Module-1 Viscous Flow · 2020. 3. 30. · Module-1 Viscous Flow 3 Darshan Institute of Engineering & Technology, Rajkot Sr. No. Questions Exam Year Marks 1 Explain the terms: Hydraulic

Last Updated on:

Module-1

Viscous Flow

[email protected]

98795 10743

Applied Fluid Mechanics (2160602)

10-02-2020

Prof. Mehul Pujara

2Module-1 Viscous Flow Darshan Institute of Engineering & Technology, Rajkot

Sr. No.

Contents Teaching hours Weightage %

1 • Introduction• Continuity equation• Energy equation• Momentum equation• Major and minor energy losses,• Hydraulic gradient and total energy line• Pipes in series and parallel• Pipe networks• Hydraulic transmission of power• Navier-Stokes equation of motion• Initial conditions and boundary conditions• Viscous flow-Couette flow• Hagen-Poiseuille equation-flow between parallel plates• Turbulent flow in pipes• Prandtl’s mixing length theory• Velocity distribution• Smooth and rough boundaries• Water hammer phenomenon

12 30

SYLLABUS


Sr.No.

Questions Exam Year Marks

1 Explain the terms: Hydraulic gradient line and Total energy line. OR Define:(i) Critical depth (ii) Total energy line (iii) Hydraulic gradient line.

May-2019Nov-2017

33

2 Derive an expression for the loss of head due to sudden enlargement of apipe.

May-2019Oct-2016May-2016

477

3 Prove that the velocity distribution for viscous flow between two parallelplates when both plates are fixed across a section is parabolic in nature.

May-2019 7

4 Explain the terms: Pipes in parallel and Equivalent pipe. 3

5 Derive an expression for the loss of head due to friction in pipe. OR DeriveDarcy Weisbach formula for the loss of head due to friction in pipe line.

May-2019Nov-2018Nov-2017Oct-2016

4777

6 Enlist the major and minor loses in pipes. Derive the expression for loss ofhead due to sudden contraction. OR What are the minor losses? Under whatcircumstances will they be negligible? Derive the expression for loss ofhead due to sudden contraction.

Nov-2018April-2017

77

GTU MIMP QEUSTIONS LIST


Sr.No.

Questions ExamYear

Marks

7 Explain hydraulically smooth and rough pipes Nov-2018Nov-2017

33

8 Calculate the head loss due to friction using Darcy Equation and powerrequired to maintain 60 liters per second of liquid flow through a steelpipe 0.08 m radius and 900 m long. Take Sp. Gravity of the liquid =0.85 and co-efficient of friction f=0.0025.

Nov-2018 7

9 Enlist the important applications of Navier-stoke equations Nov-2018April-2017

43

10 Define (i) Cavitation (ii) Prandtl Mixing length (iii) Water Hammer ORDescribe water hammer phenomenon in pipes. OR Explain brieflyPrandtl’s mixing layer theory.OR (i) Shear Velocity

Nov-2018Nov-2017April-2017

333

11 Explain Hagen-Poiseuille theory. OR Derive the Hagen-Poiseuilleequation and state the assumptions made. OR Derive an expression forthe velocity distribution of viscous flow through a circular pipe andprove that the ratio of maximum velocity to average velocity is 2.

April-2018Nov-2017April-2017Oct-2016May-2016

37777



Sr.No.


12 Describe relation between shear stress and velocity gradient. April-2018 4

13 What is couette flow? Derive an expression of velocity and shear stressfor couette flow.

7

14 Explain different types of shear theories for turbulent flow. 3

15 Write advantages and disadvantages of shear theories. 4

16 A rough pipe of 30 cm diameter carries water. If the mean pointvelocity and the velocity gradient at a distance of 3 cm from pipe wallare 2 m/sec and 12.5 sec-1 respectively, determine the average heightof roughness projection, wall shear stress, friction factor and meanvelocity of flow. Take ρ =1000 kg/m3 and κ(kappa) = 0.4

7

17 What do you mean by pipes in series and pipes in parallel? How theloss of head is determined in both systems

3

18 Write a brief note on major and minor losses in pipes. Nov-2017 4



Sr.No.


