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Thermodynamics Heating oil Problem

  1. Dec 16, 2009 #1
    I have problem in calculating heating time of Lube Oil ( SAE 40 ). There is a tank of Lube Oil ( SAE 40) of 19000 liters at 20° C. this oil has to be heated up from 20° C to 40° C with help of a heater of 85 kW through which oil flows at 4800 liter/hour and back to tank.

    How much time it will take to heat up complete tank from 20° C to 40° C. it is a closed loop circuit and small distance. just oil from tank flows at 4800 liter/hour through 85kW heater and then goes back to tank.

    any suggestions for finding time ??


  2. jcsd
  3. Dec 16, 2009 #2


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    Really. This is pretty basic, first year physics stuff. In the easiest sense, there is a basic relation to the mass flow rate and the delta T you wish to achieve. What do you think that is? Hint: You may see it as just mass, but substitute mass flow rate (HINT HINT) to include the time factor.
  4. Dec 17, 2009 #3
    Thanks a lot .

    I used the formula Q = m* Cp * delta T and calculated for every minute. but answer was a bit unbelievable as 20 minutes.

    also i have a doubt to make it in a single step calculation as it is a iterative process and gradual increase in temp. will take place. so finding an answer in single step a bit confusing.
  5. Dec 17, 2009 #4


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    85 kW is a fair amount of heat. 20 minutes doesn't sound unreasonable to me at all considering the low amount of mass you're dealing with.

    I have done these types of problems using a finite difference method in which I set the delta T to a small value and then calculate the energy required for that delta T (I also update the Cp value for variations in temp with each step). I do this for a broad range and sum up the energy requirements until it equals the energy I have. It has worked pretty well for me as a good first estimate.
  6. Dec 17, 2009 #5


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    Nowadays (chuckle) we have FLUID elements and can couple them to surface effect elements for really good heat transfer predictions. We use a method like this to predict oil/fuel cooled bearings....being quite conservative of course.
  7. Dec 18, 2009 #6
    but when i look at system ..19000 liters is total volume to be heated from 20 degrees to 40 degrees.
    it flows through heat exchanger at 4800 liters per hour. and each time temp is increased by 35 degrees in this flow.

    so virtually if 20 minutes it takes , then we can say that only 1600 liters have crossed through heat exchanger in 20 minutes ..and can this 1600 liters make the temp of remaining 17400 liters raised by 20 degrees.

    thats the confusion ..i used the same normal equation and a temp. of delts T as 20 degrees. but end up with confusion ..
  8. Dec 18, 2009 #7


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    What's your value for Cp?
  9. Dec 18, 2009 #8
    Let me throw in my 2cents on this one...
    you can approach this from two (very) different perspectives, one will fit you better than the other (you'll figure out which works for you): Theoretical or Real.
    On the theoretical side (say you were in college and this was on a test/homework), you have far too little information to come up with the exact value since the only thing you are considering is the overly simplified scenario that you are given. That said, take a closer look at the equation you are using. If you were to replace the time rate of energy (Q) with just energy and likewise the time rate of mass flow (dm/dt) with just mass you would find that it takes somewhere in the order of 855 MJ of energy to increase the temperature by 20C. If you are adding this heat at a rate of 85 kJ/s, it would take about 2.8h. This is ignoring the pump and heat exchanger. What do these two do? Well, they simply move the fluid around and add heat to it. Will the total amount of energy change if the pump moves the fluid faster or slower? Not really. Does 2.8h sound appropriate? In my experience it does.
    So what about the pump? Well if you take your equation and plug in the (initial) values you will find that the instantaneous output will have raised the temperature by 28.3K (assuming Cp to be about 2500 J kg^-1 K^-1). You could take some different approach here and use a difference method to arrive at your number through successive approximation, but I think the easiest is to use the total amount of energy.

    Now for the 'real' scenario...
    In the real world, things will not work as nicely. Partially because you have thermal gradients that occur within the fluid (and will therefore change the heat capacity) but mainly because there will be heat lost to the environment (through convection, conduction and radiation). Due to this, you will need to consider additional points in your model: what are the surface temperatures (and hence the losses mentioned above), what are safe ranges that the oil may be heated to (you may need to heat it slower so that there are no dangerous byproducts generated if the heat exchanger doesn't heat the fluid uniformly). So in the 'real' case, the time will definitely be longer than in the ideal situation since there will be system losses that need to be accounted for.

