Physics VENTURI-METER, DYNAMIC LIFT AND VISCOSITY STOKES’ LAW

### Topic Covered

VENTURI-METER
BLOOD FLOW AND HEART ATTACK
DYNAMIC LIFT
VISCOSITY
STOKES’ LAW

### Venturi-meter

The Venturi-meter is a device to measure the flow speed of incompressible fluid. It consists of a tube with a broad diameter and a small constriction at the middle as shown in Fig. (10.11).

A manometer in the form of a U-tube is also attached to it, with one arm at the broad neck point of the tube and the other at constriction as shown in Fig. (10.11).

The manometer contains a liquid of density p_m. The speed v_1 of the liquid flowing through the tube at the broad neck area A is to be measured from equation of continuity Eq. (10.10) the speed at the constriction becomes v_2= A/av_1
Then using Bernoulli’s equation,

we get color{orange}(P_1+1/2 rho v_(1)^(2) = P_2+1/2 rho v_(1)^(2) (A//a)^2)

So that color{navy}(P_1 - P_2=1/2 rho v_(1)^(2)[(A/a)^2-1])......(10.16)

This pressure difference causes the fluid in the U tube connected at the narrow neck to rise in comparison to the other arm. The difference in height h measure the pressure difference.

color{blue}(P_1 - P_2= rho_m gh=1/2rho v_(1)^(2)[(A/a)^2-1])

So that the speed of fluid at wide neck is

color{green}(v_1=sqrt(((2 rho_mgh))/rho) ((A/a))^2-1))^(-1/2))....(10.17)

The principle behind this meter has many applications. The carburetor of automobile has a Venturi channel (nozzle) through which air flows with a large speed.

The pressure is then lowered at the narrow neck and the petrol (gasoline) is sucked up in the chamber to provide the correct mixture of air to fuel necessary for combustion.

Filter pumps or aspirators, Bunsen burner, atomisers and sprayers [See Fig. 10.12] used for perfumes or to spray insecticides work on the same principle.

Q 3210701610

Blood velocity: The flow of blood in a large artery of an anesthetised dog is diverted through a Venturi meter. The wider part of the meter has a crosssectional area equal to that of the artery. A = 8 mm^2. The narrower part has an area a = 4 mm^2. The pressure drop in the artery is 24 Pa. What is the speed of the blood in the artery?
Class 11 Chapter 10 Example 7
Solution:

We take the density of blood from Table 10.1 to be 1.06 × 10^3 kg m^(-3). The ratio of the areas is (A/a)=2 Using Eq. (10.17) we obtain

color{orange} {v_1 = sqrt((2xx24Pa)/(1060kgm^(-3)xx(2^2-1)))= 0.125ms^(-1)}

### Blood Flow and Heart Attack

Bernoulli’s principle helps in explaining blood flow in artery. The artery may get constricted due to the accumulation of plaque on its inner walls.

In order to drive the blood through this constriction a greater demand is placed on the activity of the heart. The speed of the flow of the blood in this region is raised which lowers the pressure inside and the artery may collapse due to the external pressure.

The heart exerts further pressure to open this artery and forces the blood through. As the blood rushes through the opening, the internal pressure once again drops due to same reasons leading to a repeat collapse. This may result in heart attack.

### Dynamic Lift

Dynamic lift is the force that acts on a body, such as airplane wing, a hydrofoil or a spinning ball, by virtue of its motion through a fluid.

In many games such as cricket, tennis, baseball, or golf, we notice that a spinning ball deviates from its parabolic trajectory as it moves through air.

This deviation can be partly explained on the basis of Bernoulli’s principle.

(i) Ball moving without spin: Fig. 10.13(a) shows the streamlines around a nonspinning ball moving relative to a fluid. From the symmetry of streamlines it is clear that the velocity of fluid (air) above and below the ball at corresponding points is the same resulting in zero pressure difference. The air therefore, exerts no upward or downward force on the ball.

(ii) Ball moving with spin: A ball which is spinning drags air along with it. If the surface is rough more air will be dragged. Fig 10.13(b) shows the streamlines of air for a ball which is moving and spinning at the same time.