19 A pipe system consists of three pipes arranged in series.

Transform the system to (i) an equivalent length of 30 cm diameterpipe, and (ii) an equivalent diameter for the pipe 6500 m long.

Nov-2017 7

20 Write the assumptions made in derivation of the Dynamic Equation of the Gradually varied flow.

April-2017 3

21 What are the differences between pipe flow and open channel flow?Also write the uses of pipes for hydraulic transmission of fluid.

4

22 Derive the continuity equation for one dimensional flow and discuss itsapplication.

7

23 Enlist the forces acting on Fluid in motion. 4



Sr.No.


24 Calculate the head loss due to friction using Darcy Equation and powerrequired to maintain 50.3 liters per second of liquid flow through asteel pipe 0.1 m radius and 900 m long. Take Sp. Gravity of the liquid =0.7 and co-efficient of friction f=0.0025.

April-2017 7

25 A horizontal pipe of 150 mm diameter is suddenly enlarged to 300 mmdiameter. The rate of flow of water through a pipe is 0.2 m3/sec. Thepressure intensity of smaller pipe 125 kPa. Determine (i) Loss of headdue to sudden enlargement (ii) Pressure intensity in large pipe.

Oct-2016 7

26 A pipe of diameter of 20 cm conveying water. Calculate the discharge when centre line velocity is 3 m/sec and velocity at a point 4 cm from centre is 2.5 m/sec.

7

27 A smooth pipe of diameter 400 mm and length 800 m carries water ata rate of 0.40 m3/s. determine the head lost due to friction, wall shearstress, centre line velocity and thickness of laminar sub-layer. Takekinematic viscosity of water as 0.018 stokes and coefficient of friction,f = 0.0791/ (Re)1/4

May-2016 7



Sr.No.


28 Two reservoirs are connected by two pipes in series of lengths 200 mand 300 m and of diameters 20 cm and 30 cm respectively. Thedifference of head between the two reservoir water surfaces is 10m.The friction factors for the two pipes are 0.02 and 0.015 respectively.Determine the flow rate.

May-2016 7


ALL THE BEST


It is the study of fluid motion with the forces causing flow.

The dynamic bhaviour of the fluid flow is analyzed by the Newton’s second law of motion.

F = m a

In fluid flow, the following forces are present:

Fg, gravity force

Fp, pressure force

Fv, viscous force

Ft, turbulent force

Fc, force due to compressibility

F = Fg + Fp + Fv + Ft + Fc

If Fc is neglected

F = Fg + Fp + Fv + Ft

Equation is known as Reynold’s Equation of motion.

DYNAMICS OF FLUID FLOW:


If Ft is neglected

F = Fg + Fp + Fv

Equation is known as Navier-Stokes Equation.

If is Fv neglected

F = Fg + Fp

Equation is known as Euler’s equation of motion.

DYNAMICS OF FLUID FLOW:


F = Fg + Fp + Fv

Equation is known as Navier-Stokes Equation.

Consider small element of size dx, dy and dzin x, y and z direction.

Let us consider pressure force acting onelement is

On Face ABCD = p * dy dz

On Face EFGH = [P +𝝏𝒑

𝝏𝒙dx] * dy dz

Net pressure force in x-direction is

= p * dy dz - [P +𝝏𝒑

𝝏𝒙dx] * dy dz

= -𝝏𝒑

𝝏𝒙dx dy dz …………..(1)

NAVIER-STOKES EQUATION OF MOTION:

P + 𝝏𝒑

𝝏𝒙dxp


Let us consider R is the body force per unitmass of fluid having component X, Y and Z in x,y and z dierection is

= X * ρ * dx dy dz ………….. (2)

Let Tx, Ty and Tz are the components of shearforce in x, y and z direction. ..……….. (3)

Applying Newton’s second law of motion in x –direction

Force = mass * Acceleration

X * ρ * dx dy dz -𝝏𝒑

𝝏𝒙dx dy dz - Tx = ρ * dx dy dz *

𝒅𝒖

𝒅𝒕

…………..(4)