    P.S.: I didn't check my work above so don't take this as 'correct'. But in my opinion the problem can more eloquently be solved by ignoring the pump altogether. This may just be one of those cases where the problem has added additional information that you don't need...
    Anyone else concur with my comments?
  10. Dec 18, 2009 #9
    I think Solveer summed it up pretty well. The pump can pretty much be ignored unless checking for limits like max temperature of the oil; heat addition is heat addition unless you have details on efficiencies or losses (or want to make assumptions). The pump could be used as a justification for using a lumped capacitance approach by claiming it would be a well mixed system. Using the formula Q_dot = (m*C*DELTAT)/time, I arrived at about 2.1 hours using the "unused engine oil" properties in EES (not sure how well that corresponds to SAE40). If you want to consider changing Cp, look at how much it changes from 20C to 40C. If the change is very small you could reasonably just use the Cp at 20C. You could also use the average of the two values since the Cp varies linearly, more or less, with temperature.

    For a more real world situation you have to look at the requirements and goals. You just have to remember that the simple lumped capacitance approach will underestimate the time due to heat losses of the heater and tank.
    Last edited: Dec 18, 2009
  11. Dec 18, 2009 #10
    I think Yeti08 points to an important 'additional' item here that should not be overlooked (apart from the interaction between heat capacity and temerature of course): The delta-T, the heat rise of the oil as it gets heated in the exchanger is fairly constant (other than fluid-temperature effects which should be evaluated at the fluid's film temperature). So, as time passes and the fluid starts to hit the set point, the dT is still going to be in the order of 20-30C (or more?) so the total temperature of the oil could be close to 70C. Not much of a problem for SAE40, but for more volative types of oil certainly something to consider. On top of that, the surfaces of the heat exchanger will definitely be A LOT higher, especially with 85kW going into the fluid. There is (of course) no mention as to the way the Hx works, but this will make a huge difference for any local effects... just thought I'd mention it.
  12. Dec 18, 2009 #11

    I have not checked your formula or calculations, but did just a quick calc;

    To raise the bulk (19000 L) temp from 20 C to 40 C in 20 minutes, your 20 min flow quantity (1600 L) would have to be raised to 257.5 C

    So delta T for the oil across the exchanger would need to be;
    (257.5 - 20)= 237.5 C -- in a single pass

    Even if this is possible with your equipment, you might want to consider breakdown of the lube oil within the exchanger (coking) as mentioned above.
  13. Dec 18, 2009 #12


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    I'd go a different direction and ignore the flow rate completely. He's not looking for a delta-T in the flowing fluid, he's looking for the time to heat up the tank. So if you have mass, Cp and delta-T, you can calculate it directly:

    E=Cp*dT (Which he said he used...)

    That gives total energy, so then you just divide by power and that gives you time. Simple!

    ....but make sure you have your units consistent...

    [edit: looks like someone already got there]
  14. Dec 21, 2009 #13
    Cp value is 1.802 KJ/KgK at 20° C and 1.927 KJ/KgK at 40° C.
    Density value 897 kg/m³ at 20° C and 879 Kg/m³ at 40° C

    I have got some idea from above given solutions of you all. Thankyou for that. This is not a real life peoblem , which you said that have to consider pressure loss and heat loss or some other parameters. this was just an excercise.

    I used Q = m1*Cp1*( Teq. -20° C) = m2*Cp2* ( 55° C - Teq.) for the first minute and i got the value of Teq.
    for second minute i used this value of Teq instead of 20° C of first equation. I used like heat loss from hot oil is equal to heat gained by tank oil which is at 20 ° C.

    and my answers were like on 20th minute i was getting tank temp. as 40° C which no one was able to believe as how it is possible to heat 19000 liters from 20° C to 40°C in 20 minutes with this flow rate and heater.
  15. Dec 21, 2009 #14
    I think the consensus was (towards the end of this thread at least) that the pump doesn't add anything to your problem and that the 20 minutes was not correct. I'm not sure what you are pointing out above but the finite differences method you seem to be touching on will not get you there (it would, but not with this equation). If you use the values of 1802 J kg^-1 K^-1 & 1927 J kg^-1 K^-1 for the 20C and 40C points, with density of 897 kg m^-3 & 879 kgm^-3 you will get 2.1h (the number will be 2.056h but don't even get me started on that number... way too many sig. figures). Having two data points for the heat capacity and density at the two temperatures simply means you will have to take the average value for the properties.
    I would venture a guess that you are a student in engineering... if you are a physics student, I would have expected the energy approach before the difference method you are trying to use. One word of advice that helped me during my undergrad days in ME: energy is never created nor destroyed. If you can see where energy is transformed (or added/subtracted) during a process, you will have an easier time figuring out the underlying equations.
  16. Dec 21, 2009 #15


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    I have sen it both ways. But once you mentioned that, I looked back at Chromolox's catalog and that's pretty much how they say to do it to.

    Attached Files:

  17. Dec 23, 2009 #16
    thanx a lot for your answers , the PDF file in last post was also very helpful and i have got my answer ..may be will back with some other problem ..wish you all a marry christmas and happy new year...have a gud time
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