fluid dynamics

MAEER’S MIT, Pune Civil Department

[email protected] 1

Experiment No. Date:

Study of Viscosity with Redwood Viscometer

Introduction :

Redwood viscometer is based on the principle of laminar flow through capillary tube of

standard dimension under falling head. The viscometer consists of a vertical cylinder with an

orifice at the center of the base of inner cylinder. The cylinder is surrounded by a water bath,

which can maintain temperature of the liquid to be tested at required temperature. The water bath

is heated by electric heater. The cylinder, which is filled up to fixed height with liquid whose

viscosity is to be determined is heated by water bath to the desired temperature. Then orifice is

opened and the time required to pass 50 cc of oil is noted. With this arrangement variation of

viscosity with temperature can be studied.

Object:

To study variation of viscosity of given oil with temperature.

Theory :

In case of Redwood Viscometer, the kinematic viscosity (ν) of liquid and the time (t)

required to pass 50cc of liquid are correlated by the expression

ν = 0.0026t – 1.175/t

Where,

ν - Kinematic Viscosity in stokes

t - time in seconds to collect 50 cc of oil.

Equipment:

Redwood viscometer with accessories, Measuring Flask, Thermometer,

Stopwatch etc.

Procedure :

1. Level the instrument with the help of circular bubble and leveling foot screws.

2. Fill the water bath.

3. Close the orifice with the ball valve and fill the cylinder up to the index mark with oil.

4. Record steady temperature of oil.

5. By lifting the ball valve, collect 50cc of the liquid in the measuring flask and measure

the time required for the same.


[email protected] 2

Thermometer

Inner Cylinder containing oil

Outer Cylinder containing water

Outlet valve

Ball Valve

Capillary Tube

Stirrer Fins

Stirrer wall

Index Mark

Heating Coil

Conical flask to measure 50 cc of oil

Levelling screws

Redwood Viscometerÿþýüû ùø÷ö

�� ûþ��


[email protected] 4

6. Repeat the procedure for different temperatures by heating oil with water bath.

Experimental data :

1. Diameter of cylinder = ............... mm.

2. Height of cylinder = ............... mm

3. Diameter of orifice = ............... mm

4. Length of orifice = ............... mm

Observation Table:

Sr. No. Temperature

‘ÿ’

(0C)

Time to collect 50cc of oil

‘t’

(s.)

Kinematic viscosity

‘νννν’

(stokes)

1

2

3

4

5

6

7

8

9

10

Sample calculations:


[email protected] 5

2 0 3 0 4 0 5 0 6 0 7 0 8 0 9 0 1 0 00 .0 0

0 .0 5

0 .1 0

0 .1 5

0 .2 0

0 .2 5

0 .3 0

0 .3 5

0 .4 0

K in e m a t ic v is c o s i ty ( υ ) v s . te m p e r a tu r e ( θ )

R e g io n IIIR e g io n IIR e g io n I

R e g io n I:- V e r y h ig h v is c o s ity , a ls o l a r g e v a r ia t io n inv is c o s ity w ith s m a l l c h a n g e in te m p e r a tu r eh e n c e u n s u ita b le

R e g io n II:- M o d e r a te V is c o s ity , a ls o m o d e r a te v a r ia t io nin v is c o s ity w ith c h a n g e in te m p e r a tu r eh e n c e s u ita b l e te m p e r a tu r e r a n g e f o r th e o ilto b e u s e d a s a l u b r ic a n t .

R e g io n III: - O il p o s s e s s e s v e r y l e s s v is c o s ityh e n c e u n s u ita b le

S c a leX -a x is : 1 c m = 5 0 CY -a x is : 1 c m = 0 .0 2 5 s to k e s

Kin

emat

icvi

scos

ity

υ(s

toke

s)

T e m p e r a tu r eθ ( 0 C )


[email protected] 6

Graph:

Graph of kinematic viscosity ÿ (stokes) vs. temperature � ºC

Conclusion:

1. Kinematic viscosity of given oil at 27 ºC= __________

2. Kinematic viscosity ______________ with increase in temperature.

3. Rate of decrease of kinematic viscosity ____________ with increase in temperature.


[email protected] 7


Study of Flow measuring device - Venturimeter

Introduction

Venturimeter is a device used for measurement of discharge in a pipeline and is

based on Bernoulli’s theorem. The instrument consists of a short pipe which contracts up to a

section called as throat and then enlarges up to a diameter at outlet as shown in Fig. The conical

portions joining the inlets and the throat and the outlet are called as converging cone and

diverging cone respectively.

Object

• To determine the coefficient of discharge (k) of Venturimeter

• To calibrate the Venturimeter.

Theory

By contracting the passage of flow at the throat, the velocity of flow and hence the

velocity head is increased. This increase in the velocity head causes change in pressure head.

The Pressure difference thus created is measured generally by a ‘U’ tube manometer (differential)

and the discharge through the pipe is calculated by the formula.

Qth. = hC

Qa = k Qth

Where,

Qth =Theoretical discharge through Venturimeter.

Qa = Actual discharge through Venturimeter.

k = Co-efficient of discharge of Venturimeter where,

h = Difference of head in terms of water column between inlet and throat.

C = Constant of Venturimeter =2

22

1

21 2

aa

gaa

−

Where,

a1 = Area of inlet which can be found out from inlet diameter d1.

= 214

dπ

a2 = Area of throat which can be found out from throat diameter d2.

= 224

dπ


[email protected] 9

Inverted U tube Air-Water Differential Manometer

Air Release Valve

Manometric Fluid - Air

Pipe Fluid - Water

Annular Ring

Holes of annular ring tomeasure average pressure

Inlet Converging Cone Diverging Cone Outlet

Rubber Tubes connecting pipe to manometer

Throat

Qa

@20°

Venturimeter

@6°d1=5cm

h1=p1/ν

h2=p2/ν

2.5LL

Qa

x=h1-h2

d2=1.6 cm


[email protected] 10

Actually, the coefficient of discharge k is never unity and hence it is determined

experimentally.

th

a

Q

Qk =

The above formula can also be written

Qa = Mhn

The constants M and n can be found out by plotting the graph of log Qa vs. log h.

Apparatus

1. Venturimeter

2. A Flow table with self circulating system

3. Measuring tank

4. Stopwatch

5. Differential manometer

Experimental Procedure

1. Set up the Venturimeter on the flow table and connect the inlet hose pipe.

2. The inverted U tube differential manometer is then connected to the respective pressure

tapping, making sure that no air bubble is entrapped in the tube.

3. The flow of water is then adjusted for required pressure head difference and the pressure

difference is noted.

4. The flow is then actually measured by collecting it in a measuring tank for known interval

of time.

5. The procedure is repeated for different values of pressure head difference by changing

the discharge.

Experimental Observation

1. Inlet diameter of Venturimeter = d1 = 2.6 cm.

2. Throat diameter of Venturimeter = d2 = 1.6 cm.

3. Dimensions of the measuring tank A = 50 cm. X 25 cm



Observation Table

Manometric readings Actual discharge

measurementSr. No.

h1

cm

h2

cm

h

cm

Qth.

= hC

cm3/s I.R.

cm.

F.R.

cm.

Time

Sec.

