Mud Gas Separator Calculation

DELBA III Mud Gas Separator Calculation

NOV Brandt Manufacturer

Built to the Requirements of ASME Sec VIII, Div I, 2007 Edition, 2009 Addenda Addenda

Shell Tag Nos:

Shell Type: SA-216 Gr.WCB Max Working Pressure: 200 psi

Prepared by: Erwin Gomop-as Checked by: Jeffrey Macasero

A) Minimum Required Shell Thickness per UG-27 Design parameters: Material

SA-216 Gr.WCB

Vessel inside diameter (Di)

48

in

Tangent to tangent Length (Ls)

192

in

Corrosion Allowance (CA)

0.125

in

Shell nominal thickness (tact)

0.512

in

Longitudinal Efficiency for Circular Stress (El)

1

Circular Efficiency for Longitudinal Stress (Ec)

1

Design Temperature (T) Allowable Stress at Design Temp. (from Sec.II part D table 1A) (S)

o

200 18300

F psi

Internal Design pressure at top vessel (Pi)

200

psi

External Pressure (choose Yes for FV) (Pe)

NA

Shell Outside Diameter

(Do) = Di + (2 x tn)

48

in

Shell Inside Corroded Radius

(Ri)

= (Di / 2) + CA

24.125

in

Shell Inside Radius

(Rn)

= Di / 2

24

in

Shell Corroded Thickness

(tc)

= tn - CA

0.387

in

(a) The minimum required thickness of shells under internal pressure shall not be less than that computed by the following equations. (c) Cylindrical Shells. The minimum thickness or maximum allowable working pressure of cylindrical shells shall be the greater thickness or lesser pressure as given by (1) & (2) below. (1) Circumferential Stress (Longitudinal Joints) When the thickness (t) does not exceed one-half of the inside radius (Ri), or S does not exceed 0.385 SE , equation

(Fig. 1)

(1) shall apply: t <

.5Ri

0.5 < 12.063 S =

18300 psi from ASME Section II- Part D- Table IA (As per UG-24 a casting quality factor of 80% it to be appli ed to the allowable stress) S = 18300 x 0.8 = 14640 psi P < 0.385SE 200 < 5636.4 one-half of the inside radius (Ri), and S does not exceed Since, the thickness (t) does not exceed one-half 0.385 SE. Therefore, Equation 1 is applicable.

ASME Section VIII, Division 1 2007 Edition, 2009 Addenda Addenda

Page 2 of 16

Solve for treq {(P x Ri) Ri) / (S x E) - (0.6P (0.6P)} )} treq = {(P =

Eq. 1

0.332 in

MAWP = [(S x E x t)/(Ri+ 0.6t)] =

232

psi

(2) Longitudinal Stress (Circumferential Joints)

When the thickness (t) does not exceed one-half of the inside radius (R i ) , or S does not exceed

1.25 SE , formula (2) shall apply: t <

.5Ri

0 . 5 < 1 2 .0 6 3 S = 14640 psi P < 1.25SE 200 < 18300 Since, the thickness (t) does not exceed one-half one-half of the inside radius (Ri), and S does not exceed 1.25 SE. Therefore, formula 2 is applicable. Solve for treq t = {(P x Ri) / (2S x E) + (0.4P)}

=

Eq. 2

0.165 in

MAWP = [(2 x S x E x t)/(R i+ 0.2t)] =

468

psi

(Fig. 2)

Conclusion: Since, the minimum required shell thickness of shells computed by formula (1) & (2) is less the actual thickness. Therefore, the design is satifactory. (c ) When necessary, vessels shall be provided wi th stiffeners or other additional means of support to prevent overstress or large distortions under the external l oadings.

Comment: Not Necessary. (g) Any reduction in thickness within a shell course or spherical shell shall be in accordance with UW-9 to prevent overstress or large distortions under the external l oadings.

Comment: Not Applicable.


