We require a BLDC motor which
can sustain a load torque of 25.4291 N-m (according to equation 3). A BLDC
motor with ratings 48V, 29A, 32.92 N-ms considered. The mathematical calculations for
the BLDC motor are as follows:
Pelectrical
= Pmechanical + Pcopper losses (1)
Where,
Pelectrical
is input electrical power in watts
Pmechanical
is output mechanical power in watts
Pcopper
losses is copper losses i.e. I2R losses in watts
Pelectrical
= V*I (2)
Where,
V is
supply voltage in volts (48V)
I is
current in amps (29A)
Therefore,
Pelectrical = 48*29 = 1392 W
Load
torque need to be calculated to know the amount of torque required to move the
vehicle. It is also essential in selecting a perfect motor for the desired qualities.
Tload = F*r*µ (3)
Where,
Tload is load torque in
N-m
F is the force required to spin the
wheel in Newton =251.40N (from force equation)
R is the radius of the wheel in meters
= 0.2023m
µ is the coefficient of friction = 0.5
Therefore, Tload =
251.40*0.2023*0.5 = 25.4291 N-m
Considering the BLDC motor with
torque greater than or equal to the load torque (Tload) with an
output speed of 300 rpm and output torque of about 32.62 N-m.
Pmechanical = Tm*ω
(4)
Where,
Tm is
motor torque in N-m i.e. 32.62 N-m
ω is angular velocity in rad/sec
i.e. ωrpm*(2π/60)
Therefore, Pmechanical =
(2πNTm)/60 = (2*3.14*300*32.62)/60 = 1024.268 W
Pcopper
losses = I2R (5)
Therefore,
Pcopper losses = 292*0.139 = 116.899 W
Therefore, Efficiency =
=
81.98%
Horse Power (1hp=750W)
|
1hp
|
Operating
voltage
|
48 v
|
Operating
current
|
15.62A
|
Starting /max
current
|
29A
|
Maximum torque
|
32.62 N-m
|
Maximum output
speed
|
300 rpm
|
