Problem - Offline Heating Loop for Equipment 1

[ASIT859]

Hi there,
New here and looking for some help and direction.

I have a piece of equipment (gearbox) (approx. 10,000 lbs of steel) that holds 50 gallons of oil. I want to heat up the oil and gearbox by kidney looping the oil through an off-line heater. I'm trying to determine how long it will take to bring everything up to temperature. A few bullet points and what I've done to try and attack this problem below.

- Starting temperature = 50*F (10 C)
- End target temperature = 120*F (48.8 C)
- 50 gallons gear oil
- equipment mass = 10,000 lbs (4536 kg)
- Kidney loop flow rate = 5-gpm (18.9-lpm)
- Kidney loop heater = 5-kW
- I want to avoid heating the oil above 130*F (55 C)
- specific heat steel = 0.5 kJ/kg/K
- specific heat oil = 1.8 kJ/kg/K

From the standard specific heat formula (Q = c*m*(T2-T1)) I know how much total heat I need to add to both the gearbox and the oil to bring them up to temperature.

- 86,400 kJ to heat the steel
- 12,700 kJ to heat the oil

Simply adding the above numbers and dividing by 5-kW gives me 5.5 hours to heat. This seems simple but I don't think it's accurate (obviously I didn't account for any losses) but I think that's beside the point.

I tried to calculate the single pass heat addition through the heater using the basic formula above, but dividing both sides by time.
I then get

- Q'(kJ/s) = M'(kg/s) * c * deltaT. Rearranging for deltaT yields a single pass temperature rise of ~10.2 *C.

In each pass through the heater the temperature of the oil rises. Each time that oil passes through the gearbox the temperature falls as heat is transferred into the gearbox. What I'm not sure of and what I think is important is how much heat is lost from the oil into the gearbox on each pass. I believe this is forced convection.

Formula for forced convection; Q'(J/s) = h*A*(Toil - Tbox).

I've found a very wide range for values of h (h= 50 to 2,000??) and I'm not sure what Area inside the gearbox is applicable.

What I think may happen during heating is that at some point the oil will reach the desired temperature or very close to it but there will still be heating left for the gearbox. I just want to try and come up with a good approximation for time to heat and be able to choose different flow rates, heater sizes, etc in order to optimize this process. I believe there should be an optimal method for this that doesn't just involve throwing more wattage at the problem.

Am I looking at this wrong? I'm not sure where to look for resources and help in this type of problem/application.

 

[chicopee]

This is a transient heat problem, so a time step analysis will need to be done. In this analysis there will be a temperature gradient thru the gear box walls and thru the internal gears. No simple task and some assumptions will need to be made. My approach would be to research periodicals on plant maintenance or machine design for related analysis.

 

[LittleInch]

Kind of depends also on how accurate you want / need to be.

Without out having some idea of the wetted area of the gearbox to the warmer oil you will find it difficult to move any further forward.

Your 5.5 hours is a "perfect" solution where all the 5.5 kW is going into heating the oil and gearbox. From your description I would take a factor between 2 and 3 on this time to allow for inefficiency in transferring the heat, losses into the outside, the gradual reduction (probably) in heat input if you limit oil temp to 130F and the reduction in heat transfer as the differential temperature between oil and gearbox reduces.

You could find that the oil heats up preferentially if heat transfer is poor and then the gearbox lags behind. As ever how close to your end temp do you need to be - temperature rise will reduce as time goes on.

2 if there is a lot of area available for the oil into the gearbox and 3 if not.

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