BY WADHONKAR PAVANKUMAR D. BABHALE NIKHIL C. ASWAR RAVINDRA D. RACHATTE GIRISH M. 1.

1

ANALYSIS OF PROCESSES OF MANUFACTURING OF

GLOBE VALVEBY

WADHONKAR PAVANKUMAR D.BABHALE NIKHIL C.

ASWAR RAVINDRA D.RACHATTE GIRISH M.

2

VALVE BODY

3

LOCKING RING SPRING WASHER

4VALVE PLUG

5SEAT RING

6

GUIDE BUSH GLAND BUSH

7

PACKING SPREADOR - LOWER

PLUG PIN

8

PACKING SPREADOR UPPER

VALVE STEM

9Bonnet Plain

10

Summary of machining time

COMPONENT NO. COMPONENT NAME

DVL TIME (min) PREPARED TIME (min)

1 VALVE BODY 80 70.31

2 LOCKING RING 3 2.92

3 SPRING WASHER 2.50 2.50

4 VALVE PLUG 37.50 29.47

5 SEAT RING 12.50 11.50

6 GUIDE BUSH 4 3.90

7 PACKING SPREADOR – LOWER

2.50 2.34

8 GLAND BUSH 3 2.98

9 PLUG PIN 7 3.89

10 VALVE STEM 27.50 16.65

11 PACKING SPREADOR – UPPER

2.50 2.46

12 BONNET PLAIN 65 33.51

TOTAL TIME 247 187.32

11

Assignment 1Study of ball valve

12

A ball valve is a valve that opens by turning a handle attached to a ball inside the valve

The ball in the ball valve has a hole or port, through the middle

When the valve is closed, the hole is perpendicular to the ends of the valve

Introduction

Durable and usually work to achieve perfect shutoff

They are easy to repair operate, manually or by actuators

From the point of view of sealing, the concept of ball valve is the excellent.

Importance and applications

13

14

A multitude of ball valve types and designs safely accommodate a wide variety of industrial applications. Regarding of type, all ball valves have the following basic parts-BodyBonnetTrimBall and SeatStem and SleevesActuatorPacking

Ball valve parts

15

Assignment 2Study of three different types of trims

16

The internal elements of a valve are collectively referred to as a valve's trim.

The trim typically includes a disk, seat, stem & sleeves needed to guide the stem.

A valve's performance is determined by the disk and seat interface and the relation of the disk position to the seat.

Because of the trim, basic motions and flow control are possible.

What is Trim ?

17

Aim : To calculate trim exit flow area and kinetic energy density at the outlet

To avoid the noise, erosion, cavitation, vibration in the valve

Therefore velocity must be reduced.

Introduction

18

What is flow coefficient ?The flow coefficient of a device is a relative measure of its efficiency at allowing fluid flow.

What is resistance coefficient of the valve ?A dimensionless number used in the study of flow resistance, equal to the resistance force in flow divided by one-half the product of fluid density, the square of fluid velocity, and the square of a characteristic length.

Introduction

19

Calculation of coefficient of resistance

Overall resistance coefficient

Cage trim with venturi shaped slots

20

Figure 1

21

Calculation of intersection area

Concentric cage trim with offset drilled holes

22

Figure 2

23

Introduction to trim with right angle expanding turns

Calculation of resistance coefficient

Value of resistance coefficient is more comparing to the other two methods

Trim with expanding right angle turns

24

Figure 3

25

Venturi valve trim is not compatible for higher pressure drops

Concentric cage trim requires more space to get more pressure drop

Compatible for low space and more pressure drop

Conclusion

26

RESISTANCE COEFFICIENT FOR 16

CAGE TRIM

Project 2

27

Introduction

Aim : To calculate the coefficient of resistance,

to calculate trim exit area and trim exit velocity

Type of trim used concentric cage trim with offset drilled holes with 16 cages

Significance : calculation of pressure drop in the trim

28

Figure

29

Calculation procedure

Values for area are taken proportionally with respect to given input from example

