API 650-RF Pad Calculation

Input Data

Parameter

 

Uni ts ts

N27A/B/C/D

N25 A/B

1 Nozzle Mark 2 Nozzle Size

Dn

mm

100

100

3 Nozzle Outside diameter

Dno

mm

114.3

114.3

4 Schedule

SCH

SCH

120

120

5 Thickness as per schedule

tn

 

mm mm

11.13

11.13

of circular opening 6 Finished diameter of circular

dn

 

mm mm

92.04

92.04

7 Corrosion Allowance

c

mm

3.2

3 .2

8 Nozzle wall thickness

tnc

mm

7.93

7.93

9 Nozzle MOC

MOC

MOC

ASTM A106 Gr.B Gr.B

ASTM ASTM A106 Gr.B

10 Allowable stress in nozzle

Sn

kPa

117900

117900

11 Shell Diameter

D

mm

51600

51600

12 Shell/Roof  thickness including corrosion allowance

tw

mm

25

25

13 Shell/Roof  thickness in corroded condition

twc

mm

21.8

21.8

t=(T2/(StsE)+c

mm

4.583

4.583

15 Shell/Roof  MOC 16 Allowable Stress in Shell (Table 5‐2b)

MOC S

MOC kPa

A36 159958

A36 159958

of opening 17 Total internal pressure @centre of opening

P=P1+Pq

kPa

6

6

butt joint 19 When opening is in solid plate or in Cat B butt joint 20 RF Pad MOC

E' MOC

NO UNIT MOC

. 1 A36

. 1 A36

21 RF Pad thickness

tpad

mm

10

10

22 RF Pad diameter

Dpad

mm

350

350

tr

mm

1.383

1.383

mm mm

1.383

1.383

mm

54.5

54.5

m mm m

44.825

44.825

Ln

mm

44.825

44.825

Ar

mm

136.094

136.094

A1

mm

2009.898

2009.898

A2

mm

586.975

586.975

A4

mm

2357

2357

of reinforcement provided 8 Total area of reinforcement

A=A1+A2+A4

mm

4953.874

4953.874

9 Is reinforcement provided is adequate or not  

Yes/No

YES

YES

14

 

Thk of wall of  wall plate required by CI 5.10.3 for the Latitudinal unit forces T2 (From design Calculation)

‐

Calculations as per API 620 CI 5.16.5

Net thickness required for a seamless tank wall at the centre of opening of opening (exclusive of  1 corrosion allowance) tr=t‐c 2

3

Net thickness required for nozzle neck(exclusive of corrosion of corrosion allowance) trn=Px(Dt+c)

/ 2xSxE

trn

Length of nozzle of nozzle neck within the limits of reinforcement of  reinforcement

Ln1

Ln =

smaller 2.5(tw‐C) or 2.5(tn‐C)+tw

Ln2

 

 

Area of reinforcement of reinforcement required at the vertical centreline of through of  through opening Ar=(dn+2C)(t‐C)E' 5 6 7

2

Area of reinforcement of reinforcement due to excess thickness in the tank wall AI=(dn+2C)(tw‐C‐tr)

2

Area of reinforcement of reinforcement due to excess thickness in the nozzle neck A2=2Ln(tn‐C‐trn)

2

Area of reinforcement of reinforcement due to excess thickness in the reinforcement pad A4=tpad(Dpad‐Dno)

2