Qa.

cm3/s

K

=th

a

Q

Q aQ10log h10log

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

Average k =



1 .4 8 1 .5 0 1 .5 2 1 .5 4 1 .5 6 1 .5 8 1 .6 0 1 .6 2 1 .6 4 1 .6 6 1 .6 82 .6 4

2 .6 6

2 .6 8

2 .7 0

2 .7 2

2 .7 4

2 .7 6

2 .7 8

2 .8 0

2 .8 2

2 .8 4A lw a y s d ra w to s a m e s c a le

kg ra ph ic a lly

= M /C=

M = 1 0 [log1 0

Q a -n lo g1 0

h ]

=

n = s lo pe o f line= (y

2-y

1)/(x

2-x

1)

=

lo g1 0

Q a v s . lo g1 0

h

S c a leX -a x is : 1 cm = 0 .0 2 un itsY -a ix s : 1 c m = 0 .0 2 un its

log 10

Qa

lo g1 0

h



0 1 0 2 0 3 0 4 0 5 0 6 0 7 00

1 0 0

2 0 0

3 0 0

4 0 0

5 0 0

6 0 0

7 0 0

8 0 0

C a lib ra tio n c u rv e ( Qa

v s . h )

F or head h = ( ) cm ,actua l d ischarge , Q

a= ( ) cm 3/s .

S ca leX-axis: 1cm = 5 cmY -ax is: 1cm = 50 cm 3/s .

n =M =Q

a=( )h ( ) cm 3/s.

Act

uald

isch

arge

Qa

(cm

3 /s)

D iffe re n tia l h e a dh (c m )



6 4 6 6 6 8 7 0 7 2 7 4 7 6 7 8 8 0 8 2 8 4 8 6 8 8 9 0 9 2 9 40 .0

0 .2

0 .4

0 .6

0 .8

1 .0

1 .2

1 .4

C o e f f ic ie n t o f d is c h a rg e (k ) v s . R e y n o ld 's n u m b e r (R e ) a t th ro a t

S c a leX -a x is : 1 c m = 2 0 0 0 u n itsY -a x is : 1 c m = 0 .1 u n its

( X 1 0 3 )

Coe

ffici

ento

fdis

char

ge

k

R e y n o ld 's n u m b e r a t th ro a t(R e )



Sample calculations

1. C =2

22

1

21 2

aa

gaa

−= _____________________________________ .sec/5.2cm

2. Qth. = hC =_____________________________________ .sec/cc

3. Qa. =[ ]

time

RIRFA .... −= _____________________________________ .sec/cc

4. k =th

a

Q

Q= _________________

Graphs

1. Plot graph of log Qa Vs log h to determine M and n.

2. Plot Calibration curve: Plot Qa Vs h

3. Plot graph of k vs. Re at throat.

Conclusions

1. The coefficient of discharge of the Venturimeter is k=…………… from calculation.

k=…………… from graph

2. The law of the Venturimeter is Qa = Mhn = ……………………………………cm3/s.

=…………………………………….m3/s.

3. Practical utility of calibration curve,

For ………cm pressure head difference across the Venturimeter, the discharge through

the Venturimeter is …………cm3/s.




Study of Flow measuring device - Orificemeter

Introduction

Orificemeter is yet another device used for measurement of discharge through a pipe

based on Bernoulli's theorem. It is different from the venturimeter in the sense that it provides sudden

change in a flow conditions instead of smooth transition provided by the venturimeter. As the liquid

passes through the orificemeter, lot of eddies are formed and there is a loss of energy due to which,

the measured value of discharge (Qa), is far less than the theoretical discharge. ( Qth )

Object

• To determine the coefficient of discharge (k) of orificemeter

• To calibrate the Orificemeter.

Theory

Orificemeter consists of a flat circular plate having a sharp edged hole called an orifice. The

plate is fitted in such a way that the orifice is concentric with the pipe. The diameter of the orifice is

about half the diameter of the pipe. The suitable pressure tappings one on each side of the orifice are

provided for measurement of pressure difference across the orifice.

The discharge through an orifice is

Qth = Ch1/2

Qa = k Qth

Where,

Qth =Theoretical discharge through 0rifice

Qa = Actual discharge through 0rifice

k = Co-efficient of discharge of 0rificemeter where,

h = Difference of head in terms of pipe fluid column across orifice

C = Constant of orificemeter = g2a0

Where,

a0 = Area of orifice =2

od4

∏where do = diameter of orifice opening

Actually, the coefficient of discharge k is never unity and hence it is determined

experimentally.



Orificemeter

h2=P2/?

h1=p1/?

doVena contracta

Orifice plate

x=h1-h2

Air ReliefValve

Rubber tube connecting manometer withpipe

Separation zone(eddies)

D Qa

Inverted U-tube Air-water differentialmanometer



th

a

Q

Qk =

The above formula can also be written

Qa = Mhn

The constants M and n can be found out by plotting the graph of log Qa vs. log h.

Apparatus

1. Orificemeter


3. Measuring tank

4. Stopwatch

5. Differential manometer


1. Set up the orificemeter on the flow table and connect the inlet hose pipe.

2. The inverted U tube differential manometer is then connected to the respective pressure

tapping, making sure that no air bubble is entrapped in the tube.

3. The flow of water is then adjusted for required pressure head difference and the pressure

difference is noted.

4. The flow is then actually measured by collecting it in a measuring tank for known interval

of time.

5. The procedure is repeated for different values of pressure head difference by changing

the discharge.


1. Diameter of orifice = d0 = 2.5 cm.

2. Diameter of pipe = d = 5 cm.

3. Dimensions of the measuring tank = 50 cm. X 25 cm.



Observation Table

Manometric readings Actual discharge

measurementSr. No.

h1

cm

h2

cm

h

cm

Qth.

= hC

cm3/s I.R.

cm.

F.R.

cm.

Time

Sec.

Qa.

cm3/s

K

=th

a

Q

Q aQ10log h10log

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

Average k =



1 .0 0 1 .0 5 1 .1 0 1 .1 5 1 .2 0 1 .2 5 1 .3 02 .5 5

2 .6 0

2 .6 5

2 .7 0

2 .7 5

2 .8 0

2 .8 5A lw ays draw to sam e sca le

kgraph ically

= M /C=

M = 10 [log10

Q a-n log10

h ]

=

n = slope of line= (y

2-y

1)/(x

2-x

1)

=

S ca leX-axis : 1 cm = 0.025 unitsY -axis : 1 cm = 0.025 units

lo g10

Q a v s . lo g10

hlo

g 10Q

a

lo g10

h



0 1 0 2 0 3 0 4 0 5 00

1 0 0

2 0 0

3 0 0

4 0 0

5 0 0

6 0 0

7 0 0

8 0 0

9 0 0

1 0 0 0

1 1 0 0

1 2 0 0

C a l ib r a t io n c u r v e ( Qa

v s . h )

F o r h e a d h = ( ) c m ,a c tu a l d is c h a rg e , Q

a= ( ) c m 3 /s .

S c a leX -a x is : 1 c m = 5 c mY -a x is : 1 c m = 1 0 0 c m 3 /s .

n =M =Q

a= ( )h ( ) c m 3 /s .

