Time to heat up lube oil tank 1

[nucleareng78]

We are filtering these rectangular lube oil tanks and want to use the skids to heat up the oil during cold startup.

I pulled out the old Heat Transfer book but I think I'm over thinking this. If we are taking cold oil out from the bottom of the tank, heating it (specific temp.), and discharging just below the oil level, how do I determine the time for the entire contents of the tank to reach that specific temp.?

I know the geometry of the tank, thermal resistance of walls, heat capacity of the oil. The troubling part is determining how the oil mixes when the suction and discharge are relatively close (just different elevations in the tank). I'm starting to think a general approximation would be sufficient...

 

[nucleareng78]

IRstuff. Your equation cancels delta T and specific heat and just leave total mass over mass flow rate. This may be why I am getting such a low number. Any thoughts?

 

[IRstuff]

They should be different delta_T's; if you wanted a final delta_T of 10°C, you wouldn't necessarily use a 10°C delta_T for the input flux.

TTFN
faq731-376

 

[IRstuff]

There was specific reason I used different notations for the two delta temperatures

TTFN
faq731-376

 

[nucleareng78]

If we are wanting to raise the temperature of the tank from X degrees to Y degrees, wouldn't that difference in temperatures be the same for both equations?

Our desired delta is the the difference between X and Y, then what is the other delta T represent?

 

[IRstuff]

If you wanted a 5° change in the entire tank, would you really only heat up the loop to 5°, and take the hit on warmup time?

Not saying you couldn't, and if you did, then yes, the quantities would cancel out, as you'd expect, since it strictly boils ;-) down to the mass exchange through the heating loop, and the calculation reduces to total_mass/mass_flow, as you'd expect.

TTFN
faq731-376

 

[nucleareng78]

can I replace the heat load with the kW rating of the heater in the circulation loop?? Because I need that in the equation somehow...

 

[IRstuff]

Then, the delta_temps would be different, wouldn't they? The first equation tells you how much power you need, which is why I had them separated.

TTFN
faq731-376

 

[nucleareng78]

Understood, but we know how much power we have right now.

Wait, I think I'm getting this now, lol. If I know the current power of the heater I can calculate what the delta T will be at the flow rate of the heater circulation line. This in turn can be applied to the total mass of the tank (this delta T will be the temperature rise we want) and eureka we have the time figured out...

 

[IRstuff]

That's assuming that level of power doesn't cook the oil...

which means that you may need to crank up the flow rate.

TTFN
faq731-376

 

[racookpe1978]

IRstuff:

Could he take it (the double-ended dynamic thermo-dynamic and flow mixing problem in two simpler steps?

Make the approximation that the lube oil is flowing through the (known capacity) electric heater at a small but steady rate that will heat up the oil coming out of the heater to a single almost-constant value, regardless of what the input temps into the heater are ... at least at the bigenning of the problem.


Then take this steady (but low-percentage of the total tank flowrate) amount into the larger tank.

Heatup rate of the large tank then becomes function of (Heat gain into tank from steady oil flow) - (heat loss from tank into room due to delta T between tank and room)/(heat capacity of tank & oil inside tank).

Heat problem (in the larger tank) then becomes a dynamic (logarithmic) problem of a-steady-amount-of-hot-oil-into-large-tank problem.

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Now, realistically, when I use an electrically-heat oil heater just like this during an oil flush to heat up power plant lube oil tanks, I don't really measure the heatup rates nor do I specifically measure the oil flowing through the heater. 10 to 15 hours is sufficient to heat a 1200 MegaWatt oil tank from room temperature to 130 degree F flush temperature using a kilowatt-sized in-line heater with a 2 inch lube oil hose going through the heater and back into the tank. (A cold plant in winter in the mezzanine is about 45 F to 60 F. A warm plant in the summer is 65 to 75 degrees oil temperature.)

In either case, once the oil is up to a flush temperature of around 130 degrees, we just turn each heater coil off sequentially on and off and keep it (output temperature temperature of the heater element) between 130 to 140 degrees. Tank temperature (system oil temperature) will be about 5 to 8 degrees lower. Don't really "calculate" things - just keep a watch on the heater output thermostat since each plant's oil room and pipe geometry and ambient (outside) temperature is different every time and for eery flush.

 

[IRstuff]

You can take the heater power, 4kW, and divide by the specific heat of the oil, resulting in a quantity with the units of gm*K/s. This is the mass_flow*delta_temp product that your circulation loop must be able to support. So if the specific heat is 2 J/gm*K, then the mass flow temperature product is 1910 gm*K/s. So if your temperature delta is 10°C, then you need to move almost 20kg/s through the loop. You can do that, but I'm not sure that the convective transfer of the oil is that good.

TTFN
faq731-376

 

[chicopee]

since the temperature of the oil will be stratisfied, I would suggest the installation of a mixing paddle to lessen the time to heat the oil and to get a more accurate bulk temperature of the oil.

 

[nucleareng78]

I think the confusing part is that this skid has a set flow rate and a set heater...can't be changed. So when the oil passes through one time is will not reach the deltaT desired the first time, may take 2-3 loops. But I think I can use a lot of this information posted on here, thanks for the help.