Page 3 of 16

B) Minimum Required Ellipsoidal Ellipsoidal Head Thickness per UG-32 Design parameters: Material

SA-516 Gr.70

Vessel inside diameter (Di)

48

in

Corrosion Allowance (CA)

0.125

in

Head thickness (tact)

0.394

in

Joint Efficiency (E) Design Temperature (T) Allowable Stress at Design Temp. (from Sec.II part D table 1A) (S) Internal Design pressure at top vessel (Pi)

1 200 20000 200

in F psi psi

External Pressure (choose Yes for FV) (Pe) Shell Outside Diameter Shell Inside Corroded Radius Shell Inside Radius Head Corroded Thickness

NA 48 24.125 24 0.269

in in in in

(Do) (Ri) (Rn) (tc)

= = = =

Di + (2 x tn) (Di / 2) + CA Di / 2 tn - CA

o

(a) Ellipsoidal Heads with t s /L ≥ 0.002. The minimum thickness of a dished head of semiellipsoidal form, in which half the minor axis (inside depth of the head minus the skirt) equals one-fourth of the inside diameter of the head skirt, shall be determined by: t = {(P x Di) / (2SE) - (0.2P)}

=

Eq. 3

0.241 in

MAWP = [(2 x S x E x t)/(Di+ 0.2t)] =

223

psi

Conclusion: Since, the minimum required shell thickness of shells computed by equation (3) is less the actual thickness. Therefore, the design is satifactory.

(Fig. 3)


Page 4 of 16

C) Minimum Nozzle Wall Thickness per UG-45 The minimum wall thickness of the nozzle necks shall be the l arger of the thickness determined by (a) or (b) below. Shear stressed caused by loadings shall not exceed the allowable stress in ( c) below.

SCHEDULE OF NOZZLE A

3000#

6"

B

3000#

1 ½"

C D

300# 150#

6" 4"

Coupling Female Coupling Female Flange Flange

E F G H I J K L

150# 150# 300# 3 0 0# 300# 300# 1 5 0# 150#

6" 10" 20" 20" 12" 6" 12" 10"

Flange Flange Flange Flange Flange Flange Flange Flange

Hydrotest Pressure Sensor Feed (Inlet) Low Pressure Mud Supply Inspection Port Discharge (Outlet) Inspection Door Inspection Door Diverter (Inlet) Feed (Inlet) Vent (Secondary) Vent (Primary)

(Fig. 4)

(C1) Nozzle 3" required thickness Design parameters: Material

SA-352 LCB

Nozzle inside diameter (Di)

3.068

in

Nominal thickness (tn)

0.216

in

Corrosion Allowance (Ca)

0.125

in

200

psi

Internal Design pressure at top vessel (Pi) Design Temperature (T) Allowable Stress at Design Temp. (from Sec.II part D table 1A) (S) Joint Effeciency (E)


200 18600 1

o

F psi

Page 5 of 16

o

18600 psi from ASME Section II- Part D- Table IA (SA-995 @ 190 F) (As per UG-24 a casting quality factor of 80% it to be appli ed to the allowable stress) S = 18600 x 0.8 = 14880 psi S =

(tact) = =

0.216

x

0.125

0.091 in

in corroded condition

Ri = (Di / 2) + C a Ri =

1.659 in


(C1.1) Circumferential Stress (Longitudinal Joints)

Solve for treq treq = {(P {(P x Ri) Ri) / (S x E) - (0.6P (0.6P)} )}

=

Eq. 4

0.022 in

(C1.2) Longitudinal Stress (Circumferential Joints)

Solve for treq treq = {(P x Ri) / (2S x E) + (0.4P)}

=

Eq. 5

0.011 in

Conclusion: Since, the minimum required shell thickness of shells computed by equation (4) & (5) is less the actual thickness. Therefore, the design is satisfactory.

wal l thickness to be not less than that computed for the applicable UG-45 (a) requires minimum nozzle wall loading plus corrosion allowance. from Equation 4 trn =

0.022 in

this thickness is compared with the minimum thickness provided which for pipe material would include a 12.5% undertolerance. 0.875 Since

x

tact =

0.080 in

0.080 inch is larger than than

0.022 inch, therefore satifactory.

