Re: Re: [MW:28906] Interpretation: IX-10-23 Subject: QW-433 and QW-452.1(b) Date Issued: March 14, 2011 File: 10-1918

Hi Rajesh,

Great catch to the point, lets talk in mm, why not 8.43 mm? why only 2.54 mm what is the reason? which is 5.08 mm only.

0.332 in - 8.43 mm

0.100 in - 2.54 mm

0.432 in - 10.97 mm

Thanks.

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Regards,

Mohamed Muneeb Mahaboob

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From: rajesh kaswala

Date: 2019-01-13 18:47

To: materials-welding

Subject: Re: [MW:28903] Interpretation: IX-10-23 Subject: QW-433 and QW-452.1(b) Date Issued: March 14, 2011 File: 10-1918

depositing 0.100 in. of E6010 and the balance  - 2t only as per 452.1 b , means 0.100 2t = 0.200 not 0.864 

regard

rajesh 

On Wed, Jan 9, 2019 at 7:09 PM muneebslsa@gmail.com <muneebslsa@gmail.com> wrote:

Greetings,

I have clarifications with below interpretaion for question 1 and agree with 2 and 3, does anyone has any observations on Q1?

Interpretation: IX-10-23 Subject: QW-433 and QW-452.1(b) Date Issued: March 14, 2011 File: 10-1918 

Background: A welder tests on an NPS 6 Sch. 80 (0.432 in. wall) coupon, depositing 0.100 in. of E6010 and the balance of 0.332 in. using E7018. Question (1): Using E6010, is the welder qualified to deposit 0.864 in. maximum of weld metal? Reply (1): Yes. Question (2): Using E7018, is the welder qualified to deposit 0.664 in. maximum of weld metal? Reply (2): Yes. Question (3): Is the welder qualified to deposit 0.864 in. of weld metal using E6010 plus 0.664 in. of E7018 weld metal deposit thickness for a total of 1.528 in. in the same groove? Reply (3): No. See QW-452.1(b).

Thanks.

---

Regards,

Mohamed Muneeb Mahaboob

CSWIP 3.2-89247/2 

IRCA-9001:2015 LA

ASNT-NDT-L2-RTFI-RT-UT-MT-PT.

SAP#:70021650 APs:JAPID, QCS & QCI.

E muneebslsa@gmail.com M +91-636-955-7189

''If you know any art by any means then SCULPT IT - no matter how cheap or useless it is - JUST SCULPT IT, you will never know when experts will be on demand'' 

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