ASME SECTION VIII Div.1

SHELL CALCULATIONS NOZZLE CALCULATIONS

CIRCUMFERENTIAL STRESSES CIRCUMFERENTIAL STRESSESVALUES VALUES

Maximum Allowable Working Pressure (P): 300 Maximum Allowable Working Pressure (P): 300

Maximum Allowable Stress Value (S): 17500 Maximum Allowable Stress Value (S): 17500Height: 75

Joint Efficiency: 1 Joint Efficiency: 1 Static Head 32.5 psi

Inside Diameter of Shell: 69.875 34.9375 inside R Outside Diameter of Nozzle: 6.625 2.7125 Rn

Corrosion Allowance: Corrosion Allowance:

Given total wall thickness: 0.653 0.6530 Nom-CA Given total wall thickness: 0.6 0.6000 Nom-CA

FOR CIRCUMFERENTIAL STRESS (Longitudinal Joints): FOR CIRCUMFERENTIAL STRESS (Longitudinal Joints):

t = PR/SE - 0.6P P = SEt/R + 0.6t t = PR/SE - 0.6P P = SEt/R + 0.6t

t =: 0.6052 inches P =: 323 psig t =: 0.0470 inches P =: 3417 psig

FOR LONGITUDINAL STRESS (Circumferential Joints): FOR LONGITUDINAL STRESS (Circumferential Joints):

t = PR/2SE + 0.4P P = 2SEt/R - 0.4t t = PR/2SE + 0.4P P = 2SEt/R - 0.4t

t =: 0.2984 inches P =: 659 psig t =: 0.0232 inches t =: 8493 psig 0.605150.0470

The minimum required shell thickness for a MAWP of 300 psig The minimum required nozzle thickness for a MAWP of 300 psig 0.6052

is 0.6052 inches, including CA. is 0.0470 inches, including CA. 0.0470

The MAWP with a shell 0.6530 inches, including CA The MAWP with a shell 0.6000 inches, including CA

thick is: 323 psig. thick is: 3417 psig.

NOZZLES

Required thickness of shell (tr): 0.6052 L1 = 5.4250 3.9655L2 = 1.6325 2.2500

Required thickness of nozzle (trn): 0.0470

Diameter of circular opening (d): 5.4250 Nozzle: Inserted (Inserted or Abutting)

Correction Factor (F): 1

Nominal thickness of nozzle wall (tn): 0.6000 Actual tn: 0.60000

Stress value of nozzle (Sn): 17500

Stress value of shell (Sv): 17500 1.25

Stress value of reinforcment (Sp): 12500

Inside shell diameter (D): 69.8750

Outside diameter of reinforcement (Dp): 10 Actual Dp: 10.0000 5.42505.4250

Thickness of reinforcement (te): 0.75 Actual te: 0.7500 0.751.6325

Efficiency (E1): 1.0000

Nominal thickness of shell (t): 0.6530 Actual t: 0.65300

Nozzle projection (h): 0

Corrosion allowance: 0.0000

Top nozzle weld size: 0.75 Actual Weld: 0.7500 1.5 0.75001.6325

Outer reinforcement weld size: 0.5 0.4250 0.4250

Inward nozzle weld: 0.0000

Area required: dtrF+2tntrF(1-fr1) A=: 3.2832 SQUARE INCHES

Area: sheld(Et-Ftr)-2tn(Et-Ftr)(1-fr1) A1=: 0.2593 0.2593or 2(t+tn)(Et-Ftr)-2tn(Et-Ftr)(1-fr1) 0.1198

Area: nozzle 5(tn-trn)fr2t A2=: 1.6590 1.8055or 5(tn-trn)fr2tn 1.659

Area: inward nozzle 2(tn-c)fr2h A3=: 0.0000

Area:nozzle weld ouleg*leg*fr2 A41=: 0.5625

Area: nozzle weld inleg*leg*fr2 A43=: 0.0000Nozzle is acceptable.

