Question on RPM

[kevinp]

Para sa mga Automotive Engineers: Given two equal engines, will an engine running at 1000rpm for 6minutes consume the same amount of gas as an engine running at 6000rpm for 1 minute? Naisip ko baka mas malakas ang consumption per revolution ng engine kung mas mataas ang rev.

E kung meron 1 car driving on a dyno in 1st gear at 6000rpm for 1 minute and another car in 5th gear revving at 3000rpm for 2 minutes, which one will consume more fuel all other things being equal (e.g. weight, etc.)?

[ghosthunter]

basically it take more gas to make more power.

since at higher RPMs, you engine will also produce more power.

BTW, since engine efficiency drops as you increase engine RPMs, the engine will consume more than six times more fuel at 6000rpm than at 1000rpm.

[kevinp]

ghosthunter, could you explain your point further? logic tells me that the volume of fuel that is injected into the combustion chamber per revolution of the engine will be the same if its running at 1000rpm or 6000rpm. afaik, the increase in power at higher revs is caused by the corresponding increase in revolutions of the crankshaft and its momentum.

[vermicelli]

"the volume of fuel that is injected into the combustion chamber per revolution of the engine will be the same if its running at 1000rpm or 6000rpm" is correct IF it has the same load and the engine's torque curve is flat for the given range of rpm.

Also, the losses introduced by the engine itself becomes more apparent as the rpm increases.

The amount of fuel your talking about is just enough to sustain that particular rpm (at 1000rpm and at 6000rpm) but once you put load it will drop (notice that when you try to release the clutch of your car while idling in first gear the rpm drops and you just compensate by pressing the accelerator more, which puts up more gas just to maintain the same rpm).

The load i'm talking about here is the engine losses which are different when revving at 1000 and at 6000 rpm.

[ghosthunter]

ghosthunter, could you explain your point further? logic tells me that the volume of fuel that is injected into the combustion chamber per revolution of the engine will be the same if its running at 1000rpm or 6000rpm. afaik, the increase in power at higher revs is caused by the corresponding increase in revolutions of the crankshaft and its momentum.

in theory, 1000rpm & 6000rpm should only differ by comsuming six times more fuel for the same amount of time.

in reality, there are inefficiencies to consider. Like the volumetric efficiency of the fuel system and air-intake. Example, if the air-intake is sufficient at 1000 rpm, it might not be sufficient at 6000rpm and would cause the engine to run over-rich. The opposite would happen if the fuel system is not sufficent at 6000 rpm, the engine would run very lean and might lead to detonation.

Also as mentioned, fuel consumption under no-load and loaded conditions are different. Thats why there is such a thing as "brake-horsepower" rating. A free-reving engine (with no load) reving at its redline is not producing its maximum power, its just reving very fast.

[yebo]

kunin natin sa basic physics.

there will be higher loses at higher rpm for a reciprocating piston engine. remember that momentum is 1/2 x M x V x V. the higher the engine rpm the higher will be the momentum of the piston because the crankshaft rotates faster, and this momentum of the piston is loss when the piston reverses direction. a piston travelling at say 6 meters per second and then reverses direction will have 36 times more momentum loss than a piston travelling at 1 meter per second that reverses direction. this is due to the momentum equation, the value of velocity being raised to the 2nd power (squared). 1 x 1 = 1, 6 x 6 = 36. even though the total number of revolutions is the same (6000 x 1 minute = 1000 x 6 minutes) the mechanical losses are higher on an engine at 6000 rpm due to this higher energy losses from the piston momentum. thus the consumption of fuel will be higher.

the thermodynamic losses (heat, volumetric efficiency, flue gas) will also be more on the engine running at 6000 rpm (assuming they are both at the same temperature) due to the higher energy input.

this applies for an engine irregardless of load, whether zero load or any value.

[ghosthunter]

yebo i think thats more than kevinp would need. :wink:

btw, I think thermodynamics isn't basic physics, well at least its not taught in high school (yet).

[yebo]

the momentum equation is basic high school physics.

thermodynamics is basic, err, college physics, hehehehe! basic pa rin, college nga lang! :mrgreen: i just mentioned thermodynamics kasi baka may kumontra na mas malaki heat losses on an engine running for a longer time.

[kevinp]

the mechanical losses are higher on an engine at 6000 rpm due to this higher energy losses from the piston momentum. thus the consumption of fuel will be higher.

so your point is the volume of fuel injected into the combustion chamber per cycle at 6000rpm is higher than at 1000rpm for the engine to be able to maintain its speed?

actually may naisip pa ako. paki-check na lang kung tama or mali yung pagka-intindi ko:

So here's the question: If your foot is steady on the gas, say at 2000 rpm, then you floor the gas pedal and the rev reaches 3000 rpm, what causes the acceleration in rpm?

