Work order: Parameterize the frequency inverter
auxiliary aid: Technical Data Manual
In order to enable a smooth start and stop, the drive for the conveyor belt is controlled by a VFD. Your work order is to understand the information of the motor rating plate and to determine the correct parameters for the VFD.
A1: Explain the information given on the rating plate.
A2: How should the motor be switched on our three-phase network 400 V / 230 V if you would connect it directly (Star or Delta)? Please give a short explanation!
A3: Determine slip, slip speed and efficiency for nominal operation and operating frequency f = 50 Hz.
A4: So that the belt starts and stops smoothly, choose a frequency converter. According to the manufacturer, the following motor parameters must be entered for initial commissioning:
P0304: nominal voltage
P0305: nominal current
P0307: nominal power
P0308: nominal power factor
P0310: nominal frequency
P0311: nominal speed
Enter the specific values. Missing values you have to calculate.
A5: The workpiece carriers (workpiece carriers) should be conveyed at a maximum speed of 0.4 m/s. For starting the acceleration for the belt should be 0.10 m/s2. When stopping, the deceleration should be 0.25 m/s2. To which values do you have to set the VFD parameters P0320 (ramp-up time) and P0321 (ramp-down time)?
A2: If two voltages are specified on the motor rating plate, the lower voltage stands for the permissible phase voltage (voltage at the motor coils). That´s why the motor must be connected in star to a 400V supply voltage.
A3: nslip = ns-nN = 1500min-1 – 1400 min-1 = 100 min-1
slip s = (ns – nN) / ns * 100% = 10 %
η = Pmech / Pelectr = 0,16 kW / 0,2106 kW = 0,76
(Pzu = √3 U * I *cosφ = √3 * 400V * 0,4A * 0,76 = 0,2106 kW)
A4: Just watch the motor plate
A5: Within the v(t)-diagram, the ramps represent a uniform acceleration.
=> v(t) = a * t remark: v stands for „velocity“ and a for „acceleration“
=> tStart = vmax / a1 = 0,4 m/s / 0,1 ms2 = 4 s
=> tStopp = vmax / a2 = 0,4 m/s / 0,25 ms2 = 1,6 s