Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Automotive Fuel efficiency

  1. Mar 28, 2012 #1
    The qwustion is:
    2 identical cars with their engine rev 2000rpm. The only difference is that car num #1 is in 4th gear and car num #2 is in 3rd.
    Which car will use more fuel per minute?
    Considering they are driving on leveled ground.
    And does it Make a difference if they are going uphill or downhill instead?

    Was arguing with a friend and I would be glad if you could help out

    Thanks alot
     
  2. jcsd
  3. Mar 28, 2012 #2
    We need to know the speeds of the two cars before we can answer that properly. For example, are they both going the same speed or is one going 30 mph in 3rd and the other at 40 mph on 4th?
     
  4. Mar 28, 2012 #3

    Averagesupernova

    User Avatar
    Gold Member

    I don't agree that we need to know the speed since the question is how much fuel per minute and not fuel per distance. There should be more vacuum on the vehicle in the lower gear since it is more lightly loaded. Engine vacuum is just about ALWAYS a relative indicator of fuel economy. So maybe I have given you enough clues to figure it out yourself?
     
  5. Mar 28, 2012 #4

    jack action

    User Avatar
    Science Advisor
    Gold Member

    The car in 4th gear will most likely take more fuel per minute than the one in 3rd gear.

    First, if the cars are identical, the one in 4th gear will necessarily go faster at the same rpm. More speed means more power required. If we assume that the 3rd gear ratio is 1.26:1 and the 4th gear ratio is 1:1, the speed will be 1.26 times faster in the 4th gear and it will require 2 times the power to fight the aerodynamic drag (= 1.26³).

    The fuel consumption rate (r) per power produced (P) by an engine is the BSFC. So:

    r = P * BSFC

    In a typical engine - at 2000 rpm - the BSFC will probably be lower in 4th gear, around 15% less than in 3rd gear. So the fuel consumption rate should be around 1,70 times greater in 4th gear than in 3rd gear (= 2 times the power * 0.85 times the BSFC).

    This is assuming no acceleration and neglecting rolling resistance, which should lower the difference between the two because it is not related to speed and mostly constant. If the speeds are small, it might leads to a lower fuel consumption rate in 4th gear because the power needed will be nearly the same in both gears (0.85 ~ 1 times the power * 0.85 times the BSFC).

    Going uphill or downhill will simulate a constant acceleration or deceleration. Going uphill will further lower the difference between the two - just like rolling resistance - because the power needed is also independent of speed. Going downhill should have the reverse effect.

    I had this nice discussion in another thread that might interest you.
     
  6. Mar 29, 2012 #5
    So the assumption is that the speeds are different; reasonable assumption but I wasn't going to make that. Better to have a clear description of what the proposed scenario is before answering, but maybe that's just me.:smile:
     
  7. Mar 29, 2012 #6

    Averagesupernova

    User Avatar
    Gold Member

    2 identical cars each in different gears and both engines spinning 2000 RPM, how could you assume anything BUT they are traveling at different speeds?
     
  8. Mar 29, 2012 #7
    Easily, given the way people ask questions. Seen it quite a lot, have stopped assuming that what they type is what they mean.

    "Oh, I meant identical other than the final drive gear ratio! My buddy and me were wondering whether having a numerically higher differential gear with the transmission in direct drive would be more efficient than a lower diff gear and the transmission in 3rd (same car speed for 2000 rpm). In other words, how much difference is there between the efficiency of bevel gears and hypoid gears.":smile:
     
    Last edited: Mar 29, 2012
  9. Apr 16, 2012 #8
    The answer has been given here already, but as a little addition it is worth noting that VERY low loads can cause poorer specific fuel consumption due to immense pumping losses experienced due to an almost completely closed butterfly valve on the throttle. Though in this example I'm sure this doesn't need to be taken into account as traveling along at 2000RPM in either 3rd or 4th gear should provide ample butterfly valve opening for this phenomena to be irreverent. I have attached a little JPEG of an experiment I performed recently demonstrating this effect.
     

    Attached Files:

    • SFC.jpg
      SFC.jpg
      File size:
      18.8 KB
      Views:
      113
  10. May 16, 2012 #9
    Actually, the pumping loss is most extreme at this engine speed in a lightly loaded cruise mode- it's ALL about throttle angle- so if both were at WOT up hill the loss would be insignificant. If you were to take a vacuum gauge and run it at idle then at 2,000, I think you will see what I mean.

    At 2k only the load will be different for this test- they will both use very nearly the same amount of fuel, with the 4th gear engine using slightly more.
     
  11. May 17, 2012 #10

    Ranger Mike

    User Avatar
    Science Advisor

    You are almost there...but...I suspect hanging a vacuum gage on an engine in the dyno room is not telling the whole story.


    Leaving aside air resistance for the moment, the answer is the lowest RPM in the highest gear.

    As for the specific question you posed, air resistance also comes into play Big time. For any vehicle, the higher the speed, the greater the air resistance. But some vehicles have far less drag than others, and so the effect of air resistance for those vehicles is less than it is for others.
    Waddel Wilson ( famous Nascar Crew chief) one said “ ifin you're moving , you're moving air” or something like that
    Basic mathematical formula for the amount of power it takes to move your car at certain speeds.

