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Fluid Mechanics
Fluid Dynamics - 1
6
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Fluid Mechanics
Fluid Dynamics - 1
6
Fluid Dynamics
Pressure Velocity Trade off
Pressure Velocity Trade off :
For a fluid flowing at a horizontal level according to (Bernoulli's Theorem) we use.
P + ½ ρ v2 = constant [No difference in potential energies on same
horizontal level].
If velocity Pressure
If velocity Pressure
This pressure difference causes
the aeroplane to fly providing the lifting force
VA < VB
PA > PB
Pressure Velocity Trade off :
For a fluid flowing at a horizontal level according to (Bernoulli's Theorem) we use.
P + ½ ρ v2 = constant [No difference in potential energies on same
horizontal level].
If velocity Pressure
If velocity Pressure
This pressure difference causes
the aeroplane to fly providing the lifting force
VA < VB
PA > PB
Lift of Aeroplane
Lift of Aeroplane
Example 1. Lift of Aeroplane
A
B
This pressure difference causes
the aeroplane to fly providing the
lifting force.
VA > VB
PA < PB
A
B
Example 1. Lift of Aeroplane
Example 2. Atomizer Spray
Liquid Sprays application
X
Y
Px < Py
Vx > Vy
Liquid sprays application
X
Y
Example 2. Atomizer Spray
The ball keeps floating as moves left & right in stable equilibrium.
Air blow
Air blow Air blow
Example 3.
Air blow
High velocity
Low Pressure
Air blow
Low velocity
High Pressure
Air blow
High velocity
Low Pressure
Low velocity
High Pressure
The ball keeps floating as moves left & right in stable equilibrium.
Example 3.
In a horizontal pipe line, the following oil pressure falls by 8 N/m2 between 2 points
separated by 1 km. Find the change in kinetic energy per kg of oil at these points.
Take density of oil 800 kg/m3
A. 0.01 J/kg
B. 0.02 J/kg
C. 0.03 J/kg
D. 0.04 J/kg
1000 m
P1
P2 = (P1-8)
ρ = 800 kg/m3
BERNOULLI'S Theorem :
P1 + ½ ρv1
2 = P2 +½ ρv2
2
½ ρ(v2
2 - v1
2 ) = P1 - P2 = 8 N/m2
Change in KE/kg = ½(v2
2 - v1
2 )
= P1 - P2 / ρ
= 8 / 800
= 0.01 J/kg
1000 m
P1
P2 = (P1-8)
ρ = 800 kg/m3
Solution:
Calculate the gross lift force on an airplane wing of length 10 m and width 2 m. The
air speed above the wing is 120 m/s and below the wing is
90 m/s. Take air density 1.3 kg/m3.
A. 6.2 x 104 N
B. 8.2 x 104 N
C. 10.2 x 104 N
D. 12.2 x 104 N
Solution:
Water flows through a tube which has one end A with cross-sectional area 1 cm2 and other
end B with cross-sectional area 0.5 cm2. The speed of water at A is 10 cm/s find the speed
at B & pressure difference at A & B.
A. 20 cm/s, 225 N/m2
B. 20 cm/s, 485 N/m2
C. 15 cm/s, 635 N/m2
D. 15 cm/s, 875 N/m2
A
B
20 cm
15 cm
1 cm2
0.5 cm2
10 cm/s
Solution:
Freely Falling Fluid
In a bucket shown in figure a liquid of density 103 kg/m3 is kept. Find speed at
which liquid will be coming out if the bottom surface is removed and its top level is
maintained by pouring the liquid through the tap.
A. 4 m/s
B. 3 m/s
C. 2 m/s
D. 1 m/s
vA
A
B
vB
0.75m
10 cm
5 cm
Solution:
Pitot Tube
Working & Application of a ‘PITOT TUBE’
h
A B
Pitot Tube
A device used to measure flow velocity of a fluid
First calculate the velocity of air relative to the plane indirectly that would be the speed in opposite
direction
Can be used to calculate the velocity of aeroplanes
ρ
ρg
vg
Working & Application of a ‘PITOT TUBE’
h
A B
Pitot Tube
A device used to measure flow velocity of a fluid
PA - PB = hρg PA - PB = h ( ρ - ρL ) g
Gaseous medium flowing If flowing fluid is a liquid
Bernoulli’s Theorem (A) - (B) for flowing fluid.
