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  ![](media/image1.png){width="0.875in"   **Fermilab**
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**Particle Physics Division**

**Mechanical Department Engineering Note**

# Number: MD-ENG- 079 Date: June 20, 2005

Project Internal Reference:

Project: **ILC**

Title: Vacuum Vessel Engineering Note for 1.3 GHz, β=1, Cryomodule

Author(s): Edward Chi

Reviewer(s):

Key Words:

Crymodule, Vacuum, Pressure, Operating Temperature, Vessel Shell,

Stiffening Rib, Reinforcement, Flange, Weldment, Up & Downstream

Abstract Summary:

The vacuum vessel of 1.3 GHz, β=1, Cryomodule has 38" O. D, \~ 448"

Length with multiple openings for different connections and

instrumentation. This engineering note is presented extensive

discussion, analysis and calculations for the vessel and nozzle shells,

stiffening rings, multiple openings and reinforcements, different

flanges, welds and others per Fermilab, ASME and other applicable

codes.

Applicable Codes:

ES&H Manual Chapter 5033, Fermilab.

"Boiler & Pressure Vessel Code" ASME VIII, Div.1

Addenda July 1, 2003

"Allowable Stress Design", AISC, 9^th^ edition

"Structural Welding Code-Steel", AWS D1.1-90

**[CONTENTS]{.underline}**

*Section Descriptions Pages*

1\. Calculation for the Vacuum Vessel Cylindrical Shell Thickness 4

2\. The Permissible Out of Roundness of the Vessel Cylindrical Shell 6

3\. Analysis and Calculations for the Stiffening Rings

3.1 To Define the Stiffening Ring for the Vessel Shell Under the 7

> External Pressure and the Calculations of the Weld

3.2 To Define the Downstream End Flange as the Stiffening Ring for the
10

Vessel Shell Under the External Pressure

3.3 To Define the Upstream End Flange as the Stiffening Ring for the 12

Vessel Shell Under the External Pressure

4\. Calculations and Analysis for the Vessel Shell Openings

4.1. Calculations for the Opening of Port Coldmass for the Reinforcement
13

> and Weld Size

4.2. Calculations for the Opening of MC Port (#9) for the Reinforcement
and 18

Weld Size

4.3 Calculations for the Opening of Instrumentation Port for the 22

> Reinforcement and Weld Size

4.4 Calculations for the Opening of Port #16 for the Reinforcement and
25

Weld Size

4.5 Calculations for the Opening of Port #22 for the Reinforcement and
32

Weld Size

4.6 Calculations for the Opening of Port #24 for the Reinforcement and
36

Weld Size

7.  Analysis for the Opening Port #14 for the Reinforcement and 40

> Weld Size

8.  Calculations for the Opening of Port #20 for the Reinforcement and
    44

> Weld Size

5\. Calculations and Analysis for the Flanges, Bolts and Welds

5.1 Upstream End Flange, Sliding Flange and the Bellow Flange 48

5.2 Downstream End Flange and the Bellow Flange 58

5.3 Calculations for Flanges #8, #13, #15, #19, #21 and #23 61

6\. Calculations and Analysis for the Saddle Support 63

7\. Appendix 66

![](media/image2.png){width="6.752777777777778in"
height="5.545833333333333in"}

Figure i.1, [An overall view of the vacuum vessel weldment of 1.3
MHz,]{.underline}

[β=1,Cryomodulue]{.underline}

Reference drawing: ME -- 443027

Overall length = 448.23 in

W~ws~ = 7,745 lbs. the weight of the vessel shell weldment.

Assuming the vessel shell is operating under the room temperature.

Vessel shell material: SA 516-60

Materials for flanges, ports and nozzles: see respective notes and
drawings.

**1. [Calculation for the Vacuum Vessel Cylindrical Shell
Thickness]{.underline}**

*[The Vacuum Vessel Cylindrical Shell:]{.underline}*

Given:

P: 14.7 psi, the external design pressure,

D~o~: 38.00 in, outside diameter of the vessel shell

D~i~: 37.25 in, inside diameter of the vessel shell

t~n~: The nominal vessel shell thickness.

F~a~: 15,000 psi, the max. allowable stress of the shell material at the
operating temp,

also know as S~v~. (See Page 1-1).

Vessel shell material: SA 516-60, carbon steel, see dwg. #E-443054

L~v~: Design length of the vessel between lines of the support.

M.A: Material corrosion and other misc. allowance, \~ 1/16"

Assuming:

L~v~ = 228.30 in (per drawing #E-443027-01)

Find out the required vessel shell thickness t~n~:

1.  Try t~n~ = 0.375"

L~v~/D~o~ = 228.30" / 38.00" = 6.0079

D~o~/t~n~ = 38.00" / 0.375" = 101.3333

Per section UG-28, also Fig. G and Fig. CS-2 in Subpart 3 of Section II,
part D,

ASME VIII, Div. 1:

It is found out that:

A = 0.0002

B = 2850

Then:

P~a~ = (4B) / (3D~o~/t~n~)

= (4 x 2850) ÷ (3 x 101.3333)

= 37.50 psi

where:

P~a~ : The max. allowable external working pressure, psi

A: Factor determined from Fig. G

B: Factor determined from the applicable material chert in Fig. CS-2.

2.  Try t~r2~ = 0.25

> L~v~/D~o~ = 6.053
>
> D~o~/t~r2~ = 38"/0.25" = 152
>
> Find:
>
> A ≈ 0.000115
>
> Since the value of A falling to the left of the applicable
> material/temperature
>
> curve, so:
>
> P~a2~ = (2AE/3(Do/t~r2~)) = 15.13 psi
>
> So, **t~r~** **= 0.250"**
>
> So the vessel shell thickness is required 0.25", using 3/8" as vessel
> shell thickness is above the shell thickness required and it is
> satisfactory.
>
> **2. [The Permissible Out of Roundness of the Vessel Cylindrical
> Shell]{.underline}**

Given:

P: 14.7 psi, the external design pressure,

D~o~: 38.00 in, outside diameter of the vessel shell

D~i~: 37.25 in, inside diameter of the vessel shell

t~n~: 0.375", the nominal vessel shell thickness.

F~a~: 15,000 psi, the max. allowable stress of the shell material at the
operating temp,

also know as S~v~. (See Page 1-1).

Vessel shell material: SA 516-60, carbon steel, see dwg. #E-443054

L~v~: \~230", design length of the vessel between lines of the support.

The difference between the max. dia and the mini. dia. of the vessel
shell at any cross

section shall not exceed:

1.  x 38 in = 0.38 in

> (per section UG-80 (b)(1) of ASME VIII, Div.1)

Per section UG-80(b)(2) and figure UG-80 of ASME VIII, Div.1, it is
found out that

the maximum plus or minus deviation from the true circular form,
measured radially on

the outside or inside of the cylindrical vessel shall not exceed the
maximum permissible

deviation e:

e ≈ 0.85t

= 0.85 x 0.375 in

= 0.319 in

where:

D~o~ / t = 101

L~v~ / D~o~ = 6.052

Per Figure UG-29.2, section UG-29 of ASME VIII, Div.1, it is found out
that:

Arc length ≈ 0.36 D~o~

> = 0.36 x 38 in
>
> = 13.68 in
>
> Chord length = 2 x Arc length
>
> = 2 x 13.68 in
>
> = 27.36 in

So, in a chord length of 27.36 in, the maximum plus or minus deviation
from the true

circle form shall not exceed 0.32 in.

**3. [Analysis and Calculation for the Stffening Rings]{.underline}**

1.  **[To Define the Stiffening Ring for the Vessel Shell Under the
    External Pressure]{.underline}**

**[and the Calculations of the Weld]{.underline}**

Reference drawings: ME-443027

Given:

> E = 29 x 10^6^ psi, Modules of elasticity of the stiffening materials
>
> l~e~ = 1.1 (D~o~t)^0.5^ = 1.1 (38" x 0.375")^0.5^
>
> = 4.152", the effective regional length of the vessel shell
>
> F~y~: 38 ksi, min. yield stress of the stiffening material SA516-70
> under the operating
>
> temperature.

[*To find the required moment of inertia of the combined stiffening ring
-- shell cross section about the neutral axis parallel to the axial of
the shell, in^4^, I~s~'*:]{.underline}

I~s~' = \[D~o~^2^L~s~(t + A~s~/L~s~)A\] / 10.9 (eq. x-1)

(per UG-29, ASME VIII, Div.1)

Where:

A~s~: Cross-sectional area of the stiffening area, in^2^

= 1.50 in^2^, see figure 1.

L~s~: Distance between the stiffening ring and support, \~230.0 in

( See drawing ME-443027)

B: Factor determined from the applicable chart,

= 0.75 \[PD~o~ ÷ (t + As/Ls)\]

= 0.75 \[14.7 psi x 38 in ÷ (0.375 in + 1.50 in^2^/230 in)\]

= 1098

Per figure CS-2 in Subpart 3 of section II, Part D, where B = 1098
falling below the left end of the material/temperature curve, so:

A = 2B/E

= 2 x 1098 / (29 x 10^6^)

= 0.0000757

(per step 5 of UG-29, ASME VIII, Div. 1)

Then the Eq. x-1 becomes:

I~s~' = {\[38^2^ x 230 x (0.375 + 1.50/230) x 0.0000757\] / 10.9}in^4^

= 0.88 in^4^

[*To find the combined moment of inertia of the cross section area of
the* stiffening ring and shell]{.underline}

Figure 3.1 is the cross-section view of the stiffening ring with the
vessel shell, also refer to view S-S of drawing ME-443027, MC-443052

![](media/image3.png){width="6.585416666666666in"
height="2.842361111111111in"}

> Figure 3.1, [Cross-section view of the vessel shell with the
> stiffening rings]{.underline}

It is found that from the figure 3.1:

A~1~ = 4.152 in x 0.375 in = 1.557 in^2^

A~2~ = 0.5 in x 3.0 in = 1.500 in^2^

> y~1~ = 0.1875 in
>
> y~2~ = 1.8125 in
>
> y = (A~1~y~1~ + A~2~y~2~) ÷ (A~1~ + A~2~)
>
> = 0.9848 in
>
> dy~1~^2^ = (y -- y~1~)^2^ = 0.6357 in^2^
>
> dy~2~^2^ = (y -- y~2~)^2^ = 0.6851 in^2^
>
> I~1~ = 0.0182 in^4^

I~2~ = 1.125 in^4^

> I = ∑I~i~ + ∑A~i~dy~i~^2^
>
> = [3.1507 in^4^]{.underline} \> I'= 0.88 in^4^

Since the requirement moment of inertia I' is less than the value of I
provided by the combined moment inertia of stiffening ring and vessel
shell, so the stiffening ring design is satisfactory and it can be
treated as a support of the vessel.

*[Calculations of the attachment welds of the stiffening
ring]{.underline}:*

Per UG-30(e), ASME VIII, Div. 1:

Radial pressure load PL~s~ = 14.7 psi x 230 in

= 3381 lb/in

Radial shear load V = 0.01 PL~s~D~o~ = 0.01 x 14.7 psi x 230 in x 38 in

= 1,285 lb

Weld shear flow due to radial shear load: VQ/I

where Q is the first moment of area, per figure 3.1,

Q = 4.152 x 0.375 (0.9848 -- 0.375/2)

= 1.24 in^3^

VQ/I = 1,285 lb x 1.24 in^3^/ 3.1507 in^4^

= 505.73 lb/in

The combined weld load (required)

P~r~ = (3,381^2^ + 506^2^)^1/2^ lb/in

= 3,419 lb/in

The calculated welding load from the ring welds:

The most weaker material is the shell:

S~v~ = 15 ksi (SA 516-60)

The allowable fillet weld stress:

F~af~ = 0.55S~v~ = 0.55 x 15,000 psi

= 8,250 psi (per UW-18(d), ASME VIII, Div.1)

The minimum fillet weld leg size is: 0.25" for the attachment.

(per UG-30(f), ASME VIII, Div.1)

Since there is weld on both sides of the ring (see view S-S, ME-443027)

The allowable load from the weld is:

P~aw~ = 2 x 0.25 in x 8,250 psi

= 4,125 lb/in \> P~r~ = 3,419 lb/in

**So the** **min. fillet size of 0.25" is acceptable for the current
stiffening ring design.**

2.  **[To Define the Downstream End Flange as the Stiffening Ring for
    the Vessel]{.underline}**

**[Shell Under the External Pressure]{.underline}**

Reference drawings: ME-443027, MD-443039

Givien:

> E = 28 x 10^6^ psi, Modules of elasticity of the flange materials
>
> l~e~ = 1.1 (D~o~t)^0.5^ = 1.1 (38" x 0.375")^0.5^
>
> = 4.152", the effective regional length of the vessel shell
>
> F~y~: 25 ksi, min. yield stress of the stiffening material A182-F304L
> under the
>
> operating temperature.

![](media/image4.png){width="4.708333333333333in"
height="4.0680555555555555in"}

Figure 3.2, [The cross-section view of the end flange with the vessel
shell, also refer to]{.underline}

[detail L of drawing ME-443027, and drawing MD-443039]{.underline}..

