home heating oil change in volume due to temperature change

There are two good answers to this question.  Both are equally valid, and
actually come out to be the same for you.

1.  31.2 inches
2.  31.2 inches * (1.000165)

Answer 1 is probably the easiest.  You simply aren't going to be able to
measure the difference if you can only measure to one digit past the
decimal.

Michael

On Tue, 06 Jun 2006 03:26:20 GMT, "Herman Family"


Thank you for the reply.
So if I have a tank with the dimensions of 60"x27"x17" (17" depth)fill
with heating oil and its 60 degrees out, I can calculate it to have
(60x27x17)/2319.2207792 gallons of heating oil. If the temperature
rise is 5 degrees I'll get (5 x 1.000165)x17….014025,
(85.014025"x27"x60)/231Y6.202 gallons. This doesn't seem correct. Am
I doing something wrong?
Ken

ken wrote:

  ...


No, it's not correct. Yes, you're doing something wrong.

Assuming that the new volume of oil after a change of 1 degree F really
is 1.000165 times the old volume -- I haven't checked, but it seems
reasonable -- then a better way to express it is as a volume expansion
coefficient: .000165/degree F The _change_ for a rise of 5 degrees is
then 5 x .000165 = .000825. Multiply that by the original volume, and
you get 0.09835714284, the change in volume due to temperature. Or
multiply by 1.000165 to get the new volume, 119.31913634284.

It's silly to use so many significant figures. How much oil do you wipe
off the dipstick with each measurement? How accurately can you read the
dipstick?

What the figures show is a change of as close to 0.1 gallon as you can
hope to measure. To substantiate it, you need to read the oil's
temperature when you measure it. Stir the oil first.

Jerry
--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Hi Jerry and thanks for that clarification. Your right its
insignificant, I was just curious.
Thanks again.
Ken

It's a reasonable question.  Home heating oil might not change much in
volume due to temperature, but lighter fractions most certainly do.

* Petrol for your car changes enough to overflow the tank if you fill it
right to the top on a cold morning and leave the car parked in the hot sun
later.
* Jet fuel changes so much that leaving ESD valves at either end of a
pipeline closed for more than a few minutes on a hot day can burst the
pipe..

HTH,
Cameron:-)

ken wrote:


Try (1 + 5 * (0.000165)).

But you are obviously doing this calculation on a PC or hand calculator.
  119.2207792 gallons?  A mouse turd is about 1mm in diameter and 3mm
long, or 0.003cc, or about 0.0000007 gallons -- so should a mouse crap
in your tank the volume of actual heating oil is now 119.2207785
gallons.  This ruins your precision.

Heat on the metal of the tank, wind pressing against the side of the
tank, kids boinking balls against the side of the tank -- all of these
may change your measurements far more than that mouse looking for a
place to go.  And unless your measurement is 31.2000000 inches, with a
really _expensive_ ruler, your calculations are moot anyway.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Posting from Google?  See http://cfaj.freeshell.org/google/

"Applied Control Theory for Embedded Systems" came out in April.
See details at http://www.wescottdesign.com/actfes/actfes.html

wrote:


That's a funny analogy and with those numbers I'll be sure to keep the
mice away from my oil tank.
Ken

Tim Wescott wrote:

  ...


Think big. Mouse turds are seldom produced one at a time. :-)


And you would need to set the dipstick down extremely slowly so as not
to make waves, and then subtract the submerged volume of the dipstick
(probably the greatest of the individual sources of offset).

Jerry
--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

You are talking very small (negligible) peanuts--However, for my 2 cents
Density=Wt/Vol
The weight of the Fluid doesn't change, so for the two conditions:
D1V1ÒV2
V2=(D1/D2)*V1
OR for the change in the volume:
V2-V1= V1*(D1/D2-1)
Solve for an absolute number or just look at the percent change in Volume.
Knowing the configuration of your tank figure out the change in height.
Obviously, you need to know how the fluid density varies with temperature.
MLD

What you are doing in principle is determined using the "thermal coefficient
of expansion".
It's published in tables of physical properties.  Mine only shows the
coefficient of linear epansion for solids and nothing for liquids.  What you
want is the thermal coefficient of volumetric expansion which is 3 times the
linear value.

I'd go to the local library to find physical constants like this.

See:

http://www.ajdesigner.com/phpthermalexpansion/thermal_expansion_equation_volumetric_coefficient.php

Fred