Fluid Dynamics

MAEER’S MIT, Pune

Experiment No.

Civil Department

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]

1

MAEER’S MIT, Pune

Civil Department

Thermometer Inner Cylinder containing oil Outer Cylinder containing water

Index Mark Stirrer wall Stirrer Fins Heating Coil Ball Valve Capillary Tube Outlet valve ûþ

Conical flask to measure 50 cc of oil Levelling screws ÿþýüû ùø÷ö

Redwood Viscometer

[email protected]

2

MAEER’S MIT, Pune

Civil Department

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

Time to collect 50cc of oil

Kinematic viscosity

‘ÿ’

‘t’

‘ν ν’

(s.)

(stokes)

0

( C) 1 2 3 4 5 6 7 8 9 10

Sample calculations:

[email protected]

4

MAEER’S MIT, Pune

Civil Department

K i n e m a t ic v i s c o s i t y ( υ ) v s . t e m p e r a t u r e ( θ ) 0 .4 0 S c a le 0 X - a x is : 1 c m = 5 C Y - a x is : 1 c m = 0 . 0 2 5 s t o k e s

0 .3 5

R e g io n I:- V e r y h ig h v is c o s ity , a l s o l a r g e v a r ia tio n in v is c o s it y w it h s m a l l c h a n g e in t e m p e r a tu r e h e n c e u n s u ita b l e R e g io n II:- M o d e r a t e V is c o s ity , a l s o m o d e r a t e v a r ia tio n in v is c o s ity w ith c h a n g e in te m p e r a tu r e h e n c e s u it a b l e t e m p e r a tu r e r a n g e f o r t h e o il to 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 it y h e n c e u n s u ita b l e

0 .2 5

υ (stokes)

Kinematic viscosity

0 .3 0

0 .2 0 0 .1 5 0 .1 0 0 .0 5 R e g io n I

R e g io n II

R e g io n III

0 .0 0 20

30

40

50

60

70

80

90

100

T e m p e ra tu re 0 θ ( C )

[email protected]

5

MAEER’S MIT, Pune

Civil Department

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]

6

MAEER’S MIT, Pune

Civil Department

Experiment No.

Date:

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. =

C h

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 =

a1a 2 2 g a1 − a 2 2

2

Where, a1 = Area of inlet which can be found out from inlet diameter d 1. =

π 4

d1

2

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

π 4

d2

2

[email protected]

7

MAEER’S MIT, Pune

Civil Department

Air Release Valve Inverted U tube Air-Water Differential Manometer

Manometric Fluid - Air

x=h1-h2

Holes of annular ring to measure average pressure

h1=p1/ν

Annular Ring

h2=p2/ν

Pipe Fluid - Water

Qa d1=5cm Inlet

@20°

Throat

Converging Cone

Qa

@6°

d2=1.6 cm

Diverging Cone

Outlet

2.5L

L

Rubber Tubes connecting pipe to manometer

Venturimeter [email protected]

9

MAEER’S MIT, Pune

Civil Department

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

k=

Qa Q th The above formula can also be written n

Qa = Mh

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

[email protected]

10

MAEER’S MIT, Pune

Civil Department

Observation Table

Manometric readings Sr. No. h1

h2

h

cm

cm

cm

Qth.

Actual discharge

=C h

measurement

cm3/s

K =

I.R.

F.R.

Time

Qa.

cm.

cm.

Sec.

cm3/s

Qa Qth

log10 Qa

log10 h

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Average k =

[email protected]

11

MAEER’S MIT, Pune

Civil Department

lo g 1 0 Q a v s . lo g 1 0 h A lw a y s d r a w t o s a m e s c a le

2 .8 4 S c a le X -a x is : 1 c m = 0 .0 2 u n its Y -a ix s : 1 c m = 0 .0 2 u n its

2 .8 2 2 .8 0 2 .7 8

log10Qa

2 .7 6 n = s lo p e o f lin e = (y 2 -y 1 )/(x 2 -x 1 ) =

2 .7 4 2 .7 2

M = 10 =

2 .7 0

[lo g 1 0 Q a - n lo g 1 0 h ]

2 .6 8 k g r a p h ic a lly = M /C =

2 .6 6 2 .6 4 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 8

lo g 1 0 h

[email protected]

12

MAEER’S MIT, Pune

Civil Department

C a lib ra tio n c u rv e ( Q a v s . h ) 800 S c a le X -a x is : 1 c m = 5 c m 3 Y -a x is : 1 c m = 5 0 c m /s .