P + 𝝏𝒑

𝝏𝒙dxp


Let us consider shear stress in x direction is

𝝉 = 𝝁𝒅𝒖

𝒅𝒙

The shear force acting on face ABCD is

= 𝝁 𝒅𝒚 𝒅𝒛𝒅𝒖

𝒅𝒙

The shear force acting on face EFGH is

= 𝝁 𝒅𝒚 𝒅𝒛𝝏

𝝏𝒙(𝒖 +

𝒅𝒖

𝒅𝒙dx)

= 𝝁 𝒅𝒚 𝒅𝒛 (𝝏𝒖

𝝏𝒙+

𝒅𝟐𝒖

𝒅𝒙𝟐dx)

The net shear force along x–direction on facesABCD and EFGH is

= 𝝁 𝒅𝒚 𝒅𝒛𝒅𝒖

𝒅𝒙- 𝝁 𝒅𝒚 𝒅𝒛 (

𝝏𝒖

𝝏𝒙+

𝒅𝟐𝒖

𝒅𝒙𝟐dx)

= − 𝝁𝒅𝟐𝒖

𝒅𝒙𝟐dx 𝒅𝒚 𝒅𝒛 ………….(5)


𝝁𝝏

𝝏𝒙(𝒖 +

𝒅𝒖

𝒅𝒙dx)𝝁

𝒅𝒖

𝒅𝒙


The net shear force along x–direction on facesBCGF and ADEH is

= − 𝝁𝒅𝟐𝒖

𝒅𝒛𝟐dx 𝒅𝒚 𝒅𝒛 ………. (6)

The net shear force along x-direction on faceCDGH and ABEF is

= − 𝝁𝒅𝟐𝒖

𝒅𝒚𝟐dx 𝒅𝒚 𝒅𝒛 .………(7)


𝝁𝝏

𝝏𝒛(𝒖 +

𝒅𝒖

𝒅𝒛dz)

𝝁𝒅𝒖

𝒅𝒛

The net shear force along x–direction on all faces is

Tx = − 𝝁 [𝒅𝟐𝒖

𝒅𝒙𝟐+𝒅𝟐𝒖

𝒅𝒚𝟐+𝒅𝟐𝒖

𝒅𝒛𝟐] dx 𝒅𝒚 𝒅𝒛 ……………… (8)

The net shear force along y–direction on all faces is

Ty= − 𝝁 [𝒅𝟐𝒗

𝒅𝒙𝟐+𝒅𝟐𝒗

𝒅𝒚𝟐+𝒅𝟐𝒗

𝒅𝒛𝟐] dx 𝒅𝒚 𝒅𝒛 ……………… (9)

The net shear force along z–direction on all faces is

Tz= − 𝝁 [𝒅𝟐𝒘

𝒅𝒙𝟐+𝒅𝟐𝒘

𝒅𝒚𝟐+𝒅𝟐𝒘

𝒅𝒛𝟐] dx 𝒅𝒚 𝒅𝒛 …………… (10)


Remember equation (4)….


𝝏𝒙dx dy dz - Tx = ρ * dx dy dz *

𝒅𝒖

𝒅𝒕

Let’s put the value of Tx from equation (8) in belowequation (4)


𝝏𝒙dx dy dz + 𝝁 [

𝒅𝟐𝒖



𝒅𝒛𝟐] dx

𝒅𝒚 𝒅𝒛 = ρ * dx dy dz *𝒅𝒖

𝒅𝒕

Dividing on both the sides with ρ * dx dy dz

X -𝟏

𝝆

𝝏𝒑

𝝏𝒙+𝝁

𝝆[𝒅𝟐𝒖



𝒅𝒛𝟐] =

𝒅𝒖

𝒅𝒕

X -𝟏

𝝆

𝝏𝒑

𝝏𝒙=𝒅𝒖

𝒅𝒕-𝝁

𝝆[𝒅𝟐𝒖



𝒅𝒛𝟐]