UG-45 (b) requires requires determining the one applicable wall thickness from (b)(1), (b)(2), or (b)(3), comparing that with the thickness from (b)(4) and then choosing the smaller of those two values. wal l thickness to be not less than the thickness required for UG-45 (b)(1) requires minimum nozzle wall internal pressure of the head or shell where the nozzle is located but in no case l ess than that thickness required by UG-16(b). tr =

0.332 in

from Eq. 1 1

and in UG-16(b) minimum is /16 in. Therefore, ASME Section VIII, Division 1 2007 Edition, 2009 Addenda Addenda

0.332 inch thickness governs. Page 6 of 16

UG-45 (b)(2) applies to vessels designed for external ex ternal pressure only and therefore not applicable. UG-45 (b)(3) applies to vessels designed for both external and internal pressure and therefore not applicable. UG-45 (b)(4) requires minimum nozzle wall thickness of standard wall pipe accounting for undertolerance plus the thickness added for corrosion allowance. Undertolerance for pi pe manufactured in accordance with ASME B36.10M is 12.5% is 0.375 inch. Thus, the minimum wall thickness is 0.375 x (1.0 - 0.125) = 0.328 in Therefore, the minimum nozzle wall thickness required by UG-45(b) is the smaller of (b)(1) or (b)(4), or 0.328 inch.

(C2) Nozzle 4" required thickness Dimensions: Material

SA-672 Gr-B60


3.826

in


0.337

in


0.125

in

200

psi

Internal Design pressure at top vessel (Pi) Design Temperature (T) Allowable Stress at Design Temp. (from Sec.II part D table 1A) (S) Joint Effeciency (E) (tact) = =

0.337

-

200 17100 1

o

F psi

0 .1 2 5

0.212 in


Ri = (Di / 2) + C a Ri =

2.038 in



Solve for treq {(P x Ri) Ri) / (S x E) - (0.6P (0.6P)} )} treq = {(P =

Eq. 6

0.024 in



=

Eq. 7

0.012 in

Conclusion: Since, the minimum required shell thickness of shells computed by formula (6) & (7) is less the actual thickness. Therefore, the design is satisfactory. ASME Section VIII, Division 1 2007 Edition, 2009 Addenda Addenda

Page 7 of 16

UG-45 (a) requires minimum nozzle wall wal l thickness to be not less than that computed for the applicable loading plus corrosion allowance. from Eq. 6 trn =

0.024 in

this thickness is compared with the minimum thickness provided which for pipe material would include a 12.5% undertolerance. 0.875

x

tact =

0.186 in


Since


UG-45 (b) requires requires determining the one applicable wall thickness from (b)(1), (b)(2), or (b)(3), comparing that with the thickness from (b)(4) and then choosing the smaller of those two values. UG-45 (b)(1) requires minimum nozzle wall wal l thickness to be not less than the thickness required for internal pressure of the head or shell where the nozzle is located but in no case l ess than that thickness required by UG-16(b). tr =

0.332 in

from Eq. 1 1

and in UG-16(b) minimum is /16 in. Therefore,

0.332 inch thickness governs.



SA-524 Gr-I


6.065

in


0.280

in


0.125

in

200

psi

Internal Design pressure at top vessel (Pi) Design Temperature (T) Allowable Stress at Design Temp. (from Sec.II part D table 1A) (S) Joint Effeciency (E) ASME Section VIII, Division 1 2007 Edition, 2009 Addenda Addenda

200 17100 1

o

F psi

Page 8 of 16

(tact) = =

0.280

-

0 .1 2 5

0.155 in


Ri = (Di / 2) + C a Ri = 3.15 .1575 in




=

Eq. 8

0.037 in



=

Eq. 9

0.018 in

Conclusion: shells computed by formula (3) & (4) is less the the Since, the minimum required shell thickness of shells actual thickness. Therefore, the design is satisfactory.

wal l thickness to be not less than that computed for the applicable UG-45 (a) requires minimum nozzle wall loading plus corrosion allowance. from formula 5 trn =

0.037 in


x

tact =

0.136 in



UG-45 (b) requires requires determining the one applicable wall thickness from (b)(1), (b)(2), or (b)(3), comparing that with the thickness from (b)(4) and then choosing the smaller of those two values. wal l thickness to be not less than the thickness required for UG-45 (b)(1) requires minimum nozzle wall internal pressure of the head or shell where the nozzle is located but in no case l ess than that thickness required by UG-16(b). tr =