*[To find the required moment of inertia of the combined end flange --
shell cross section about the neutral axis parallel to the axia of the
shell, in^4^, I~s~':]{.underline}*

I~s~' = \[D~o~^2^L~s~(t + A~s~/L~s~)A\] / 10.9 (eq. x-1)

(per UG-29, ASME VIII, Div.1)

Where:

A~sd~: Cross-sectional area of the downstream end flange stiffening
area, in^2^

= (1.18 x 2.26) in^2^ = 2.6668 in2, see figure 3.2

L~sd~: Distance between the stiffening ring and downstream end flange,
\~230.0 in

( See drawing ME-443027)

B: Factor determined from the applicable chart,

= 0.75 \[PD~o~ ÷ (t + As/Ls)\]

= 0.75 \[14.7 psi x 38 in ÷ (0.375 in + 2.6668 in^2^/230 in)\]

= 1084

Per figure CS-2 in Subpart 3 of section II, Part D, where B = 1084
falling below the left end of the material/temperature curve, so:

A = 2B/E

= 2 x 1084 / (28 x 10^6^)

= 0.0000774

(per step 5 of UG-29, ASME VIII, Div. 1)

Then the Eq. x-1 becomes:

I~s~' = {\[38^2^ x 230 x (0.375 + 2.6668/230) x 0.0000774\] / 10.9}in^4^

= ((332,120 x 0.3866 x 0.0000723) /10.9) in^4^

= 0.912 in^4^

*[To find the combined moment of inertia of the cross section area of
the stiffening ring and shell]{.underline}*

It is found that from the figure 3.2:

A~1~ = 2.67 in x 0.375 in = 1.001 in^2^

A~2~ = 1.18 in x 2.26 in = 2.6668 in^2^

> y~1~ = 0.1875 in
>
> y~2~ = 1.505 in
>
> y = (A~1~y~1~ + A~2~y~2~) ÷ (A~1~ + A~2~)
>
> = 1.1454 in
>
> dy~1~^2^ = (y -- y~1~)^2^ = 0.9176 in^2^
>
> dy~2~^2^ = (y -- y~2~)^2^ = 0.1293 in^2^
>
> I~1~ = 0.0117 in^4^

I~2~ = 1.1351 in^4^

> I = ∑I~i~ + ∑A~i~dy~i~^2^
>
> = (1.1468 + 1.2633) in^4^
>
> = [2.4101 in^4^]{.underline} \> I'= 0.912 in^4^

Since the requirement moment of inertia I' is less than the value of I
provided by the combined moment inertia of the end flange(downstream)
and vessel shell, so the downstream end flange is satisfactory and it
can be treated as a support of the vessel.

3.  **[To Define the Upstream End Flange as the Stiffening Ring for the
    Vessel Shell]{.underline}**

**[Under the External Pressure]{.underline}**

Reference drawings: Detail M of drawing ME-443027-2,

MD- 443040

L~su~: Distance between the stiffening ring and upstream end flange, \~
45.0 in

(See figure 4-4.3 of Page 32)

A~su~: ≈ 0.5 (45.67 - 38.12) x 1.57 in^2^ = 5.927 in^2^

Since L~su~ \< L~sd~; A~su~ \> A~sd~

From the calculation steps of section x.x, we can conclude that the
requirement moment of inertia I' is less than the value of I provided by
the combined moment inertia of the end flange(upstream) and vessel
shell, so the upstream end flange is satisfactory and it can be treated
as a support of the vessel.

**4. [Calculations and Analysis for the Vessel Shell
Openings]{.underline}**

1.  **[Calculations for the Opening of Coldmass Port for the
    Reinforcement and]{.underline}**

> **[Weld Size]{.underline}**

*[The main vessel shell]{.underline}*:

P: 14.7 psi, external design pressure

D~o~: 38.00 in, outside dia. Of the vessel shell.

D~i~: 37.25, inside dia. of the vessel shell.

t: 0.375 in, the nominal pipe wall thickness.

F~a~: 15,000 psi, the max. allowable stress of the shell material at the
operating temp,

also know as S~v~. at the operating temp., -20^0^ F to 100^0^ F (per
Table 1A, part D,

Section II, ASME Boiler & Pressure Vessel Code 1995 edition).

Vessel shell material: SA 516, grade 60, carbon steel (see ME-443054)

L~v~: Design length of the vessel section between lines of support (see
Page UG-28(b)),

≈ 230"

Find out the required vessel shell thickness t~r~:

Try t~r1~ = 0.375"

> L~v~/D~o~ = 230"/38" = 6.053
>
> D~o~/t~r~ = 38/0.375 = 101.333
>
> From section UG-28, also Fig. G & Fig. CS-2 in Subpart 3 of section
> II, part D,
>
> ASME VIII, Div I:
>
> Find:
>
> A = 0.0002
>
> B = 2850
>
> P~a1~= (4B)/(3(D~o~/t~r~)) = 37.50 psi
>
> where:
>
> A: Factor determined from Fig. G.
>
> B: Factor determined from applicable material chart in Fig. CS-2
>
> P~a~: Max. allowable external working pressure, psi

3.  Try t~r2~ = 0.25

> L~v~/D~o~ = 6.053
>
> D~o~/t~r2~ = 38"/0.25" = 152
>
> Find:
>
> A ≈ 0.000115
>
> Since the value of A falling to the left of the applicable
> material/temperature
>
> curve, so:
>
> P~a2~ = (2AE/3(Do/t~r2~)) = 15.13 psi
>
> So, **t~r~** **= 0.250"**

[The coldmass port:]{.underline}

d~o~: 25.197", outside dia. of the port.

D~i~: 15.512", inside dia . of the port.

t~n~: 4.842", wall thickness of the port.

t~rn~: required thickness of the port.

L: length of projection defining the thickness portion of integral
reinforcement of the

port beyond the outside surface of the vessel wall.

E: 28 x 10^6^ psi, modulus of elasticity of the material at the opening
temp.

Material: SA 182- 304L stainless steel pipe, F~a~ = S~n~ = 16,700 psi at
the operating

temp., -20^0^ F to 100^0^ F (per Table 1A, part D, Section II, ASME
Boiler &

Pressure Vessel Code 1995 edition).

Reference drawings: MD-443049, MD-443050, ME-443027, and also as shown
from

figure 4.1.

![](media/image5.png){width="5.615972222222222in"
height="4.533333333333333in"}

Figure 4.1, [3D view of the coldmass port and the vessel
shell]{.underline}

Find out the thickness required:

1.  Assuming L~1~ = 6.64 in, t~r1~ = 0.375 in, then:

> L~1~/d~o~ = 6.64/25.197 = 0.2635
>
> d~o~/t~r1~ = 25.197/0.375 = 67
>
> From Fig. G and Fig. HA-3 of subpart 3 of Sect. II, part D:
>
> A = 0.013
>
> B = 13300
>
> P~a~ = (4B)/(3(d~o~/t~r1~)) = (4 x 13300)/(3 x 67) = 264 psi \> P

2.  Try: L~2~ = 6.64 in, t~r2~ = 0.0625 in

> L~2~/d~o~ = 0.2635
>
> d~o~/t~r2~ = 403
>
> get:
>
> A = 0.00073
>
> B = 8000
>
> P~a~ = (4B)/(3(d~o~/t~r2~)) = (4 x 8000)/(3 x 403) = 26.47 psi \> P

3.  Try: L~3~ = 6.64 in, t~r3~ = 0.046 in

> L~3~/d~o~ = 0.2635
>
> d~o~/t~r3~ = 25.197/0.046 = 548
>
> get:
>
> A = 0.000445
>
> B = 6100
>
> P~a~ = (4B)/(3(d~o~/t~r2~)) = (4 x 6100)/(3 x 548) = 14.84 psi ≈ P
>
> So: **t~rn~** **= t~r3~ = 0.046 in**

*[Size of the weld required]{.underline}*:

Per UG-37, UW-16 and Fig. UW-16.1(p) of ASME VIII, Div.1.

t~o~: the throat dimension of the outer attachment weld, =0.5 t~min~.

t~i:~ the throat dimension of the inside of the shell cutout, ≥
0.7t~min~

t~o~ = 0.5 t~min~ = 0.5 x 0.375" = 0.1875", **so the fillet leg size is
0.27"**

t~i~ = 0.7 t~min~ = 0.7 x 0.375" =0.26", **so the fillet leg size is
0.37"**

*[Area of the reinforcement required A~r~]{.underline}*:

Per section UG-37(d)(1):

A~r~ = 0.5(dt~r~F + 2t~n~t~r~F(1-f~r1~)

= 0.5 x (16.299" x 0.25" x 1.0 +2 x 4.842" x 0.25" x 1.0 x 0)

= **2.0374 in^2^**

Where:

d: Finished diameter of circle opening, see Fig. UG-37.1 of UG-37.

D = 16.299 in (per drawing ME-443054)

F: Correction factor

f~r~ : Strength reduction factor, not greater than 1.0

f~r1~ =1.0 for nozzle wall abutting the vessel wall as shown in figure
UG-40(j),

(k), (n) and (o).

f~r2~ = S~n~ / S~v~

*[Area of reinforcement available A~a~]{.underline}:*

A~1~: Larger of the following calculated value: (area available in
shell):

= d(E~1~t -- Ft~r~)-2t~n~(E~1~t -- Ft~r~) x (1-f~r1~)

= 16.299" x (1 x 0.375" -- 1.0 x 0.25") -- 0

= 2.0374 in^2^

or = 2(t + t~n~)(E~1~t --Ft~r~)-2t~n~(E~1~t -- Ft~r~) x (1-f~r1~)

= 2 (0.375" + 4.842") x (1.0 x 0.375" -- 1.0 x 0.25") -- 0

= 1.30425 in^2^

So pick **A~1~ = 2.0374 in^2^**

A~2~: Smaller of the following:

= 5 (t~n~-t~rn~)tf~r2~

= 5 x (4.842 in -- 0.046 in) x 0.375 in x 1.0

= 8.9925 in^2^

or = 5 (t~n~-t~rn~) f~r2~t~n~

= 5 x (4.842 in -- 0.046 in) x 1.0 x 4.842 in

= 116.11 in^2^

So pick **A~2~ = 8.9925 in^2^**

A~3~: Area available in inward nozzle, use the smallest value:

Since there is no inward nozzle

So pick **A~3~ = 0.0 in^2^**

A~41~: Outward nozzle weld

= leg^2^ f~r2~

= 0.27^2^ in^2^ x 1.0

=**0.0729 in^2^**

A~43~: Inward nozzle weld

= leg^2^ f~r2~

= 0.37^2^ in^2^ x 1.0

= **0.1369 in^2^**

Then:

A~a~ = A~1~ + A~2~ + A~3~ + A~41~ + A~43~

= (2.0374 + 8.9925 + 0 + 0.0729 + 0.1369) in^2^

= **11.2397 in^2^** \> A~r~ = 2.0374 in^2^

# **Per Fig. UG-37.1, section UG-37, the opening is adequately reinforced under the current designated boundary conditions.**

The analysis and conclusion should be also true for openings of #2 (have
2 openings @ both ends, see drawing ME-443027), this is because: the
value of L~v~ for #2 opening is equal or smaller than the value for #3
opening, the smaller value of L~v~ will lead to smaller value of t~r~,
accordingly, it will lead to the smaller value of A~r~.

**4.2 [Calculations for the Opening of MC Port (#9) for the
Reinforcement and ]{.underline}**

**[Weld Size]{.underline}**

[The main vessel shell]{.underline}:

From section 4.1, it is found that **t~r~** **= 0.250"**

[The MC port:]{.underline}

d~o~: 11.339", outside dia. of the port.

D~i~: 10.945", inside dia . of the port.

t~n~: 0.197", wall thickness of the port.

t~rn~: required thickness of the port.

L: length of projection defining the thickness portion of integral
reinforcement of the

port beyond the outside surface of the vessel wall.

E: 28 x 10^6^ psi, modulus of elasticity of the material at the opening
temp.

Material: SA 249-TP304L stainless steel pipe, F~a~ = S~n~ = 14,200 psi
at the operating

temp., -20^0^ F to 100^0^ F (per Table 1A, part D, Section II, ASME
Boiler &

Pressure Vessel Code 1995 edition).

Reference drawings: ME-443027, and also as shown from figure 4.2.

![](media/image6.png){width="2.9055555555555554in"
height="2.595138888888889in"}![](media/image7.png){width="3.0006944444444446in"
height="2.6166666666666667in"}

Figure 4.2, [The partial 2D and 3D views of the MC port from
ME-443027]{.underline}

Find out the wall thickness required:

1.  Assuming L~1~ = 8.54 in, t~r1~ = 0.125 in, then:

> L~1~/d~o~ = 8.54/11.339 =0.7532
>
> d~o~/t~r1~ = 11.339/0.125 = 91
>
> From Fig. G and Fig. HA-3 of subpart 3 of Sect. II, part D:
>
> A = 0.00225
>
> B = 10500
>
> P~a~ = (4B)/(3(d~o~/t~r1~)) = (4 x 10500)/(3 x 91) = 153 psi \> P

2.  Try: L~2~ = 8.54 in, t~r2~ = 0.0625 in

> L~2~/d~o~ = 0.7532
>
> d~o~/t~r2~ = 181
>
> get:
>
> A = 0.00075
>
> B = 8050
>
> P~a~ = (4B)/(3(d~o~/t~r2~)) = (4 x 8050)/(3 x 181) = 59.3 psi \> P

Try: L~3~ = 8.54 in, t~r3~ = 0.033 in

> L~2~/d~o~ = 0.7532
>
> d~o~/t~r3~ = 344
>
> get:
>
> A = 0.000280
>
> B = 3950
>
> P~a~ = (4B)/(3(d~o~/t~r2~)) = (4 x 3950)/(3 x 344) = 15.31 psi ≈ P

So: **t~rn~** **= t~r3~ = 0.033 in**

*[Size of the weld required]{.underline}*:

Per UG-37, UW-16 and Fig. UW-16.1© of ASME VIII, Div.1.

t~min~: 0.197"

t~c:~ ≥ 0.7t~min~ = 0.138", **so the fillet leg size is 0.195"**

Full penetration weld plus the fillet weld t~c.~

*[Area of the reinforcement required A~r~]{.underline}*:

Per section UG-37(d)(1):

A~r~ = 0.5(dt~r~F + 2t~n~t~r~F(1-f~r1~)

= 0.5 x (11.339" x 0.25" x 1.0 +2 x 0.197" x 0.25" x 1.0 x 0.053)

= **1.4226 in^2^**

Where:

d: Finished diameter of circle opening, see Fig. UG-37.1 of UG-37.

