Mud Gas Separator Calculation

DELBA III

Mud Gas SeparatorCalculation

NOV Brandt

Manufacturer

Built to the Requirements

of

ASME Sec VIII, Div I, 2007 Edition, 2009 Addenda

Tag No: N-H21717

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:

=

=

=

=

(a) The minimum required thickness of shells under internal

pressure shall not be less than that computed by the follo-

wing 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

(1) shall apply:

t <

<

S = from ASME Section II- Part D- Table IA

(As per UG-24 a casting quality factor of 80% it to be applied to the allowable stress)

S = x

=

P <

<

Since, the thickness (t) does not exceed one-half of the inside radius (Ri), and S does not exceed

0.385 SE. Therefore, Equation 1 is applicable.

psi

in

in

in

in

24

0.236

oF

in

in

in

in

psi

1

SA-216 Gr.WCB

48

192

0

0.236

150

20000

125

NA

1

48

24

psi

.5Ri

Longitudinal Efficiency for Circular Stress (El)

16000 psi

Vessel inside diameter (Di)

Tangent to tangent Length (Ls)

Corrosion Allowance (CA)

125

External Pressure (choose Yes for FV) (Pe)

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

Design Temperature (T)

Di + (2 x tn)

(Di / 2) + CA

12

Internal Design pressure at top vessel (Pi)

Material

(Ri)

20000

0.2

Circular Efficiency for Longitudinal Stress (Ec)

20000

(Rn)

Shell nominal thickness (tact)

Shell Outside Diameter

Shell Inside Corroded Radius

Shell Inside Radius

Shell Corroded Thickness (tc)

0.385SE

Di / 2

tn - CA

(Do)

(Fig. 1)

0.8

6160

ASME Section VIII, Division 1

2007 Edition, 2009 Addenda Page 2 of 16

Solve for treq

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

= in

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

= 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 <

<

S = psi

P <

<

Since, the thickness (t) does not exceed 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)}

= in

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

= psi

Conclusion:

Since, the minimum required shell thickness of shells computed by formula (1) & (2) is less the

actual thickness. Therefore, the design is satifactory.

(d ) When necessary, vessels shall be provided with stiffeners or other additional means of support

to prevent overstress or large distortions under the external loadings.

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 loadings.

Comment: Not Applicable.

157

MAWP

314

MAWP

125 20000

16000

Eq. 1

Eq. 2

1.25SE

treq

0.188

0.094

.5Ri

0.2 12

(Fig. 2)



B) Minimum Required Ellipsoidal Head Thickness per UG-32

Design parameters:

=

=

=

=

(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)}

= in

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

= psi

Conclusion:

Since, the minimum required shell thickness of shells computed by equation (3) is less the

actual thickness. Therefore, the design is satifactory.

MAWP

oF

125

NA

148

150

20000

0

0.177

1

48

24

48

SA-516 Gr.70

0.177

in

in

in

in

psi

(tc) tn - CA

psi

in

in

in

in

24

(Di / 2) + CA

Shell Inside Radius (Rn) Di / 2

Head Corroded Thickness


External Pressure (choose Yes for FV) (Pe)

Shell Outside Diameter (Do) Di + (2 x tn)

Corrosion Allowance (CA)


Eq. 3

0.150

Material

(Fig. 3)

MinimumHead thickness (tact)

Joint Efficiency (E)

Vessel inside diameter (Di)


Shell Inside Corroded Radius (Ri)



C) Minimum Nozzle Wall Thickness per UG-45

The minimum wall thickness of the nozzle necks shall be the larger of the thickness determined by

(a) or (b) below. Shear stressed caused by loadings shall not exceed the allowable stress in ( c) below.

(a) Nozzle 3" required thickness

Design parameters:

3.068

0

H 300#

300#

300#

150

125

K

oF

in

in

in

psi

psi

6"150#

6"

6"

18600

1

300#

0.216

SA-352 LCB

Diverter (Inlet)

Nozzle inside diameter (Di)

J


Joint Effeciency (E)

Inspection Door

Inspection Door

I



Nominal thickness (tn)

Corrosion Allowance (Ca)

Material

Discharge (Outlet)

Flange

SCHEDULE OF NOZZLE

Hydrotest

10"150#

Inspection Port

L

6"

12"150# Flange

Flange

10"

20"

20"

12"

Flange

Flange

Flange

Flange

Coupling

1 ½"

4"

Female

150#

E

F

G

Female

Coupling

Flange

Flange

3000#

300#

B

C

D 150# Flange

3000#

(Fig. 4)

Feed (Inlet)

Pressure Sensor

Vent (Secondary)

Vent (Primary)

Mud Supply

Low Pressure

Feed (Inlet)

A



S = from ASME Section II- Part D- Table IA (SA-995 @ 150o F)

(As per UG-24 a casting quality factor of 80% it to be applied to the allowable stress)

S = x

=

tact = -

= in in corroded condition

Ri = (Di / 2) + Ca

Ri = in corroded condition

(a.1) Circumferential Stress (Longitudinal Joints)

Solve for treq

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

= in

(a.2) Longitudinal Stress (Circumferential Joints)

Solve for treq

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

= 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.