    Driving load = av + bv2 + cv3

    The "av + bv2" part has to do with how much power it takes to overcome the resistance due parasitic drag, internal friction of the engine, like ring drag, bearing rotating friction etc.. plus tire

    resistance, disc brake drag, drive shaft friction, transaxle gear friction etc..all to get the car rolling at a given speed.

    The "cv3" part has to do mostly with power required to overcome wind resistance (or "drag") while driving.

    When doing this formula, your replace a, b, and c with the speed you're driving at. Let's assume this speed is in miles per hour (mph).

    The biggie is the "power of 3" aspect of this formula..It means that wind drag will be the largest force your car will have to fight. And believe me..it is HUGE.

    Ifin we are going 60 mph:

    Driving load = 60 +3600 + 216,000 = 219660 units

    These "units" could be anything - horsepower, Newtons, Joules ,cans of beer??

    Based on this formula, the amount of power required to drive at 60 mph is 219660 units. Based on the same formula, the amount of power required to drive at 65 mph is 278915 units. The difference is 59255! This is an increase of about 27% more effort your car has to make just to drive 5 mph faster

    not being a degreed professional engineer...i can not explain it more but recommend you check out this link which is very in detail reagrding the formula...
    http://autopedia.com/stuttgart-west/Physics/StuttPhysics06.html
     
    Last edited: May 17, 2012
  12. May 17, 2012 #11
    /

    Did you check the picture I posted? Clearly, pumping losses are NOT at the most extreme at mid-cruising load, throttle angle is further reduced at zero load under tickover situations, therefore pumping losses are the most extreme here.
     
  13. May 17, 2012 #12
    I don't care what your graph shows, really, anyone can make a graph to represent anything. Use a vacuum gauge- or even a scanner, on a properly running engine and get back to me.

    There is a reason EGR valves were introduced on vehicle engines besides the blending of a neutral gas. The introduction of the EGR necessitates the increase of the throttle angle to maintain the same power level- reducing the pumping loss.
     
  14. May 18, 2012 #13

    Ranger Mike

    User Avatar
    Science Advisor

    mylar..not cool...no need to get personal...no one is going to dummie up a graph to thwart you..



    .the Exhaust Gas Reg valve is on the engine for one reason..so cars can pass the smog test ..i.e. stupid regulations for car emissions...makes engines run super lean and super hot...
     
  15. May 18, 2012 #14
    The tests were conducted within an enclosed engine test cell, which includes vast amounts of digital measurement equipment (yes, including digital measurement of intake manifold pressure). I'm not here to get into conflict, only share knowledge with like minded people, if you have data to prove your stance on this issue then please, I'd love to hear it. But currently all you have is your opinion - and that doesn't really account for much.
     
  16. May 18, 2012 #15
    Chase, my deepest apologies if you thought I was assaulting you in any way, shape, or form. I am too new here and didn't know your credentials, my apologies. But after 35+ years of working on cars, I still have to disagree with your tests in a real world platform. Oh my, I just looked again at your picture and noted that it's based on DIESEL performance- NO WONDER I disagree! Diesels don't have a throttle plate to induce pumping loss.

    Ranger, that is a very popular misconception about Exhaust Gas Recirculation- it was ORIGINALLY used in that way, but was found years later to have the added benefit of reduction in pumping losses. As EGR flow is introduced into the cylinder, a wider angle of throttle opening is required to maintain the desired power level- thus the reduction in pumping losses. Believe me, it took a while for me to wrap my head around it as well. There are many things we, as technicians, were taught incorrectly and have been forced to "unlearn"!
     
  17. May 18, 2012 #16

    Ranger Mike

    User Avatar
    Science Advisor

    thank you mylar..i will look into this..maybe ya ca n teach old dog new tricks??
     
  18. May 18, 2012 #17
    Traditionally no they don't, modern designs are trending to do so however in order to achieve suitable intake manifold vacuum for EGR operation and to avoid excessive lean out and therefore NO and NOx production at higher loads.

    Now please, let this lie here and allow the thread to die like it should have done several posts ago.
     
  19. May 20, 2012 #18
    Ok, consider it done, I'll not discuss anything diesel with someone who thinks they can run "lean". Have a nice weekend.

    BTW, EGR TPs are gone already.
     
  20. May 20, 2012 #19
    *Sigh*

    Yet another test on an older diesel engine with direct injection and no throttle valve. Note how NO and NOx production increases as air to fuel ratio becomes more LEAN.

    Really, all I have heard so far is how your experience as a technician for 35 years is a far more valid proof than any actual data collected in the real world with scientific tests.
     

    Attached Files:

  21. May 20, 2012 #20

    brewnog

    User Avatar
    Science Advisor
    Gold Member

    Hmm, your graph shows NO2 to NO ratio; not NOx production. Just saying...
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Fuel efficiency
  1. Fuel System (Replies: 16)

  2. Fuel Cell Efficiency? (Replies: 3)

Loading...