Solution VFD

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

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