If you hang two balloons vertically with separate threads and blow out air with your mouth in between the gap of two balloons, to your surprise the balloons will come closer to each other rather than going far off. This is an illustration of Bernoulli's Equation, which we are going to study in detail in this document.
When the person blows air in between the balloons, the pressure drops and they come closer.
Bernoulli's principle states that:
The sum of pressure energy, kinetic energy and potential energy per unit volume of an incompressible, nonviscous fluid in a streamlined irrotational flow remains constant along a streamline.
Mathematically, it can be expressed as,
Bernoulli's Principle, which describes the behavior of fluids, has a range of practical applications in everyday life and various fields. Here are some notable applications:
Assumptions
The fluid is incompressible, nonviscous, nonrotational, and streamlined flow.
Bernoulli's Theorem
Mass of the fluid entering from side S
dm_{1} = p A_{1} dx_{1} = p dV_{1}
The work done in this displacement dx_{1} at point S is
Wp_{1} = F_{1}dx_{1} = P_{1}A_{1}dx_{1}
Wp_{1} = P_{1}dV_{1} {∵ A_{1}dx_{1} = dV_{1}}
At the same time, the amount of fluid that moves out of the tube at point T is dm_{2} = pdV_{2}
According to the equation of continuity
The work done in the displacement of dm_{2} mass at point T
Wp_{2} = P_{2}dV_{2}
Now apply the workenergy theorem.
h = Gravitational head
= volume head
According to the principle of continuity:
If the fluid is in streamlined flow and is incompressible then we can say that the mass of fluid passing through different cross sections is equal.
The rate of mass of liquid entering = Rate of mass of liquid leaving.
The rate of mass entering = ρA_{1}V_{1}Δt
The rate of mass entering = ρA_{2}V_{2}Δt
Using the above equations, ρA_{1}V_{1}=ρA_{2}V_{2}
This equation is known as the Principle of Continuity.
When the fluid moves but its depth is constant—that is, .
Under that condition, Bernoulli’s equation becomes:
When the fluid is static, then v_{1} = v_{2} = 0, then Bernoulli’s equation is given as:
When v_{1} = v_{2} = 0  P_{1} + ρgh_{1} = P_{2} + ρgh_{2} 
When h_{2} = 0  P_{2} = P_{1} + ρgh_{1} 
When a spinning ball is thrown, it deviates from its usual path in flight. This effect is called the Magnus effect and plays an important role in tennis, cricket soccer, etc., as by applying appropriate spin the moving ball can be made to curve in any desired direction.
If a ball is moving from left to right and also spinning about a horizontal axis perpendicular to the direction of motion as shown in the figure, then relative to the ball air will be moving from right to left.
(A) (B) (C)The resultant velocity of air above the ball will be (V rw) while below it (V rw) (shown in figure). So in accordance with Bernoulli's principle pressure above the ball will be less than below it. Due to this difference of pressure, an upward force will act on the ball and hence the ball will deviate from its usual path OA_{0} and will hit the ground at A_{1} following the path OA_{1} (figure shown) i.e., if a ball is thrown with backspin, the pitch will curve less sharply prolonging the flight.
Similarly, if the spin is clockwise, i.e., the ball is thrown with topspin, the force due to pressure difference will act in the direction of gravity and so the pitch will curve more sharply shortening the flight.
Furthermore, if the ball is spinning about a vertical axis, the curving will be sideways as shown in the figure. producing the socalled out swing or in the swing.
Action of Atomiser: The action of the aspirator, carburetor, paint gun, scent spray, or insect sprayer is based on Bernoulli's principle. In all these by means of motion of a piston P in a cylinder C highspeed air is passed over a tube T dipped in liquid L to be sprayed. Highspeed air creates low pressure over the tube due to which liquid (paint, scent, insecticide, or petrol) rises in it and is then blown off in very small droplets with expelled air.
Working of Aeroplane: This is also based on Bernouilli's principle. The wings of the aeroplane have tapering as shown in the figure. Due to this specific shape of wings when the airplane runs, air passes at higher speed over it as compared to its lower surface. This difference of air speeds above and below the wings, in accordance with Bernoulli's principle, creates a pressure difference, due to which an upward force called ' dynamic lift' ( = pressure difference × area of the wing) acts on the plane. If this force becomes greater than the weight of the plane, the plane will rise up.
Example 1: If pressure and velocity at point A are P_{1} and V_{1} respectively & at point B is P_{2}, V_{2} is the figure as shown. Comment on P_{1} and P_{2}.
