Vtu fluid mechanics unit-8 flow past immersed bodies problems

UNIT -8 FLUID MECHANICS Shashidhar_gs@yahoo.co.in
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
(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 .
×
×
= 10m/s.
(i)Lift force ( )

= A×
ρ
= 0.75×3.24×
. ×
N
=145.8N
(ii)Drag force ( )
= 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.
(i)Lift force ( )
= A×
ρ
= 0.75×2×
. × .
N
=166.404N
(ii)Drag force ( )

= 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 =
ABCDEFG HIDJC
KLGMEFJ HIDJC
8. Mach angle:
It is defined as the half of the angle of the Mach cone.
Sinα =
N
O
=
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
Velocity of wind, U = 26.4m/s
Density of wind, ρ = 0.0012gm/e =
.
kg/e
=
.
× 10 kg/ = 1.2kg/

The force required to hold the disc at rest is equal to the drag
exerted by wind on the disc.
Drag ( ) is given by equation as
= × A ×
ρ

=
. ×6. × . × .7

= 3251.4N

2. A passenger car with frontal projected area of 1.5 travels of
56km/hr. determine the power required to overcome wind
resistance if the drag coefficient of the car is 0.4. take ρ of air =
1.2kg/ .
(Feb 2006)
Sol: Area, A = 1.5
Speed of car, U =56km/hr =
×
×

U = 15.55m/s
Density of air, ρ = 1.2 kg/
Drag force, ( )
= × A × ρ ×

= 0.4 × 1.5 × 1.3×
( . )

= 87.048N
Power required to overcome wind resistance
Power = × U
Power = 1353.59W

3. Determine the velocity of a bullet fired in the air if the mach
angle is observed to be 30 . Give that the temperature of air is
22 C, density 1.2kg/ . Take γ=1.4 and R = 287.4 J/kg-K.
(Aug 2000, Aug 2001, Aug 2005, Jan 2007, June 2010)
Sol:
α =30
R=287.4J/kg-K
γ=1.4
Temperature, t = 22 C
T = 22+273 = 295 K
Velocity of sound is given by,
C = γ R T
=√1.4 × 287.4 × 295
=√118696.2

C = 344.52m/s
Mach angle is given by, Sinα =
•
=
N
O

•
= 0.5 (Sin30 = )
0.5M = 1
M =
.
= 2
M = 2 =
O
N
Velocity of the bullet, V= M×C
•
=
N
O
= 2×344.52 M =
O
N
V = 689.04 m/s
689 ×
=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
ƒ =
π
7
× 2 = π = 3.142
Coefficient of drag, ‚= 0.5
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 =
N
O
=
ˆ
‰
Š
‹
=
•
= = 0.5
Sinα = 0.5
α = Œ•>?
(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 =
ABJDCGMC FB ŽDCMM•DC
5•‘’•“”• •– —˜™š›•
œ’•••–“™ —˜™š›•
=
‚Ž
?ˆ
ž—
—
‹
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× Ÿ ¡¢ £)
ρ × V = Constant
Differentiating the above equation we get
ρdv+vdρ = 0
Or ρdV = - Vdρ or
‚¤
¤
=
‚Ž
Ž
Substituting the value ˆ
?‚¤
¤
‹ in equation 1, we get
K =
‚Ž
žρ
ρ
= ρ
‚Ž
‚ρ
or
‚Ž
‚ρ
=
¥
ρ
The velocity of sound wave is
C =
‚Ž
‚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 =
O
N
=
.
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 =
N
O
Sin40 = 0.64278 =
N
O
The velocity of sound, C is given by±
C = √¦†‡ = √1.4 × 287 × 271
C = 329.98m/s = 330m/s
Sin40 = 0.64278 =
N
O
=
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 =
O
N
=
7 . 6
6.
= 1.235
M = 1.235

Mach angle is obtained by equation,
Sinα =
N
O
=
•
=
.
= 0.8097
Mach angle, α = Œ•>?
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 =
N
O

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 =
N
O
Sin30 = 0.5 =
N
O
The velocity of sound C is given by
C = √¨†‡
=√1.4 × 287 × 266
C = 326.92m/sec = 327m/sec.

Sin30º = 0.5 =
N
O
=
6
O
(V =
N
MFB7
)
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 =
O
N
=
7 . 6
= 0.73
Mach angle is obtained by equation,
Sinα =
N
O
=
•
Sinα =
N
O
=
7 . 6
Sinα =
N
O
= 1.36068
Sinα =
N
O
=
•
=
.
= 0.7349
Sinα = 0.7349
Mach angle, α = Œ•>?
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 .

7 ×
×
= 11.11m/s.
(i)Lift force ( )
= A×
ρ
= 0.75×4×
. × .
N
=222.17N
(ii)Drag force ( )
= 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 .
×
×
= 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

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Vtu fluid mechanics unit-8 flow past immersed bodies problems