# A pressure question

#### Greg

##### Well-Known Member
Here's one for the engineers in here. What is the formula for determining pressure in a given space that changes. Example: 1 cubic foot sealed box has 10 psi, the box volume is reduced to 1/2 cubic foot, what is the pressure now?? Disreguard heat. anyone? anyone?

Question 2, what is the final "spring rate" on a 6" travel air-bump with say 100ps??i

Greg #### Kritter

##### Krittro Campbell
In an ideal situation it would be PV=mRT. R being the gas constant of the gas being used, P pressure, V volume, m is mass and T is absolute temp.

For #2 I need some dimensions of the airbump or intial and final volumes.

Kris

<A target="_blank" HREF=http://www.dmsrace.com>www.dmsrace.com</A>

#### rdc

##### - users no longer part of the rdc family -
considering the same temperature you can use:
p1v1=p2v2 or in another form you have (p1/p2)=(v2/v1)

For the 2nd question. If you give me some dimensions of the piston area and volums etc I can then figure it out.
-Chris

2000 v6 4x4 TRD 5-Speed Tacoma<P ID="edit"><FONT SIZE=-1>Edited by PismoTaco on 05/06/02 02:49 PM (server time).</FONT></P>

##### Well-Known Member
Hey Greg,

From the nature of your question, I will assume that you are talking about a known, uniform gas in this space. If so, the ideal gas law should do it for you. PV=nRT

P = pressure
V = volume
n = number of moles of gas = mass * Mole number
R = ideal gas constant
T = temperature

It can also be taken as Pv=RT where v=specific volume (Volume/mass)

If there is more to it or you need help with any of the values, send me a PM or email.

Note: You can not discount the temperature other than to say that there is no heat added to the system. Anytime you do work on a closed system it adds energy and there will be an increase in temperature.

#### Kritter

##### Krittro Campbell
make sure you use absolute temp or you will be off by a factor of 273 or 459 depending on units.

Kris

<A target="_blank" HREF=http://www.dmsrace.com>www.dmsrace.com</A>

#### MNotary

##### Well-Known Member
For one change in volume and using Nitrogen, heat would be negligible. If the tempature increases, the pressue increases. But the pressure will increase at P1 and P2 so the difference is close to the same (some more heat for each cycle).

P1= 10 psi
V1= 1 cu. ft.
V2= 0.5 cu. ft.

P1V1=P2V2

P2=P1V1/V2

P2= 10(1)/0.5

P2= 20 psi

2.5 x 10 resevoir at 250 psi. 7/8 shaft 14 in. stroke

0.875(0.875)(14)= 10.72 ci

2.5(2.5)10 =62.5 ci

P2= 250(62.5)/(62.5-10.72)

P2= 302 psi

Didn't Fox have the spring rates for their air shocks in the old catalog?

#### BIG_FAT_LOSER

##### Well-Known Member
1st question = about a hundred

2nd question = 2 bizillion

<font color=red>PAT KAPKO</font color=red>
<font color=yellow>Self appointed King of ghetto fab</font color=yellow>