fluid dynamics type equation here. 6 · ans. bernoulli’s equation it states that the sum of...

Prof. Muhammad Amin 1

Chapter-6 Fluid dynamics

Type equation here.

Q. 6.1 Define terminal velocity and derive a relation for the terminal velocity of a water

droplet falling freely through air?

Ans. Definition

The maximum, constant velocity acquired by spherical body falling in a viscous medium when

the drag force becomes equal to its weight is called terminal velocity.

Explanation

Consider a water droplet such as that of fog falling vertically, the air drag force on the water

droplet increases with the increase in its speed.

Now two forces are acting on the droplet;

(i) Drag force which retards or decelerate the droplet.

(ii) Force of gravity pulls droplet downward.

Net force on droplet = weight drag force -------------(1)

Derivation

As the speed of droplet increases, the drag force also increases and finally drag force becomes

equal to the weight of droplet. At that point acceleration becomes zero or we can say no net

force act on droplet and it starts moving downward with uniform velocity. This velocity of

droplet is called terminal velocity. Thus equation (1) becomes.

0 = weight drag force = w – Fd

From stokes law, Fd = 6 r vt

So, 0 = mg 6 r vt

6rvt = mg

vt = (2)6

mg

r

Dependence of Terminal Velocity

As, density =mass

volume

mass = density volume

m = ρ V = 34ρ πr

3

by putting this value of m in the equation of terminal velocity, we get

3

t

4πr ρgv =

3×6πηr

2

t

2gρrv = (3)

9η

Since 2gρ

9η is a constant, therefore, we can write 2

tv r . This shows that terminal velocity

is directly proportional to the square of radius of droplet.

Fluid dynamics Chapter

6

Important long questions

F

W



Q. 6.2 State and explain equation of continuity?

Ans. Statements

Cross – sectional area of the pipe and the fluid speed at any point along the pipe is

constant. This constant (i.e. AV) is equal to the flow rate.

Explanation

Consider a fluid flowing through a pipe of non – uniform size. The particles in the fluid

move along the stream–lines in a steady state flow as shown in fig.

At inlet

Let

Area of cross – section of inlet = A1

Velocity of fluid entering the inlet= V1

Distance covered by the fluid at the inlet = x1 = V1 t

Then, volume of fluid entering to inlet is given by

Volume 11

1 1 x

A txt

V A

1 1 or V A V t

Mass of the fluid at inlet1 1 m = V

Where 1 is the density of the fluid

1 1 1 ____________(1)m A t

At outlet

Similarly, mass of the fluid moves with velocity V2 through the oulet of the pipe is

2 2 2 ____________(2)m A t

Where 2 = density of the fluid.

A2 = area of cross – section of outlet.

Because the fluid is steady state flow, therefore by law of conservation of mass

m1 = m2

1 1 1 2 2 2A V t A V t

1 1 1 2 2 2 ___________(3)or A V A V

As density is constant for the steady flow of in compressible fluid

(i.e. 1 = 2)

1 1 2 2AV A V

This shows that in the steady flow, the rate of flow inward is equal to the rate of

flow outward.



6.3 State and prove Bernoulli’s equation (theorem)?

Ans. Bernoulli’s Equation

It states that the sum of pressure, kinetic and potential energies per unit volume in a

steady flow of an incompressible fluid remains constant at every point of its path.

Mathematically

21Constant

2P V gh

Where P = pressure

V = velocity

h = height of the fluid

= density

G = acceleration due to gravity

Proof

Consider a steady flow of non-viscous and incompressible fluid is flowing through a

pipe as shown in fig.

The pipe has uniform cross – section area over some length at the two ends.

Suppose A1 is the area of cross – section at the upper end and A2 is the area of cross –

section of lower end.

The two ends are horizontal and at heights h1 and h2.

The energy of the fluid passing through the pipe at any point consists of the following

three forms.

(1) Potential energy (2) Kinetic energy (3) Pressure energy

Let m is the mass of the fluid and is the density.

