- #1
gpsimms
- 30
- 1
Thanks in advance to any help on this:
I am working in a combustion lab and we are designing a high pressure tube reactor. My background is only in math and I have never taken a materials or design course before, so I apologize that this is probably a very basic application and a very classical problem--but I need help.
We want to pass tubing through the chamber wall, and to do that we are going to tap female 3/8 NPT threads into the wall. Inside the chamber, we plan to go up to pressure of at least 50 atm.
The wall in which I want to tap the hole is 3/8" thick, and I want to be sure that this thickness allows enough threads to engage and secure the male NPT piece so that the threads do not strip.
It is easy to use the pressure in the chamber and the area of the tapped hole to determine the force (F) against the male NPT.
Then I divide that total force by the tensile stress area, which I got from a formula A=pi/4(d - .9743/n)^2, where d is the same diameter I used to calculate F, and n for 3/8 NPT is 18.
Finally, I divide the total force I calculated F by the tensile stress area to get tensile stress.
I compared this result to the yield strength of stainless steel and find that the yield strength is approximately 34 times the tensile stress, and the design is 'safe.'
OK here are my questions: In the above calculations, there is nothing about how many of the threads need to be engaged. Is the tensile stress formula assuming all threads are engaged? Is there a "rule of thumb" about the minimum number of threads needed for this formula to be applied? Also, I think the formula is probably meant to be used to for straight, not NPT threads. Is there a significant difference? The threaded section of the NPT piece I am planning to use is longer than 3/8", which is the wall thickness I am hoping to use. If only 3/8" worth of threads (about 5-6 threads) are in the wall, should that be sufficient?
Of course, my feeling is that I am safe, since the number I calculated has a safety factor of 34, but in the interest of not blowing up our lab and everyone in it, I'd like to be a little more sure on how to do the proper analysis. : )
Thanks!
simms
I am working in a combustion lab and we are designing a high pressure tube reactor. My background is only in math and I have never taken a materials or design course before, so I apologize that this is probably a very basic application and a very classical problem--but I need help.
We want to pass tubing through the chamber wall, and to do that we are going to tap female 3/8 NPT threads into the wall. Inside the chamber, we plan to go up to pressure of at least 50 atm.
The wall in which I want to tap the hole is 3/8" thick, and I want to be sure that this thickness allows enough threads to engage and secure the male NPT piece so that the threads do not strip.
It is easy to use the pressure in the chamber and the area of the tapped hole to determine the force (F) against the male NPT.
Then I divide that total force by the tensile stress area, which I got from a formula A=pi/4(d - .9743/n)^2, where d is the same diameter I used to calculate F, and n for 3/8 NPT is 18.
Finally, I divide the total force I calculated F by the tensile stress area to get tensile stress.
I compared this result to the yield strength of stainless steel and find that the yield strength is approximately 34 times the tensile stress, and the design is 'safe.'
OK here are my questions: In the above calculations, there is nothing about how many of the threads need to be engaged. Is the tensile stress formula assuming all threads are engaged? Is there a "rule of thumb" about the minimum number of threads needed for this formula to be applied? Also, I think the formula is probably meant to be used to for straight, not NPT threads. Is there a significant difference? The threaded section of the NPT piece I am planning to use is longer than 3/8", which is the wall thickness I am hoping to use. If only 3/8" worth of threads (about 5-6 threads) are in the wall, should that be sufficient?
Of course, my feeling is that I am safe, since the number I calculated has a safety factor of 34, but in the interest of not blowing up our lab and everyone in it, I'd like to be a little more sure on how to do the proper analysis. : )
Thanks!
simms