solutions of examples for practice - · pdf file1 - 2 fluid mechanics properties of fluids w g...

Solutions of Examples for Practice

Example 1.29 :

Solution : Given data : u =32

y y 3� 2 , � = 8 Poise = 0.8 Pa-s

To find : Shear stress �.

Step - 1 : Calculate the shear stress at various points

It is given that,

u =32

y y 3� 2

�dudy

=32

y1�32

2

As per Newton's law of viscosity,

� = �dudy

= 0.832

32

y1 2� ��

��

At y = 9 cm = 0.09 m

� = 0.832

32

0.091 2� � ��

��

= 0.84 N m 2 … Ans.

Example 1.30 :

Given data : d1 = 6 mm = 6 10 3� � m, d2 = 12 mm = 12 10 3� � m,

= 0.0725 N/m, � = 0º.

To find : Difference of water levels.

Step - 1 : Calculate the difference of levels in the tubes

Capillary effect in smaller limb is,

h 1 =4 �

� �

cosd

=4 0.0725 cos(0)

9810 6 10 3

� �

� � �

� h 1 = 4.9269 10 m3� � = 4.9269 mm

Capillary effect in larger limb is,

h 2 =4 �

� �

cosd

=4 0.0725 cos (0)

9810 12 10 3

� �

� � �

1 - 1 Fluid Mechanics

Chapter - 1

PROPERTIES OF FLUIDS

Unit - I

� h 2 = 2.4634 10 m3� � = 2.4634 mm

Difference in the levels of capillary tubes is,

h = h h1 2� = 4.9269 – 2.4634 = 2.4635 mm … Ans.

Example 1.31

Solution :

Given data : Side of cube b = 0.5 m, m = 50 kg, � � 45º,

� = 2.5 10 3� � N - s m2 , t = 0.05 mm = 0.05 10 m3� �

Step - 1 : Calculate the shear force on the bottom surface of cube

Weight of cube W = mg = 50 9.81� = 490.5 N

Component of weight W acting along the plate is equal to the shear force on the bottom

surface of cube.

� Shear force F = W � sin � = 490.5 sin (45)

F = 346.8358 N

Step - 2 : Calculate the speed of the block


� = �dudy

=FA

� F = �dudy

� A = �Ut

� A … (A = Contact area)

� 346.8358 = 2.5 � ��

� ��

10 3 U

0.05 10(0.5 0.5)

3

� U = 27.7468 m/sec … Ans.


Properties of Fluids

W

G

θ

W sin θ

0.2

m

0.2 m

θ = 45º

W cos θ

t = 0.05 mm

Contactarea

Fig. 1.1

Exmaple 1.32 :

Solution :

Given data : d = 0.15 m, t = 0.05 m, N = 80 rpm, T = 6 10 4� � N-m.

To find : Viscosity (�)

Step - 1 : Calculate the viscosity of oil

Refer Fig. 1.2.

Velocity of the disc is,

V =� dN60

� V =� � �0.15 80

60

� V = 0.6283 m/sec.

We know that,

T = Force � Radius = Fd2

�

� 6 10 4� � = F0.152

� � F = 8 10 N3� �


� = �dudy

=FA

� F = �dudy

A� = � �� Ut

d t

� 8 10 3� � = � �� 0.6283 0.15 0.05

� � = 0.5403 Pa-s … Ans.

Example 1.33

Solution :

Given data : rd = 1.5 cm, � dd = 1.5 2� = 3 cm, rj = 0.8 cm

� dj = 0.8 2� = 1.6 cm, patm = 100 kN m2 , = 0.05 N/m

To find : pgauge and pabs

Step - 1 : Calculate the gauge pressure and absolute pressure for a droplet and for ajet of water



80 rpm

0.15 m

t = 0.05 m

Fig. 1.2

i) For a droplet

Gauge pressure is,

p =4 d =

4 0.05

3 10–3�

�= 66.6667 N/m2 … Ans.

Absolute pressure is,

pabs = p patm� = 66.6667 + 100 � 10 3

� pabs = 100.0667 kN/m2 …Ans.

ii) For a jet of water

Gauge pressure is,

p =2 d =

2 0.05

1.6 10–3�

�= 62.5 N/m2 … Ans.

Absolute pressure is,

pabs = p patm� = 62.5 + 100 � 10 3

� pabs = 100.0625 kN/m2 …Ans.

Properties of Fluids ends....




Example 2.11

Solution :

Given data :

Sm = 0.8, h 1 = 100 cm = 1m, h = 30 cm = 0.3 m, h 2 = 50 cm = 0.5 m

To find : Pressure difference (p p )M N�

Step - 1 : Calculate the pressure difference between M and N

We know that,

p pM N� = � � �1 1 m 2 2h h h� �

= (9810 1) (9810 0.8 0.3) (9810 0.5)� � � � � �

p pM N� = 2550.6 kN m 2 … Ans.

Example 2.12

Solution :

We know that,

Pressure in left column = Pressure in right column

� p h h1 1 2 2� �� = 0

� p = � � � � �[(9810 0.2) (9810 13.6 0.2)]

� pM = � �28.6452 10 N m3 2 = � 28.6452 kN m 2 … Ans.

Example 2.13

Solution :

Given data : Z = 0.4 m

To find : Pressure p

Step - 1 : Calculate the pressure due to 0.4 m columni) For water,

p = � � �g Z = 1000 9.81 0.4� �


Chapter - 2

Fluid pressure and its measurement

Unit - I

� p = 3924 N/m 2 … Ans.ii) For oil,

Pressure is given by,

p = � � �g Z = (0.9 1000) 9.81 0.4� � �

� p = 3531.6 N/m 2 … Ans.iii) For mercury,

p = � � �g Z = (13.6 1000) 9.81 0.4� � �

� p = 53.3664 103� N/m 2 … Ans.

Example 2.14

Solution :

Pressure in left column above datum = Pressure in right column above datum

� p hA 1� � 1 = � �2 2 3 3h h�

� pA = � � � �1 1 2 2 3 3h h h� �

� pA = � � � � � � � �9810 0.6 13.6 9810 0.45 0.88 9810 0.3

� pA = 56.7410 103� N/m 2 … Ans.

Fluid Pressure and its Measurement ends…


Fluid Pressure and its Measurement


Example 3.23

Solution :

Given data : d = 1.2 m, Distance of C.P. from liquid surface = 2 m, � � 90º

To find : Total pressure (P) and center of pressure (h)

Step 1 : Calculate the total pressure and center of pressure

Refer Fig. 3.1.

Total pressure on the plate is,

P = �A x = � ��4

� d 2 � x

� P = 9810 ��4

� 1.22 � 2

� P = 22.1896 � 103 N = 22.1896 kN ... Ans.Depth of center of pressure vertically below liquid surface is,

� h = x +IAx

G = x +

�

�64

42

�

� �

d

d x

4


Chapter - 3

Hydrostatic forces

Unit - I

2 m

1.2 mCG

CP

CG

CP

hx

Fig. 3.1

� h = 2 +

�

�64 (1.2)

4 (1.2) 2

4

2

�

� �

� h = 2.045 m below free liquid surface. … Ans.

Example 3.24

Solution :

Given data : b = 3 m, Distance of top edge from liquid

surface = 1.5 m.

To find : Total pressure (P) and centre of pressure

Step - 1 : Calculate the total pressure and position ofcentre of pressure

Refer Fig 3.2.