The answer that I came to after thinking (not researching) about it: During the time that the engine was steady at 2000 rpm, the engine needed a certain amount of fuel per cycle of the engine to maintain that rpm. let's say 1 mL of fuel. To increase revs to 3000 rpm requires more power and more power requires more fuel. so the moment you depress the gas pedal beyond the steady 2000 rpm level, the stroke immediately after you depress the pedal (assuming the engine is still at 2000 rpm) will require more fuel (say 1.001 mL) so that the next cycle will produce more power to bring up the revs to 2001rpm. Then the 2001 rpm stroke that will require again more power and more fuel (due to the losses at higher rpm accdg to yebo) to bring it up to 2002 rpm and so on. Tama ba yan? I also assume that once the engine reaches 3000rpm and holds steady, it will theoretically consume less fuel per cycle than during the period of acceleration from 2000-3000rpm but more fuel than when the engine was running steady at 2000 rpm.

Thanks for the replies mga pards. :D Di kasi ako engineer and curious lang ako to learn as much about cars as i can.

[ghosthunter]

Accelerating a car takes more power than to keep a steady speed. Its not just a matter of increasing the revs of the engine. Hence your basic understanding is correct.

Increased velocity would also increase certain forces that will try to slow down your car. Example is air drag, tire friction, internal engine friction, internal engine windage losses. And some of these forces increases exponentially (as against linearly) as you increase speed like air drag. That's why to maintain a faster speed requires more fuel than at lower speeds.

In ideal circumstances, like in a vaccuum and frictionfree environment, the force required to maintain a speed of 30kph & 100kph should be the same.

[yebo]

In ideal circumstances, like in a vaccuum and frictionfree environment, the force required to maintain a speed of 30kph & 100kph should be the same.

pare, the force required to maintain a steady velocity (meaning it is now a vector, velocity = speed and direction) in space where it is almost a perfect vacuum and thus no friction/drag will be zero. kaya off ang engines ng space shuttle after it attains it's orbital velocity.

outer space, btw, is not a perfect vacuum. free electrons, neutrons, "space dust", leftover particles from comets, solar wind material, not to mention the theoretical "dark matter" permeates outer space.

clarification lang, di ako kumokontra ha. :mrgreen:

peace.

[ghosthunter]

well, kaya nga di space yung example ko... i did say "vaccuum & frictionfree environment" lang. Ideal circumstance lang. Just like some physics problem solving we got during high school & college. :mrgreen:

[kinyo]

just have an idea on this topic, gumawa ako calculation:

given a car consumes 10km/liter on highway speed 100km/hr, how much fuel is injected on each chamber?

assuming typical data for my civic esi, the engine is at 3000rpm (approx) when running 100 kph at 5th gear, and noting there are 2 injections per revolution for a 4-cylinder engine,

the car is consuming 1 liter in 6 minutes (10km/liter)/(100km/hr)=(1/10)hr
in 6 minutes, the engine made 18,000 revolutions (6 x 3000rpm)
the car consumed 1 liter in 36,000 injections
hence, the amount of fuel per injection is 1/36,000 liters or 0.02777 ml.

yun lang ...

[ghosthunter]

the problem is your assumptions are wrong. It nearly impossible to measure fuel consumption based on time unless your car is on a threadmill (trying to get close to ideal conditions). It doesn't take into considerations that air drag, road inclines, acceleration rate, car weight and other friction inducing forces play a significant role in a car's fuel consuption.

A better measure would be km/L.

[yebo]

korek. drag alone may take up to 90% of an engine's power output when moving at a steady speed (depending on speed, ok). transmission losses, traction loss, road friction, alternator, power steering and and a/c takes up the rest.

[kinyo]

A better measure would be km/L.

i used 10km/L in my calculations ... purposely to take care of every imaginable losses ... your insight is correct that a better measure is km/L that's why i used it ... i'm not interested in calculating losses, rather just the amount of fuel injected on every injection and starting with km/L value is what i thought to be most sensible

[kevinp]

nabasa ko rin somewhere that each car uses only approximately 15% of its total energy output to turn the wheels. dami kasing moving parts and other variables kaya daming energy lost sa isang auto due to friction and heat.

On another topic, nabasa ko rin pala na ang drag is exponential. forgot the exact formula but its something like <some variable> x velocity squared. So ibig sabihin the drag that a vehicle encounters at 80kph is four times the drag at 40kph. I guess to achieve maximum fuel efficiency, one must balance it between drag and travel time (since the car travelling at 80kph experiences 4x as much drag but for only half the time as a car travelling at 40kph for the same distance).

[yebo]

F(drag) = A x Cd x V x V x m,

where A = cross sectional area of the object perpendicular to the direction of motion
Cd = coefficient of drag
V = velocity
m = density of the fluid through which the object is moving. in the case of a car, that will be the density or air. for a submarine it will be the density of sea water.

you will find this formula in basic fluid mechanics.

[yebo]

oh **** ngayon ko lang napansin, tricycle drayber na ako! funyeta kaya pala parang magaan ang paa ko di na ko padyak drayber!

are you taking up engineering kevinp? looks like it eh. well, welcome to the club. :mrgreen:

[kevinp]

are you taking up engineering kevinp? looks like it eh. well, welcome to the club.

actually, nope. i'm more of a computer guy. hobby ko lang yung interest ko sa cars. di ako nag fluid mechanics nung college e (or di ako nakinig :mrgreen: ). i wouldn't mind though combining my interest in cars and computers, like making traction control software for F1 cars (in my dreams siguro) :D