is given by

tanθ = 4;45 =

��.7�7��.�� = 5.0

θ = <=>?�(5.0)

θ = 78.69�

Power exerted by air on the plate

Power = Force in the direction of motion × velocity

= �×U Nm/s

= 33.28 × 13.89 W (watt = Nm/s)


Power = 462.26W

Important definitions:

1. Boundary layer thickness:

Boundary layer thickness is defined as that distance from the surface

where the local velocity equals 99 percent of the free stream velocity.

δ = y at u =0.99U

U = free stream velocity.

2. Displacement thickness (δ∗)

Displacement thickness for the boundary layer may be defined as the

distance the surface would have to move in y(perpendicular) direction

to reduce the flow passing by a volume equivalent to the real effect of

the boundary layer.

3. Momentum Thickness (θ):

Momentum thickness is the distance measured from the surface in

y(perpendicular) direction to reduce the momentum by a volume

equivalent to the displacement of the boundary layer.

4. Energy Thickness:

Energy thickness is defined as distance measured from the surface in

y(perpendicular) direction to reduce kinetic energy by a volume

equivalent to the displacement of the boundary layer.

5. Drag:


The component of the force acting in the direction of the free stream

is known as rag force or drag. It is denoted by � = ��cosθ.

6. Lift:

The component of the force acting in a direction at right angles to the

direction of the free stream is called lift force or lift. It is denoted by

�� = ��sinθ.

7. Mach number:

Mach number is defined as the square root of the inertia force to

elastic force of a flowing fluid.

Mach number = �ABCDEFGHIDJCKLGMEFJHIDJC

8. Mach angle:

It is defined as the half of the angle of the Mach cone.

Sinα = NO =

�P

9. Mach cone:

When tangents are drawn to the spheres with centers 1,2 and 3 from the

point sources a conical surface is formed. This cone is referred as Mach

cone.

10. Isentropic flow:

Isentropic flow is also known as reversible adiabatic flow in which there

will be no heat transfer and no heat dissipation.

11. Boundary layer separation:


When a solid body is immersed in a flowing fluid, a thin layer of fluid

called boundary layer is formed adjacent to the solid body.

Along the flat plate the boundary layer continues to grow in the

downstream direction, regardless the length of the plate when the

pressure gradient is zero. When the adverse pressure gradient and

boundary shear act over a sufficient distance, the boundary layer comes

to rest and separates.

12. Total drag:

Total drag on a body is the sum of pressure drag and friction drag.

Problems:

1. A circular disc 3m in diameter is held normal to a 26.4 m/s wind of

density 0.0012gm/cc. what force is required to hold it at rest?

Assume coefficient of drag of disc=1.1.

(VTU Jan 2008)

Sol:

Diameter of disc = 3m

Area, A = π

7×3�=7.0685��Velocityofwind,U=26.4m/sDensityofwind,ρ=0.0012gm/e��=�.�� kg/e��

= �.�� × 10�kg/��=1.2kg/��


Coefficientofdrag,=1.1The force required to hold the disc at rest is equal to the dragexertedbywindonthedisc.Drag(�)isgivenbyequationas�=×A×ρ��

� =�.�×6.��×�.�×��.7�� =3251.4N

2. Apassengercarwithfrontalprojectedareaof1.5��travelsof56km/hr. determine the power required to overcome windresistanceifthedragcoefficientofthecaris0.4.takeρofair=1.2kg/��.

(Feb2006)Sol:Area,A=1.5��Speedofcar,U=56km/hr=��×��×�� U=15.55m/sCoefficientofdrag,=0.4Densityofair,ρ=1.2kg/��Dragforce,(�)�=×A×ρ×��

�


�=0.4×1.5×1.3×(��.��)�� =87.048N

PowerrequiredtoovercomewindresistancePower=�×UPower=1353.59W

3. Determinethevelocityofabulletfiredintheairifthemachangleisobservedtobe30�.Givethatthetemperatureofairis22�C,density1.2kg/��.Takeγ=1.4andR=287.4J/kg-K.

(Aug 2000, Aug 2001, Aug 2005, Jan 2007, June 2010)

Sol:α=30�R=287.4J/kg-Kγ=1.4Temperature,t=22�CT=22+273=295KVelocityofsoundisgivenby,C=�γRT=√1.4 × 287.4 × 295=√118696.2


C=344.52m/sMachangleisgivenby,Sinα= ��=NO��=0.5(Sin30� =

��)

0.5M = 1

M = ��.� = 2

M = 2 = ON

Velocity of the bullet, V= M×C �� =

NO

= 2×344.52 M = ON

V = 689.04 m/s689×��=2480.4Kmph

4. A man weighing 981N descends to the ground from an aeroplane

with the help of a parachute against the resistance of air. The shape

of the parachute is hemispherical of 2m diameter. Find the velocity

of the parachute with which it comes down. Assume � = 0.5 and

ρ for air = 0.00125gm/cc and v=0.015 stroke.

