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

2. VALVE BODY 2

3. LOCKING RING SPRING WASHER 3

4. 4

5. 5

6. GUIDE BUSH GLAND BUSH 6

7. PACKING SPREADOR - LOWER PLUG PIN 7

8. PACKING SPREADOR UPPER VALVE STEM 8

9. 9

10. COMPONENT NO.COMPONENT NAME DVL TIME (min)PREPARED TIME (min) 1VALVE BODY8070.31 2LOCKING RING32.92 3SPRING WASHER2.50 4 VALVE PLUG37.5029.47 5SEAT RING12.5011.50 6GUIDE BUSH43.90 7PACKING SPREADOR – LOWER 2.502.34 8GLAND BUSH32.98 9PLUG PIN73.89 10VALVE STEM27.5016.65 11PACKING SPREADOR – UPPER 2.502.46 12BONNET PLAIN6533.51 TOTAL TIME247187.32 10

11. Study of ball valve 11

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 12

13.  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. 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-  Body  Bonnet  Trim  Ball and Seat  Stem and Sleeves  Actuator  Packing 14

15. Study of three different types of trims 15

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. 16

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. 17

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. 18

19.  Calculation of coefficient of resistance  Overall resistance coefficient 19

20. Figure 1 20

21.  Calculation of intersection area 21

22. 22

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 23

24. 24

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 25

26. RESISTANCE COEFFICIENT FOR 16 CAGE TRIM Project 2 26

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 27

28. Figure 28

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 29

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) 30

31. Calculation (Continued) Trim exit velocity is Calculated value for trim exit velocity is 94.63 ft/sec 31

32. Conclusion Exit flow velocity is allowable (less than 100 ft/s) Trim can not fit into the valve 32

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 33

34. Tortuous flow disc is type of trim which is used for controlling velocity and reducing pressure across the valve 34

35.  Cross section is ( 5 X 3) and ( 6 X 3) alternately  Value of resistance coefficient (K o ) = 121.99  Outlet flow area (a o ) = 7.2685 in 2  Exit flow velocity (V O ) calculation = 99.33 ft/s  Disc diameters (ID X OD) = (50φ X 140φ)  Outlet flow area for 14 discs = 7.03 in 2 Hence, outlet flow area for 14 discs is approximately equal with the calculation of outlet flow area from ‘K o ’. So, it is safe. 35

36.  Cross section at inlet is ( 5 X 3) and increasing to ( 5.1 X 3)  Value of resistance coefficient (K o ) = 121.99  Outlet flow area (a o ) = 7.2685 in 2  Exit flow velocity (V O ) = 99.33 ft/s  Disc diameters (ID X OD) = (50φ X 150φ)  Outlet flow area for 15 discs = 6.2 in 2 Hence, outlet flow area for 15 discs is not equal with the calculation of outlet flow area from ‘K o ’. So it is not recommended. 36

37.  Cross section at inlet is ( 5 X 3) and increasing to ( 5.2 X 3)  Value of resistance coefficient (K o ) = 121.99  Outlet flow area (a o ) = 7.2685 in 2  Exit flow velocity (V O ) = 99.33 ft/s 37

38.  Disc diameters (ID X OD) = (50φ X 153φ) Outlet flow area for 15 discs = 6.3 in 2  Disc diameters (50φ X 160φ) Outlet flow area for 12 discs = 7.03 in 2  Disc diameters (50φ X 155φ) Outlet flow area for 15 discs = 6.29 in 2  Disc diameters (50φ X 163φ) Outlet flow area for 15 discs = 7.55 in 2  Disc diameters (65φ X 178φ) Outlet flow area for 10 discs = 7.16 in 2 38

39. CASE I  Considering rectangular cross section  P 2 – P 1 = [ + h L ] h L =  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. 39

40.  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. 40

41.  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 41

42.  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. 42

43.  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. 43

44. DESIGN OF 5 CAGE TRIM WITH GAP BETWEEN TWO CAGES AND PRESSURE DROP CALCULATION 44

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 valve 3) To avoid the vibration  Use of the baffle plate 45

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 46

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 47

48. Input data  Flow rate (Q) = 265000 l/hr  Density of liquid (ρ) = 720 to 860 kg/m 3  Valve size = 4” 900 class  Required pressure drop – calculate maximum achievable pressure drop 48