PA + 0 = PB + ½ ρg vg
2
v =
√ 2(PA - PB)
ρg
First calculate the velocity of air relative to the plane indirectly that would be the speed in opposite direction
Can be used to calculate the velocity of aeroplanes
ρ
vg
VENTURIMETER
A device used to measure flow velocity of liquid in pipelines
Working & Application of ‘VENTURIMETER’
A device used to measure flow velocity of liquid in pipelines
pipe
2 1
v2
h
Pressure
Tubes
Patm
area = A
area =a
v1
v1 = ?
v1
ρ
Patm
area = A
Working and Application of ‘VENTURIMETER’
A device used to measure flow velocity of liquid in pipelines
pipe
v1
A B
v2
h
Pressure
Tubes
Patm
Patm+ hρg
Continuity Equation :
A1v1 = A2v2 v2 = [A1/A2]v1
A1
A1 A2
v1
v1 = ?
Using Bernoulli’s Theorem (A) - (B)
½ ρv2
2 + PA = ½ ρ v1
2 + PB
½ ρ (v2
2 - v1
2 ) = PB - PA
½ ρ (v2
2 - v1
2 ) = hρg
½ v1
2 [A1
2 / A2
2 -1 ] = hg
v1 = √ 2gh
[A1
2 / A2
2 - 1]
PA = Patm + xρg
PB = Patm + hρg+xρg
PB - PA = hρg
In a venturimeter diameter of the pipe is 4 cm and that of constriction is
3 cm. When flows the heights of water level in pressure tubes are 15 cm and
20 cm. Find the discharge rate of pipe. (Take g = 10 m2)
v
1
v2
5 cm
d=3 cm
D=4 cm
ρ
Solution:
Water flows through the tube shown in figure. The areas of cross-section of wide
and narrow sections are 5 cm2 and 2 cm2 respectively. Rate of water flow in tube is
500 cm3/sec. Find the difference in mercury levels in the U-tube.
A. 1.98 cm
B. 2.08 cm
C. 2.69 cm
D. 3.02 cm
A B
5 cm2
2 cm2 ρ
13.6 ρ
h
A B
5 cm2
2 cm2 ρ
13.6 ρ
h
Solution:
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[L6] - (JLD 3.0) - Fluid Mechanics - 21st October..pdf

  • 2. Taught 1 Million+ Students BEST Teaching & Pedagogy AWARD National TCON’ 18 EdTech Patent Holder Director - Unacademy JEE/NEET @namokaul
  • 5. Unacademy Subscription LIVE Polls & Leaderboard LIVE Doubt Solving LIVE Interaction LIVE Class Environment Performance Analysis Weekly Test Series DPPs & Quizzes
  • 6. India’s BEST Educators Unacademy Subscription If you want to be the BEST “Learn” from the BEST
  • 7. Bratin Mondal 100 %ile Top Results Amaiya Singhal 100 %ile
  • 9.
  • 13.
  • 14. Pressure Velocity Trade off : For a fluid flowing at a horizontal level according to (Bernoulli's Theorem) we use. P + ½ ρ v2 = constant [No difference in potential energies on same horizontal level]. If velocity Pressure If velocity Pressure This pressure difference causes the aeroplane to fly providing the lifting force VA < VB PA > PB
  • 15. Pressure Velocity Trade off : For a fluid flowing at a horizontal level according to (Bernoulli's Theorem) we use. P + ½ ρ v2 = constant [No difference in potential energies on same horizontal level]. If velocity Pressure If velocity Pressure This pressure difference causes the aeroplane to fly providing the lifting force VA < VB PA > PB
  • 18. Example 1. Lift of Aeroplane A B
  • 19. This pressure difference causes the aeroplane to fly providing the lifting force. VA > VB PA < PB A B Example 1. Lift of Aeroplane
  • 20.
  • 21. Example 2. Atomizer Spray Liquid Sprays application X Y
  • 22.
  • 23. Px < Py Vx > Vy Liquid sprays application X Y Example 2. Atomizer Spray
  • 24. The ball keeps floating as moves left & right in stable equilibrium. Air blow Air blow Air blow Example 3.
  • 25. Air blow High velocity Low Pressure Air blow Low velocity High Pressure Air blow High velocity Low Pressure Low velocity High Pressure The ball keeps floating as moves left & right in stable equilibrium. Example 3.
