Diesel storage over night

[mabbas72]

I am trying to calculate the temperature change with time for diesel left in a storage tank overnight in cold weather.

What do I need to know? and is there a standard formula or program?

The tank is made of Alminum, volume stored is about 6000 liters, tank dimensions are 6.75 m long, 2.3 m wide and 1.92 m high.

 

[25362]

Here are some thoughts:

Primus. One deals here with transient heat transfer, since it is assumed the tank surfaces and contents cool down as time passes.

Secundus. If the prismatic tank is not thermally insulated and is exposed to the elements one must consider heat lost by convection (wind) and radiation, this is without taking into consideration rain and snow effects, that would add a conductivity element to the estimation of heat losses.

Tertius. Since the tank is only partly filled (6 m³ out of 29.8 m³) the "dry" heat conductive aluminum walls and roof would also participate in the heat transfer and losses. It is assumed the tank bottom plate doesn't exchange heat with the ground.

Quartus. Convective heat losses of the vertical walls would be different than those from the horizontal roof. As an example on natural convection:

If the temperature difference between the tank plates and the surroundings is, say, 50^oF the heat loss from the roof would be 20 Btu/(h*sqft), and from the vertical walls it would be about 32 Btu/(h*sqft).

Quintus. Forced convection includes wind effects. A change of wind velocity from 5 to 20 mph may increase the heat transfer coefficient by a factor of 3.

Sextus. Radiation from aluminum metal walls is about 1/10 that from dark paints; anyway the effect is there and would have to be estimated.

Septimus. There are simplified nomographs based on delta T between the tank surfaces and the surroundings, which could help in reaching an estimate.

Octavus. One has to consider that the cooling in the diesel itself would be progressive and would consist in sensible heat transferred from the bulk to the walls based on its conductivity of about 0.14 W/(m.^oC). All these meaning that the diesel oil in the tank wouldn't have an equal temperature all through and would show a profile that may induce some kind of layering.

Nonus. Even before the first morning light, the process may start reversing, a fact that may add complexity to the estimation.

Decimus. One is lead to think that due to the variability of the factors affecting the temperature in the tank, the most appropriate thing to do, is to take diesel oil temperatures every hour through the night, recording the weather conditions at the same time, i/o to find some kind of practical correlation.

 

[mabbas72]

Thanks for the response, that makes sense. Let me add more information to clarify the above.

1) The 6000 L is an example of part of the load. The tank has a cylinderical type tank, which can carry between 18-22 m3 fuel, where 6000 L is one compartment.

2) This compartment is left overnightstill filled with fuel, hence, you have some empty areas from the remaining space.

3) The temperature drops to - 5 °C, and assume the tank was at 5 °C. How long would it take it reach zero degrees? or alternatively, what would be the delta T/delta time? Assume wind is not a factor

4) heat loss by conduction would be minimal, then the main two factors would be the Qconv and QRad.

 

[25362]

As a ballpark estimate of the time for "isothermal" cooling see the following.

time=(MCp/UA)*ln(t₁-t_o]/[t₂-t_o)​

The external temperature is constantly t_o=-5^oC; the combined (convection + radiation) HT coefficient U assumed roughly at 10 kcal/(m²*h*^oC); the mass of diesel M=5,000 kg; the specific heat capacity Cp=0.5 kcal/(kg*^oC); taking the exposed area A at, say, 50 m²; t₁=5^oC; t₂=0^oC.

time=[(5000*0.5)/(50*10)](ln 2)=5*0.693=3.5 hours​

I repeat, this is a very rough estimate that may serve as an example. From the formula above and the right thermal and geometric paramaters you'd be able to estimate the relation of delta T with delta time. This is quite a simple formula because the process is isothermic for the surroundings. Actual measurements may help in adjusting the assumed values.

 

[unclesyd]

25362,
We used to call this the "Bottle of Wine Formula" or “always purchase you gasoline in the morning formula”.

 

[mabbas72]

Thanks again for the information, I have looked at the calculations and the values and I could trace almost all back through formulas and tables, but I could not get how you estimated the 10 kcal/m2.hr.°C overall U. Can you please explain a bit further?

 

[IRstuff]

That's just an estimated heat transfer coefficient.


Looks like about 1/3 from natural convection and about 2/3 from radiative loss.

TTFN

 

[srkangas]

there is a standard program... found at

http://www.thermoanalytics.com