Example for a voltage drop calculation
We calculate for you the voltage drop which occurs when you start your motor/machine combination, if we know the configuration of your feeding grid in addition to the technical data of motor and driven machine. Below you can find a typical example for a calculation. 2 voltage levels (40 kV and 6 kV) are given. The 40 kV grid has a minimal short circuit power of Sk" = 250 MVA. The 40 / 6 kV transformer has the following features: 10.000 kVA power and uk = 8 % short circuit voltage. Further motorical loads of 4.000 kVA run on the 6 kV level beside the motor, which will be started by us. We calculate the voltage drop from these data when starting your machine to 1,5 % on the 40 kV level and 4,5 % on the 6 kV level. The diagrams show the corresponding curves of time, torque, current and voltage during acceleration.
motor rated current | In | 250 | A | |
motor rated voltage | P | 2.000 | kW | |
motor start current DOL | Ia | 5 | x In | |
motor speed (synchronous) | nn | 1.500 | rpm | |
start time | 10 | s | ||
min. short circuit power at 40 kV | Sk'' | 250 | MVA | |
voltage drop at 40 kV | ΔU | 1,4 | % | |
feeding transformer 40 → 6 kV | ||||
power | S | 10.000 | kVA | |
short circuit voltage | uk | 8 | % | |
parallel motorical load | S | 4.000 | kVA | |
voltage drop at 6 kV | ΔU | 4,5 | % |
network configuration