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Unit 5

Internal & External Flow

Q1) Define Renolds number.

A1) Reynolds Numbers (Re)

It is defined as the ratio of the inertia force to the viscous force.

Re = =

Re =

For pipe flow Re =

In a pipe, a laminar flow occurs when Reynolds number is less than 2000 & turbulent flow occurs when Reynolds number is more than 4000.

In turbulent flow fluid motion is irregular & there is complete mixing of fluid due to collision of fluid masses with one another.

The shear in turbulent flow is mainly due to momentum transfer.

The contribution of fluid viscosity to the total shear is small.

In case of laminar flow definite relationship exists between shear stress due to viscosity & velocity.

Q2) Explain the concept of entrance region and fully developed flow with the help of diagram?

A2)

Consider a flow entering a pipe.

The entering flow being uniform, so having negligible viscosity

As soon as the flow hits the pipe many changes take place.

The most important of these is that viscosity imposes itself on the flow and the "No Slip" condition at the wall of the pipe comes into effect.

Consequently the velocity components are each zero on the wall,u = v = 0.

The flow adjacent to the wall decelerates continuously.

A layer close to the body where the velocity builds up slowly from zero at wall to a uniform velocity towards the centre of the pipe.

This layer is what is called the Boundary Layer.

Viscous effects are dominant within the boundary layer. Outside of this layer viscous effects are negligible or absent.

The boundary layer is not a static phenomenon; it is dynamic, it grows meaning that its thickness increases as flow move downstream.

From Fig. it is seen that the boundary layer from the walls grows to such an extent that they all merge on the centreline of the pipe.

Once this takes place, inviscid core terminates, and the flow is all viscous.

The flow is now called a Fully Developed Flow.

The velocity profile becomes parabolic. Once the flow is fully developed the velocity profile does not vary in the flow direction.

In this region the pressure gradient and the shear stress in the flow are in balance.

The length of the pipe between the start and the point where the fully developed flow begins is called the Entrance Length.

Q3) Explain the flow of viscous fluid in circular pipes.

A3) Flow of viscous fluid in circular Pipes – Hagen Poiseuille Law

(Velocity & shear stress distribution for laminar flow in pipe)

I 2.png

Fig: Shows a horizontal circular pipe of radius R, having laminar flow of fluid through it.

Consider a small concentric cylinder of radius r and length dx as a free body.

If is the shear stress, the shear force F is given by

F = x 2 r x dx

Let P be the intensity of pressure at left end and the intensity of pressure at the right end be (

Thus, the force acting on the fluid element are:

i) the shear force, x 2r x dx on the surface of fluid element.

ii) The pressure force, P x on the left end.

iii) The pressure force, ( on the right end.

For steady flow, the net force on the cylinder must be zero.

[ p x - (] –

- -

Shear stress is zero at the centre of pipe (r = 0) and maximum at the pipe wall given by

I 1.png

()

From Newton’s Law of viscosity.

In this equation, the distance y is measured from the boundary. The radial distance r is related to distance y by the relation.

y = R-r ordy = -dr

Comparing two values of

du = ()r.dr

Integrating the above equation w. r. t. ‘r’ we get

u = . r2 + C --------------- (ii)

Where C is the constant of integration and its value is obtained from the boundary conditions.

At r = R, u = 0

0 = . R2 + Cor C = . R2

Substituting this value of C in eqn. (ii) we get,

u = . r2 . R2

u = - . (R2 – r2)

Shows that velocity distribution curve is a parabola.

The maximum velocity occurs, at the centre and is given by.

umax = - R2

Q4) Explain the Flow of viscous fluid between two parallel plates when one plate is moving & other of Rest – Coutte flow.

A4) Flow of viscous fluid between two parallel plates when one plate is moving & other of Rest – Coutte flow

Consider laminar flow between two parallel flat plates located at a distance b apart such that lower plate is at rest & the upper plates moves uniformly with a constant velocity U as shown in fig.

I 4.png

A small rectangular element of fluid of length dx, thickness dy & unit width is

Considered as a free body.

The forces acting on the fluid element are

1) The pressure force p dy x 1 on left end

2) The pressure force (p +. dx) dy x1 on the right end.