700

500

3

Qa (cm /s)

Actual discharge

600

n = M = Q a= (

400

)h

(

)

3

c m /s .

300 For head h = ( ) cm , a c tua l d is c ha rge , Q a = (

200

3

) c m /s .

100 0 0

10

20

30

40

50

60

70

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

[email protected]

13

MAEER’S MIT, Pune

Civil Department

C o e f fic ie n t o f d is c h a r g e ( k ) v s . R e y n o ld 's n u m b e r ( R e ) a t th r o a t 1 .4 S c a le X -a x is : 1 c m = 2 0 0 0 u n its Y -a x is : 1 c m = 0 .1 u n its

1 .0

0 .8

k

Coefficient of discharge

1 .2

0 .6

0 .4

0 .2

0 .0 64

66

68

70

72

74

76

78

80

82

84

86

88

90

92

94

3

( X 10 )

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

[email protected]

14

MAEER’S MIT, Pune

Civil Department

Sample calculations

a1a 2 2 g

1. C =

a1 − a 2 2

2

= _____________________________________ cm

2.5

/ sec .

2. Qth. =

C h =_____________________________________ cc / sec .

3. Qa. =

A[F .R. − I .R.] = _____________________________________ cc / sec . time

4. k =

Qa = _________________ Qth

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

n

3

= Mh = ……………………………………cm /s. =…………………………………….m3/s.

3.

Practical utility of calibration curve, For ………cm pressure head difference across the Venturimeter, the discharge through 3

the Venturimeter is …………cm /s.

[email protected]

16

MAEER’S MIT, Pune

Civil Department

Experiment No.

Date:

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 1/2

Qth = Ch

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 =

a 0 2g

Where, a0 = Area of orifice =

∏ 4

do

2

where do = diameter of orifice opening

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

[email protected]

17

MAEER’S MIT, Pune

Civil Department Air Relief Valve

Inverted U-tube Air-water differential manometer

x=h1-h2

h1=p1/? Orifice plate

h2=P2/?

do Vena contracta

D

Qa

Separation zone (eddies) Rubber tube connecting manometer with pipe

Orificemeter [email protected]

18

MAEER’S MIT, Pune

k=

Civil Department

Qa Q th The above formula can also be written

Qa = Mh

n

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

Apparatus 1. Orificemeter 2. A Flow table with self circulating system 3. Measuring tank 4. Stopwatch 5. Differential manometer

Experimental Procedure 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.

Experimental Observation 1. Diameter of orifice = d0 = 2.5 cm. 2. Diameter of pipe = d = 5 cm. 3. Dimensions of the measuring tank

[email protected]

= 50 cm. X 25 cm.

19

MAEER’S MIT, Pune

Civil Department

Observation Table

Manometric readings Sr. No. h1

h2

h

cm

cm

cm

Qth.

Actual discharge

=C h

measurement

cm3/s

K =

I.R.

F.R.

Time

Qa.

cm.

cm.

Sec.

cm3/s

Qa Qth

log10 Qa

log10 h

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Average k =

[email protected]

20

MAEER’S MIT, Pune

Civil Department

lo g 1 0 Q a v s . lo g 1 0 h

A lw a y s d ra w to s a m e s c a le

2 .8 5 S c a le X-a x is : 1 c m = 0 .0 2 5 units Y -a x is : 1 c m = 0 .0 2 5 u nits

2 .8 0

log10Qa

2 .7 5 n = s lo pe o f line = (y 2 -y 1 )/(x 2 -x 1 ) =

2 .7 0

2 .6 5

M = 10 =

2 .6 0

2 .5 5 1 .0 0

[lo g 1 0 Q a -n lo g 1 0 h ]

k g r a ph ic a lly = M /C =

1 .0 5

1 .1 0

1 .1 5

1 .2 0

1 .2 5

1 .3 0

lo g 1 0 h

[email protected]

21

MAEER’S MIT, Pune

Civil Department

C a lib r a t io n c u r v e ( Q

a

vs. h )

1200 S c a le X - a x is : 1 c m = 5 c m 3 Y - a x is : 1 c m = 1 0 0 c m /s .

1100 1000 900

700

n = M = Q a= (

600

)h

(

)

3

c m /s .

