I've got an interesting problem in thermal expansion. We have a 304
stainless steel shaft running through a High Density Polyethylene bearing
block, with about 0.020" clearance. Turns like a greased doorknob at room
temperature. But the thing is in a cryo chamber that gets -200C, and at
some temperature in the middle the bearing latches onto the shaft and quits
turning.
My old CRC handbook does not see fit to list thermal expansion coefficients
of either material, so I'm off on a search for that info.
Once found, I'm trying to figure out what to do with it. Would the inside
of the plastic bearing contract at the:
1. Coefficient of expansion rate or
2. Coefficient of expansion times Pi or something like that, since it is a
circle?
Hmmm. Shouldn't have slept through Physics.
-- Lawrence Lile
Sr. Project Engineer
Salton inc. Toastmaster Div.
573-446-5661 Voice
573-446-5676 Fax
Lawrence Lile wrote:
>
> I've got an interesting problem in thermal expansion. We have a 304
> stainless steel shaft running through a High Density Polyethylene bearing
> block, with about 0.020" clearance. Turns like a greased doorknob at room
> temperature. But the thing is in a cryo chamber that gets -200C, and at
> some temperature in the middle the bearing latches onto the shaft and quits
> turning.
Lawrence Lile wrote, in part:
>I've got an interesting problem in thermal expansion. We have a 304
>stainless steel shaft running through a High Density Polyethylene bearing
>block, with about 0.020" clearance. [...]
>Once found, I'm trying to figure out what to do with it. Would the inside
>of the plastic bearing contract at the:
>
>1. Coefficient of expansion rate or
>
>2. Coefficient of expansion times Pi or something like that, since it is a
>circle?
>
>Hmmm. Shouldn't have slept through Physics.
As a former physics professor, I've seen more sleeping engineering
students than I can throw a stick (or a chalkboard eraser) at. But
now you see why your physics professor begged you to pay attention
when you were a student (and you, like all engineering students the
world over, scoffed, "I'll never need to know this stuff.").
At any rate. To determine if a disk will fit in a circular hole
(approximate both as 2-dimensional), you need the coefficient of
area expansion. You may not find this listed for your materials,
but fortunately you don't need to, because, as a simple argument
shows (I'll spare you the physics details), the coefficient of
area expansion is 2 times the coefficient of linear expansion, which
you *will* find listed (this only applies to isotropic materials,
of course). So, you measure the shaft cross-sectional area Ao at
a reference temperature, compute the area A at your working
temperature via A = Ao(1 + g Delta-T), where g is the coefficient
of area expansion and delta-T is the temperature difference. Do
the same thing for the hole, and the difference gives you the
clearance (or overlap) in area units (sq. cm, for example). Special
note: When you cool an object so that the material contracts, if
it has a hole in it, does the hole get bigger (as the material
recedes away from it) or smaller (since everything is shrinking)?
The answer is: it gets smaller. So when you cool down your assembly,
*both* the shaft and the hole are shrinking, but, evidently, the
hole is shrinking faster. Do the calculation.
Wake up! Class is over!
Michael Vinson
Thank you for reading my little posting.
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We must have had the same professor. Mine tried to boil alcohol out of a
plastic beaker on a bunsen burner. What beautiful flames!
Lawrence Lile
<llile@TOASTMA To: .....PICLISTKILLspam.....MITVMA.MIT.EDU
STER.COM> cc:
Sent by: pic Subject: Re: +AFs-OT+AF0-: Coefficient of Thermal Expansion
microcontrolle
r discussion
list
<PICLIST@MITVM
A.MIT.EDU>
09/27/01 12:54
PM
Please respond
to Lawrence
Lile
Thanks, Michael! Actually, I did pay quite a lot of attention in Physics,
and never for a minute thought I would never use this stuff. I note the
volume coeff. of expansion is three times the linear coeff., but the area C
of E is not something most people mention. Thanks for clarifying this.
Now, If I could just find a reference that actually gives the C of E for
these materials I am using ...
Nobody ever threw an eraser at me, although I did have a Chem professor who
musta slept through Chem I, because he put a non-pyrex gallon beaker of
water on the bunsen burner at the beginning of a class, saying he'd show us
an experiment once it reached a boil. When it burst, it soaked all his
papers and most of the front row!
Uh... by dimensional analysis I get the area coefficient of expansion as
the square of the linear coefficient. If you are looking for square
inches of hole minus square inches of shaft cross section your answer
has to be in square inches.
On the other hand if you assume they are both round and just look at the
diameter of the hole minus the diameter of the shaft the whole area
thing is moot.
I loved my high school physics teacher, but I think I annoyed him.
