Thursday, January 17, 2019

Re: Re: [MW:28925] Interpretation: IX-10-23 Subject: QW-433 and QW-452.1(b) Date Issued: March 14, 2011 File: 10-1918

yes because 7018 f4 so welder qualify f 1 to 4 and 6010 f no. 3 so qualify

your answer no , why ?

regard
rajesh

On Wed, Jan 16, 2019 at 3:17 PM muneebslsa@gmail.com <muneebslsa@gmail.com> wrote:
Hi Rajesh,

Greetings,

I hope its F3, so when you are restricted with F3, you cannot pull and trigger the F4 mode, which is what i have observed and you have observed, hence Question (1): Using E6010, is the welder qualified to deposit 0.864 in. maximum of weld metal? Reply (1): YES, but answer should be NO.

I want to see more experts view on this.

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.
muneebslsa@gmail.com +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 ITyou will never know when experts will be on demand'' 
 
Date: 2019-01-15 16:09
Subject: Re: Re: [MW:28909] Interpretation: IX-10-23 Subject: QW-433 and QW-452.1(b) Date Issued: March 14, 2011 File: 10-1918
your  all question and answer is correct 

1) total thickness 0.432 so qualified 2t = 0.864 not 1.528
2) 6010 f no. 2 and 7018 f no. 4 so qualified 6010, allow total thickness 0.864

regard
rajesh


On Tue, Jan 15, 2019 at 6:01 AM muneebslsa@gmail.com <muneebslsa@gmail.com> wrote:
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.

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.
muneebslsa@gmail.com +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 ITyou will never know when experts will be on demand'' 
 
Date: 2019-01-13 18:47
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.
muneebslsa@gmail.com +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 ITyou will never know when experts will be on demand'' 

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