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

Which prime mover do you think will win?

  1. May 18, 2016 #1
    I already know the answer to this easy question. Just curious what people know on the internet.

    A 36 volt DC electric motor, model P66SR274 is rated for 1865 watts (2.5 HP) at 2000 RPM, and its output shaft is to be direct coupled to a shaft that is 25mm in diameter. The power supply is 3X, 12 volt car batteries wired in series that are brand new and fully charged. The 25mm shaft will have a rope going through a drilled hole in the shaft and knotted. This rope will lift a mass X as shown in the list:

    80Kg
    100Kg
    135Kg

    After the testing is done with the electric motor, a Honda GX100 gas engine with clutch to engage at 3600 RPM is rated for 2100 watts (2.8 HP) at 3600 RPM and will be geared down to 1800 RPM to a pulley that is attached to the 25mm diameter shaft and the gas engine will be pulling up the same mass X as the electric motor did.


    Now the question is:

    Which prime mover will stall?
     
  2. jcsd
  3. May 18, 2016 #2

    berkeman

    User Avatar

    Staff: Mentor

    It's a bit rude for you to ask us to put out lots of throw-away effort when you have a solution. How about you show us the solution, and ask for comments instead? That would be a much more polite approach to this question, IMO...
     
  4. May 19, 2016 #3
    Sorry, not trying to be rude. The answer is the DC electric motor will win and lift even the highest load of 135Kg. PM DC motors have inverted parabolic power curves that max at 50% efficiency and linear increasing torque curves that max at stall where max current is reached when the current is limited to the resistance of the motor's armature windings and other associated parts of the circuit with the voltage applied. When the load increases the armature will slow down (rotor part), which in turn causes the back EMF generated to go down resulting in the differential voltage going up that is applied to the resistance of the armature circuit, which causes current to increase (Vsupply - BEMF), and since the magnetic field is proportional to the current (there will be magnetic saturation as current gets very high), the magnetic force will increase more and more that is multiplied with rotor radius to give more and more torque.

    If I need to explain max power transfer theorem to explain why peak power occurs at 50% efficiency in a brushed DC motor then let me know.

    Internal combustion engines cannot increase in torque and power when you overload them. The reason is there is no mechanism in a combustion engine to ram in more fuel/air mixture once the power stroke has commenced. Thus, if load where to increase during the power stroke phase, then the explosive force produced will not be able to over come the force required to move the piston. So you get a stall condition. I don't think its even possible to design a combustion engine that can add in more fuel/air mix "on-the-fly" during a power stroke. It would be very difficult to do without losing compression. Lost compression during a power stroke will mean less explosive force against the piston face. The problem is exacerbated more if the the load demand increases on the other cycles where the fly wheel is having to release the stored energy to carry the piston through those cycles where power stroke does not occur.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted