Find total length of cylinder (springs, natural frequency)

[Engineer101]

Homework Statement

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

Homework Equations

[/B]

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

 

[BvU]

How big is L ?

 

[Engineer101]

How big is L ?

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

 

[gneill]

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

 

[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?

 

[BvU]

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

 

[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?

 

[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?

 

[BvU]

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?

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)

 

[Engineer101]

Thank you very much for replying.

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

 

[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

 

[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

 

[Engineer101]

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

 

[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 !

 

[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!

 

[Engineer101]

I'd really appreciate it if you could take a look at my other thread, if you have the time so close to Christmas!
Thanks! https://www.physicsforums.com/threa...-crankshaft-of-an-engine.788785/#post-4953803

 

[BvU]

I'd really appreciate it if you could take a look at my other thread, if you have the time so close to Christmas!
Thanks! https://www.physicsforums.com/threa...-crankshaft-of-an-engine.788785/#post-4953803

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 :)

 

[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!

 

[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..

 

[BvU]

pi (0.025)^2

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