Fluid Dynamics

Steady Flow (Stream Line Flow)

It is the flow in which the velocity of fluid particles crossing a particular point is the same at all the times. Thus, each particle takes the same path as taken by a previous particle through the point.

 

Enquiry: Can we study fluids in motion?

             Motions of smoke from chimney, flow of water in river, flow of any gas in a tube are examples of flow of fluids. This is generally a complex stream of mechanics. But, if we restrict ourselves to a simpler type of motion, i.e. steady, streamline or laminar flow and not turbulent flow then we can study the behavior of such fluids.

 

Enquiry: What is steady, streamline or laminar flow?

                The flow of fluid is said to be streamline if the velocity at any point in the fluid remains constant with time (in magnitude as well as in direction) in this case energy needed to drive the fluid is used up in overcoming the viscous force between its layers. All particles passing through a point in a steady flow follow the same path. The paths are known as streamlines> An example is water slowly flowing through a pipe.

 

Enquiry: What is turbulent flow?

             When the motion of a particle varies rapidly in magnitude and direction, the flow is said to be turbulent. In other words, when the velocity exceeds beyond the critical velocity, the paths and velocities of liquid change continuously and haphazardly. This flow is called turbulent flow. An example is water coming out of a fountain.

 

Enquiry: What is critical velocity?

             If in the case of a steady flow of fluids the velocity of flow is gradually increased it is found that the motion remains steady (streamline or laminar) upto a certain limit. If the velocity of flow crosses a certain limit the fluid particles do not follow the path of the preceding one and the motion becomes turbulent. The maximum velocity upto which fluid motion remains steady is called critical velocity. According to Reynold, in case of motion of fluids in narrow tubes, critical velocity depends on the density r and coefficient of viscosity η of the fluid as well as radius of the tube. i.e.

                Vc  = η/rρ     or   Vc = R η/ρr

             Here R is a dimensionless constant called Reynold's number. For steady flow R < 2000. For 2000<R<3000, flow is transitional and for R>3000 flow is turbulent.

 

Pause:   We shall be studying about viscosity later on in the course of this chapter.

Note:      In further discussions, we shall be considering only the streamline motion of a non-viscous incompressible fluid.

 

Enquiry: Will the velocity of a fluid remain constant even if the tube through which it flows has constrictions?

critical velocity

             In case of a fluid flowing through a tube of non-uniform cross-section, the product of the area of cross-section, density and the velocity of flow is same at every point in the tube. This is known as the principle of continuity, which actually means that amount of mass remains constant. For a fluid, mass flowing in through end B=mass flowing out from C. If the fluid is compressible

                                        A1V1ρ1 = A2V2ρ2.

If the fluid is incompressible then ρ1 = ρ2.

Therefore, A1V1 = A2V2 = Constant.

             Therefore, we can infer that if the cross-sectional area of a tube changes, the velocity of flow will also change.

 

Illustration:

                Let us analyse the problem above. Why does the fluid flow through the hole? We know that fluid exerts, force on the walls of container and walls of container produce equal and opposite reaction. But at the hole there is no wall to produce a reaction. Hence the volume element near the hole experiences a net downward force and accelerates out of the hole. In terms of pressure, we may say that pressure inside the container (near the hole) is greater than pressure outside the container and fluid flows towards lower pressure. Also observe that in the container, velocity of an infinitesimal volume element is less than its velocity when it emerges out.

             Similarly, consider a waterfall. Water gains speed as it reaches the bottom. Let us think why it falls down. The answer is its tendency to reduce its potential energy. Its potential energy gets reduced but kinetic energy is increased. Note that pressure was same at both ends i.e. atmospheric pressure. (This can also be explained by Newton's laws of motion)

 

Line of Flow

It is the path taken by a particle in flowing liquid. In case of a steady flow, it is called streamline. Consider an area S in a fluid in steady flow. Draw streamlines from all the points on the periphery of S. These streamlines enclose a tube, on which S in a cross-section.

line of flow

No fluid enters or leaves across the surface of this tube.

 

Equation of Continuity

In a time Dt, the volume of liquid entering the tube of flow in a steady flow is (A1v1 Δt). The same volume must flow out as the liquid is incompressible. The volume flowing out in time Δt is A2 v2 Δt.

coninuity

=> A1 v1 - A2 v2.