Act

uald

isch

arge

Qa

(cm

3 /s)

D i f fe r e n t ia l h e a dh ( c m )



Sample calculations

1. C = g2a0 =_____________________________________ .sec/5.2cm

2. Qth. = hC =_____________________________________ .sec/cc

3. Qa. =[ ]

time

RIRFA .... −= _____________________________________ .sec/cc

4. k =th

a

Q

Q= ___________

Graphs

1. Plot graph of log Qa vs log h to determine M and n.

2. Plot Calibration curve : Plot Qa vs h

Conclusions

1. The coefficient of discharge of the Orificemeter is k=…………… from calculation.

k=…………… from graph

2. The law of the Orificemeter is Qa = Mhn = ……………………………………cm3/s.

=…………………………………….m3/s.

3. Practical utility of calibration curve,

For ………cm pressure head difference across the Orificemeter, the discharge through the Orificemeter

is …………cm3/s.




Study of Flow Through Pipe Fittings

Introduction

In a pipe flow there are different types of losses. The losses due to change of section or obstruction or

change of direction of flow are called minor losses. The major loss is frictional loss of head which is significant

for longer pipes (L/d > 500). For short pipes, (L/d < 500), minor losses become significant.

Coupling

Coupling is a connection between two pipes of either same diameter or of different diameters. In the

present apparatus, the coupling is in the form of gradual expansion and gradual contraction.

Object

• To determine the loss of energy due to gradual expansion and to determine the coupling coefficient (k1)

for given area ratio.

• To determine the loss of energy due to gradual contraction and to determine the coupling coefficient (k2)

for given area ratio.

Theory

(A) The loss of energy due to gradual expansion

In gradually diverging coupling, the change in the cross sectional area causes the change in the

magnitude of velocity of fluid and large scale turbulance is generated due to formation of eddies. Some

portion of kinetic energy is utilized in this and it is to be considered as loss. The loss of energy due to gradual

expansion is given as

hLa. = k1 hLth.

Where,

hLa = Actual loss of energy.

k1 = Coefficient which depends on angle of divergence and area ratio

hLth. = Theoretical loss of energy.

= (V1-V2)2/2g

Where, V1 = Velocity of flow at inlet.

V2 = Velocity of flow at outlet.

(B) The loss of energy due to gradual contraction

In the gradually converging coupling, the pressure energy is converted into kinetic energy and flow gets

accelerated. Gradually accelerated flow has an inherent stability and since it is free from separation,

energy loss is very small. In converging coupling, Vena-contracta is formed in narrower pipe after which

the stream of fluid widens again to fill the pipe completely. In between vena-contracta and the wall of the

pipe, eddies are formed which cause considerable loss of energy. The loss of energy due to gradual

contraction is

hLa. = k2 hLth.

Where,

hLa = Actual loss of energy.



K2 = Coefficient which depends on area ratio

hLth. = Theoretical loss of energy.

= V22/2g

Apparatus

1. Coupling forming gradual expanding section and gradually contracting section


3. Measuring tank

4. Stopwatch

5. Manometer


1. Obtain the flow with maximum discharge through the pipe.

2. The manometer is then connected to the respective pressure tapping, making sure that no air

bubble is entrapped in the tube.

3. Take the manometric readings.

4. The flow is then actually measured by collecting it in a measuring tank for known interval of time.

5. The procedure is repeated for different values of pressure head difference by changing the discharge.


(A) Gradually expanding section

1. Upstream Diameter of pipe = d1 = 2.5 cm.

2. Downstream diameter of pipe = d2 = 5 cm.

(B) Gradually contracting section

1. Upstream Diameter of pipe d1 = 5 cm.

2. Downstream diameter of pipe d2 = 2.5 cm.

3. Dimensions of the measuring tank A = 50 cm. X 25 cm.



Observation Table :-

Gradually Expanding Section.

Sr.

No.

Manometric

readings

Actual discharge measurement

h1

cm

h2

cm

I.R

cm

F.R

cm

Time

sec.

Qa.

cm3/s

V1

cm/s

V2

cm/s

E1

cm

E2

cm

hLa

cm

hLth

cm

K1

cm

1

2

3

4

5

6

7

8

9

10

Average k 1 =

Observation Table :-Gradually Contracting Section.



Sr.

No.

Manometric

readings

Actual discharge measurement

h1

cm

h2

cm

I.R

cm

F.R

cm

Time

sec.

Qa.

cm3/s

V1

cm/s

V2

cm/s

E1

cm

E2

cm

hLa

cm

hLth

cm

K2

cm

1

2

3

4

5

6

7

8

9

10

Average k 2 =



Total Energy line (TEL) and Hydraulic grade line (HGL) in pipe fittings

Expansion Contraction

datum level

Z

h1= __P1

?

__V1

2g

2

E1

1 Z

h2=__P2

?

V2

2g

2

E2

2 Z

h1= __P1

?

V1

2g

2

E1

1 Z

h2=__P2

?

V2

2g

2

E2

2

____

__

Qa Qa

hla= E1 E2_ hla= E1 E2

_T.E.L. (Ideal fluid)

T.E.L. (Real fluid)

T.E.L. (Ideal fluid)

T.E.L. (Real fluid)

H.G.L.

H.G.L.

Eddies



Sample calculations

(A) Gradually expanding section

5.g2

)vv(h

221

Lth

−= =_____________________________________cm

6. =+= g2/vhE 2111 _____________________________________cm

7. =+= g2/vhE 2222 _____________________________________cm

8. =−= 21 EEhLa _____________________________________cm

9. ==Lth

La

h

hk1 _________

(B) Gradually contracting section

1.g2

vh

22

Lth = =_____________________________________cm

2. =+= gvhE 2/2111 _____________________________________cm

3. =+= gvhE 2/2222 _____________________________________cm

4. =−= 21 EEhLa _____________________________________cm

5. ==Lth

La

h

hk2 __________

Conclusions

3. Average Coefficient 1K for gradual expanding section is = ______

4. Average Coefficient 2K for gradual contracting section is =______

5. The loss of energy in gradual contacting section is ______ than that through gradual expanding

section




Study of Laminar and Turbulent Flow Through Pipes

Introduction:

Depending upon the value of Reynolds number, the flow of viscous liquid in the pipe can be laminar or

turbulent. For Reynolds number less than 2000 the flow is laminar while for Reynolds number more than 2500

it is likely to be turbulent. In general, it is definitely turbulent for Reynolds number more than 4000. The friction

factor ‘f’ and hence the loss of head in the pipe varies greatly with the nature of flow.

Object:

1. To study how head loss due to friction ‘hf’ varies with the velocity ‘v’ in laminar and turbulent

flow

2. To study variation of friction factor ‘f ’ in laminar and turbulent flow.

Theory:

In general, the frictional loss of head in pipe is give by Darcy weisbach formula.

5

22

1.122 d

flQ

gd

flvhf == in SI units

With usual notations, for laminar flow, Re64=f where Re is the Reynolds number. In case of laminar flow

‘hf’, the loss of head proportional to ‘v’ while it is proportional to v2 in case of turbulent flow. Thus we can write.

nf kvh =

Where,

K=gd

fl

2

Experimentally the loss of head hf for known length of pipe ‘l’ is actually measured. After finding out

the corresponding discharge through the pipe and knowing the diameter of the pipe, the velocity of flow is

found out. With the help of a graph between fh and the velocity ‘v’, the nature of relation between the loss of

head and the type of flow can be obtained.

