
Torque Capabilities of DC and AC Drives in the
Constant Horsepower Range Using a Wound Rotor Motor
Replacement as a Typical Application Example
Copyright © 2002 Francis J. Martino
A copper tube drawing machine was powered by a 250 HP, 1725 RPM wound
rotor motor. A new drive was desired that would eliminate the seventy year old
controls and motor.
The replacement would be required to increase the speed of the machine so
that the maximum possible production would be achieved. A replacement DC
motor could be run in armature control from zero to base speed of 1750 RPM and
above base speed by field weakening. An AC motor could be run in the constant
torque range from 0 to 60 HZ and in the constant HP range from 60 to 90 HZ.
Using a nominal base speed of 1750 RPM, either of the two types of motors
could achieve the desired maximum speed of 2300 RPM.
The Wound Rotor Motor
There are two major advantages of a wound rotor motor: the high starting
torque capability and the ability to run at continuously varying speeds.
The maximum starting torque capability of a wound rotor motor is determined
at the time of design by the characteristics of both the motor rotor winding and
the resistors that vary the current in the rotor winding.
A wound rotor motor can be designed to develop a starting torque up to 270%
of its rated full load torque. If a high starting torque is not required on an
application then the system might have been designed for a starting torque of
less than 200% of full load torque.
Any replacement drive must be capable of matching or exceeding the starting
torque capability of the wound rotor motor. The motor and controls of the copper
tube drawing machine developed a starting torque between 150% and 175% of
full load torque. The starting torque capability can be determined by measuring
the stator and rotor amps on starting inrush or by evaluating the wattage rating of
the rotor control resistors.
To use a wound rotor motor as an adjustable speed drive, the rotor control
resistors must be rated for continuous current. If the motor is used only for a slow
acceleration or high starting torque but then operates at its maximum speed for
the duration of the work cycle, then the resistors will be removed from the circuit
when at motor rated speed. In that case they will have been duty cycle rated for
starting duty only.
Torque Calulations for the Wound Rotor Motor
A copper tube drawing machine is a constant torque application,
therefore it will require a constant torque throughout it speed range
regardless of speed. Only upon starting will the machine require a
high starting torque in order to “break away” the machine and
load inertias.
The full load torque (T) capability of the 250 HP, 1725 RPM motor is:
T = (HP x 5252) / RPM
250 HP x 5252) / 1725 RPM = 761 lbsft
The starting torque capability of the tube drawing machine has been
determined to be a maximum of 175% of running torque, or
1.75 x 761 = 1331 lbsft
Torque Characteristics of DC and AC Drives
A typical DC drive or AC drive will develop a constant torque from zero
speed to motor rated base speed. In that range the DC motor is
controlled by a linear increase in the armature voltage from zero to the
motor rated voltage. The AC drive will maintain a constant volts per hertz
ratio as both voltage and hertz are raised from zero volts at zero hertz to
motor rated voltage at rated hertz.
Startup torque allowed by most drives on the market today is 200% of
the rated full load torque of the motor.
When exceeding rated base speed in the field weakening range of a DC
shunt wound motor, the torque developed will be inversely proportional
to the speed, which is a true constant horsepower. Torque is therefore
proportional to 1/RPM.
When an AC motor is operated in the typical 60 to 90 Hz range of a
variable frequency drive, the motor will be operating with an approximate
constant horsepower characteristic. In the constant horsepower range,
motor voltage is held constant as the frequency increases. Thus, the
volts per hertz (V/Hz) ratio decreases as hertz increases.
The torque capability of an AC motor drops off with the square of the volts
per hertz ratio. * Consequently, the torque developed is inversely
proportional to the square of the speed since the speed is proportional to
the volts per hertz ratio. Torque is therefore proportional to 1/(V/Hz)^{2}, or, 1/RPM^{2}.
See Figure 1 for a torque versus speed graph of DC and AC drives.
Torque Calculations for a DC and AC Drive
We will assume that both the DC and AC motors that are candidates
to replace the wound rotor motor have base speeds of 1750 RPM
and maximum speeds that exceed 2300 RPM.
The maximum rated full load torque available from both the DC and
AC motors will be the torque rating at base speed:
T = (HP x 5252) / RPM
For a 250 HP motor:
(250 HP x 5252) / 1750 RPM = 750 lbsft
The torque available from the DC motor in the field weakening range
will be proportional to 1 / RPM. See Figure 2.
T = Maximum Rated Torque x (1750 / RPM)
At 1750 RPM, torque capability is 100%:
T = 750 lbsft x (1750 / 1750) = 750 lbsft
At 2000 RPM, torque capability is 87.5%:
T = 750 lbsft x (1750 / 2000) = 656 lbsft
At 2300 RPM, torque capability is 76.1%:
T = 750 lbsft x (1750 / 2300) = 570 lbsft
At 150% of base speed, torque capability is 66.7%:
T = 750 lbsft x (1750 / 2625) = 500 lbsft
The torque available from an AC motor in the constant horsepower
range from 60 to 90 Hertz will be inversely proportional to the square of
the V/Hz ratio. A variable frequency drive, when operating a motor at
60 Hz and 460 Volts, will have a V/Hz ratio of 7.666:
T = Maximum Rated Torque x [ (V/Hz) / 7.666 ]^{2}
Calculations below are given for both V/Hz and RPM for a 250 HP
motor operating in the constant horsepower range of a variable
frequency drive. See Figure 3.
At 1750 RPM, 60 Hz, torque capability is 100%:
T = 750 lbsft x [ (460/60) / 7.666 ]^{2} = 750 lbsft
T = 750 lbsft x (1750/1750)^{2} = 750 lbsft
At 2000 RPM, 68.6 Hz, torque capability is 76.5%:
T = 750 lbsft x [ (460/68.6) / 7.666 ]^{2} = 574 lbsft
T = 750 lbsft x (1750/2000)^{2} = 574 lbsft
At 2300 RPM, 78.9 Hz, torque capability is 57.8%:
T = 750 lbsft x [ (460/78.9) / 7.666 ]^{2} = 434 lbsft
T = 750 lbsft x (1750/2300)^{2} = 434 lbsft
At 150% of base speed, 90 Hz, torque capability is 44.4%:
T = 750 lbsft x [ (460/90) / 7.666 ]^{2} = 333 lbsft
T = 750 lbsft x (1750/2625)^{2} = 333 lbsft
Selection of Motor Horsepower Based on Torque Requirements
The starting torque requirement of the tube drawing machine was
determined to be no more than 175% of the wound rotor full load
torque requirement, as discussed above, which is 175% of 761 lbsft,
or, 1332 lbsft.
A 250 HP DC or AC motor with 750 lbsft of running torque will
develop 200% of 750 lbsft on startup, or 1500 lbsft, as noted
above, which exceeds the ability of the wound rotor system to
be replaced.
The full load torque capability of 761 lbsft developed by the wound
rotor motor must also be equaled or exceeded by the replacement motor
unless it can be determined that the torque requirement of the
machine is less than 761 lbsft. Measuring the maximum running amps of
the wound rotor motor will determine what per cent of the maximum load
capability of the motor is required by the machine.
A tube drawing machine will present a constant torque requirement
throughout the speed range provided that no mechanical difficulties arise
as the machine is driven into the extended speed range developed by the
new replacement motor.
See the Tables below for a comparison of torque versus speed of
the DC and AC drives.
Torque Capability of 250, 300, 400 and 500 HP Motors,
Operating at 1750, 2000, 2300, 2625 RPM
Torque vs. RPM of 250 HP DC and AC Motors






