# Agitator Design

July 17, 2017 | Author: Deepika | Category: Screw, Turbine, Physical Quantities, Continuum Mechanics, Classical Mechanics

#### Short Description

Agitator Design...

#### Description

Project :

PEDC

Design Calculations for : Calc by N M Patel Checked by

500 KL SS Storage Tank

Job No. Date Date

22/11/10

DESIGN OF 500 KL SS STORAGE TANK QTY.

3

Design Calculation No:

FZD- 01

DATA Sr

Parameter

Nom

Unit

Value

1 Content

Ethylene glycol

2 Working Capacity

VL

KL/MT

15

3 Vessel Diameter

D

M

2.6

4 Height

H

M

2.9

5 Specific Gravity

G

6 Corrosion Allowance

C

1.2 mm

0

7 MOC

SS 316

8 Viscosity

u

9 Ratio ,H/D

Ni

CP

30 1.12

10 Sweep Diameter

Da

M

0.9

11 TIP Speed

Ts

M/s

2

Nb

14 Distance Between Impeller

Si

3 M

1.3

15 Ratio of Si/Da

1.50

17 Type of impeller 16 Nos of impeller

multiimpeller

17 Type of Seal

Mechanical seal LS

M

3.4

19 Weight of 1st impeller

W a1

Kg

18

20 Weight of 2nd impeller

W a2

Kg

18

18 Length of Shaft

18

Page 1 of 9

Nos of Impeller =

2 3

1 Speed of impeller:

N=

Where Da = Sweep Diameter,M

60 τ S

= D/3

ΠD a N=

44.1 33.0

RPM

2 Pumping Rate, Q in M3/sec

Q=N Q ND

a3

Q=

Impeller Type Nos of Baffles Propeller 0 Propeller 3 to 8 Turbine,VT blade 0 Turbine,VT blade 4 Pitch Turbine,45◦ 0 Pitch Turbine,45◦ 4 Anchor 0

0.9 Np 0.3 0.33 -0.37 0.93 -1.08 3 to 5 0.7 1.3-1.4 0.28

Where NQ = Pumping Number

12.9

NQ 0.3 0.4-0.55 0.33-0.34 0.7-0.85 0.3 0.6-0.87 0.28

0.6

M3/min

3 Mixing Time,in Min for 100 volume to be change

QS =100

( QV ) Qs =

116

min

4 Reynold Number,Re

ND Re =

a2

ρ

u =Viscosity in Kg.M/hr

μ =

108

16524

Hence,Re >10000 Flow is Turbulent Power Number for Reynold and For 3 bladed pitch turbine is Np = 5

0.8

Power Required : PL =Power loss , =

H . P=1. 1

[

FN P ρN 3 D S

735

= Provided H.P=

a

5

+ PL

]

0.41

H.P

1

H.P

0.2 0.5

For Mechanical Seal For Stuffing Box

F= Multi impellerFactor Pi / Da

0.1 to 0.5 0.5 to 1.0 1.0 to 1.5 1.5 to 2.0 2.0 to 3.0 3.0 to 4.0 >4

Page 2 of 9

1.6

Multi impeller factor

F 1.4 1.5 1.6 1.7 1.8 1.9 2

Page 3 of 9

40

mm

7 Diameter of Shaft,mm

d s =100

[

J m ( H . P ) ΣL D a NN i

ds = Provided ds =

]

0. 33

Drive length =

0.5

Ls=Length of Shaft ,M =

3.8

Where,Jm =Jamming Factor EL =Overhung length=

58.2 60

M

1.5 to 2 2

Jm = L1+L2

5.55 2.8 L1 =Distance from Centre of Bottom First Impeller to

mm mm

to Vessel Top Surface

=

3.4

M

L2 =Distance from Centre of BottomSecond Impeller to Vessel Top Surface =

2.15

M

H.P t b =555 NN i N b W

(

tb =

0. 5

)

+2CA

5.1

mm

3

mm

Provided tb = 9 Critical Speed,RPM Nc=

[

1 1 . 92 W S L

EI S3

2. 67

+Wa1 L

a1 2

( La 1 + L 1 ) +W a2 L a2 2 ( La 2 + L 1 ) +W c L c2 ( Lc + L 1 )

]

0. 5

¿ ¿ ¿¿ ¿ ¿

Where E=Modulas of Elasticity = I=Moment of Inertia,cm4 = W S=Weight of Shaft,Kg = = Bearing Span,L1