FIRST CALCULATIONS Reinforcement is required.More reinforcement is required

A1 + A2 + A3 + A41 + A43=: 2.4808

Reinforcement is required.

REINFORCEMENT CALCULATIONS

Area required: dtrF+2tntrF(1-fr1) A=: 3.2832 SQUARE INCHES

Area: shell: d(Et-Ftr)-2tn(Et-Ftr)(1-fr1 A1=: 0.2593 0.2593or 2(t+tn)(Et-Ftr)-2tn(Et-Ftr)(1-fr1) 0.1198

Area: nozzle: 5(tn-trn)fr2t A2=: 1.8055 1.8055or 2(tn-trn)(2.5tn+te)fr2 2.4885

Area: inward nozzle 2(tn-c)fr2h A3=: 0.0000

Area nozzle out wel leg*leg*fr3 A41=: 0.4018 fr1 = 1.0000 1 1fr2 = 1.0000 1

Area out rein. weld: leg*leg*fr4 A42=: 0.1290 fr3 = 0.7143 0.7143fr4= 0.7143 0.7143

Area in nozzle weld leg*leg*fr2 A43=: 0.0000

Area in repad (Dp-d-2tn)tefr4 A5=: 1.8081

A1 + A2 + A3 + A41 + A42 + A43 + A5 =: 4.4037

Nozzle is acceptable.

High Temp 650 High Stress 18.8Low Temp 600 Low Stress 19.4Difference 50 Difference 0.6

MAWT 620 2040

Actual Stress 19,160 0.24

Inspection Intervals:

Time: 7 years Corrosion Rate: 0.0171 in/yr

Previous thickness: 0.5 in. Corrosion Allowance: 0.2029 in.

Present thickness: 0.38 in Remaining Life: 11.836 years

Required thickness: 0.1771 in.

Metal Loss: 0.12 in.

Semi-Elliptical Heads

HEAD CALCULATIONS

VALUES

Maximum Allowable Working Pressure (P): 150

Maximum Allowable Stress Value (S): 17500

Joint Efficiency: 1

Inside Diameter of Head: 69.875 34.9375 (inside radius minus corrosion allowance)

Corrosion Allowance:

Mill tolerance:

Given total wall thickness: 1.125 1.1250 (Nominal Thickness minus corrosion allowance)

FOR SEMI-ELLIPTICAL HEADS:

t = PD/2SE - 0.2P P = 2SEt/D + 0.2t

t =: 0.2997 inches P =: 562 psig

Critical Thickness: 0.315 in. 0.7124

Flat Circular Heads

HEAD CALCULATIONS

VALUES

Maximum Allowable Working Pressure (P): 250

Maximum Allowable Stress Value (S): 20000

Joint Efficiency: 1

Inside Diameter of Head: 32 16.0000 (inside radius minus corrosion allowance)

Corrosion Allowance: 0.0000C Factor 0.17Given total wall thickness: 2.6883 2.2313 (Nominal Thickness minus corrosion allowance)

FOR SEMI-ELLIPTICAL HEADS:

t =: 1.4751 inches

Critical Thickness: 1.549 in. 1.8532

t = D√(CP/SE)

Flat Circular Heads (raised face)

HEAD CALCULATIONS

Maximum Allowable Working Pressure (P):

Maximum Allowable Stress Value (S):

Joint Efficiency:

(inside radius minus corrosion allowance) Inside Diameter of Head:

Corrosion Allowance:

C Factor

(Nominal Thickness minus corrosion allowance) Given total wall thickness:WhG

FOR SEMI-ELLIPTICAL HEADS:

t =: 1.9596 inches

Critical Thickness: 2.058

t = D√(CP/SE)

VALUES

250

20000

1

32 16.0000 (inside radius minus corrosion allowance)

