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Induktiomoottorin käynnistäminen sarjainduktansin avulla
DC-moottorin käynnistäminen sarjavastuksen avulla
Example. The nominal values of a constantly magnetised DC-motor are: UN = 220 VDC, IN = 220 A and nN = 950 rpm and the resistance in the armature circuit is 0.05 x UN/IN = 0.05 Ω. Choose the step count and resistance values for the starter, when the start current is allowed to vary in the range 330 A…440A.
Solution. The voltage drop with the nominal current is: 220 A x 0.05 Ω = 11 VDC, which is 5 % of the nominal voltage. Thus the rest 95 % remains for internal back voltage. Now we can calculate the idling speed nS from the equation nN = 0.95nS. We get n0 = 1000 rpm. The slip speed ΔnN in the nominal point is thus 50 rpm.
It is easy to see that the last start-up step is left with the slip speed Δn0 = 100 rpm, for the slip voltage is thus double the final value 11 VDC, which corresponds to the current 220A, and the resistance is the same. The last start-up step is connected on with the speed Δn1 = (440/330) Δn0 = (Imax/Imin) Δn0. This is seen from the equation, which equates the ratio of the slip voltages in the opposite ends of the same start-up step with the currents in those same ends. This follows from the fact that the resistance in the circuit is constant during any start-up step. It is obvious that a step is connected on with the slip speed:
In our example the current ratio is 4/3 and we get the sequence: 100, 133, 177, 237, 316, 421, 561, 749, 998 and 1331, k gets the successive values 0, 1, 2, … k. We see that in practice we need 8 steps. The sequencing of the resistors is got as follows:
The resistor value of the first step is seen to be R1 = (Imax/Imin)RA – RA. In general the rule is:
,
From which the step resistors can be calculated, when k gets the values 0, 1, 2, and so on. This equation follows from the fact that the slip voltage is constant during the change of the start-up step, but the current changes at that time from the minimum value to the maximum.