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Distance for an IC engine for a given load

  1. Sep 21, 2009 #1

    I have this question that, how do we calculate the distance an IC engine can propel a vehicle for given certain conditions? And what would be those required certain conditions?

    As a test case, i was looking at this engine:
    http://www.yanmar.com.au/industrial/la_series/l48ae.htm [Broken]

    now, from the performance curve given, i took the RPM approx 3200, for which the bsfc is approx 0.26 Kg/KWh and Power is 3kW.
    Fuel is gasoline (density=0.74616 Kg/L , CV=42900 KJ/kg)

    so, the brake thermal efficiency = 1 / (bsfc x CV ) [units adjusted] = 0.32 = 32 %
    now, as the engine is giving 3 KW of (assuming continuous) power, the input energy from fuel is = 3 / 0.32 = 9.1KW = 9.1 KJ/s

    as CV = 42900 KJ/kg, the fuel that comes in, or the mass flow rate of fuel is = 9.1/42900 = 0.0002 kg/s.

    Now, firstly, are the calculations uptil now correct? Are the assumptions (like constant power for a vehicle) valid?

    If yes, how do i proceed from here?

    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 21, 2009 #2
    You cant really proceed to work out how far a car will travel until you know the wheel diameter, and drive ratios.

    Are you talking about how far it could go on a set amount of fuel, assuming steady conditions?
  4. Sep 21, 2009 #3
    right...what else is needed?

    wheel dia is 19"
    gear ratios, i know the one for the 4th is 1:1, and is compounded in the driveline by a factor of 3.4

    main question is, how do these factors link up with what i need? Yes, i am talking about how far it ll go on a set amount of fuel (1 L to be precise).
  5. Sep 21, 2009 #4
    You work out the distance and speed travelled from:

    engine rpm*drive ratio*wheel circumfrence.

    From speed and distance you get a time and power. you can then find how long the fuel will last by bsfc (EDIT: which you appear to have done already).

    You can then work it back to find total distance.

    You could also use your calcualtion of fuel used per seconf to find out how many seconds the engine will run. And work out speed and distance in the same way.

    EDIT: This is only valied for steady load and conditions.
  6. Sep 21, 2009 #5

    Ranger Mike

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    don't forget road surface, paved ? mud/ dirt?
    is ti flat? any hills in this scenario.
    also wind, temperature, humidity
    aero dynamics too...
    one more thing is diameter of drive wheels with driver and fuel.
    what does this pig weigh?
  7. Sep 21, 2009 #6
    umm...you sure this is applicable here? I mean it is useful, as i have tried, to set a limit to the max speed the vehicle will travel for a set rpm and wheel dia, by altering the drive ratio? No?
    However, i was not able to get anything with dimensions of distance (apart from the wheel radius) in my calculations...am i doing it all wrong here?

    Here is what i tried after the relations in the original post:
    mass flow rate of fuel = 0.0002 kg/s
    (Volume) flow rate = mass flow rate / density = 0.0002 / .74616 = 0.000289 L/s
    Total number of liters = 1
    so time all the fuel lasts = 1/0.000289 = 3462.3s = 57.7 mins

    is this right?

    hi ranger mike, and thanks!

    i was wondering if these things come into play. For these calculations (which are pretty crude), i am assuming a lot of things just to get an idea of what mileage i can get, so these are the assumed values:
    Coefficient of Drag = 0.2
    Coefficient of Rolling Resistance = 0.03
    Mass of the vehicle = 200 kg
    And neglecting the grade, and other natural factors like wind, humidity, temperature.

    with these values, the power requirements at around 35-40 km/h (which is the desired max speed) come out to be somewhere between 0.8-1 kW. (the engine i have mentioned has significantly higher max power, but i am just making approx calcs)
    Last edited: Sep 21, 2009
  8. Sep 21, 2009 #7
    Before we go any futher, what is the application?
  9. Sep 21, 2009 #8