UG-45 (a) requires minimum nozzle wall thickness to be not less than that computed for the applicable

loading plus corrosion allowance.

from Equation 4

trn = in

this thickness is compared with the minimum thickness provided which for pipe material would include

a 12.5% undertolerance.

x = in

Since inch is larger than than 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.

UG-45 (b)(1) requires minimum nozzle wall 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 less than that thick-

ness required by UG-16(b).

tr = in from Eq. 1

and in UG-16(b) minimum is 1/16 in. Therefore, inch thickness governs.

treq Eq. 4

0.006

0.013

0.013

0.875

0.8

in

Eq. 5

0.013

treq

0.188

18600

18600 psi

0.216

0.189

1.534

14880 psi

0.216

tact

0

0.189

0.188



UG-45 (b)(2) applies to vessels designed for external 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 under-

tolerance plus the thickness added for corrosion allowance. Undertolerance for pipe manufactured

in accordance with ASME B36.10M is 12.5% is 0.375 inch.

Thus, the minimum wall thickness is

x (1.0 - 0.125) = in

Therefore, the minimum nozzle wall thickness required by UG-45(b) is the smaller of (b)(1) or (b)(4),

or inch.

(b) Nozzle 4" required thickness

Dimensions:

tact = -


Ri = (Di / 2) + Ca

= in corroded condition

(b.1) Circumferential Stress (Longitudinal Joints)

Solve for treq

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

= in

(b.2) Longitudinal Stress (Circumferential Joints)

Solve for treq

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

= in

Conclusion:



oFDesign Temperature (T) 150

psi

3.826

in


SA-672 Gr-B60

psi

in

inNominal thickness (tn)


Material

0.328


0.337 0

0.337

1.913

1

0.337

0

125

Allowable Stress at Design Temp. (from Sec.II part D table 1A) (S) 17100

0.328

treq Eq. 6

0.375


in

0.014

treq Eq. 7

0.007





from Eq. 6

trn = in



x = in







tr = in from Eq. 1




applicable.





x (1.0 - 0.125) = in


or inch.

(c) Nozzle 6" required thickness

Dimensions:

0

125

oF150

17100

1

in

psi

psi

6.065

0.2950.875 tact

0.295 0.014

0.188

0.188

0.014

0.328

Material

0.375 0.328

SA-524 Gr-I


Nominal thickness (tn) 0.280

in

in








tact = -


Ri = (Di / 2) + Ca


(c.1) Circumferential Stress (Longitudinal Joints)

Solve for treq

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

= in

(c.2) Longitudinal Stress (Circumferential Joints)

Solve for treq

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

= in

Conclusion:





from formula 5

trn = in



- = in







tr = in from Eq. 1



0.875 tact

0.245

0.188

3.0325 in

0.280 0

0.28

treq Eq. 8

0.022

treq Eq. 9

0.022

0.188

0.011

0.022

0.245




applicable.





x (1.0 - 0.125) = in


or inch.

(d) Nozzle 10" required thickness

Dimensions:

tact = -


Ri = (Di / 2) + Ca

= in corroded condition

(d.1) Circumferential Stress (Longitudinal Joints)

Solve for treq

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

= in

(d.2) Longitudinal Stress (Circumferential Joints)

Solve for treq

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

= in

Conclusion:



0

in

in

in

SA-524 Gr-I

10.020

0.365

125

150

17100



1

oF

psi

psi

Material

0.328

0.375 0.328






0.365 0

0.365

5.01 in

treq Eq. 10

0.037

treq Eq. 11

0.018





from formula 5

trn = in



x = in







tr = in from Eq. 1




applicable.





x (1.0 - 0.125) = in


or inch.

(e) Nozzle 20" required thickness

Dimensions:

18.814 in

0.593

0

125

13700

1

150

in

in

psi

psi

oF

SA-106 Gr-A

0.875 tact 0.319

0.188

0.188

0.319 0.037

0.037

0.328

Material

0.375 0.328










tact = -


Ri = (Di / 2) + Ca


(e.1) Circumferential Stress (Longitudinal Joints)

Solve for treq

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

= in

(e.2) Longitudinal Stress (Circumferential Joints)

Solve for treq

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

= in

Conclusion:





from formula 5

trn = in



x = in







tr = in from formula 1




applicable.