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/v  Internal circumference of cage 1  No of holes per cage = Internal circumference of cage pitch  Total 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 49

50. Calculation Procedure  Pressure drop calculation: 1) head loss due to friction = k v o 2 / (2*g n ) 2) ΔP = ρ* g n [(z1-z2) + (v o 2 - v i 2 )/ (2*g n ) + h L ]  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 50

51. Case I  Flow is from inside to outside  Cage thickness = 4 mm  Gap between every two cages = 2 mm  No. of rows = 3 51

52. Diameters of cages Cage Internal Diameter (mm) Cage Thickness (mm) Outer Diameter (mm) Gap Between Cages (mm) Cage 19741052 Cage 210941172 Cage 312141292 Cage 413341412 Cage 51454153- 52

53. Result Sr.No. Abbrevi ation Cage 1 Gap 1- 2 Cage 2 Gap 2- 3 Cage 3 Gap 3- 4 Cage 4 Gap 4- 5 Cage 5 Total pressur e 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-- 3Pressure drop(psi) 112.85-117.05-116.93-104.85-100.97552.65 53

54. Case II  Flow is from outside to inside.  Cage thickness = 4 mm  Gap between every two cages = 2 mm  No. of rows = 3 54

55. Diameter of cages (case II) Cage Internal Diameter (mm) Cage Thickness (mm) Outer Diameter (mm) Gap Between Cages (mm) Cage 59741052 Cage 410941172 Cage 312141292 Cage 213341412 Cage 114541532 55

56. Result Sr.N o. Abbreviat ion Cage 1 Gap 1- 2 Cage 2 Gap 2- 3 Cage 3 Gap 3- 4 Cage 4 Gap 4- 5 Cage 5 Total press ure 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-- 3Pressure drop(psi) 96.41-107.51-116.93-117.08-117.40553.32 56

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 57

58. Diameter of cages Cage Internal Diameter (mm) Cage Thickness (mm) Outer Diameter (mm) Gap Between Cages (mm) Cage 59741052 Cage 410941172 Cage 312141292 Cage 213341412 Cage 11454153- 58

59. Result Sr. No. Abbre viation Cage 1 Gap 1-2 Cage 2 Gap 2-3 Cage 3 Gap 3-4 Cage 4 Gap 4-5 Cage 5 Total pres sure drop 1 Velocit y / hole in the cage( m/s) 30- - - - - 2 Velocit y in gap(m/ s) -1.58-1.72-1.89-2.11-- 3Pressu re drop(p si) 112.8 7 117.3 2 -117.3-117.2 6 -117.2 2 581.9 7 59

60. Baffle plate  To increase the pressure drop  Calculation of diameter of one hole 60

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. 61

62. 8” X 8” 62

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 63

64. Pressure drop  Total pressure drop across Total pressure drop across valve including baffle plate 593.28 (41.71 Kg/cm 2 ) 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 - 5Total pressure drop across trim 1Velocity / hole in the cage(m/s) 30- - - - - 2Velocity in gap(m/s) -0.46 25 -0.492 6 -0.527-0.566-- 3Pressure drop(psi) 97.64999.49- - - 495.63 (34.85 Kg/cm 2 ) 64

65. Comparison of results Sr. No.DescriptionCustomer ValuesDesigned values 1 Condition of Critical Flow for LiquidThe flow of liquid is critical The flow of liquid is subcritical 2 Liquid Critical pressure ratio factor (F F )0.09 3 Liquid pressure recovery factor (F L )0.940.52 4 Valve Flow Coefficient (C v )61.29106.84 5 Valve outlet velocity6.49 m/s 6 Trim exit velocity30 m/s 7 Rated valve flow coefficient (C V )85 8 Percentage opening of valve72.10 %125.69 % 9 Pressure drop ratio (x)0.810.24 10 Cavitation index (σ) based on upstream pressure of valve 0.0040.014 11 Cavitation index (σ) based on downstream pressure of valve -0.97-0.98 65

66. Conclusion  Calculated pressure drop 41.71 Kg/cm 2 (495.63 psi) does not satisfy the required value of pressure drop 137.78 kg/cm 2 (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”). 66