500

a

3

Q (cm /s)

Actual discharge

800

400 F or head h = ( ) cm , a c tu a l d is c h a r g e , Q a = (

300

3

) c m /s .

200 100 0 0

10

20

30

40

50

D if f e r e n t ia l h e a d h (c m )

[email protected]

22

MAEER’S MIT, Pune

Civil Department

Sample calculations

a 0 2g =_____________________________________ cm 2.5 / sec .

1. C = 2. Qth. =

C h =_____________________________________ cc / sec .

3. Qa. =

A[F .R. − I .R.] = _____________________________________ cc / sec . time

4. k =

Qa = ___________ Qth

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

n

3

= Mh = ……………………………………cm /s. 3

=…………………………………….m /s. 3. Practical utility of calibration curve, For

………cm

pressure head difference across the Orificemeter, the discharge through the Orificemeter

3

is …………cm /s.

[email protected]

23

MAEER’S MIT, Pune

Civil Department

Experiment No.

Date:

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. 2

= (V1-V2) /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.

[email protected]

24

MAEER’S MIT, Pune

Civil Department

K2 = Coefficient which depends on area ratio hLth. = Theoretical loss of energy. 2

= V2 /2g

Apparatus 1. Coupling forming gradual expanding section and gradually contracting section 2. A Flow table with self circulating system 3. Measuring tank 4. Stopwatch 5. Manometer

Experimental Procedure 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.

Experimental Observation (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

[email protected]

A = 50 cm. X 25 cm.

25

MAEER’S MIT, Pune

Civil Department

Observation Table :Gradually Expanding Section. Sr.

Manometric

No.

readings

Actual discharge measurement

h1

h2

I.R

F.R

Time

Qa.

cm

cm

cm

cm

sec.

cm3/s

V1

V2

E1

E2

hLa

hLth

K1

cm/s

cm/s

cm

cm

cm

cm

cm

1 2 3 4 5 6 7 8 9 10

Average k1 =

Observation Table :Gradually Contracting Section. [email protected]

26

MAEER’S MIT, Pune

Sr.

Manometric

No.

readings

Civil Department

Actual discharge measurement

h1

h2

I.R

F.R

Time

Qa.

cm

cm

cm

cm

sec.

cm3/s

V1

V2

E1

E2

hLa

hLth

K2

cm/s

cm/s

cm

cm

cm

cm

cm

1 2 3 4 5 6 7 8 9 10

Average k2 =

[email protected]

27

MAEER’S MIT, Pune

Civil Department

E1 2

V1 __ 2g

T.E.L. (Ideal fluid) T.E.L. (R

eal fluid)

. H.G.L

P1 h1 = __ ?

E2 hla= 2 V 2 __ 2g

E1 E_E 1

P2 h2= __ ?

Qa

2 1

V __ 2g

2

E2

T.E.L. (Ideal fluid) T.E.L. (Real fluid)

H.G .L

.

P1 h1 = __ ?

hla= E1_ E2 2 V2 __ 2g

P2 h2= __ ?

Qa Z1

Eddies

Z2

Z1

Z2

datum level

Expansion

Contraction

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

[email protected]

28

MAEER’S MIT, Pune

Civil Department

Sample calculations (A) Gradually expanding section

(v1 − v 2 ) 2 =_____________________________________cm = 2g

5.

h Lth

6.

E 1 = h 1 + v1 / 2g = _____________________________________cm

7.

E 2 = h 2 + v 2 / 2g = _____________________________________cm

8.

hLa = E1 − E 2 = _____________________________________cm

9.

k1 =

2

2

hLa = _________ hLth

(B) Gradually contracting section 2

v2 =_____________________________________cm 2g

1.

h Lth =

2.