0.00000.3

2.6883 1.8818 (Nominal Thickness minus corrosion allowance)

in. 1.9207

HEAD CALCULATIONS

VALUES

Maximum Allowable Working Pressure (P): 750

Maximum Allowable Stress Value (S): 19100

Joint Efficiency: 0.85

Inside Diameter of Head: 72 36.1250 (inside radius minus corrosion allowance)

Corrosion Allowance: 0.1250

Mill tolerance:

Given total wall thickness: 0.2722 0.1472 (Nominal Thickness minus corrosion allowance)

FOR HEMISPERICAL HEADS:

t = PL t = 0.8383 inches P = 2SEt P =

2SE - 0.2P L + 0.2t

Critical thickness: 0.8802 inches or 0.4927 inches

(inside radius minus corrosion allowance)

(Nominal Thickness minus corrosion allowance)

245 psig

VALUES

Maximum Allowable Working Pressure (P):

Maximum Allowable Stress Value (S): 20000

Joint Efficiency: 1

Outside Diameter of Pipe: 4.5 1.99

Corrosion Allowance: 0.0500Mill tolerance: 0Given total wall thickness: 0.26 0

For Straight Run Pipe:

t = PD/2(SE + Py) P =: 2tSE/D-2tyt =: 0.0000 inches P =: 2423 psig

t required = 0.0500

d = 168.25C = 0.3P = 147S = 20,000E = 1W = 3,000,000hg = 4.625

t = 8.3817

API 650 Tanks

D = 50 ft. t nom = 0.5 in.

H = 48 ft.

G = 1.3

Sd = 20000

CA = 0.0625 in.

t = 2.6D(H-1)G + CASd

min t = 0.3972 in. critical t = 0.421 in.

t = 0.4597 in. Flag t = 0.4375 in.

ASME SECTION I

DRUM CALCULATIONS

VALUES

Maximum Allowable Working Pressure (P): 2500

Maximum Allowable Stress Value (S): 18800

Joint Efficiency (E): 1

Ouside Diameter of Drum (D): 84 Inside Radius of Drum (R):

Temperature Coefficient (y): 0.4000

Threading Allowance (C: 0.0000

Given total wall thickness: 5.303

For Outside Diameter:

t = (PD/2SE + 2yP) + C P = 2SE(t - C)/D - 2y(t - C)

t =: 5.3030 inches P =: 2500 psig

For Inside Radius

t = (PR/SE (1 - y)P) + C P = SE(t-C)/R + (1 - y)(t - C)

t =: 5.3210 inches P =: 2492 psig

Inside Radius of Drum (R): 36.8212

Boiler Tubes

P = 1949

D = 2.5

S = 15000

E = 1

e = 0

t = PD + 0.005D + e2S + P

t = 0.1650

0.2050

P = (2St)/D * 1000 * F CSA Z662

P = Design PressureS = Specified Minimum Yield Strength (SMYS)t = Nominal Wall ThicknessD = Outside DiameterF = Combined Factor (F, L, J & T)CA = Corrosion Allowance

S = 359 Mpat = 5.56 mmD = 114.3 mmF = 0.6CA = 1.61 mm

P = 20956 kPa

Pit Depth Evaluation Pit Length Evaluation

CA % = 29% L = 1.12Bsqrt(Dt)

Pit Depth = 1 mm L = Maximum Length of Corroded Area18% B = 4.0 for depths >10% & <=17.5%, calculated value for >17.5% & <= 80%

Acceptable Yes D = Outside Diametert = Nominal Wall Thickness

L = 102.32 mm

P = 2F(s/D)t ASME B31G

P = Design PressureF = Design Factor, usually 0.72S = Specified Minimum Yield Strength (SMYS)t = Nominal Wall ThicknessD = Outside Diameter

S = 48000 psit = 0.159 in

D = 4.5 inF = 0.72

P = 2,442 psig16,827 kPa

4.0 for depths >10% & <=17.5%, calculated value for >17.5% & <= 80%

B = 3.624

t = Pd/2F-P

P = 586 kPag

D = 1372 mm

F = 165000 kPa

t = 2.440673 mm

0.09609 in