67. 4” X 6” 67

68. Introduction  Aim : to check whether five cage trim design is recommended for given valve size of 4” X6”  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 68

69. Pressure drop  Total pressure drop across valve including baffle plate = 594.88 (41.82 Kg/cm 2 ) 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 1Velocity / hole in the cage(m/s) 30- - - - - 2Velocity in gap(m/s) -0.66-0.72-0.78-0.85-- 3Pressure drop(psi) 96.40 7 100.28 7 -100.281-100.27 4 -100.26 6 497.517 (34.98 Kg/cm 2 ) 69

70. Comparison of results Sr. No.DescriptionCustomer ValuesDesigned values 1 Condition of Critical Flow for LiquidThe flow of liquid is critical The flow of liquid critical 2 Liquid Critical pressure ratio factor (F F ) 0.41 3 Liquid pressure recovery factor (F L )10.83 4 Valve Flow Coefficient (C v )33.01 5 Valve outlet velocity3.5 m/s 6 Trim exit velocity30 m/s 7 Rated valve flow coefficient (C V )60 8 Percentage opening of valve55.02 % 9 Pressure drop ratio (x)0.590.49 10 Cavitation index (σ) based on upstream pressure of valve 0.0310.037 11 Cavitation index (σ) based on downstream pressure of valve -0.97-0.96 70

71. Conclusion  A) Calculated pressure drop (41.82 Kg/cm 2 ) does not satisfy the required value of pressure drop (50.22 kg/cm 2 )  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”) 71

72. VALVE SIZE 4” X 6” (WITH MAXIMUM 33 M/S VELOCITY) 72

73. Introduction  Aim : to check whether five cage trim design is recommended for given valve size of 4” X 6”  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. 73

74. Pressure drop  Total pressure drop across valve including baffle plate is 724.76 psi (50.95 Kg/cm 2 ) Sr. No. Abbreviat ion 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- - - - - 2 Velocity in gap(m/s) -0.66-0.72-0.78-0.85-- 3Pressure drop(psi) 118.43121.38-121.37- -121.36603.91 (42.46 Kg/cm 2 ) 74

75. Comparison of results Sr. No. DescriptionCustomer ValuesDesigned values 1Condition of Critical Flow for LiquidThe flow of liquid is critical 2Liquid Critical pressure ratio factor (F F )0.41 3Liquid pressure recovery factor (F L )11 4Valve Flow Coefficient (C v )33.01 5Valve outlet velocity3.5 m/s 6Trim exit velocity30 m/s 7Rated valve flow coefficient (C V )60 8Percentage opening of valve55.02 % 9Pressure drop ratio (x)0.590.61 10Cavitation index (σ) based on upstream pressure of valve 0.031 11Cavitation index (σ) based on downstream pressure of valve -0.97 75

76. Conclusion  Calculated pressure drop 50.95 Kg/cm 2 (724.76 psi) satisfies the required value of pressure drop 50.22 kg/cm 2 (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) 76

77. VALVE SIZE 3” X 4” 77

78. Introduction  Aim : to check whether five cage trim design is recommended for given valve size of 3” X 4”  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 78

79. Pressure drop  Total pressure drop with baffle plate 753.11 psi (52.94 kg/cm 2 ) Sr. No Abbreviatio n 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 1Velocity / hole in the cage(m/s) 30- - - - - 2Velocity in gap(m/s) -0.85-0.93-1.02-1.14-- 3Pressure drop(psi) 119.58128.5-128.49-128.4 8 -128.46633.53 (44.54 kg/cm 2 ) 79

80. Comparison of results Sr. No.DescriptionCustomer ValuesDesigned values 1Condition of Critical Flow for LiquidThe flow of liquid is critical 2Liquid Critical pressure ratio factor (F F )0.95 3Liquid pressure recovery factor (F L )0.980.47 4Valve Flow Coefficient (C v )15.4029.11 5Valve outlet velocity12.52 m/s 6Trim exit velocity30 m/s 7Rated valve flow coefficient (C V )25 8Percentage opening of valve61.62 %116.44 % 9Pressure drop ratio (x)0.960.22 10Cavitation index (σ) based on upstream pressure of valve 1.034.40 11Cavitation index (σ) based on downstream pressure of valve 0.0353.40 80