The formula to calculate the resistance coefficient is modified

30

Calculation (Continued)

Calculation of resistance coefficient

Value of resistance coefficient calculated from our calculation is 104.22

Outlet flow area

Calculated value for outlet flow area is 6.72 (in * in)

31

Calculation (Continued)

Trim exit velocity is

Calculated value for trim exit velocity is 94.63 ft/sec

32

Conclusion

Exit flow velocity is allowable (less than 100 ft/s)

Trim can not fit into the valve

33

Calculation for 5 cages

Calculation of resistance coefficient with same procedure ( Ko = 9.5)

Trim exit velocity was not in the safe range (Vo = 313.18 ft/s)

Both cases are not recommended

34

DESIGN OF TRIM WITH 10 EXPANDING RIGHT ANGLE TURNS

Tortuous flow disc is type of trim which is used for controlling

velocity and reducing pressure across the valve

35

CASE I Cross section is ( 5 X 3) and ( 6 X 3)

alternately Value of resistance coefficient (Ko) = 121.99 Outlet flow area (ao) = 7.2685 in2 Exit flow velocity (VO) calculation = 99.33

ft/s Disc diameters (ID X OD) = (50φ X 140φ) Outlet flow area for 14 discs = 7.03 in2

Hence, outlet flow area for 14 discs is approximately equal with the calculation of outlet flow area from ‘Ko’. So, it is safe.

36

CASE II Cross section at inlet is ( 5 X 3) and

increasing to ( 5.1 X 3) Value of resistance coefficient (Ko) =

121.99 Outlet flow area (ao) = 7.2685 in2 Exit flow velocity (VO) = 99.33 ft/s Disc diameters (ID X OD) = (50φ X 150φ) Outlet flow area for 15 discs = 6.2 in2

Hence, outlet flow area for 15 discs is not equal with the calculation of outlet flow area from ‘Ko’. So it is not recommended.

37

CASE III Cross section at inlet is ( 5 X 3) and

increasing to ( 5.2 X 3) Value of resistance coefficient (Ko) =

121.99 Outlet flow area (ao) = 7.2685 in2 Exit flow velocity (VO) = 99.33 ft/s

38

SUMMERY OF CASE III Disc diameters (ID X OD) = (50φ X 153φ)

Outlet flow area for 15 discs = 6.3 in2

Disc diameters (50φ X 160φ)Outlet flow area for 12 discs = 7.03 in2




39

CALCULATION OF PRESURRE DROPCASE I Considering rectangular cross section P2 – P1 = [ + hL]

hL =

Pressure drop = 159.36 psi Pressure drop for 10 turns = 10 X 159.36 =

1593.6 psi We required pressure drop of 1820 psi in 10

turns and calculated is 1593.6 psi.

40

CASE II In this case input flow rate value is

changed. Pressure drop = 140 psi Pressure drop for 10 turns = 10 X 140 =

1400 psi We required pressure drop of 1820 psi in

10 turns and calculated is 1400 psi.

41

CASE III To calculate pressure drop for second

right angle turn Section 3 – (5.2 X 3) Here, width – 5.2 mm and depth – 3 mm Section 6 – (6.24 X 3) Pressure drop = 114.84 psi

42

CASE IV With the same cross sections as case III

Keep velocity less than 30 m/s in the trim

Pressure drop = 46.4 psi Pressure drop for 10 turns = 10 X 46.4 =

464 psi We required pressure drop of 1820 psi in

10 turns and calculated is 464 psi.

43

CONCLUSION Any case is not recommended as

pressure drop required is less than required.

This type of trim design can be used for valves requiring less pressure drops.

