Ellipsoidal head calculations in accordance with part 4

Liquid static head at ellipsoidal head

Max. liquid level in vessel (acting on ellip. head) LL = 1659 mmDensity of liquid ς = 849.3 kg/m3

Equivalent pressure at ellipsoidal head Ps = ς * g * LL

= 0.01382 N/mm2

= 13.82 KPa

Ellipsoidal head under internal pressure

Material of ellip. headDesign temperature T = 60 CInside diameter of ellip. head D = 477.8 mmInside diameter of ellip. head in corroded condition to be used in calculation D = 481 mmInternal design pressure P = 600 KPaWeld joint factor E = 1Internal corrosion allowance CAi 1 6 mm

SA-234 WPB

Note 1: This pressure is already included in the design pressure, as specified on vessel datasheet, so it shall not added to design pressure

Internal corrosion allowance CAi 1.6 mmExternal corrosion allowance CAo 0 mm

Step-1

Height of ellipsoidal head h = 120.25 mm

coefficient k = D/(2h)= 2

Since 1.7 ≤ k ≤ 2.2 The rules in paragraph 4.3.7.1 are applicable

Inside knuckle radius r = D(0.5/k-0.08)= 81.77 mm

Inside crown radius L = D(0.44k+0.02)= 432.90 mm

Assumed Min. thickness of ellip. head t = 15.09 mm

Nominal thickness of straight flange tn = 15.09 mm

Thickness of ellip. head in corroded condition to be used in calculation t = 13.49 mm

Step-2

L/D ratio L/D = 0.9r/D ratio r/D = 0.17L/t ratio L/t = 32.09

conditions check0.7 ≤ L/D ≤ 1.0 TRUEr/D ≥ 0.06 TRUE20 ≤ L/t ≤ 2000 TRUE

Since all above three conditions are true, hence we can proceed to step-3

Step-3

Geometric constant ßth = arccos ((0.5D-r)/(L-r))= 1.1017

Geometric constant øth = (√L*t)/r= 0.9346

For øth < ßth Rth = (0.5D-r)/cos(ßth-øth)+rFor øth ≥ ßth Rth = 0.5D

Since øth < ßth, hence corresponding value of Rth Rth = 242.74 mmcorresponding value of Rth Rth 242.74 mm

Step-4

For r/D ≤ 0.08 C1 = 9.31(r/D)-0.086For r/D > 0.08 C1 = 0.692(r/D)+0.605

corresponding value of coefficient C1 C1 = 0.7226

For r/D ≤ 0.08 C2 = 1.25For r/D > 0.08 C2 = 1.46-2.6(r/D)

Since r/D > 0.08, hence corresponding value of coefficient C2 C2 = 1.0180

Step-5

Modulus of elasticity at max. design temperature ET = 200000000 KPa

Internal pressure expected to produce elastic buckling of knuckle Peth = (C1*ET*t2)/(C2*Rth*(Rth/2-r))

= 2687582.70 KPaStep-6

Min. specified yield strength at design temperature (from annex 3.D) Sy = 229800 KPaAllowable stress value at design temperature (see annex 3.A) S = 153000.00 KPaIf allowable stress at design temperature is governed by time-independent properties then C3 = SyIf allowable stress at design temperature is governed by time-dependent properties & allowable stress is based on 90% yield criterian, then C3 C3 = 1.1*SIf allowable stress at design temperature is governed by time-dependent properties & allowable stress is based on 67% yield criterian, then C3 C3 = 1.5*SHence, choosen value of C3 C3 = 229800

Internal pressure producing max. stress in knuckle equal to material yield strength Py = (C3*t)/(C2*Rth*(Rth/2r-1))

= 25902 51 KPa= 25902.51 KPaStep-7

Constant G = Peth/Py= 103.76

For G ≤ 1.0 Pck = 0.6*PethFor G > 1.0 Pck = Py (0.77508G-0.20354G2+0.019274G3)

(1+0.19014G-0.089534G2+0.0093965G3)

Since G > 1.0, hence corresponding value of pressure Pck Pck = 52653.16 KPa

Step-8

Allowable pressure based on buckling failure of knuckle Pak = Pck/1.5

= 35102.11 KPaStep-9

Allowable pressure based on rupture of crown Pac = (2SE)/(L/t+0.5)

= 9389.26 KPaStep-10

The maximum allowable internal pressure Pa = min (Pak,Pac)

= 9389.26 KPa

Step-11

Total internal pressure plus liquid static head on ellip. head Ptotal = 613.82 KPa

Since Ptotal <= Pa, hence ellipsoidal head thickness for internal pressure is safe

Ellipsoidal head under external pressure

Design external pressure Pe = 0 KPa

Step-1

Assumed initial thickness of ellip. Head in corroded condition t = 13.49 mm

Step-2

Outside radius of ellip. head Ro = 253.99 mmModulus of elasticity at design temperature (see part 3) Ey = 200000000 KPaPredicted elastic buckling stress Fhe = 0.075Ey(t/Ro)

= 796684.91 KPa

Step-3

Ratio Fhe/Sy Fhe/Sy = 3.47

Fhe/Sy ≥ 6.25 FALSE corresponding value of Fic = 229800.001.6 < Fhe/Sy < 6.25 TRUE corresponding value of Fic = 65204.030.55 < Fhe/Sy ≤ 1.6 FALSE corresponding value of Fic = 246813.28Fhe/Sy ≤ 0.55 FALSE corresponding value of Fic = 796684.91

Hence choosen value of predicted buckling stress Fic = 65204.03 KPa

Step-4

Fic ≤ 0.55Sy TRUE 20.55Sy < Fic < Sy FALSE 2.1967Fic = Sy FALSE 1.6670

Hence, choosen value of design factor FS = 2

Step-5

Allowable hoop compressive membrFha = Fic/FS= 32602.01 KPa

Allowable external pressure Pa = 2Fha(t/Ro)

corresponding value of FS =corresponding value of FS =corresponding value of FS =

= 3463.14 KPaStep-6

Since Pe <= Pa, hence ellipsoidal head thickness for external pressure is safe

Minimum thickness requirement check

The min. thickness of head shall be = 1.6+CAi+CAo= 3.2 mm

The min. thickness requirement is fulfilled

Hence the adopted thickness of ellipsoidal head is safe & sufficient.

Extreme fibre elongation ( see paragraph 6.1.2.3 & table 6.1)

Final mean radius Rf = 89.315 mmOriginal mean radius Ro = infinityNominal thcikness of plate t = 15.09 mm

calculated forming strain Ef = (75t/Rf)*(1-Rf/Ro)

Ef = 12.67144377 %

Since head is ASME Standard Pressure Part (ASME B16.9), it needs not be heat treated due to fiber elongation if exceeds 5%.