3) The shear force

4) The shear force [ . dy] dx x 1 on the surface.

For steady & uniform flow the resultant force in the direction of flow is zero.

p. dy – (p + . dx) dy – + (+ . dy) dx = 0

-dx. dy + dy. dx = 0

Dividing by the volume of element dx. dy we get

= -------------------(1)

Above equation shows the interdependence of shear & pressure gradients

According to Newton’s law of viscosity for laminar flow the shear stress.

= .

put in equation (1) we get,

Integrating above equation twine w. r. t. y

u = y2 + C1 y + C2 -------------(2)

At y = 0, u = 0 & at y = b, u = U

We get C2 = 0 & C1 = - () b

Put values of C1& C2 in eqn. (2)

u = y - (by – y2)

q = = (by – y2)] dy

U. -

The distribution of shear stress across any section may be determined by using

Newton’s Law of viscosity

= u. = [ - () (b – 2y)]

= - (b – 2y)

Q5) Explain the Flow of viscous fluid between two parallel plates when both plates are at the rest– Coutte flow.

A5) Both Plates at Rest:

In this case the equations for velocity, discharge q and the shear can be obtained from similar equation for generalise Coutte flow by putting U = 0

Thus, for flow between two stationary parallel plates.

I 5.png

u = - . (by – y2)

Discharge per unit width,

q = .

Shear stress . (b –2y)

Q6) Explain the concept of hydro dynamically smooth and rough boundaries.

A6)

If K is the arrange height of the irregularities of the surface of a boundary, then in general the boundary is said to be rough if value of K is high and smooth if K is low.

For proper classification of smooth and rough boundaries, besides the boundary characteristics, the fluid and fluid characteristics need to be considered.

As shown in fig: when the average height K of the irregularities is much less than the thickness of the laminar sublayers

the flow outside the laminar sublayer is turbulent; the eddies of various sizes present, try to penetrate the laminarsublayer.

These eddies cannot reach the surface irregularities due to greater thickness of the laminar sub layer and so the boundary acts as a smooth boundary.

As the Reynolds numbers (Re) increases the thickness of the sublayer decreases the irregularities will then project through laminar sublayer and eventually the laminas sublayer is destroyed completely.

Subsequently the eddies will come in contact with the surface irregularities and there will be a large amount of energy loss.

This type of boundary is known as hydrodynamically rough boundary.

Through experiment Nikuradse found that the boundary behaves as:

i) Hydrodynamically smooth boundary, when () < 0.25

ii) Hydrodynamically rough boundary, when () > 6.0

iii) Boundary in transition, when 0.25 < () < 6.0

Q7) Explain velocity distribution for turbulent flow in smooth pipes.

A7) Velocity Distribution for turbulent flow in smooth pipes.

The velocity distribution for turbulent flow in smooth or rough pipe is given by equation as

u =logey + c

It may be seen that at y = 0, the velocity u at wall is +

This means that velocity u is positive at some distance away from the wall and - (minus Infinity) at the wall.

Hence at some finite distance from wall the velocity will be equal to zero.

Let this distance from pipe wall is y

Now, the constant C is determined from the boundary condition.

at y = y, u = 0.

Hence above equation become as

0 = logey’+ c

c =- logey’

Substituting the value of c in above eqn.

u = logey - logey’

= loge

Substituting the value of K = 0.4, we get

u = logeloge

= log10

= log10---------(i)

From Nikuradse’s experiment the value of y’ is given as

y’ =

Where = , where = Kinematic viscosity of fluid.

y = x =

Substituting this value of y’ in equation (i), we obtain

= 5.75 log10 ( )

= 5.75 log10 ( ) = 5.75 log10 (

= 5.75 log10 + 5.75 log10 9.259

= 5.75 log10 + 5.55

The above equation is known as Karman Prandtl equation.

Q8) Explain Velocity Distribution for Turbulent flow in Rough Types

A8) In case of rough boundaries, the thickness of Laminar sub – layer is very small.

The surface irregularities are above the laminar sub- layer and hence the laminar sub layer is completely destroyed.

Thus y’ can be considered proportional to the height of protrusions k.