Mass flow rate = ρAV (where ρ is the density of the liquid).

 

Bernoulli's Theorem

Consider a tube of flow, ABCD. In a time Dt, liquid moves and the liquid element becomes A'B'C'D'. In other words, we can also interpret that ABB'A' has gone to DCC'D'.

        Δm = ρA1v1 Δt = ρ A2v2 Δt

flow

Work done by fluid pressure at end 1 = (P1A1)v1 Δt = P1 Δm/r

Work done by fluid pressure at end 2 = -(P2A2)v2 Δt = -P2 Δm/r

Work done by gravity = -(Δm)g(h2 - h1)

Change in kinetic energy = (1/2) Δm [v22 - v12]

Using, work-energy theorem (W = ΔK),

        P1 Δm/ρ - P2 Δm/ρ - Δmg (h2 - h1) = 1/2 Δm [V22 - V12]

Or     p1/ρ + gh1 + V12/2 = P2 + gh2 + V22/2

Or     P1 + ρgh1 + ρV12/2 = P2 + ρgh2 + ρV22/2

Or     P + ρgh + ρV2/2 = constant

In a streamline flow of an ideal fluid, the sum of pressure energy per unit volume, potential energy per unit volume and kinetic energy per unit volume is always constant at all cross sections of the liquid.

  • Bernoulli's equation is valid only for incompressible steady flow of a fluid with no viscosity.

 

Illustration:

 

The figure shows how the stream of water emerging from a faucet necks down as it falls. The area changes from A0 to A through a fall of h. At what does the water flow the top tap?

faucet

Solution:

        Equation of continuity: A0 V0 = AV

        Bernoulli equation:       P0 + ρgh + 1/2 ρ v02 + ρg(0) + 1/2 ρv2

        Solving for v0, v0 = √((2ghA2 A02)/(A02-A2 ))

                                = (AA0 √2gh)/√(A02-A2 ).

Note: As the jet is going into the atmosphere, the pressure at A and A0 are equal to the atmospheric pressure.

 

Illustration:

        Water enters a house through a pipe with inlet diameter of 2.0 cm at an absolute pressure of 4.0 × 105 Pa (about 4 atm). A 1.0 cm diameter pipe leads to the second floor bathroom 5.0 m above. When flow speed at the inlet pipe is 1.5 m/s, find the flow speed, pressure and volume flow rate in the bathroom.

 

Solution:

Let points 1 and 2 be at the inlet pipe and at the bathroom, then from continuity equation

        a1v1 = a2v2     =>       v2 = 6.0 m/s

Now, applying Bernoulli's equation at the inlet (y = 0) and at the bathroom (y2 = 5.0 m).

As p + 1/2 σv2 + σgy = constant

Hence, p2 = p1 - 1/2 ρ(v22-v12 ) - ρg(y2-y1 )

which gives p2 = 3.3 × 105 Pa

The volume flow rate = A2v2 = A1v1 = Π/4 (0.1)26 = 4.7 × 10-4 m3/s.

 

Illustration:

Water coming out of jet having across sectional area a, with a speed v strikes a stationary plate and stops after striking. Find the force exerted by the water jet on the plate.

 

Solution:

The change of momentum of water in time dt = 0 - ρav2dt î = -ρav2dt î where is a unit vector in the direction of the velocity of the jet. The rate of change of momentum of water jet = -ρav2 î

        Thus the force exerted on the water jet by the plate = -ρav2 î

        The force exerted on the plate by the water jet = ρav2 î .

Enquiry: Can the in-swing and out-swing of a Cricket ball be explained by this principle?

             Yes, when a spinning ball is thrown, it deviates from its usual path in flight. This effect is called Magnus effect and plays an important role in tennis, cricket etc. By applying appropriate spin the moving ball can be made to deviate in any desired direction.

             If a ball is moving from left to right and is also spinning about a horizontal axis perpendicular to the direction of motion, the resultant velocity of air above the ball will be (V + r ω) while below the ball (V-rω). So in accordance with the Bernoulli's principle pressure above the ball will be lesser than that below it. Due to this difference of pressure upward force will act on the ball and hence, the ball will deviate from its usual path and will touch the ground following a path which curves less sharply, prolonging the flight.

anti-clock-wise-spin

             Similarly if the spin is clockwise i.e. the ball is thrown with topspin, the force due to pressure difference will act in the direction of gravity and the ball will pitch more sharply shortening the flight.

