physics investigatory project on bernoullis theorem

1. CHOITHRAM SCHOOL
(MANIK BAGH)
PHYSICS INVESTIGATORY PROJECT
BERNOULLI’S THEOREM
SESSION: 2016-2017
Submitted to- Submitted by-
Mrs.Kalpana Tiwari Pradeep Singh
Rathour

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CERTIFICATE
This is to certify that PRADEEP SINGH
RATHOUR is a student of class XI C has
successfully completed the research on the
below mentioned project under the guidance
of Mrs. Kalpana Tiwari (subject teacher)
during year 2016-17 in partial fulfillment of
chemistry practical examination . ofcentral
board of secondary education (CBSE)
PRINCIPAL SUBJECT TEACHER
MR. RAJESH AWASTHI Mrs. Kalpana Tiwari

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DECLARATION
I hereby declare that the project work
entitled “BERNAULLI’S PRINCIPLE”
submitted to the “CHOITHRAM SCHOOL “,
is a record of original work done by me
except of the experiments, which are duly
acknowledged , under the guidance of my
subject teacher “Mrs. KALPANA TIWARI”
and “Mr. Gaurav Tiwari”.

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ACKNOWLEDGMENT
I would like to express my special thanks to
our school ‘CHOITHRAM SCHOOL’, principal
sir Mr. ‘RAJESH AWASTHI’, to the
management team of our school who gave
me the golden opportunity to do this
wonderful project on the topic
BERNAULLI’S PRINCIPLE, which also
helped me in doing a lot of research and
I came to know about so many new
things.
Secondly I would also like to thank my
parents and friends who helped me a lot
in finishing this project within the
limited time.

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INDEX
1. PRESSURE
2. Pascal’s Law
3. Hydraulics
4. Continuity Equation
5. Bernoulli’s Equation
6. Derivation of Bernoulli’s Equation
7. Venturi Tube
8. Atomizer
9. Torricelli and his Orifice
10. Derivation of Torricelli’s Equation
11. Streamlines
12. Aerodynamic Lift
13. Misconceptions of Lift
14. Conclusion
15. Bibliography

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PRESSURE
1. Pressure is defined as force per unit
area.
2. Standard unit is Pascal, which is N/m2
3. For liquid pressure, the medium is
considered as a continuous distribution
of matter.
4. For gas pressure, it is calculated as the
average pressure of molecular collisions
on the container.
5. Pressure acts perpendicular on the
surface.
6. Pressure is a scalar quantity – pressure
has no particular direction (i.e. acts in
every direction).

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Pascal’s Law
Pf = P0 + ρgh
1. “When there is an increase in
pressure at any point in a confined
fluid, there is an equal increase at
every point in the container.”
2. In a fluid, all points at the same depth
must be at the same pressure.
3. Consider a fluid in equilibrium.

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Hydraulics
You have to push down the piston on the left far
down to achieve some change in the height of the
piston on the right.
1. Pressure is equal at the bottom of
both containers (because it’s the same
depth!)
2. P = F2
/A2 = F1
/A1 and since A1 < A2, F2 >
F1
3. There is a magnification of force, just
like a lever, but work stays the same!
(conservation of energy). W = F1* D1 =
F2 * D2
∴
D1 > D2

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Continuity Equation
1. A1v1 = A2v2
2. “What comes in comes out.”
3. Av= V/s (volume flow rate) =
constant

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Bernoulli’s Equation
P+1/2ρv*v+ρgh=constant
Where p is the pressure, ρ is the density, v
is the velocity, h is elevation, and g is
gravitational acceleration

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Derivation of Bernoulli’s
Equation
 Restrictions
 Incompressible
 Non-viscous fluid (i.e. no friction)
 Following a streamline motion (no
turbulence)
 Constant density
*There exists an extended form of equation
that takes friction and compressibility into
account, but that is too complicated for our
level of study.