44

PROJECT 4

DESIGN OF 5 CAGE TRIMWITH GAP BETWEEN TWO CAGESAND PRESSURE DROP CALCULATION

45

Introduction

Aim of the project : Design the five cage trim to get the pressure drop of 700 psi at the outlet of trim

Significance :1) To reduce the noise 2) To avoid wear and tear of the valve3) To avoid the vibration

Use of the baffle plate

46

Design Aspects

The velocity through the one hole of the cage should be maximum 30 m/s

Pressure drop decides the no. of cages and there is a limit for maximum no of cages.

Direction of the flow should be decided

47

Sequence of Calculation

Calculate velocity in the valve Calculate diameters of cages Calculate diameter of one hole in the cage Calculate area of the gap between the cages &

velocity at that section Calculate pressure drop in every cage of trim &

Hence pressure drop across the trim Calculate pressure drop across the baffle plate

connected Calculate total pressure drop across the valve

48

Input data

Flow rate (Q) = 265000 l/hr Density of liquid (ρ) = 720 to 860

kg/m3

Valve size = 4” 900 class Required pressure drop – calculate

maximum achievable pressure drop

49

Calculation Procedure

Inlet velocity

v1 = Q/A1 Diameter of the cages Calculation of diameter of one hole in the cage

Total area provided in cage 1 = Q/vInternal circumference of cage 1No of holes per cage = Internal circumference of

cage pitchTotal no of holes in one cage

= No of holes per cage * No. of rows Area of one hole in first cage = Total area provided in

cage 1

Total no of holes in one cage

50


Pressure drop calculation:

1) head loss due to friction = k vo2/ (2*gn)

2) ΔP = ρ* gn [(z1-z2) + (vo2- vi

2)/ (2*gn) + hL]

Velocity in the gap:

1) Area of gap between two cages

2) Flow through gap = total flow

no. Of holes in the cage 1

3) Velocity in the gap = flow through gap

area of gap

51

Case I

Flow is from inside to outside Cage thickness = 4 mm Gap between every two cages = 2 mm No. of rows = 3

52

Diameters of cages Cage Internal

Diameter (mm)Cage Thickness

(mm)Outer Diameter

(mm)Gap Between Cages (mm)

Cage 1 97 4 105 2

Cage 2 109 4 117 2

Cage 3 121 4 129 2

Cage 4 133 4 141 2

Cage 5 145 4 153 -

53

ResultSr.No. Abbrevia

tionCage

1Gap 1-

2Cage 2 Gap 2-

3Cage 3 Gap 3-

4Cage 4 Gap 4-

5Cage 5 Total

pressure drop

1 Velocity / hole (m/s)

30 - 29.97 - 29.69 - 28.36 - 27.82 -

2 Velocity in gap(m/s)

- 1.71 - 0.76 - 1.39 - 0.63 - -

3 Pressure drop(psi)

112.85 - 117.05 - 116.93 - 104.85 - 100.97 552.65

54

Case II

Flow is from outside to inside. Cage thickness = 4 mm Gap between every two cages = 2 mm No. of rows = 3

55

Diameter of cages (case II)

Cage Internal Diameter

(mm)

Cage Thickness

(mm)

Outer Diameter

(mm)

Gap Between

Cages (mm)

Cage 5 97 4 105 2

Cage 4 109 4 117 2

Cage 3 121 4 129 2

Cage 2 133 4 141 2

Cage 1 145 4 153 2

56

Result

Sr.N

o.

Abbreviati

on

Cage

1

Gap 1-

2

Cage

2

Gap 2-

3

Cage

3

Gap 3-

4

Cage

4

Gap 4-

5

Cage

5

Total

pressu

re drop

1 Velocity /

hole (m/s)

27.82 - 28.39 - 23.69 - 29.97 - 30 -

2 Velocity in

gap(m/s)

- 1.24 - 0.70 - 1.5 - 0.84 - -

3 Pressure

drop(psi)

96.41 - 107.51 - 116.93 - 117.08 - 117.40 553.32

57

Case III

Flow is from outside to inside. All cage (1-5) hole diameters are same

(constant) & its velocity is limited to 30 m/sec (max)

Cage thickness = 4 mm Gap between every 2 cages = 2 mm Pitch of the hole in first cage (Horizontally) =

23.43 mm Pitch of the hole (Vertically) = 13 mm No. of rows = 4 No. of holes in each cage = 52

58

Diameter of cages

Cage Internal Diameter

(mm)

Cage Thickness

(mm)

Outer Diameter

(mm)

Gap Between Cages (mm)

Cage 5 97 4 105 2

Cage 4 109 4 117 2

Cage 3 121 4 129 2

Cage 2 133 4 141 2

Cage 1 145 4 153 -

59

Result

Sr.No.

