# Calculations for the Opening of Rough Port Pipe for Pipe Thickness, Weld Size and 

**Reinforcement-Numi Decay Pipe Downstream End.**

By: Edward Chi -- PPD/ME 03/31/2003

This edition superseded the edition dated on 05/28/2002

(For Chapter 3 only)

[The main decay pipe]{.underline}:

P: 15 psi, external design pressure

D~o~: 78.75 in, outside dia. Of the decay pipe.

D~i~: 78 in, inside dia. Of the decay pipe.

t: 0.375 in, the nominal pipe wall thickness.

F~a~: 13,800 psi, the max. allowable stress of the pipe material at the
operating temp,

also know as Sv. (See Page 1-1).

Pipe material: SA 285, grade C, carbon steel.

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

=81.875" (see Page 1-1)

Find out the reguired pipe shell thickness t~r~:

1.  Try t~r1~ = 0.35"

> L~v~/D~o~ = 81.875"/78.75" = 1.04
>
> D~o~/t~r~ = 78.75/0.35 =225
>
> From section UG-28, also Fig. G & Fig. CS-2 in Subpart 3 of section
> II, part D,
>
> ASME VIII, Div I:
>
> Find:
>
> A = 0.0004
>
> B = 5750
>
> P~a1~= (4B)/(3(D~o~/t~r~)) = 34.07 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

2.  Try t~r2~ = 0.250"

> L~v~/D~o~ = 1.04
>
> Do/tr2 = 78.75/0.25 = 315
>
> Find:
>
> A = 0.00025
>
> B = 3550
>
> Pa2 = (4B/3(Do/tr2)) = 15.026 psi
>
> So, **t~r~** = 0.250"

[The rough prot pipe]{.underline}:

d~o~: 6.625", outside dia. Of the rough-port pipe.

d~i~: 6.357", inside dia . of the rough-port pipe.

t~n~: 0.134", wall thickness of the rough-port pipe.

t~rn~: required thickness of a seamless nozzle wall.

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

nozzle neck beyond the outside surface of the vessel wall.

E: 29 x 10^6^ psi, modulus of elasticity of the material at the opening
temp.

Material: SA 312 304 stainless steel pipe, Fa = Sn = 16,700 psi at the
operating temp.

-20^0^ F to 200^0^ F (per talbe 1A, part D, Section II, ASME Boiler &
Pressure

Vessel Code 1995, 1997 Addenda).

Reference drawings: MC-406148, MD-406870 & ME-406873

1.  Try: L~1~ = 30 in, t~r1~ = 0.125 in, then:

> L~1~/d~o~ = 30/6.625 = 4.5283
>
> d~o~/t~r1~ = 6.625/0.125 = 53
>
> From Fig. G and Fig. HA-1 of subpart 3 of Sect. II, part D:
>
> A = 0.00069
>
> B = 7800
>
> P~a~ = (4B)/(3(d~o~/t~r1~)) = (4 x 7800)/(3 x 53) = 196 psi \> P

2.  Try: L~2~ = 40 in, t~r2~ = 0.0625 in

> L~2~/d~o~ = 6.0377
>
> d~o~/t~r2~ = 106
>
> get:
>
> A = 0.00021
>
> B = 2800
>
> Pa = (4B)/(3(d~o~/t~r2~)) = (4 x 2800)/(3 x 106) = 35.22 psi \> P

3.  Try: L~3~ = 40 in, t~r3~ = 0.045 in

> L~3~/d~o~ = 6.0377
>
> d~o~/t~r3~ = 6.625/0.045 = 147
>
> get:
>
> A = 0.000115
>
> Since the value of "A" falls to the left side of the curve of the Fig.
> HA-1, so:
>
> P~a~ = 2AE/3(d~o~/t~r3~) = (2 x 0.000115 x 29 x 10^6^) / (3 x 147) =
> 15.12 psi
>
> So: **t~rn~** = t~r3~ = 0.045 in

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

Per UG-37, UW-16 and Fig. UW-16.1(i) of ASME VIII, Div. I of 1998.

t~1~ = t~2~ = Not less than the smaller of ¼ in or 0.7 t~min~

t~1~ = t~2~ = 0.134" x 0.7 = 0.094 in

t~1~ + t~2~ = 0.094 in + 0.094 in =0.188 in \> 1.25 t~min~ = 0.1675 in

So, the weld size t~1~ = t~2~ = 0.094 in, the fillet size is about 0.13
in

[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 (6.357" x 0.250" x 1.0 +2 x 0.134" x 0.250" x 1.0 x 0)

= 0.7946 in^2^

Where:

d: Finished diameterof circle opening, see Fif. UG-37.1

F: Correction factor

F~r1~: S~n~/S~v~, for nozzle wall inserted through the vessel wall.

(not greater than 1.0)

Area of reinforcement available A~a~:

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

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

= 6.357" x (1 x 0.375" -- 1.0 x 0.25") -- 0

= 0.7946 in^2^

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

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

= 0.15625 in^2^

So pick **A~1~ = 0.7946 in^2^**

A~2~: Smaller of the following (area available in feedthru pipe
projected outwards):

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

= 5 x (0.134 in -- 0.045 in) x 0.375 in x 1.0

= 0.1669 in^2^

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

= 5 x (0.134 in -- 0.045 in) x 1.0 x 0.134 in

= 0.05963 in^2^

So pick **A~2~ = 0.05963 in^2^**

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

= 5tt~i~f~r2~

= 5 x 0.375 in x 0.134" x 1.0

= 0.25125 in^2^

or =5t~i~t~i~f~r2~

= 5 x 0.134 in x 0.134 in x 1.0

= 0.08978 in^2^

or = 5ht~i~f~r2~

= 5 x 0.94 in x 0.134 in x 1.0

= 0.6298 in^2^

So pick **A~3~ = 0.08978 in^2^**

where:

t~i~: Nominal thickness of the internal projection of the nozzle wall
(see Fig. UG-37.1).

A~41~: Outward nozzle weld

= leg^2^ f~r2~

= 0.13^2^ in^2^ x 1.0

=0.0169 in^2^

A~43~: Inward nozzle weld

= leg^2^ f~r2~

= 0.13^2^ in^2^ x 1.0

= 0.0169 in^2^

Then:

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

= (0.7946 + 0.05963 + 0.08978 + 0.0169 + 0.0169) in^2^

= 0.9778 in^2^ \> A~r~ = 0.7946 in^2^

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