Experimental equipment:

A set of 4 pipes fitted with control valves for varying the flow and provided with pressure tapping,

manometer board, flow collection vessel, and measuring cylinders, stopwatch etc.



Overflow

Air-water invertedU-tube differentialmanometer

Upstream FlowControl Valve

Rubbertube

Constant head supply water tank

1

air

h1 h2

Air reliefValve

Darcy Weisbach equation

Outlet

hf

2

D=0.32 cm

0

10

20

30

40

50

Water

gD

fLVhf 2

2

=

Study of flow through pipes

L = 70 cm

Downstream FlowControl Valve

MeasuringCylinder



Experimental Procedure:

1. Set the experimental device on the flow table and mark the inlet and outlet connection. With the help

of constant head water supply tank ensure steady conditions throughout the experiment.

2. Connect first inverted U tube differential mercury manometer to the pressure tapping on the pipe in

which the loss of head is to be measured.

3. While making the manometer connections, see that no air is locked anywhere in the system.

4. Start the flow through the pipe and adjust it gradually till you are able to measure the loss of about

10mm.

5. Measure the discharge Q in the measuring flask. Note the time for it.

6. Repeat the procedure for different discharges. Note that the variation in the discharge should be done

very gradually so that the differenence between the two consecutive readings is sufficiently small.

7. Measure the distance ‘l’ between the pressure tappings on pipe.

Experimental Data :

1. Nominal diameter of pipe (brass) d = 0.32 cm

2. Length between the pressure tapping l = 70 cm

3. Kinematic viscosity of water at room

Temperature (to be measured) ν = ………….stokes



Observation Table:

a) For Increasing flow:

Actual Discharge

measurement

(Qa) Log10hf Log10v Log10Re Log10f

Sr.No Head

loss (hf)

cm of

water Vol.

cm3

Time

Sec.

Qa

cm3/s

Velocity

V

=2

4d

Qa

π

cm/s

Reynolds

Number

Re

=vd/ν

Friction

factor

f

=2

2

vl

gdhf ×

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15



Calculations:--

1. ==time

VolQ ________________________cc/sec

2. ==A

Qv __________________________cm/sec.

3. ==νvd

Re ________________

4. ==l

h

v

gdf f2

2_____________________________________

Graphs:--

1. Graph of Log10 Re on X-axis and log10 f on Y-axis. i.e. Moody diagram.

2. Graph of log10 v on X-axis and log 10 hf on Y-axis.

Conclusions:--

1. Laminar and turbulent are two types of flow having transitional state in between them.

2. In the laminar flow ....vαfh from graph whereas in turbulent flow, ....vαfh from graph.

3. From Moody diagram plotted, it is seen that friction factor ‘f’ varies linearly with Re in the laminar

flow.

4. Practical utility of Moody diagram:---

For a flow rate of ………cc/sec., the velocity is ……cm/sec.; hence Re is ……….

For this value of Re=……..from the Moody diagram, friction factor f=………

Assuming the pipe of 3.2 mm diameter, for a flow rate of ……….cc/sec., the velocity v =…….cm/sec.

For a length of say 1 m. of the above pipe head loss due to friction fh is estimated using Darcy-

Weisbach equation as

==gd

flvhf 2

2

_____________________________________

= ……… cm.




Trial and Error Solution for a given Flow Problem

Theory

To determine the friction factor ‘f’ for steady incompressible fluid flow through a pipe so as to use itin Darcy-Weisbach equation to estimate frictional head loss through the pipe.

gD

flVhf 2

2

=

Friction factor ‘f’ is not constant, but it varies with the type of flow whether laminar or turbulent andin turbulent flow, whether the boundary is hydro dynamically smooth or rough or is in transition betweensmooth and rough.

In general for the turbulent flow,

ÿ��

��→

D

kff s

n Re,

Where,

ks = Nikuradse’s Equivalent sand grain roughness

D = Pipe diameter

Re= Reynold’s Number

To estimate this friction factor ‘f’ Colebrook-White suggested an implicit equation in ‘f’ valid for entire

range of Reynold’s number.

��

�

�

��

�

��

��

+−=−f

kR

k

R

fs

s Re

7.181log274.1log2

11010

This equation being implicit was required to be solved by trial and error. Hence Lewis Moody gave an

approximate Explicit equation as

��

�

�

��

�

��

��

�++=

31

6

Re

102000010055.0

D

kf s

which gave the result within ± 5% variation of Colebrook-White equation.

Swami-Jain further gave an accurate explicit equation within ± 1% accuracy of Colebrook’s equation.

��

��

� +−=9.010 Re

25.21log214.1

1

D

k

fs

Aim

1. To solve the Colebrook-White implicit equation in f by trial and error method using a computer

program, for the given flow rate.

2. To determine f from standard Moody diagram, Moody equation and Swami-Jain equation and to

compare the value of f obtained from these equations with the value obtained from Colebrook-

White Equation.



3. To find head loss due to friction (hf) in the pipe using the correct estimated value of f by putting

it in the Darcy-Weisbach equation

Flow Problem

Using Colebrook-White equation, estimate the friction factor f for steady incompressible flow of water

through a galvanized iron pipe of ____ mm diameter when the flow rate through the pipe is ____ lps.

Also compute the value of friction factor f for same flow conditions from the

a. Moody Diagram

b. Moody Equation

c. Swami-Jain Equation

Compare these values of f with the value obtained from Colebrook-White equation.

Estimate the frictional head loss in meter of water for the same flow through the pipe assuming that

the pipe is connected from underground water tank to overhead water tank, traveling a total distance

of 20 m. Write a program to solve the above problem in any programming language.

Given Data

1. Fluid flowing through pipe – Water

a. mass density of water (ρ) = 1000 kg/m3

b. kinematic viscosity of water at 270C (ν) = 0.0085 cm2/s.

2. Pipe diameter (D)= ___ mm= ____ m

3. Pipe length (L)= 20 m.

4. Pipe material = Galvanized iron (G.I.) with equivalent sand grain roughness (ks)= 0.15mm

5. Flow rate through pipe (Q) = ____ lps

Calculation Steps

1. Calculate average velocity of flow =A

QVavg =

2. Calculate flowν

DVavg=Re

3. Calculate relative roughness ratio Dks

4. Assume any value of f in between 0.01 to 0.1 and solve Colebrook-White equation to get Re

��

�

�

��

�

��

��

+−=−f

kR

k

R

fs

s Re

7.181log274.1log2

11010

5. To make trial and error easier, select approximate value of ‘f ’ obtained from Moody equation.

For this selected ‘f ’, obtain value of Re by Colebrook-White equation.

6. If calculated flow Re is less than actual flow Re, then increase assumed value of ‘f ’ by

appropriate amount and use it for the next trial.

7. Repeat above steps till calculated value of Re matches with actual flow Re.

8. The friction factor corresponding to this value of Re is the solution of Colebrook-White equation.



9. Substitute the flow Re computed earlier in Swami-Jain explicit equation to get another value of ‘f

’.