Total pressure is,

P = �Ax = � � � ��

��

�12

BC AD x

� P = 0.9 981012

3 3 cos 30 (1.5 + 1.5)� � � ��

��

�

� P = 103.2228 10 N3� = 103.2228 kN … Ans.The position of centre of pressure is,

h = x +IAxG = x +

I12

BC AD x

G

� � �

But IG =bh36

3=

BC (AD)36

3�=

3 (3 cos (30))36

3�= 1.4614 m4

� h = (1.5 1.5)1.4614

12

3 cos (30) (1.5 1.5)� �

� � � �3

� h = 3.125 m below free liquid surface. … Ans.

Example 3.25

Solution :

Given data : Diagonal AC = 4 m � Side of square b42

� = 2.8284 m

To find : P and h

Step - 1 : Calculate the hydrostatic force and the depth of centre of pressure


Hydrostatic Forces

1.5 m

3 mCG

CP

30ºA

B

C

D

Fig. 3.2

Refer Fig. 3.3.

Hydrostatic force is,

P = �A x = 0.85 9810 (2.8284 ) (1 2)� � � � �2 8284.

� P = 200.1201 10 N 200.1201 kN3� � … Ans.

Depth of centre of pressure is,

h = xIAxG�

But, IG =b12

4=

(2.8284)12

4= 5.3331 m4

� h = (1 2)5.3331

(2.8284 2.8284) (1 2)� �

� � �

� h = 3.2222 m below free liquid surface … Ans.

Example 3.26

Solution :

Given data : b = 5 m, d = 6 m, � � 45º

To find : Total pressure and centre of pressure.

Step - 1 : Calculate the total pressure and the position of centre of pressure

Refer Fig. 3.4.

x = 5 + OA sin 45 = 5 + 3 sin 45 = 7.1213 m

Total pressure is,

P = �A x = 9810 (5 6) 7.1213� � �

� P = 2.0958 10 N6� … Ans.

Position of centre of pressure is,


Hydrostatic Forces

P

1 m

x = 1 + 2 = 3 mh

4 m

b b

A

B

C

DO

CG

CP

Fig. 3.3

h = xI sin

AxG

2�

�= x

bd12

sin

(b d) x

32

��

�

�

= 7.1213

5 612

(sin 45)

5 6 7.1213

32

�

��

� �


Example 3.27

Solution :

Given data : d = 4 m, b = 3 m, � = 90º

To find : Total pressure (P) and centre of pressure

Step - 1 : Calculate the total pressure and position of centre of pressure

Refer Fig. 3.5.

Total pressure is,

P = � A x

But, A = 2 3 6 m2� � and x = 3.5 m

� P = 9810 � 6 � 3.5

� P = 206.01 � 103 N … Ans.The position of centre of pressure is,

h = xIA x

G�

But, IG =bd12

3 412

16 m3 3

4��

�


Hydrostatic Forces

hx 5 m

B

45º

45º AB

O

3 cos45

3 m3 sin45

45º A

O

CP

CG

5 m

6 m

Upper edge

Fig. 3.4

CGCP

3.5 m

4 m

3 m

P

h

Fig. 3.5

� h = 3.516

6 3.5�

�


Hydrostatic Forces ends....


Hydrostatic Forces


Example 5.27

Solution :

u = 6x2 , v = 8xyz, w = � �4 xz 12 xz2

Now, differentiate the velocity components with respect to x, y and z��ux

= 12x��uy

= 0��uz

= 0

��vx

= 8yz��vy

= 8 xz��vz

= 8xy

��wx

= � �4z 12 z2 ��wy

= 0��wz

= – 8xz – 12x

i) To satisfy continuity equation (for continuous flow)

��

��

��

ux

vy

wz

� � = 0

� 12x + 8xz – 8xz – 12x = 0

Hence, the flow is continuous. … Ans.

ii) For rotational flow, the rotational component about z-axis should not be zero. The

component of rotation about z-axis is,

� z =12

��

��

vx

uy

��

��

� � z =12 (8yz 0)� = 4 yz 0

As � z 0, the flow is rotational … Ans.

iii) Rotation components at point (1, 1, 1),

�x =12

��

��

wy

vz

��

��

� �x =12 (0 8 xy) 1

2 (0 8 1 1)� � � � �


Chapter - 5

Kinematics of fluid motion

Unit - II

� �x = – 4 … Ans.

�y =12

��

��

uz

wx

��

��

� �y =12 (0 12z))� � �( 4 2z

� �y =12 (0 12 1)� � � � �( 4 12

�y = 8 … Ans.

� z =12

��

��

vx

uy

��

��

� � z =12 (8 yz )� 0 =

12 (8 1 1 )� � � 0

� z = 4 … Ans.

Example 5.28

Solution : To satisfy the continuity equation,��

��

��

ux

vy

wz

� � = 0 … (i)

u = x y z2 2 2� � 2 ��ux

= 2x

v = � � �x y xy yz2 ��vy

= � � �x x z2

Substituting these values in equation (i),

� 2x x x zwz

2� � � ��

= 0

��wz

= x x z2 � �

� �w = (x x z) z2 � � �

Integrating on both sides,

� w = x z xz +z2

C22

� �

But, constant of integration cannot be a function of z.

Hence, it is a function of x and y i.e. C = f(x, y).

� w = x z xz +z2

f (x, y)22

� � … Ans.


Kinematics of Fluid Motion

Example 5.29

Solution :

Stream function � = 2xy

i) As per defination of stream function,

��x

= v = 2y and��y

= – u = 2x

� u = – 2x and v = 2y

We know that,

V = u v2 2�

at point (2, 3),

u = – 2x = � �2 2 = – 4

v = 2y = 2 3� = 6

V = ( )� �4 62 2 = 7.211 m/s … Ans.

ii) As per defintion of potential function,��x

= – u = – (– 2x) = 2x … (i)

and��y

= – v = – (2y) = – 2y … (ii)

From equation (i),

�� = 2x x� �


� =2x2

C2

� = x C2 �

But, constant of integration is a function of y i.e. C = f (y).

� � = x f (y)2 � … (iii)

To find f (y) we will use equation (ii). Initially differentiate equation (iii) with respect to yi.e.

��y

= 0 � �f (y)

But,��y

= – 2y … [From equation (ii)]

� f (y)� = – 2y

or f(y) =�

�2y2

Constant2



f(y) = � �y constant2

Substituting this value in equation (iii),

� � = x y Constant2 2� � … Ans.

Example 5.30

Solution :

Velocity potential � = 2xy – x

As per definition of velocity potential �,��x

= – u = (2y – 1) � u = 1 – 2y

and��

�y

= – v = 2x � v = – 2x

Now, as per definition of stream function �,

��x

= v = – 2x … (i)

and��y

= – u = 2y – 1 … (ii)

From equation (i),��x

= v = – 2x

� �� = – 2x � �x


� =�

�2 x2

C2

� � = – x C2 �

But, the constant of integration cannot be a function of x.

Hence, it is a function of y i.e. C = f(y).

� � = � �x f(y)2 … (iii)

Initially differentiate equation (iii) with respect to y i.e.

��y

= 0 f (y)� �

But,��y

= 2y – 1

� 2y – 1 = f (y)�



� f(y) =2 y

2y

2�

� f(y) = y y2 �

Substituting this value in equation (iii),

� = � � �x y y2 2 … Ans.

