# Thread: I am not sure if this is calculus or not

1. ## I am not sure if this is calculus or not

I have to come up with a way to plug in a number (in this case measured in Kilowatts, but that is less important) and calculate how many hours it would take before we run out of fuel. Here is the information I have available:

Fuel amount available
Fuel burn rates for the diesel generators at specific load %'s

Example: a 1500KW generator burns 29.2 gallons per hour at 25% (1500 * .01945), 47.6 gallons at 50% (1500 * .03175), 68.1 Gallons at 75% (1500 * .0454), and 87.3 Gallons at 100% (1500 * .0582)

What I need to do is be able to figure out at any given percentage load how much it will burn instead of just at those specific loads. From that I would be able to take fuel amounts and divide it by the fuel burn at the given load. I am not sure how to figure out if I have a 18% load what the multiplier would be for each percentage point. (assume a resolution of no more than 1 decimal point)

2. ## Re: I am not sure if this is calculus or not

You can do it crudely by working out the gradient between each %load and then interpolating between them by having set ranges and domains between 0%, 25%, 50%, 75%, and 100%. Another way you could do it is to plot the graph of these points, and then draw a curve/line of best fit, which you could use to estimate graphically any load/burn rate combination.
If it works out to be a fairly linear graph, you could use linear regression to calculate a line graph equation, which can be used as a "plug and play" formula for your results on burn/load.

If you need any more information on any of these suggestions, just shout.

3. ## Re: I am not sure if this is calculus or not

Using 25% and 29.2 Gallons, etc. for the four points I get a linear regression of 9.35+.775x

So now I have to talk this out. If I use the first equation I need to take the total load in KW and turn it into a percentage, then I would need to plug that percentage into X, this would tell me how many gallons I am using per hour (note I have more than 1 generator so this could be up to 400%). EG 4500KW total load = 4500/1500 (this would take a full 3 1500KW generators to run) = 3.0 or 300%, that means 9.35 + .775*(3) = 241.85 Gallons/Hr. Assuming I have 19624 Gallons of Fuel that would be 81.14 hours of run time.

If that run time is correct the calculated run time I had prior to this (which I am sure is slightly off) was 92.8 hours @ 75% load. This is a change of about 14% less than previous. Does my math make sense right now?

4. ## Re: I am not sure if this is calculus or not

Your linear regression doesn't make sense because I would expect at 0% load or 0KW power output to have a fuel burn of near 0 gallons / hour. I expect in you regression formula you have not accounted for a zero burn at 0% load or something as it comes out to 9.35 gallons per hour? Anyway, I done a quick regression, and it comes out to 3.74+0.854x which gives a slightly more realistic value of 3 gallons at 0% load.
The rest of your calculations do make sense but this whole mathematical model takes a big assumption that this is a linear system of power output vs fuel burn which is often not the case in real life. I have plotted this as a graph though and it seems fairly linear, I just don't see how it will be like this due to working friction and efficiency ratios etc etc but i guess that's the nature of mathematical modelling of these things on a basic level.
For your own information, it would be more difficult to model power generators if they have any sort of exponent type relationship for what you are doing as you'd need a curve of best fit, equations to represent that, and also a model of each generator separately (without saying its 4500kw = 300% load) as each generator would have a fuel burn that doesn't increase linearly.
In summary, yes your maths works in this basic model, with a tweak to the regression formula!

hope this helps.