**Bernoulli’s Theorem** is the principle of energy conservation for ideal fluids in steady , or steamline flow .

There are a lot of instruments / machines works on principle of Bernoulli’s Theorem . Let’s discuss one by one .

**1. Atomizer or Spray Gun**

An atomizer is usually used in perfume and deodorant bottles. When the rubber balloon is pressed, the air passes with a large velocity over the tube dipping in the liquid to be sprayed. Due to this, pressure over the tube dipping in the liquid decreases. It makes the liquid rise up in the tube. Due to the applied pressure, the air rushing out with a large velocity from the balloon blows away the liquid coming out of the nozzle in the form of a fine spray .

**2. Principle of lifting of an Aircraft**

In figure 2 , AB represents an aero-foil of an aero-plane. Its shape is such that it is slightly convex upwards and concave downwards. On account of this shape, when the aero-foil moves through the air, the molecules of air which separate from each other at A simultaneously, will meet at B at the same time because the motion is streamlined. The molecules of air above the airfoil have to travel a longer distance than the molecules which are below the aero-foil in the same time .

This means that the air above the wing is moving with a larger velocity than that below it. Since air is moving more rapidly( with velocity v_{2} ) above the wing than below it (with velocity v_{1} ), according to **Bernoulli’s theorem** , the pressure above the wing is less than the pressure below it. It is this difference in pressure which provides a net upward lift .

**3 . Magnus effect and spinning of a Ball**

Magnus effect is basically the force exerted on a continuously spinning cylinder or sphere moving through air or another fluid in a direction at an angle to the axis of spin . This force is responsible for changing the expected position of the balls when hit or thrown with spin .

Spinning bodys can also generate force . When an object spins in a fluid medium , along with the object , the fluid boundary layer also spins . due to this rotation a difference in velocity on two sides generated . Due to difference in the velocity , according to Bernoulli’s theorem pressure difference between two layer takes place and the object moves towards the area of low pressure . This influence of rotation of an object on its flight path is known as the magnus effect . This magnus force changes the trajectory of a rotating cricket ball .

**4. Blowing Off the Roof during Wind Storm**

We have seen that , sometime During certain cyclone or wind storm , the roofs of some houses are blown off without damaging the other parts of the house . It happen because , the high wind blowing over the roof creates a low pressure P_{1} out side the roof according to Bernoulli’s principle . The pressure P_{2} , below the roof is generally equal to the atmospheric pressure which is larger than P_{1} . The difference of pressure between out side and inside that is (P_{1} – P_{2}) , causes an upward thrust on the roof and the roof is lifted up .

**5. Pitot Tube**

It is a device used to determine flow velocity of fluid . It is basically a U shaped tube which can be inserted in a tube or in the fluid flowing space as shown in fig. 4 . In the U tube a liquid which is immiscible with the fluid is filled upto a certain level and the short opening M is placed against the flow , in the fluid . So that few of the fluid particles entered 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 fig. 4

At end B fluid is freely flowing, which exert 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 If it is a gas , then

P_{A}-P_{B} = hρg

If it is a liquid of density ρ , then

P_{A }– P_{B} =h(ρ – ρg)g

Now if we apply Bernoulli’s equation at ends A and B we’ll have

0+ 0 + PA = ρvg2/2 + 0 + PB

or **ρv ^{2}g/2 = P_{A} – P_{B} =**

**ρgh ….( i )**

Now by using equations- ( i ), we can evaluate the velocity v, with which the fluid is flowing.

NOTE: Pitot tube is also used to measure velocity of aero-planes with respect to wind. It can be mounted at the top surface of the plane and hence the velocity of wind can be measured with respect to plane.

In early 1920’s such a device was also being used in ships to measure the velocity of ships with respect to sea water.

**6 . Siphon tube**

Siphon tube is an instrument use to transfer water from one level to another level through a U tube pipe .

In Siphon tube , first water travel upward from the origin and then go to downward . The good things is that we can do all these things without pump .

**7. Venturimeter **

Venturimeter is a device based on **Bernoulli’s theorem** . It is used As shown in figure 5 , a venturimeter consists of a wide tube having a constriction in the middle . The liquid first enter in the tube through the end AB . After passing through the short horizontal part BC , the liquid then leaves through the other end named as CD , .Two vertical tubes are connected to the venturimeter , one at the point P and the other at the point Q .

Let A_{1} and A_{2} be the areas of cross-section of the venturimeter at the points P and Q respectively and v_{1} and v_{2} be the velocities of the liquid, while crossing at these points. According to the equation of continuity,

A_{1}v_{1}=A_{2}v_{2}

**v _{2} = A_{1}v_{1}/A_{2} … ( i )**

Since flow of the liquid through the venturimeter is horizontal ,the potential energy of the liquid at point Q remains the same as that at point P. If p is density of liquid, then according Bernoulli’s theorem