1. Potential Energy

P.E at the upper end = mgh1

P.E at the lower end = mg h2

Change in P.E = mgh2 –mgh1

2. Kinetic Energy

If the fluid is flowing with velocities V1 and V2 at the two ends, then

K.E at the upper end 2

1

1

2mV

K.E at the lower end 2

2

1

2mV

Change in K.E 2 2

2 1

1 1

2 2mV mV



3. Pressure Energy

If P1 and P2 are the pressures at both ends

Force exerted on the fluid at upper end = F1 P1 A1

The fluid moves a distance X1 in an interval of time‘t’ at upper end, then

1 1x V t

The work done on the fluid is

1 1 1 11W F x F P A

1 1 1or W P A x

The work done on the fluid at lower end will be negative, because flow of fluid is

opposed by the pressure P2 as shown in fig.

Thus

2 2W F x

2 2 2 2 2 2 2or W P A x F P A

Where x2 is the distance moved by the fluid in the same time t.

The net work done is given by

1 2W W W

1 1 1 2 2 2 ________(1)W P A x P A x

From equation of continuity

1 1 2 2A V A V

1 1 2 2or A V t A V t V Volume

1 1 2 2A x A x V

Density = Mass / Volume

or Volume = Mass / Density

mV

1 1 2 2 _________(2)m

A x A x V

Putting in equation (1)

1 2W P V PV

1 2

mW (P P ) (3)

According to the law of conservation of energy, the net work done on the fluid must be

equal to the change in its P.E. and K.E.

(𝑃1 − 𝑃2)𝑚

𝜌=

1

2𝑚𝑣2

2 −1

2𝑚𝑣1

2 + 𝑚𝑔ℎ2 − 𝑚𝑔ℎ1

(𝑃1 − 𝑃2)𝑚

𝜌= 𝑚 [

1

2𝑣2

2 −1

2𝑣1

2 + 𝑔ℎ2 − 𝑔ℎ1]



𝑃1 − 𝑃2 = 𝜌 [1

2𝑣2

2 −1

2𝑣1

2 + 𝑔ℎ2 − 𝑔ℎ1]

𝑃1 − 𝑃2 =1

2𝜌𝑣2

2 −1

2𝜌𝑣1

2 + 𝜌𝑔ℎ2 − 𝜌𝑔ℎ1

𝑃1 +1

2𝜌𝑣1

2 + 𝜌𝑔ℎ1 = 𝑃2 +1

2𝜌𝑣2

2 + 𝜌𝑔ℎ2

Hence 𝑃 +1

2𝑣2 + 𝜌𝑔ℎ = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

This equation is called Bernoulli’s equation



Q. 6.1 Explain what do you understand by the term viscosity?

Ans. Viscosity

An internal frictional force between different layers of a flowing fluid is called viscosity

of fluid.

Viscosity measures, how much force is required to slide one layer of the liquid over another

layer. Substances that do not flow easily, such as thick tar and honey etc. have large

coefficients of viscosity, usually denoted by Greek letter ‘’. Substances which flow easily,

like water, have small coefficients of viscosity. Since liquid and gases have non zero

viscosity.

S.I unit of viscosity is Kg m-1 s-1

Q. 6.2 What is meant by drag force? What are the factors upon which drag force acting

upon a small sphere of radius r, moving down through a liquid, depend?

Ans. The resistive force experienced by a body moving through a viscous medium is called

drag force.

Accourding to stoke’s law, the drag force on a small sphere is given by

F = 6 𝜂 r v

So the drag force depends upon

o Viscosity of the medium

o Radius of sphere

o speed of sphere

Q. 6.3 Why fog droplets appear to be suspended in air?

Ans. As the relation for the terminal velocity is given by

2

t

2gρ rv =

9η

⇒ 𝑣𝑡 ∝ 𝑟2

Since size of droplet is very small so it gets terminal velocity very soon so the drag force

becomes equal to the weight of droplet, then net force acting on the droplet is zero. Hence

the fog droplet appears to be suspended in air as it falls with a very small constant

velocity.