Abbreviation

Cage 1

Gap 1-2

Cage 2

Gap 2-3

Cage 3

Gap 3-4

Cage 4

Gap 4-5

Cage 5

Total pressure drop

1 Velocity / hole in the cage(m/s)

30 - 30 - 30 - 30 - 30 -


- 1.58 - 1.72 - 1.89 - 2.11 - -


112.87

117.32

- 117.3 - 117.26

- 117.22

581.97

60

Baffle plate

To increase the pressure drop Calculation of diameter of one hole

61

Conclusion

For case I flow is from inside to outside which is not suitable for liquids as it may cause damage to valve plug and also less pressure drop so this case is not recommended for liquids.

In case II flow is from outside to inside but the design is not convenient to manufacture also less pressure drop so this case is not recommended.

Pressure drop gain is good in this case & it is convenient to manufacture. So this case is recommended.

62

ASSIGNMENT 48” X 8”

63

Introduction

Aim : to check whether five cage trim design is recommended for given valve size of 8” X 8”

Design of five cage trim Valve flow coefficient Calculation sheet for

design data Valve flow coefficient Calculation sheet for

customer data Comparison between customer data and

design data

64

Pressure drop

Total pressure drop across Total pressure drop across valve including baffle plate 593.28 (41.71 Kg/cm2)

Sr. No.

Abbreviation

Cage - 1

Gap 1-2

Cage - 2

Gap 2-3

Cage – 3

Gap 3-4

Cage - 4

Gap 4-5

Cage - 5 Total pressure

drop across trim


30 - 30 - 30 - 30 - 30 -


- 0.4625

- 0.4926

- 0.527 - 0.566 - -


97.649 99.49 - 99.49 - 99.49 - 99.49 495.63 (34.85

Kg/cm2)

65

Comparison of resultsSr. No. Description Customer Values Designed values

1 Condition of Critical Flow for Liquid The flow of liquid is critical

The flow of liquid is subcritical

2 Liquid Critical pressure ratio factor (FF) 0.09 0.09

3 Liquid pressure recovery factor (FL) 0.94 0.52

4 Valve Flow Coefficient (Cv) 61.29 106.84

5 Valve outlet velocity 6.49 m/s 6.49 m/s

6 Trim exit velocity 30 m/s 30 m/s

7 Rated valve flow coefficient (CV) 85 85

8 Percentage opening of valve 72.10 % 125.69 %

9 Pressure drop ratio (x) 0.81 0.24

10 Cavitation index (σ) based on upstream pressure of valve

0.004 0.014

11 Cavitation index (σ) based on downstream pressure of valve

-0.97 -0.98

66

Conclusion

Calculated pressure drop 41.71 Kg/cm2 (495.63 psi) does not satisfy the required value of pressure drop 137.78 kg/cm2 (593.28 psi)

Calculated valve flow coefficient (106.84) is greater than rated valve flow coefficient (85)

Hence 5 cage trim design is not recommended for case I (For valve size 8” X 8”).

67

ASSIGNMENT 54” X 6”

68

Introduction

Aim : to check whether five cage trim design is recommended for given valve size of 4” X6”




design data

69

Pressure drop

Total pressure drop across valve including baffle plate = 594.88 (41.82 Kg/cm2)

Sr.No.

