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I need help with Potential Energy compared to energy used.

  1. Jul 29, 2015 #1
    your going up hill in a big truck ,18 wheeler at a (*10 % Grade) or (* 5.7106 DEGREES)
    your fully loaded at (*80,000 lbs) or (*36287kg) or (*40 short tons)
    your traveling at (*30 mph) or (*13.41120m/s)
    you travel for about (*10 miles) or (*1609.344 Meters) the actual distance traveled would be 9.9499 miles

    Based on real world data you would use about 2.5 gallons of diesel fuel.
    That is (4mpg: miles per gallon) companies whould have you belive this number was higher
    For this exercise lets say 3 gal of fuel.
    thats about (3.33mpg)

    now here is the senerio
    1) you get to the top of this hill and stop on a bridge that is old and about to fall. The bottom just happens to be the exact height that you started. 1 mile down.
    your Gravitational Potential Energy in Jouels
    PE=Mass (kg)*G(9.8m/s)*H(m)
    PE would be 572,303,004.13 or about 572 mega jouels or fairly close
    now a gallon of fuel has a potential of anywhere from 45.3 to 141 mega jouels of energy based on where you look my pysics text book gave 45.3 while a web search turned up many numbers in that range, all different. so we will use the largest of 141
    141 * 3 (actual gallons used) = 423 mega jouels of energy to get a potential of 572 mega jouels

    Where did I go wrong?
     
  2. jcsd
  3. Jul 29, 2015 #2
    I seamed to have missed adding the problem. I am trying to show an example of entropy using a real world example. This example should have been an easy one. I figured that showing how much energy was used to get to the top of this hill compared to how much energy could be extracted on the way down would work. Now either my logic is off, my math is off, or just driving to the top of this hill will provide more energy then it took to get there. That is also assuming a 100 percent efficiency rate of the engine. I haven't gotten that far in my math, but I need to figure this out first. I should also note that the amount of fuel used was experimentally tested and the grad is the average.
     
  4. Jul 29, 2015 #3

    jbriggs444

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    3-4 miles per gallon is correct for an 18 wheeler travelling on level ground. Climbing a 10 percent grade would result in considerably lower fuel economy. I seriously doubt that this was experimentally verified.
     
  5. Jul 29, 2015 #4

    CWatters

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  6. Jul 30, 2015 #5
    You are absolutely correct if we were living 20 years ago. Newer models can general get 8 to 9 mpg on flat ground, especially if they are equipped with trailer tails. That is according to the trip computer attached to the main brain of the truck. As far as the experiment, as stated above, the magical hill, cliff, and bridge are completely hypothetical. To keep the post from becoming a full paper I gave the results based on observations made of the equipment installed in the truck. the odometer gives the distance; the gps shows altitude from sea level; and the trip computer shows the amount of fuel used. It was a simple matter of subtracting the starting number, at the bottom of a hill, from the ending number, at the top of the hill. Do this enough times (and if you ever spent 11 hours a day for months, behind the wheel of a truck you know there is plenty of time.) you get enough step hills to add together to make a mile in altitude. Then you simple do some basic sums.

    I was and still am not trying to prove any type of great revelation. It was simple observations that were documented to come up with the numbers above. It was not supposed to be a lab controlled experiment with radars, lasers, surveying equipment, or metered fuel. I was generous on the side of energy spent in all of my numbers and rounding. I posted asking for help because the answer didn't seam to be correct to me. I figured that the most likely possibility was there was an error in my logic. Am I trying to compare apples to oranges type of thing. Maybe the formulas or math are just wrong. that is what I am asking help with. If there is no error then it could be that the energy density of low sulfur diesel fuel is actually higher than what I am looking at and I need to find a much more reputable source for that information. And yes the equipment used could be off or just simply misleading.

    And yes this article is completely true. New trucks are being advertised with as high as 12 mpg in the U.S. So based on the above article the 20 to 25% difference is reasonable.
     
    Last edited by a moderator: Jul 30, 2015
  7. Jul 30, 2015 #6

    NascentOxygen

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    The mathematics will have to accord with actual measurements. Where your estimates fall short is in the fuel used. Few drivers would be able to find a local highway with a 1:10 grade over 10 miles for testing. That gradient means you would travel around 10 miles while rising 5,000 ft; that is going to be a very fuel-gulping ear-popping crawl for a loaded transporter.

    One reasonable way to look at this is to ask how much additional fuel would be needed to travel on a 1:10 gradient compared with travelling under the same conditions (same speed, same gear) on the level. Suppose the vehicle consumes X gallons to travel 10 miles on a level road, then to lift its laden weight vertically by 1 mile would need the extra 572MJ you calculated. If the power source and drive train were 100% efficient then you'd need an extra Y gallons, I make it around 4 extra gallons, meaning fuel to negotiate that steep incline will total (X+Y) gallons, at least.
     
  8. Jul 30, 2015 #7

    CWatters

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    I rechecked your calculations and they appear correct based on 2.5 gallons consumed. I got 573MJ of PE gained for 339MJ burned. So something is clearly wrong with the original estimate of 2.5 Gallons burned.

    I doubt this is the cause but.. If you measured fuel consumed going up a large number of very small hills did you factor in any loss in KE? For example if you start up a hill at 50mph (22m/s) and slow to 10mph then you loose about 8.7MJ energy equivalent to a height error of about 24m per hill.
     
  9. Aug 1, 2015 #8
    Thank you C Watters. I did not account for the KE loss. And as it turns out I wouldn't be able to make it up the hypothetical hill with those losses. Makes a whole lot more sense now.
     
  10. Aug 1, 2015 #9
    At a true steady 30 mph up a 10% grade, at 40% efficiency, this 80,000 lb rig would get closer to 1 mpg (US) using about 640 hp to lift the weight.
     
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