RPM: 
1750 
2000 
2300 
2625 
RPM 
DC 
750 
656 
570 
500 
lbsft 
AC 
750 
574 
434 
333 
lbsft 
Torque vs. RPM of 300 HP DC and AC Motors






RPM: 
1750 
2000 
2300 
2625 
RPM 
DC 
900 
788 
685 
600 
lbsft 
AC 
900 
689 
520 
400 
lbsft 
Torque vs. RPM of 400 HP DC and AC Motors






RPM: 
1750 
2000 
2300 
2625 
RPM 
DC 
1200 
1050 
913 
800 
lbsft 
AC 
1200 
918 
694 
533 
lbsft 
Torque vs. RPM of 500 HP DC and AC Motors






RPM: 
1750 
2000 
2300 
2625 
RPM 
DC 
1500 
1313 
1142 
1000 
lbsft 
AC 
1500 
1148 
867 
666 
lbsft 
From the Tables we see that, for operation at 2000 RPM, a 300 HP DC
drive or a 400 HP AC drive will be required to exceed the 761 lbsft
capability of the wound rotor motor.
For operation at 2300 RPM, as required by the replacement motor
specifications as was noted above, either a 400 HP DC drive or a 500 HP
AC drive will be required.
If a 250 HP, 1750 RPM DC motor is used, the maximum torque
requirement of the machine must not exceed the capability of the DC drive
at 2000 RPM, which is 656 lbsft. Thus, a 250 HP motor will be
sufficient only if the maximum horsepower required by the machine at
1725 RPM is:
(656 lbsft maximum machine torque requirement x 1725 RPM) / 5252 = 215 HP
For 2300 RPM operation, the maximum HP required of the machine at
1725 RPM must not exceed:
(570 lbsft maximum machine torque requirement x 1725 RPM) / 5252 = 187 HP
Thus, to prevent oversizing of the replacement motor, the full load
running amps of the existing wound rotor motor must be measured to
determine the actual horsepower requirement of the machine.
Reference:
* Mark M. Hodowanec, “Proper Application of Motors Operated on
Adjustable Frequency Control,” IEEE Industry Applications Magazine,
Vol. 6, No. 5 September/October 2000, pp 4046.
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