Q. 6.4 Explain the difference between laminar flow and turbulent flow?

Ans.

Streamline or Laminar Flow or Steady Flow

The flow is said to be streamline or laminar (or steady) flow, if every particle that

passes a particular point, moves along exactly the same path, as followed by particles

which passed that point earlier.

Important short questions



Turbulent Flow

The irregular or unsteady flow of the fluid is called

turbulent flow.

In turbulent flow, the velocity of the fluid changes

suddenly and the path of the particles of the fluid is

disturbed.

Q. 6.5 State Bernoulli’s relation for a liquid in motion and describe some of its

applications?

Ans. Bernoulli’s Equation

It states that the sum of pressure, kinetic and potential energies per unit volume in a

steady flow of an incompressible fluid remains constant at every point of its path.

Mathematically

21Constant

2P V gh

Bernoulli’s theorem is according to law of conservation of energy

Applications

Swing of ball

Aerodynamic lift

Working of carburetor

Blood flow

Torricell,s theorem

Q. 6.6 A person is standing near a fast moving train. Is there any danger that he will fall

towards it?

Ans. When a train is moving fast, the velocity of air between the person and the train also

increases because it is dragged by the train.

According to Bernoulli’s equation

“When the velocity of a fluid increases, its pressure decreases”

Thus the pressure of air between the person and strain decreases. The air behind the

person pushes him towards train. So there is a danger that he may fall towards it.

Q. 6.7 Two row boats moving parallel in the same direction are pulled towards each

other. Explain.

Ans. When two boats are moving parallel in the same direction, the velocity of water flowing

through them will be large.



Due to this decrease of pressure row boats are pulled towards each other because of the

high pressure on the other sides of the boats.

Q. 6.8 Explain, how the swing is produced in a fast moving cricket ball.

Ans.





Due to the spin of the ball, the velocity of the air above the ball is greater than that

below the ball. As the pressure of the air below the ball is greater than that above the

ball. This push gives and extra curvature to the ball known as swing. Such a b all

produces difficulties for a batsman.

Q. 6.9 Explain the working of a carburetor of a motorcar using by Bernoulli’s principle.

Ans. According to Bernoulli’s equation


The carburetor of a car uses a venture duct which feeds the correct mixture of air and

petrol to the cylinders. Air is drawn through duct and along a pipe to the cylinders.

The air through duct moves at a greater speed, producing low pressure. So the petrol is

pushed along with air stream.

Q. 6.10 In an orbiting space station, would the blood pressure in major arteries in the leg

ever be greater than the blood pressure in major arteries in the neck?

Ans. On earth, gravity is pulling the fluid in the body towards the feet. In orbiting space

station, the blood balance inside arteries is disturbed. In weightless environment it is

easier to get the blood back up from the legs.

Hence the blood pressure in neck arteries rises as compared to that in leg arteries.

Q. 6.10 What are the properties of an ideal fluid?

Ans: Properties

The fluid is non-viscous i.e there is no internal frictional force between adjacent

layers of fluid.

The fluid is incompressible, i.e its density is constant.

The fluid motion is steady

Q. 6.11 What is Venturi’s effect?

Ans: The effect of decrease in pressure with increase in speed of the fluid in a horizontal

pipe is known as Venturi’s effect

Mathematically 2

2

1 2

1

2P P v



Q. 6.12 What effect is used in chimney?

Ans: According to Bernoulli’s equation


Chimney works on the principle of Bernoulli’s theorem. When air passes above the

chimney, it reduces the pressure above the chimney. The smoke rises up in the

chimney. The air and smoke together move away into the atmosphere.

Q. 6.13 How perfume bottles and paint spray work.

Ans: Perfume bottles and paint sprays also works on the principle of

Bernoulli’s equation by which


A stream of air passing over a tube dipped in a liquid will cause

the liquid to rise in the tube as shown in fig. This is because of the

fact that pressure of fast moving air becomes lesser than the air

above the paint level. Hence paint rises through pipe due to

difference of pressure.