0.875

0.593 0

0.593

9.407 in

tact 0.519

0.188

0.188

treq Eq. 12

0.086

treq Eq. 13

0.519 0.086

0.043

0.086







x (1.0 - 0.125) = in


or inch.

D) 12" Class 150 Effective gasket seating & Total required bolt area.

Dimensions:

(a) Effective gasket seating.

bo = N/2 from ASME Section-VIII Div-I Table 2-5.1 N= 1

= in

b = 0.5√ bo

= in

G = God - 2b

= in effective gasket seating

(b) Total bolt area required.

H = 0.785 x G2

x P

= in

= 2 x b x π x G x m x Pi

= lbs

(Fig. 4)

SA-516 Gr-70

SA-193Gr-B7

15

14

12

20000

20000

2.500

10000

25000

23750

13700

0.125

125

1

in

in

in

psi

psi

psi

psi

psi

psi

in

psi

0.328

12" ASME B16.5 Blind Flange 150 lbs

0.375 0.328

Raised face outside diameter (Do)

Gasket outside diameter (God)

Gasket inside diameter (Gid)

Flange allowable stress @ ambient (Sa)

Gasket factor (m)

Bolt material

Bolt allowable stress @ ambient (Sa)

17339

0.5

Flange allowable stress @ operating (So)

0.354

13.293



Minimum design seating stress (y)

Bolt allowable stress @ operating (So)



Hp

9227.9



Wm1 = H + Hp

= lbs

Wm2 = π x b x G x y

= lbs

(c) Calculate the total bolt area and determine the number and size of bolts.

Am1 = Wm1 / So

= in

Am2 = Wm2 / Sa

= in2

The largest required area Am = in2.

Use 16 - M24 bolts, Ab= 7.79 in2.

(d) Determine gasket seating design load.

A = .5 (Am + Ab)

= in2

Wa = A x Sa

= lbs

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

c = d = G C = flange O.D.

= in

hg = 0.5 (C-G)

= in

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

t = d√( (1.9Whg/SEd3)

= in

(g) For operating conditions.

t = d√(cP/SE) + (1.9Whg/SEd3)

= in

treq = t + CA

= in

Therefore, the gasket seating condition controls and the minimum required thickness, including corrosion

allowance is 1.7423 inches.

1.7101

7.59

0.3

17.00

1.85

1.5851

0.8266

26567

147647

7.3823

189669

7.38

1.1186



D) 20" Class 150 Effective gasket seating & Total required bolt area.

Dimensions:

(a) Effective gasket seating.bo = N/2 from ASME Section-VIII Div-I Table 2-5.1 N= 1

= in

b = 0.5√ bo

= in

G = God - 2b

= in effective gasket seating

(b) Total bolt area required.

H = 0.785 x G2

x P

= in

= 2 x b x π x G x m x Pi

= lbs

Wm1 = H + Hp

= lbs

Wm2 = π x b x G x y

= lbs

(c) Calculate the total bolt area and determine the number and size of bolts.

Am1 = Wm1 / So

= in2

1

in

125

SA-193Gr-B7

23

22

20

20000

20000

2.5

10000

SA-516 Gr-70

psi

in

25000

23750

13700

0.125

psi

psi

psi

psi

in

in

psi

psi

12" ASME B16.5 Blind Flange 150 lbs

Raised face outside diameter (Do)

Gasket outside diameter (God)

Gasket inside diameter (Gid)

Flange allowable stress @ ambient (Sa)

Flange allowable stress @ operating (So)

Gasket factor (m)

Minimum design seating stress (y)

Bolt material

Bolt allowable stress @ ambient (Sa)

Bolt allowable stress @ operating (So)

2.4956


0.5

14782

59270

236505

Hp



44489

0.354

21.293




Am2 = Wm2 / Sa

= in2

The largest required area Am = in2.

Use 20 - M24 bolts, Ab= 14.02 in2.

(d) Determine gasket seating design load.

A = .5 (Am + Ab)

= in2

Wa = A x Sa

= lbs

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

c = d = G C = flange O.D.

= in

hg = 0.5 (C-G)

= in

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

t = d√( (1.9Whg/SEd3)

= in

(g) For operating conditions.

t = d√(cP/SE) + (1.9Whg/SEd3)

= in

treq = t + CA

= in

Therefore, the gasket seating condition controls and the minimum required thickness, including corrosion

allowance is 1.58 inches.

1.4621

1.1577

1.5871

12.92

11.825

11.8

1.85

258493

0.3

25.00



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Mud Gas Separator Calculation