A CIM motor is attached to a small dynamometer which exerts a fixed constant load torque of 100 oz-in on the motor shaft. The CIM is directly attached to a 12VDC source (assume no internal resistance) with heavy gauge wire and connectors (assume no resistance). Assume the CIM’s specs are exactly as specified in the attached screenshot.
Questions:
What is the CIM’s speed?
What is the CIM’s current?
How much power (watts) is being input to the CIM?
How much power (watts) is the CIM delivering into the load (dynamometer)?
How do you account for the difference between 4 and 3 above?
Maybe
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~3763.7 RPM
~40.64 Amperes
487.68 Watts in
~256 Watts out
The inefficiencies of the motor (mostly friction) prevent power from being transfered without loss
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What is the CIM’s speed?
If free speed is 5310 RPM and stall torque is 343.4 and the calculated speed is the free speed - reduction in speed due to load then speed = free speed - (rated torque * (free speed/max torque)) or
5310 - (100 oz.in. * (5310/343.4)) = 3763.69 RPM
What is the CIM’s current?
If stall torque is 343.4 oz. in. and stall current is 133, then the current constant is .3873A/oz.in. therefore current at 3763.69 RPM is the constant * RPM + I min
I= (3763.69RPM * .3873A/oz.in.) + 2.7A = 38.73 + 2.7A = 41.43A
How much power (watts) is being input to the CIM?
P IN = I * V = 41.43A * 12V = 497.16 W
How much power (watts) is the CIM delivering into the load (dynamometer)?
The conversion factor for oz. in and speed is 0.00074 and Pout = Torque * RPM * conversion factor,
Therefore P out = 100 oz. in. * 3763.69 * 0.00074 = 278.51W
How do you account for the difference between #4 and #3 above?
As the lost electrical power is the current squared * motor winding resistance, and R motor is 12V/133A (stall current) = .0902 ohms, then electrical power loss is 41.43A^2 * .0902ohms = 154.87W.
If the input power is 497.16W and electrical loss is 154.87 and power out is 278.51W then all other losses amount to 497.16-154.87-278.51=63.78W
These losses are attributable to bearing friction, brush friction, back EMF, friction of armature with the air, etc.
A CIM motor is attached to a small dynamometer which exerts a fixed constant load torque of 100 oz-in on the motor shaft. The CIM is directly attached to a 12VDC source (assume no internal resistance). Assume the CIM’s specs are exactly as specified here.
Questions:
What is the CIM’s speed?
Solution: Assume that the speed vs torque curve is linear, and use the given load torque (100 oz-in) to interpolate the rpm:
rpm = 5310*(1-100/343.4) = 3763.7 rpm
What is the CIM’s current?
Solution: Assume that the current vs torque curve is linear, and use the given load torque (100 oz-in) to interpolate the current:
How do you account for the difference between #4 and #3 above?
Solution: The total power loss is 487.7-278.3 = 209 watts. Most of this loss is due to I^2R heat loss in the motor windings. The motor winding resistance is 12/133 = 0.09023 ohms, so the I^2R loss is (40.64)^2*(0.09023) = 149 watts. The remaining loss (60 watts) is due to a combination of motor bearing friction, windage, and motor core losses (hysteresis losses and induced eddy currents).
Now, suppose I open the circuit and insert a 0.1 ohm resistor in series with the motor. Assume that the load torque remains the same, at 100 oz-in, and that the resistor is large enough that it does not overheat.
Questions:
What is the CIM’s speed?
What is the CIM’s current?
How much power (watts) is being input to the CIM?
How much power (watts) is the CIM delivering into the load (dynamometer)?
Explain any assumptions you had to make to get the answers.
I didn’t calculate the output current, I drew a picture of the curve and guessed. That would be why I got it wrong. We haven’t learned (much) about angular motion in physics yet
Well, lets see… Assuming the motor is mostly linear in performance, The No-Load Current is 2.7 amps, the constant (lets use this) D determining the ratio of current to torque is (64-0)/(27-2.7)= 2.633 oz-in per amp, the no-load 12v speed is 5310 rpm, with it losing (5310-4320)/(64) rpm per oz-in, or 5310-4320/27-2.7 per amp.
Oh, and 5310 RPM per 12 volts equates to 442.5 RPM’s per volt.
Based on this:
5310 RPM no load
2.7 amps no load
2.633 oz-in per amp of torque
-15.46 rpm per oz-in torque, or 40.74 rpm per amp
442.5 RPM per volt, no load.
So, lets answer number two first. In order for there to be 100 oz-in of torque, the current applied (which in turn causes the force) must be the same, and I’m hoping this math was done correct before, 40.64 amps.
Which in turn brings us back to #1. The voltage loss across a .1 ohm resistor at 40.64 amps would be equal to V=IR, or .1*40.64= 4.064 volts (can’t believe I used my TI-89 on that.)
12-4.064 volts = 7.936 volts left for the CIM motor.
A no load speed would be 3511.68 rpm. Subtracting the torque losses would result in an actual speed of -15.46 * 100 + 3511.68 = 1965.68 RPM.
Number 3: Power input to the CIM = VI= 7.936 V * 40.64 amps = 322.5 watts.
Numbah 4: Power output, mechanical: Speed * Force. 1966 RPM * 100 oz-in ~ using a random converter = 100 oz .278N per oz (32.76 revs * 2 pi inches * .0254m/in) per sec = OMG = 145.3 ? Watts.
Number 5: I believe I had to answer the second question first. You don’t have to, but it eliminates a few extra steps of variable making.
In the first scenario, the motor is generating 100 oz-in of net torque (against the external load) plus torque losses at 3764 rpm (due to friction and windage).
In the second scenario, the motor is generating 100 oz-in of net torque (against the external load) plus torque losses at 1965 rpm (due to friction and windage).
In the second scenario the torque losses are less (due to the lower rpm). Therefore the total torque is less than the first scenario, and the current required should be slightly less. How much less ?
speed: go faster!
heat: it makes to much of this
torque: enough to work on accessories, and great for pulling or driving stuff if you put a 12.75:1 gearbox on
current: work with the battery, when run through the tan device
Being a mechanical subteam is no excuse to not know this; personally, anyone on my team that is working design MUST know at least half of this stuff to do any work.
Being a mechanical subteam is no excuse to not know this; personally, anyone on my team that is working design MUST know at least half of this stuff to do any work.
Being on mechanical means that you should know this. You can never design anything well if you don’t understand the big picture.
True, but in the context of the FRC, knowing exactly how back-emf and inductance losses play into a motors performance isn’t critical to design, although I do agree that knowing the effects that voltage, torque and load play into the operation of the motor are necessary.
Please note that back emf does not cause any “losses”. It simply limits the amount of current that the source can deliver to the load.
The losses in the motor are due to I^2R heat in the motor windings, motor bearing and brush friction, windage, and core losses (hysteresis and eddy currents).