D = 11.339 in (per drawing ME-443054)

F: Correction factor

f~r~ : Strength reduction factor, not greater than 1.0

f~r1~ = S~n~ / S~v~ for nozzle wall inserted through the vessel wall

f~r2~ = S~n~ / S~v~ = 14.2 / 15 =0.947

*[Area of reinforcement available A~a~]{.underline}*:

A~1~: Larger of the following calculated value: (area available in
shell):

= d(E~1~t -- Ft~r~)-2t~n~(E~1~t -- Ft~r~) x (1-f~r1~)

= 11.339" x (1 x 0.375" -- 1.0 x 0.25") -- 2 x 0.197 (1.0 x 0.375 -- 1.0
x 0.25) x

0.053

= (1.4174 -- 0.0026) in^2^

= 1.4148 in^2^

or = 2(t + t~n~)(E~1~t --Ft~r~)-2t~n~(E~1~t -- Ft~r~) x (1-f~r1~)

= 2 (0.375" + 0.197") x (1.0 x 0.375" -- 1.0 x 0.25") -- 2 x 0.197 (1.0
x 0.375 -- 1.0 x

0.25) x 0.053

= (0.143 -- 0.0026) in^2^

= 0.1405 in^2^

So pick **A~1~ = 1.4148 in^2^**

A~2~: Smaller of the following:

= 5 (t~n~-t~rn~)tf~r2~

= 5 x (0.197 in -- 0.033 in) x 0.375 in x 0.947

= 0.2912 in^2^

or = 5 (t~n~-t~rn~) f~r2~t~n~

= 5 x (0.197 in -- 0.033 in) x 0.947 x 0.197 in

= 0.153 in^2^

So pick **A~2~ = 0.153 in^2^**

A~3~: Area available in inward nozzle, use the smallest value:

Since there is no inward nozzle

So pick **A~3~ = 0.0 in^2^**

A~41~: Outward nozzle weld

= leg^2^ f~r2~

= 0.195^2^ in^2^ x 0.947

=**0.036 in^2^**

A~43~: Inward nozzle weld

= leg^2^ f~r2~

= 0.0^2^ in^2^ x 1.0

= **0.0 in^2^**

Then:

A~a~ = A~1~ + A~2~ + A~3~ + A~41~ + A~43~

= (1.4148 + 0.153 + 0 + 0.036 + 0.0) in^2^

= **1.6038 in^2^** \> A~r~ = 1.4226 in^2^

# **Per Fig. UG-37.1, section UG-37, the opening is adequately reinforced under the current designated boundary conditions.**

The analysis and conclusion should be true for all 8 openings (#9
openings, see drawing ME-443027 & ME-443054). Any issue related the
multiple openings will discuss in different section.

> **4.3 [Calculations for the Opening of Instrumentation Port (#10
> opening)]{.underline}**
>
> **[for the Reinforcement and Weld Size]{.underline}**

*[The main shell (Tube #9)]{.underline}*

From section 4.2, it is found that: **t~r~ = 0.033 in**

*[The instrumentation port:]{.underline}*

d~o~: 3.50", outside dia. of the port.

D~i~: 3.334", inside dia . of the port.

t~n~: 0.083", wall thickness of the port.

t~rn~: required thickness of the port.

L: length of projection defining the thickness portion of integral
reinforcement of the

port beyond the outside surface of the vessel wall.

E: 28 x 10^6^ psi, modulus of elasticity of the material at the opening
temp.

Material: SA 249-TP304L stainless steel pipe, F~a~ = S~n~ = 14,200 psi
at the operating

temp., -20^0^ F to 100^0^ F (per Table 1A, part D, Section II, ASME
Boiler &

Pressure Vessel Code 1995 edition).

Reference drawings: section G-G, ME-443027, and also as shown from
figure 4.2.

Find out the wall thickness requires:

> Assuming L~1~ = 2.0075 in, t~r1~ = 0.083 in, then:

where: L~1~ = 7.48 -- (11.339/2 -- 0.197) = 2.0075"

> L~1~/d~o~ = 2.0075/3.50 =0.5736
>
> d~o~/t~r1~ = 3.50/0.083 = 42
>
> From Fig. G and Fig. HA-3 of subpart 3 of Sect. II, part D:
>
> A = 0.009
>
> B = 13200
>
> P~a~ = (4B)/(3(d~o~/t~r1~)) = (4 x 13200)/(3 x 42) = 419 psi \> P

Try: L~2~ = 2.0075 in, t~r2~ = 0.015625 in

> L~2~/d~o~ = 0.5736
>
> d~o~/t~r2~ = 224
>
> get:
>
> A = 0.0007
>
> B = 8000
>
> P~a~ = (4B)/(3(d~o~/t~r2~)) = (4 x 8000)/(3 x 224) = 47.62 psi \> P
>
> Try: L~3~ = 2.0075 in, t~r3~ = 0.0094 in
>
> L~3~/d~o~ = 0.5732
>
> d~o~/t~r3~ = 372
>
> get:
>
> A = 0.0003
>
> B = 4250
>
> P~a~ = (4B)/(3(d~o~/t~r2~)) = (4 x 4250)/(3 x 372) = 15.23 psi ≈ P

So: **t~rn~** **= t~r3~ = 0.0094 in**

*[Size of the weld required]{.underline}*:

Per UG-37, UW-16 and Fig. UW-16.1© of ASME VIII, Div.1.

t~min~: 0.083"

t~c:~ ≥ 0.7t~min~ = 0.056", **so the fillet leg size is 0.08"**

Full penetration weld plus the fillet weld t~c.~

*[Area of the reinforcement required A~r~]{.underline}*:

Per section UG-37(d)(1):

A~r~ = 0.5(dt~r~F + 2t~n~t~r~F(1-f~r1~)

= 0.5 x (3.5" x 0.033" x 1.0 + 2 x 0.083" x 0.033" x 1.0 x 0)

= **0.0578 in^2^**

Where:

d: Finished diameter of circle opening, see Fig. UG-37.1 of UG-37.

D = 11.339 in (per drawing ME-443054)

F: Correction factor

f~r~ : Strength reduction factor, not greater than 1.0

f~r1~ = S~n~ / S~v~ for nozzle wall inserted through the vessel wall

f~r2~ = S~n~ / S~v~ = 14.2 / 14.2 = 1.0

*[Area of reinforcement available A~a~]{.underline}*:

A~1~: Larger of the following calculated value: (area available in
shell):

= d(E~1~t -- Ft~r~)-2t~n~(E~1~t -- Ft~r~) x (1-f~r1~)

= 3.50" x (1 x 0.197" -- 1.0 x 0.033") -- 2 x 0.083 (1.0 x 0.197 -- 1.0
x 0.033) x 0

= 0.574 in^2^

or = 2(t + t~n~)(E~1~t --Ft~r~)-2t~n~(E~1~t -- Ft~r~) x (1-f~r1~)

= 2 (0.197" + 0.083") x (1.0 x 0.197" -- 1.0 x 0.033") -- 2 x 0.197 (1.0
x 0.375 -- 1.0

x 0.25) x 0.0

= (2 x 0.28 x 0.164) in^2^

= 0.0918 in^2^

So pick **A~1~ = 0.574 in^2^**

Since A~1~ = 0.574 in^2^

Also: A~a~ = A~1~ + A~2~ + A~3~ + A~41~ + A~43~

That leads to:

A~a~ \> A~r~ = 0.0578 in^2^

# **Per Fig. UG-37.1, section UG-37, the opening is adequately reinforced under the current designated boundary conditions.**

The analysis and conclusion should be true for all 8 instrumentation
ports (#10 openings, see section G-G, drawing ME-443027).

4.  **[Calculations for the Opening of Port #16 for the Reinforcement
    and]{.underline}**

**[Weld Size]{.underline}**

*[The main vessel shell]{.underline}*:

From section 4.1, it is found that **t~r~** **= 0.250"**

[The port #16:]{.underline}

d~o~: 11.811", outside dia. of the port.

D~i~: 11.417", inside dia . of the port.

t~n~: 0.197", wall thickness of the port.

t~rn~: required thickness of the port.

L: length of projection defining the thickness portion of integral
reinforcement of the

port beyond the outside surface of the vessel wall.

E: 30 x 10^6^ psi, modulus of elasticity of the material at the opening
temp.

Material: SA 249-TP304L stainless steel pipe, F~a~ = S~n~ = 14,200 psi
at the operating

temp., -20^0^ F to 100^0^ F (per Table 1A, part D, Section II, ASME
Boiler &

Pressure Vessel Code 1995 edition).

Reference drawings: Section H-H of ME-443027, and also as shown

from figure 4-4.1 and figure 4-4.2..

![](media/image8.png){width="3.532638888888889in"
height="4.092361111111111in"}

Figure 4-4.1 [Multiple openings of the upstream end of the
vessel]{.underline}

![](media/image9.png){width="6.620138888888889in"
height="3.9159722222222224in"}

Figure 4-4.2, [The 2D view of the port #16 from drawing
ME-443027]{.underline}

Find out the wall thickness required:

> Assuming L~1~ = (23.62 -- 18.625) in = 5.00 in, t~r1~ = 0.035 in,
> then:
>
> L~1~/d~o~ = 5.00/11.811 = 0.4233
>
> d~o~/t~r1~ = 11.811/0.035 = 337
>
> From Fig. G and Fig. HA-3 of subpart 3 of Sect. II, part D:
>
> A = 0.00057
>
> B = 7700
>
> P~a~ = (4B)/(3(d~o~/t~r1~)) = (4 x 7700)/(3 x 337) = 30.46 psi \> P

1.  Try: L~2~ = 5.00in, t~r2~ = 0.025 in

> L~2~/d~o~ = 0.4233
>
> d~o~/t~r2~ = 472
>
> get:
>
> A = 0.00033
>
> B = 4650
>
> P~a~ = (4B)/(3(d~o~/t~r2~)) = (4 x 4650)/(3 x 472) = 13.13 psi \< P
>
> Try: L~3~ = 5.00 in, t~r3~ = 0.026in
>
> L~2~/d~o~ = 0.4233
>
> d~o~/t~r3~ = 454
>
> get:
>
> A = 0.00035
>
> B = 5000
>
> P~a~ = (4B)/(3(d~o~/t~r2~)) = (4 x 5000)/(3 x 454) = 14.68 psi \> ≈ P

So: **t~rn~** **= t~r3~ = 0.026 in**

*[Size of the weld required]{.underline}*:

Per UG-37, UW-16 (d.1) and Fig. UW-16.1(k) of ASME VIII, Div.1.

t~o~: the throat dimension of the outer attachment weld, ≥ 0.7t~min~.

t~i:~ the throat dimension of the inside of the shell cutout, ≥
0.7t~min~

t~o~ = 0.7 t~min~ = 0.7 x 0.197" =0.138", **so the fillet leg size is
0.19"**

t~i~ = 0.7 t~min~ = 0.7 x 0.197" =0.138", **so the fillet leg size is
0.19"**

t~o~ + t~i~ = 0.138" + 0.138" = 0.276" ≥ 1.25 t~min~ =0.246"

*[Area of the reinforcement required A~r~]{.underline}*:

Per section UG-37(d)(1):

A~r~ = 0.5(dt~r~F + 2t~n~t~r~F(1-f~r1~))

= 0.5 x (11.417" x 0.25" x 1.0 + 2 x 0.197" x 0.25" x 1.0 x 0.053)

= (1.4271 + 0.0026) in^2^

= **1.4297 in^2^**

Where:

d: Finished diameter of circle opening, see Fig. UG-37.1 of UG-37.

D = 11.417 in (per drawing ME-443054)

F: Correction factor

f~r~ : Strength reduction factor, not greater than 1.0

f~r1~ = S~n~ / S~v~ for nozzle wall inserted through the vessel wall

f~r2~ = S~n~ / S~v~ = 14.2 / 15 =0.947

*[Area of reinforcement available A~a~]{.underline}:*

A~1~: Larger of the following calculated value: (area available in
shell):

= d~16~(E~1~t -- Ft~r~)-2t~n~(E~1~t -- Ft~r~) x (1-f~r1~)

= 4.90" x (1 x 0.375" -- 1.0 x 0.25") -- 2 x 0.197" x 1.0 x .125" x
0.053

= (0.6125 -- 0.0026) in^2^

= 0.6099 in^2^

or = 2(t + t~n~)(E~1~t --Ft~r~)-2t~n~(E~1~t -- Ft~r~) x (1-f~r1~)

= 2 (0.375" + 0.197") x (1.0 x 0.375" -- 1.0 x 0.25") -- 2 x 0.197" x
1.0 x .125" x

0.053

= (0.143 -- 0.0026) in^2^

= 0.1404 in^2^

So pick **A~1~ = 0.6099 in^2^**

where:

d~16~ ≈ 2 x 51% x \[(Π d~o~ x 13.3^o^) ÷ 360^o^ + 2d~n~\] = 4.90", per
UG-42 (a)(1) and section

C-C of drawing ME-443054. 51% is the portion of the cross section can be

considered between two openings (#16 & #9) by the ratio of the
diameters.

A~2~: Smaller of the following:

= 5 (t~n~-t~rn~)tf~r2~

= 5 x (0.197 in -- 0.026 in) x 0.375 in x 0.947

= 0.3036 in^2^

or = 5 (t~n~-t~rn~) f~r2~t~n~

= 5 x (0.197 in -- 0.026 in) x 0.947 x 0.197 in

= 0.1595 in^2^

So pick **A~2~ = 0.1595 in^2^**

A~3~: Area available in inward nozzle, use the smallest value:

Since there is no inward nozzle

So pick **A~3~ = 0.0 in^2^**

A~41~: Outward nozzle weld

= leg^2^ f~r2~

= 0.19^2^ in^2^ x 1.0

=**0.036 in^2^**

A~43~: Inward nozzle weld

= leg^2^ f~r2~

= 0.19^2^ in^2^ x 1.0

= **0.036in^2^**

Then:

A~a~ = A~1~ + A~2~ + A~3~ + A~41~ + A~43~

= (0.6099 + 0.1595 + 0 + 0.036 + 0.036) in^2^

= **0.8414 in^2^** \< A~r~ = 1.4297 in^2^

The opening is not adequate, either to reduce the t~r~ value or adding
reinforcing element.

Assuming to add a stiffening rib like #26 of drawing ME-443027 at the
location as shown in figure 4-4.3

![](media/image10.png){width="4.245138888888889in"
height="4.404861111111111in"}

Figure 4-4.3, [New version upstream end of the vessel with new
stiffening rib]{.underline}.

If going through the similar procedures as shown on section 4.1, it was
found that:

**t~r~ ≈ 0.13"** when L~v~ = 45"

then:

*[Area of the reinforcement required A~r~]{.underline}*:

Per section UG-37(d)(1):

A~r~ = 0.5(dt~r~F + 2t~n~t~r~F(1-f~r1~))

= 0.5 x (11.417" x 0.13" x 1.0 + 2 x 0.197" x 0.13" x 1.0 x 0.053)

= (0.7421 + 0.0014) in^2^

= **0.7435 in^2^**

*[Area of reinforcement available A~a~]{.underline}:*

A~1~: Larger of the following calculated value: (area available in
shell):

= d~16~(E~1~t -- Ft~r~)-2t~n~(E~1~t -- Ft~r~) x (1-f~r1~)

= 4.90" x (1 x 0.375" -- 1.0 x 0.13") -- 2 x 0.197" x 1.0 x .245" x
0.053

= (1.2005 -- 0.0051) in^2^

= 1.1954 in^2^

or = 2(t + t~n~)(E~1~t --Ft~r~)-2t~n~(E~1~t -- Ft~r~) x (1-f~r1~)

= 2 (0.375" + 0.197") x (1.0 x 0.375" -- 1.0 x 0.13") -- 2 x 0.197" x
1.0 x .245" x

0.053

= (0.2803 -- 0.0051) in^2^

= 0.2752 in^2^

So pick **A~1~ = 1.1954 in^2^**

where:

d~16~ ≈ 2 x 51% x \[(Π d~o~ x 13.3^o^) ÷ 360^o^ + 2d~n~\] = 4.90", per
UG-42 (a)(1) and section

C-C of drawing ME-443054. 51% is the portion of the cross section can be

considered between two openings (#16 & #9) by the ratio of the
diameters.

A~2~: Smaller of the following:

= 5 (t~n~-t~rn~)tf~r2~

= 5 x (0.197 in -- 0.026 in) x 0.375 in x 0.947

= 0.3036 in^2^

or = 5 (t~n~-t~rn~) f~r2~t~n~

= 5 x (0.197 in -- 0.026 in) x 0.947 x 0.197 in

= 0.1595 in^2^

So pick **A~2~ = 0.1595 in^2^**

A~3~: Area available in inward nozzle, use the smallest value:

Since there is no inward nozzle

So pick **A~3~ = 0.0 in^2^**

A~41~: Outward nozzle weld

= leg^2^ f~r2~

= 0.19^2^ in^2^ x 1.0

=**0.036 in^2^**

A~43~: Inward nozzle weld

= leg^2^ f~r2~

= 0.19^2^ in^2^ x 1.0

= **0.036in^2^**

Then:

A~a~ = A~1~ + A~2~ + A~3~ + A~41~ + A~43~

= (1.1954 + 0.1595 + 0 + 0.036 + 0.036) in^2^

= **1.4269 in^2^** \> A~r~ = 0.7435 in^2^

# **Per Fig. UG-37.1, section UG-37, the opening is adequately reinforced under the modified boundary conditions.**

Applying the similar approach, it also will find out that the opening #9
at the upstream end is adequately reinforced under the modified boundary
condition.

5.  **[Calculations for the Opening of Port #22 for the Reinforcement
    and]{.underline}**

**[Weld Size]{.underline}**

*[The main vessel shell]{.underline}*:

From section 4.4, it is found that **t~r~** **= 0.13"**

[The port #22:]{.underline}

d~o~: 8.780", outside dia. of the port.

D~i~: 8.386", inside dia . of the port.

t~n~: 0.197", wall thickness of the port.

t~rn~: required thickness of the port.

L: length of projection defining the thickness portion of integral
reinforcement of the

port beyond the outside surface of the vessel wall.

E: 28 x 10^6^ psi, modulus of elasticity of the material at the opening
temp.

Material: SA 249-TP304L stainless steel pipe, F~a~ = S~n~ = 14,200 psi
at the operating

temp., -20^0^ F to 100^0^ F (per Table 1A, part D, Section II, ASME
Boiler &

Pressure Vessel Code 1995 edition).

Reference drawings: Section K-K and detail L of ME-443027, and also as
shown

from figure 4-5.1

Find out the wall thickness required:

1.  Assuming L~1~ = 14.533 in, t~r1~ = 0.0625 in, then:

> L~1~/d~o~ = 14.533/8.780 =1.6552

d~o~/t~r1~ = 8.780/0.0.0625 = 140

From Fig. G and Fig. HA-3 of subpart 3 of Sect. II, part D:

> A = 0.00053
>
> B = 7400

P~a~ = (4B)/(3(d~o~/t~r1~)) = (4 x 7400)/(3 x 140) = 70.48 psi \> P

2.  Try: L~2~ = 14.533 in, t~r2~ = 0.03125 in

> L~2~/d~o~ = 1.6552

d~o~/t~r2~ = 281

> get:

A = 0.00017

B = 2300

P~a~= (4B)/(3(d~o~/t~r2~)) = (4 x 2300)/(3 x 281)

= 10.91 psi \< P

![](media/image11.png){width="4.911111111111111in"
height="5.497916666666667in"}

Figure 4-5.1, [The 2D view of the ports of #22, #24 & #14]{.underline}

> Try: L~3~ = 14.533 in, t~r3~ = 0.035 in
>
> L~2~/d~o~ = 1.6552
>
> d~o~/t~r3~ = 251
>
> get:
>
> A = 0.00021
>
> B = 2800
>
> P~a~ = (4B)/(3(d~o~/t~r2~)) = (4 x 2800)/(3 x 251) = 14.87 psi ≈ P

So: **t~rn~** **= t~r3~ = 0.035 in**

*[Size of the weld required]{.underline}*:

Per UG-37, UW-16 (d.1) and Fig. UW-16.1(k) of ASME VIII, Div.1.

t~o~: the throat dimension of the outer attachment weld, =0.5 t~min~.

t~i:~ the throat dimension of the inside of the shell cutout, ≥
0.7t~min~

t~o~ = 0.7 t~min~ = 0.7 x 0.197" =0.138", **so the fillet leg size is
0.19"**

t~i~ = 0.7 t~min~ = 0.7 x 0.197" =0.138", **so the fillet leg size is
0.19"**

t~o~ + t~i~ = 0.138" + 0.138" = 0.276" ≥ 1.25 t~min~ =0.246"

*[Area of the reinforcement required A~r~]{.underline}*:

Per section UG-37(d)(1):

A~r~ = 0.5(dt~r~F + 2t~n~t~r~F(1-f~r1~))

= 0.5 x (8.386" x 0.13" x 1.0 +2 x 0.197" x 0.13" x 1.0 x 0.053)

= 0.5 x (1.0902 + 0.0027)

= **0.5465 in^2^**

Where:

d: Finished diameter of circle opening, see Fig. UG-37.1 of UG-37.

D = 8.386 in (per drawing ME-443054)

F: Correction factor

f~r~ : Strength reduction factor, not greater than 1.0

f~r1~ = S~n~ / S~v~ for nozzle wall inserted through the vessel wall

f~r2~ = S~n~ / S~v~ = 14.2 / 15 =0.947

*[Area of reinforcement available A~a~]{.underline}:*

A~1~: Larger of the following calculated value: (area available in
shell):

= d~22~(E~1~t -- Ft~r~)-2t~n~(E~1~t -- Ft~r~) x (1-f~r1~)

= 3.243"x (1 x 0.375" -- 1.0 x 0.13") -- 2 x 0.197 x (1.0 x 0.375"-1.0 x
0.13) x

0.053

= (0.7945 -- 0.0051) in^2^

= 0.7894 in^2^

or = 2(t + t~n~)(E~1~t --Ft~r~)-2t~n~(E~1~t -- Ft~r~) x (1-f~r1~)

= 2 (0.375" + 0.197") x (1.0 x 0.375" -- 1.0 x 0.13") -- 2 x 0.197 x
(1.0 x 0.375"-1.0

x 0.13) x 0.053

= 0.28 -- 0.0051

= 0.2749 in^2^

So pick **A~1~ = 0.7894 in^2^**

where:

d~22~ ≈ 2 x 77% x \[(Π d~o~ x 5.4^o^) ÷ 360^o^ + d~n22~ + d~n24~ \] =
3.243", per UG-42 (a)(1) and

section C-C of drawing ME-443054. 51% is the portion of the cross
section can

be considered between two openings (#22 & #24) by the ratio of the
diameters.

A~2~: Smaller of the following:

= 5 (t~n~-t~rn~)tf~r2~

= 5 x (0.197 in -- 0.035 in) x 0.375 in x 0.947

= 0.2877 in^2^

or = 5 (t~n~-t~rn~) f~r2~t~n~

= 5 x (0.197 in -- 0.046 in) x 0.947 x 0.197 in

= 0.1408 in^2^

So pick **A~2~ = 0.1408 in^2^**

A~3~: Area available in inward nozzle, use the smallest value:

Since there is no inward nozzle

So pick **A~3~ = 0.0 in^2^**

A~41~: Outward nozzle weld

= leg^2^ f~r2~

= 0.19^2^ in^2^ x 1.0

=**0.0361 in^2^**

A~43~: Inward nozzle weld

= leg^2^ f~r2~

= 0.19^2^ in^2^ x 1.0

= **0.0361in^2^**

Then:

A~a~ = A~1~ + A~2~ + A~3~ + A~41~ + A~43~

= (0.7894 + 0.1408 + 0 + 0.0361 + 0.0361) in^2^

= **0.9979 in^2^** \> A~r~ = 0.5465 in^2^

# **Per Fig. UG-37.1, section UG-37, the opening is adequately reinforced under the current designated boundary conditions.**

The analysis and conclusion should be also true for openings of #2 (have
2 openings @ both ends, see drawing ME-443027), this is because: the
value of L~v~ for #2 opening is equal or smaller than the value for #3
opening, the smaller value of L~v~ will lead to smaller value of t~r~,
accordingly, it will lead to the smaller value of A~r~.

**4.6 [Calculations for the Opening of Port #24 for the Reinforcement
and ]{.underline}**

**[Weld Size]{.underline}**

*[The main vessel shell]{.underline}*:

From section 4.4, it is found that **t~r~** **= 0.130"**

[The port #24:]{.underline}

d~o~: 2.756", outside dia. of the port.

d~i~: 2.52", inside dia . of the port.

t~n~: 0.118", wall thickness of the port.

t~rn~: required thickness of the port.

L: length of projection defining the thickness portion of integral
reinforcement of the

port beyond the outside surface of the vessel wall.

E: 28 x 10^6^ psi, modulus of elasticity of the material at the opening
temp.

Material: SA 249-TP304L stainless steel pipe, F~a~ = S~n~ = 14,200 psi
at the operating

temp., -20^0^ F to 100^0^ F (per Table 1A, part D, Section II, ASME
Boiler &

Pressure Vessel Code 1995 edition).

Reference drawings: Section K-K and detail L of ME-443027, and also as
shown

from figure 4-4.3 and figure 4-5.1.

Find out the wall thickness required:

> 1\. Assuming L~1~ = (22.44 -- 18.625) in = 3.815 in, t~r1~ = 0.0625
> in, then:
>
> L~1~/d~o~ = 3.815/2.756 =1.3843
>
> d~o~/t~r1~ = 2.756/0.0.0625 = 44
>
> From Fig. G and Fig. HA-3 of subpart 3 of Sect. II, part D:
>
> A = 0.0035
>
> B = 11500
>
> P~a~ = (4B)/(3(d~o~/t~r1~)) = (4 x 11500)/(3 x 44) = 348 psi \> P

3.  Try: L~2~ = 3.815 in, t~r2~ = 0.03125 in

> L~2~/d~o~ = 1.3843
>
> d~o~/t~r2~ = 88
>
> get:
>
> A = 0.0012
>
> B = 8950
>
> P~a~ = (4B)/(3(d~o~/t~r2~)) = (4 x 8950)/(3 x 88) = 135.6 psi \> P

3\. Try: L~3~ = 3.815 in, t~r3~ = 0.013 in

> L~2~/d~o~ = 1.3843
>
> d~o~/t~r3~ = 212
>
> get:
>
> A = 0.00033
>
> B = 4550
>
> P~a~ = (4B)/(3(d~o~/t~r2~)) = (4 x 2800)/(3 x 251) = 28.62 psi \> P

4\. Try: L~3~ = 3.815 in, t~r3~ = 0.011 in

> L~2~/d~o~ = 1.3843
>
> d~o~/t~r3~ = 251
>
> get:
>
> A = 0.000215
>
> B = 2900
>
> P~a~ = (4B)/(3(d~o~/t~r2~)) = (4 x 2900)/(3 x 251) = 15.41 psi ≈ P

So: **t~rn~** **= t~r3~ = 0.011 in**

*[Size of the weld required]{.underline}*:

Per UG-37, UW-16 (d.1) and Fig. UW-16.1(k) of ASME VIII, Div.1.

t~o~: the throat dimension of the outer attachment weld, =0.5 t~min~.

t~i:~ the throat dimension of the inside of the shell cutout, ≥
0.7t~min~

t~o~ = 0.7 t~min~ = 0.7 x 0.118" =0.0826", **so the fillet leg size is
0.117"**

t~i~ = 0.7 t~min~ = 0.7 x 0.197" =0.0826", **so the fillet leg size is
0.117"**

t~o~ + t~i~ = 0.0826" + 0.0826" = 0.165" ≥ 1.25 t~min~ =0.148"

*[Area of the reinforcement required A~r~]{.underline}*:

Per section UG-37(d)(1):

A~r~ = 0.5(dt~r~F + 2t~n~t~r~F(1-f~r1~))

= 0.5 x (2.52" x 0.13" x 1.0 + 2 x 0.118" x 0.13" x 1.0 x 0.053)

= 0.5 x (0.3276 + 0.0016) in^2^

= **0.1646 in^2^**

Where:

d: Finished diameter of circle opening, see Fig. UG-37.1 of UG-37.

d = 2.52 in (per drawing ME-443054)

F: Correction factor

f~r~ : Strength reduction factor, not greater than 1.0

f~r1~ = S~n~ / S~v~ for nozzle wall inserted through the vessel wall

f~r2~ = S~n~ / S~v~ = 14.2 / 15 =0.947

*[Area of reinforcement available A~a~]{.underline}:*

A~1~: Larger of the following calculated value: (area available in
shell):

= d~24~(E~1~t -- Ft~r~)-2t~n~(E~1~t -- Ft~r~) x (1-f~r1~)

= 0.9686" x (1 x 0.375" -- 1.0 x 0.13") -- 2 x 0.118" x 1.0 x .245" x
0.053

= (0.237 - 0.0031) in^2^

= 0.2339 in^2^

or = 2(t + t~n~)(E~1~t --Ft~r~)-2t~n~(E~1~t -- Ft~r~) x (1-f~r1~)

= 2 (0.375" + 0.118") x (1.0 x 0.375" -- 1.0 x 0.13") -- 2 x 0.118" x
1.0 x .245" x

0.053

= (0.1208 -- 0.0031) in^2^

= 0.1177 in^2^

So pick **A~1~ = 0.2339 in^2^**

where:

d~24~ ≈ 2 x 23% x \[(Π d~o~ x 5.4^o^) ÷ 360^o^ + d~n22~ + d~n24~ \] =
0.9686", per UG-42 (a)(1)

and section C-C of drawing ME-443054. 51% is the portion of the cross
section

can be considered between two openings (#22 & #24) by the ratio of the

diameters.

A~2~: Smaller of the following:

= 5 (t~n~-t~rn~)tf~r2~

= 5 x (0.118 in -- 0.011 in) x 0.375 in x 0.947

= 0.190 in^2^

or = 5 (t~n~-t~rn~) f~r2~t~n~

= 5 x (0.118 in -- 0.011 in) x 0.947 x 0.118 in

= 0.0598 in^2^

So pick **A~2~ = 0.0598 in^2^**

A~3~: Area available in inward nozzle, use the smallest value:

Since there is no inward nozzle

So pick **A~3~ = 0.0 in^2^**

A~41~: Outward nozzle weld

= leg^2^ f~r2~

= 0.11^2^ in^2^ x 1.0

=**0.012 in^2^**

A~43~: Inward nozzle weld

= leg^2^ f~r2~

= 0.11^2^ in^2^ x 1.0

= **0.012in^2^**

Then:

A~a~ = A~1~ + A~2~ + A~3~ + A~41~ + A~43~

= (0.2339 + 0.0598 + 0 + 0.0121 + 0.0121) in^2^

= **0.3179 in^2^** \> A~r~ = 0.1646 in^2^

# **Per Fig. UG-37.1, section UG-37, the opening is adequately reinforced under the current designated boundary conditions.**

**4.7 [Analysis for the Opening of Port #14 for the Reinforcement and
Weld Size]{.underline}**

*[The main vessel shell]{.underline}*:

From section 4.1, it is found that **t~r~** **= 0.250"**

[The port #14:]{.underline}

d~o~: 6.625", outside dia. of the port.

d~i~: 6.187", inside dia . of the port.

t~n~: 0.219", wall thickness of the port.

t~rn~: required thickness of the port.

L: length of projection defining the thickness portion of integral
reinforcement of the

port beyond the outside surface of the vessel wall.

L = 22.44"- 18.625 = 3.815"

E: 28 x 10^6^ psi, modulus of elasticity of the material at the opening
temp.

Material: SA 249-TP304L stainless steel pipe, F~a~ = S~n~ = 14,200 psi
at the operating

temp., -20^0^ F to 100^0^ F (per Table 1A, part D, Section II, ASME
Boiler &

Pressure Vessel Code 1995 edition).

Reference drawings: Section K-K and detail L of ME-443027, and also as
shown

from figure 4-4.3 and figure 4-5.1.

Find out the wall thickness required:

> 1\. Assuming L~1~ = (22.44 -- 18.625) in = 3.815 in, t~r1~ = 0.0625
> in, then:
>
> L~1~/d~o~ = 3.815/6.625 =0.5758
>
> d~o~/t~r1~ = 6.625/0.0625 = 106
>
> From Fig. G and Fig. HA-3 of subpart 3 of Sect. II, part D:
>
> A = 0.0023
>
> B = 10500
>
> P~a~ = (4B)/(3(d~o~/t~r1~)) = (4 x 10500)/(3 x 106) = 132 psi \> P

2.  Try: L~2~ = 3.815 in, t~r2~ = 0.02 in

> L~2~/d~o~ = 0.5758
>
> d~o~/t~r2~ = 331
>
> get:
>
> A = 0.0004
>
> B = 7000
>
> P~a~ = (4B)/(3(d~o~/t~r2~)) = (4 x 7000)/(3 x 331) = 28.2 psi \> P

3\. Try: L~3~ = 3.815 in, t~r3~ = 0.0170 in

> L~3~/d~o~ = 0.5758
>
> d~o~/t~r3~ = 390
>
> get:
>
> A = 0.00031
>
> B = 4350
>
> P~a~ = (4B)/(3(d~o~/t~r2~)) = (4 x 4350)/(3 x 390) = 14.87 psi ≈ P

So: **t~rn~** **= t~r3~ = 0.017 in**

*[Size of the weld required]{.underline}*:

Per UG-37, UW-16 (d.1) and Fig. UW-16.1(k) of ASME VIII, Div.1.

t~o~: the throat dimension of the outer attachment weld, =0.5 t~min~.

t~i:~ the throat dimension of the inside of the shell cutout, ≥
0.7t~min~

t~o~ = 0.7 t~min~ = 0.7 x 0.219" =0.153", **so the fillet leg size is
0.216"**

t~i~ = 0.7 t~min~ = 0.7 x 0.219" =0.153", **so the fillet leg size is
0.216"**

t~o~ + t~i~ = 0.153" + 0.153" = 0.306" ≥ 1.25 t~min~ =0.274"

t~i22~ = t~o22~ = 0.19" \< t~i14~ = t~o14~ = 0.21"

*[Area of the reinforcement required A~r~]{.underline}*:

Per section UG-37(d)(1):

A~r~ = 0.5(dt~r~F + 2t~n~t~r~F(1-f~r1~))

= 0.5 x (6.187" x 0.25" x 1.0 + 2 x 0.219" x 0.25" x 1.0 x 0.053)

= (0.7734 + 0.0058) in^2^

= **0.7792 in^2^**

Where:

d: Finished diameter of circle opening, see Fig. UG-37.1 of UG-37.

d = 6.187 in (per drawing ME-443054)

F: Correction factor

f~r~ : Strength reduction factor, not greater than 1.0

f~r1~ = S~n~ / S~v~ for nozzle wall inserted through the vessel wall

f~r2~ = S~n~ / S~v~ = 14.2 / 15 =0.947

*[Area of reinforcement available A~a~]{.underline}:*

A~1~: Larger of the following calculated value: (area available in
shell):

= d~14~(E~1~t -- Ft~r~)-2t~n~(E~1~t -- Ft~r~) x (1-f~r1~)

= 5" x (1 x 0.375" -- 1.0 x 0.25") -- 2 x 0.219" x 1.0 x .125" x 0.053

= (0.625 - 0.0058) in^2^

= 0.6192 in^2^

or = 2(t + t~n~)(E~1~t --Ft~r~)-2t~n~(E~1~t -- Ft~r~) x (1-f~r1~)

= 2 (0.375" + 0.219") x (1.0 x 0.375" -- 1.0 x 0.25") -- 2 x 0.219" x
1.0 x .125" x

0.053

= (0.1485 -- 0.0058) in^2^

= 0.1427 in^2^

So pick **A~1~ = 0.6192 in^2^**

where:

d~14~ ≈ 2 x 71% x (Π d~o~ x 10.9^o^) ÷ 360^o^ = 5.08", per UG-42 (a)(1)
and section C-C of

drawing ME-443054. 71% is the portion of the cross section can be
considered

between two openings (#24 & #14) by the ratio of the diameters.

A~2~: Smaller of the following:

= 5 (t~n~-t~rn~)tf~r2~

= 5 x (0.219in -- 0.017 in) x 0.375 in x 0.947

= 0.3587in^2^

or = 5 (t~n~-t~rn~) f~r2~t~n~

= 5 x (0.219 in -- 0.017 in) x 0.947 x 0.219 in

= 0.2095 in^2^

So pick **A~2~ = 0.2095 in^2^**

A~3~: Area available in inward nozzle, use the smallest value:

Since there is no inward nozzle

So pick **A~3~ = 0.0 in^2^**

A~41~: Outward nozzle weld

= leg^2^ f~r2~

= 0.21^2^ in^2^ x 1.0

=**0.044 in^2^**

A~43~: Inward nozzle weld

= leg^2^ f~r2~

= 0.21^2^ in^2^ x 1.0

= **0.044in^2^**

Then:

A~a~ = A~1~ + A~2~ + A~3~ + A~41~ + A~43~

= (0.6192 + 0.2095 + 0 + 0.044 + 0.044) in^2^

= **0.9167 in^2^** \> A~r~ = 0.7792 in^2^

**Per Fig. UG-37.1, section UG-37 and UG-42, the opening is adequately
reinforced under the boundary conditions.**

Notes:

1.  The actual t~r~ = 0.13" in the upstream end area after adding the
    stiffening rib as

> shown in figure x.1, that leads even smaller value of A~r~ and larger
> value of A~a~,
>
> so the additional reinforcing element is not necessary.

2.  The analysis of the above is based on the opening in the upstream
    end of the

> Vessel. Per sections A-A and C-C of drawing ME-443054, sections A-A
> and K-K of drawing ME-443027, #14 opening in the downstream end of the
> vessel has
>
> larger space distance to its neighboring opening, to this end, we
> conclude that #14 opening at the downstream end is adequately
> reinforced under the current design configuration.

**4.8 [Calculations for the Opening of Port #20 for the Reinforcement
and Weld Size]{.underline}**

*[The main vessel shell]{.underline}*:

From section x.x, it is found that **t~r~** **= 0.130"**

[The port #20:]{.underline}

d~o~: 4.50", outside dia. of the port.

d~i~: 4.26", inside dia . of the port.

t~n~: 0.120", wall thickness of the port.

t~rn~: required thickness of the port.

L: length of projection defining the thickness portion of integral
reinforcement of the

port beyond the outside surface of the vessel wall.

E: 28 x 10^6^ psi, modulus of elasticity of the material at the opening
temp.

Material: SA 249-TP304L stainless steel pipe, F~a~ = S~n~ = 14,200 psi
at the operating

temp., -20^0^ F to 100^0^ F (per Table 1A, part D, Section II, ASME
Boiler &

Pressure Vessel Code 1995 edition).

Reference drawings: Section J-J of ME-443027, and also as shown

from figure 4-4.3 and figure 4-8.1..