Abbreviation

Cage - 1

Gap 1-2

Cage - 2

Gap 2-3

Cage - 3

Gap 3-4

Cage - 4

Gap 4-5

Cage - 5

Total pressure drop across

trim


30 - 30 - 30 - 30 - 30 -


- 0.66 - 0.72 - 0.78 - 0.85 - -


96.407

100.287

- 100.281 - 100.274

- 100.266

497.517 (34.98 Kg/cm2)

70



The flow of liquid critical

2 Liquid Critical pressure ratio factor (FF)

0.41 0.41

3 Liquid pressure recovery factor (FL) 1 0.83








0.031 0.037


-0.97 -0.96

71

Conclusion

A) Calculated pressure drop (41.82 Kg/cm2) does not satisfy the required value of pressure drop (50.22 kg/cm2)

B) Hence 5 cage trim design does not meet required pressure drop.

Hence 5 cage trim design is not recommended for case II. (Valve size 4” X 6”)

72

ASSIGNMENT 6

VALVE SIZE 4” X 6”(WITH MAXIMUM 33 M/S VELOCITY)

73

Introduction


Design of five cage trim Valve flow coefficient Calculation sheet for design

data Valve flow coefficient Calculation sheet for customer

data Comparison between customer data and design data As calculated pressure drop in assignment 5 is

nearly equal to required pressure drop, we are making velocity

33 m/s to meet that pressure drop.

74

Pressure drop

Total pressure drop across valve including baffle plate is 724.76 psi (50.95 Kg/cm2)

Sr.

No.

Abbreviati

on

Cage -

1

Gap

1-2

Cage -

2

Gap

2-3

Cage -

3

Gap

3-4

Cage

- 4

Gap

4-5

Cage -

5

Total pressure

drop across trim

1 Velocity /

hole in the

cage(m/s)

33 - 33 - 33 - 33 - 33 -

2 Velocity in

gap(m/s)

- 0.66 - 0.72 - 0.78 - 0.85 - -

3 Pressure

drop(psi)

118.43 121.38 - 121.37 - 121.3

7

- 121.36 603.91 (42.46

Kg/cm2)

75

Comparison of results

Sr. No. Description Customer Values Designed values


The flow of liquid is critical


3 Liquid pressure recovery factor (FL) 1 1








0.031 0.031


-0.97 -0.97

76

Conclusion

Calculated pressure drop 50.95 Kg/cm2 (724.76 psi) satisfies the required value of pressure drop 50.22 kg/cm2 (714.29 psi)

Calculated valve flow coefficient (33.01) is less than rated valve flow coefficient (60)

Hence 5 cage trim design is recommended (For valve size 4” X 6” with velocity 33 m/s maximum)

77

ASSIGNMENT 7VALVE SIZE 3” X 4”

78

Introduction





design data

79

Pressure drop

Total pressure drop with baffle plate 753.11 psi (52.94 kg/cm2)

Sr.No

Abbreviation Cage - 1

Gap 1-2

Cage - 2

Gap 2-3

Cage - 3

Gap 3-4

Cage - 4

Gap 4-5

Cage - 5

Total pressure drop


30 - 30 - 30 - 30 - 30 -


- 0.85 - 0.93 - 1.02 - 1.14 - -


119.58 128.5 - 128.49 - 128.48

- 128.46 633.53 (44.54 kg/cm2)

80













1.03 4.40


0.035 3.40

81

Conclusion

Calculated pressure drop (52.94 Kg/cm2) does not satisfy the required value of pressure drop (225.44 kg/cm2)

Value of calculated flow coefficient (29.11) is greater than rated value of flow coefficient (25)

Hence 5 cage trim design is not recommended for valve size 3” X 4”

82

ASSIGNMENT 8VALVE SIZE 4”

83

Introduction

Aim : to check whether five cage trim design is recommended for given valve size of 4”




design data

84

Pressure drop

Total pressure drop across valve including baffle plate = 989.01 psi (69.53 Kg/cm2)

Sr.No.

