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Time to cool down oil Pool

Time to cool down oil Pool

Time to cool down oil Pool

(OP)
Is there any way to calculate how much time it will take to cool down a oil pool on the ground if have following information:

1. Temp change of Oil 150 Deg F to 100 deg F
2. Ambient air & Soil Temp 60 deg F
3. Wind vel 5 MPH
4. Sp. heat of oil 0.55 BTU/lb -deg F
5. Sp Gr of oil - 0.85
5. Soil Conductivity 0.3 BTU/Ft2 - Hr deg F
6. Area of oil pool 5000 Sq. Ft
7. Volume of oil pool 20000 Bbls

Any help in pointing to correct direction will be appriciated.

RE: Time to cool down oil Pool

You apply Newton's Law of Cooling.  

The only numbers you're missing for a first order calculation are soil thermal conductivity and oil thermal conductivity.

TTFN

FAQ731-376: Eng-Tips.com Forum Policies

RE: Time to cool down oil Pool

You can find here some worked examples similar to your problem (see more particularly page 30).
The heat transfer coefficient to air can be estimated as a function of wind speed from published tables. In the absence of such data I would use 15 to 20 W/m²°C (sorry, don't know how this translates to US units).
For the heat transfer to soil, by patiently googling 'heat conduction in a semi infinite solid' (or similar) you should find values of an equivalent heat transfer coefficient from a small circular area (tank) to a semi infinite body (soil) as a function of soil conductivity.
And IRstuff oil conductivity is not necessary to solve this problem and in fact it is purposedly not supplied in the data.
BTW gasoperations, how this problem is stated strangely resembles a school homework problem: if you come back please describe the context in which the problem is posed, it will help in suggesting how to go on with it.

prex
http://www.xcalcs.com : Online engineering calculations
http://www.megamag.it : Magnetic brakes and launchers for fun rides
http://www.levitans.com : Air bearing pads

RE: Time to cool down oil Pool

If you look at how data are provided, they suggest (or even force) a solution for a single lumped mass representing the pool as a whole, thus no need for conductivity. If the temperature distribution in the fluid had to be accounted for, this would be a really strong problem (only fluid FEA could be used) and more data would be required (e.g. viscosity).

prex
http://www.xcalcs.com : Online engineering calculations
http://www.megamag.it : Magnetic brakes and launchers for fun rides
http://www.levitans.com : Air bearing pads

RE: Time to cool down oil Pool

Well, there are lumped mass, and there are lumped mass.  What's the point of doing the calculation if there's no chance that it'll reflect reality?  And why would "they" want that?

At the very minimum, one should be able to bound the answer, and say, "the cooldown time will be between T1 and T2."

TTFN

FAQ731-376: Eng-Tips.com Forum Policies

RE: Time to cool down oil Pool

If I may throw in my 2cents on this, it appears that the problem is indeed somewhat simplified in that radiation seems to be ignored even though this is a very large pool of oil which is likely to have a high emissivity/absorption.
As a second point, it makes sense to look at the applicability of lumped system equations through the Biot number (Bi=h*V*k^-1*A^-1)... is this larger than 0.1 (=not lumped)? ...so again, you would need to know the termal conductivity (k) to even make the judgement call as to whether this can or cannot be treated as a lumped system. Oil has a k vaule of approx. 0.15 (W*m^-1*K^-1) depending on type.
The value of h would be obtained through forced & natural convection calcs.

http://www.solveering.com

RE: Time to cool down oil Pool

(OP)
I am sorry for not being able to log in and read all the good replies as there was so problem with my login account.

Now this is not a school problem.

It is a real world problem. There is a high pour point oil (+100 degF) stored in tank that is continuously heated and circulate.

Now the estimates are being made if the oil will flow to another location if there is big breach in the tank before it is cooled down due to heat lost to the ground and ambient air.

This is how the original question was raised. Anyway the problem is still unresolved and if some of the original reply may still want to open this subject and enlighten please feel most welcome.
 

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