![](media/image12.png){width="5.247916666666667in"
height="4.465277777777778in"}

Figure 4-8.1, [The 2D view of the port #20 from drawing
ME-443027-2]{.underline}

Find out the wall thickness required:

> 1\. Assuming L~1~ = 2.771 in, t~r1~ = 0.035 in, then:
>
> L~1~/d~o~ = 2.771/4.50 = 0.6158
>
> d~o~/t~r1~ = 4.50/0.035 = 129
>
> From Fig. G and Fig. HA-3 of subpart 3 of Sect. II, part D:
>
> A = 0.0015
>
> B = 9500
>
> P~a~ = (4B)/(3(d~o~/t~r1~)) = (4 x 9500)/(3 x 129) = 98 psi \> P

3.  Try: L~2~ = 2.771 in, t~r2~ = 0.025 in

> L~2~/d~o~ = 0.6158
>
> d~o~/t~r2~ = 180
>
> get:
>
> A = 0.00089
>
> B = 8400
>
> P~a~ = (4B)/(3(d~o~/t~r2~)) = (4 x 8400)/(3 x 180) = 62.2 psi \> P

3\. Try: L~3~ = 2.771 in, t~r3~ = 0.013 in

> L~2~/d~o~ = 0.6158
>
> d~o~/t~r3~ = 375
>
> get:
>
> A = 0.00032
>
> B = 4400
>
> P~a~ = (4B)/(3(d~o~/t~r2~)) = (4 x 4400)/(3 x 375) = 15.64 psi \> ≈ P

So: **t~rn~** **= t~r3~ = 0.013 in**

*[Size of the weld required]{.underline}*:

Per UG-37, UW-16 (d.1) and Fig. UW-16.1(k) of ASME VIII, Div.1.

t~o~: the throat dimension of the outer attachment weld, ≥ 0.7t~min~.

t~i:~ the throat dimension of the inside of the shell cutout, ≥
0.7t~min~

t~o~ = 0.7 t~min~ = 0.7 x 0.120" =0.084", **so the fillet leg size is
0.119"**

t~i~ = 0.7 t~min~ = 0.7 x 0.120" =0.084", **so the fillet leg size is
0.119"**

t~o~ + t~i~ = 0.084" + 0.084" = 0.168" ≥ 1.25 t~min~ =0.15"

*[Area of the reinforcement required A~r~]{.underline}*:

Per section UG-37(d)(1):

A~r~ = 0.5(dt~r~F + 2t~n~t~r~F(1-f~r1~))

= 0.5 x (4.26" x 0.13" x 1.0 + 2 x 0.120" x 0.13" x 1.0 x 0.053)

= 0.5 x (0.5538 + 0.0017) in^2^

= **0.2778 in^2^**

Where:

d: Finished diameter of circle opening, see Fig. UG-37.1 of UG-37.

d = 4.26 in (per drawing ME-443054)

F: Correction factor

f~r~ : Strength reduction factor, not greater than 1.0

f~r1~ = S~n~ / S~v~ for nozzle wall inserted through the vessel wall

f~r2~ = S~n~ / S~v~ = 14.2 / 15 =0.947

*[Area of reinforcement available A~a~]{.underline}:*

A~1~: Larger of the following calculated value: (area available in
shell):

= d~20~(E~1~t -- Ft~r~)-2t~n~(E~1~t -- Ft~r~) x (1-f~r1~)

= 3.346" x (1 x 0.375" -- 1.0 x 0.13") -- 2 x 0.120" x 1.0 x .245" x
0.053

= (0.8198 - 0.0031) in^2^

= 0.8167 in^2^

or = 2(t + t~n~)(E~1~t --Ft~r~)-2t~n~(E~1~t -- Ft~r~) x (1-f~r1~)

= 2 (0.375" + 0.120") x (1.0 x 0.375" -- 1.0 x 0.13") -- 2 x 0.120" x
1.0 x .245" x

0.053

= (0.2426 -- 0.0031) in^2^

= 0.2395 in^2^

So pick **A~1~ = 0.8167 in^2^**

Where: d~20~ ≈ 2 x 27% x 6.1975 = 3.346", per UG-42 (a)(1) and front
view of drawing

ME-443054. 27% is the portion of the cross section can be considered
between

two openings (#16 & #20) by the ratio of the diameters.

6.1975" = 14.036" (ctr. to ctr. distance of the openings #20 & #16) --
0.5 x

15.677"

A~2~: Smaller of the following:

= 5 (t~n~-t~rn~)tf~r2~

= 5 x (0.120 in -- 0.013 in) x 0.375 in x 0.947

= 0.190 in^2^

or = 5 (t~n~-t~rn~) f~r2~t~n~

= 5 x (0.120 in -- 0.013 in) x 0.947 x 0.120 in

= 0.0608 in^2^

So pick **A~2~ = 0.0608 in^2^**

A~3~: Area available in inward nozzle, use the smallest value:

Since there is no inward nozzle

So pick **A~3~ = 0.0 in^2^**

A~41~: Outward nozzle weld

= leg^2^ f~r2~

= 0.12^2^ in^2^ x 1.0

=**0.014 in^2^**

A~43~: Inward nozzle weld

= leg^2^ f~r2~

= 0.12^2^ in^2^ x 1.0

= **0.014in^2^**

Then:

A~a~ = A~1~ + A~2~ + A~3~ + A~41~ + A~43~

= (0.8167 + 0.0608 + 0 + 0.014 + 0.014) in^2^

= **0.9055 in^2^** \> A~r~ = 0.2778 in^2^

# **Per Fig. UG-37.1, section UG-37, the opening is adequately reinforced under the designated boundary conditions.**

**5. [Analysis and Calculations for the Flanges,Their Bolts and
Welds]{.underline}**

**5.1 [Upstream End Flange, Sliding Flange and Bellow
Flange]{.underline}**

References: ME-443027-1, MD-443039, MD-443038,

TTF-CRY3-01.01.00, TTF-CRY3-01.03.00,

and Figure 5.1

![](media/image13.png){width="5.995138888888889in"
height="5.2652777777777775in"}

Figure 5.1, [Upstream end flanges with the below flange]{.underline}

Figure x.x.1 is the assembly configuration of the upstream flange
connecting with the below flange, where the upstream flange is assumed
as the combination of the flange MD-443038 and MD-443039.

It was found that:

A = 45.669" (dia.)

B~1~ = 42.677" (dia.)

B~2~ = 37.25" (dia.)

C = 44.488" (dia.)

G = 43.15" (41.85) (dia.)\*

T~1~ = 1.654"

T~2~ = 1.181"

P = 14.7 psi

n = 24

E~1~ = E~2~ = 2.8 x 10^7^ psi

\*: There are 2 o'ring grooves, conservatively pick the larger one.

Material: SA 182-F304L stainless steel, F~a~ = S~f~= 16,700 psi at the
operating

temp., -20^0^ F to 100^0^ F (per Table 1A, part D, Section II, ASME
Boiler &

Pressure Vessel Code 1995 edition).

assuming bolt spec: ½ - 13 UNC

A~b~ = 24 x 0.1257 in^2^ = 3.0168 in^2^

The cross-sectional area of the bolts using the root dia. (0.4001") of
the

thread in sq. in.

Per section Y-5.1, Y-5.2, figure Y-5.1.2 of Appendix Y,

Section 2-4, figure 2-4 of Appendix 2 of ASME VIII:

B~1~ \> ½ C,

So it is defined as: Class 2, Category 3 flange

*[Flange Analysis]{.underline}*:

The moment due to the pressure M'~p~:

M'~p~ = H~D~'h~D~' + H~T~'h~T~' (per Y-6.2(3), Appendix Y, ASME VIII)

H~D~'= 0.785 B~2~^2^ P

= 0.785 x 37.25^2^ in^2^ x 14.7 psi

= 16,012 lbs.

H~T~'= 0.785P (B~1~^2^ --B~2~^2^)

= 0.785 x 14.7 psi (42.677^2^ -- 37.25^2^) in^2^

= 5,006 lbs.

h~D~' = (B~1~ -- B~2~) ÷ 2

= (42.677 -- 37.25) in ÷ 2

= 2.7135 in

h~T~' = (B~1~ -- B~2~) ÷ 4

= (42.677 -- 37.25) in ÷ 4

= 1.3568 in

So: M'~p~ = (16,012 x 2.7135) lbs-in + (5,006 x 1.3568) lbs-in

= 50,241 lbs-in

= C~5~

(per section Y-6.2(4)(a), Appendix Y, ASME VIII, Div.1)

Let: C~6~ = 0.829 / log(B~1~/B~2~)

= 0.829 ÷ 0.059

= 14.051

(per section Y-6.2(4)(b), Appendix Y, ASME VIII, Div.1)

C~1~ = \[1 -- 2.095J~s~ log(A/B~1~)\] ÷ (- C~6~ -- 1.738 J~s~)

(per section Y-6.2(4)(c), Appendix Y, ASME VIII, Div.1)

Where:

J~s~ = (1/B~1~) \[2h~D~/β + h~C~/α\] + π ґ~B~

h~D~ = (C -- B) / 2

= (44.488 -- 42.677) / 2 in

= 0.9055 in

h~C~ = (A -- C) / 2

= (45.669 -- 44.488) / 2 in

= 0.5905 in

β = ( C + B)/ 2B

= (44.488 + 42.677) / (2 x 42.677)

= 1.021

α = ( A + C)/ 2B

= (45.669 + 44.488) / (2 x 42.677)

= 1.056

ґ~B~ = (1/n)\[4/(1 -- ($\overline{A}\overline{R}$)^2^)^1/2^$$tan^-1^((1+
$\overline{A}\overline{R}$)/(1 - $\overline{A}\overline{R}$))^1/2^ -- π
-2$\overline{A}\overline{R}$\]

where:

$\overline{A}\overline{R}$= nD/ πC

= (24 x 0.551") / π 44.488"

= 0.0946

D: the bolt hole diameter

ґ~B~ = (1/24)\[4/(1 -- (0.0946)^2^)^1/2^$$tan^-1^((1+ 0.0946)/(1 --
0.0946))^1/2^ -- π --

2 x 0.0946\]

= 0.041667 x (4.018 x 47.7141- π -- 0.1892)

= 7.849

J~s~ = (1/42.677") \[2 x 0.9055/1.021 + 0.5905/1.056\] + π x 7.849

= 0.0234 x (1.774 + 0.559) + 24.658

= 24.713

So:

C~1~ = \[1 -- 2.095 x 24.713 x log(45.669/42.677)\] ÷ ( - 14.051 --
1.738 x 24.713)

= ( 1 -- 1.5236 ) ÷ (-57.0022)

= 0.0091

C2 = (1.738J~p~M~p~ -- C~5~C~6~) ÷ (- C~6~ -- 1.738 J~s~)

(per section Y-6.2(4)(d), Appendix Y, ASME VIII, Div.1)

where:

J~p~ = (1/B~1~) \[h~D~/β + h~C~/α\] + π ґ~B~

= (1/42.677") \[0.9055/1.021 + 0.5905/1.056\] + π x 7.849

= 0.0234 x (0.8869 + 0.559) + 24.658

= 24.692

M~p~ = H~D~h~D~ + H~T~h~T~ + H~G~h~G~ (per sectionY-3, Appendix Y, ASME
VIII)

H~D~ = 0.785 B~1~^2^ P

= 0.785 x 42.677^2^ in^2^ x 14.7 psi

= 21,017 lbs.

H = 0.785G^2^P

= 0.785 x 43.15^2^ in^2^ x 14.7 psi

= 21,486 lbs.

H~T~ = H - H~D~

= 21,486 -- 21,017

= 469 lbs.

h~G~ = (C -- G)/2

= (44.488 -- 43.15) in / 2

= 0.669 in

h~D~ = 0.9055 in

h~T~ = (h~D~ + h~G~) / 2

= 0.7873 in

H~G~ = W -- H, gasket load

(per section 2-3, appendix 2, ASME VIII)

W = H + H~p,~

flange design load for the operating condition

(per section 2-5(c)(1), appendix 2, ASME VIII)

which leads to:

H~G~ = W -- H

= H + H~p~ -- H

= H~p~

= 2b x 3.14GmP

= 2 x 0.315 in x 3.14 x 43.15 in x 6.00 x 14.7 psi

= 7,529 lbs.

