Starting Methods and Starter Calculations

Full-load conditions first, since everything on this page is quoted against them:

The efficiency, as a check that the data are self-consistent:

Reasonable for a 15 kW machine. Note that \(I_{FL}\) is the line current; the delta-connected windings each carry \(30/\sqrt3 = 17.3\) A.

The starting torque from the current ratio. Torque is proportional to \(I_2'^2R_2'/s\), and the rotor resistance is the same at both conditions, so the ratio is

Exactly full-load torque — which sounds satisfactory until the current is looked at. Note also how the slip enters: a machine with a low-resistance rotor has a small \(s_{FL}\) and therefore a poor starting torque, and vice versa.

The starting kVA, which is what the supply actually sees:

Against a full-load input of \(\sqrt3(400)(30) = 20.8\) kVA. Supply authorities set limits in exactly this form — a typical rule permits direct-on-line starting only up to a few kW on a domestic supply and a few tens of kW on an industrial one.

Why the exchange is so poor. At standstill the power crossing the air gap is \(P_g\) and all of it becomes rotor copper loss, since \(P_m = (1-s)P_g = 0\). Five times the current produces twenty-five times the heat in the rotor, and the shaft sees no more torque than at full load. A motor started direct on line spends the whole run-up dumping into the rotor an energy equal to the kinetic energy it finally acquires.

Fix what does not change. The starter reconnects the windings but does not alter them, so the standstill impedance per phase, \(Z\), is the same in both connections. The supply line voltage \(V_L\) is also the same. Only the voltage across each winding changes.

Delta connection. Each winding sees the full line voltage, and the line current is \(\sqrt3\) times the winding current:

Star connection. Each winding sees \(V_L/\sqrt3\), and the line current is the winding current:

Divide, and the factor of three appears as two separate factors of \(\sqrt3\):

One \(\sqrt3\) comes from the reduced winding voltage and the other from the change in the line-to-phase current relation. Both act in the same direction, which is why the reduction is three and not \(\sqrt3\).

The torque ratio uses winding current, not line current. Torque comes from the ampere-turns of each winding, so \(T \propto I_{ph}^2\) at a given slip:

The two ratios come out equal, but by different routes: the current ratio is \(1/3\) because of the two \(\sqrt3\) factors, and the torque ratio is \(1/3\) because a single \(\sqrt3\) is squared. Confusing the two is the classic error here.

Now the numbers for this machine:

The equivalent tapping. A star connection puts \(1/\sqrt3\) of the line voltage across each winding, so it does exactly what an autotransformer set to

would do, and the arithmetic agrees: \(x^2 = 1/3\) for the torque and \(x^2 = 1/3\) for the supply current. The star–delta starter is an autotransformer starter with one fixed tap and no autotransformer.

Two practical points the arithmetic does not show. First, the machine must be designed to run in delta at the supply voltage, and all six winding ends must be brought out. Second, the changeover from star to delta briefly disconnects the motor; when delta is re-made the machine is running but the currents are not in step with the supply, and a transient of several times full-load current can occur. Closed-transition starters exist to suppress it.

The motor side. The motor sees \(xV\) across an unchanged standstill impedance, so

Torque goes as the square because it goes as the square of the current at a fixed slip.

The supply side, and the reason for the second factor of \(x\). The autotransformer steps the voltage down by \(x\), so it steps the current down by \(x\) as well. Neglecting the starter's own losses and exciting current, volt-amperes in equal volt-amperes out:

This is the whole case for the autotransformer starter. A series resistance or reactance in the stator also gives \(xV\) at the motor, but the supply current then equals the motor current, \(xI_{sc}\), and none of the second reduction is obtained.

At the 60% tapping, with \(I_{sc} = 150\) A and \(T_{st,\text{DOL}} = 99.5\) N·m:

1.8 times full-load current from the supply for 36% of full-load torque. The motor terminals are at \(0.6\times400 = 240\) V during this period.

The tapping for half full-load torque:

The tapping is the square root of the torque ratio required, a relation worth remembering: to double the starting torque, multiply the tapping by \(\sqrt2\).

The stator-resistance alternative. To get the same torque it must also apply \(x = 0.707\) at the motor, so the motor current is again 106.1 A — but that current now comes straight from the supply:

41% more supply current for the same torque, and the difference is burnt as heat in the starting resistors. The autotransformer costs more to buy and is worth it on any machine large enough for the question to arise.