X -𝟏

𝝆

𝝏𝒑

𝝏𝒙=𝒅𝒖

𝒅𝒕- 𝜗 [

𝒅𝟐𝒖



𝒅𝒛𝟐]

Similarly in y and z direction,

Y -𝟏

𝝆

𝝏𝒑

𝝏𝒚=𝒅𝒗

𝒅𝒕- 𝜗 [

𝒅𝟐𝒗



𝒅𝒛𝟐]

Z -𝟏

𝝆

𝝏𝒑

𝝏𝒛=𝒅𝒘

𝒅𝒕- 𝜗 [

𝒅𝟐𝒘



𝒅𝒛𝟐]


P + 𝝏𝒑

𝝏𝒙dxp

The equations are known as Navier-Stokes equation.


After rearranging the equation𝒅𝒖

𝒅𝒕= X -

𝟏

𝝆

𝝏𝒑

𝝏𝒙+ 𝜗 [

𝒅𝟐𝒖



𝒅𝒛𝟐]

𝒅𝒗

𝒅𝒕= Y -

𝟏

𝝆

𝝏𝒑

𝝏𝒚+ 𝜗 [

𝒅𝟐𝒗



𝒅𝒛𝟐]

𝒅𝒘

𝒅𝒕= Z -

𝟏

𝝆

𝝏𝒑

𝝏𝒛+ 𝜗 [

𝒅𝟐𝒘



𝒅𝒛𝟐]

In vector form

𝑫𝒗

𝑫𝒕= R -

𝟏

𝝆grad p + 𝝑 𝜵𝟐 𝒗


P + 𝝏𝒑

𝝏𝒙dxp


The important applications of Navier-Stokes equations are Laminar flow in circular pipe

Laminar flow between concentric rotating cylinders

Laminar flow between two fixed plates

Laminar flow between parallel plates having relative motion

The deign of aircraft and cars

Study of blood flow

The design of power station

NAVIER-STOKES EQUATION APPLICATION:


FLOW OF VISCOUS FLUID THROUGHCIRCULAR PIPE:

The flow through the circular pipe will beviscous or laminar, if the Reynoldsnumber (Re) is less than 2000.

Re =𝜌𝑣𝑑

𝜇

Let’s consider pipe of radius R. Considerfluid element of radius r sliding incylinder fluid element of radius r+dr.

Consider length of fluid element is ∆x.

1. p = intensity of pressure on the face AB

2. (p+𝜕𝑝

𝜕𝑥∆𝑥) = intensity of pressure on the

face CD

Hagen Poiseuille Formula:


The value of pressure force…

1. P * π r2 on face AB

2. (p+𝜕𝑝

𝜕𝑥∆𝑥) * π r2 on face CD

The value of shear force is

3. τ 2π𝑟∆𝑥 on the surface of fluidelement.

Summation of all these forces must beequal to zero as no acceleration.

P * π r2 - (p+𝜕𝑝

𝜕𝑥∆𝑥) * π r2 - τ 2π𝑟∆𝑥 =0

∴ -𝜕𝑝

𝜕𝑥∆𝑥π r2 - τ 2π𝑟∆𝑥 =0

∴ -𝜕𝑝

𝜕𝑥r - 2 τ = 0

So, τ= -𝜕𝑝

𝜕𝑥

𝑟

2…………(1)



The shear force and velocity distributionis shown in second figure.

1. Velocity Distribution:

The value of shear stress…

τ = μ𝑑𝑢

𝑑𝑦

y is measured from pipe wall so,

y = R – r

dy = -dr

τ = - μ𝑑𝑢

𝑑𝑟

Put the above value in equation (1)

