Dinghy Pressure due to temp?

[CAELESTIS]

Ok, engineers, if I have a 10 RIB inflated to 3.5psi at 70*F, what is the psi at 90*? What about 115*F?

Thanks!

[Wotname]

Originally Posted by CAELESTIS

Ok, engineers, if I have a 10 RIB inflated to 3.5psi at 70*F, what is the psi at 90*? What about 115*F?

Thanks!

If you want accurate numbers, we need far more information about the RIB construction etc but for a rule of thumb you could likely use 1 psi increase for every 10*F increase.

So 3.5psi at 70*F becomes 5.5psi at 90*F and 8psi at 115*F.

See https://en.wikipedia.org/wiki/Cold_inflation_pressure

[JPA Cate]

Great question, Caelestis. I have to admit I relate to it because of leaving a previously sewn and patched Achilles on the beach in the sun, having gone for a hike, and returned to find it self-disinflated. Dangies, hate it when that happens!

Now, back to V1 P1 over T1 equals V2 P2 over T2, and allied concerns.

Ann

[AndyEss]

Originally Posted by CAELESTIS

Ok, engineers, if I have a 10 RIB inflated to 3.5psi at 70*F, what is the psi at 90*? What about 115*F?

Thanks!

Try avoiding dark colored inflatable dinghies - or dark dinghy chaps. Right now my nearly 5 y o inexpensive PVC West Marine dinghy has a coat of blindingly white, rubberized roof paint. It makes a world of difference in how hot my dinghy gets in the strong Mexican sun.

[Ded reckoner]

Ann's equation is the combined law of Boyle, Charles, and Gay-Lussac. It is for an ideal gas (which we don't have, but will think good enough). V1 is approximately V2 (contrary to Ann's anecdote about what happens with over-inflation, volumes will be assumed not much different unless you have experimental volumetric measurements). So, it's P1/T1 = P2/T2. So, P2= P1*T2/T1. This needs to be solved with temperatures converted to Kelvin and then back. Pressures need to be converted to pascals and then back.

Going from 70F to 90F, pressure rises to 3.63 psi from 3.5psi. Going from 70 to 115, the pressures rises to 3.8psi.

My understanding is dinghy tubes can handle up to something like 10psi.

[sanibel sailor]

Originally Posted by Wotname

If you want accurate numbers, we need far more information about the RIB construction etc but for a rule of thumb you could likely use 1 psi increase for every 10*F increase.

So 3.5psi at 70*F becomes 5.5psi at 90*F and 8psi at 115*F.

See https://en.wikipedia.org/wiki/Cold_inflation_pressure

conversely, 2.5 psi at 60, 1.5 at 50 and essentially completely deflated at 35. And that's not true.


For this rule of thumb, from the wiki article...
Hence, for a tire filled to 32 psi, the approximation usually made

if you look at the article's table, the increase decreases as initial pressure decreases.
I would extrapolate an inflatable goes up 0.1 psi for 10F.
I am too lazy to calculate using Kelvin and Boyle's Law.

[StuM]

Originally Posted by Ded reckoner

So, it's P1/T1 = P2/T2. So, P2= P1*T2/T1. This needs to be solved with temperatures converted to Kelvin and then back.

FIFY

Pressures need to be converted to pascals and then back.

Why?
3.5 * 319.3 / 294.3 = 3.8


However, you are ignoring atmospheric pressure. 3.5 PSI in the tube is an absolute pressure of 18.2 PSI.

(3.5 + 14.7) * 319.3 /294.3 = 19.7
19.7- 14.7 = 5

[StuM]

Originally Posted by sanibel sailor

I would extrapolate an inflatable goes up 0.1 psi for 10F.
I am too lazy to calculate using Kelvin and Boyle's Law.

Gay-Lussac's Law actually

[zemurray]

Inflate it with nitrogen. problem solved. ��

[wsmurdoch]

Here is my take...

The problem has to be worked in absolute pressure and in absolute temperature.
Fahrenheit + 460 = Rankine
psig + 14.7 = psia

70F = 530R
3.5 psig = 18.2 psia

PV = nRT
P/T = nR/V and assume V like n and R is constant
nR/V = 18.2/530 = 0.03434

P = (nR/V) * T

if T = 70 F or 530 R, then P = 18.2 psia or 3.5 psig
if T = 90 F or 550 R, then P = 18.9 psia or 4.2 psig
if T = 115 F or 575 R, then P = 19.7 psia or 5.0 psig
if T = 260 F or 720 R, then P = 24.7 psia or 10.0 psig

[StuM]

Originally Posted by zemurray

Inflate it with nitrogen. problem solved. ��

Gas Laws still apply regardless of what gas it is filled with

[StuM]

Originally Posted by wsmurdoch

Fahrenheit + 460 = Rankine

OMG, do people still use °R ?

[wsmurdoch]

Originally Posted by StuM

OMG, do people still use °R ?

I do, but I am an old fart.

[Ded reckoner]

Originally Posted by StuM

FIFY
Why?
3.5 * 319.3 / 294.3 = 3.8


However, you are ignoring atmospheric pressure. 3.5 PSI in the tube is an absolute pressure of 18.2 PSI.

(3.5 + 14.7) * 319.3 /294.3 = 19.7
19.7- 14.7 = 5

The reference to converting pressure to pascals and back should more completely have been phrased "...and back to psi". Similarly, temperature is converted to kelvin and back to fahrenheit (or celsius, if preferred; kelvin is not a commonplace temperature reference).

Thank you for the correction about pressure. Absolute, not gauge pressure, must be used (as you indicated, in the calculations, 14.7psi from atmospheric pressure must be added to the gauge pressure to get absolute pressure). Doing the math correctly this time, at 90F, the gauge pressure will be 4.2psi. At 115F, the gauge pressure will be 5.0psi.

[AndyEss]

Originally Posted by StuM

Gas Laws still apply regardless of what gas it is filled with

👍🏼

Heck, if you filled it with helium, you would be a very small way into being a dirigible, and not just a mere dinghy