0.332 in

from Eq. 1 1



ex ternal pressure only and therefore not applicable. UG-45 (b)(2) applies to vessels designed for external


Page 9 of 16

UG-45 (b)(3) applies to vessels designed for both external and internal pressure and therefore not applicable. UG-45 (b)(4) requires minimum nozzle wall thickness of standard wall pipe accounting for undertolerance plus the thickness added for corrosion allowance. Undertolerance for pi pe manufactured in accordance with ASME B36.10M is 12.5% is 0.375 inch. Thus, the minimum wall thickness is 0.375 x (1.0 - 0.125) = 0.328 in Therefore, the minimum nozzle wall thickness required by UG-45(b) is the smaller of (b)(1) or (b)(4), or 0.328 inch.


SA-524 Gr-I


10.020

in


0.365

in


0.125

in

200

psi

Internal Design pressure at top vessel (Pi) Design Temperature (T) Allowable Stress at Design Temp. (from Sec.II part D table 1A) (S) Joint Effeciency (E) (tact) =

0.365

x

=

0.24

in

200 17100 1

o

F psi

0 .1 2 5 in corroded condition

Ri = (Di / 2) + C a Ri =

5.135 in




=

Eq. 10

0.060 in



=

Eq. 11

0.030 in



Page 10 of 16

UG-45 (a) requires minimum nozzle wall wal l thickness to be not less than that computed for the applicable loading plus corrosion allowance. from formula 5 trn =

0.060 in

this thickness is compared with the minimum thickness provided which for pipe material would include a 12.5% undertolerance. 0.875

x

tact =

0.210 in


Since



0.332 in

from Eq. 1 1





SA-106 Gr-A


18.814

in


0.593

in


0.125

in

200

psi

Internal Design pressure at top vessel (Pi) Design Temperature (T) Allowable Stress at Design Temp. (from Sec.II part D table 1A) (S) Joint Effeciency (E) ASME Section VIII, Division 1 2007 Edition, 2009 Addenda Addenda

200 13700 1

o

F psi

Page 11 of 16

(tact) = =

0.593

x

0 .1 2 5

0.468 in


Ri = (Di / 2) + C a Ri =

9.532 in




=

Eq. 12

0.140 in



=

Eq. 13

0.069 in


wal l thickness to be not less than that computed for the applicable UG-45 (a) requires minimum nozzle wall loading plus corrosion allowance. from formula 5 trn =

0.140 in


x

tact =

0.410 in




0.332 in

from formula 1 1



ex ternal pressure only and therefore not applicable. UG-45 (b)(2) applies to vessels designed for external

UG-45 (b)(3) applies to vessels designed for both external and internal pressure and therefore not applicable. ASME Section VIII, Division 1 2007 Edition, 2009 Addenda Addenda

Page 12 of 16

UG-45 (b)(4) requires minimum nozzle wall thickness of standard wall pipe accounting for undertolerance plus the thickness added for corrosion allowance. Undertolerance for pi pe manufactured in accordance with ASME B36.10M is 12.5% is 0.375 inch. Thus, the minimum wall thickness is 0.375 x (1.0 - 0.125) = 0.328 in Therefore, the minimum nozzle wall thickness required by UG-45(b) is the smaller of (b)(1) or (b)(4), or 0.328 inch.

D) 12" Class 150 Effective gasket seating & Total required bolt area. Dimensions: 12" ASME B16.5 Blind Flange 150 lbs

SA-516 Gr-70

Raised face outside diameter (Do)

15

in

Gasket outside diameter (God)

14

in

Gasket inside diameter (Gid)

12

in

Flange allowable stress @ ambient (Sa)

20000

psi

Flange allowable stress @ operating (So)

12000

psi

Gasket factor (m) Minimum design seating stress (y) Bolt material

2.500 10000 psi SA-193Gr-B7

Bolt allowable stress @ ambient (Sa)

25000

psi

Bolt allowable stress @ operating (So)

21000

psi

Allowable Stress at Design Temp. (from Sec.II part D table 1A) (S)

13700

psi


0.125

in

200

psi

Internal Design pressure at top vessel (Pi) Joint Effeciency (E)

