# Find total length of cylinder (springs, natural frequency)

1. Dec 22, 2014

### Engineer101

1. The problem statement, all variables and given/known data

A simple pressure relief device is shown in Figure Q6.1 below. It consists of a
piston of mass 0.5 kg and thickness 3 cm sliding without leakage or friction into
a cylinder of diameter 5 cm. The bottom section of the cylinder can be exposed
to a varying source of pressure, while the upper section is open to the
atmosphere.
A spring is attached to the upper face of the piston at point A and to a fixed
frame at its other extremity. Without the piston attached, the spring rest
position of A is located up 15 cm from the bottom of the cylinder, as shown in
Figure Q6.1. When attached to the piston, A drops by 3 cm as shown in
Figure Q6.1. Assume that g = 9.8 m/s

1 -
Calculate the total length of the cylinder so that the bottom face of the
piston reaches the top of the cylinder (and therefore vents the pressure out)
when the internal pressure reaches 10 kPa gauge. Assume a spring
constant of 150 N/m for this question and a position of the piston point A at
rest of 12 cm from the bottom of the cylinder.

2 -
Consider the cylinder base open to atmosphere with a free flow of air. The
system of piston and spring is at rest. Calculate the natural frequency of
this system. Assume a spring constant of 150 N/m for this question.
Arrow on right is total length of cylinder, but the image was too big so I had to crop it.
Consider the cylinder base open to atmosphere with a free flow of air. The
system of piston and spring is at rest. Calculate the natural frequency of
this system. Assume a spring constant of 150 N/m for this question.w
2. Relevant equations

3. The attempt at a solution
I don't even know where to begin to be honest.
Found Pressure upwards using F=PA, F=10,000 x pi (0.025)^2 = 19.63N which seems very low.
Then found force on hanging spring to beh 450N, F=kL = 150 x 3 = 450N
So I'm stuck because there isn't enough force to overcome the weight already on the spring so I've gone wrong somewhere.

Haven't even covered natural frequency to be honest.
Thanks

2. Dec 22, 2014

### BvU

How big is L ?

3. Dec 22, 2014

### Engineer101

Ah. Of course! Silly me. I'll give it another shot!
Thanks for pointing out that stupid mistake!

4. Dec 22, 2014

### Staff: Mentor

Watch your units. The length dimensions are given in cm but the spring constant is in Newtons per meter.

5. Dec 22, 2014

### Engineer101

I don't know how I missed that to be honest. I'm not entirely sure how I go about solving this problem could you give me a clue?

6. Dec 22, 2014

### BvU

Spring deals with weight of piston. How far can your 19.6 N compress the spring ?

7. Dec 23, 2014

### Engineer101

It can deliver 19.6 - (150 x 0.03)N upwards? = 19.6 - 4.5 = 15.1N upwards?
So then would I use F=kL, rearrange to L=F/k and that will find the length it pushes upwards?
In that case L = 15.1/150 = 0.1m
Therefore total length of cylinder = 19cm?

8. Dec 23, 2014

### Engineer101

Then for natural frequency, is the equation f= 1/2pi x SQRT(k/m)?
Therefore, f = 1/2pi x SQRT(150/0.5) = 2.76Hz?

9. Dec 23, 2014

### BvU

I agree about the natural frequencyin post #8.

But I would expect the 19.63 N to be 19.63 N, all of it available to push up the piston from bottom side 9 cm to 9 + 150/19.63.

(the 4.5 N you mention is required when the bottom side is at 12 cm, not when the spring is holding all the weight, at 9 cm)

10. Dec 23, 2014

### Engineer101

Thank you very much for replying.

Why wouldn't the pressure force have to overcome the weight of the piston though?

11. Dec 23, 2014

### Engineer101

Also, I think you made a mistake it should be 19.63/150 not the other way around? Unless I slipped up somewhere?
then in that case 0.09+19.63/150 = 0.22m, to 2 dp

12. Dec 23, 2014

### BvU

It has to overcome weight of piston AND force of spring. So the first 4.5 N would bring it back to a situation (12 cm) where the spring is at rest (force 0). Then there's 15.1 N left to compress the spring and you end up with a required length of the cylinder 3 cm more than you found in post #7

13. Dec 23, 2014

### Engineer101

Oh okay I see thank you, so the answer will be 0.22m?

14. Dec 23, 2014

### BvU

Hehe, in PF we are not allowed to give answers ! Would get us into all kinds of trouble with teachers, lawyers and what have you. But we are allowed to inspire self-confidence in students (and engineers?), so: I think you're doing fine !

15. Dec 23, 2014

### Engineer101

Ah I see haha, well that the answer you HELPED me achieve feel promising! Thank you!
I'm studying aeronautical engineering, so a bit of both!

16. Dec 23, 2014

### Engineer101

17. Dec 25, 2014

### BvU

I was looking at that already, but you were in very good hands, so I didn't intrude (didn't have much to contribute either, I must confess -- certainly didn't like the ambiguity and confusion)

Congrats with the 80% and Merry Christmas :)

18. Dec 25, 2014

### Engineer101

I was CWatters helped me so much! I know I feel really bad about the confusion, I probably could have worded it better but I was pretty confused myself!
Thank you I'm pleased with that! Merry Christmas have a great day!

19. Jan 11, 2016

### xavierengineering

Hi there. I am currently attempting to complete the same question. But i a stuck. Right at the start where you are working out F=PA. how did you come across the A=0.025^2..

20. Jan 11, 2016

### BvU

Don't forget the $\pi$ !
And read the problem statement carefully ...