Motovario M Serie Manual De Instrucciones página 21

efficiency
moment of inertia
acceleration time
braking time
ratio between mechanical power delivery and electrical power absorption
h = P / P
h once we know the efficiency, the power delivered to the shaft can be calculated as follows:
asynchronous three-phase motor
asynchronous single-phase motor
Product of rotating mass m [kg] and the square of the equivalent radius of rotation r [m]: J = mr
In practice one uses PD
J [kg·m ]
2
lows that: PD = 4J
2
[kgp·m2]
Note that the weight in the practical system corresponds (numerically) to the mass in the SI system
In evaluating the acceleration and braking times we must sum the motor's moment of inertia J
obtain the total moment of inertia: J
and analogously: PD
2
t
Furthermore, to the torque delivered by the motor M
resisting torque Mr, to obtain, as a first approximation:
during acceleration, the accelerating torque: M
during braking, the braking torque: M
As a first approximation we can use for M
tion, given the load curve, can be obtained by integrating from 0 to the nominal speed.
The acceleration time, for a speed variation of Dw (or Dn), is:
in the SI system
t [s]
in the practical system
a
t [s]
f The same formulas apply to the braking time, with M
If the external loads are connected by gear reducers or speed multipliers, the respective moments of inertia must be referred
to the motor axis by multiplying them by the square of the ratio between the load speed n
J (n /n ) and analogously for PD2.
2
ext c m
To refer the inertia to a load of mass M drive in a linear motion by the motor to the motor's shaft, we must know the ratio
between the linear speed v and the corresponding speed n (or w) of the motor; the corresponding moment of inertia will be:
in the SI system
in the practical system
where P is the weight of the moving part.
Motori elettrici / Electric motors / Moteurs électriques / Elektromotoren / Motores eléctricos / 电动机
h% = P / P
a a
P = √3V I
hcosw
[W] [V] V[A]
P = E I
hcosw
[W] [V] E[A]
, the product of the weight [kgp] and the square of the equivalent diameter of rotation D [m]; it fol-
2
[kg·m2]
= J + J
t m ext
= PD + PD
2 2
m txt
, which may be accelerating or braking, we must subtract or add the
m
= M - M
a m r
= M + M
f m r
the value of the starting torque as given in the catalogue; a more precise calcula-
m
t = [J / M ]·Dw [kg·m ]
2
a t a
t = [2.67 PD / M ]·Dn·10
2 -3
a t a
replaced by M
a
J = M (v /w )
ext [kg] [m/s] m[rad/s]
PD = 365 P (v /nm
2
[kgp] [m/s]
·100
2
to that of the load J
m
[kgp·m ]
2
and bearing in mind that M and Dn are negative.
f a
and the motor speed n
c
2
)
2
[rpm]
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M
, to
ext
:
m
19