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...

 

[Latexman]

This is an unsteady state heat transfer problem that requires a differential equation (heat balance) to be solved (integrated). Some editions of Perry's Handbook and Kern's Process Heat Transfer have several cases already solved.

Good luck,
Latexman

 

[nucleareng78]

If I know the kW rating of the heater outside the tank (through which the oil is passing), would this simplify the analysis. I want to treat the tank as a insulated body.

An approximation is satisfactory because an exact time to reach a specific temperature is unecessary.

 

[Latexman]

If you Google "batch heating time calculation" you can probably find the equations and/or a website with a calculator. The equations are not too difficult. Yes, you can also approximate this numerically on Excel. I'm pretty sure if you use Search (between Forum and FAQs) in this forum you can find a description how to do that in an old thread.

Good luck,
Latexman

 

[nucleareng78]

I'll search around. If I know the specific heat of the oil and the kW rating of the heater, it should be a pretty straight forward equation.

Thanks for pointing me in the right direction.

 

[Latexman]

No problem. Let us know how it turns out, or if you have more questions.

Good luck,
Latexman

 

[ApplicationENG]

why dont you put an RTD in the tank, run a loop for 30 min., plot T vs. t, and correlate that to a curve.

 

[dvd]

This probably will get you to third base...

http://www.chromalox.com/catalog/re...ions-and-Heater-Selection-Heating-Liquids.pdf

 

[TD2K]

If it's an electric heating element, I'd be willing to bet that the heat transfer is only weakly a function of oil temperature. I'd use the mass of oil, its heat capacity and temperature change to calculate the duty you need. The kW rating of the heater should give you the time.

 

[nucleareng78]

This is what I did..

Converted 4 kW to BTU/hr. Multiplied this by the specific heat of the oil to get X.

Divided X by the total weight of the oil to get Y.

Divided Y by the temperature difference I am trying to achieve to get Z.

This left me with Z in units of 1/hr

Just took the inverse of this and got it in hours. Again this is simplifying the problem drastically. But I was conservative in the temperature range based on the ambient conditions. I could in theory add some sort of safety factor to this to account for the mixing of different temperature fluids in the tank and the convection heat transfer from the air.

Does my methodology seem correct? I can post exact numbers and units if that helps.

 

[IRstuff]

That would be the perfectly ideal time, which requires simultaneous heating and mixing of every molecule in the tank. Since you are not doing that, your actual time will be higher.

TTFN
faq731-376

 

[nucleareng78]

Yes I understand, but due to the reasoning behind heating the oil, an exact time is not crucial. Would it be wise to take this time and double it, or even triple it? Just to add some margin

 

[IRstuff]

But the way you describe it, the time is determined by the amount of changeover of the circulating oil. You obviously cannot dump the entire heat load instantaneously into the circulated oil, so your calculation is off. Your equation should be based on the mass transport of the circulation loop.

TTFN
faq731-376

 

[nucleareng78]

Understood. The inlet and outlet of the circulating loop are 1-2 feet from each other in the top of the tank, but are at different elevations in the tank. Due to mixing in the oil I was trying to stay away from doing an unsteady state calculation since the exact time is not urgent, but rather an estimate.

If I know the flow rate of the cirulation loop and the volume of the tank, is it possible to somehow include this in my calculation? Just doesn't seem to be a spot to include this...

 

[IRstuff]

Yes, you replace the mass with the mass flow, multiply by the specific heat, and multiply by the temperature change you can achieve in this side loop, which gives you a power that is divided into the total mass to get a time. Again, this ignores any transport delays needed for mixing.

TTFN
faq731-376

 

[IRstuff]

Oh, that was just pitiful

mass_flow*specific_heat*delta_temp = heat_flow_in (Watts)

total_mass*specific_heat*desired_delta_temp = thermal_mass (Joules)

thermal_mass / heat_flow_in = time (seconds)

TTFN
faq731-376

 

[nucleareng78]

If I know the kW rating of the heater would that factor into the equation above?

 

[nucleareng78]

I ended up with 15 hours using the equation above and 9 hours using the original method I posted, so the 15 hours makes more sense since this is a circulation loop.

Thanks IRstuff for the input!

 

[IRstuff]

The heater's power output is relevant in needing to be large enough to provide the heat flow back into the tank.

Note that the above analysis is EXTREMELY superficial; there are lots of parasitic losses that are not accounted for.



TTFN
faq731-376

 

[nucleareng78]

After re-working the problem in Mathcad I ended up getting a much lower heating time using IRstuff's method versus my original method...this doesn't quite make sense to me since IRstuff's method is more accurate and takes into account the flow rate of the circulation loop.

Or am I just overthinking this?