home heating oil change in volume due to temperature change

Hello, I hope someone can help me out. I am looking for a formula that I can use to calculate the change in volume due to temperature change, home
heating oil is the medium. For example if I have 175 gallons in a 275 gallon tank on a 60 degree day and the depth of the fuel oil measures 31.2 inches, what will the depth be at 63 degrees.
I have two variables to work with and they are temperature and depth of fuel.
I saw a correction factor of (.000055)-1 per degree F for kerosene in my search and I was hoping for something as simple as that.
Thank you for any and all help. Regards. Ken
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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
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On Tue, 06 Jun 2006 03:26:20 GMT, "Herman Family"

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

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:-)
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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
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wrote:

Hi Tim, That's a funny analogy and with those numbers I'll be sure to keep the mice away from my oil tank. Ken
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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
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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: D1V1V2 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
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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
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