E1 = h1 + v1 / 2 g = _____________________________________cm

3.

E2 = h2 + v2 / 2 g = _____________________________________cm

4.

hLa = E1 − E 2 = _____________________________________cm

5.

k2 =

2

2

hLa = __________ hLth

Conclusions 3. Average Coefficient

K 1 for gradual expanding section is = ______

4. Average Coefficient

K 2 for gradual contracting section is =______

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

[email protected]

29

MAEER’S MIT, Pune

Civil Department

Experiment No.

Date:

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.

hf =

flv 2 flQ 2 = in SI units 2 gd 12.1d 5

With usual notations, for laminar flow,

f = 64

Re

where Re is the Reynolds number. In case of laminar flow 2

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

h f = kv n Where, K=

fl 2 gd

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

h f 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.

[email protected]

30

MAEER’S MIT, Pune

Civil Department Constant head supply water tank

Overflow Darcy Weisbach equation

Air-water inverted U-tube differential manometer

fLV 2 hf = 2gD

Air relief Valve

air

Upstream Flow Control Valve

hf

Rubber tube

h1

h2 Water

Downstream Flow Control Valve

Outlet 1

2 D=0.32 cm 50 40

Measuring Cylinder

30

L = 70 cm

20 10 0

Study of flow through pipes

[email protected]

31

MAEER’S MIT, Pune

Civil Department

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)

[email protected]

ν = ………….stokes

32

MAEER’S MIT, Pune

Civil Department

Observation Table: a) For Increasing flow: Sr.No

Head

Actual Discharge

Velocity

Reynolds

Friction

loss (hf)

measurement

V

Number

factor

cm of

(Qa)

Qa

Re

f

water

=

Vol.

Time

Qa

π d2 4

cm3

Sec.

cm3/s

cm/s

=vd/ν

=

Log10hf

Log10v

Log10Re

Log10f

hf ×2gd l v2

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

[email protected]

33

MAEER’S MIT, Pune

Civil Department

Calculations:--

1.

Q=

Vol = ________________________cc/sec time

2.

v=

Q = __________________________cm/sec. A

3.

Re =

4.

f =

vd

ν

= ________________

2 gd h f = _____________________________________ v2 l

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

h f α v .... from graph whereas in turbulent flow, h f α v .... 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

h f is estimated using Darcy-

Weisbach equation as

flv 2 hf = = _____________________________________ 2 gd = ……… cm.

[email protected]

34

MAEER’S MIT, Pune

Civil Department

Experiment No.

Date:

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 it in Darcy-Weisbach equation to estimate frictional head loss through the pipe.

hf =

flV 2 2 gD

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

f → f n Re,

ks Dÿ

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.

1

R − 2 log10 = 1.74 − 2 log10 1 + ks f

18.7 R Re

ks f

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

k 10 6 f = 0.0055 1 + 20000 s + D Re

1

3

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.

1 f

= 1.14 − 2 log10

k s 21.25 + D Re 0.9

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 ColebrookWhite Equation.

[email protected]

35

MAEER’S MIT, Pune

Civil Department

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/m

b.

kinematic viscosity of water at 27 C (ν) = 0.0085 cm /s.

3

0

2

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 = Vavg

2. Calculate flow

Re =

=

Q A

Vavg D

ν

3. Calculate relative roughness ratio

ks

D

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

1

R − 2 log10 = 1.74 − 2 log10 1 + k f s

18.7 R

ks

Re f

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.

[email protected]

36

MAEER’S MIT, Pune

Civil Department

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.

Vavg =

2.

Re =

Q =__________________________ A

Vavg D

ν

=_______________________________

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

R − 2 log 10 = 1.74 − 2 log 10 1 + ks f

1

18.7 R

ks f

Re

Re1 =_____________________________________

4. f from Moody equation

k 10 6 f m = 0.0055 1 + 20000 s + D Re

1

3

=____________________________________

5. Accuracy of Moody equation in percentage =

fm − f × 100 fm

=________________________ 6. f from Swami-Jain equation

1 f

= 1.14 − 2 log10

k s 21.25 + D Re 0.9

f =_________________________________________ 7.