10. Compare the two friction factors ‘f’ from Colebrook-White equation and Swami-Jain equation.

11. Use estimated accurate value of ‘f ’ in Darcy-Weisbach equation to find head loss due to friction

of pipe.

Sample Calculations

1.A

QVavg = =__________________________

2.ν

DVavg=Re =_______________________________

3. Substituting f from Moody Equation, Re from Colebrook-White Equation

��

�

�

��

�

��

��

+−=−f

kR

k

R

f

s

s Re

7.181log274.1log2

11010

Re1 =_____________________________________

4. f from Moody equation

��

�

�

��

�

��

��

�++=

31

6

Re

102000010055.0

D

kf s

m =____________________________________

5. Accuracy of Moody equation in percentage = 100×−

m

m

f

ff

=________________________

6. f from Swami-Jain equation

��

��

� +−=9.010 Re

25.21log214.1

1

D

k

fs

f =_________________________________________

7. Accuracy of Swami-Jain equation in percentage = 100×−

s

s

f

ff

=_______________________

8. Head loss,5

2

1.12 D

flQhf = =_____________________________________



/*Program for Trial and error solution of a given flow problem*/

#include<math.h>

#define pi 3.1415926535897932384626433832795

void main()

{

long double hf,q,d,l,k,ks,v,nu,r,d1,c1;

long double re,f,fm,fs,a,b,c,m,e,re1,diff;

clrscr();

printf("--------------------------------------------------------------------------------");

printf("\t\t\t\t INPUT DATA FOR FLUID FLOW\n\n");

printf("Enter Discharge flowing through pipe (Q) in lps:");

scanf("%Lf",&q);

printf("Enter kinematic viscosity of water in cm2/sec. :");

scanf("%Lf",&nu);

printf("--------------------------------------------------------------------------------");

printf("\t\t\t\t INPUT DATA FOR PIPE \n\n");

printf("Enter Diameter of pipe (D) in mm:");

scanf("%Lf",&d1);

printf("Enter Length of pipe (L) in m. :");

scanf("%Lf",&l);

printf("Enter Equivalent sand grain roughness (ks) in mm:");

scanf("%Lf",&k);

printf("--------------------------------------------------------------------------------");



printf("\t\t\t\t OUTPUT DATA \n");

d=d1/(pow(10,3));

r=d/2;

ks=k*1/(pow(10,3));

v=q/((pow(10,3)*(d*d*pi/4)));

re=v*d/(nu*1/(pow(10,4)));

printf("\n\t Average Velocity of the flow Vavg in m/s= %15.8Lf\n\t\t\t\t Actual Flow Re= %15.8Lf", v, re);

fm=0.0055*(1+pow((20000*(ks/d))+(pow(10,6)/re),1/3));

f=fm;

do

{

a=1/(sqrtl(f));

b=2*log10(r/ks);

c=1.74-(a-b);

c1=c/2;

m=pow(10,c1);

e=m-1;

re1=(18.7*(r/ks))/(e*sqrtl(f));

if ((re-re1)<=1/pow(10,9))

break;

if ((re-re1)<=1/pow(10,9))

break;

if (re1<re)



{

diff=((re-re1)/re);

f=f*(1-diff/1000);}

else

{

diff=((re1-re)/re1);

f=f*(1+diff/1000);

}

if (re1<=0)

f=(f+0.05);

}

while(re1!=re);

fs=1/pow(1.14-2*log10((ks/d)+(21.25/pow(re,0.9))),2);

hf=(f*l*pow(v,2))/(2*9.81*d);

printf("\n\t\t Re from Colebrook-White Equation= %15.8Lf\n Friction factor f from Colebrook-White Eq.= %15.8Lf ",re1,f);

printf("\n\t Friction Factor f from Moody Equation= %15.8Lf",fm);

printf("\n\t Friction f from Swami-Jain Equation= %15.8Lf",fs);

printf("\nFrictional Head Loss in given pipe in m. of water= %15.8Lf",hf);

printf("\n--------------------------------------------------------------------------------");

getch();

}



INPUT DATA

Water::

Discharge flowing through pipe (Q) in lps = 2

Kinematic viscosity of water in cm2/sec. = 0.0085

Pipe::

Diameter of given pipe section (D) in mm =25

Length of given pipe section (L) in m =15

Eq. sand grain roughness coeff.(ks) in mm = 0.15

--------------------------------------------------------------------------------

OUTPUT DATA

Average Velocity of the flow (Vavg) in m/s= 4.07436654

Re calculated from actual flow conditions = 119834.31009272

Re obtained from Colebrook-White Equation = 119834.31009272

Friction factor f from

Colebrook-White Eq.= 0.03280198

Moody Equation = 0.03324313

Swami-Jain Equation= 0.03300098

Accuracy of Moody Equation in percentage = -1.34488866

Accuracy of Swami-Jain Eq. in percentage = +0.60300172

Frictional Head Loss (hf) in m. of water = 16.65223552



--------------------------------------------------------------------------------

INPUT DATA

Water::



Pipe::




--------------------------------------------------------------------------------

OUTPUT DATA














Accuracy of Moody Equation in percentage = +0.62173993



--------------------------------------------------------------------------------

INPUT DATA

Water::



Pipe::




--------------------------------------------------------------------------------



OUTPUT DATA














Flow Through Packed Bed

Aim

1. To estimate void fraction for different packings.

2. To use the void fraction, thus calculated to find packed bed Reynolds number for given data.

3. To use Ergun equation to find the packed bed friction fp.

4. To determine the pressure drop across a packed bed for given data.

Theory

In a packed bed. the fluid, liquid or gas flows through solid packings. The solid packings could

be spherical, cylindrical, solids or specially prepared packings such as Raschig ring, Berl Saddle,

Intalox Saddle etc. During chemical reaction, the packings provide a large contact area between the

liquids and gases especially during two phase flow. Sometimes, the packings also act as catalyst in

different reactions. Because of the packings, as the fluid moves through them, a very large pressure

drop across the depth of packed bed which is required to be evaluated.

Common practical applications involving packed bed include:

i) Oil removal from porous rocks

ii) Filtration process.

iii) Catalytic cracking process.

iv) Chemical process such as distillation, humidification etc.

v) Polyester fibre drawn from molten polymers.

Equation

To evaluate the pressure drop, across a packed bed, packed bed friction factor fp is defined

as:

L

p

V

EDf p

p

∆−

=∞ )1(2

3

ερWhere,

pD = Particle diameter



Case I Case II Case IIIPackings only Packings + Water Water

Packed BedDetermining void fraction (e)

1000 cc 1000 cc

700 cc



E = void fraction =bedpackedofvolumeTotal

FluidofVolume

pf depends on the manner of packing whether dumped randomly or stacked properly by

hand. (In this experiment., the packings are randomly dumped).

∞V = Superficial velocity which is the velocity that would exist in absence of solid particles.

A

QV =∞

where:

A=c/s area of packed bed.

L = depth of bed.

� = fluid mass density.