Douglas Butler wrote:
>
> Uh... by dimensional analysis I get the area coefficient of expansion as
> the square of the linear coefficient. If you are looking for square
> inches of hole minus square inches of shaft cross section your answer
> has to be in square inches.
>
> On the other hand if you assume they are both round and just look at the
> diameter of the hole minus the diameter of the shaft the whole area
> thing is moot.
>
> I loved my high school physics teacher, but I think I annoyed him.
I know the feeling. I frequently shrink steel parts into
aluminum bores - Said Al bores vary in diameter by the coeff
of linear expansion, in this case .00001244 in/deg F, within
the limits of my measuring equipment.
1. Tried the formulae, after I found the right coefficients on 2 different
repudable sources on the net (you can't be too careful)
At -200C, my parts theoretically have 0.001" clearance, zip for practical
purposes. UHMW has an annoying characteristic of being as slick as telfon,
until you squeeze it a little and then it sticks. I've respecified the
holes so I have a comfortable 0.020" clearance at COLD temperatures,
downright sloppy at room temperature.
2. Thanks to all you guys for the help
3. this is NOT a toaster - (do toasters get cold? Do toasters use liquid
nitrogen? Answer: Does the Pope program PICS? Is Ossama Bin Laden a Nice
Guy? ) It is a cryogenic processing system that will be used to make the
mirrors on the new outriggers at the Keck observatory.
www2.keck.hawaii.edu:3636/realpublic/gen_info/kiosk/index.html#Future
What else do I have to do when toasters get boring?
Douglas Butler wrote, in part:
>Uh... by dimensional analysis I get the area coefficient of expansion as
>the square of the linear coefficient. If you are looking for square
>inches of hole minus square inches of shaft cross section your answer
>has to be in square inches.
Careful! Recall the definition of the coefficient of linear expansion:
delta-L = a Lo delta-T, (1)
where delta-L is the change in length, "a" (usually written as
alpha, but I can't find the alpha key on my keyboard) is the
coefficient of linear thermal expansion, Lo is the original
length, and delta-T is the change in temperature.
Dimensional analysis of this equation implies that the dimensions
of "a" are inverse-temperature, i.e., the units might be 1/kelvin
or 1/fahrenheit or whatever. No length dimension, since "a" describes
the *fractional* change in length. For example, if (a delta-T) had the
value 0.1, then that would mean the object increased in length
by ten percent, so if it were originally 1 cm long, it would now
be 1.1 cm long. The key point here is that thermal expansion is
described as a fractional change in size, not an absolute change.
Turning to the question of area expansion, dimensionally we have
A = L^2. (2)
Therefore a small change dL in length will cause a small change dA
in area given by
dA = d(L^2)
= 2L dL. (3)
Thus if we define g, the coefficient of area expansion via
delta-A = g Ao delta-T, (4)
then:
delta-A = 2Lo delta-L (from (3) above)
= 2 Lo (a Lo delta-T) (using equation (1) above)
= (2a) Ao delta-T (Rearrange and use equation (2))
and by comparing with equation (4), we see that g = 2a, as I asserted
earlier. Go ahead and track the dimensions throughout this argument,
you will see it all works out.
Again, there are no dimensional issues here because the
coefficients of expansion are defined in terms of *fractional* change
in length or area. It's like saying, if I have a square, and increase
the length of both sides by 1%, then how does the area change? You
don't square the 1%, as Douglas thought, because it doesn't have
dimension of length. If you work it out, you'll see the change in area
is about 2%.
Michael
Thank you for reading my pedantic little posting.
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What you said is certainly correct for infinitesimal (differential) changes
in length, but not exactly true for large changes.
If we have deltaL=a*Lo*deltaT, and if A=L^2 (as you said), then
deltaA=(deltaL+Lo)^2-Lo^2
which is equal to deltaL^2 +2*deltaL*Lo, which is the same as
(a*Lo*deltaT)^2+2*a*Ao*deltaT where Ao=Lo^2. Note that this is not EXACTLY
equal to 2*a*Ao*deltaT, as you suggested. For cases where there is a
significant linear expansion (probably only important in VERY high accuracy
calculations or very strange materials), we might not be able to neglect
the (a*Lo*deltaT)^2 term. Normally, though, it would be so much smaller
than the other term (because a is much less than 1 so a^2 is much less than
2*a). However, we also have deltaT^2 in the neglected term, so if deltaT is
extreme, it might also come back into play. Note, too, that this actually
shows area expansion to be a nonlinear function of deltaT, even if linear
expansion is a linear (no pun intended) function of deltaT.