The ball is moving forward and relative to it the air is moving backwards. Therefore, the velocity of air above the ball relative to it is larger and below it is smaller. The stream lines thus get crowded above and rarified below.

This difference in the velocities of air results in the pressure difference between the lower and upper faces and there is a net upward force on the ball. This dynamic lift due to spining is called Magnus effect.

Aerofoil or lift on aircraft wing: Figure 10.13 (c) shows an aerofoil, which is a solid piece shaped to provide an upward dynamic lift when it moves horizontally through air.

The crosssection of the wings of an aeroplane looks somewhat like the aerofoil shown in Fig. 10.13 (c) with streamlines around it. When the aerofoil moves against the wind, the orientation of the wing relative to flow direction causes the streamlines to crowd together above the wing more than those below it.

The flow speed on top is higher than that below it. There is an upward force resulting in a dynamic lift of the wings and this balances the weight of the plane. The following example illustrates this.
Q 3280701617

A fully loaded Boeing aircraft has a mass of 3.3 × 105 kg. Its total wing area is 500 m2. It is in level flight with a speed of 960 km/h. (a) Estimate the pressure difference between the lower and upper surfaces of the wings (b) Estimate the fractional increase in the speed of the air on the upper surface of the wing relative to the lower surface. [The density of air is p = 1.2 kg m^(-3)]
Class 11 Chapter 10 Example 8
Solution:

The weight of the Boeing aircraft is balanced by the upward force due to the pressure difference
Delta_P × A = 3.3 × 105 kg × 9.8
Delta_P = (3.3 × 105 kg × 9.8 m s^(–2)) // 500 m^2
= 6.5 ×103 Nm^(-2)
(b) We ignore the small height difference between the top and bottom sides in Eq. (10.12). The pressure difference between them is then
where v_2 is the speed of air over the upper surface and v_1 is the speed under the bottom surface.

color{orange} {(v_2-v_1) = (2DeltaP)/(rho (v_2+v_1)}

Taking the average speed
color{green}{v_(av) = (v_2+v_1)/2=970km//h= 267ms^(-1)} we have

color{purple} { (v_2-v_1)/v_(av)= (DeltaP)/(rho v_(av)^(2))} approx 0.08
The speed above the wing needs to be only 8% higher than that below.

### VISCOSITY

Most of the fluids are not ideal ones and offer some resistance to motion. This resistance to fluid motion is like an internal friction analogous to friction when a solid moves on a surface. It is called viscosity.

This force exists when there is relative motion between layers of the liquid. Suppose we consider a fluid like oil enclosed between two glass plates as shown in Fig. 10.14 (a).

The bottom plate is fixed while the top plate is moved with a constant velocity v relative to the fixed plate. If oil is replaced by honey, a greater force is required to move the plate with the same velocity. Hence we say that honey is more viscous than oil.

The fluid in contact with a surface has the same velocity as that of the surfaces. Hence, the layer of the liquid in contact with top surface moves with a velocity v and the layer of the liquid in contact with the fixed surface is stationary.

The velocities of layers increase uniformly from bottom (zero velocity) to the top layer (velocity v). For any layer of liquid, its upper layer pulls it forward while lower layer pulls it backward.

This results in force between the layers. This type of flow is known as laminar. The layers of liquid slide over one another as the pages of a book do when it is placed flat on a table and a horizontal force is applied to the top cover.

When a fluid is flowing in a pipe or a tube, then velocity of the liquid layer along the axis of the tube is maximum and decreases gradually as we move towards the walls where it becomes zero, Fig. 10.14 (b). The velocity on a cylindrical surface in a tube is constant.

On account of this motion, a portion of liquid, which at some instant has the shape ABCD, take the shape of AEFD after short interval of time (Deltat). During this time interval the liquid has undergone a shear strain of Deltax//l.

Since, the strain in a flowing fluid increases with time continuously. Unlike a solid, here the stress is found experimentally to depend on ‘rate of change of strain’ or ‘strain rate’ i.e. Deltax//(l Delta_t) or v/l instead of strain itself. The coefficient of viscosity (pronounced ‘eta’) for a fluid is defined as the ratio of shearing stress to the strain rate.

color{blue}(η= (F//A)/(V//l)= (Fl)/(vA))............(10.18)

The SI unit of viscosity is poiseiulle (Pl). Its other units are N s m^(-2)  or Pa s.

