Robin RGV13000T Service Manual page 23

NOTE 2 : Motor-driven appliances mentioned in lets (3) and (4) vary in their required motor starting
power depending on the kind of motor and start-up load. If it is difficult to determine the
optimum generator capaciv, select a generator with a larger capacity.
.-.
(5) Appliances without any indication as to power consumption
Some appliances have no indication as to power consumption; but instead the work load
(output) is indicated. In such a case, power consumption is to be worked out according to
the numerical formula mentioned below.
(Output of electrical appliance) = (Power consumption~
(Efficiency)
Efficiencies of some electrical appliances are as follows:
Single-phase motor
l l l l l l
-
l l
0.6 to 0.75
II-
The smaller the motor, the lower
Three-phase motor
l l l l
*
l l l l l
0.65 to 0.9
the efficiency.
Fluorescent lamp
l l l l
9
l l
-
l l l
0.7 to 0.8
Example 1
: A 40 W fluorescent lamp means that its luminous output is 40W. Its efficiency is
0.7 and accordingly, power consumption will be 40+0.7=57W. As explained in
item(2),
multiply this power consumption value of 57 W by 1.2 to 2 and you will
get the figure of the necessary capacity of a generator. In other words, a genera-
tor with a rated output of 1000 W capacity can light nine to fourteen 40W fluores-
cent lamps.
,-
Example
2 : Generally speaking, a 400 W motor means that its work load is 400 W.
Efficiency of this motor is 0.7 and power consumption will be 400+0.7=570 W.
When this motor is used for a motor-driven tool, the capacity of the generator
should be multiple of 570W by 1.2 to 3 as explained in the
Item (3).
570
(W) x I .2 to 3=684 (W) to1 710 (W)
-.
Applications
Applicable Wattage on/)
RGV12000 (60 Hz)
I
Incandescent Lamp, Heater
I
10000
I
I
Fluorescent Lamp, Electric Tool
I
4950
I
1 Pump, Compressor
I
2500
I
-2o-