Nikuradse’s experiment shows the value of y’ for pipes coated with uniform sand (rough pipes) as

y’ =

Substituting this value of y’ in equation

= 5.75 log (y/y’)

= 5.75 log10 () = 5.75 [ log10 (y/k) x 30]

= 5.75 log10 (y/k) + 5.75 log10 (30.0)

= 5.75 log10 (y/k) + 8.5

Q9) Explain the Laminar flow through circular pipe - Hagen Poiseuille Law.

A9)

I 2.png

Fig: Shows a horizontal circular pipe of radius R, having laminar flow of fluid through it.

Consider a small concentric cylinder of radius r and length dx as a free body.

If is the shear stress, the shear force F is given by

F = x 2 r x dx

Let P be the intensity of pressure at left end and the intensity of pressure at the right end be (

Thus, the force acting on the fluid element are:

i) the shear force, x 2r x dx on the surface of fluid element.

ii) The pressure force, P x on the left end.

iii) The pressure force, ( on the right end.

For steady flow, the net force on the cylinder must be zero.

[ p x - (] –

- -

Shear stress is zero at the centre of pipe (r = 0) and maximum at the pipe wall given by

I 1.png

()

From Newton’s Law of viscosity.

In this equation, the distance y is measured from the boundary. The radial distance r is related to distance y by the relation.

y = R-r or dy = - dr

Comparing two values of

du = ()r.dr

Integrating the above equation w. r. t. ‘r’ we get

u = . r2 + C --------------- (ii)

Where C is the constant of integration and its value is obtained from the boundary conditions.

At r = R, u = 0

0 = . R2 + Cor C = . R2

Substituting this value of C in eqn. (ii) we get,

u = . r2 . R2

u = - . (R2 – r2)

Shows that velocity distribution curve is a parabola.

The maximum velocity occurs, at the centre and is given by.

umax = - R2

Q10) Explain the concept of Transition from laminar to turbulent flow

A10) Transition from laminar to turbulent flow

Turbulent flows can often be observed to arise from laminar flows as the Reynolds number is increased1.

The transition to turbulence happens because small disturbances to the flow are no longer damped by the flow, but begin to grow by taking energy from the original laminar flow.

Laminar flows are solutions of the Navier- Stokes equations in the limit when nonlinearities can be neglected.

This situation corresponds to the Stokes approximation; the flow is given by a balance between external forcing (mechanical or thermodynamically) and dissipation.

The corresponding velocity field is entirely predictable and has the same regularity properties as the applied external forces.

As the Reynolds number is increased, nonlinearities are no longer negligible.

There is a critical Reynolds number, above which the system bifurcates toward completely new solutions.

These solutions correspond to breaking of symmetries of the original flow.

When the Reynolds number is increased even further, the flow that emerged after the primary bifurcation can itself become unstable to new modes breaking new symmetries.

The set of bifurcation continues until the global dynamics of the flow becomes very complicated, i.e. turbulent.

Q11) Explain the Development of boundary layer on a flat plate Laminar, transitional and turbulent boundary.

A11) Laminar Boundary Layer

Consider the flow of fluid having free stream velocity(U) over a smooth thin plate which is flat and placed parallel to the direction for free stream of fluid as shown in the figure.

Let us consider the flow with zero pressure gradient on one side of the plate which is stationary.

The velocity of the fluid on the surface of the plate should be equal to the velocity of the plate. But plate is stationary and hence velocity of fluid on the surface of the plate is zero. But at a distance away from the plate, the fluid is having certain velocity.

Thus, a velocity gradient is setup in the fluid near the surface of the plate. This velocity gradient develops shear resistance, which retards the fluid.

Thus, the fluid with a uniform free stream velocity(u) is retarded in the vicinity of the solid surface of the plate and the boundary layer region begins at the sharp leading edge.

At subsequent points downstream the leading edge, the boundary layer region increases because the retarded fluid is further retarded.

This is also referred as the growth of boundary layer.

Near the leading edge of the surface of the plate, where the thickness is small, the flow in the boundary layer is laminar though the main flow is turbulent.

This layer of the fluid is said to be laminar boundary layer. This is shown by AE in figure.