��




Example 6.3

Solution :

Given data : d1 = 60 cm = 0.6 m � a 1 =�4

d12 =

�4

� (0.6) 2 = 0.2827 m2

d2 = 20 cm = 0.2 m � a 2 =�4

d22 =

�4

� (0.2) 2 = 0.0314 m2

Z1 = 20 cm, = 0.2 m, S = 0.9, p1 = 180 kPa = 180 10 3� Pa

p2 = 100 kPa = 100 10 3� Pa, h L = 2 % of h, Cd = 0.97.

To find : Discharge Q.

Step - 1 : Calculate the flow rate through the venturimeter

Applying Bernoulli's equation between inlet (section 1) and

throat (section 2) by considering section 2 as a datum,

p V2g

Z12

11

�� =

p V2g

Z h22

2 f2

�� … (i)

But, for vertical venturimeter differential head is

h =p p

(Z Z )1 21 2

��

�

� h =180 10 1 0 10

0.9 9810(0.2 0)

3 3� � ��

� �0

… S f

w�

��

��

� h = 9.2610 m … (ii)

Head lost, h L = 2 % h = 0.02 9� .2610 = 0.1852 m

Also by continuity equation,

Q = a V1 1 = a V2 2

� 0.2827 V1� = 0 0314. � V2

� V1 = 0.1110 V2


Chapter - 6

FLUID DYNAMICS

Unit - III

1

2

Inlet

20 cm

Throat

Outlet

Fig. 6.1

Equation (i) can be written as,

p p1 2��

�(Z Z )1 2 =

V V2g

h22

12

L�

�

9.2610 =V (0.1110 V )

2 9.810.18522

22

2��

�

� 9.0758 = 0.0503 V22

� V2 = 13.4325 m/s

Discharge is,

Q = a V2 2 = 0 13.4325.0314 �

� Q = 0.4217 m sec3 = 421.7820 lps … Ans.

Example 6.34

Solution :

Given data : d0 = 20 cm = 0.20 m � a 0 =�4

d02 =

�4

� (0.20)2 = 0.0314 m2

d1 = 40 cm = 0.4 m � a 1 =�4

d12 =

�4

� (0.4) 2 = 0.1256 m2

x = 0.8 m, S = 0.9, Cd = 0.64

To find : Flow rate of oil Q.

Step - 1 : Calculate the flow rate of oil through the orifice meterPressure head through orifice meter is,

h = xSS

1m ��

� �

= 0.813.60.9

1��

� �

� h = 11.2888 m

Discharge through the orifice meter is,

Q = C a a2gh

a ad 0 112

02

��

� Q = 0.64 0.0314 0.12562 9.81 1 .2888

(0.1256) (0.03142� � �

� �

�

1

)2

� Q = 0.3088 m sec3 = 308.8853 lps … Ans.

Example 6.35

Solution :

Given data : x = 200 mm = 0.2 m, Sm = 1, � air = 16 N m 3 , Cv = 0.98


Fluid Dynamics

To find : Velocity of aeroplane (V).

Step - 1 : Calculate the velocity of aeroplanePressure head is,

h = xSS

1m

air�

��

= x 1m

air

��

��

� �

� h = 0.2981016

1��

� �

= 122.425 m of air

Velocity of aeroplane is,

V = C 2ghv = 0.98 2 9.81 122.425� � �

� V = 48.0297 m/sec = 172.9069 km/hr … Ans.Example 6.36

Solution :

Given data : d = 200 mm = 0.2 m �a =�4

d2 =�4

� (0.2) 2 = 31.4159 10 m2� �3

dc = 16.4 cm = 0.164 m �a c =�4

dc2 =

�4

� (0.164)2 = 21.1240 10 m3 2� �

h = 5.2 m, Q = 180 lps = 180 10 3� � m sec3

To find : Hydraulic coefficients (C , C , C and C )c d v r

Step - 1 : Calculate the values of hydraulic coefficientsCoefficient of contraction is,

Cc =aac =

21.1240 10

31.419 10

3

3�

�

�

�= 0.6723 … Ans.

Theoretical discharge is,

Qth = a Vth� = 31.4169 10 2 9.81 5.23� � � �� … (V 2gh)th �

� Qth = 0.3173 m sec3

Coefficient of discharge is,

Cd =QQ

act

th=

180 100.3173

3� �= 0.5672 … Ans.

We know that,

Cd = C Cc v� � Cv =CC

d

c

� Cv =0.56720.6723

= 0.8437 … Ans.


Fluid Dynamics

Coefficient of resistance is,

Cr =1

Cv2

– 1 =1

(0.8437)2�1 = 0.4046 … Ans.

Example 6.37

Solution :

Given data : � = 30º, S = 0.9, h m = 600 mm = 0.6 m, Cd = 0.98

Inlet diameter d1 = 300 mm = 0.3 m

� Area of inlet a 1 =�4

d12 =

�4

� (0.3) 2 = 0.0706 m2

Throat diameter d2 = 200 mm = 0.2 m

� Area of throat a 2 =�4

d22 =

�4

� (0.2) 2 = 0.0314 m2

To find : Flow rate (Q)

Step - 1 : Calculate the flow rate of oil

Applying Bernoulli's equation between section 1 and 2 by considering horizontal through

section 1 as a datum,

p V2g

Z12

11

�� =

p V2g

Z22

22

��

�V V

2g22

12�

=p p1 2�

��

� ��

(Z Z )2 1 = h

But, h = xSS –1m

o

�

� �


Fluid Dynamics

1

2

0.6 m

0.7 × sin 30 = 0.35 m

30º

0.7 m

Mercury

Fig. 6.2

� h = 0.6 13.60.9 –1

��

� h = 8.4666 m

Discharge through venturimeter is ,

Q =C a a 2gh

a a

d 1 2

12

22

�

�

� Q =0.98 0.0706 0.0314 2 9.81 8.4666

(0.0706) (0.0314)2

� � � � �

� 2

� Q = 0.4428 m sec3 … Ans.

Fluid Dynamics ends …


Fluid Dynamics


Example 7.25

Solution :

Given data : B = 200 mm = 0.2 m, umax = 2 m/sec, � = 2.5 poise = 2.5 Pa-s.

To find : i) � max ii) dp ii)dudy

iv) u

Step - 1 : Calculate the shear stress at the plates

For laminar flow between parallel plates,

u avg =23

u max =23

� 2 = 1.3333 m/sec

But, u avg =B2

12��

��

�

��

px

1.3333 =(0.2)

2.5

2

12 ��

�

��

��

px

��

�

��

��

px

= 1000Pam

Shear stress at the plates is maximum, hence

�max =��

�

��

��

px

B2

= 1000 �0.22

�max = 100 N m 2 or Pa … Ans.

Step - 2 : Calculate the difference in the pressure and velocity gradient at the platesPressure difference between two points which are 20 m apart is,

p p1 2� = dp =12� u L

B

avg2

=12 2.5 1.3333 30

(0.2) 2� � �

p p1 2� = dp = 29.9995 10 N m3 2� … Ans.


� = �dudy


Chapter - 7

LAMINAR FLOW

Unit - IV

It is important to note that, y is the distance measured from the plates and shear stress (�)at the plates is maximum.