Accuracy of Swami-Jain equation in percentage =

fs − f × 100 fs

=_______________________ 8.

[email protected]

Head

loss,

hf =

flQ 2 12.1D 5

=_____________________________________

37

MAEER’S MIT, Pune

Civil Department

/*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("--------------------------------------------------------------------------------"); [email protected]

38

MAEER’S MIT, Pune

Civil Department

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
39

MAEER’S MIT, Pune

Civil Department

{ 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(); }

[email protected]

40

MAEER’S MIT, Pune

Civil Department

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.= Moody Equation

=

Swami-Jain Equation=

0.03280198 0.03324313 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 = [email protected]

16.65223552 41

MAEER’S MIT, Pune

Civil Department

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

INPUT DATA

Water:: Discharge flowing through pipe (Q) in lps = 3 Kinematic viscosity of water in cm2/sec. = 0.0085 Pipe:: Diameter of given pipe section (D) in mm =20 Length of given pipe section (L) in m

=18

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

OUTPUT DATA

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

9.54929659

Re calculated from actual flow conditions = 224689.33142385 Re obtained from Colebrook-White Equation = 224689.33142385 Friction factor f from Colebrook-White Eq.= Moody Equation [email protected]

=

0.03477678 0.03500932 42

MAEER’S MIT, Pune

Civil Department

Swami-Jain Equation=

0.03489697

Accuracy of Moody Equation in percentage =

-0.66866007

Accuracy of Swami-Jain Eq. in percentage =

+0.34439360

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

145.47074637

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

INPUT DATA

Water:: Discharge flowing through pipe (Q) in lps = 4 Kinematic viscosity of water in cm2/sec. = 0.0085 Pipe:: Diameter of given pipe section (D) in mm =15 Length of given pipe section (L) in m

=20

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

OUTPUT DATA

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

22.63536968

Re calculated from actual flow conditions = 399447.70030907 [email protected]

43

MAEER’S MIT, Pune

Civil Department

Re obtained from Colebrook-White Equation = 399447.70030907 Friction factor f from Colebrook-White Eq.= Moody Equation

=

Swami-Jain Equation=

0.03803432 0.03779784 0.03809669

Accuracy of Moody Equation in percentage =

+0.62173993

Accuracy of Swami-Jain Eq. in percentage =

+0.16373547

Frictional Head Loss (hf) in m. of water = 1324.31266246 --------------------------------------------------------------------------------

INPUT DATA

Water:: Discharge flowing through pipe (Q) in lps = 5 Kinematic viscosity of water in cm2/sec. = 0.0085 Pipe:: Diameter of given pipe section (D) in mm =30 Length of given pipe section (L) in m

=10

Eq. sand grain roughness coeff.(ks) in mm = 0.15 -------------------------------------------------------------------------------ssnimbalkar@mitpune.com

44

MAEER’S MIT, Pune

Civil Department

OUTPUT DATA

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

7.07355303

Re calculated from actual flow conditions = 249654.81269317 Re obtained from Colebrook-White Equation = 249654.81269317 Friction factor f from Colebrook-White Eq.= Moody Equation

=

Swami-Jain Equation=

0.03073918 0.03136514 0.03086229

Accuracy of Moody Equation in percentage =

-2.03634316

Accuracy of Swami-Jain Eq. in percentage =

+0.39888324

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

[email protected]

26.13047561

45

MAEER’S MIT, Pune

Civil Department

Experiment No.

Date:

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:

fp =

∆p ρV∞ (1 − ε ) L Dp E 3 2

Where,

D p = Particle diameter

[email protected]

46

MAEER’S MIT, Pune

Civil Department

1000 cc

1000 cc

700 cc

Case I

Case II

Packings only

Packings + Water

Case III Water

Determining void fraction (e)

Packed Bed

[email protected]

47

MAEER’S MIT, Pune

Civil Department

E = void fraction =

fp

Volume of Fluid Total volume of packed bed

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.

V∞ =

Q A where: A=c/s area of packed bed. L = depth of bed. = fluid mass density.