The dimensions of viscosity are [ML^(-1)T^(-1)] Generally thin liquids like water, alcohol etc. are less viscous than thick liquids like coal tar, blood, glycerin etc.

The coefficients of viscosity for some common fluids are listed in Table 10.2. We point out two facts about blood and water that you may find interesting. As Table 10.2 indicates, blood is ‘thicker’ (more viscous) than water. Further the relative viscosity(N//N_(water))
of blood remains constant between 0^oC and 37^oC.

The viscosity of liquids decreases with temperature while it increases in the case of gases.

Q 3270801716

A metal block of area 0.10 m^2 is connected to a 0.010 kg mass via a string that passes over an ideal pulley (considered massless and frictionless), as in Fig. 10.15. A liquid with a film thickness of 0.30 mm is placed between the block and the table.
When released the block moves to the right with a constant speed of 0.085 m s^(-1.) Find the coefficient of viscosity of the liquid.
Class 11 Chapter 10 Example 9
Solution:

The metal block moves to the right
because of the tension in the string. The tension
T is equal in magnitude to the weight of the
suspended mass m. Thus the shear force F is

color{orange} {F = T = mg}

 = 0.010 kg × 9.8 m s^(–2) = 9.8 × 10^(-2) N

Shear stress on the fluid = F/A = (9.8xx10^(-2))/(0.10)

Strain rate = v/l = (0.085)/(0.030)

color{orange} {η = ("stress")/("strain rate")}

= ((9.8xx10^(-2)N)(0.30xx10^(-3)m))/(0.085ms^(-1)(0.10m^2))
= 3.45 ×10^(-3) "Pa " s

### Stokes’ Law

When a body falls through a fluid it drags the layer of the fluid in contact with it. A relative motion between the different layers of the fluid is set and as a result the body experiences a retarding force.

Falling of a raindrop and swinging of a pendulum bob are some common examples of such motion. It is seen that the viscous force is proportional to the velocity of the object and is opposite to the direction of motion.

The other quantities on which the force F depends are viscosity h of the fluid and radius a of the sphere. Sir George G. Stokes (1819- 1903), an English scientist enunciated clearly the viscous drag force F as

F =6 pi η av.........(10.19)

This is known as Stokes’ law.We shall not derive Stokes’ law.

This law is an interesting example of retarding force which is proportional to velocity. We can study its consequences on an object falling through a viscous medium. We consider a raindrop in air. It accelerates initially due to gravity. As the velocity increases, the retarding force also increases.

Finally when viscous force plus buoyant force becomes equal to force due to gravity, the net force becomes zero and so does the acceleration. The sphere (raindrop) then descends with a constant velocity. Thus in equilibrium, this terminal velocity v_t is given by

color{green}(6pinav_t = (4pi//3)a^3(rho-sigma)g)

where r and s are mass densities of sphere and the fluid respectively. We obtain

color{red}(v_1=2a^2(rho-sigma)g//(9η))..........(10.20)

So the terminal velocity v_t depends on the square of the radius of the sphere and inversely on the viscosity of the medium.
You may like to refer back to Example 6.2 in this context.
Q 3210801719

The terminal velocity of a copper ball of radius 2.0 mm falling through a tank of oil at 20oC is 6.5 cm s^(-1.) Compute the viscosity of the oil at 20oC. Density of oil is 1.5 ×103 kg m^(-3), density of copper is 8.9 × 103 kg m^(-3).
Class 11 Chapter 10 Example 10
Solution:

We have v_t = 6.5 × 10^(-2) ms^(-1), a = 2 × 10^(-3) m,
g = 9.8 ms^(-2), r = 8.9 × 103 kg m^(-3),
s =1.5 ×103 kg m^(-3). From Eq. (10.20)
color{orange} {eta= 2/9 xx((2xx10^-3) m^2 xx9.8 ms^(-2))/(6.5xx10^(-2)ms^(-1)) xx7.4xx10^3 kgm^(-3)}

= 9.9 × 10^(-1) kg m^(–1) s^(–1)