I am in a muddle here, so thanks for making the effort to have a read.
I understand that electrical power going into motor is given by:
\$P_e=I^2R\$
I understand that mechanical power out is given by:
\$P_m=\tau\cdot\omega\$
Where \$\tau\$ = Torque at motor output shaft
Where \$\omega\$ = Speed at output shaft
I understand that for a DC motor, torque is proportial to current:
\$\tau=k_E\cdot I\$
Therefore if current is proportional to voltage via resistance, and electrical power in is proportional to current, power is proportional to current.
Then imagine that a motor is reacting a fixed torque value, and the supply voltage is increased to raise the output speed. A fixed torque and a rising speed means a rising output power. However input power is fixed, since torque is fixed because current is fixed. Graphing this (x-supply voltage, y-Power) gets you a flat line for electrical power in and an angled line for mechanical power out. Therefore the lines must intersect and therefore on one side of the intersection point, output power must be greater then input power, which is impossible.
I am clearly overlooking something so any pointers would be greatly appreciated.
Cheers now, all the best.
Best Answer
Some of your basic premise is correct : torque is proportional to current, and power DISSIPATED IN THE MOTOR ITSELF is a constant I^2*R, where R is the (constant) DC resistance of the motor, as measured across its terminals with the motor stationary.
Now let's run the motor at current I. The V required is not IR. (If it is, the motor is stalled so that the back EMF = 0.)
Instead, V = IR + back-EMF.
Now, I * back-EMF is the electrical power delivered to the load as mechanical power, and I * IR is the power wasted in the motor as heat.
Let's increase V and increase speed keeping I constant. Now, input power has increased (IV) but the motor's resistance hasn't changed : therefore IR is the same and I*IR is the same. But what HAS changed is the back-EMF - obviously, since it is proportional to speed (which has increased).
So the power dissipated in the motor as heat is constant; but the electrical power delivered to the load (I * back-EMF) has increased, exactly as the mechanical output power (torque * speed) has.
No magic, and it all adds up correctly.
But what IS interesting is that the efficiency has increased because the wasted power is constant but the useful power has increased. So a general rule is that electrical efficiency is higher in a lightly loaded motor running fast, than a heavily loaded motor running slow and drawing high current.
(There are limits to this : the less you load a motor and the faster you run it, the higher a proportion of power lost to friction in bearings and especially brushes. Bearings (ballraces) are easy : brushes are not, so brushless motors have a big advantage at high speed and high efficiency)