Q. 6.14 What is the difference between systolic and diastolic blood pressure.

Ans: Systolic Pressure

The upper limit of range (of about 120 torr) of human blood pressure is called systolic

pressure.

Diastolic Pressure

The lower limit of range (of about 75 to 80 torr) of human blood pressure is called

diastolic pressure. The human blood pressure can be measured by an instrument called

sphygmomanometer.

Q. 6.15 State and explain Torricelli’s Theorem?

Ans: The speed of efflux is equal to the velocity gained by the fluid in falling through the

distance (h1-h2) under the action of gravity.

Explanation:

Suppose a large tank of fluid has two orifices A and B on it.

Since orifices are so small, the efflux speed V 2 and V3 will be much larger than 1V

Therefore V1 ≈ 0

Hence Bernoulli’s equation can be written as

2

1 1 2 2 2

1

2P gh P V gh

1 2P P = Atmospheric Pressure

Hence V2 = 1 22g h h



Q. 6.16 Explain lift of an aero-plane?

Ans: lift of an aero-plane also works on the principle of Bernoulli’s equation by which

“Where streamlines are force close together, speed is height and pressure will be low”

The wings of an aero-plane are designed to deflect the air so that the streamlines are

close together above the wings than below it. Thus, the air is faster on the upper side of

the wing as compared to the lower wing. This creates a pressure difference since

pressure gets low above the wing, which helps in lift ing the aero-plane upward.

6.1 Certain globular protein particle has a density of 1246 kg m-3. It falls through pure water

(=8.0 x 10-4 Nm-2s) with a terminal speed of 3.0 cm h-1. Find the radius of the particle.

Solution:

Data: Density of protein particle = 1246 Kg m-3

Co-efficient of viscosity = 8.0 x 10-4Nm s-3

Terminal speed vt =3cm h-1

-2-1

-6 -1

t

3.0×10= ms

3600

v =8.3×10 ms

To find: Radius r = ?

Calculation:

As the expression of terminal velocity of the particle is

2

t

2gr ρv =

9η

or 2 tv ×9η

r = ............(1)2gρ

Substituting the values in equation (1)

-6 -42 -10

2 -10 12

-12 -6

8.3×10 ×9×8.0×10 597.6r = = ×10

2×9.8×1246 24421.6

r = 0.02447×10 2.447 10

r= 2.447×10 =1.6 10 m

r = 1.6 10-6 m Ans

Numerical Problems



6.2 Water flows through a hose, whose internal diameter is 1cm at a speed of 1ms-1. What

should be the diameter of the nozzle if the water is to emerge at 21ms-1?

Solution:

Data: Internal diameter of hose = d1 = 1cm = 10-2m

Speed of water in the hose v1 = 1m/s

Speed of emerging water v2 = 21m/s

To find: Diameter of nozzle = d2 = ?

Calculation:

As the equation of continuity is;

1 1 2 2A v = A v

2 2

1 1 2 2

dv (πr )=v (πr ) r =

2

2 2

1 21 2

d dv =v

2 2

2 2

1 21 2

2 2

1 1 2 2

2 212 1

2

d dor v × =v ×

4 4

v d =v d

vd = ×d ................(1)

v

Substituting the values in (1)

2 -2 2 -4

2

-4 -2

2

1d = ×(10 ) =0.0476×10

21

d = 0.0476×10 =0.218×10 =0.22cm

d2 = 0.22 cm Ans

6.3 Water is flowing smoothly through a closed pipe system. At one point the speed of water is

3.0ms-1, while at another point 3.0 m higher, the speed is 4.0 ms-1. If the pressure is 80 kPa at

the lower point, what is pressure at the upper point?

