A concept problem in overpressure

[johns123]

A "concept" problem in "overpressure"

Here's the kind of physics problem that really drives
me crazy:

The overpressure in a tire is 2.4 atm on a day
when the atmospheric pressure is .95 atm. So,
on a day when the atmospheric pressure reads 1.01 atm,
what will the overpressure be ?

This is a trick question. It makes me want to strangle
a certain physics professor. The problem seems simple,
but it is not. It is a definition problem, plus it is
a real world problem. There is a "concept" worth
learning here. The professor's answer is 2.34 atm :-)

 

[sankalpmittal]

Here's the kind of physics problem that really drives
me crazy:

The overpressure in a tire is 2.4 atm on a day
when the atmospheric pressure is .95 atm. So,
on a day when the atmospheric pressure reads 1.01 atm,
what will the overpressure be ?

This is a trick question. It makes me want to strangle
a certain physics professor. The problem seems simple,
but it is not. It is a definition problem, plus it is
a real world problem. There is a "concept" worth
learning here. The professor's answer is 2.34 atm :-)

Hint : How do you calculate overpressure ? Can you come up with proportionality relationship between atmospheric pressure and overpressure ?

You can use the formula stated here : http://en.wikipedia.org/wiki/Overpressure#Overpressure_calculation_.28explosive.29

 

[johns123]

The correct answer ( according to him ) is
the tire gage goes out of calibration by
1.01 - .95 = .06 , so the overpressure
( which doesn't change from 2.4 ) now
reads 2.34 .. Geeze! He also says the
2.4 answer is not exactly right, because
a tire gage is calibrated at 1 atm. And
what do you think is the right answer now?
I'm guessing 2.37 atm .. because less
calibration error is subtracted ?

 

[johns123]

In other words, the original 2.4 measure is also in error, because the reading was taken at .95 atm. He gives that so the "overpressure error" of .06 can be calculated. I still find it difficult to determine the +/- sign of the calibration error. I do realize that the pressure inside the tire has nothing to do with the pressure outside the tire .. and that is the concept. The pressure outside the tire simply controls the length of the stick in the tire gage. And it is specifically calibrated at 1 atm to read "0" overpressure. Isn't this fun ... NOT!

 

[cjl]

He also says the
2.4 answer is not exactly right, because
a tire gage is calibrated at 1 atm.

This part is wrong (or at the very least misleading). Tires are not specified by absolute pressure - rather, they are specified by gauge pressure (or what you are calling "overpressure"), which is the difference between their internal pressure and ambient. This is what a tire pressure gauge measures.

 

[AlephZero]

Isn't this fun ... NOT!

Maybe it's not "fun" if your interest is theoretical physics, but it's the sort of thing that experimentalists have to deal with (and get right!) every day of the week.

Reading a "number" off ANY measurement device is worthless, unless you know how that number relates to the real world situation.

 

[johns123]

That's exactly what I said through the entire problem ? The overpressure is a calibrated pressure, and therefore the tire gage measures it wrong if used at other than 1 atm. If you notice in the problem, the professor only gives a way to calculate the tire gage error of .06 in the tire gage stick. I agree it is confusing, but wouldn't call it misleading. I think he is trying to get us to grasp the "concept" of incompressible fluids.

He just gave us another problem using sand in a large barrel ... find the pressure at the bottom of the barrel. Naturally all of us added in the 1.01x10^5 pascal atmospheric pressure plus the overpressure of the sand in the barrel ( WRONG ). Sand is pourous, so there is no overpressure. The whole business of "pressure" is nuts! And now he's telling us you can have pressure and overpressure at the same time. You can't win.