∴ - μ𝑑𝑢

𝑑𝑟= -

𝜕𝑝

𝜕𝑥

𝑟

2

∴𝑑𝑢

𝑑𝑟=

1

2μ𝜕𝑝

𝜕𝑥r



1. Velocity Distribution:𝑑𝑢

𝑑𝑟=

1

2μ𝜕𝑝

𝜕𝑥r

Integrating above equation w.r.t. r

We get,

u =1

4μ𝜕𝑝

𝜕𝑥r2 + C

To find out the value of C let’s applyboundary condition,

At r = R , u = 0

∴ 0 =1

4μ𝜕𝑝

𝜕𝑥R2 + C

∴ C = -1

4μ𝜕𝑝

𝜕𝑥R2

So, u =1

4μ𝜕𝑝

𝜕𝑥r2 -

1

4μ𝜕𝑝

𝜕𝑥R2




u =1

4μ𝜕𝑝

𝜕𝑥r2 -

1

4μ𝜕𝑝

𝜕𝑥R2

u = -𝟏

𝟒μ𝝏𝒑

𝝏𝒙[R2 - r2]

The above equation of u is a equation ofparabola.

The velocity distribution in pipe isparabola as shown in figure.

2. Ratio of Maximum velocity to AverageVelocity:

The velocity is maximum when r=0,

so Umax = -𝟏

𝟒μ𝝏𝒑

𝝏𝒙R2



Average Velocity:

The average velocity can be obtained bydividing discharge by area of pipe.

The fluid flow through elementary ring is

dQ = velocity at radius r * area

= u * 2π r dr

= -1

4μ𝜕𝑝

𝜕𝑥[R2 - r2] * 2π r dr

The total discharge is….

Q = 0

𝑅𝑑𝑄

= 0𝑅

−1

4μ𝜕𝑝

𝜕𝑥[R2 − r2] ∗ 2π r dr

= −1

4μ𝜕𝑝

𝜕𝑥2π 0

𝑅[R2 − r2] r dr

= −1

4μ𝜕𝑝

𝜕𝑥2π 0

𝑅[R2r− r3] dr


Q =−1

4μ𝜕𝑝

𝜕𝑥2π

0

𝑅[R2r− r3] dr

= −1

4μ𝜕𝑝

𝜕𝑥2π [

R4

2−

R4

4]

=1

4μ(−

𝜕𝑝

𝜕𝑥) 2π

R4

4

Q=𝟏

𝟖μ(−

𝝏𝒑

𝝏𝒙) π R4


Average Velocity:

𝑢 =𝑄

𝐴

=

π8μ(−

𝜕𝑝

𝜕𝑥)R4

π R2

𝒖=1𝟖μ

(−𝝏𝒑

𝝏𝒙) R2

Ratio of Umax to 𝒖 :

Umax 𝒖

=− 1

4μ𝜕𝑝

𝜕𝑥R2

18μ(−

𝜕𝑝

𝜕𝑥)R2

Umax 𝒖

= 2



3. Drop of pressure for a given length ofpipe:

The average velocity is…

𝑢=18μ(−

𝜕𝑝

𝜕𝑥) R2

Or

(−𝜕𝑝

𝜕𝑥) =

8μ 𝑢R2

Integrating above equation w.r.t. x

2

1−𝑑𝑝 =

2

1 8μ 𝑢R2

𝑑𝑥

- [p1-p2] =8μ 𝑢R2

[x1-x2]

[ p1-p2] =8μ 𝑢R2

L {L = x2-x1

[p1-p2] =8μ 𝑢𝐿

(D/2)2


1

x1

2

x2

P1 P2

d

L

[p1-p2] =32μ 𝑢𝐿

D2

Now,

Loss of pressure head =p1−p2𝜌𝑔

p1−p2𝜌𝑔

= hf =32μ 𝑢𝐿𝜌𝑔D2

The equation is Hagen PoiseuilleFormula.


Example:


Consider two parallel fixed plates kept at adistance ‘t’ apart as shown in fig.

If the width of the element in the directionperpendicular to the paper is unity then theforces acting on the fluid element are:

The value of pressure force are

1. P * ∆𝑦 ∗ 1on face AB

2. (p+𝜕𝑝

𝜕𝑥∆𝑥) * ∆𝑦 ∗ 1 on face CD

The value of shear force are

3. τ ∗ ∆𝑥 ∗ 1 on face BC

4. (τ +𝜕τ

𝜕𝑦∆𝑦) * ∆𝑥 ∗ 1 on face AD

Summation of all these forces must beequal to zero as no acceleration.