Solution:

Data: Speed of water at one point v1 = 3ms-1

Speed of water at 2nd point v2 = 4 ms-1

Height of lower point h1 = 0

Height of upper point h2 = 3 m

Pressure at lower point P1 = 80kPa

= 80 x 103 Pa

To find: Pressure at upper point P2 = ?



Calculation:

According Bernoulli’s theorem

2 2

1 1 1 2 2 2

2 2

2 1 1 1 2 2

2 2

2 1 1 2 1 2

1 1P + ρv +ρgh =P + ρv +ρgh

2 2

1 1P =P + ρv +ρgh - ρv -ρgh

2 2

1P =P + ρ(v -v )+ρg(h -h )...........(1)

2

For water, Density = 1000 Kg m-3

Putting the values in eq (1)

P2 = 80000 + 1

21000[(3)2 – (4)2] + 1000 x 9.8 (-3)

P2= 80000 + 500 (9 - 16) + 9800(-3)

P2= 80000 + 500 (-7) + 9800(-3)

P2= 80000 - 3500 -29400

P2= 80000 – 32900 = 47100

P2 = 47.1 kPa

P2 = 47kPa Ans

6.4 An airplane wing is designed so that when the speed of the air across the top of the wing is

450 ms-1, the speed of air below the wing is 410 ms-1. What is the pressure difference between

the top and bottom of the wings? (Density of air = 1.29kgm-3)

Solution:

Data: Speed of air at the top of wing v1 = 450ms-1

Speed of air below the wing = v2 = 410 ms-1

Density of air = 1.29 Kg m-2

To find: Pressure difference = P2 – P1 = ?

Calculation:

Using Bernolli’s eq,

2 2

1 1 1 2 2 2

2 2

2 1 1 2 1 2

2 2

2 1 1 2 1 2

2 2

2 1 1 2 1 2

1 1P +ρgh + ρv =P +ρgh + ρv

2 2

1 1P -P = ρv - ρv +ρgh -ρgh

2 2

1P -P = ρ(v -v )+ρg(h -h )

2

1P -P = ρ(v -v ) (h =h )

2



Substituting the values

2 2

2 1

2 1

2 1

3

2 1

1P -P = ×1.29× 450 - 410

2

1P -P = ×1.29× 202500-168100

2

1P -P = ×1.29×34400=22188

2

P -P =22.188×10 Pa = 22.2kPa

Hence, the pressure difference

P2 – P1 = 22kPa Ans

6.5 How large must a heating duct be if air moving 3.0ms-1 along it can replenish the air in a

room of 300m3 volume every 15 min? Assume the air’s density remains constant.

Solution:

Data: Speed of air in the duct = v = 30 ms-1

Volume of air = V = 300 cm3

Heating time = t = 15 min

t = 15 x 60s = 900 sec

To find: Size of heating duct = r = ?

Calculation:

As we know that

Volume of air replenished per sec = A1v1

2V= πr v

t

2

2

V=r

t×πv

300 300r = = =0.0354

900×3.14×3 8478

r = 0.188m=0.19m=19cm

Hence, the size of duct;

r = 19cm. Ans

6.6 What gauge pressure is required in the city mains for a stream from a fire hose connected to

the mains to reach a vertical height of 15.0m?

Solution:

Data: Vertical height = h = (h1 – h2) = 15m

Density of water = = 1000 Kgm–3

Gauge pressure = pressure difference



To find: P = P2 – P1 = ?

Calculation:

As the Bernoulli’s eq

P1+ 1

2 v1

2 + gh1 = P2 + 1

2 v2

2+ gh2

P2 – P1 =g(h1 – h2) + 1

2 (v1

2 – v22)

The speed of streamline is same through out the motion then v1 = v2 therefore

P2 – P1 = g (h1 – h2)

P = gh

P = 1000 × 9.8 × 15 = 147000 = 1.47 × 105 Pa

Hence, Gauge pressure.

P = 1.47 × 105 Pa Ans

fluid dynamics type equation here. 6 · ans. bernoulli’s equation it states that the sum of...

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