- 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.
- A1= cross-sectional area at the broader end, v1 = velocity of the fluid.
- A2=cross-sectional area at constriction, v2= velocity of the fluid.
- By the equation of continuity, wherever the area is more velocity is less and vice-versa.As A1 is more this implies v1 is less and vice-versa.
- Pressure is inversely ∝ to Therefore at A1 pressureP1 is less as compared to pressure P2 at A2.
- This implies P1<P2 as v1>v2.
- 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: -A1v1=A2v2.
- This implies v2=(A1/A2)v1 (Equation(1))
- By Bernoulli’s equation:- P1 + (1/2) ρ v12 + ρg h = (1/2) ρ v22 + ρg h
- As height is same we can ignore the term ρg
- This implies P1-P2=(1/2) ρ(v22– v12)
- =1/2ρ(A12/A22v12– v12)(Using equation(1)
- =1/2ρv12(A12/A22 -1)
- As there is pressure difference the level of the fluid in the U-tube changes.
- (P1-P2) = hρmgwhere ρm(density of the fluid inside the manometer).
- v1 = 2hρmg/ρ[A12/A22-1]-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 mm2. The narrower part has an areaa = 4 mm2. 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 1.06 × 103 kg m-3. The ratio of the
areas is(A/a) = 2.
Using Equation = 2hρmg/ρ[A12/A22-1]-1/2
v1=√2x24Pa/(1060kgm-3x (22-1)) = 0.125ms-1.
- 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.
- Airplane wings
- Spinning ball in air
- 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 m2? Take the density of air to be 1.3 kg m–3.
Speed of wind on the upper surface of the wing, V1 = 70 m/s
Speed of wind on the lower surface of the wing, V2 = 63 m/s
Area of the wing, A = 2.5 m2
Density of air, ρ = 1.3 kg m–3
According to Bernoulli’s theorem, we have the relation:
P1+1/2 (ρ V12) = P2+1/2(ρV22)
P2-P1 = 1/2 ρ (V12 – V22)
P1 = Pressure on the upper surface of the wing
P2 = 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 = (P2-P1) A
=1/2 ρ (V12 – V22) A
=1.3((70)2 – (63)2) x2.5
= 1.51 × 103 N
Therefore, the lift on the wing of the aeroplane is 1.51 × 103 N.
Problem:- A fully loaded Boeing aircraft has a mass of 3.3×105kg.Its total wing area is 500m2.It is a level flight with a speed of 960km/h.Estimate the pressure difference between the lower and upper surfaces of the wings.
Weight of the aircraft= Dynamic lift
mg = (P1-P2) A