FLOW OF VISCOUS FLUID BETWEEN TWO PARALLEL PLATES:


The value of pressure force are

1. p * ∆𝑦 ∗ 1on face AB

2. (p+𝜕𝑝

𝜕𝑥∆𝑥) * ∆𝑦 ∗ 1 on face CD

The value of shear force are

3. τ ∗ ∆𝑥 ∗ 1 on face BC

4. (τ +𝜕τ

𝜕𝑦∆𝑦) * ∆𝑥 ∗ 1 on face AD

Summation of all these forces must be equal to zero as noacceleration.

p ∆𝑦 ∗ 1 - (p+𝜕𝑝

𝜕𝑥∆𝑥) ∆𝑦 ∗ 1 - τ ∆𝑥 ∗ 1 + (τ +

𝜕τ

𝜕𝑦∆𝑦) ∆𝑥 ∗ 1 = 0

∴ -𝜕𝑝

𝜕𝑥∆𝑥 ∆𝑦 +

𝜕τ

𝜕𝑦∆𝑦 ∆𝑥 = 0

Dividing by ∆𝑥 ∆𝑦

-𝜕𝑝

𝜕𝑥+𝜕τ

𝜕𝑦= 0

𝝏𝒑

𝝏𝒙=𝝏𝝉

𝝏𝒚



𝝏𝒑

𝝏𝒙=𝝏𝝉

𝝏𝒚


The value of shear stress using Newton’s lawof viscosity is

τ = μ𝑑𝑢

𝑑𝑦

Putting the value in equation𝝏𝒑

𝝏𝒙=𝝏𝝉

𝝏𝒚

𝝏𝒑

𝝏𝒙=𝝏(μ 𝑑𝑢

𝑑𝑦)

𝝏𝒚𝝏𝒑

𝝏𝒙= μ

𝝏2u𝝏𝒚𝟐

𝝏2u𝝏𝒚𝟐

=𝟏

μ𝝏𝒑

𝝏𝒙

Integrating above equation w.r.t. y weget,


𝑑𝑢

𝑑𝑦=𝟏

μ𝝏𝒑

𝝏𝒙y + c1

Integrating again

u =𝟏

μ𝝏𝒑

𝝏𝒙

y2

2+ c1y + c2

c1 and c2 are constant ofthe equation.



u =𝟏

μ𝝏𝒑

𝝏𝒙

y2

2+ c1y + c2

c1 and c2 are find out by putting theboundary condition.

I. y = 0 , u = 0II. y = t , u = 0

Applying first boundary condition

0 = 0 + 0 + c2∴ c2 = 0

Applying second boundary condition

0 =𝟏

μ𝝏𝒑

𝝏𝒙

t2

2+ c1t

c1 = -𝟏

2μ𝝏𝒑

𝝏𝒙t


Substituting the value inequation

u =𝟏

μ𝝏𝒑

𝝏𝒙

y2

2+ c1y + c2

u =𝟏

μ𝝏𝒑

𝝏𝒙

y2

2+ (-

𝟏

2μ𝝏𝒑

𝝏𝒙t) y

u = -𝟏

𝟐μ𝝏𝒑

𝝏𝒙[ty - y2]



u = -𝟏

𝟐μ𝝏𝒑

𝝏𝒙[ty - y2]

The above value shows that velocityprofile is parabola type.