Sean
At 08:01 AM 9/28/01 -0700, you wrote:
Again, there are no dimensional issues here because the {Quote hidden}
>coefficients of expansion are defined in terms of *fractional* change
>in length or area. It's like saying, if I have a square, and increase
>the length of both sides by 1%, then how does the area change? You
>don't square the 1%, as Douglas thought, because it doesn't have
>dimension of length. If you work it out, you'll see the change in area
>is about 2%.
>
>Michael
>
>Thank you for reading my pedantic little posting.
>
>
>_________________________________________________________________
>Get your FREE download of MSN Explorer at http://explorer.msn.com/intl.asp
>
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Sean H. Breheny wrote, in part:
>[much deleted for brevity]
>Note, too, that this actually
>shows area expansion to be a nonlinear function of deltaT, even if linear
>expansion is a linear (no pun intended) function of deltaT.
By definition, all the coefficients of thermal expansion (whether for
length, area, or volume) are for linear changes. Same idea as
resistance, where the definition is V = I R, whether or not the
potential actually varies linearly with the current; in cases where
the variation *is* (to a good enough approximation) linear, R is a
constant and is called the resistance. The derivation I gave is
correct, because the coefficients are *defined* for the linear regime
(in which change in size is proportional to change in temperature).
In that regime, the coefficient of area expansion is exactly 2 times
the coefficient of length expansion.
Nonlinear effects can also be treated, of course, to as high an
order as you need to go to get the precision that you need.
Michael
Thank you for reading my little posting.
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>Again, there are no dimensional issues here because the
>coefficients of expansion are defined in terms of *fractional* change
>in length or area. It's like saying, if I have a square, and increase
>the length of both sides by 1%, then how does the area change? You
>don't square the 1%, as Douglas thought, because it doesn't have
>dimension of length. If you work it out, you'll see the change in area
>is about 2%.
>Michael
Works for me, and my non-math-intensive mind. If I have something
that's 1 x 1 and it grows by 1% now it's 1.01 x 1.01 and the
area, according to my Pentium, is now 1.0201.
Unfortunately, I haven't been following this thread. Did the original
poster want to know how to compute area expansion from linear expansion, or
did they want to know what the relationship between the defined linear and
area coefficients of expansion? I had guessed (perhaps incorrectly) that it
was the former that they wanted. In other words, they had a practical
application where they knew alpha for a material and wanted to know how
much it would expand in area. (I admit, though, that in most circumstances
this is splitting hairs since the alpha^2 term would be so small).
Sean
At 01:00 PM 9/28/01 -0700, you wrote:
By definition, all the coefficients of thermal expansion (whether for {Quote hidden}
>length, area, or volume) are for linear changes. Same idea as
>resistance, where the definition is V = I R, whether or not the
>potential actually varies linearly with the current; in cases where
>the variation *is* (to a good enough approximation) linear, R is a
>constant and is called the resistance. The derivation I gave is
>correct, because the coefficients are *defined* for the linear regime
>(in which change in size is proportional to change in temperature).
>In that regime, the coefficient of area expansion is exactly 2 times
>the coefficient of length expansion.
>
>Nonlinear effects can also be treated, of course, to as high an
>order as you need to go to get the precision that you need.
>
>Michael
>
>Thank you for reading my little posting.
>
>
>_________________________________________________________________
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>
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The thread originated with a steel shaft going through a plastic
bearing. The shaft turned fine at room temp, but bound up when very
cold. That lead to discussion of the diameter expansion vs the area of
the hole expansion.
COE(area) = 2 * COE(linear) is a good enough approximation for real
values, but it can not be completely correct. If we start with a 1" x
1" piece and expand it to 1.5" x 1.5", the Linear expansion is
(1.5"-1")/1" = 0.5. The Area expansion is (2.25""-1"")/1"" = 1.25 which
is not 2*0.5.
Doug, most mech engineering catalogues will
rate bearings in ranges of operating temperature,
you don't need calcs, just ring your local bearing
supplier and buy bearings suited for cyro
temperature use. :o)
-Roman
>
> The thread originated with a steel shaft going through a plastic
> bearing. The shaft turned fine at room temp, but bound up when very
> cold. That lead to discussion of the diameter expansion vs the area of
> the hole expansion.
>
> COE(area) = 2 * COE(linear) is a good enough approximation for real
> values, but it can not be completely correct. If we start with a 1" x
> 1" piece and expand it to 1.5" x 1.5", the Linear expansion is
> (1.5"-1")/1" = 0.5. The Area expansion is (2.25""-1"")/1"" = 1.25 which
> is not 2*0.5.