M~p~ = H~D~h~D~ + H~T~h~T~ + H~G~h~G~

= (21,017 x 0.9055 + 489 x 0.7873 + 7,529 x 0.669) lbs-in

= (19,031 + 385 + 5,037) lbs-in

= 24,453 lbs-in

C2 = (1.738J~p~M~p~ -- C~5~C~6~) ÷ (- C~6~ -- 1.738 J~s~)

= (1.738 x 24.692 x 24,453 -- 14.051 x 52,041) lbs-in ÷ (-14.051 --
1.738 x

24.713)

= 318,165 lbs-in / (- 57.002)

= - 5,582 lbs-in

For the rigid body rotation of the flange times E\*:

E~1~\* θ~rbi~ = X(C~4~ -- C~2~) ÷ (1.206 log(A/B~1~) --XC~3~ -- (1 --
X)C~1~)

where:

C~3~ = - (0.575 -- 1.206 Js log(A/B~1~)) ÷ (Js + t~1~^3^/F~1~')

$\because$ F~1~'= 0 (per section Y-3(6c), Appendix Y, ASME VIII, Div.1)

$\therefore$ C~3~ = - (0.575 -- 1.206 Js log(A/B~1~)) ÷ (Js)

= - (0.575 -- 1.206 x 24.713 x 0.0294) / 24.713

= 0.0122

C4 = - (J~p~M~p~) ÷ (Js + t~1~^3^/F~1~')

= - (24.692 x 24,453 lbs-in) / 24.713

= - 24,432 lbs-in

X = E~1~\* / (E~1~\* + E~2~\*)

E~1~\* = E~1~ t~1~^3^

E~2~\* = E~2~ t~2~^3^

$\because$ E~1~ = E~2~

$\therefore$ X = E~1~ t~1~^3^ / (E~1~ t~1~^3^ + E~2~ t~2~^3^ )

= t~1~^3^ / (t~1~^3^ + t~2~^3^ )

= 4.525 / (4.525 + 1.647)

= 0.733

E~1~\* θ~rbi~ = 0.733(-24,432 -- (-5,582)) ÷ \[(1.206 x 0.0294) --(0.733
x 0.0122) -- (1 --

0.733) x 0.00919)\]

= 0.733 x (-18,850) lbs-in ÷ (0.0355 -- 0.0089 -- 0.00245)

= - 13,817 lbs-in ÷ 0.02415

= - 572,133 lbs-in

E~2~\* θ~rbii~ = - E~1~\* θ~rbi~ (E~2~\* / (E~1~\*)

= - (-572,133 x 0.3635) lbs-in

= 207,970 lbs-in

*Total Flange Moment at Diameter B~1~*:

M~s1~ = C~3~ (E~1~\* θ~rbi~) + C~4~

= 0.0122 x (- 572,133 lbs-in) + (- 24,432 lbs-in)

= - 31,412 lbs-in

M~s2~ = C~1~ (E~2~\* θ~rbii~) + C~2~

= 0.00919 x 207,970 lbs-in + (- 5,582 lbs-in)

= - 3,671 lbs-in

*Unbalanced Flange Moment at Diameter B~1~:*

M~u1~ = 1.206 (E~1~\* θ~rbi~) log(A/B~1~)

= 1.206 x (- 572,133 lbs-in) x 0.0294

= - 20,286 lbs-in

M~u2~ = 1.206 (E~2~\* θ~rbii~) log(A/B~1~)

= 1.206 x (207,970 lbs-in) x 0.0294

= 7,374 lbs-in

*Balanced Flange Moment at Diameter B~1~:*

M~b1~ = M~s1~ - M~u1~

= - 31,412 lbs-in - (- 20,286 lbs-in)

= - 11,126 lbs-in

M~b2~ = M~s2~ - M~u2~

= - 3,671 lbs-in -- 7,374 lbs-in

= - 11,045 lbs-in

*Slope of Flange at Diameter B~1~ times E*:

E~1~ θ~Bi~ = (5.46/π t~1~^3^) (J~s~M~b1~ + J~p~M~p~) + E~1~\* θ~rbi~ /
t~1~^3^

= 0.384 in^-3^ x (24.713 x (-- 11,126 lbs-in) + 24.692 x 24,453
lbs-in) + (-

572,133 lbs-in / 4.525 in^3^)

= (126,273 + (-126,438)) psi

= -165 psi

E~2~ θ~Bii~ = (5.46/π t~2~^3^) (J~s~M~b2~ + J~p~M~p~) + E~2~\* θ~rbii~ /
t~2~^3^

(per section Y-6.2(e), Appendix Y, ASME VIII, Div.1)

= \[1.055 x (24.713 x (-11,045) + 24.692 x 24,453) + 207,970 / 1.647\]
psi

= (349,035 + 126,272) psi

= 475,307 psi

*Contact Force between Flange at h~c~:*

H~c~ = (M~p~ + M~b1~) / h~c~

= (24,453 +(- 11,126)) lbs-in / 0.5905 in

= 22,569 lbs.

*Bolt Load at Operating Conditions:*

W~m1~ = H + H~G~ + H~C~

=( 21,486 + 7,529 + 22,569) lbs

= 51,584 lbs.

*Operating Bolt Stress f~b~*:

f~b~ = W~m1~ /A~b1~

= 51,584 lbs / 3.0168 in^2^ (See page 52, section 5.1 of the note)

= [17,099 psi]{.underline}

*Design Pre-stress in Bolts:*

S~i~ = f~b~ -- 1.159 h~c~^2^ (M~p~ + M~b1~) / (2(1 -- X)a t~1~^3^ l
r~e1~B~1~)

where:

r~e1:~ ≈ 1.0, elasticity factor, the ratio between E~flange~ and E~bolt~

a = (A + C) / 2B~1~

= (45.669 + 44.488) in / 2 x 42.677 in

= 1.056 (shape factor)

l: = t~1~ + t~2~ + 0.5d~b~ (calculated strain length of the bolt)

= (1.654 + 1.181 + 0.25) in

= 3.085 in

S~i~ = 17,099 psi-- 1.159 x 0.5905^2^ in^2^ x (24,453 -- 11,126)lbs-in /
(2(1 -- 0.733) x

1.056 x 1.654^3^ in^3^ x 3.085 in x 1.0 x 42.677 in)

= 17,099 psi -- 5,386 lbs-in^3^ / 336 in^5^

= 17,083 psi

*Redial Stress in Flange I @ Bolt Circle*:

S~R1~ = 6(M~p~ + M~S1~) / t~1~^2^(πC --nD)

= 6 x ( 24,453 -31,412) lbs-in / 1.654^2^ in^2^ x (π 44.488 in -- 24 x
0.551 in)

= - 41,754 lbs-in / 346.175 in^3^

= \| -121 \| psi \< S~n~ = 16,700 psi

*Tangential Stress in Flange I @ Inside Diameter*:

S~t1~ = t~1~ E~1~ θ~Bi~ / B~1~

= 1.654 in x -165 psi / 42.677 in

= \|-6.39 \| psi \< S~n~ = 16,700 psi

*Tangential Flange Stress Adjacent to Central Nozzle*:

S~t2n~ = Y(M'~p~ + M~S2~) / t~2~^2^ B~2~

where:

Y = (1/(K-1))\[0.66845 + 5.71690 K^2^log~10~K/(K^2^ -1)\]

(per figure 2-7.1, Appendix 2, ASME VIII, Div. 1)

K = A/B

= 45.669 / 37.25

= 1.226

Y = (1/(1.226-1))\[0.66845 + 5.71690 x 1.5031 x log~10~K/(K^2^ -1)\]

= 4.4248 x (0.6685 + 5.7169 x 0.2643)

= 9.6437

$\therefore$ S~t2~ = 9.6437 x (50,241 -- 3,671) lbs-in / (1.181^2^ x
37.25) in^3^

= 8,644 psi \< S~n~ = 16,700 psi

*Longitudinal Hub Stress in Flange 1:*

S~h1~ = 0

(per equation (34c), section Y-6.3, appendix Y, ASME VIII, Div.1)

*Radial Stress in Flange 2 @ bolt circle*:

S~r2b~ = 6(M~p~ + M~S2~) / t~2~^2^(πC --nD)

**=** 6 x ( 24,453 -- 3,671) lbs-in / 1,181^2^ in^2^ x (π 44.488 in --
24 x 0.551 in)

= 124,692 lbs-in / 176.492 in^3^

= 707 psi \< S~n~ = 16,700 psi

*Radial Stress in Flange 2 @ Diameter B~1~*:

S~r2B1~ = 6 M~S2~ / t~2~^2^ πB~1~

= 6 x (-3,671) lbs-in / 1.181^2^ in^2^ x π 42.677 in

= \|- 118 \| psi \< S~n~ = 16,700 psi

*Tangential Stress in Flange 2 @ Diameter B~1~*:

S~t2B1~ = t~2~ E~2~ θ~Bii~ / B~1~ -- 1.8 M~S2~ / t~2~^2^ πB~1~

= (1.181 in x 475,307 psi / 42.677 in) -- (1.8 x (-- 3,671 lbs-in)) /
1.181^2^ in^2^

π 42.677 in

= 13,153 psi + 35 psi

= 13,188 psi \< S~n~ = 16,700 psi

*Radial and Tangential Stress at Center of Flange 2*:

S~r2~ = S~t2~ = (0.3094 PB~1~^2^ / t~2~^2^) -- (6 M~S2~ / πB~1~ t~2~^2^)

= (0.3094 x 14.7 psi x 42.677^2^ in^2^ / 1.181^2^ in^2^ )

-   ( 6 x (-3,671 lbs-in) / π 42.677 in x 1,181^2^ in^2^ )

= 5,939 psi + 118 psi

= 6,057 psi \< S~n~ = 16,700 psi

*[Welds Analysis and Calculations:]{.underline}*

See UW-13 (e)(2), Fig. UW-13.2, section 2-3 and Fig. 2-4(4a), Appendix 2
of

ASME VIII, Div.1. for the related codes and the physical representation
of the

nomenclatures.

Per above codes, it is required that the throat of the fillet c =
0.7t~min~. = 0.26",

So the fillet leg size is \~ 0.37".

However, per the assembly configuration of the upstream flange
connecting with

the below flange (see drawing TTF-CRY3-01.01.00 & TTF-CRY3-01.03.00), it
is

found that there only has \~0.19 clearance between flange MD-443038 and

MD-443039, so it is suggested to use 0.37" bevel groove instead of 0.37"
fillet weld to

meet the related code required and to avoid the parts interference.

It is found that:

t~p~= 0.32" \> 0.25"

where is the smallest of t~n,~ t~x~ and 0.25" is 0.25"

c = t~n~ = 0.375" = leg of the fillet or bevel groove

b = 0.16" + 2 x tan 15^o^ x 0.43"

= 0.39"

a = 0.43" + 0.37"- 0.06"

= 0.74"

$\therefore$ a + b = 1.13" ≥ 3t~n~ = 1.125"

The weld design for the end flange is ok.

*[Also run a weld strength calculation for the noted location as the
2^nd^ analysis]{.underline}*

*[approach]{.underline}*:

Radial pressure load F~rp~:

F~rp~ = P L~sd~

= 14.7 psi x 230 in

= 3,381 lbs/in

Shear flow due to radial shear load F~sr~:

F~sr~ = VQ/I

where:

Radial shear load V:

V = 0.01 P L~sd~ D~o~

= 0.01 x 14.7 psi x 230 in x 38,0 in

= 1,285 lbs.

The 1^st^ moment of area Q:

Q = 2.67 in x 0.375 in x (1.1454 -- 0.1875) in

= 0.959 in^3^

I = 2.4101 in^4^

F~sr~ = VQ/I

= 1,285 lbs. x 0.959 in^3^ / 2.4101 in^4^

= 511 lbs/in

The required resultant weld load F~rw~:

F~rw~ = ( 3,381^2^ + 511^2^)^1/2^ lbs/in

= 3,419 lbs/in

The welds for connecting the flange and the vessel shell is single
groove weld inside

and intermittent fillet outside, the lowest allowable weld stress F~aw~:

F~aw~ = 0.60 S~n~ = 0.60 x 16.7 ksi

= 10.02 ksi

The requires calculated minimum weld size C~m:~

C~m~ = F~rw~ / F~aw~

= 3,419 / 10,020 (in)

= 0.341 in \< 0.43 in (designated single groove weld)

If adding the value of the outside intermittent bevel groove weld
(0.0.37"), the weld

size even can go further smaller size.

So, the design of the weld for the upstream end flange is satisfactory.

Per the above calculationsand analysis, the upstream end flange and the
bellow flange have met the codes of ASME VIII, Div.1.

**5.2. [Downstream End Flange and Bellow Flange]{.underline}**

References: ME-443027-1, MC - 443040,

TTF-CRY3-01.01.00, TTF-CRY3-01.03.00

Since there some information are missing at this moment, so it is to
make the best assumption as followings:

1\. The downstream end flange MC-443040 connect to the upstream end
flange of the

bellow (see drawing TTF-CRY3-01.03.00).

> 2\. The above two flanges are connected by the clamps, the center line
> of the clamp
>
> C = 43.78" + (45.47 -- 43.78") / 2 = 44.625".
>
> 3\. A = 45.669", G = 44.488", T~1~ = 1.57", T~2~ = 1.57", B~1~ =
> 42.677", B~2~ = 37.37.25".

4\. n = 12, D = 0.55"

Material: SA 182-F304L stainless steel, F~a~ = S~f~= 16,700 psi at the
operating

temp., -20^0^ F to 100^0^ F (per Table 1A, part D, Section II, ASME
Boiler &

Pressure Vessel Code 1995 edition).

Per section Y-5.1, Y-5.2, figure Y-5.1.2 of Appendix Y,

Section 2-4, figure 2-4 of Appendix 2 of ASME VIII:

B~1~ \> ½ C,

So it is defined as: Class 2, Category 3 flange

In order to reduce the amount of work required as shown on section 5.1,
per section Y-9, of Appendix Y, ASME VIII, Div.1, we apply the method of
estimating flange thickness and bolting for the downstream flanges.

t~a~ = 2.45 \[Mp / ((πC --nD)S~f~)\]^1/2^

(per eq. 39 of Y-9, appendix Y, ASME VIII, Div.1)

where:

M~p~ = H~D~h~D~ + H~T~h~T~ + H~G~h~G~ (per sectionY-3, Appendix Y, ASME
VIII)

H~D~ = 0.785 B~1~^2^ P

= 0.785 x 42.677^2^ in^2^ x 14.7 psi

= 21,017 lbs.

H = 0.785G^2^P

= 0.785 x 44.488^2^ in^2^ x 14.7 psi

= 22,839 lbs.

H~T~ = H - H~D~

= 22,839 -- 21,017

= 1,822 lbs.

h~G~ = (C -- G)/2

= (44.625 -- 44.488)in / 2

= 0.0685 in

h~D~ = (C -- B) / 2

= ( 44.625 -- 42.677) / 2

= 0.974 in

h~T~ = (h~D~ + h~G~) / 2

= 0.5213 in

H~G~ = W -- H, gasket load

(per section 2-3, appendix 2, ASME VIII)

W = H + H~p,~

flange design load for the operating condition

(per section 2-5(c)(1), appendix 2, ASME VIII)

which leads to:

H~G~ = W -- H

= H + H~p~ -- H

= H~p~

= 2b x 3.14GmP

= 2 x 0.315 in x 3.14 x 44.488 in x 6.00 x 14.7 psi

= 7,766 lbs.

