Finding force due to pressure on Pipe Outlet

Tanjina · February 26, 2015, 17:43

Hello everyone,

Is there any way for find the force on pipe outlet due to pressure? From Fluent manual, I came to know that Fluent gives force on wall only..... so I was wondering if anyone has any idea about it ?

Regards,
Tanjina

jsm · February 26, 2015, 17:52

Hi,

I can not understand the intention of your question.

Pressure = Force/Area. You can find out the area from Fluent and multiply with pressure.

Tanjina · February 26, 2015, 20:46

Sorry, jsm. I forgot to mention that it is partially filled pipe at outlet, so pressure distribution is like attached figure (First line)...... so in my thinking I have to integrate the area and pressure to find out the force. That's why I am looking for some way in fluent which can do this work for me. Correct me if I a wrong.

Regards,
Tanjina

jsm · February 28, 2015, 10:27

Ok. Now I understood your question.

You can do it in another way. Create iso surface in outlet for liquid phase and find out the area of liquid surface and then multiply with pressure.

If your outlet is complex shape, you need to do trial and error by using iso clip .

February 26, 2015, 17:43	Finding force due to pressure on Pipe Outlet	#1
Tanjina Senior Member Tanjina Afrin Join Date: May 2013 Location: South Carolina Posts: 169 Rep Power: 12	Hello everyone, Is there any way for find the force on pipe outlet due to pressure? From Fluent manual, I came to know that Fluent gives force on wall only..... so I was wondering if anyone has any idea about it ? Regards, Tanjina

February 26, 2015, 17:52		#2
jsm Senior Member JSM Join Date: Mar 2009 Location: India Posts: 192 Rep Power: 20	Hi, I can not understand the intention of your question. Pressure = Force/Area. You can find out the area from Fluent and multiply with pressure. __________________ With regards, JSM

February 28, 2015, 10:27		#4
jsm Senior Member JSM Join Date: Mar 2009 Location: India Posts: 192 Rep Power: 20	Ok. Now I understood your question. You can do it in another way. Create iso surface in outlet for liquid phase and find out the area of liquid surface and then multiply with pressure. If your outlet is complex shape, you need to do trial and error by using iso clip . __________________ With regards, JSM

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