�max = �dudy y 0

��

��

… (At the plate y = 0)

100 = 2.5dudy y 0

��

��

dudy y 0

��

��

= 40 sec 1� … Ans.

Step - 3 : Calculate the velocity at 20 mm from the plate

Velocity for the fixed parallel plates is,

u =12�

��

��

��

px

(By y )2

u =1

2 2.51 0 (0.2 0.03 0.03 )2

�� 00

u = 1.02 m/sec … Ans.Example 7.26

Solution :

Given data : � = 0.08 N - s m2 , Width Z = 2 m, B = 20 mm = 0.02 m

At midway u = u max = 6 m/sec

To find : i)��px

��

��

ii) uavg iii) Q

Step - 1 : Calculate the pressure gradient along the flow

For laminar flow through fixed parallel plates,

umax =18�

��

��

��

px

B2 6 =1

8 0.08(0.02)2

��

�

��

��

px

��

�

��

��

px

= 9600

or��px

= � 9 00 N m m26 orPam

… Ans.

Step - 2 : Calculate the average velocity and discharge of an oil

Average or mean velocity is,

uavg =23

u max =23

6� = 4 m/sec … Ans.

Discharge of an oil is,

Q = Area Mean velocity� = (Z B) u avg� �

Q = (2 0.02) 4� � = 0.16 m sec3 … Ans.


Laminar Flow

Example 7.27

Solution :

Given data : D = 400 mm = 0.4 m R =D2

0.42

� = 0.2 m, umax = 2 m/sec.

To find : i) umean or uavg ii) Radius at uavg iii) u

Step - 1 : Calculate the mean velocity and its radiusFor laminar flow through a pipe,

uavg =u

2max =

22

uavg = 1 m/sec … Ans.

The radius for mean velocity is,

r = 0.707 R = 0.707 0.2�

r = 0.1414 m … Ans.Step - 2 : Calculate the velocity at 60 mm from the pipe wallVelocity for the pipe flow is,

u = u 1rRmax

2� �

��

�

��

�

��

… [From equation (7.5)]

But, r = R – y = 0.2 – 0.06 = 0.14 m [y is measured from the pipe wall]

u = 2 10.140.2

2� �

��

�

��

�

��

u = 1.02 m/sec … Ans.Example 7.28

Solution :

Given data : � = 1.5 poise = 0.15 Pa-s, S = 0.9, D = 50 mm = 0.05 m R = 0.025 m,

��

��

��

px

= 10 kN m2 = 10 10 3� N m2

To find : i) Mass flow rate �m ii) � iii) Re iv) Power P

Step - 1 : Calculate the mass flow rate and shear stress at pipe wallFor laminar flow through a pipe, mass flow rate is,

�m = � � �A uavg = � �oil2

avgR u� � … (i)

But, pressure drop is,

p p1 2� =32 � u L

D

avg2


Laminar Flow

or��

�

��

��

px

=32� u

D

avg2

… (� L = dx)

10 10 3� =32 0.15 u

(0.05)

avg2

� �

u avg = 5.2083 m/sec

Substituting in equation (i),

�m = 0.9 1000 (0.025) 5.20832� � � �� … ( )� � �oil water= S

�m = 9.2038 kg/sec

�m = 9 60.2038 � = 552.2295 kg/min … Ans.

Shear stress is maximum at the pipe wall i.e.

�max =��

�

��

��

px

R2

= (10 10 )0.025

23� �

�max = 125 N m 2 … Ans.

Step - 2 : Calculate the Reynolds number of flow and the required powerReynold's number of flow is,

Re =�

�

u Davg=

0.9 1000 5.2083 0.050.15

� � �

Re = 1562.49 … Ans.

Re < 2000, it means flow is laminar.

Power required to maintain the flow is,

P = � Q h f = ��

� � ��

��

Area Mean velocityp p1 2

P = � R u (p p )2avg 1 2� � �

P = ��

R uu L

D2

avgavg2

� �32

P =� � � � � � �(0.025) 5.2083 32 0.15 5.2083 60

(0.05)

2

2… (� L = 60 m)

P = 6135.8446 W = 6.1358 kW … Ans.Example 7.29 :

Solution :

Given data : D = 80 mm = 0.08 m, R =D2

0.082� = 0.04 m, L = 500 m,


Laminar Flow

Slope = 1 in 50, � = 0.8 N-s/m2, S = 0.9, Q = 6 lps = 6 10 m s–3 3�

To find : i) Flow is laminar or not ? ii) Ppump iii) u anddudy

Step - 1 : Calculate the Reynolds number to find the type of flow

Discharge through the pipe is,

Q = A u avg� =�4

� �D u2avg

6 10 3� – =�4

2� �(0.08) u avg

u avg = 1.1936 m/sec

Reynolds number is,

Re =�

��

�f avg w avgu D S u D

��

Re =0.9 1000 1.1936 0.08

0.8� � �

Re = 107.424 < 2000 Flow is laminar … Ans.Step - 2 : Calculate the required power of pumpLoss of head due to friction is,

h f =32 u L

Davg

f2

�

�=

32 0.8 1.1936 50.9 9810 (0.08)2

� � �

� �

00

h f = 270.3816 m

A slope of 1 in 50 indicates that, for 50 m length there is 1 m height (inclined pipe) of the

pipe. It means, for 500 m length of pipe there is height of 10 m. Hence, the pump has to

overcome frictional resistance in the pipe and the extra 10 m height of the pipe.

Total head against which the pump works is,

H = h f + Head due to slope = 270.3816 + 10 = 280.3816 m

Power required by the pump is,

P = � f QH = 0.9 9810 6 10 2 0.3816–3� � � � 8

P = 1 10 W34 8529. � …Ans.

Step - 3 : Calculate the centre line velocity and velocity gradient at the pipe wallCentre line (maximum) velocity is,

umax = 2 u 2 1.1936avg� � � = 2.3872 m/sec … Ans.

Pressure drop is given as,

p p1 2– =128 QL

D4�

�

p – pL

px

128 QD

1 24

��

��

– ��

�

�


Laminar Flow

– ��

px

��

�� =

128 0.8 6 100.08

10 N m m–3

3 2� � �

��

� ( ).

447746

Negative sign indicates pressure drop.


� = �dudy

dudy =

��

But at the pipe wall, y = 0 and shear stress � is maximum,

dudy y 0

��

��

=�

�max =

– ��

�

px

R2

��

��

dudy y 0

��

��

=4.7746 10 0.04

20.8

3� �

dudy y 0

��

��

= 119.3662 sec–1 … Ans.

Example 7.30

Solution :

Given data : � = 16 poise = 1.6 P a-s, S = 0.9, D = 60 mm = 0.06 m R = 0.03 m

p p1 2� = 10 kN/m2 = 10 � 10 N3 /m2 , L = 1.5 m

To find : i) Q in lpm iii) umax iii) FD

Step - 1 : Calculate the rate of flow of oil

For laminar flow through pipe,

p p1 2� =128�

�

QLD4

10 � 10 3 =128 1 6

0 064� � �

�

.

.

Q 1.5�

Q = 1.3253 � 10 3� m3 /s ... Ans.Step - 2 : Calculate the center line velocity

Discharge through pipe is ,

Q = Area � Mean velocity = �R2 � u avg

1 3253 10 3. � � = � � �( . )0 03 2 u avg

u avg = 0.4687 m/s

Maximum velocity in the pipe is,


Laminar Flow

u avg =u

2max 0.4687 =

u2max

u max = 0.9375 m/s ... Ans.