81. Conclusion  Calculated pressure drop (52.94 Kg/cm 2 ) does not satisfy the required value of pressure drop (225.44 kg/cm 2 )  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” 81

82. VALVE SIZE 4” 82

83. Introduction  Aim : to check whether five cage trim design is recommended for given valve size of 4”  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 83

84. Pressure drop  Total pressure drop across valve including baffle plate = 989.01 psi (69.53 Kg/cm 2 ) Sr. No. AbbreviationCage 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 1Velocity / hole in the cage(m/s) 30- - - - - 2Velocity in gap(m/s) -0.18 65 -0.42-0.47-0.52-- 3Pressure drop(psi) 163.7165.3-165.39- - 825.17 (58.01 Kg/cm 2 ) 84

85. Comparison of results Sr. No.DescriptionCustomer ValuesDesigned values 1Condition of Critical Flow for LiquidThe flow of liquid is critical 2Liquid Critical pressure ratio factor (F F )0.79 3Liquid pressure recovery factor (F L )2.052.03 4Valve Flow Coefficient (C v )41.65 5Valve outlet velocity4.62 m/s 6Trim exit velocity30 m/s 7Rated valve flow coefficient (C V )34 8Percentage opening of valve122.48 % 9Pressure drop ratio (x)0.890.87 10Cavitation index (σ) based on upstream pressure of valve 0.002230.0022 11Cavitation index (σ) based on downstream pressure of valve -0.99 85

86. Conclusion  Calculated pressure drop 988.94 psi (69.53 Kg/cm 2 ) satisfies the required value of pressure drop 1004.44 psi (70.62 kg/cm 2 )  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”) 86

87. DESIGN OF AGITATOR SHAFT IN A REACTOR 87

88.  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 88

89. 89

90. Calculate 1. Power (P) 2. Continuous average rated torque (T c ) 3. Maximum torque (T m ) 4. Polar modulus of section of shaft cross section (Z p ) 5. Transient force (F m ) 6. Bending moment (M) 7. Equivalent bending moment (M e ) 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 (N c ) 90

91.  Diameter of agitator (D a ) = 1440 mm  Maximum speed (N)= 56 rpm = 0.933 rps  Density (ρ)= 1600 kg/m 3  Viscosity of liquid in vessel (μ)= 3000 cP  Gravitational acceleration (g c ) = 9.81 m/s 2  Tensile strength= 517 N/mm 2  Yield strength= 207 N/mm 2  Allowable tensile stress (f a )=172 N/mm 2  Shear stress of A182 F316 (forging) (f S )= 46.53 N/mm 2  Radius of blade (R b )= 0.58 m  Length of shaft (l)= 5.2 m  Power function (Φ or N p )= 0.56 91

92.  Power (P) P =  Continuous average rated torque (T c ) T c =  Maximum torque (T m ) T m = 1.5 * T c 92

93.  Polar modulus of section of shaft cross section (Z p ) Z p = = Π * d 3 /16  Transient force (F m ) F m =  Bending moment (M) M = F m * l 93

94.  Equivalent bending moment (M e ) M e = 0.5 [M + ]  Stress due to equivalent bending moment (f) f = M e * 1000 / Z Z = Π d 3 /32  Deflection of shaft (δ) (Considered as cantilever beam) δ = =  Deflection of shaft (δ) (Supported at both the ends) δ = =  Critical speed (N c ) N c = 94

95. Sr no. Diamete r (mm) Calculated Stress Allowa ble stress Deflecti on(cantil ever) (cm) Deflection (simply supported )(cm) Allowable deflection (cm) Status 159.441115.62172--1.42Not recommend ed 2110172 14.90.9331.42Recommen ded (for simply supported only) 3200301721.42- Recommen ded (for both ) 95

96. Sr. No. Diameter of shaft (mm) Stress due to equivalent bending moment (N/mm 2 ) Allowable tensile stress (N/mm 2 ) Deflection of shaft (mm) Remark 180.68721.5117231.6 Not recommended 2100378.9217212.90 Not recommended 3130172 4.51Recommended 4154.12103.51722.3Recommended 96