The starter's own rating. During starting the autotransformer carries 106.1 A at 283 V on the motor side, which is \(\sqrt3(283)(106.1) = 52\) kVA — but only for a few seconds. Starting autotransformers are therefore rated for a short-time duty and a stated number of starts per hour, and are physically far smaller than a 52 kVA continuous unit.

Set up the two working expressions. With \(R\) the total rotor resistance per phase,

Differentiating the torque with respect to \(s\) gives the standard condition \(R = sX_2\) at the maximum.

Maximum torque with the rings shorted:

To bring that peak to standstill, put \(s_{\max T} = 1\):

Evaluate the three torques from \(T \propto sR/(R^2+(sX_2)^2)\), working in units of \(E_2^2\) which cancel in the ratios:

The rotor currents, which is where the method earns its reputation:

Adding the resistance raised the torque from 1.31 to 3.41 times full load and at the same time lowered the current from 6.61 to 4.77 times full load. No stator-side starter can do that.

Put the two together:

A factor of 3.6 improvement in torque per ampere. It comes from the power factor: at \(R = X_2\) the rotor circuit is at 0.707 lagging instead of 0.196, so far more of the rotor current is in phase with the rotor emf and therefore producing torque.

A starter step at 50% slip. As the machine accelerates the resistance is cut out in stages, each stage chosen so the torque stays high. For the peak to sit at \(s = 0.5\):

And the torque there is \(0.5(0.30)/(0.09+0.09) = 0.8333\) in the same units — identical to the standstill maximum, as it must be. The peak value of the torque is \(3E_2^2/(2\omega_sX_2)\) and does not contain \(R\) at all; the resistance only chooses where the peak sits.

Full-load quantities:

Direct-on-line figures, from the current ratio:

Apply the one-third factor to both:

Test both requirements:

Acceptable on both counts — but the torque margin is thin. A 20% margin over breakaway torque means the machine accelerates slowly, and the starter must be timed to hold star long enough for the motor to reach a speed at which the transition to delta does not produce a large current surge.

The limiting slip. Setting the star–delta torque exactly equal to the requirement:

A motor of the same current ratio but 3% slip — a slightly more efficient design — would not start this load on a star–delta starter. The better machine is the worse starter, because low slip means low rotor resistance and low rotor resistance means poor standstill torque.

The heavier compressor. With 0.60 pu of torque needed and 1.44 pu available direct on line:

Exactly at the supply limit, and it had to be: the torque and the supply current both scale as \(x^2\), so their ratio is a constant of the machine. Asking for \(0.60/0.48 = 1.25\) times the star–delta torque necessarily costs 1.25 times the star–delta current, and \(1.25\times2.0 = 2.5\).

Where that leaves the design. The 65% tapping is standard, so the job can be done — but there is now no margin at all on current. Any further increase in breakaway torque forces a change of machine: a higher-slip cage design, a deep-bar or double-cage rotor with better standstill torque, or a slip-ring motor with rotor resistance as in Problem 4.

Recall the base quantities for the cage machine: \(I_{FL} = 30\) A, \(I_{sc} = 150\) A, \(T_{FL} = 99.5\) N·m and \(T_{st,\text{DOL}} = 99.5\) N·m, so \(I_{sc}/I_{FL} = 5.0\) and \(T_{st}/T_{FL} = 1.00\).

The comparison, every entry obtained from \(I_{\text{motor}} = xI_{sc}\) and \(T = x^2T_{st,\text{DOL}}\), with the supply current \(x^2I_{sc}\) for the transforming starters and \(xI_{sc}\) for the resistance starter:

\(^\dagger\) Figures for the slip-ring machine of Problem 4, quoted in per unit of its own full-load values; a cage motor cannot be started this way at all.

Read the last column. The four stator-side transforming methods give identical torque per supply ampere, 0.200 pu, because both quantities scale as \(x^2\):

The stator-resistance starter is worse by exactly the factor \(x\), and the rotor-resistance method is in a different class altogether.

The other considerations, which the table of numbers cannot show:

Choosing for this machine. Suppose the load needs 0.35 pu of breakaway torque and the supply allows 2 pu of current. Star–delta gives 0.333 pu — just short. The 60% autotransformer tap gives 0.360 pu at 1.80 pu of current, and is the correct answer. The 70.7% tap would work too but at 2.50 pu, over the limit; direct on line is far over it; and a stator resistance set for 0.36 pu of torque would draw \(0.6\times5.0 = 3.0\) pu.

That is the whole design procedure: convert the load's torque requirement to a value of \(x\), read off the supply current, and check it against the limit. If no \(x\) satisfies both, the machine is wrong for the duty.