Further more, if the ball is spinning about a vertical axis, the curving will be side ways producing the so-called out-swing or in-swing.

             The following phenomena can also be explained using Bernoulli's equation.

(a)          Drop of pressure when a fluid from broader to a narrower horizontal pipe. We have Av = constant.

             \    When area is large, velocity is small and vice-versa.

             But by Bernoulli's equation

             P + 1/2ρv2 = constant for a horizontal pipe. i.e. When velocity is large pressure is small and vice-versa. Thus, pressure at smaller cross-area is lesser.

(b)          Blowing off, of roofs of windstorms

             During a tornado or a hurricane, when a high-speed wind blows over a straw or tin roof, it creates low pressure above it, in accordance with Bernoulli's principle. However the pressure under the roof is still atmospheric. So due to this difference of pressure the roof is lifted up and is then blown off by the wind.

(c)          Attraction between two closely parallel moving boats.

             When two boats are moving in the same direction the water or air between them move faster than that on the remote sides. Consequently, in accordance with the Bernoulli equation, the pressure between them is reduced. Hence due to pressure difference they are pulled towards each other creating the so-called attraction.

 

Enquiry: What is the use of Bernoulli's principle in practice? 

             Bernoulli's principle is used in the following and many other ways.

(a)          Action of Atomizer

             The action of aspirator, carburetor, paint gun, scent spray or insect sprayer is based on Bernoulli's principle. In all these, by means of motion of piston r in a cylinder C, high speed air is passed over a tube T, dipped in a liquid L to be sprayed.

atomizer

             High speed air creates low pressure over the tube due to which liquid rise in it and is then blown off with expanded air.

 

(b)          Taking off of an Aeroplane

             This is also based on Bernoulli's Principle. The wings of aeroplane are made tapering. Due to this specific shape of wings when aeroplane runs, air pases at high speed over it as compared to lower surface. This difference of air speeds above and below the wings, in accordance with the Bernoulli's principle, creates a pressure difference, which increases with the speed of the aeroplane due to which an upward force called dynamic lift (= pressure difference x area of wing) acts on the plane. When this lift becomes greater than the weight of the plane, the plane rises up.

 

efflux

 

(c)          Velocity of Efflux

             Liquid is filled in a vessel up to a height H and a hole is made at a depth h below the free surface of the liquid as shown. Taking the level of hole as reference level and applying Bernoulli's principle to the liquid just inside and outside the hole (assuming the liquid to be at rest inside), we get                                                     P0 - hpg - 0 = P0 - 1/2ρv2

                   Or  v = √2gh [Which is also known as Toricelli's Theorem].

 

Enquiry: Where do we see the application of Bernoulli's theorem? 

Bernoulli-theorem

              Yes, Bernoulli's theorem finds application in many devices, two of which are described below.

 

(a)          Venturimeter

             This is an instrument for measuring velocity of flow of liquids & gases [fluids]

             The instrument is connected horizontally in the tube in which the rate of flow is to be measured. If PA is pressure at A & PB is pressure at B. PA-PB = hrg [h is difference of heights of liquids of density r is vertical tubes].

Let         V1 = Velocity at A

             V2 = Velocity at B.

Now, A1 V1 = A2 V2.

Let us first consider the case of liquids. If the tube is held horizontal, then by application of Bernoulli's theorem,

Bernoulli-theorem1

 

          PA/ρ+ (v12)/2 = PB/ρ + (v22)/2 

          => v22 – v12 = 2/ρ (PA - PB )= 2/ρ hρg = 2hg 

          => Q2/(A22 ) - Q2/(A12 ) = 2gh, [∵Q=Av]

where Q = amount of liquid flowing in or flowing out per unit time.

          => Q = A1 A2 √(2gh/(A12 - A22 ))

As Q = A1V1 = A2V2,

Hence V1 and V2 can be calculated

Explanation the velocity is measured using this formula.