Abbreviation Cage 1

Gap 1-2

Cage 2

Gap 2-3

Cage 3

Gap 3-4

Cage 4

Gap 4-5

Cage 5

Total pressure drop across trim

1 Velocity / hole in the

cage(m/s)

30 - 30 - 30 - 30 - 30 -


- 0.1865

- 0.42 - 0.47 - 0.52 - -


163.7 165.3 - 165.39 - 165.39

- 165.39

825.17 (58.01 Kg/cm2)

85













0.00223 0.0022


-0.99 -0.99

86

Conclusion

Calculated pressure drop 988.94 psi (69.53 Kg/cm2) satisfies the required value of pressure drop 1004.44 psi (70.62 kg/cm2)

Value of calculated flow coefficient (41.45) is nearer to rated flow coefficient value (34 approximate)

Hence, five cage trim design is recommended for (Valve size 4”)

87

Project 5

DESIGN OF AGITATOR SHAFT IN A REACTOR

88

Introduction

Aim : To design the shaft such that the deflection in the shaft is within range

Design includes calculation of power of the motor

Critical speed calculation

89

Figure

90

Sequence of Calculation

Calculate1. Power (P)2. Continuous average rated torque (Tc)3. Maximum torque (Tm)4. Polar modulus of section of shaft cross section (Zp)5. Transient force (Fm)6. Bending moment (M)7. Equivalent bending moment (Me)8. Stress due to equivalent bending moment (f)9. Deflection of shaft (δ) (Considered as cantilever beam)10. Deflection of shaft (δ) (Supported at both the ends)11. Critical speed (Nc)

91

INPUT DATA Diameter of agitator (Da) = 1440

mm Maximum speed (N) = 56 rpm =

0.933 rps Density (ρ) = 1600 kg/m3 Viscosity of liquid in vessel (μ) = 3000

cP Gravitational acceleration (gc) = 9.81

m/s2 Tensile strength = 517 N/mm2 Yield strength = 207 N/mm2 Allowable tensile stress (fa) =172 N/mm2

Shear stress of A182 F316 (forging) (fS) = 46.53 N/mm2

Radius of blade (Rb) = 0.58 m Length of shaft (l) = 5.2 m Power function (Φ or Np) = 0.56

92


Power (P)P =

Continuous average rated torque (Tc)Tc =

Maximum torque (Tm)Tm = 1.5 * Tc

93

Calculation Procedure (continued)

Polar modulus of section of shaft cross section (Zp)Zp = = Π * d3/16

Transient force (Fm)

Fm =

Bending moment (M)M = Fm* l

94

Calculation Procedure (continued) Equivalent bending moment (Me)

Me = 0.5 [M + ] Stress due to equivalent bending moment (f)

f = Me * 1000 / ZZ = Π d3/32

Deflection of shaft (δ) (Considered as cantilever beam)δ = =

Deflection of shaft (δ) (Supported at both the ends)δ = =

Critical speed (Nc)Nc =

95

Conclusion Sr no.

Diameter (mm)

Calculated Stress

Allowable stress

Deflection(cantilever) (cm)

Deflection(simply supported)(cm)

Allowable deflection(cm)

Status

1 59.44 1115.62 172 - - 1.42 Not recommended

2 110 172 172 14.9 0.933 1.42 Recommended (for simply supported only)

3 200 30 172 1.42 - 1.42 Recommended (for both )

96

For 16 HP Power

Sr. No. Diameter

of shaft

(mm)

Stress due to

equivalent bending

moment (N/mm2)

Allowable

tensile stress

(N/mm2)

Deflection of

shaft (mm)

Remark

1 80.68 721.51 172 31.6 Not

recommended

2 100 378.92 172 12.90 Not

recommended

3 130 172 172 4.51 Recommended

4 154.12 103.5 172 2.3 Recommended

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BY WADHONKAR PAVANKUMAR D. BABHALE NIKHIL C. ASWAR RAVINDRA D. RACHATTE GIRISH M. 1.

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