M~p~ = H~D~h~D~ + H~T~h~T~ + H~G~h~G~

= (21,017 x 0.974 + 1,822 x 0.5213 + 7,766 x 0.0685) lbs-in

= (20,471 + 950 + 532) lbs-in

= 21,953 lbs-in

t~a~ = 2.45 \[Mp / ((πC --nD)S~f~)\]^1/2^

= 2.45 x \[ 21,953 lbs-in / ((π 44.625 in -- 24 x 0.551 in) x 16,700
psi)\]^1/2^

= 2.45 x 0.102 in

= 0.250 in

t~b~ = 0.56 B~1~ (P/S~f~)^1/2^

= 0.56 x 42.677 in x (14.7 psi / 16,700 psi)^1/2^

= 0.56 x 42.677 in x 0.0297

= 0.71 in

Since t~c~ is the greater of t~a~ or t~b~, so:

t~c~ = t~b~ = 0.71"

Per equation 42 of A-7, Appendix Y, ASME VIII, Div. 1:

A'~b~ = (H + 2 M~p~ / (A -- C)) ÷ S~b~

= (22,839 lbs + 2 x 21,953 lbs-in / (45.669 -- 44.625) in ) ÷ 20,000 psi

= (22,839 lbs + 42,056 lbs) / 20,000 psi

= 3.245 in^2^

Per Table Y-9.1, Appendix Y, ASME VIII, Div.1, it is found:

t~1~ = 1.1t~a~ = 0.275" \< T~1~ = 1.57"

t~2~ = 1.1t~c~ = 0.781" \< T~2~ = 1.57"

So the thickness of the flanges are ok for the downstream end flanges.

*[Welds Analysis and Calculations:]{.underline}*

The weld configurations are the same as the upstream flange, it designed
per section UW-13 (e)(2), Fig. UW-13.2, section 2-3 and Fig. 2-4(4a),
Appendix 2 of ASME VIII, Div.1

Also, since the valus of L~sd~ for downstream end flange is \~45"
(\~230" for upstream end flange), I value for downstream end flange is
large than the upstream flange, all these lead to a smaller values of
F~rw~ and C~m~ than the upstream end flange requires, however, the
designated welds of the downstream end flange are the same as the
upstream end flange, so the welds design are satisfactory.

Per the above calculations, the upstream end flange and the bellow
flange have met the codes of ASME VIII, Div.1.

**5.3. [Flanges #8, #11, #13, #15, #19, #21 & #23]{.underline}**

In order to reduce the amount of work required as shown on section 5.1,
per section Y-9, of Appendix Y, ASME VIII, Div.1, we apply the method of
estimating flange thickness and bolting for the above flanges.

Since the information of the most mating flanges are not available at
this moment, I made some engineering judgments per the best of my logic
knowledge.

Table 5.3, [Estimated Calculated Flange thickness and bolting VS. the
designated data]{.underline}

  ---------------- -------- ------------ ----------- ----------- ----------- ----------- ----------- -----------
                            Flanges (see                                                             
                            drawing                                                                  
                            ME-443027)                                                               

  nomenclature     unit     #8           #11         #13         #15         #19         #21         #23

                            MC-443037    MC-443034   MC-443036   MC-443031   MC-443033   MC-443035   MC-443032

  P                psi      14.7         14.7        14.7        14.7        14.7        14.7        14.7

  Sf               psi      18,800       18,800      18,800      18,800      18,800      18,800      18,800

  A                in       10.945       5.512       8.67        13.78       6.535       11.22       4.472

  B1               in       8.386        3.334       6.194       11.418      4.26        8.386       2.522

  B2               in       3.75         3.334       0           0           0           0           0

  C                in       9.646        4.882       7.874       12.99       5.748       10.236      3.628

  G                in       8.793        4.012       7.034       12.204      5.004       9.311       3.024

  Classification            3            1           3           3           3           3           3

  Categorization            3            3           3           3           3           3           3

  n                         24           4           8           8           6           12          8

  D                in       0.3125       0.394       0.394       0.394       0.394       0.335       0.335

  m                         6            6           6           6           6           6           6

  b                         0.21         0.21        0.21        0.21        0.21        0.21        0.21

  HD               lbs      812          128         443         1504        209         812         73

  H                lbs      892          186         571         1719        289         1000        106

  HT               lbs      81           57          128         214         80          189         32

  hG               in       0.4265       0.435       0.42        0.393       0.372       0.4625      0.302

  hD               in       0.63         0.774       0.84        0.786       0.744       0.925       0.553

  hT               in       0.5283       0.6045      0.6300      0.5895      0.5580      0.6938      0.4275

  HG               lbs      1023         467         818         1420        582         1083        352

  MP               lbs-in   990          337         796         1867        417         1383        161

  ta               in       0.1178       0.0885      0.1086      0.1258      0.0921      0.1253      0.0767

  tb               in       0.1313       0.0522      0.0970      0.1788      0.0667      0.1313      0.0395

  tc               in       0.1313       0.0885      0.1086      0.1788      0.0921      0.1313      0.0767

  A\'b             in\^2    0.1285       0.0668      0.1368      0.3428      0.0717      0.2027      0.0258

  t1               in       0.1296       0.0885      0.1194      0.1384      0.1013      0.1378      0.0844

  t2               in       0.1444       0.0885      0.1194      0.1967      0.1013      0.1444      0.0844

  Ab               in\^2    0.1350       0.0668      0.1436      0.3599      0.0753      0.2128      0.0271

  t1d              in       0.787        0.512       0.709       0.79        0.51        0.984       0.68

  t2d              in       N/A          0.512       N/A         N/A         N/A         N/A         N/A

  Abd              in\^2    1.088        0.271       0.542       0.542       0.4065      TBA         TBA
  ---------------- -------- ------------ ----------- ----------- ----------- ----------- ----------- -----------

The most representation of nomenclature of the Table can be found from
section 5.1 of this note, also from the appendix 2 and appendix Y of
ASME VIII, Div.1, except:

t~1~, Estimated calculated flange 1 thickness

t~2~, Estimated calculated flange 2 thickness

A~b~, Estimated calculated root diameter area of the bolts.

t~1d,~ Designated flange 1 thickness

t~2d~, Dersignated flange 2 thickness

A~bd~, Designated root diameter area of the bolts.

Per table 5.3, it is found that all t~1~ \> t~1d~; t~2~ \> t~2d~ and
A~b~ \> A~bd~ for all respective flanges, so the flange design are
satisfactory.

Per figure UW-13.2, section UW-13, appendix 2 of ASME VIII, Div. 1 and
drawing ME-443027, it is found that all flange welds are designed per
the related codes.

**6. [Calculation and Analysis of the Saddle support]{.underline}**

Reference drawing: ME-443027

Figure: 6.1 and Figure 6.2

Per drawing ME-443027 and figure x.x.1, it is found that:

L~c~ = 163.50 x 2 = 327 in (the ctr. distance between 2 saddle support.)

W~ws~ = 7,745 lbs. the weight of the vessel shell weldment.

W~other~ = 8,846 lbs.,

the weight of coldmass shields & pipe, 8 cavity string w/mech. turners,

individual dressed cavity w/turner, main coupler.

W~tot~ = 1.05 (W~ws~ + W~tot~ ) ≈ 17,421 lbs

(Total weight including 5% mice. parts weight applying to two saddle
supports)

![](media/image14.png){width="6.627083333333333in"
height="3.642361111111111in"}

Figure 6.1, [The weight and the location of the ctr. of gravity of the
vessel shell assembly]{.underline}.

The force distribution diagram on figure x.x.2 is created conservatively
based on the information from figure x.x.1 and drawing ME- 443027.

It is found that:

Q~1~ + Q~2~ = 17,421 lbs = W~tot~

Q~2~ = (W~tot~ x 148.13) / 327

the reaction force on the downstream saddle support.

= 7,892 lbs

Q~1~ = 9,529 lbs.

the reaction force on the upstream saddle support.

A = 76.854 in

the distance from the end of the upstream end to the ctr, of the 1^st^
saddle support.

L~tot~ ≈ 448 in

Θ = 120^o^, the value of the assuming contact angle of the saddle
support.

Figure 6.2, [The force distribution diagram of the vacuum
vessel]{.underline}.

*To find the longitudinal stress on the saddle support* 1:

f~1s~ = QA \[ 1 -- (1 -- A/L + (R^2^ -- H^2^) / 2AL) / (1 + 4H/3L)\] /
KR^2^t~s~

where H = 0,

R = 19.0"

K = constant, from page 88 of Pressure Vessel Handbook, 7^th^ edition,

by E. Megyesy

f~1s~ = 9,529 lbs x 76.854" x \[1 - (1 -- 76.854 / 448 + (19^2^ / 2 x
76.854 x 448))

÷ (1 + 0)\] / (0.335 x 19^2^ x 0.375 in^3^)

= 732,342 lbs-in x \[ 1- (1 -- 0.1715 + 0.0052)\] ÷ 45.351 in^3^

= 2,686 psi

*To find the longitudinal stress on the middle of the span*:

f~1m~ = (QL/4) \[(1 + 2 (R^2^ -- H^2^) /L^2^)) / (1 + 4H/3L) -- 4A/L\] /
KR^2^t~s~

= 1,067,248 lbs-in x ( 1.0036 -- 0.6862) ÷ 45.351 in^3^

= 7,469 psi

*Stress due to the pressure*:

f~p~ = PR / 2t~s~

= 14.7 psi x 19.0 in / 2 x 0.375 in

= 373 psi

So, the total longitudinal stress at the middle span of the vessel:

f~bm~ = f~1m~ + f~p~ = 7,469 psi + 373 psi

= 7,842 psi \< 0.80 F~a~ = 12 ksi

where:

F~a:~ allowable stree, see page x.

0.80, joint efficiency, see Table UW-12, section UW-12, ASME VIII,
Div.1.

*To find the tangential stress on the vessel shell:*

f~2t~ = (K^2^Q / Rt~s~ )( L -2A) / (L + 1.333H)

= ((1.171^2^ x 9,529 lbs./ 19 in x 0.375 in) x (448 -- 2 x 76.854) in) ÷
448 in

= 1,834 psi x 294.29 in / 448 in

= 1,197 psi \< 0.80 F~a~ = 12 ksi

*To find the circumferential stress on the vessel shell:*

f~4c~ = (Q / (4t~s~(b + 1.56(Rt~s~)^1/2^) -- 3K~6~Q / 2t~s~^2^

where: b, the width of the saddle support, assuming 3/4 inch, so:

f~4c~ = \[9,529 lbs / 4 x 0.375 in x (0.75 + 1.56 x (19 x 0.375)^1/2^
in\] -- (3 x 0.053 x 9,529

lbs / (2 x 0.375^2^ in^2^))

= -1,293 psi -- 5,387 psi

= \|- 6,680\| psi \< 1.50 F~a~ = 22.5 ksi

So, all the stresses are less than the allowable stress respectively, so
the saddle support design is satisfactory (Since the design hasn't
finished yet, the calculation and analysis is

based on the assumption of: the saddle support has al least 120^o^
degree contact angle with 3/8 in thick wear plate).

7.  **[Appendix]{.underline}**

    1.  **[The Lists of the Design Drawings of 1.3 MHz, β = 1,
        Cryomodule Vacuum]{.underline}**

> **[Vessel (Fermilab)]{.underline}**
>
> 1\. Vacuum Vessel Weldment Assy. ME -- 443027 -- 1 (2)
>
> 2\. Vessel Shell ME -- 443054
>
> 3\. Port End Coldmass MD -- 443050
>
> 4\. Port Middle Coldmass MD -- 443049
>
> 5\. Pick Point Weldment MC -- 443028
>
> 6\. Pad Weld Upper MC -- 443043
>
> 7\. Vessel Support Weldment MC -- 443029
>
> 8\. Pad Weld Lower MC -- 443042
>
> 9\. Flange MC Port MC -- 443037

10\. Instrumentation Flange MC -- 443034

11\. Flange Side Rib MB - 443051

12\. Flange -- K1, K3 MC -- 443036

13\. Flange D MC -- 443031

14\. MC Bracket Weldment -- LH MC -- 443058

15\. MC Bracket Weldment -- RH MC -- 443059

16\. Flange E.O.D. 166 MC -- 443033

17\. Flange G.O.D. 285 MC -- 443035

18\. Flange DN 63 DF MC -- 443032

19\. Ring Flange Stop MC -- 443053

20\. Stiffening Rib MC -- 443052

21\. Pipe Support Plate MC -- 443047

22\. Plate Welded Outrigger MC -- 443048

23\. Gusset Welded Outrigger MC -- 443041

24\. Flange, Fixed Downstream MC -- 443040

25\. Flange, Fixed Upstream MC -- 443039

26\. Flange, Bellow Sliding MD -- 443038

27\. Flange Weld Ring MC -- 443060

2.  **[The Lists of the Design Drawings of 1.3 MHz, β = 1, Cryomodule
    Vacuum]{.underline}**

> **[Vessel (ARES - TTF)]{.underline}**
>
> 1\. Vacuum Vessel Shell Assy. TTF -- CRY3 -- 01.01.00
>
> 2\. Support Post Cover TTF -- CRY3 -- 01.01.01
>
> 3\. Vacuum Vessel Shell Details TTF -- CRY3 -- 01.01.02
>
> 4\. Adjustable Support Assy. TTF -- CRY3 - 01.02.00
>
> 5\. Adjustable Support Details TTF -- CRY3 -- 01.02.01
>
> 6\. Universal Expansion Joint TTF -- CRY3 -- 01.03.00