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Drivetrain efficiency is a function of what?(gear ratio,eng.speed )

  1. Mar 21, 2014 #1
    By the conservation of energy, P_in=P_out

    So,no matter what gear ratio 'i' I choose, P_out=P_in, since the gear ratio 'i' term would get cancelled at the numerator and denominator.

    $$P_{eng}= ω_{eng} M_{eng}= i . ω_{out} \frac{M_{out}}{i}$$
    hence P_in=P_out

    But, what is the drivetrain efficiency a function of? Is it a function of gear ratio and engine speed and engine torque? Or is it a constant always?

    2.And why is the drivetrain efficiency of automatic transmissions 10% lesser than manual? [source page 417:Advanced Direct Injection Combustion Engine Technologies and Development ...
    edited by H Zhao]
  2. jcsd
  3. Mar 21, 2014 #2
    Making the assumption that P_out = P_in

    Its P_out = P_in - P_losses

    The key is what are the losses? Off the top of my head:

    Bearing drag
    Oil windage
    Transmission error


    Obviously friction related losses will increase with speed.

    Traditionally the majority of automatic efficiency loss is due to slip in the torque converter. Modern autos lock up at very low rpm so are much better. Autos also require oil pumps for hydraulics.
  4. Mar 22, 2014 #3

    Ranger Mike

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    Chrisxx is spot on as usual. I would add that unless you are working in a vacuum ( sick pun) you also have to look at Aerodynamic Drag.

    There is a basic equation for the force it takes to push something through air:

    Aerodynamic drag = 1/2 D x A x V squared

    In this equation, D is the density of the air, A is the frontal area of the moving shape, and V is its velocity relative to the air

    For real body shapes, air at standard conditions, A in square ft., V in mph, and drag in pounds of force, this equation becomes:

    Drag = 1/391 x Cd x A x V squared

    This equation shows that to calculate drag you need to know three things: Cd, the drag coefficient; A, the frontal area of whatever you’re driving through the air; and the speed of air past it. This equation shows an important point—aerodynamic forces are proportional to the square of the speed. That means you quadruple the drag or lift when you double the speed.

    Note that the power needed to push an object through a fluid increases as the cube of the velocity. A car cruising on a highway at 50 mph (80 km/h) may require only 10 horsepower (7.5 kW) to overcome air drag, but that same car at 100 mph (160 km/h) requires 80 hp (60 kW). With a doubling of speed the drag (force) quadruples per the formula. Exerting four times the force over a fixed distance produces four times as much work. At twice the speed the work (resulting in displacement over a fixed distance) is done twice as fast. Since power is the rate of doing work, four times the work done in half the time requires eight times the power. It's important to value the rolling resistance in relation to the drag force. Rotational weight of drive line components and parasitic drag must be considered too.

    If you think about the manual transmission vs automatic trans. thing the manual trans set up is straight line to the differential. Usually is lighter than the automatic. The automatic trans is converting mechanical power to fluid power than back to mechanical power with heat being the culprit for power loss by thee conversion process. It is a trade off for smoother drive and automatic gear changing but is heavier and required additional oil cooler, more trans fluid, adds heat and weight.
  5. Mar 24, 2014 #4
    To add to Ranger_Mike's automatic info...even with a lock-up converter, you're still driving the transmission pump.
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