Turbulent Boundary Layer

The thickness of the boundary layer will go on increasing in the downstream direction.

Then the laminar boundary layer becomes unstable and motion of fluid within it is disturbed and irregular which leads to a transition from laminar to turbulent boundary layer.

This short length over which the boundary layer flow changes from laminar to turbulent is called transition zone. This is shown by distance BC in figure.

Further downstream the transition zone, the boundary layer is turbulent and continues to grow in thickness. The layer of boundary is called turbulent boundary layer, which is shown by the portion FG in figure.

Q12) Explain the Nominal, displacement, momentum & energy thicknesses

A12) Boundary Layer Thickness ()

It is defined as the distance from the boundary of the solid body measured in the y-direction to the point, where the velocity of the fluid is approximately equal to 0.99 times the free stream velocity(u) of the fluid.

It is denoted by the symbol

1. = Thickness of laminar boundary layer,

2. = Thickness of turbulent boundary layer, and

3. = Thickness of laminar sublayer.

Displacement Thickness ()

It is defined as the distance, measured perpendicular to the boundary of the solid body, by which the boundary should be displaced to compensate for the reduction in flow rate on account of boundary layer formation.

It is denoted by

It is also defined as “The distance perpendicular to the boundary, by which the free stream is displaced due to the formation of boundary layer”.

Momentum Thickness (

Momentum thickness is defined as the distance, measured perpendicular to the boundary of the solid body, by which the boundary should be displaced to compensate for the reduction in momentum of the flowing fluid on account of boundary layer formation.

It is denoted by

Energy Thickness ()

It is defined as the distance measured perpendicular to the boundary of the solid body, by which the boundary should be displaced to compensate for the reduction in kinetic energy of the flowing fluid on account of boundary layer formation.

It is denoted by

Q13) Define Local and mean drag coefficients.

A13) Local and mean drag coefficients

Local coefficient of drag (CD*)

It is defined as the ratio of the shear stress to the quantity

It is denoted by CD*

Average coefficient of Drag (CD)

It is defined as the ratio of total drag force to the quantity

It is also called coefficient of drag and is denoted by CD.

where A= area of the surface (or plate)

V = free stream velocity

= mass density of fluid

Q14) Explain the concept of boundary layer separation and its control.

A14) Boundary Layer separation and its control.

Separation of boundary layer

When a solid body is immersed in a flowing fluid, a thin layer of fluid called the boundary layer is formed adjacent to the solid body.

Along the length of the solid body the thickness of the boundary layer increases.

The fluid layer adjacent to the solid surface has to do work against surface frictional at the expense of its kinetic energy.

This loss of kinetic energy is recovered from the immediate fluid layer in contact with the layer adjacent to solid surface through momentum exchange process.

Thus, the velocity of the layer goes on decreasing along the length of the solid body.

At a certain point a stage may come when the boundary layer may not be able to keep sticking to the solid body if it cannot provide kinetic energy to overcome the resistance offered by the solid body. The boundary layer will be separated from the surface.

This phenomenon is called boundary layer separation.

The point on the body at which the boundary layer is on the verge of separation from the surface is called point of separation.

Q15) Explain the Methods of preventing the separation of boundary layer.

A15) Methods of preventing the separation of boundary layer

The following are the methods of preventing the separation of boundary layer:

Motion of solid boundary:

By rotating a circular cylinder lying in a stream of fluid so that the upper side of cylinder where the fluid as well as the cylinder move in the same direction, the boundary layer does not form.

2. Acceleration of fluid in the boundary layer:

This method of controlling separation consists of supplying additional energy to particles of fluid which are being retarded in the boundary layer. This may be achieved either by injecting the fluid into the region of boundary layer from the interior of the body with the help of some available device as shown in the figure or by diverting a portion of fluid of the mainstream from the region of high pressure to the retarded region of boundary layer through a slot provided in the body.

3. Suction of fluid from the boundary layer:

In this method, the slowmoving fluid in the boundary layer is removed by suction through slots or through a porous surface as shown in the figure.

