VLT ® 2000 Series How to size your VLT I Selection of frequency converter Usually the size of frequency converter is chosen on the basis of the shaft output, as this may be the only value known. However, if the data are known for both the application, the motor and the frequency converter, it is recommended to make a more accurate sizing. I Sizing on the basis of the apparent power S M [kVA] consumed by the motor Exercise A motor must yield constant torque at continous operation. Usually the necessary values will appear from the motor plate or from the motor catalogue. The values must be based on the rated motor speed.
Solution From the table on the next page you can see that VLT type 2025 can yield 2.0 kVA (415 V) at continuous operation. The solution is therefore VLT type 2025. VLT 2000 only operates according to the CT (constant torque) characteristic I Sizing when you know the motor current Exercise Conveyor plant with a 1.1 kW, 3 Ч 380 V motor. At continuous operation the motor current is 2.5 A ( 3 Ч 415 V). Read values Motor current = 2.5 A (3 Ч 415 V) s M = U Ч I Ч √ 3 [kVA] 1000 = 415 Ч 2.5 Ч √ 3 [kVA] 1000 Solution From the table on the next page you can see that a VLT type 2025 can yield 2.8 A at continuous opera- tion. The right solution is therefore VLT 2025.
= 1.8 kVA I Sizing on the basis of the power requirement P VLT [kW] of the motor Exercise A machine tool is driven by a 3 kW motor. The power requirement is indicated to be 2.4 kW. The motor efficiency η is 0.80, cos ϕ = 0.81 and the motor volt- age is 3 x 415 V. η and cos ϕ are measured at 3 kW output. We estimate η and cos ϕ to be approximately the same at 80 % load. Solution From the table on the next page you can see that VLT type 2040 can yield 4.0 kVA (415 V) at continuous operation. The right solution is therefore VLT type 2040. S VLT = P m η Ч cos ϕ = 2.4 kW 0.80 Ч 0.81 = 3.7 kVA 30 MG.20.B6.02 – VLT is a registered Danfoss trademark