    Ranger Mike

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    good start! and yes..anytime you are moving..you are moving air..imagine a curtain that weight 14 pounds per square inch, stick your hand out the car window when moving 25 MPH. what kind of ground clearance can you get away with....lower the better to keep air out from under the vehicle.

    being old ..ancient school..what the H??? is KM/H...
    how much horsepower is that???

    your CDs seems a little low..check out the attached..
    a 440 pound car to me should not have that much rolling resistance..are you running disc brakes on all four corners?
    with that low a weight you can loose the disc brake pad drag and go with drum brake and shoes ifin you want to really go for the MPG numbers..ifin you really want to trick out the set up, go super light on the rotating mass, wheels, tires, drive train..what kind of transmission...lot of parasitic drag on components..flywheel ..what diameter? drive shaft or half shafts?
    can you go carbon fiber? gun drill them?

    also regarding the IC,,can you u duct cold air to the intake?
    can you up the water temperature to 240 degrees F..more heat makes more HP..can you run synthetic oil and differential grease in it.. all these will increase HP or fuel mileage.

    Attached Files:

    • cd.jpg
      File size:
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  10. Sep 21, 2009 #9
    Hang on 200kg with a driver 40odd kph, lawnmower engine... you're building a kart arent you? (seems a tad heavy for a road kart so maybe one of those offroad jobbies?)

    Also, to find the distane you will travel you have to use.


    Rpm * ratio gives the amount of time the wheel spins for each eagine revolution.

    So for example a drive ratio of 1:1 and engine speed of 3600 rpm, the wheel will rotate 3600times per min.
    The circumfrence of the tyre then lets you convert his angualr speed to a linear speed.

    So say we have a circumfrence of 1m.

    With 1m circumfrence, and a 1:1 drive ratio and 3600rpm.

    We travel 1 meter for every revolution. Or 3.6 km per min, which is a whopping 216 kmh.
    This means you will travel 205 km on 1 l of fuel in an hour.

    Now obviosly the figures are rubbish because you arent going to be using a 1:1 drive ratio or have a circumfrence of 1m. It just demonstrates the resoning behind the calc.

    Now if you've limited it to 40kph you'll go 0.95*40 miles on the fuel.

    Before you go complicating it with losses aich as drag and rolling resistance. Work it out as an ideal case, then make it more complicated. Then you can work out if you've got the power to do that speed.
    Last edited: Sep 21, 2009
  11. Sep 21, 2009 #10
    ok...the application is an energy-efficient mini-car for an event called eco marathon organized by Shell. In fact, we have talked about this here, and have been consulting with a book or two (as was told to do) since then.

    Ranger Mike, I sometimes find it difficult to understand certain portions of your messages, but always find them the most encouraging and helpful. Anyhow,
    - Minimum ground clearance the rules allow is 10 cm
    - Cd, 0.2, is a bit on the lower (and optimistic) side, but i did a small research on teams from previous years, and they did as low as 0.18
    - Rolling resistance; wikipedia says 0.03 for 'ordinary tires on asphalt'. I guess this can be lowered (but i havent reached the tires chapter yet, so can't say for sure :-) ) There are supposedly some Michelin tires specifically for the event, but i haven't managed to find a tangible link. The e-shop from the organizers isn't up yet, so there's time still for that i guess
    - Disc brakes on all wheels necessary (the rules)
    - super-light rotating parts. Actually, we intend to use a 'branded' engine. The power required (assumed at the moment) to overcome drag and rolling resistance comes out to be around 0.8-1 kW. The track gradient is max 5%, which puts around 0.15 kW more required on certain sections. If we manage to arrange a carbon-fibre supplier, this could drop marginally (due to weight). So my teacher recommended to look up (not necessarily finalise) en engine twice that much power. So, there are a few i could locate:
    very high rpm, if my basics are correct, this would require a compound transmission ration of around 20 for speed limited to 36kph!
    would require a compound transmission ratio of around 10 for the same speed.
    Now, i am also unable to get the complete performance curves (including the bsfc) for each.

