10mm_
Mechanical
- Jan 6, 2017
- 3
Hello,
I am new to the forum so if this belongs somewhere else I apologize.
I am working on a product built to contain an arc flash explosion in a metal box with spring loaded latches on the door. This allows the door to "burp" gasses out during the arc flash event which reduces stress on the overall structure.
Customers ask for door mounted accessories, which can blow out of the door during an arc event.
What I would like to do is calculate an equivalent force that I can test door mounted devices to see if they will remain in place during the arc blast without running an expensive dangerous arc flash test.
Is there a way to calculate the force/pressure on the interior wall of a box that contains an explosion? Could I use burst panels on the door to measure the pressure during an actual arc flash and then solve for an equivalent static force. The idea would be to build a jig that holds the doors and press on door mounted devices with the equivalent load and if they break/push out of the door then they can't be used.
I hope I explained it well, your thoughts are welcome!
I am new to the forum so if this belongs somewhere else I apologize.
I am working on a product built to contain an arc flash explosion in a metal box with spring loaded latches on the door. This allows the door to "burp" gasses out during the arc flash event which reduces stress on the overall structure.
Customers ask for door mounted accessories, which can blow out of the door during an arc event.
What I would like to do is calculate an equivalent force that I can test door mounted devices to see if they will remain in place during the arc blast without running an expensive dangerous arc flash test.
Is there a way to calculate the force/pressure on the interior wall of a box that contains an explosion? Could I use burst panels on the door to measure the pressure during an actual arc flash and then solve for an equivalent static force. The idea would be to build a jig that holds the doors and press on door mounted devices with the equivalent load and if they break/push out of the door then they can't be used.
I hope I explained it well, your thoughts are welcome!