For gases, let d be the density of the gas

             PA - PB = hrg

from Bernoulli's Equation

Bernoulli-theorem3

 

                  PA/d+ (V12)/2 = PB/d + (V22)/2 

                 => v22 – v12 = 2/d (PA - PB ) 

                 => v22 – v12 = 2/d (PA - PB )hρg 

                => Q2/(A22 ) - Q2/(A12 ) = 2/d hρg 

                => Q = A1 A2 √(2hρg/(d(A12 - A22)))

 

b)            Pilot Tube

             Like Venturimeter, a pilot tube is also used for measuring flow rates of fluids.

pilot-tube

For liquids, if A & B are in the same horizontal plane explain the figure. Applying Bernoulli's Equation.

 PA/ρ +  1/2 v2 =  PB/ρ + 0

or 2(PB - PA )/P = v2

or v = √(2(PB - PA )/ρ) = √2gh

[∵ (PB - PA ) = hρg]

                   Q = Av = A√2gh.

For gases, Let r be the density of the liquid in vertical tube

                   PB - PA = hrg.

Applying Bernoulli's theorem at A & B

 

PA + 1/2 dv2= PB

=>  v2 =  2(PA - PB )/d

=>  v = √(2hρg/d)

Then      Q = A√(2hρg/d).

 

Enquiry: Considering the motion of fluids, do the fluid streamlines encounter any opposition?

              If you recall, while studying critical velocity we talked about something called viscosity.

             The property of a fluid due to which it opposes the relative motion between its consecutive layers is called viscosity and the force between the layers opposing the motion is called viscous force.

             According to Newton, the viscous force F acting on any layer of a fluid is directly proportional to its area and the velocity gradient across the layer i.e.

        F = Adv/dy

or      F = ηAdv/dy,

             where A is the area of the layers, dv/dy is velocity gradient and η is called the coefficient of viscosity. The negative sign shows that viscous force acts in a direction opposite to that of the flow of fluid. Coefficient of viscosity depends only on the nature of the fluid and is independent of area considered. Its dimensions are [ML-1 T-1]

Regarding viscosity the following are worth noting.

(i)          Viscosity of liquids is much greater than gases.

(ii)         With rise in temperature the viscosity of the liquids decreases while that for gases increases.

(iii)         In case of liquids viscosity increases with density while for gases it decreases with increase in density.

(iv)        With increase in pressure, the viscosity of liquids (except water) increases while that of gases is practically independent of pressure.

(v)         From kinetic theory, viscosity represents transport of momentum, i.e. viscosity is due to the difference in the momenta of two layers.

(vi)        In case of steady flow of a liquid of viscosity η in a cylindrical tube of length L and radius r under a pressure difference r across it, the velocity of flow at a distance y from the axis is given by v = ρ/4ηL (r2 - y2 ) The velocity profile is a parabola. On x-axis y=0

                        => v = pr2/4ηL

(vii)        In case of a steady flow of a viscous liquid in a cylindrical tube of length L and radius r under a pressure difference p across it, the volume of liquid flowing per second is given by Q = (πpr4)/8ηL

             This is Poiseuille's Formula

  

Think:    When we throw a dense object in a fluid the force of buoyancy cannot balance the force of gravity and then the object accelerates downwards. Had it been an ideal fluid, then it would have accelerated indefinitely till it reached the bottom but due to relative motion between object and fluid, resistive (viscous) forces come into play.

 

Stokes Law

              When a body moves through a fluid, the fluid in contact with the body is dragged with it. This establishes relative motion in fluid layers near the body, due to which viscous starts operating. The fluid exerts viscous force on the body to oppose its motion. The magnitude of the viscous force depends on the shape and size of the body, its speed and viscosity of the fluid. As the viscous force depends on velocity of the body it keeps on increasing with the increase in velocity. If it balances the net downward force then body falling freely in a viscous medium attains a constant velocity, which is called terminal velocity. If a shape of radius r moves with a velocity v through a fluid of viscosity η, the viscous force opposing the motion of the sphere is

                        F = 6pηrv.

This law is called Stokes law. The sphere dropped in the fluid moves down with a decreasing acceleration until the net force on it becomes zero. At this point it is said to have attained terminal velocity.

Viscous force = Weight - Upthrust

             or,  F = 6pηrvT. = (4/3 πr3 )ρg - (4/3 πr3 σ)g,

             Where r is the density of the solid and σ is the density of the fluid.

                   VT = (2r2 g)/9  [(ρ-σ)/η].

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