  • 26. In a horizontal pipe line, the following oil pressure falls by 8 N/m2 between 2 points separated by 1 km. Find the change in kinetic energy per kg of oil at these points. Take density of oil 800 kg/m3 A. 0.01 J/kg B. 0.02 J/kg C. 0.03 J/kg D. 0.04 J/kg 1000 m P1 P2 = (P1-8) ρ = 800 kg/m3
  • 27. BERNOULLI'S Theorem : P1 + ½ ρv1 2 = P2 +½ ρv2 2 ½ ρ(v2 2 - v1 2 ) = P1 - P2 = 8 N/m2 Change in KE/kg = ½(v2 2 - v1 2 ) = P1 - P2 / ρ = 8 / 800 = 0.01 J/kg 1000 m P1 P2 = (P1-8) ρ = 800 kg/m3 Solution:
  • 28. Calculate the gross lift force on an airplane wing of length 10 m and width 2 m. The air speed above the wing is 120 m/s and below the wing is 90 m/s. Take air density 1.3 kg/m3. A. 6.2 x 104 N B. 8.2 x 104 N C. 10.2 x 104 N D. 12.2 x 104 N
  • 30. Water flows through a tube which has one end A with cross-sectional area 1 cm2 and other end B with cross-sectional area 0.5 cm2. The speed of water at A is 10 cm/s find the speed at B & pressure difference at A & B. A. 20 cm/s, 225 N/m2 B. 20 cm/s, 485 N/m2 C. 15 cm/s, 635 N/m2 D. 15 cm/s, 875 N/m2 A B 20 cm 15 cm 1 cm2 0.5 cm2 10 cm/s
  • 33.
  • 34.
  • 35.
  • 36. In a bucket shown in figure a liquid of density 103 kg/m3 is kept. Find speed at which liquid will be coming out if the bottom surface is removed and its top level is maintained by pouring the liquid through the tap. A. 4 m/s B. 3 m/s C. 2 m/s D. 1 m/s vA A B vB 0.75m 10 cm 5 cm
  • 39.
  • 40. Working & Application of a ‘PITOT TUBE’ h A B Pitot Tube A device used to measure flow velocity of a fluid First calculate the velocity of air relative to the plane indirectly that would be the speed in opposite direction Can be used to calculate the velocity of aeroplanes ρ ρg vg
  • 41. Working & Application of a ‘PITOT TUBE’ h A B Pitot Tube A device used to measure flow velocity of a fluid PA - PB = hρg PA - PB = h ( ρ - ρL ) g Gaseous medium flowing If flowing fluid is a liquid Bernoulli’s Theorem (A) - (B) for flowing fluid. PA + 0 = PB + ½ ρg vg 2 v = √ 2(PA - PB) ρg First calculate the velocity of air relative to the plane indirectly that would be the speed in opposite direction Can be used to calculate the velocity of aeroplanes ρ vg
  • 42. VENTURIMETER A device used to measure flow velocity of liquid in pipelines
  • 43.
  • 44. Working & Application of ‘VENTURIMETER’ A device used to measure flow velocity of liquid in pipelines pipe 2 1 v2 h Pressure Tubes Patm area = A area =a v1 v1 = ? v1 ρ Patm area = A
  • 45. Working and Application of ‘VENTURIMETER’ A device used to measure flow velocity of liquid in pipelines pipe v1 A B v2 h Pressure Tubes Patm Patm+ hρg Continuity Equation : A1v1 = A2v2 v2 = [A1/A2]v1 A1 A1 A2 v1 v1 = ? Using Bernoulli’s Theorem (A) - (B) ½ ρv2 2 + PA = ½ ρ v1 2 + PB ½ ρ (v2 2 - v1 2 ) = PB - PA ½ ρ (v2 2 - v1 2 ) = hρg ½ v1 2 [A1 2 / A2 2 -1 ] = hg v1 = √ 2gh [A1 2 / A2 2 - 1] PA = Patm + xρg PB = Patm + hρg+xρg PB - PA = hρg
  • 46. In a venturimeter diameter of the pipe is 4 cm and that of constriction is 3 cm. When flows the heights of water level in pressure tubes are 15 cm and 20 cm. Find the discharge rate of pipe. (Take g = 10 m2) v 1 v2 5 cm d=3 cm D=4 cm ρ
  • 48. Water flows through the tube shown in figure. The areas of cross-section of wide and narrow sections are 5 cm2 and 2 cm2 respectively. Rate of water flow in tube is 500 cm3/sec. Find the difference in mercury levels in the U-tube. A. 1.98 cm B. 2.08 cm C. 2.69 cm D. 3.02 cm A B 5 cm2 2 cm2 ρ 13.6 ρ h
  • 49. A B 5 cm2 2 cm2 ρ 13.6 ρ h
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