Delta REG2000 Series, Руководство пользователя

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10 Power Regenerative Unit Selection | REG2000
10-3
Model selection method 3
(based on the application load characteristics)
Crane/Hoist Application
The equipment weighs 10 tons, drive model: VFD075CH43A, with a 5.5 kW motor.
From the crane's specification we know
1. In high speed operation, the time it takes from top to bottom is 9 (m)/3.6 (m/min) = 2.5 (min) =
150 (sec).
2. Assuming the motor efficiency is 85%, the mechanical efficiency is 85%, and the drive and the
REG2000's efficiencies are both 95%, the useful power output would be 5.5 kW*0.85^2*0.95^2 =
3.57 kW.
3. When using the REG2000, and the mains voltage is 380 V, the current would be 3.57
kW/(sqrt(3)*380 V) = 5.4 A.
4. From the table below, at 40% ED, and a working duration of 150 seconds, the output current must
be lower than 90% of the rated current to avoid overload.
**
REG Performance Curve
60
70
80
90
100
110
120
130
140
150
160
1 10 100 1000
Time of Regeneration (sec.)
10 ED
%
25 ED
%
50 ED
%
60 ED
%
80 ED
%
Continuous
10 ED
%
25 ED
%
40 ED
%
50 ED
%
60 ED
%
80 ED
%
Continuous
※
The diagram above shows the testing result by combining REG and a DC choke.
 Definition of ED%: See Chapter 03 Specifications.
5. From the specification sheet, the REG075A43A-21 rated current is 10.5 A, and 10.5 A*90% = 9.45
A > 5.4 A; therefore, in this case, you can select the REG075A43A-21.