What causes the rpm difference between peak brushed motor torque and peak brushed motor efficiency?
typical brushed motor performance http://members.toast.net/joerger/pic2/motorcurve.gif source: http://members.toast.net/joerger/AskAaron/motors.html
page 15 of this brushed & brushless motor performance summary talks about efficiency but doesn't mention a connection between torque and efficiency. http://www.dtic.mil/dtic/tr/fulltext/u2/a577582.pdf
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Peak efficiency means best ration of mechanical power (velocity * torque) to electrical power (I * (IR + BEMF)).
The ability to keep high torque is lost towards highest RPM, this is why there is a peak instead constant growth.
The DC motor performance curves posted in the question are drawn for a fixed voltage and variable load torque. At zero speed, the load has been increased to the point that the motor is stalled. The motor is producing its maximum torque, but the shaft will not turn because the load is so high. Since mechanical power is speed multiplied by torque, the output mechanical power is zero. 100% torque X 0% speed = 0% power. However, the motor current is high, so electrical power is going into the motor even though no mechanical power is coming out. All of the power going into the motor is producing heat inside the motor. Efficiency is output mechanical power divided by input electrical power. No output divided by some input = zero efficiency.
At 100% speed, the load has been reduced to zero so that the motor has nothing holding it back and runs at the maximum possible speed. Here again, there is no mechanical power being produced. 0% torque X 100% speed = 0% power. Here again, there is some current going into the motor, so power is going into the motor, but no power is coming out. Here again, the efficiency is zero.
The speed at which efficiency is maximum is someplace between zero and maximum speed. The losses in the motor are mostly losses due to the resistance of the winding. Losses in resistance are proportional to the current squared, and increase rapidly as the current increases in proportion to the torque. The output power of the motor rises as the torque increases, but reach a peak when the speed has decreased as much as the torque has increased. At that point, the losses have risen to equal the output power of the motor. Half of the input power to the motor is going to losses and half to output, so the efficiency has fallen to 50%.
Put simply, the rpm difference is caused by current causing a voltage drop across the motor's internal resistances (brushes, commutator, armature windings).
Torque is proportional to current. When the motor is free running it speeds up until the back-emf (almost) equals the supply voltage, and it draws just enough current to overcome internal losses. As loading increases the current must increase to supply a torque to match the load. This current causes a voltage drop across the motor's internal resistance. The motor slows down because the driving voltage (supply voltage - voltage drop across internal resistance) is lower.
Peak torque occurs at maximum current. This occurs at zero rpm because at that point there is no back-emf so current is limited only by resistance. However power output = speed * torque, so efficiency is zero.
Efficiency improves at higher rpm because then the current is lower so resistive ('copper') loss reduces. However as speed increases so too do frictional and magnetic ('iron') losses. rpm is at a maximum when free running, but then torque is zero so power output and efficiency are also zero.
Peak efficiency occurs at the point where copper loss = iron loss, typically at 80~90% of no-load rpm.