Well, I started this mess ... ah.. thread. Here's the results: Last
Saturday we sis a shakedown cruise of our cryogenic processor. As you may
recall, I have some fans on Stainless Steel shafts, with the motors outside
the 'fridge, the fan blades inside the 'fridge, and HDPE bearing blocks.
The fans would run until the system reached about -600C, then the fans would
run slow, stop, motors overheat, and so on. So the problem is to find the
coeefficients of expansion of Stainless and HDPE, and compute the cleaqrance
required so the bearings will not sieze at -1950C.
I settled on 1/16" clearance at room temperature, and this squeezes down to
a few tens of thousanths at cryogenic temperatures. Fans kind of wobble at
room temperatures, but behave nicely once the bearings squeeze down. No
observable leakage of LN past the bearing blocks, for some reason.
We ran the system all the way down, without having any fans sieze up. It
was (pun intended) pretty cool. Also got to goof around with a bucket of
Liquid Nitrogen. When I got done testing the calibration of my sensors in
it, and then testing the brittleness of several plastics in it (I was plased
to find Nylon 66 was quite flexible at -1950C - this was not what I had read
in books.)
We also discovered that immersing a can of root beer (opened, to let off any
pressure) will result in a pleasing root beer float, almost instantly. The
carbonated beverage also foams up, the foam spilling out of the can, which
makes a kind of instant ice cream when it hits the LN. Quite tasty, once it
warmed up to freezing. Don't try this at home. Dumping a cup of LN into a
bucket of water freezes the water 2" thick in a minute. We were using
distilled water/ice slush as a calibration standard, so we happily had to do
this a couple of times to make more ice.
All quite a lot of fun. They pay us to do this kind of stuff?
>The fans would run until the system reached about -600C, then the fans
would
>run slow, stop, motors overheat, and so on. So the problem is to find the
>coeefficients of expansion of Stainless and HDPE, and compute the
cleaqrance
>required so the bearings will not sieze at -1950C.
Hmm, I can see our cryogenics people are going to have to try a lot harder
when testing our spacecraft components. They only approach -273 degrees C :)
Somehow I get the feeling you wrote this in HTML with superscript "o" for
degree symbols :) but when converted to plain text it does look rather
"cool"
Sounds neat! Glad it is working well. I assume you slipped a decimal
point on your temperatures though, -1950C would be well below absolute
zero! If the wobble at room temperature is too great you might try a
spring loaded bushing.
Do you have a good reference for materials for cryogenic use? I, a EE
sonar and firmware guy, have recently been tasked with designing a LN2
recirculation system for another part of the company testing cryogenic
gear. I am fumbling with where to start, other than getting my boss a
new psychiatrist.
> Well, I started this mess ... ah.. thread. Here's the results: Last
> Saturday we sis a shakedown cruise of our cryogenic processor.
> We also discovered that immersing a can of root beer (opened, to let off any
> pressure) will result in a pleasing root beer float, almost instantly. The
> carbonated beverage also foams up, the foam spilling out of the can, which
> makes a kind of instant ice cream when it hits the LN. Quite tasty, once it
> warmed up to freezing. Don't try this at home. Dumping a cup of LN into a
> bucket of water freezes the water 2" thick in a minute. We were using
> distilled water/ice slush as a calibration standard, so we happily had to do
> this a couple of times to make more ice.
>
> All quite a lot of fun. They pay us to do this kind of stuff?
>
> --Lawrence Lile
>
>
Many years ago at Liverpool University, a humble banana was immersed in LN,
and then unfortunately dropped by the undergraduate who tried to fish it out
of the dewar. The banana slid across the lab floor, out of the open door,
across the corridor and tumbled down the stairwell smashing into thousands
of slippery slithers. The stairwell was out of bounds for several hours
cleaning up the last traces of cryogenic banana.
Even more off topic, when I was a college in 85, a stair carpet was thrown
down a hall of residence stairwell, to deter some marauders. It missed the
intruders but smashed the main cast iron water pipe which ran from the 2000
gallon tank on the roof.
Again, another stairwell and most of the ground floor off the residence were
out of action for several hours. The basement took days to dry out!
I missed all this fun, because I had my nose in some Z80 assembler at the
time. Had PICs been widely available then, life may have had a more
exciting turn!
At a NASA open house around the time of the moon landings, there was a
demonstration of cryogenics.
Live goldfish were dropped into LN2, then back into their bowl. In a few
minutes the frozen goldfish would thaw out and begin swimming about as if
nothing had happened.
An individual goldfish would make the trip several times before showing
signs of wear.
All went well until the demonstrator dropped a frozen fish onto the floor
and it shattered into tiny shards.
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