1

(D1) Effective gasket seating bo = N/2 =

from ASME Section-VIII Di Div-I Ta Table 2-5.1 N= 1

0.5 in

b = 0.5√ bo =

0.354 in

G = God - 2b = 13.293 in

effective gasket seating

(D2) Total bolt area required 2

H = 0.785 x G x P = 27742 in Hp = 2 x b x π x G x m x P i =

14765 lbs (Fig. 4)


Page 13 of 16

Wm1 = H + Hp =

42507 lbs

Wm2 = π x b x G x y = 147647 lbs

(c) Calculate the total bolt area and determine the number and size of bolts. bolts. Am1 = Wm1 / So = 2.0241 in Am2 = Wm2 / Sa = 7.3823 in

2

2

The largest required area A m = 7.38 in . 2

Use 12 - M24 bolts, A b= 8.414 in .

(d) Determine gasket seating design load. A = .5 (Am + Ab) =

7.90

in

2

Wa = A x Sa =

197460

lbs

( e)Determine the minimum thickness using Equation 2 in UG-34. c =

0.3

d = G

C = fl flange O.D. = 17.00 in

hg = 0.5 (C-G) =

1.85

in

(f) For gasket seating only, P=0 and W=W a therefore. 3

t = d√( (1.9Whg/SEd ) t = 1.6173 in

(g) For operating conditions. 3

t = d√(cP/SE) + (1.9Whg/SEd ) = 1.3498 in treq = t + CA = 1.7423 in Therefore, the gasket seating condition controls and the minimum required thickness, including corrosio allowance is 1.7423 inches. ASME Section VIII, Division 1 2007 Edition, 2009 Addenda Addenda

Page 14 of 16

D) 20" Class 150 Effective gasket seating & Total required bolt area. Dimensions: 12" ASME B16.5 Blind Flange 150 lbs

SA-516 Gr-70

Raised face outside diameter (Do)

23

in

Gasket outside diameter (God)

22

in

Gasket inside diameter (Gid)

20

in

Flange allowable stress @ ambient (Sa)

20000

psi

Flange allowable stress @ operating (So)

12000

psi

Gasket factor (m) Minimum design seating stress (y) Bolt material

2.5 10000 psi SA-193Gr-B7

Bolt allowable stress @ ambient (Sa)

25000

psi

Bolt allowable stress @ operating (So)

21000

psi

Allowable Stress at Design Temp. (from Sec.II part D table 1A) (S)

13700

psi


0.125

in

200

psi

Internal Design pressure at top vessel (Pi) Joint Effeciency (E)

1

(D1) Effective gasket seating bo = N/2 =

from ASME Section-VIII Di Div-I Ta Table 2-5.1 N= 1

0.5 in

b = 0.5√ bo =

0.354 in

G = God - 2b = 21.293 in

effective gasket seating

(D2) Total bolt area required 2

H = 0.785 x G x P = 71182 in Hp = 2 x b x π x G x m x P i =

23650 lbs

Wm1 = H + Hp =

94832 lbs

Wm2 = π x b x G x y = 236505 lbs

(c) Calculate the total bolt area and determine the number and size of bolts. bolts. Am1 = Wm1 / So 2

= 4.5158 in ASME Section VIII, Division 1 2007 Edition, 2009 Addenda Addenda

Page 15 of 16

Am2 = Wm2 / Sa = 11.825 in

2

2

The largest required area A m = 11.8 in . 2

Use 20 - M30 bolts, A b= 21.91 in .

(d) Determine gasket seating design load. A = .5 (Am + Ab) =

16.87 in

2

Wa = A x Sa =

337352

lbs

( e)Determine the minimum thickness using Equation 2 in UG-34. c =

0.3

d = G

C = fl flange O.D. = 25.00 in

hg = 0.5 (C-G) =

1.85

in

(f) For gasket seating only, P=0 and W=W a therefore. 3

t = d√( (1.9Whg/SEd ) t = 1.6703 in

(g) For operating conditions. 3

t = d√(cP/SE) + (1.9Whg/SEd ) = 1.8905 in treq = t + CA = 2.0155 in Therefore, the gasket seating condition controls and the minimum required thickness, including corrosio allowance is 2.0155 inches.


Page 16 of 16

Mud Gas Separator Calculation

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