= W C=Weight of Coupling,Kg=

1800000

Kg/cm2

63.6

cm4

[dS2LS/159] 86.0

Kg

0.3

M

20

Kg

Lc =Distance from Centre of Coupling to tanks Top height = NC =

97.6

0.8

M

RPM

Hence,Operating Speed Should be

1.3*NC

=

126.9

RPM

OR 0.7*N ≤ C

=

68.3

RPM

Page 4 of 9

18

Wa1

18

Wa2

M

10

Deflection,mm:

A Deflection due to 1st Impeller Weight Weight of 1st impeller W a1 =

[ ][

δ 1=

Wa1 X 6 EI

=

2 L a1 L1 +3 L a1 X− X 2 ] 2.24225

= 0.0224225

18

Kg

X=Distance from centre of lower bearing to distance at which deflection is to be measured = 3.4 M 340 Deflection at the end of 1st impeller ,X=La1

cm

cm M

=

3.4

Bearing Span,L1 = =

M

340

0.3 30

M cm

cm

B Deflection due to 2nd Impeller Weight Weight of 2nd impeller W a2 =

2 L a 2 L1 + 3 Lalignl ¿ ¿¿ Wa2 X [¿ X −X 2] δ 2= 6 EI

]

=

0.59370

cm

=

0.005937

M

= 2.15 M 215 Deflection at the end of 2nd impeller ,X=La2

=

2.15

Bearing Span,L1 = = C Deflection due to Shaft Weight

[

δ s=

WsL

s

8 EI

3

]

=

3.69

cm

=

0.03693

M

D Deflection due to coupling weight

δ3=

[ ][ WcX 6 EI

=

2 Lc L1 +3L c X− X 2 ] 0.00000

M

E Critical Speed at Overhung,Nc

[

30 g N C= Π δ 1 +δ 2 +δ 3 +δ s =

Kg

X=Distance from centre of lower bearing to distance at which deflection is to be measured

a2

[

18

117

0.5

]

Rpm

Page 5 of 9

M

215

0.3 30

M cm

cm

cm

11 Critical speed of shaft when shaft supported at two end 1 Deflection due to Shaft Weight

δ S=

5WS L

S3

5.371

384 EI

mm

2 Deflection due to 1st Impeller Weight

δ 1=

W1 L

2

a 12

( Ls −La 1 )

0.255163

3 EIL S

mm

3 Deflection due to 2nd Impeller Weight

δ 2=

W2 L

2

a2 2

( LS −La 2 )

3 EIL s

1.736142

mm

4 Deflection due to 3rd Impeller Weight

δ3=

W3 L

2

a3 2

( L S −La 3 )

3 EIL S

mm

0.882918

5 Critical Speed ,Nc

[

30 g N C= Π δ 1 +δ 2 +δ 3 +δ s NC =

329.55

0.5

]

RPM

Hence,Operating Speed Should be

1.3*NC

=

428.4

RPM

OR ≤ 0.7*NC

=

230.7

RPM

Page 6 of 9

12 Design of Hub and Key Part Description

Yield stress Shear stress

MOC

σy in MPA

τ in MPA

Crushing stress σc in MPA

Shaft

SS-304

175

43.75

87.5

Hub

CS

180

45

90

Key

CS

180

45

90

Coupling

CS

180

45

90

bolt

CS

180

45

90

A Design of Hub 1 Outer Diameter of Hub,Dh in mm

Dh =2 d s

120

mm

2 Length of Hub,Lh in mm

Lh =1 .5 d s

90

mm

3 Induced shear stress in Hub 16TD h τ s= π D 4 −d 4

(

h

s

Torque

)

τs =

T = 215.981

0.679

Mpa

1 Width of key,W

15

mm

2 Thick of Key,t

15

mm

<

45

ok

B Design of Key For Square Key

3 Shaft strength reductin factor,e

e=1−0 .2

W h −1. 1 d d e=

0.8125

4 Length of Key,Lk

94

mm

5 Induced shear stress

τ is =

Torque

2T LK Wd s τis =

T=

4.14

215.98

Mpa

<

45

ok

Mpa

<

90

ok

6 Induced Crushing Stress

σ ic =

2T LK tds σic =

4.14

Page 7 of 9

N.M

N.M

Page 8 of 9

13 Design of Rigid coupling 1 Design of flange Hub Diameter,D

120

MM

Thick of flange,tf

30

mm

Checking for induced shear stress in flange

τ if =

2T πD 2 t f τif = 0.318444 Mpa

<

45

ok

2 Design for Bolts PCD of bolts,D1 = 3d

180

OD of Flange,D2 = 4d

mm

240

mm

3.36

mm

Nos of bolts n=

[

8T db= 4 πτ b nD 1

]

0. 5

db =

Bolt size 4nos of M6 size

Page 9 of 9

6