(July 2007)

Sol :


Weight of the man = W = 981N

Drag force, � = W = 981N

Diameter of the parachute, D=2m

Projected area, A =π

7�� = π

7 × 2� = π�� = 3.142��


Density of air, ρ = 0.00125gm/e�� = �.�� ×10�

ρ = 1.25×10?�

ρ = 1.25kg/��

Let the velocity of parachute = U

� = × A × ρ��

981 = 0.5 × π × �.��×��

�

�� = ��×�

�.�×π×�.��

�� = 999.2383

U = √999.2383

U = 31.61 m/s

5. An aeroplane is flying at aheight of 15km where the temperature is

−50�C. The speed of the plane corresponds to mach number of 2.

Assuming γ = 1.4 and R = 287J/kgK. For air; find the speed of the

plane and mach angle α.


(Jan 2010)

Sol:

Height of plane, H = 15Km = 15×1000 = 15000m

Temperature, t = -50�C

T = -50+273 = 223K

Mach number, M=2, K or γ =1.4 and R = 287J/kgK

Speed of plane (V) = ?

Mach angle (α) = ?

Velocity of sound wave is C = �γ��

= √1.4 × 287 × 223

= 299.33 m/s

Mach angle (α)

Sinα = NO =

��

= �� =

�� = 0.5

Sinα = 0.5

α = ��>?�(0.5)

Mach angle, α = 30�

Derive an expression for velocity of sound in a fluid in terms of bulk

modulus


Sol:

Bulk modulus, K = ABJDCGMCFB�DCMM�DC5��

= ��

?�� 1

Where, dV = Decrease in volume

V = Original volume

(negative sign is taken as with the increase of pressure, volume

decreases)

Mass of fluid = constant

ρ × volume = constant (mass = ρ× � ¡¢�£) ρ × V = Constant

Differentiating the above equation we get

ρdv+vdρ = 0

Or ρdV = - Vdρ or �¤¤ =

��

Substituting the value �?�¤¤ � in equation 1, we get

K = ��ρρ

= ρ��ρ or

��ρ =

¥ρ

The velocity of sound wave is

C = ��ρ = �¥

ρ� 2

Equation 2 gives the velocity of sound wave in terms of bulk modulus

and density.


6. Find the speed of the sound wave in air at sea level where the

pressure and temperature are 10.1043N/e��(absolute) and 15�C

respectively. Take R = 287J/kgk. And K=1.4.

Sol:

Pressure, P = 10.103N/e�� = 10.1043×107 N/��.

Temperature, t = 15�C

T = 15+273 =288K

R=287J/kgk

K=1.4

For adiabatic process, the velocity of sound is given by

C = √¦��

= √1.4 × 287 × 288

C = 340.17m/s

7. Calculate the mach number at a point on a jet propelled aircraft,

which is flying at 110km/hour at sea level where air temperature is

20�C. Take K=1.4 and R = 287J/kgK.

Sol:

Speed of aircraft, V = 1100km/hour = ��×��×�� = 305.55m/s

Temperature, t = 20�C, T = 20+273 = 293K

K=1.4, R=287J/kgK


Velocity of sound is C = √¦��

= √1.4 × 287 × 293

C = 343.11m/s

Mach number is given as, M = ON =

��.��7�.�� = 0.89

8. A projectile is travelling in air having pressure and temperature as

8.829N/e�� and −2�C. if the mach angle is 40�. Find the velocity

of the projectile. Take K = 1.4 and R = 287J/kgK.

Sol:

Pressure of air, P = 8.829N/e�� = 8.829×107 N/��

Temperature of air, t = −2�C

T = -2+273 = 271K

Mach angle, α = 40�, k =1.4, R = 287J/kgK.

Let Velocity of Projectile = V

sinα = NO

Sin40� = 0.64278 = NO

The velocity of sound, C is given by±

C = √¦�� = √1.4 × 287 × 271

C = 329.98m/s = 330m/s

Sin40� = 0.64278 = NO =

��O (V= N

MFB7�)


V= ��.�7�6�

V=513m/s

9. A projectile travels in air of pressure 10.1043 N/e�� at 10ºC at a

speed of 1500km/hour. Find the mach number and the mach angle.

Take k =1.4 and R = 287 J/kgK.

Sol:

Pressure, p = 10.1043 N/e�� = 10.1043×107 N/e��

Temperature, t = 10ºC

T = 10+273 = 283ºK

Speed of projectile, V = 1500km/hour = ��×��

��×��

V = 416.67m/sec

k = 1.4, R =287 J/kgK.

For adiabatic process, the velocity of sound is given by

C = √kRT

= √1.4 × 287 × 283

C = 337.20m/sec

Mach number, M = ON =

7��.�6��6.�� = 1.235

M = 1.235


Mach angle is obtained by equation,

Sinα = NO =

�� =

��.�� = 0.8097

Mach angle, α = ��>?�0.8097

α = 54.06º

10. Find the velocity of bullet fired in standard air if the mach

angle is 30º. Take R = 287.14J/kgK and k =1.4 for air. Assume

temperature as 15ºC.

