Delta REG2000 Series, Руководство пользователя

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10 Power Regenerative Unit Selection | REG2000
4
Elevator Application
For this example, take an elevator with two tons of working load, a speed of 60 m/min, floors from
B1–4F, using 22 kW motor, and a counterweight of 48%.
1. The elevator takes 30 seconds to reach floor 4F from B1, and the whole trip takes 100 seconds,
then ED = 30/100 = 30%.
2. Assuming the motor efficiency is 85%, mechanical efficiency is 85%, and the drive and the
REG2000's efficiencies are both 95%, the useful power output would be 22 kW * 85% * 85% * 95%
* 95% = 14.3 kW.
3. When using the REG2000, and the mains voltage is 380 V, the current would be 14.3
kW/(sqrt(3)*380 V) = 21.8 A.
4. From the table below, at 30% ED, and a working duration of 30 seconds, the output current must be
lower than 115% 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 REG110A43A-21 rated current is 17 A, 17 A*115% = 19.55 A < 21.8 A, and
the REG150A43A-21 rated current is 20 A, 20 A*115% = 23 A > 21.8 A;
therefore, in this case, you can select the REG150A43A-21.