2. Ratio of Maximum velocity to Averagevelocity:

The velocity is maximum when y = t/2,applying the value in above equation,

Umax = -𝟏

𝟐μ𝝏𝒑

𝝏𝒙[t (

𝒕

𝟐) - (

𝒕

𝟐)2 ]

= -𝟏

𝟐μ𝝏𝒑

𝝏𝒙[𝒕𝟐

𝟐-𝒕𝟐

𝟒]

= -𝟏

𝟐μ𝝏𝒑

𝝏𝒙*𝒕𝟐

𝟒

Umax= -𝟏

𝟖μ𝝏𝒑

𝝏𝒙t2




Average Velocity:

The average velocity can be obtained bydividing discharge by area of pipe.

dQ = velocity at distance y * area of strip

= -1

2μ𝜕𝑝

𝜕𝑥[ty - y2] * dy *1

Q = 0𝑡

dQ = 0𝑡

−1

2μ𝜕𝑝

𝜕𝑥[ty − y2] ∗ dy ∗1

= -1

2μ𝜕𝑝

𝜕𝑥[𝒕𝟑

𝟐-𝒕𝟑

𝟑]

= -1

2μ𝜕𝑝

𝜕𝑥

𝒕𝟑

𝟔

Q= -𝟏

𝟏𝟐μ𝝏𝒑

𝝏𝒙t3




Average Velocity:

The value of discharge,

Q= -𝟏

𝟏𝟐μ𝝏𝒑

𝝏𝒙t3

The average velocity is given by,

𝑢 =𝑄

𝐴

=− 1

12μ𝜕𝑝

𝜕𝑥t3

𝑡∗1

𝒖 = -𝟏

𝟏𝟐μ𝝏𝒑

𝝏𝒙t2




Maximum velocity is

Umax= -𝟏

𝟖μ𝝏𝒑

𝝏𝒙t2

Average velocity is

𝒖 = -𝟏

𝟏𝟐μ𝝏𝒑

𝝏𝒙t2

So ratio is

Umax 𝒖

=− 𝟏

𝟖μ𝝏𝒑

𝝏𝒙t2

− 𝟏𝟏𝟐μ

𝝏𝒑

𝝏𝒙t2

Umax 𝒖

=𝟑

𝟐



3. Drop of pressure head for a givenlength:

The value of average velocity is given by,

𝒖 = -𝟏

𝟏𝟐μ𝝏𝒑

𝝏𝒙t2

Rewriting the above equation𝝏𝒑

𝝏𝒙= -

𝟏𝟐μ 𝒖t2

Integrating above equation w.r.t. x

21dp = 2

1−

12μ 𝑢t2

dx

[p1-p2] =−12μ 𝑢

t2[x1-x2]

[p1-p2] =12μ 𝑢

t2[x2-x1]

[p1-p2] =−12μ 𝑢𝐿

t2



3. Drop of pressure head for a givenlength:

[p1-p2] =−12μ 𝑢𝐿

t2

If hf is drop of pressure head then,

hf =p1−p2𝝆𝒈

= −𝟏𝟐μ 𝒖𝑳𝛒𝐠t2

4. Shear stress Distribution:

The shear stress value is

τ = μ𝑑𝑢

𝑑𝑦

And velocity is given by,

u = -𝟏

𝟐μ𝝏𝒑

𝝏𝒙[ty - y2]



4. Shear stress Distribution:

Putting the value of u in shear stressequation then,

τ = μ𝑑

𝑑𝑦[−

1

2μ𝜕𝑝

𝜕𝑥(ty − y2)]

τ = μ[−1

2μ𝜕𝑝

𝜕𝑥(t − 2y)]

τ = −1

2

𝜕𝑝

𝜕𝑥(t − 2y)

The value of τ with y only, at y=0 or tshear stress is maximum,

τ0 = −1

2

𝜕𝑝

𝜕𝑥t



Example:


References:

1. Fluid Mechanics and Fluid Power Engineering by D.S. Kumar, S.K.Kataria & Sons

2. Fluid Mechanics and Hydraulic Machines by R.K. Bansal, LaxmiPublications

3. Fluid Mechanics and Hydraulic Machines by R.K. Rajput, S.Chand & Co

4. Fluid Mechanics; Fundamentals and Applications by John. M. CimbalaYunus A. Cengel, McGraw-Hill Publication

Documents

Module-1 Viscous Flow · 2020. 3. 30. · Module-1 Viscous Flow 3 Darshan Institute of Engineering & Technology, Rajkot Sr. No. Questions Exam Year Marks 1 Explain the terms: Hydraulic