Pressure drop across packed bed can be given as:

f p LρV∞ (1 − ε ) 2

∆p =

Dp E 3

The packed bed friction factor

fp =

150 Re p

for Re ≤ 10

f p = 1.75 fp =

f p can be evaluated by using Ergun equation:

for Re ≥ 1000

150 + 1.75 for Re between 10 and 1000 Re p

Where,

Re p = Packed bed Reynolds number =

D p ρV∞ (1 − ε ) µ

The above equation Indicates that the void fraction is very instrumental in calculations of pressure drop ∆p .

[email protected]

48

MAEER’S MIT, Pune

Civil Department

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 D p . 9. Assume same flow rate through the packed bed Q. 10. Find the superficial velocity.

V∞ =

Q

π D2 4

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 o

4. Kinematic viscosity of water at 27 c ( ) 5.

Dynamic viscosity of water ( )

3

kg/m

2

=0.0085 cm /s =…………

6. Volumetric rates of flow (Q) -7

3

-5

3

-5

3

i. Q1=____ X 10 m /s ii. Q2=____ X 10 m /s iii. Q3=____ X 10 m /s

[email protected]

49

MAEER’S MIT, Pune

Civil Department

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

Void Fraction

Avg.

(ml) 1 1

2

Raschig Ring

3 1 Bearl Saddle

2

2

3 1 Intalox Saddle

2

3

3

Sample Calculation: For --------------------------------------------------- packing 3

Q= ______________ m /sec.

V∞ =

Q

π D2 4

=________________________________

D p ρV∞

Re p =

(1 − ε ) µ

For the

=_____________________________

Re p obtained, using Ergun equation,

f p = _______________=_____________________________________ f p LρV∞ (1 − ε ) 2

∆p =

Dp E 3

[email protected]

=_____________________________________

50

MAEER’S MIT, Pune

Civil Department

Conclusion 1. The void fraction changes with the type of packing and manner of packings. 2. From values of packed bed friction factor

f p , 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 .

[email protected]

51

MAEER’S MIT, Pune

Civil Department

#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;

[email protected]

52

MAEER’S MIT, Pune

Civil Department

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); Rep2=Dp1*v2*rho/((1-e1)*mue); Rep3=Dp1*v3*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); Rep4=Dp2*v4*rho/((1-e2)*mue); Rep5=Dp2*v5*rho/((1-e2)*mue); Rep6=Dp2*v6*rho/((1-e2)*mue);

[email protected]

53

MAEER’S MIT, Pune

Civil Department

fp4=150/Rep4; fp5=1.74+(150/Rep5); fp6=1.75; temp=rho*(1-e2)*l/(Dp2*e2*e2*e2); 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); Rep7=Dp3*v7*rho/((1-e3)*mue); Rep8=Dp3*v8*rho/((1-e3)*mue); Rep9=Dp3*v9*rho/((1-e3)*mue); fp7=150/Rep7; fp8=1.74+(150/Rep8); fp9=1.75; temp=rho*(1-e3)*l/(Dp3*e3*e3*e3); 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 printf("\n\tRep1=%f

[email protected]

Berl Saddle

Rep1=%f

Intalox Saddle");

Rep1=%f",Rep1,Rep4,Rep7);

54

MAEER’S MIT, Pune printf("\n\tRep2=%f printf("\n\tRep3=%f

Civil Department

Rep2=%f Rep3=%f

Rep2=%f",Rep2,Rep5,Rep8); Rep3=%f",Rep3,Rep6,Rep9);

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\tfp2=%f

fp3=%f

fp2=%f",fp2,fp5,fp8);

printf("\n\tfp3=%f

fp3=%f

fp3=%f",fp3,fp6,fp9);

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);

printf("\n\tp2=%f

p2=%f

p2=%f",p2,p5,p8);

printf("\n\tp3=%f

p3=%f

p3=%f",p3,p6,p9);

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

[email protected]

55

MAEER’S MIT, Pune Rasching Ring

Civil Department

Berl Saddle

Rep1=1.544693

Intalox Saddle

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

fp2=2.516853

fp3=1.858679

fp2=2.053473

fp3=1.750000

fp3=1.750000

fp3=1.750000

fp1=39.184181

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

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

[email protected]

p3=298.529938

56

MAEER’S MIT, Pune

Civil Department

Experiment No.