Step - 3 : Calculate the total friction drag over 1 km of the pipe

Total friction drag for 1500 m length is,

FD = �max � Contact area = �max � �DL

FD =��

��

��

��

px

R2 DL =

10 101000

0 032

0 063��

�

��

..� 1.5 10 3

FD = 42.4115 N ... Ans.

Laminar Flow ends …


Laminar Flow


Example 8.11

Solution :

Given data : D = 0.2 m, R =0.22

= 0.1 m, u max = 5 m/sec,

At 60 mm from the centre, y = 100 � 60 = 40 mm = 0.04 m, u = 4.4 m/sec

To find : � 0

Step 1 : Calculate the shear stress at the wall

For turbulent flow,u u

umax

*

�= 5.75 log

Ry10

��

��

5 4.4

u *

�= 5.75 log

0.10.0410

��

��

u * = 0.2622 m/sec

Shear velocity is,

u * =��0

0.2622 =�0

1000

�0 = 68.7593 N m 2 … Ans.

Example 8.12

Solution :

Given data : f = 0.04, u avg = 0.6 m/sec, Radial distance = 0.5 r

r = Pipe radius y = r 0.5 r� = 0.5 r

To find : i) u ii) umax

Step 1 : Calculate the local velocity at a radial distance of 0.5 r


Chapter - 8

TURBULENT FLOW

Unit - IV

For rough pipe,1f

= 2 logRk

1.7410��

��

�

10.04

= 2 logRk

1.7410��

��

�

logRk10

��

��

= 1.63

But,u

uavg

*= 5.75 log R

k10��

��

� 475.

0.6u *

= �5.75 1.63 4.75� �

u * = 0.0424 m/sec

For smooth and rough pipe,u u

uavg

*

�= 5.75 log

yR

3.7510��

��

�

u 0.60.0424

�= 5.75 log

0.5rr

3.7510� ��

��

� … (R = r)

u = 0.6857 m/sec … Ans.Step 2 : Calculate the velocity at the centre of pipe

At the centre of pipe, y =D2 = r and u = u max

For smooth and rough pipe,

u u

uavg

*

�= 5.75 log

yR

3.7510��

��

�

u 0.6

0.0424max �

= 5.75 log rr10

��

��

� 375.

umax = 0.759 m/sec … Ans.

Example 8.13

Solution :

Given data : D = 0.5 m, R =0.52 = 0.25 m, umax = 3.5 m/sec.

u = 3 m/sec, y = 0.25 � 0.15 = 0.1 m

To find : i) Q ii) f iii) k

Step 1 : Calculate the discharge through the pipe

For turbulent flow,


Turbulent Flow

u uu

max

*

�= 5.75 log

Ry10

��

��

3.5 3u *

�= 5.75 log

0.250.110

��

��

u * = 0.2185 m/sec

For smooth and rough pipe,u u

umax avg

*

�= 3.8

3.5 u

0.2185avg�

= 3.8

uavg = 2.6696 m/sec

Discharge through the pipe is,

Q = Area � Velocity =�4

D u2avg�

Q = ��4

6696� �0.5 22 . = 0.5241 m3 /sec … Ans.

Step 2 : Calculate the friction factor and height of roughness projection

We know that,

u * =f8

u avg�

0.2185 =f8

.6696� 2

f = 0.0535 … Ans.Height of roughness projection for turbulent flow,

1f

= 2 logRk

1.7410��

��

�

1

0.0535= 2 log

0.25k

1.7410��

��

�

log0.25k10

��

��

= 1.2898

0.25

k = 19.4908

k = 12.8265 � �10 3 m = 12.8265 mm … Ans.

Turbulent Flow ends …


Turbulent Flow


Example 9.30

Solution :

Given data : i) Pipes are connected in parallel :

d1 = 30 cm = 0.3 m, d2 = 40 cm = 0.4 m,

Q = 200 lps = 0.2 m sec3 , L L1 2� = 350 m, f f1 2� = 0.02

ii) Pipes are connected in series :

L = 600 m, Q = 0.2 m sec3 , f f1 2� = 0.02

To find : i) Q1 and Q2 ii) H

Step - 1 : Calculate the discharge through the each pipe when pipes are connected inparallelRefer Fig. 9.1 (a)

For pipes in parallel,

Q = Q Q1 2� = 0.2 … (i)

and h f1 = h f2

f L Q

12.1d1 1 1

2

15

=f L Q

12.1d2 2 2

2

25

But, f1 = f2 and L1 = L2

�Q

(0.3)12

5=

Q

(0.4)22

5

� Q12 = 0.2373 Q2

2

� Q1 = 0.4871 Q2Substituting this value in equation (i),

� 0.2 = 0.4871 Q Q2 2�

� 0.2 = 1.4871 Q2

� Q2 = 0.13445 m sec3 = 134.48 lps … Ans.


Q1

Q2

QQ

φ 0.3 m

φ 0.4 mL = 350 m

Fig. 9.1 (a)

Chapter - 9

FLOW THROUGH PIPES

Unit - V

and Q1 = 0.4871 Q2 = 0.4871 � 0.1345

� Q1 = 0.0655 m sec3 = 65.5 lps … Ans.

Step - 2 : Calculate the water level difference between the tanks

Refer Fig. 9.1 (b).

For pipes in series,

Q1 = Q2 = Q = 0.2 m sec3

Total head is given by,

H = h hf1 f2�

� H =f L Q

12.1d1 1 1

2

15

+f L Q

12.1d2 2 2

2

25

� H =0.02 350 (0.2)

12.1 (0.3)

0.02 350 (0.2)

12.1 (0.

2

5

2� �

��

� �

� 4)5

� H = 9.5228 + 2.2598

� H = 11.783 m … Ans.

Example 9.31

Solution :

Given data : H = 18 m, d = 100 mm = 0.1 m, L = 300 m, hc = 4 m,

L1 = 130 m, f = 0.02.

To find : i) Discharge Q ii) Pressure at summit (pc)

Step - 1 : Calculate the discharge through the siphon

If minor losses are neglected, then the loss of head is,

h f = H =f LQ

12.1 d

2

5

� 18 =0.02 300 Q

12.1 (0.1)

2

5� �

�

� Q = 0.019 m sec3 … Ans.


Flow Through Pipes

φ 0.2 m

H

φ 0.3 m

250 m 250 m

Fig. 9.1 (b)

Step - 2 : Calculate the pressure at the summit

Pressure at the summit is,

pc�

= � � �

��

�

��

�

��h

V2g

1+fLdc

21

But, Q = AV � V =QA =

0.019

(0.1) 2�4

�= 2.4191 m/sec

�pc�

= � ��

��

��

�

��

�

��4 (2.4191)

2 9.811 0.02 130

0.1

2

�pc�

= – 12.0533 m of water … Ans.

or pc = � �12.0533 9810 = – 118.243 10 Pa (Vacuum)3� … Ans.

Example 9.32

Solution :

Given data : d = 15 cm = 0.15 m, L = 4 km = 4 103� m, H = 26 m

L1 = 2 km = 2 10 m3� , f = 0.025

To find : Increases in discharge.

