Fuel Volume Formula

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.
Mark HinesAsked:
Who is Participating?
I wear a lot of hats...

"The solutions and answers provided on Experts Exchange have been extremely helpful to me over the last few years. I wear a lot of hats - Developer, Database Administrator, Help Desk, etc., so I know a lot of things but not a lot about one thing. Experts Exchange gives me answers from people who do know a lot about one thing, in a easy to use platform." -Todd S.

d-glitchCommented:
You need to know the density of the fuel as a function of temperature.
0
Mark HinesAuthor Commented:
.82 (Jet A)

Can I email you directly..???
0
Mark HinesAuthor Commented:
Somewhere you wrote what factor to multiply the gallons by for very degree over 60 Deg F.
but I can't seem to find it now.
0
Get your problem seen by more experts

Be seen. Boost your question’s priority for more expert views and faster solutions

Mark HinesAuthor Commented:
I have a 10,000 gallon tank that I am measuring the height and calculating the volume.
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.
0
d-glitchCommented:
The formula on page 20 is not relevant, but here is the data:
Jet-Fuel-for-ExEx.png
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?
0
Mark HinesAuthor Commented:
Im trying to calculate my net and gross gallons..

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..???
0
d-glitchCommented:
Try

Gallons(T)  =. Gallons(60) * ( 1 + (T-60)*Factor)
0
d-glitchCommented:
I don't remember the earlier post, but I will get back to this tomorrow
0
d-glitchCommented:
What units are you using?
0
Mark HinesAuthor Commented:
Volume = US Gallons
Temperature = Degrees F
Max Tank Size = 50,000 Gallons
0
Mark HinesAuthor Commented:
FYI...

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).
0
Mark HinesAuthor Commented:
I apologize if Im making this harder than it is, but I need to make sure I understand exactly what I need to do.

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
0
d-glitchCommented:
Deriving the Formula
0
d-glitchCommented:
The final result would be:
                                         771.1
     Net Gallons  =  Std Gallons ( ------------------ )
                                    832 - 1.015*Temp

Open in new window

I still haven't found the earlier result you mentioned, but this is probably more accurate.
0
d-glitchCommented:
Here is an Excel plot of the relative volume versus temperature.
Note that is not quite linear with temperature since it is proportional to the density (which is linear).

You could fit a line to this curve.  That may be close enough. but it does introduce additional and unnecessary errors.
One final graph
0
Mark HinesAuthor Commented:
The last formula is hard to read because it did not seem to displayu correctly..sorry.

So IF I had 25000 gallons at 80 degrees F ...

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

Is this correct..???
0
Mark HinesAuthor Commented:
X = Volume (Gallons) Value that is Temperature Corrected (25,675.94)
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.
0
Mark HinesAuthor Commented:
Well...I can now see that I am interpreting your answer incorrectly,
so if you would please let me know what I am doing wrong.
0
d-glitchCommented:
>>  So IF I had 25000 gallons at 80 degrees F ...

The terms Net, Standard, Corrected, Gross, and Actual are all kind of confusing.  60 deg F seems to be the standard temperature, but I am going to use the number not the word.

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 starts at 60 deg F and warms up or cools down.  The final temperature here is 80 deg F.

     V_80  =  V_60 * [(771.10) / (832 - 1.015*80)]

Calculate the factor in the square brackets first.  You should get a number near one: 1.0270 in this case.    You can check it on the Excel chart or plot.

The final expression is:      V_80  =  V_60 * 1.0270

And the answer:                  V_60  =  25000 / 1.0271  =  24343 gallons
0
Mark HinesAuthor Commented:
Hi Again..

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..???
0
d-glitchCommented:
Sorry for the delay and the confusion.
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
New DerivationThese 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
0
Mark HinesAuthor Commented:
Can you make this simpler to understand..??

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

Im limited in my controller what I can do.

Thanks
0
Mark HinesAuthor Commented:
Example:
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..??
0
d-glitchCommented:
This would be the PLC formula:            V_60  =  V_T * (832.5 - 0.4359*T_F) / (806.3)

V_T is the volume of fuel in gallons at some temperature T_F in deg F.

(832.5 - 0.4359*T_F)  is the density of the fuel at T_F.

(806.3)  is the density of the fuel at 60 deg F.

V_60  is the value you want, the volume of your fuel at 60 deg F.
0
d-glitchCommented:
As a check, this is what the formula gives at a few temperatures:
      
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

Open in new window

0
Mark HinesAuthor Commented:
All the FAA guys here say I must use a value of .82 for the density value of Jet A fuel.

That value of 806.3 is confusing us..
0
d-glitchCommented:
Apparently there are two varieties of Jet A fuel:
     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:            V_60  =  V_T * (845.8 - 0.4359*T_F) / (820.0)

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

Open in new window

0

Experts Exchange Solution brought to you by

Your issues matter to us.

Facing a tech roadblock? Get the help and guidance you need from experienced professionals who care. Ask your question anytime, anywhere, with no hassle.

Start your 7-day free trial
Mark HinesAuthor Commented:
This final answer is not even close to being correct..!!!!
0
It's more than this solution.Get answers and train to solve all your tech problems - anytime, anywhere.Try it for free Edge Out The Competitionfor your dream job with proven skills and certifications.Get started today Stand Outas the employee with proven skills.Start learning today for free Move Your Career Forwardwith certification training in the latest technologies.Start your trial today
Math / Science

From novice to tech pro — start learning today.

Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.