    Obviously, these are not finalised, just on the table. We d be also looking at engines from the junk as well. If anyone has an idea of better ones, that ll be welcome. We d definitely have to make some changes to them, and i think ill be getting in touch with you soon:smile:if thats ok with you

    the relations are helpful...thanks!
    why did you multiply the final thing by 0.95??? couldn't get that part
  12. Sep 21, 2009 #11
    Oh yeah, I remember now.

    Sorry that was just a quick in the head calc, it came from the fact you said you had 57.7 mins of fuel (from your calcualtion) at that engine speed. So I just took the 57 mins out of the hour.

    so 57/60 is .95

    However, for the ultra ecomomy, the way we;ve calcualted the fuel economy is going to be way way wrong for your needs. As we are calcualting the engine running at a constant 3600 rpm, which is putting out far too much power for your needs and is wasting tons of power and therefore vital fuel. 3 kw as opposed to the 1 you said you needed. If you are stuck with that motor, you may want to gear it so that you operate lower in the rev range (at the power output you need). Sacrificing specific fuel economy for less combustion events, giving an overall gain.

    Better yet pick a smaller motor.

    As I remember you are in a road going class, so its stop start. Is there a miminum accelration up to top speed required?
  13. Sep 21, 2009 #12

    Ranger Mike

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    you find it difficult to understand me..??? so did my ex wife///ifin you are intending on winnin this
    let me know
    if some futile homework bs
    ok too
    let me know cause this whole post can kick the dog do do out of the competition ifin you let um
    guys like fred and danger an xxchris etc...
  14. Sep 22, 2009 #13
    hey chris!
    No, we aren't stuck with that motor. in fact, its what one of the teams used 2 years back. They managed 91 km to the Litre. We are looking in the market (the net for makers like honda etc) and the local junks for smaller ones. As i posted above, the 50cc Honda one giving 1.6 kW at 7000 rpm comes the closest in the ones we have seen upto now.

    Secondly, there is no max speed specified. The event does say, that the vehicle must complete 6 legs of a 2.7 km track in a certain time (including about 3 pit stops of 10 secs each) the average speed for which should be 30kph at least. There is no mention of a minimum accelaration up to top speed, but obviously, all of it has to fit in the given time.

    Ranger Mike,
    winning this? thats going to be a long shot. here are the results from this year:
    as you can see, the highest is a whopping 589 km/L, but many of these teams might not participate this year in asia. but that doesn't mean the bar will be much lower coz there are going to be good teams from here too. We ll try our best
    Last edited by a moderator: Apr 24, 2017
  15. Sep 22, 2009 #14

    Ranger Mike

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    it looks challenging..are you the main wrench bender or are you the design wizard on this team?
    how many crew do you have?
    give us some background
  16. Sep 22, 2009 #15
    hmm...thats an interesting one:confused:! At the moment, i am the design (not-so-wizard) person in the team.
    We have a person from electrical engineering, and he is working on the electrical systems (the lighting, horns)
    Then there is a person from mechatronics and four,five from mechanical [btw, all our going to start our third year]. At the moment, we have one person researching the previous years of this competition, trying to get as much data as available for the better teams. One-two guys will start on the chassis this week, with our teacher.
  17. Sep 22, 2009 #16
    So what basically, what this means is, and thats the rresult i have been getting in my calcs btw, that whatever mileage i will get will be:
    mileage = vehicle speed x time it takes for the engine to burn 1 L

    1. if we make the max speed go to, say hypothetically, 200 kph, for the engine we have discussed, i get 200 x 0.95 = 190 km/L?!!!!
    2. i actually think there would be more to this. What i mean here is that the organizers have put a limit to the min. speed (30 kph (average)) so i infer from this is that mileage is inversely related to speed: i.e more speed=>less mileage and vice-versa. Umm???

