@ John Hill
>> Double the voltage and half the time gives the same watt hours per hour.
Not True !!!
Power is related to the square of the current. So double the voltage gives 4X the power.
Using Power=Amps X Volts or Power = Amps X Amps X Resistance. (P=IV, P=I squared R)
Example : A 1 Ohm load with a 1Volt supply will dissipate 1 watt of power. A 1 Ohm load with a 2V supply will dissipate 4 watts of power.
So a 50% pulse width at twice the voltage will double the power in a resistive load.
However, in a motor, things are more complex. When supplied with DC, a motor will generate a voltage almost equal and opposite to the voltage supplied. The almost bit is where the motor losses (winding resistance etc) get their power from.
When a motor is pulse width modulated the peak current is determined by the pulse frequency, motor inductance, motor resistance, and the back emf (motor generated voltage).
Its quite possible that at double the voltage some significant problems could occur. Likewise, just connecting the motor across the supply, at turn on when its at rest, the inrush current flowing will probably exceed the 20 Amps your PSU can supply (Assuming the winding resistance is <1 Ohm).
IMHO Use a PWM controller with a frequency high enough to limit the current in the motor windings. (by means of the motor inductnce)
Alternately (and more simply

) get a controller with a built in current limit.
Hope this helps a bit.
picclock
PS 1HP is about 32 Amps+ at 24V.