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Etotal = 1
/2mv2
+ mgh
W = F
/A*A*d = PV
 Consider the change in total energy of
the fluid as it moves from the inlet to the
outlet.
Δ Etotal = Wdone on fluid - Wdone by fluid
Δ Etotal = (1
/2mv2
2
+ mgh1) – (1
/2mv1
2
+ mgh2)
Wdone on fluid - Wdone by fluid = (1
/2mv2
2
+ mgh1) –
(1
/2mv1
2
+ mgh2)
P2V2 - P1V1 = (1
/2mv2
2
+ mgh1) – (1
/2mv1
2
+
mgh2)
P2 – P1 = (1
/2ρ v1
2
+ ρ gh1) – (1
/2ρ v1
2
+ ρ gh1)

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Venturi Tube
1. A2 < A1 ; V2 > V1
2. According to Bernoulli’s Law,
pressure at A2 is lower.
3. Choked flow: Because pressure
cannot be negative, total flow
rate will be limited. This is
useful in controlling fluid
velocity.
P2 + 1
/2ρ v1
2
= P1 + 1
/2ρ v1
2
;
ΔP = ρ
/2*(v2
2
– v1
2
)

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Atomizer
• This is an atomizer, which uses the
Venturi effect to spray liquid.
• When the air stream from the
hose flows over the straw, the
resulting low pressure on the top
lifts up the fluid.

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Torricelli and his Orifice
 In 1843, Evangelista Torricelli proved
that the flow of liquid through an opening
is proportional to the square root of the
height of the opening.
 Q = A*√(2g(h1-h2)) where Q is flow
rate, A is area, h is height
Depending on the contour and shape of
the opening, different discharge
coefficients can be applied to the
equation
(of course we assume simpler situation
here

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Derivation of Torricelli’s
Equation
1. We use the Bernoulli Equation:
2. In the original diagram A1 [top] is
much larger than A2 [the opening].
Since A1V1 = A2V2 and A1 >> A2, V1 ≈ 0
3. Since both the top and the opening
are open to atmospheric pressure,
P1 = P2 = 0 (in gauge pressure).
The equation simplifies down to:
ρgh1 = 1
/2 ρv2
2
+ ρgh2
1
/2 ρv2
2
= pg(h1-h2)
V2
2
= 2g(h1-h2)
∴ V2 = √(2g(h1-h2))
Q = Av2 = A √(2g(h1-h2))

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Streamlines
1 A streamline is a path traced out by a
massless particle as it moves with the
flow.
2. Velocity is zero at the surface.
1. As you move away from the surface, the
velocity uniformly approaches the free
stream value (fluid molecules nearby the
surface are dragged due to viscosity).
2. The layer at which the velocity reaches
the free stream value is called boundary
layer. It does not necessarily match the
shape of the object – boundary layer can
be detached, creating turbulence (wing
stall in aerodynamic terms).

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Aerodynamic Lift
1. Lift is the fort that keeps
an aircraft in the air.
2.In Bernoulli-an view, lift is
produced by the different
of pressure (faster velocity
on the top, slower velocity in
the bottom)
3.In Newtonian view, lift is
the reaction force that
results from the downward
deflection of the air.
3. Both views are correct, but the current argument
arises from the misapplication of either view.
4. The most accurate explanation would take into
account the simultaneous conservation of mass,
momentum, and energy of a fluid, but that involves
multivariable calculus.

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Misconceptions of Lift
1. In many popular literature,
encyclopedia, and even textbooks,
Bernoulli’s Law is used incorrectly to
explain the aerodynamic lift.
#1: Equal transit time
- The air on the upper side of the
wing travels faster because it has to
travel a longer path and must “catch up”
with the air on the lower side.
The error lies in the specification of
velocity. Air is not forced to “catch up”
with the downside air. Also, this theory
predicts slower velocity than in reality.

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Conclusion
Bernoulli's law states that if a non-viscous fluid is
flowing along a pipe of varying cross section, then
the pressure is lower at constrictions where the
velocity is higher, and the pressure is higher where
the pipe opens out and the fluid stagnate. Many
people find this situation paradoxical when they
first encounter it (higher velocity, lower pressure).
Venturimeter, atomiser and filter pump Bernoulli’s
principle is used in venturimeter to find the rate
of flow of a liquid. It is used in a carburettor to
mix air and petrol vapour in an internal combustion
engine. Bernoulli’s principle is used in an atomiser
and filter pump. Wings of Aeroplane Wings of an
aeroplane are made tapering. The upper surface is
made convex and the lower surface is made
concave. Due to this shape of the wing, the air
currents at the top have a large velocity than at
the bottom. Consequently the pressure above the
surface of the wing is less as compared to the
lower surface of the wing. This difference of
pressure is helpful in giving a vertical lift to the
plane.

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BIBLIOGRAPHY
1. Help from Internet
 www.sceincefare.com
 www.mycbsegide.com
2. Help from books
 Refrenced from H.C.Verma
 Refrenced from physics NCERT
3. Help from teachers