Step - 1 : Calculate the discharge through the single pipeline


Flow Through Pipes

c

Inlet

leg Outlet leg

A

BH

hc

ZAZB

Zc

Reservoir

Reservoir

Summit

Datum

Fig. 9.2

Refer Fig. 9.3 (a).

Discharge through the single pipe line

is,

H = h f =f LQ

12.1 d

2

5

� 26 =0.025 4 10 Q

12.1 (0.15)

3 2

5� � �

�

� Q = 0.015 m sec3

Step - 2 : Calculate the increase in discharge due to installation of new pipe

Refer Fig. 9.3 (b).

As pipes 1 and 2 are in parallel,

h f1 = h f2

�f L Q

12.1 d1 1

2

15

=f L Q

12.1 d2 2

2

25

But, d1 = d2 , L1 = L2

� Q1 = Q2Total discharge is,

Q3 = Q Q1 2� = 2 Q1For pipes in series (pipes 1 and 3),

H = h hf1 f3� =f L Q

12.1 d1 1

2

15

+f L Q

12.1 d3 3

2

35

� 26 =0.025 2 10 Q

12.1 (0.15)0.025 2 10 (2Q )

12.1

312

5

31

2� � �

��

� � �

� (0.15)5

� 26 = 54.4162 10 Q + 217.6648 10 Q312 3

12� � = 272.0810 10 Q3

12�

� Q1 = 9.7754 10 3� � m sec3

� Q3 = 2 Q1 = 2 9.7754 10 = 0.019553� � � m sec3

Increase in discharge is

Q Q3 � = 0.01955 – 0.015

� Q Q3 � = 4.55 � �10 3 m sec3 … Ans.


Flow Through Pipes

3

1

2

2 km2 km

26 m

φ 15 cm

Fig. 9.3 (b)

φ 15 cm

4 km

26 m

Fig. 9.3 (a)

Example 9.33

Solution :

Given data : d1 = 17 cm = 0.17 m, L1 = 8 m, d2 = 33.5 cm = 0.335 m,

L2 = 18 m, H = 8 m, Coefficient of friction = 0.04,

To find : Discharge (Q) and draw EGL

Step - 1 : Calculate the discharge through the pipe line

Refer Fig. 9.4 (a).

Friction factor,

f = 4 � Coefficient of friction

� f = 4 0.04� = 0.16

As per continuity equation,

Q = A V1 1 = A V2 2

� V1 =AA

V2

12

� V1 =d

dV2

2

12 2 =

(0.335)

(0.17)V

2

2 2� = 3.88 V2

Total head loss through the pipe is,

� H =Loss of headat entrance

Loss of headdue to f

�

��

�

��

rictionLoss of head due

to sudden expansion�

��

�

��

1

�

��

�

��

��

��

�

��

Loss of headdue to friction

Loss of headat2 exit

�

��

�

��

� H = h h h h hen f1 e f2 ex� � � �

� H = 0.5V2g

f L V2gd

(V V )2g

f L V2gd

V2g

12

1 12

1

1 22

2 22

2

22

� ��

� �

� 8 =0.5(3.88 V )

2 9.810.16 (3.88 V )

2 9.81 0.17(2

22

2

��

� ��

�8 3.88 V V )

2 9.812 2

2��

+0.16 16 V2 9.81 0.335

V2 9.81

22

22� �

� ��

�

� 8 = 0.3836 V22 + 5.77 V2

2 + 0.4227 V22 + 0.3895 V2

2 + 0.051 V22

� 8 = 7.0168 V22

� V2 = 1.0677 m/sec

and V1 = 3.88 V2 = 3.88 1.0677� = 4.1426 m/sec


Flow Through Pipes

8 m18 m

d = 0.17 m1

d = 0.335 m2

8 m

1

2

Fig. 9.4 (a)

Discharge through the pipe line is,

Q = A V2 2 =�4

d V22

2� =�4

� �(0.335) 1.06772

� Q = 0.0941 m sec3 … Ans.

� h en = 0.3836 V22 = 0.3836 (1.0677)2� = 0.4372 m

h f1 = 5.17 V22 = 5.17 (1.0677)2� = 5.8937 m

h e = 0.4227 V22 = 0.4227 (1.0677)2� = 0.4818 m

h f2 = 0.3895 V22 = 0.3895 (1.0677)2� = 0.4440 m

h ex = 0.051 V22 = 0.051 (1.0677)2� = 0.05813 m

Refer Fig. 9.4 (b) for Energy Gradient Line (EGL).

Example 9.34

Solution :

Given data : L = 1800 m, d = 350 mm = 0.35 m, H = 18 m, f = 0.02

To find : Discharge and length of new

pipe.

Step - 1 : Calculate the discharge

through the single pipe

Refer Fig. 9.5 (a).

Neglecting minor losses,

H = hf L Q

12.1 df

2

5�


Flow Through Pipes

A

0.05813 m

0.4096 m

EGLHGL

5.8937 m

0.4818 mEGL

0.05813 m

0.4440 mHGL

H = 8 m

BC

Fig. 9.4 (b)

1800 m

Ø 0.35 m

18 m

Fig. 9.5 (a)

� 18 =0.02 1800 Q

12.1 (0.35)

2

5� �

�

� Q = 0.1783 m sec3

Step - 2 : Calculate the length of new pipe

Refer Fig. 9.5 (b).

Increase in discharge = 15 %

�0.15 =Q Q

Q1 �

=Q 0.1783

0.17831 �

� Q1 = 0.205 m sec3

For parallel pipes,

h f2 = h f3

�f L Q

12.1 d2 2

2

25

=f L Q

12.1 d3 3

2

35

� L Q2 22 = L Q3 3

2

Also, Q1 = Q Q2 3� … (i)

Total loss of head is,

H = h hf L Q

12.1 d

f L Q

12.1 df1 f21 1

2

15

2 22

25

� � �

� 18 =0.02 900 (0.205)

12.1 (0.35)

0.022

5

� �

��

900 Q

12.1 (0.35)22

5

� �

�

� Q2 = 0.1467 m sec3

But, Q1 = Q Q2 3� � 0.205 = 0.1467 + Q3

� Q3 = 0.0583 m sec3

Substituting these values in equation (i),

900 (0.1467)2� = L (0.0583)32�

� L3 = 5698.5665 m = 5.6985 kM ... Ans.

Flow Through Pipes ends …


Flow Through Pipes

3

1

2Ø 0.35 m

900 m900 m

16 m

L3

Fig. 9.5 (b)


Example 10.26

Solution : The functional relationship between the dependent and indepenedent variables

is given by,

T = f(D, N, , )� �, V

or f (T, D, N, , )� �, V = 0 or constant

Dimensions of different variables are,

T D N � � V

[ML T ]2 2� [L] [T ]1� [ML ]3� [ML T ]1� �1 [LT– 1

]

� Total number of variables n = 6

� Number of primary dimensions m = 3

� Number of � terms = n – m = 6 – 3 = 3 ( , , )� � �1 2 3� f ( , , )� � �1 2 3 = 0 or constant … (i)

� As m = 3, the repeating variables are also 3.