    Second, you have said to 'work it out as an ideal case' before making it more complicated. What would be the ideal case here? (i thought that by eliminating all those factors like drag etc, this is the ideal case; not sure) ?

  18. Sep 22, 2009 #17
    Well... its more complicated than that. The equation we used is very very basic, and highly simplified.

    Our calculation shows that the faster you go the better milage you get. This is only valied for a linear region, where the power requirements are low.

    So for example, take the engine you showed in the link. If you geared it to go 10 miles in 1 hour we would be getting 10mpl. If you geared it to 20 miles in 1 hour, you would be getting double the fuel milage.
    This is valid becuase the engine is producing enough power for both speeds to be viable.

    Going at 160odd mph requires much much more power than the egnine can provide, so the equation we used is not valied.

    What you need to do is to find a speed you want to go, ideally the minimum speed you can get away with, to reduce power requirements. Add in losses. This gives you the minimum power requirement to perform the task.

    You then find the engine with the lowest specific consumtion at this speed.

    I'm busy atm, i'll come back and answer this more thoroghly at some point.
  19. Nov 8, 2009 #18
    hey Chris...you there? sorry for invigorating the thread after this long but i was hoping if you could remember :-)
  20. Nov 8, 2009 #19
    That's ok, i'm not 100% sure what is the best way to do this really. I've not got too much experience in doing this so this is basically my best educated guesss. Hopwfullly someone with more experience can help you more.

    1st task is to find a power requirement for an engine. For this you will need.
    An estimate of the losses (forces acting to slow the car) of the car moving forward at top speed. This gives a force.

    Then you need to specify a minimum top speed required, and an overall average race speed. This is done by finding the race distance and / maximum time alllowed. This will give you a velocity.

    These two are then used to find a 2 power ratings (cruise and max speed) by: P=W/t
    This is the power required at the wheel to move the car the race distance in the certain time.
    EDIT: You may very well want to specify a max acceleration as well, to find a force from F=ma to make sure the engine is capable of acceleratin the car in a given gear.

    As this power is at the wheels you need to factor in losses to get power needed at the motor output by estimating efficiencies.

    motor power * efficiencies = wheel power.

    This gives you the power motor you need, match this power requirement to a suitable engine. ie one with max power in roughly the correct rpm range and one with the lowest bsfc.

    Then you can calulate theoretical mileage from that.

    I'd suspect that is the best way for you to go about matching engine to purpose and determining potential fuel milage.
    Last edited: Nov 8, 2009
  21. Nov 9, 2009 #20
    thanks again chris...

    Ok..forces opposing motion
    with 36 kph (10m/s) as the max speed, and about 27 kph (7.5 m/s) as the average speed:
    1. Drag (with Cd~0.25, A=1, air density=1.2)
    = 15 N at max speed, 8.4 N at average speed

    2. Rolling resistance (Cf~0.03, W=200x9.81 N)
    = 58.9 N

    3. Gradient (max grad~5%(0.05))
    = 98 N
    (although this is a big fraction of the total opposing forces, this only acts in a portion of the track. i don't think it needs to be kept into account if have enough kinetic energy at the bottom to negate the potential energy change and losses, even if the engine is turned off) so i am not going to use this in the power calculations

    Power required to overcome
    1. Drag = Drag*velocity
    = 150 W for 36kph, 63.3 W for 27kph

    2. Rolling resistance = RR*v
    = 588.6 W for 36 kph, 441.5W for 27kph

    Total max power = 150 + 588.6 = 738.6 W
    Total average power = 63.3 + 441.5 = 504.8 W

    Now, i have in mind to use gears in two stages, firstly as the 'transmission', and then the final drive reduction by a chain-sprocket assembly. Assuming efficiencies to be 73% (85% x 85%)

    power of the engine = 738.6 / 0.73 = 1.012 kW
    so i need an engine that gives me this much power right?

    then i need to get engine charts for that engine to get the projected mileage?
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