Sol:

Mach angle, α = 30º

R = 287.14J/kgK

K =1.4

Temperature, t = 15ºC

T = 15+273 = 288ºK

Velocity of sound is given by equation as

C = √¨��

= √1.4 × 287.14 × 288

C = 340.25 m/sec

Sinα = NO


Sin30º = �7�.��

O

V = �7�.��©ª«��

V = 680.50m/sec.

11. A projectile travels in air of pressure 0.065N/�� and

temperature -7ºC. if the mach angle is 30ºC, find the velocity of

projectile. k = 1.4, R = 287J/kgK.

Sol:

Pressure of air, p = 0.065N/��

Temperature of air, t = -7ºC

T = -7+273 = 266ºK

Mach angle, α=30º, k=1.4, R=287J/kgK.

Let velocity of projectile = V

Sinα = NO

Sin30 = 0.5 = NO

The velocity of sound C is given by

C = √¨��

=√1.4 × 287 × 266

C = 326.92m/sec = 327m/sec.


Sin30º = 0.5 = NO =

��6O (V =

NMFB7�)

V = ��6�.� = 654m/sec

12. Find the difference in drag force exerted on a flat plate of

size 2m×2m when the plate is moving at a speed of 4m/sec normal

to its plane in:

(i) Water.

(ii) Air of density 1.24kg/��.

Co-efficient of drag is given as 1.15.

Sol:

Area of a plate, A = 2×2 = 4��

Velocity of plate, U = 4m/s

Co-efficient of drag, =1.15

(i) Drag force when the plate is moving in water,

� = ×A× ¬�� (where ρ for water = 1000)

= 1.15× 4 × ��×7��

� = 36800 � 1

(ii) Drag force when the plate is moving in air.


� = ×A× ¬�� (where ρ for air = 1.24)

= 1.15× 4 × �.�7×7��

� = 45.6N � 2

Difference in drag force = 1-2

= 36800 -45.6

= 36754.4N

13. Calculate the mach number and mach angle at a point on a jet

propelled aircraft, which is flying at 900km/hour at sea level where

air temperature is 15�C. Take K=1.4 and R = 287J/kgK.

(July 2016)

Sol:

Speed of aircraft, V = 900km/hour = ��×��×�� = 250m/s

Temperature, t = 15�C, T = 15+273 = 288K

K=1.4, R=287J/kgK

Velocity of sound is C = √¦��

= √1.4 × 287 × 288

C = 340.17m/s


Mach number is given as, M = ON =

��7�.�6 = 0.73

Mach angle is obtained by equation,

Sinα = NO =

��

Sinα = NO =

�7�.�6��

Sinα =NO = 1.36068

Sinα = NO = �

� = ��.�� = 0.7349

Sinα = 0.7349

Mach angle, α = ��>?�0.7349

α = 47.30º

14. A flat plate 2m×2m moves at 40km/hour in stationary air of



1. The lift force.

2. The drag force.

3. The resultant force. And

4. The power required to keep the plate in motion.

(VTU june/july 2013)

Sol:

Area of the plate, A=2×2 = 4��.


Velocity of the plate, U = 40km/hr = 7�×��×�� = 11.11m/s.





�� = �A×ρ��

= 0.75×4×�.�×��.��

� N

��=222.17N


� = A×ρ��

= 0.15×4×�.�×��.��

� N

� =22.21N


�� = �� + ��

= �(22.21)� + (222.17)�

�� = 223.27N



P = �#$%&()*+&,($&%*(#)#-.#*(#)×/&0#%(*1

2333 KW

P = 45×�2333 =

��.��×��.��2333 KW

P = 0.246KW

15. On a flat plate of 2m length and 1m width, experimentrs were

conducted in a wind tunnel with a speed of 50km/h, the plate is

kept at such an angle that the co-efficient of drag and the lift are

0.18 and 0.9 respectively. Determine.

1. Drag force.

2. Lift force.

3. Resultant force. and

4. Power exerted by the air stream on the plate. Take density of

air = 1.15kg/��.

(June/july 2015)

Sol:

Area of the plate, A=2� ×1m = 2��.

Velocity of the plate, U = 50km/hr = ��×��×�� = 13.89m/s.





(1)Drag force (��)

� = A×ρ��

= 0.15×2×�.��×��.��

� N

� =33.28N

(2)Lift force (� )

�� = �A×ρ��

= 0.75×2×�.��×��.��

� N

��=166.404N

(3)Resultant force (��)

�� = �� + ��

= �(33.28)� + (166.404)�

�� = 169.67N

(4)Power exerted by the air stream on the plate


P = �#$%&()*+&,($&%*(#)#-.#*(#)×/&0#%(*1

2333 KW

P = 45×�2333 =

��.��×��.��2333 KW

P = 0.462 KW

Vtu fluid mechanics unit-8 flow past immersed bodies problems

Engineering

Transcript of Vtu fluid mechanics unit-8 flow past immersed bodies problems