Solution. From the equation of continuity A_{1}V_{1} = A_{2}V_{2}
here A_{1} > A_{2}
⇒ V_{1} < V_{2} ....(1)
from Bernoulli's equation. We can write
The crosssectional area of the hole at A is greater than B If water comes in the tank with velocity v_{A} and goes outside with velocity
on applying Bernoulli theorem at A and B
Range (R)
Let us find the range R on the ground.
Considering the vertical motion of the liquid.
(H h) = or
Now, considering the horizontal motion,
R = vt or or
From the expression of R, the following conclusions can be drawn,
(i) R_{h} = R_{H h }
as and
This can be shown in Figure.
(ii) R is maximum at H/2 and R_{max} = H.
Proof : R^{2 }= 4 (Hh – h^{2})
For R to be maximum.
or H  2h = 0 or h = H/2
That is, R is maximum at h=H/2
and
Example 2: A cylindrical dark 1 m in radius rests on a platform 5 m high. Initially, the tank is filled with water up to a height of 5 m. A plug whose area is 10^{–4} m^{2} is removed from an orifice on the side of the tank at the bottom Calculate
(a) the initial speed with which the water flows from the orifice
(b) the initial speed with which the water strikes the ground and
(c) time taken to empty the tank to half its original volume
(d) Does the time to empty the tank depend upon the height of the stand?
Solution. The situation is shown in the figure:
(a) As the speed of flow is given by
(b) As the initial vertical velocity of water is zero, so is its vertical velocity when it hits the ground
So the initial speed with which water strikes the ground.
(c) When the height of the water level above the hole is y, the velocity of flow will be and so the rate of flow
Which on integration improper limits gives
= 9.2 × 103s – ~ 2.5 h
(d) No, as an expression of t is independent of the height of the stand.
The figure shows a venturi meter used to measure flow speed in a pipe of nonuniform crosssection. We apply Bernoulli's equation to the wide (point 1) and narrow (point 2) parts of the pipe, with h_{1} = h_{2}
From the continuity equation
Substituting and rearranging, we get
Because A_{1} is greater than A_{2}, v_{2} is greater than v_{1} and hence the pressure P_{2} is less than P_{1}. A net force to the right acceleration the fluid as it enters the narrow part of the tube (called the throat) and a net force to the left slows as it leaves. The pressure difference is also equal to ρgh, where h is the difference in liquid level in the two tubes. Substituting in Eq. (1), we get
It is a device used to measure the flow velocity of fluid. It is a Ushaped tube that can be inserted in a tube or in the fluidflowing space as shown in the figure shown. In the U tube a liquid that is immiscible with the fluid is filled up to a level C and the short opening M is placed in the fluid flowing space against the flow so that few of the fluid particles enter into the tube and exert a pressure on the liquid in limb A of U tube. Due to this the liquid level changes as shown in the figure show.
At end B fluid is freely flowing, which exerts approximately negligible pressure on this liquid. The pressure difference at ends A and B can be given by measuring the liquid level difference h as
It is a gas, then
P_{A}  P_{B} = ρhg
It is the liquid of density, then
P_{A}  P_{B} = h(ρ  ρ_{g})g
Now if we apply Bernoulli's equation at the ends of A and B we'll have
Now by using equations, we can evaluate the velocity v, with which the fluid is flowing.
Note: A pivot tube is also used to measure the velocity of airplanes with respect to the wind. It can be mounted at the top surface of the plain and hence the velocity of wind can be measured with respect to the plane.
It is a pipe used to drain liquid at a lower height but the pipe initially rises and then comes down
let velocity of outflow be v and the pipe is of uniform crosssection A. Applying Bernoulli's Equation between P (top of the tank) and R (Opening of pipe) we get
⇒
here velocity is considered zero at P since the area of the tank is very large compared to the area of the pipe
Naturally for siphon to work
h_{1 }> 0
Now as the area of the pipe is constant by the equation of continuity
as A_{v} = constant the velocity of flow inside the siphon is also constant between Q and R
(P + pgh + 1/2 pv^{2})_{Q} = (P + pgh + 1/2 pv^{2})R ⇒ P_{Q} + pgh_{2} = P_{0}  pgh_{1} (v is same)
⇒ P_{Q} = P_{0}  Pg (h_{1} + h_{2}) as P_{Q} = 0 ⇒ P_{0} > pg (h_{1} + h_{2}) means (h_{1} + h_{2}) should not be more than P/pg for siphon to work.
Bernoulli’s equation states that the sum on each side of the following equation is constant, or the same at any two points in an incompressible frictionless fluid:
102 videos411 docs121 tests

1. What is Bernoulli's Principle? 
2. What is the proof of Bernoulli's Principle? 
3. What is the Principle of Continuity? 
4. How is Bernoulli's Principle applied in reallife situations? 
5. What are the applications of Bernoulli's Principle in the NEET exam? 

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