Pressure drop across packed bed can be given as:

3

2 )1(

ED

VLfp

p

p ερ −=∆ ∞

The packed bed friction factor pf can be evaluated by using Ergun equation:

1000and10betweenRefor1.75Re

150

1000Refor75.1

10ReforRe

150

p

+=

≥=

≤=

p

p

pp

f

f

f

Where,

pRe = Packed bed Reynolds number

=µε

ρ)1( −∞VDp

The above equation Indicates that the void fraction is very instrumental in calculations of

pressure drop p∆ .



Experimental Setup

Two measuring cylinders (1000 ml capacity), beakers, water, porcelain packings such as Raschig

Ring, Berl Saddle and Intalox Saddle.

Experimental procedure

1. Select the Raschig rings and pack them randomly in the cylinders to touch the 1000 ml mark.

2. Carefully put the water up to 1000 ml mark to fill all the voids.

3. Take another measuring cylinder and pour the water from the total packed bed into it to

record the volume of water in packed bed ‘v’’.

4. Calculate the void fraction E=v’/V where V=Total volume of packed bed.

5. Change the manner of packing for new random packing and repeat the procedure for two

more readings.

6. Find the average void fraction to estimate void fraction for that packing.

7. Perform the experiment for another packing.

8. Record the total depth of a packed bed ‘L’, the diameter of packed bed D and the individual

particle diameter pD .

9. Assume same flow rate through the packed bed Q.

10. Find the superficial velocity.

2

4 D

QV π=∞

11. Find the packed bed Reynolds number for different Packings.

12. Find the packed bed friction factor for different Packings.

13. Also find the pressure drop for the packed bed ( p∆ ) for diff. Packings.

14. Write a program to generate the data and results in tabular format after processing of data.

Experimental Data

1. Dia. of packed bed (D). =______ cm.

2. Depth of packed bed (L). =______ cm=______ m.

3. Mass density of water (ρ). =1000 kg/m3

4. Kinematic viscosity of water at 27oc ( � ) =0.0085 cm2/s

5. Dynamic viscosity of water ( � ) =…………

6. Volumetric rates of flow (Q)

i. Q1=____ X 10-7 m3/s

ii. Q2=____ X 10-5 m3/s

iii. Q3=____ X 10-5 m3/s



7. individual diameter of packing.

i. Raschig Ring Dp1 =_____ cm

ii. Bearl saddle Dp2 =_____ cm

iii. Intalox saddle Dp3 =_____ cm

8. Total volume of packed bed V=1000 ml.

Observation Table

Sr. No. Type of Packing Vol. Of Fluid

(ml)

Void Fraction

�

Avg.

�

1

21 Raschig Ring3

1

22

Bearl Saddle

3

1

23

Intalox Saddle

3

Sample Calculation:

For --------------------------------------------------- packing

Q= ______________ m3/sec.

2

4 D

QV π=∞ =________________________________

pRe =µε

ρ)1( −∞VDp

=_____________________________

For the pRe obtained, using Ergun equation,

pf = _______________=_____________________________________

3

2 )1(

ED

VLfp

p

p ερ −=∆ ∞

=_____________________________________



Conclusion

1. The void fraction changes with the type of packing and manner of packings.

2. From values of packed bed friction factor pf , it is clear that with increase in superficial velocity

∞V , the friction factor decreases.

3. However with increase in superficial velocity though the friction factor decreases, the pressure

drop p∆ across the packed bed increases as seen from values of p∆ .



#include<math.h>

#include<stdio.h>

main()

{float i,Q1,Q2,Q3,rho,nu,mue,Dp1,Dp2,Dp3,A1,A2,A3,A4,A5,A6,A7,A8,A9;

float v1,v2,v3,v4,v5,v6,v7,v8,v9,Rep1,Rep2,Rep3,Rep4,Rep5,Rep6;

float Rep7,Rep8,Rep9,e1,e2,e3,fp1,fp2,fp3,fp4,fp5,fp6,fp7,fp8,fp9;

float l,p1,p2,p3,p4,p5,p6,p7,p8,p9,temp;

clrscr();

l=0.352;

rho=1000; /* kg/m3 */

nu=0.0085*pow(10,-4); /* m2/sec */

mue=nu*rho;

Q1=2*pow(10,-8); /* m3/sec */

Q2=2.5*pow(10,-6); /* m3/sec */

Q3=5.5*pow(10,-5); /* m3/sec */

printf("\n\t\t\t****Input Data****\n");

printf("\t-------------------------------------------------------------");

printf("\n\tDiameter of Rasching Ring in cm=");

scanf("%f",&Dp1);

printf("\tDiameter of Berl Saddle in cm=");

scanf("%f",&Dp2);

printf("\tDiameter of Intalox Saddle in cm=");

scanf("%f",&Dp3);

Dp1=Dp1*pow(10,-2);

Dp2=Dp2*pow(10,-2);

Dp3=Dp3*pow(10,-2);

A1=Dp1*Dp1*3.142/4;



A2=Dp2*Dp2*3.142/4;

A3=Dp3*Dp3*3.142/4;

/*For Rasching Ring*/

v1=Q1/A1;

v2=Q2/A1;

v3=Q3/A1;

printf("\n\tVoid Fraction for rasching Ring=");

scanf("%f",&e1);

Rep1=Dp1*v1*rho/((1-e1)*mue);



fp1=150/Rep1;

fp2=1.74+(150/Rep2);

fp3=1.75;

temp=rho*(1-e1)*l/(Dp1*e1*e1*e1);

p1=fp1*v1*v1*temp;

p2=fp2*v2*v2*temp;

p3=fp3*v3*v3*temp;

/*For Berl Saddle*/

v4=Q1/A2;

v5=Q2/A2;

v6=Q3/A2;

printf("\tVoid Fraction For Berl Saddle=");

scanf("%f",&e2);






fp4=150/Rep4;

fp5=1.74+(150/Rep5);

fp6=1.75;


p4=fp4*v4*v4*temp;

p5=fp5*v5*v5*temp;

p6=fp6*v6*v6*temp;

/*For Intalox saddle*/

v7=Q1/A3;

v8=Q2/A3;

v9=Q3/A3;

printf("\tVoid fraction For Intalox Saddle=");

scanf("%f",&e3);




fp7=150/Rep7;

fp8=1.74+(150/Rep8);

fp9=1.75;


p7=fp7*v7*v7*temp;

p8=fp8*v8*v8*temp;

p9=fp9*v9*v9*temp;

printf("\n\n\t\t\t****Output Data****\n");

printf("\n\tReynold`s Number For Packed Beds");

printf("\n\n\tRasching Ring Berl Saddle Intalox Saddle");

printf("\n\tRep1=%f Rep1=%f Rep1=%f",Rep1,Rep4,Rep7);





printf("\n\t------------------------------------------------------------");

printf("\n\n\tFriction Factor For Packed Beds");

printf("\n\n\tRasching Ring Berl Saddle Intalox Saddle");

printf("\n\tfp1=%f fp1=%f fp1=%f",fp1,fp4,fp7);



printf("\n\t-----------------------------------------------------------");

printf("\n\n\tPressure Drop Across Packed Beds In Pascals");

printf("\n\n\tRasching Ring Berl Saddle Intalox Saddlle");

printf("\n\tp1=%f p1=%f p1=%f",p1,p4,p7);



