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What do I multiply my gallons of jet fuel to get true volume based on temperature.??

I think somewhere it was discussed but I can not find it now.

I think somewhere it was discussed but I can not find it now.

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but I can't seem to find it now.

But I read somewhere that temperature has to be taken into effect.

For every degree over 60 you multiply the gallons by this factor to get the compensated value.

http://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&cad=rja&uact=8&ved=0CCcQFjAB&url=http%3A%2F%2Fdtic.mil%2Fdtic%2Ftr%2Ffulltext%2Fu2%2Fa132106.pdf&ei=0gUzVJmsFq2asQSTvYG4DA&usg=AFQjCNFV8HPrqky_rRaw9RKLrwZahQgX3g&sig2=NpsB0N9K7O84TzNOqdnOPg&bvm=bv.76802529,d.cWc

You can measure either volume or mass.

The volume will change with temperature, the mass won't.

So if you have a tank with a measured volume and a known temperature, you can calculate the mass or the volume at the standard temperature which appears to be 15 deg C (60 deg F).

Density is linear with temperature, but volume will not be. How accurate do you need to be?

In an earlier post you said....

For every degree above 60F, multiply by 0.00171, subtract from 1.0, that's your conversion factor from hot gallons to standard gallons. For example a factor of 0.900 means one hot gallon is actually 0.9 standard gallons.

And for every degree below 60F, multiply by 0.00171, add to 1.0, that's your conversion factor from cold gallons to standard gallons.

What do you mean by subtract from 1.0 and add to 1.0..????

How does that work into the formula..???

I measure the fuel height in Inches to calculate the volume in Gallons.

Im using this all in a PLC (AB Micrologix 1500).

Im trying to calculate the Gross Gallons and Net Gallons to display on my HMI.

Im already displaying the height (inches), temperature (degrees F).

This is what I think you are telling me, but please correct me if I am wrong.

Gross Gallons = Height measured converted to Volume (Gallons).

Net Gallons = Temperature Compensated Volume (Gallons).

Net Gallons = (Gross Gallons) x ((Temp - 60) + 1) x .00171

Please show revised formula if this is not correct.

Thank you in advance

```
771.1
Net Gallons = Std Gallons ( ------------------ )
832 - 1.015*Temp
```

I still haven't found the earlier result you mentioned, but this is probably more accurate.
So IF I had 25000 gallons at 80 degrees F ...

Volume Corrected = 25000 x 771.1/832 - (1.015 x 80)

Is this correct..???

Y = Degrees F of Fuel (80)

Z = Volume (Gallons) Value that is "NOT" Temperature Corrected (25,000)

X = (Z x 771.1) divided by 832 minus (1.015 x Y)

X = (25,000 x 771.1) divided by 832 minus (1.015 x 80)

X = (19,277,500) divided by 832 minus (81.2)

X = 19,277,500 divided by 750.8

X = 25,675.94

Please let me know if this is correct or not if you would.

IF this is not what you met, please correct me.

Thanks for all your help.

so if you would please let me know what I am doing wrong.

The terms

You have is a volume of fuel at 80 deg F, and you want to know what the volume would be at 60 deg F.

The formula tells you what happens to a volume of fuel that

V_80 = V_60 *

Calculate the factor in the square brackets first. You should get a number near one:

The final expression is: V_80 = V_60 * 1.0270

And the answer: V_60 = 25000

You say above that the density at 60 Deg F is 771.1

but everything I have says that the density of Jet A at 60 Deg F is .82

Am I missing something..???

My calculations have been bad, but my physics has been worse.

I am going to start clean, rather than try to find the earlier error.

>> You say above that the density at 60 Deg F is 771.1 but everything I have says

that the density of Jet A at 60 Deg F is .82

The Typical Fuel Density Chart (which is the basis for this analysis) indicates a density

for Jet_A at 15 deg C ( = 59 deg F) slightly less than 0.81

These six points all check out.

The equation you need is:

V_60*D_60 = V_T*D_T

V_60 = V_T * (832.5 - 0.4359*T_F) / (806.3)

For 25,000 gallons at 80 deg F:

V_60 = 0.9892 * 25,000 = 24,730

V_60 = V_80*(D_80/D_60) = V_80*(797.6/806.3) = 0.9892*V_80

V_60 = 0.9892 * 25,000 = 24,730

Just a simple formula that I can substitute the different Temps and Gallons..!!

Im limited in my controller what I can do.

Thanks

X = Temperarure

Y = Measured Gallons

What formula can I use that when the X and Y varies it will

give me the true "Temperature Compensated" gallons..??

```
20.00 102.17
30.00 101.63
40.00 101.09
50.00 100.55
60.00 100.01 <== Small round-off error
70.00 99.47
80.00 98.92
90.00 98.38
100.00 97.84
```

That value of 806.3 is confusing us..

http://en.wikipedia.org/wiki/Jet_fuel#Typical_physical_properties_for_Jet_A_and_Jet_A-1

One of them does have the 0.82 (or 820 specification). My chart used the other one.

Since all the density curves seem to have the same slope, we can add 13.7 to the coefficients.

This would give a new PLC formula:

This is what the formula gives at a few temperatures:

```
20 102.08
30 101.55
40 101.02
50 100.49
59 100.01
60 99.96
70 99.43
80 98.89
90 98.36
100 97.83
```

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densityof the fuel as a function of temperature.