� As T is dependent variable, it cannot be taken as repeating variable. The repeating

variables are selected such that,

i) Geometric property (D)

ii) Flow property (N)

iii) Fluid property (�)

� As per Buckingham � theorem the � terms can be written as,

� 1 = D N Tx1 y1 z1� … (ii)

� 2 = Dx2 y2 z2� � � … (iii)

� 3 = Dx3 y 3 z 3� � V … (iv)

1) Consider equation (ii) and substitute the dimensions,

[M L T ]0 0 0 = [L] [T ] [ML ] [ML T ]x1 y1 3 z1 2� � �1 2

Equating the powers of MLT on both sides,


Chapter - 10

DIMENSIONAL ANALYSIS

Unit - V

For M : 0 = z1 �1 � z1 = – 1

For L : 0 = x 3z 22 1� � � x1 = – 5

For T : 0 = � �y 21 � y1 = – 2

Substituting these values in equation (ii),

� � 1 = D N 1� � �5 2 � T

� �1 =

�� 2D5

2) Consider equation (iii) and substitute the dimensions,

[M L T ]0 0 0 = [L] [T ] [ML ] [ML T ]x2 y2 3 z2 1� � � �1 1


For M : 0 = z2 �1 � z2 = – 1

For L : 0 = x 3z2 2� �1 � x2 = – 2

For T : 0 = � �y 12 � y2 = – 1

Substitute these values in equation (iii),

� � 2 = D N2 1� � �1 � �

� � 2 =�

�ND2

3) Consider equation (iv) and substitute the dimensions,

[M L T ]0 0 0 = [L] [T ] [ML ] [LT ]x3 y 3 3 z 3 1� � �1


For M : 0 = z 3 � z 3 = 0

For L : 0 = x 3z3 3� �1 � x 3 = – 1

For T : 0 = � �y 13 � y3 = – 1

Substitute these values in equation (iv),

� � 3 = D N1� �1 0� V

� � 3 =V

DN

Now, put the values of �, � 2 in equation (i),

� fT

N D,

ND,

VDN2 5 2�

�

�

�

�

�

�� = 0 or constant

orT

N D2 5�= �

�

� ND,

VDN2

�

�

�

��


Dimensional Analysis

� T = �N D2 5 ��

�ND DN

V

2,

�

�

�

�� …Ans.

Example 10.27

Solution : The functional relationship between the dependent and indepenedent variables

is given by,

� = f( , V, D, )� �, K

or f( , , V, D, )� � �, K = 0 or constant

Dimensions of different variables are,

� � V D � K

[ML T ]2� �1 [ML ]3� [LT– 1

] [L] [ML T ]1� �1 [L]

� Total number of variables n = 6

� Number of primary dimensions m = 3

� Number of � terms = n – m = 6 – 3 = 3 ( , , )� � �1 2 3� f ( , , )� � �1 2 3 = 0 or constant … (i)

� As the number of primary dimensions m = 3, the repeating variables are also 3.

� As � is a dependent variable, it cannot be taken as repeating variable. The

repeating variables are selected such that,

i) Geometric property (D)

ii) Flow property (V)

iii) Fluid property (�)

� As per Buckingham � theorem the � terms can be written as,

� 1 = D Vx1 y1 z1� � … (ii)

� 2 = D Vx2 y2 z2� � … (iii)

� 3 = D Vx3 y 3 z 3� K … (iv)

1) Consider equation (ii) and substitute the dimensions,

[M L T ]0 0 0 = [L] [LT ] [ML ] [ML T ]x1 y1 3 z1 1� � � �1 2


For M : 0 = z1 �1 � z1 = – 1

For L : 0 = x 3z1 1� � �y1 1 � x1+ y1 = – 2

For T : 0 = � �y 21 � y1 = – 2

and x1 = 0

Substituting the values of x1, y1 and z1 in equation (ii),

� � 1 = D V 10 2� ��



� �1 =�

�V 2

2) Consider equation (iii) and substitute the dimensions,

[M L T ]0 0 0 = [L] [LT ] [ML ] [ML T ]x2 y2 3 z2 1� � � �1 1


For M : 0 = z2 �1 � z2 = – 1

For L : 0 = x y2 2� � �3 12z � x y2 2� = – 2

For T : 0 = � �y 12 � y2 = – 1

and x2 = – 1

Substituting the values of x , y , z2 2 2 equation (iii),

� � 2 = D V1 1� � �1 � �

� � 2 =�

�VD

3) Consider equation (iv) and substitute the dimensions,

[M L T ]0 0 0 = [L] [LT ] [ML ] [L]x3 y 3 3 z 3� �1

In this case, as the dimensions of D and K are same,

We can write by observation,

� � 3 =KD

Now, put the values of �, � 2 , � 3 in equation (i),

� f�

�

��V

,VD

,KD2

�

�

�

�� = 0 or constant

or�

�V2= �

��VD K

D,�

� ��

…Ans.

Example 10.28

Solution :

Given data : Scale ratioLLm

p=

110 , Hm = 0.14 m, Qm = 0.1 m

3/sec

To find : i) HP ii) QP

As it is a spillway model, Froude law must be applicable.

1. Lr =LL

m

p=

110

� LP = 10 Lm

� HP = 10 Hm = 10 � 0.14 = 1.4 m …Ans.



2. Qr = Ar � Vr = L Lr2

r� = Lr5/2

� Qr =110

5 2��

��

/=

QQ

m

p=

0.1Qp

� QP = 31.6227 m3/sec …Ans.

Example 10.29

Solution :

Given data : dp = 2.5 cm = 25 mm, dm = 6 mm, Vm = 3 m/s, � p = 825 kg/m3

�m = 995.7 kg/m3, �p = 0.02 Pa-S, �m = 0.801 Pa-S

To find : Vp

For flow through pipe, Reynolds law is applicable.

As�

�VD

p

��

��

=�

�VD

m

��

��

�825 V 25

0.02� ��

� ��

=995.7 3 6

0.801� ��

� ��

� VP = 0.0216 m/s …Ans.

Example 10.30

Solution :

Given data : Scale ratioLLm

p=

116 ,

HP = 7.2 m, Lp = 150 m, Qp = 2150 m3/s, hp = 4 m

As it is a spillway model, Froude law must be applicable.

1. Lr =LL

m

p=

116

� Lm =L

16p

� Lm =15016

= 9.375 m …Ans.

2. Qr = Lr5/2

� Qr = (1/16)5/2 =QQ

m

P=

Q2150

m

� Qm = 2.0996 m3/sec …Ans.

3.hh

m

p= Lr =

116

�h4m =

116

� hm =416

= 0.25 m …Ans



4. Lr =116

=LL

m

p� Lp = 16 Lm

� Hp = 16 Hm � 7.2 = 16 Hm

� Hm = 0.45 m …Ans

Dimensional Analysis ends …




Example 11.21

Solution : Momentum thickenss is,

� =uU

uU

dy10

��

��

�=

uU

u

Udy

2

2�

�

��

�

�

0

�

=32

12

32

12

2

0

y y y y3

3

3

3� � � �

��

�

��

�

� � �

�

��

�

�

�

��

�

�

��

=3y y 9y 3y y y3

3

2

2

3

3

6

62 42

0� ��

��

�

��

�

��

� � � � ��

��

�

��

=3y y 9y 3y y3

3

2

2

4 6

62 4 40

� ��

��

�

��

�

��

=3 y

21 y

49 y 32

3

4

2

3

2 4� ��

�

��

�

� �

�

��

�

� �

�

��

�

� �

4

5

4

7y5

1 y7

�

��

�

� �

�

��

�

�

�

��

�

��

�

0

=34 8

9

12 10 28

2 4

3

3

2

5

4

7

6��

�

�

�

�

��

�

�

�� =

34 8

912 10

128

��

��

� ��

��

=39280

…Ans.

