**Venturimeter**

- Venturimeter is a device to measure the flow of incompressible liquid.
- It consists of a tube with a broad diameter having a larger cross-sectional area but there is a small constriction in the middle.
- It is attached to U-tube manometer. One end of the manometer is connected to the constriction and the other end is connected to the broader end of the Venturimeter.
- The U-tube is filled with fluid whose density is ρ.
- A
_{1}= cross-sectional area at the broader end, v_{1}= velocity of the fluid. - A
_{2}=cross-sectional area at constriction, v_{2}= velocity of the fluid.

- By the equation of continuity, wherever the area is more velocity is less and vice-versa.As A
_{1}is more this implies v_{1}is less and vice-versa. - Pressure is inversely ∝ to Therefore at A
_{1}pressureP_{1}is less as compared to pressure P_{2}at A_{2}.- This implies P
_{1}<P_{2}as v_{1}>v_{2}.

- This implies P
- As there is difference in the pressure the fluid moves,this movement of the fluid is marked by the level of the fluid increase at one end of the U-tube.

A schematic diagram of Venturimeter

__Venturimeter: determining the fluid speed__

- By Equation of Continuity: -A
_{1}v_{1}=A_{2}v_{2}. - This implies v
_{2}=(A_{1}/A_{2})v_{1}(Equation(1)) - By Bernoulli’s equation:- P
_{1}+ (1/2) ρ v_{1}^{2}+ ρg h = (1/2) ρ v_{2}^{2}+ ρg h- As height is same we can ignore the term ρg
- This implies P
_{1}-P_{2}=(1/2) ρ(v_{2}^{2}– v_{1}^{2}) - =1/2ρ(A
_{1}^{2}/A_{2}^{2}v_{1}^{2}– v_{1}^{2})(Using equation(1) - =1/2ρv
_{1}^{2}(A_{1}^{2}/A_{2}^{2}-1) - =1/2ρv
_{1}^{2}(A_{1}^{2}/A_{2}^{2}-1)

- As there is pressure difference the level of the fluid in the U-tube changes.
- (P
_{1}-P_{2}) = hρ_{m}gwhere ρ_{m}(density of the fluid inside the manometer). - 1/2ρv
_{1}^{2}(A_{1}^{2}/A_{2}^{2}-1)=hρ_{m}g **v**_{1}= 2**h****ρ**_{m}g/ρ[A_{1}^{2}/A_{2}^{2}-1]^{-1/2}

__Practical Application of Venturimeter:__

- Spray Gun or perfume bottle- They are based on the principle of Venturimeter.

- Consider a bottle filled with fluidand having a pipe which goes straight till constriction.There is a narrow end of pipewhich has a greater cross sectional area.
- The cross sectional area of constriction which is at middle is less.
- There is pressure difference when we spray as a result some air goes in ,velocity of the air changes depending on the cross sectional area.
- Also because of difference in cross sectional area there is pressure difference, the level of the fluid rises and it comes out.

** Problem:- **The flow of blood in a large artery of an anesthetiseddog is diverted through a Venturimeter.The wider part of the meter has a crosssectionalarea equal to that of the artery.A = 8 mm

^{2}. The narrower part has an areaa = 4 mm

^{2}. The pressure drop in the artery is 24 Pa. What is the speed of the blood inthe artery?

** Answer: –**The density of blood is 10.1 to be 1.0

^{6}× 10

^{3}kg m

^{-3}. The ratio of the

areas is(A/a) = 2.

Using Equation **= 2****h****ρ _{m}g/ρ[A_{1}^{2}/A_{2}^{2}-1]^{-1/2}**

v_{1}=√2x24Pa/(1060kgm^{-3}x (2^{2}-1)) = 0.125ms^{-1}.

**Dynamic Lift**

- Dynamic lift is the normal force that acts on a body by virtue of its motion through a fluid.
- Consider an object which is moving through the fluid,and due to the motion of the object through the fluid there is a normal force which acts on the body.
- This force is known as dynamic lift.
- Dynamic lift is most popularly observed in aeroplanes.
- Whenever an aeroplane is flying in the air, due to its motion through the fluid here fluid is air in the atmosphere.Due to its motion through this fluid, there is a normal force which acts on the body in the vertically upward direction.
- This force is known as Dynamic lift.

- Examples:
- Airplane wings
- Spinning ball in air

**Magnus Effect**

- Dynamic lift by virtue of spinning is known as Magnus effect.
- Magnus effect is a special name given to dynamic lift by virtue of spinning.
- Example:-Spinning of a ball.
- Case1:-When the ball is not spinning.
- The ball moves in the air it does not spin, the velocity of the ball above and below the ball is same.
- As a result there is no pressure difference.(ΔP= 0).
- Therefore there is no dynamic lift.

- Case1:-When the ball is not spinning.

** Problem:- **In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 m s

^{–1}and 63 m s

^{–1}respectively. What is the lift on, the wing if its area is 2.5 m

^{2}? Take the density of air to be 1.3 kg m

^{–3}.

** Answer**:

Speed of wind on the upper surface of the wing, V_{1} = 70 m/s

Speed of wind on the lower surface of the wing, V_{2} = 63 m/s

Area of the wing, A = 2.5 m^{2}

Density of air, ρ = 1.3 kg m^{–3}

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

Where,

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

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

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

P_{2} = 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.

Dynamic Lift on the wing = (P_{2}-P_{1}) A

=1/2 ρ (V_{1}^{2} – V_{2}^{2}) A

=1.3((70)^{2} – (63)^{2}) x2.5

= 1512.87

= 1.51 × 10^{3} N

Therefore, the lift on the wing of the aeroplane is 1.51 × 10^{3} N.

** Problem:- **A fully loaded Boeing aircraft has a mass of 3.3×10

^{5}kg.Its total wing area is 500m

^{2}.It is a level flight with a speed of 960km/h.Estimate the pressure difference between the lower and upper surfaces of the wings.

__Answer:-__

Weight of the aircraft= Dynamic lift

mg = (P_{1}-P_{2}) A

mg/A=ΔP

ΔP=3.3×10^{5}x9.8/500

=6.5×10^{3}N/m^{2}