getch();}

****Input Data****

-------------------------------------------------------------

Diameter of Rasching Ring in cm=6.4

Diameter of Berl Saddle in cm=1.25

Diameter of Intalox Saddle in cm=2.5

Void Fraction for rasching Ring=0.697

Void Fraction For Berl Saddle=0.763

Void fraction For Intalox Saddle=0.687

****Output Data****

Reynold`s Number For Packed Beds



Rasching Ring Berl Saddle Intalox Saddle

Rep1=1.544693 Rep1=10.111289 Rep1=3.828075

Rep2=193.086670 Rep2=1263.911133 Rep2=478.509430

Rep3=4247.906738 Rep3=27806.046875 Rep3=10527.208008

------------------------------------------------------------

Friction Factor For Packed Beds

Rasching Ring Berl Saddle Intalox Saddle

fp1=97.106644 fp1=14.834904 fp1=39.184181

fp2=2.516853 fp3=1.858679 fp2=2.053473

fp3=1.750000 fp3=1.750000 fp3=1.750000

-----------------------------------------------------------

Pressure Drop Across Packed Beds In Pascals

Rasching Ring Berl Saddle Intalox Saddlle

p1=0.000018 p1=0.005919 p1=0.000884

p2=0.007479 p2=11.586688 p2=0.723758

p3=2.516854 p3=5280.054199 p3=298.529938




Verification of Stoke’s law

Aim of experiment:

1. To find the terminal settling velocity of the sphere in glycerine – U

2. To verify the stoke’s faw.

Theory:

Stoke’s law state that

FD=3� � UDP

Where,

FD= Drag resistence offered by fluid

� = Absolute viscosity of fluid

U = Terminal or uniform settling velocity

DP = Diameter of spherical body immersed in fluid

Stoke’s law is applicable only for small spherical freely suspended particle moving at

extremely small velocity in an infinite expanse of a very viscous fluid such that the flow Reynold’s

No. Re � 1.

Initially the solid body accelerates downward in the fluid under its weight W, whereas the

fluid offers upward drag resistance FD.

Then the body stops accelerating and moves down in the fluid with uniform velocity U called

terminal or settling velocity.

At equilibrium

W � =FD�

(� p- � f) � Dp3/6=3 �� UDP

Where, � p =specific weight of the solid particle.

� f =specific weight of the fluid.

Common applications based on Stoke’s law

1. Design of grit chambers and sedimentation tanks in water treatment plants

2. In sewage treatment works

3. In air clarifiers during air pollution control

4. Ground water flow through porous media

5. Determination of fluid viscosity



PSteel ball accelerate underits own weight

FD

WEquillibrium Condition

Steel ball moving withterminal velocity U

L

Wooden board

Stoke's apparatus

glycerine

D

D

Stoke's cylinder



Experimental setup: Stoke’s law apparatus, steel balls, beakers, stopwatch etc.

Experimental Procedure :

1. Fill the fluid upto the top of vertical transperent glass cylinder. Ensure that there are no air bubbles

and the fluid is static before dropping the ball.

2. Select a particular diameter steel ball and after carefully placing at the center of the tube, lightly drop

it into the fluid.

3. Record the time ‘t’ seconds required to pass the ball in the fluid between the permanent marks made

on the setup covering a distance ‘L’.

4. Take the average of the three readings to accurately record time.

Find U’=L/t

5. Repeat the procedure for another ball of different diameter.

6. Find the terminal velocity U after applying the end correction as

��

�

��

��

��

�++=

2

'

4

9

4

91

D

D

D

DUU pp

7. Find the submerged unit weight and drag resistance and compare to verify stoke’s law.

Experimental data :

1. Diameter of steel balls

Dp1=4.7mm; Dp2=4.0mm; Dp1=2.3mm

2. Mass of steel balls

m1= 0.42g. ;m2= 0.25g. ; m3= 0.11g.

3. Diameter of stoke’s apparatus cylinder D = 9.6cm

4. Mass density of glycerine � f= 1266.67 kg/m3.

5. Absolute viscosity of glycerine at room temperature � =0.77Ns/m2.

Observation Table :

Sr.

No.

Dia. Of

steel

ball

Length

L (cm)

Time

T (sec.)

U�

=L/T

cm/s.

Avg U’

cm/s

Corrected

U

cm/s

FD

(N)

W

(N)

K

W/FD

1

2

3



4

5

6

7

8

9

Calculations:

1. Calculate mass density of steel balls as � P=m/ (� D3 p/6)=____________________________

2. Calculate Weight of each ball as W= ( � D3 p/6)g( � p – � f) =_____________________________

3. Calculate FD=3� � UDP=_________________________

4. Calculate K=W/FD=____________________________

5. Calculate Re= � f U Dp/ � =_____________ to check whether Re � 1

Conclusion :

As FD � W at Re <=1, Stoke’s law is Verified.




Study of Operating Characteristics of centrifugal Pump

Aim:

To study variation of pump characteristics such as Manometeric head (Hm), power input (Pin), Overall

efficiency ( � ) with the discharge (Q).

Apparatus:

Centrifugal pump with suction and delivery pipes, control valves. Pressure gauges, energy meter and

90o V-notch.

Theory:

A pump is a device which converts any other form of energy into hydraulic ÷energy. Centrifugal pump

is a type of Rotodynamic machinery which is characterised by the rotary motion of the impeller, run at a

constant RPM. This rotary motion of impeller induces an additional centrifugal head on the fluid, thereby

increasing both its kinetic energy and pressure energy. Thus centrifugal pumps are commonly used to lift and

drag fluids.

A centrifugal pump is designed for maximum overall efficiency(� o), corresponding to the efficiency (�

o), the design discharge (Qo), maximum power input required (Pino) and maximum design manometric head

(Hmo). But during actual operation of the pump, it is not necessary that it is always at the design values. In fact,

frequently the centrifugal pump gets operated at different discharges than the design discharge. Hence the

operating curves of a centrifugal pump show graphically behavior and performance of the pump under various

conditions of operation; for a const rpm. of the impeller. These curves in the non-dimensional form help to

predict the pump performance when run at any other discharge, than the design discharge.

Procedure:

1. Prime the pump to remove all the air, gas from the suction pipe, centrifugal pump body and section of

the pipe up to delivery valve, with delivery valve closed.

2. Start the pump to its design speed. Adjust the discharge and speed by controlling the delivery valve

and measure it with the help of triangular notch by measuring the head above the crest of the notch

with the help of piezometer.

3. Record the suction head, delivery head and input power on the electronic device.

4. Repeat the procedure for 8 to 10 observations.



Component parts of a Centrifugal Pump

Suction HeadHS

Delivery HeadHD

Foot Valve

Suction pipe

Strainer

Sump

Eye

Impeller

Volute Casing

Delivery Valve

Delivery pipe



Observation Table :-

Sr.

No.

