Requirements: 3 hp | 600rpm
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BLDC Motor Assumptions: 20kv motor constant, 0.02ohm resistance (2 phase measurement - lead to lead)
Controller Assumptions: BLDC >80a motor current limit
Battery Assumptions: 12S * 3.7v nominal per cell = 44.4v nominal
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3 horsepower = 2.2371 kW mechanical
(600rpm * 2 * pi) / 60 = 62.83185rad/sec
2.2371 kW mechanical = 2237.1w mechanical
2237.1w mechanical / 62.83185rad/sec = 35.60455405976427560226Nm
Torque required @ 600rpm = 35.60455405976427560226Nm
KT motor constant (torque per amp)= 60/(2 * pi * 20kv) = 0.4774648292756860073067 Nm / amp
35.60455Nm Required / 0.4774648292756860073067 KT = 74.56a motor current
Back EMF Voltage @ 600rpm = 600rpm / 20kv = 30v
Pack Voltage 44.4v * XX.XXX% duty cycle bldc = BLDC effective voltage
(XX.XXXv effective - 30v back emf) / 0.02ohm resistance = 74.56a motor current
2347.98w electrical = 74.56a * (30v bemf + (0.02ohm * 74.56a))
2347.98w electrical / 74.56a motor current = 31.49v effective
31.49v effective / 44.4v pack voltage = 70.92% duty cycle
74.56a motor current^2 * 0.02ohm = 111.183872w copper loss
2347.98w electrical / 44.4v pack voltage = 52.88a battery current
111.183872w copper loss + 2237.1w mechanical = 2347.98w electrical------------------------
Conclusions:
With a 20KV 0.02ohm BLDC motor & 44.4v battery, you can achieve 3 horsepower at 600rpm (35.60Nm @ 62.38rad/sec) with a 74.56a motor current limit controller setting, 2347.9w electrical power consumption (52.88a battery current @ 44.4v), & 111.18w copper loss assuming the stator is physically large enough that it won't magnetically saturate with less than ~74.5a stator current.