4. Streamlining of the body shapes:

By the use of suitable shaped bodies, the point of transition of the boundary layer from laminar to turbulent can be moved downstream which results in the reduction of the skin friction drag. Furthermore, by streamlining of the body shapes the separation may be eliminated.

5. Supplying additional energy from a blower.

6. Rotating boundary in the direction of flow.

7. Providing small divergence in a diffuser.

8. Providing guide blades in a bend.

9. Providing a trip wire ring in the laminar region for the flow over a sphere.

Q16) A Lubricating oil of viscosity 1 poise & sp. gr 0.9 is pumped through a 30mm diameter pipe. If the pressure drop per meter length of pipe is 20KN/m2. Determine

i) The mass flow rate in kg/min

ii) The shear stress at the pipe wall

iii) The Reynold number of flow

iv) The power required per some length of the pipe to maintain the flow

A16) Given,

= 1 poise = 0.1 N. S/m2

0.9

= 0.9 x 1000 = 900kg/m3

= 0.03,

= (P1 – P2) = 20 x N.S/m2

(P1 – P2) =

20 x =

= 5.625 m/s

A = x = x= 7.068 X

Q = A. = 7.068 x x 5.625= 3.97 x /sec

Mass flow rate =

= 900 x 3.97 x x 60

= 214.65 kg/min

= 20 x x

= 150

Reynolds Number

Re = = = 1518.7

Loss of head hf = = =2.265 m

Power = hf

= (0.9 x 9810 x 3.97 x x 2.265)

= 79.49 watts

For 50m length power required

P = 79.49 x 50

= 3974.5 watts.

Q17) A 0.2 m diameter pipe carries liquid in laminar region. A pilot tube placed in the flow at a radial distance of 6 cm from the axis of the pipe indicates velocity of 0.5 m/s. Calculate the maximum velocity, mean velocity & discharge in the pipe.

A17) Given,

D = 0.2 m, R = 0.1 m.

u(6cm) = 0.5 m/s

r = 6 cm = 0.06m

u = - ()

0.5 = - ()

- = 78.125

Umax = -

= 78.125 x

= 0.7812 m/s

= Umax= 0.7812

= 0.3906 m/s

A = x = X = 0.0314

Q = A.u

= 0.0314 x 0.3906

= 0.01227 /s

Q18) Water is flowing over a thin smooth plate of length 4 m and breadth 2m at a velocity of 1.0m/s. If the boundary layer flow changes from laminar to turbulent at a Reynolds number 5× 105. Find

i) the distance from leading edge up to which boundary layer is laminar.

ii) The thickness of the boundary layer at the transition point.

iii) The drag force on one side of the plate

u = 9.81× 10-4 NS /m2

A18) L=4m b=2m U=1.0m/s Re = 5×105 u = 9.81×10-4 Ns/m2

(i)

(ii)

(iii) Drag force due to laminar boundary

Drag force due to turbulent boundary layer from F to G.

(FFG)turb = Drag force due to turbulent

boundary layer E to G – Drag force due to turbulent flow from E to F.

=(FEG)turb – (FEF)turb

Drag force on one side of plate = Drag force due to laminar boundary layerupto F + Drag force due to turbulent boundary layer from F to G.

=0.92 + 11.163 = 12.083 N

Q19) Find the displacement thickness, the momentum thickness and energy thickness for the velocity distribution in the boundary layer given by , where u is the velocity at a distance y from the plate and u=U at y=, where is boundary layer thickness. Also calculate the value of .

A19) Velocity distribution

(i) Displacement thickness is given by the equation

= (Since )

= = = =

(ii) Momentum thickness is given by the eqn

Substituting the value of ,

(iii) Energy thickness is given by the equation:

= = = - =

Q20) Find the displacement thickness and energy thickness for the velocity distribution in the boundary layer given by 2.

A20) Velocity distribution 2.

(i) Displacement thickness is given by the equation

= (Since )

(ii) Momentum thickness is given by the eqn

(iii) =

Q21) A jet plane which weighs 29.43kN and having a wing area of 20 flies at a velocity of 950km/hr when the engine delivers 7357.5kW power. 65% of the power is used to overcome the drag resistance of the wing. Calculate the coefficients of lift and drag for the wing. The density of the atmospheric air is 1.21kg/