Suction

Head

Delivary

Head

Manometric

Head Hm

Actual discharge

Measurement

Pin

Kw

Pout

Kw

� H/Ho Q/Qo Pin/Pin0 � /� o

Hs in

mm

of Hg

Hs in

m. of

water

Hd in

Kg/cm2

Hd in

m. of

water

=Hs+Hd

m. of water

h1

cm

h

cm

Q

m3/s

1

2

3

4

5

6

7

8

9

10



Formulae to be used:

1. Hs (m. of water) = SH×1000

6.13(mm of mercury)

2. Hd (m. of water) = Hd (kg/cm2) X 10

3. Zero of the notch = 5 cm.; head over notch, h=h1-5

4. Discharge, Q = 1.417(h)2 .5 cm3/s.

5. Pout =1000

HQ mγ=9.81QHm kW

6. Efficiency = η =in

out

P

P

Sample Calculations

1. Discharge over 900 triangular notch

Q=1.417 X h2 . 5 m3/s.--------------------------------

2. Manometric head= Hm = Hs + Hd-----------------------------------

3. Output power = γ Q Hm =-----------------------------

4. Efficiency = η = P0/Pin X 100

Graphs:-

On a single graph plot,

H/Ho , Pin /Pin(0) , � /� o on y axis against Q/Qo on x axis.

Conclusion

1. At maximum efficiency � o=--------- for the centrifugal pump the design discharge Qo =----------- m3/sec.

2. Minimum power input Pin(0) corresponding to design discharge=---------kw is necessary to provide

maximum efficiency.

3. The design manometric head is Hm(0) = -------------m

4. At zero discharge the maximum value of shut off head=-------------m.

5. The characteristic curves indicate that with increase in the flow rate Q, the input power Pin

increases, the manometric head decreases and the efficiency increases initially but after the

design discharge, it again decreases.




Pump and Blower Specification writing in a format routinely used by Process

Industry

Data sheets, consisting of pumps, blower specifications are extremely useful in providing a summary

of information to the bidder and provides the basis for evaluating after comparison, the different bid offered

and provided by various manufactures. For relatively simple or inexpensive pumps, blower or for either

replacement of duplication only, specifications are not written but a direct quotation is asked for generally,

after specifying minimum requirements. For costly machinery, with the involvement of many trade makes and

manufacturers advancements in design, manufacturing technology, legalities involved etc; specifications

become very useful. Also they help the client to include special requirements than normal ones.

The specification may be broadly classified as

i) Construction specifications which gives details about the materials used, methods used during pump

design, construction, installations, maintenance etc.

ii) Performance specifications which establish the performance which the pump, blower must achieve

during its operation (Service requirement).

Sub-classifications include technical, commercial specifications. Specification writing is based on use of

different codes and standards for reference such as say.

ANSI: American National Standards Institute.

ASTM: American Standard for Testing, Manufacture.

ASME: American Society for Mechanical Engineers.

BS: British Standards.

BIS: Bureau of Indian Standards.

API: American Petroleum Institute.

ISO: International Standard Organizations, Standards of the Hydraulic Institute

NFPA: National Fire Protection Association.

FM: Factory Manual for Insurance etc.

The specifications are written to cover the following points. (only a few very important Points are

listed).

i) Duty:- Purpose of use; also consist of the duty cycle and a diagram Indicating lengths, elevation,

gradients, valves etc, necessary for piping.

ii) Fluid:- Type, Physical, Chemical characteristics.

iii) Material of construction (For the main body as well as the parts) In this the materials that are

recommended for use based on particular purpose are specified.

iv) Flow paths, flow rates, pressures, temperatures for various system operating conditions; NSPH for

pumps.

v) Graphic representation by diff. Characteristic curves.



vi) Consideration towards alternate modes of operation such as whether in continuous use or intermittent

use, operations under constant head or varying head, operations under large temperature gradients

or small ones etc.

vii) Assess impact of wear of equipment during use on operating characteristics and allow for permissible

margins over and above the normal rating.

viii) Scope for future system changes.

ix) Driver type:- Whether electric, steam, gas, etc whether variable speed; or constant speed, voltage

fluctuations, etc.

x) Code requirements to be satisfied, tolerances permissible.

xi) Performance testing.

A typical centrifugal pump data sheet may be as shown below:-

Details about plant, location, layout, reference drawing, requisition number, process involved.,

Manufacture job no, data sheet no, data etc. are mentioned on the data sheet

Technical specifications:-

1) Liquid pumped., specific gravity

2) Viscosity— Ns/m2.

Vapour pressure Pa

3) Temperature in F/0C/Max/Min

4) PH value

5) Solids concentrations, size

6) Flow ratings/Min/Max GPM/cumecs, LPS/Min.

7) Suction pressure. Pa

8) Discharge pressure. Pa

9) Differential pressure rating/shut—off value

10) NPSH(R)

11) Type of pump/model/no of stages.

12) RPM.

13) Efficiency, HP at rating, HP maximum.

14) Impeller diameter: Bid/Min/Mix.

15) Impeller eye: area/Peripheral velocity.

16) Working pressure: Max/Hydro /Test.

17) Clearance: Wear Ring/Bearing/Impeller.

18) Hydro Thrust: Rating/Maximum.

19) Torque: Rating/Max.

20) Suction: Size/Rating.

21) Discharge: Size/Rating.



22) Base Plate:

23) Coupling: Type/ manufacturer

24) Strainer: Size/diameter.

25) Bearing Lubricant Max. Pressure.

26) Shaft seal: Type.

27) Materials:

for

i) Case

ii) Shaft

iii) Impeller

iv) Wear ring

v) Liners

28) Driver: Type of motor, RPM, HP, furnished by weight,

drawing reference, bearing description.

thrust rating.

29) Drawing No: Outline/Sectional/performance curves.

30) Net.Wt. Rotating elements.

31) Inspection: Std. No. ASMEIII

32) TESTING: Ultrasonic/Eddy current/magnetic Part/

performance field.

33) Quality Assurance Mfr std.

34) Seismic Design Requirements Class I/ClassII.

Data sheet for centrifugal Blowers:

1) Fluid blown, specific gravity

2) Viscosity- NS/M2

3) Vapour pressure Pa

4) Flow rating Min/Max.

GPM/Cumecs/LPS/Min.

5) Type of Blower/Model no.

6) RPM

7) Efficiency, Hp at Rating, Hp Maximum.

8) Impeller dia area/Bid/Max/Min

Impeller eye area/Peripheral velocity.

9) Compression Ratio (P2?P1)

10) Working Pressures

Inlet P1 Max/Min/Rated

Outlet P2 Max/Min/Rated



11) temperatures in 0F/0C

Inlet P1 Mix/Min

Outlet P2 Mix/Min

12) Head developed

13) Whether coolant used

14) Whether single/Multiple stages.

15) Inlet size.

16) Discharge size.

17) Coupling Type. MFR.

18) Bearing Lubricant Max. Pressure.

19) Shaft seal type.

20) Materials Case and Size

Shaft

Impeller

Wear ring

Liners.

21) Driver: Type of motor, RPM, HP, furnished by wt, drawing reference, bearing description, thrust rating.

22) Drawing No. Outline/sectional/performance curve.

23) Net wt. Blower/Removable/Rotating parts

24) Inspection STD no. ASME III

25) Testing Ultrasonic/Eddy Current/magnetic

part of performance field

26) Quality assurance MFR Ltd.

27) Seismic design requirements Class I/Class II

fluid dynamics

Documents

e t e mp e r

e skinematicviscositystokest

e n c e s u

n g e i n t e mp e r

yh e n c e u n s u

r g e v

s o mo d e r

i n e ma t i c