Date:

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 3

( p- f) Dp /6=3 Where,

UDP

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

[email protected]

57

MAEER’S MIT, Pune

Civil Department

D DP

Steel ball accelerate under its own weight

W Equillibrium Condition Wooden board

FD

Steel ball moving with terminal velocity U glycerine

L Stoke's cylinder

Stoke's apparatus

[email protected]

58

MAEER’S MIT, Pune

Civil Department

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

U = U ' 1+

9D p 4D

+

2

9D p 4D

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=

3

1266.67 kg/m . 2

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

Observation Table: ’

Sr.

Dia. Of

Length

Time

U

Avg U

Corrected

FD

W

K

No.

steel

L (cm)

T (sec.)

=L/T

cm/s

U

(N)

(N)

W/FD

ball

cm/s.

cm/s

1 2 3

[email protected]

59

MAEER’S MIT, Pune

Civil Department

4 5 6 7 8 9

Calculations: 1. Calculate mass density of steel balls as

P=m/

3

( D p/6)=____________________________

3

2. Calculate Weight of each ball as W= ( D p/6)g( 3. Calculate FD=3

p

–

f)

=_____________________________

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.

[email protected]

60

MAEER’S MIT, Pune Experiment No.

Civil Department Date:

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 o

90 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.

[email protected]

61

MAEER’S MIT, Pune

Civil Department

Delivery pipe Delivery Valve Delivery Head HD Impeller

Eye Suction Head HS

Volute Casing

Suction pipe

Sump

Foot Valve

Strainer

Component parts of a Centrifugal Pump

[email protected]

62

MAEER’S MIT, Pune

Civil Department

Observation Table :Sr.

Suction

Delivary

Manometric

Actual discharge

No.

Head

Head

Head Hm

Measurement

Hs in

Hs in

Hd in

mm

m. of

Kg/cm

of Hg

water

2

Hd in

=Hs+Hd

h1

h

Q

m. of

m. of water

cm

cm

m /s

Pin

Pout

Kw

Kw

H/Ho

Q/Qo

Pin/Pin0

/

o

3

water

1 2 3 4 5 6 7 8 9 10

[email protected]

63

MAEER’S MIT, Pune

Civil Department

Formulae to be used: 1. Hs (m. of water) =

13.6 × H S (mm of mercury) 1000 2

2. Hd (m. of water) = Hd (kg/cm ) X 10 3. Zero of the notch = 5 cm.; head over notch, h=h1-5 4. Discharge, Q = 1.417(h) 5. Pout =

γ Q Hm 1000

2 .5

3

cm /s.

=9.81QHm kW

6. Efficiency = η =

Pout Pin

Sample Calculations 0

1. Discharge over 90 triangular notch 2.5

Q=1.417 X h

3

m /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 3

1.

At maximum efficiency

2.

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

o=---------

for the centrifugal pump the design discharge Qo =----------- m /sec.

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.

[email protected]

64

MAEER’S MIT, Pune

Civil Department

Experiment No.

Date:

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.

[email protected]

65

MAEER’S MIT, Pune vi)

Civil Department

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/ C/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:

20)

Suction:

Size/Rating.

21)

Discharge:

Size/Rating.

[email protected]

0

Rating/Max.

66

MAEER’S MIT, Pune

Civil Department

22)

Base Plate:

23)

Coupling:

Type/ manufacturer

24)

Strainer:

Size/diameter.

25)

Bearing Lubricant

Max. Pressure.

26)

Shaft seal:

Type.

27)

Materials: for

28)

i)

Case

ii)

Shaft

iii)

Impeller

iv)

Wear ring

v)

Liners

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-

3)

Vapour pressure

4)

Flow rating

2

NS/M Pa

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

[email protected]

67

MAEER’S MIT, Pune 11)

0

Civil Department 0

temperatures in F/ C 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

19)

Shaft seal type.

20)

Materials

Max. Pressure.

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

[email protected]

68