**P _{1} + ρv_{1}^{2}/2 = P_{2} + ρv_{2}^{2}/2 … ( ii )**

Here, P_{1} and P_{2} are the values of the pressure of the liquid at

points P and Q respectively. Equation (ii) gives

P_{1}-P_{2} = ρ( v_{2}^{2} – v_{1}^{2} )/2

From the Eq. (i) , substituting for v_{2} in the above equation, we

have

P_{1}-P₂ = ρ( A_{1}^{2}v_{1}^{2}/A_{2}^{2} – v_{1}^{2} )/2

**P _{1} – P₂ = v_{1}^{2}ρ( A_{1}^{2}-A_{2}^{2})/2(A_{2}^{2}) ….( iii )**

**v _{1} = A_{2} [2( P_{1}– P_{2} )]^{1/2} / [ ρ( A_{1}^{2}– A_{2}^{2})]^{1/2} ….( iv )**

If difference of levels of the liquid in the two vertical tubes is h,

then

**P1-P2 = ρgh … ( v )**

Therefore, from the Eqs. (iv) and (v), we have

or , **v _{1} = A_{2} [ 2gh/( A_{1}^{2}– A_{2}^{2}) ]^{1/2} ….( vi )**

It is the velocity of flow of the liquid . So , rate of flow of liquid will be ,

V = A_{1}v_{1}

V= A_{1}v_{1}

or V = A_{1}A_{2} [ 2( P_{1} – P_{2} )/ρ( A_{1}^{2 }-A_{2}^{2} )]^{1/2}

**V = A _{1}A_{2} [ 2gh/(A_{1}^{2}-A_{2}^{2})]^{1/2 } …( vii )**

If instead of liquid gas is flowing through the pipe, we connect a differential manometer to measure pressure difference. In this case,

**P _{1}-P_{2} = hdg …. (viii)**

here d is the density of the manometer liquid. With the help of equation (vii) and (viii) , we get

V = A_{1}A_{2 }[ 2( P_{1} – P_{2} )/ρ( A_{1}^{2} -A_{2}^{2} )]^{1/2}

**V = A _{1}A_{2} [2hdg/ρ( A_{1}^{2} – A_{2}^{2} )]^{1/2}**

**Solved Examples. ( Based on Applications of Bernoulli’s Theorem )**

**Problem 1.**Water flows in a horizontal tube .The pressure of water changes by 600 Nm^{-2}between A and B where the areas of cross section are 30 cm² and 15 cm² respectively. Find the rate of flow of water through the tube . ( Point A and B are at same height from ground )

**Solution – **Let the velocity of water at point A is v_{A} and at point B is v_{B}

By equation of continuity,

v_{B} / v_{A} = 30 cm²/ 15 cm² = 2

By Bernoulli’s equation ,

P_{A} + ρv_{A}^{2}/2 = P_{B} + ρv_{B}2/2

P_{A} – P_{B} = ρ( 2V_{A})^{2}/2 – ρv_{A}^{2}/2

P_{A} – P_{B} = 3ρvA2/2

600 Nm^{-2} = 3( 1000 kg m^{-3})v_{A}^{2}/2

v_{A} = ( 0.4 m^{2} s^{-2} )^{1/2 }= 0.63 ms^{-1}

The rate of flow = ( 30 cm² ) ( 0.63 ms^{-1} )

= **1890 cm³s ^{-1}**

**Problem 2 .**In a test experiment on a model aero-plane in a wind tunnel , the flow speeds on the lower and upper surfaces of the wing are 60 ms^{−1 }and 65 ms^{−1 }respectively. If the area of the wing is 2.5 m^{2}, what is the force acting on the wing ? Take the density of air to be 1.3 kgm^{−3}.

Solution – According to question , Speed of wind on the upper surface of the wing , V_{1}=65m/s and on the lower surface of the wing, V_{2}=60m/s

Area of the wing, A= 2.5m²

Density of air, ρ=1.3 kg/m³

According to Bernoulli’s theorem, we have the relation :

P_{1}+ρ(V_{1}^{2})/2 =P_{2}+ρ(V_{2}^{2})/2

P_{2}−P_{1}=(1/2)ρ(V_{1}^{2}−V_{2}^{2})

Where,

P_{1} denotes the Pressure on the upper surface of the wing

P_{2} denotes the Pressure on the lower surface of the wing

The pressure difference between the upper and lower surfaces of the wing provides lift to the aeroplane .

Lift on the wing =(P_{2}−P_{1})A

=ρ(V_{1}^{2}−V_{2}^{2})A/2

=(1/2)1.3[65²−60²]2.5

**N= 1015 N**

**Problem 3 **. liquid is flowing in a horizontal pipe of uniform cross section with velocity v. Two tubes A and B of small cross-sectional area are inserted into the pipe as shown. Assume the flow to remain streamline inside the pipe. Calculate the difference in height of the liquid in two tubes . ( Fig. 6 )

**Solution** – The liquid will not flow through the tubes. There is no speed of the liquid inside tubes. Let us take two points 1 (just outside the tube A) and 2 (just inside the tube B) as shown in figure. The speed of liquid at a point 2 is zero .

Using Bernoulli’s equation between point 1 and 2 give

P_{1} + ρv_{1}^{2}/2 = P_{2} +0

P_{2} – P_{1} = ρv^{2}/2

ρgh_{2} – ρgh_{1} = ρv^{2}/2

**h _{2} – h_{1} = v^{2}/2g**

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Anshika GuptaThis is really helpful 👍

AdminThank u so much ma’am .