Example 11.22

Solution :

i)uU

= 22 3y y

� ��

��

� ��

��

� u = 22 3

Uy

Uy2 3

� ��


Chapter - 11

BOUNDARY LAYER THEORY

Unit - VI

Differentiating above equation with respect to y,

dudy

=4Uy 3Uy2

� �2 3�

at y = 0

�dudy y 0

��

��

= 0

Asdudy y 0

��

��

has zero value, the flow is just on the verge of separartion. …Ans.

ii)Uu

=y y� �

��

��

� ��

��

22

� u =Uy 2Uy2

� �2�

Differentiating above equation with respect to y,

dudy

=2Uy 2U

� �2�

at y = 0

�dudy y 0

��

��

=� 2U

�

Asdudy y 0

��

��

has negative value, the flow is detached or separated. …Ans.

Example 11.23

Solution :

Given data : L = 4 m, d = 1.5 m, U = 14.4 km/hr = 4 m/sec.

� air = 1.2 kg/m 3 , � = 1.8 � �10 4 poise = 1.8 � �10 5 Pa-s

To find : Drag force on both sides of plate.

Step - 1 : Calculate the drag force when boundary layer is laminar

Reynold's number over the complete plate is,

ReL =�

�UL

=1.2 4

1.8 10 5� �

� �4

= 1.067 106�

Average coefficient of drag for laminar boundary layer is,

CDa =1.328ReL

=1.328

1.067 106�= 1.2858 � �10 3


Boundary Layer Theory

Drag force on both sides of plate is,

FD =12

2� ��

��

�C AUDa2� … (Both sides)

� FD =12

1.2858 10 1.2 4 1.5 (4) 23 2� � � � � ��

��

��

� FD = 0.1481 N … Ans.

Step - 2 : Calculate the drag force when boundary layer is turbulent

Average coefficient of drag for turbulent boundary layer is,

CDa =0.074

ReL( ) 1 5=

0.074

( . )1 067 106 1 5�= 4.6089 � �10 3

Drag force on both sides of plate is,

FD =12

2� ��

��


� FD =12

4.6089 10 1.23 2� � � � � ��

��

�� 4 15 4 2. ( )

� FD = 0.5309 N … Ans.

Example 11.24Given data : L = 3 m, b = 1 m, � = 1.75 10 Pa - s5� �

Re = 5 105� , U = 2 m/sec., � = 1.2 kg/m 3

To find : Drag force FDStep - 1 : Calculate the drag force on both sides of the plate.

CDa =0.074

(Re )

1700Re

L1 5 L

�

But, ReL =�

�UL

=1.2 3

1.75 10 5� �

� �2

= 411.43 10 3�

� CDa =0.074

(411.43 10 )

1700

(411.43 10 )3 1 5 3��

�

� CDa = 1.4446 � �10 3

Drag force on both sides of the plate is,

FD =12

2� ��

��


� FD =12

1.4446 10 1.23 2� � � � � ��

��

�� 3 1 2 2

� FD = 0.0208 N … Ans.

Boundary Layer Theory ends …


Boundary Layer Theory


Example 12.23

Solution :

Given data : A = 0.35 m2 , U = 45 km/hr = 12.5 m/sec, W = 1.7 N, � = 18º, T = 4 N,

� air = 1.15 kg/m 3

To find : CL and CD.

Step - 1 : Calculate the coefficients of lift and drag

From Fig. 12.1,

Fy� = 0 = F W T cos (18)L � �

� FL = W T cos (18)�

= C A U2L

2� �

�

�1.7 4 cos (18)� � = C 0.35 1.15(12.5)

2L

2� � �

� CL = 0.1750 … Ans.Now,

Fx� = 0 = F T sin (18)D �

� FD = T sin (18)

But, FD = C A U2D

2� �

�

� 4 sin (18)� = C 0.35

1.15 12.52D � �

� 2

� CD = 0.0393 … Ans.


FL

FDU

18º Kite

T = 4 NW = 1.7 N

Fig. 12.1

Chapter - 12

FORCES ON IMMERSED BODIES

Unit - VI

Example 12.24

Solution :

Given data : W = 38.62 kN = 38.62 � 10 3 N, U = 960 km/hr = 266.67 m/sec,

Span C = 16 m, A = 33 m2 , C 0.03D � , � air = 1.22 kg/m 3

To find : i) CL ii) P iii) �

Step - 1 : Calculate the coefficient of lift

We know that,

Lift force = Weight of plane

i.e. FL = W

� C AU2L

2� �

�= W

� C 31.22 (266.67)

2L

2� �

�3 = 38.62 � 10 3

� CL = 0.0269 … Ans.

Step - 2 : Calculate the required power and boundary layer circulation

Power given is,

P = F UD � = C AU2

UD

2� � �

�

� P = C AU2D

3� �

�= 0.03 33 1.22 (266.67)

23

� ��

� P = 11.4521 � 106 W = 11.4521 103� kW … Ans.

Theoretical boundary layer circulation is,

� = �C U sin�

As plane is flying in horizontal direction, � = 0º.

� � = � � �15 266.67 sin 0 = 0 … Ans.

Example 12.25

Solution :

Given data : D = 6 cm = 0.06 m, � S = 22.323 kN/m 3 = 22.323 � 10 3 N/m 3 ,

U = 22 m/sec

To find : Tension T and its inclination �.

Step - 1 : Calculate the tension in the string and its inclination

Drag force is,


Forces on Immersed Bodies

� FD = C AU2D

2� �

�= C D

U2D

22

� � �4

� FD = 0.64

0.061.25 20

22

2� � �

�

� FD = 0.4241 N

Weight of sphere (W) = � S � Volume of sphere

� W = 22.323 1043

0.062

33

� � � � ��

��

� W = 2.5246 N

From Fig. 12.2,

Weight W = T cos �

But, tan � =FWD =

0.42412.5246

� � = 9.5359º … Ans.

� 2.5246 = T cos (9.5359)�

� T = 2.5599 N … Ans.

Example 12.26

Solution :

Given data : L = b = 3 m, � air = 1.4 kg/m 3 , U = 60 kmph = 16.67 m/sec,

CD = 0.2, CL = 0.8

To find : i) FL ii) FD iii) FR iv) � v) P

Step - 1 : Calculate the lift force and drag force on the plate

Lift force is, FL = C AU2L

2� �

�=

0.8 3 3

1.4 16.672

2� � �

�

� FL = 1400.56 N … Ans.Drag force is,

FD = C AU2D

2� �

�=

0.2 3 3

1.4 16.672

2� � �

�

� FD = 350.14 N … Ans.Step - 2 : Calculate the resultant force and power required to keep the plate in

motion

Resultant force is,

FR = F FD2

L2� = 350.14 1400.562 2�



Airflow FD

Sphereθ

W

θ

Fig. 12.2

� FR = 1443.6 N … Ans.Direction of resultant force is,

tan � =FFL

D� tan � =

1400.56350.14

� � = 75.963º … Ans.

Power required to keep the plate in motion,

P = F UD � = 350.14 16.67�

� P = 5836.83 W … Ans.

Forces on Immersed Bodies ends …



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