Power Flow and Efficiency

Fix the direction of travel first. The 100 kW is the shaft output, at the far end of the chain, and everything asked for lies upstream of it. So the whole solution runs backwards along the power flow:

A 500 rpm synchronous speed at 50 Hz means \(P = 120f/N_s = 12\) poles, and the rotor runs at \(N = 500(1-0.018) = 491\) rpm.

Step back over the friction and windage to reach the mechanical power developed inside the machine:

This is the power the rotating field actually produces at the rotor; the bearings and the fan take 1.2 kW of it before anything reaches the coupling.

The rotor copper loss follows from the slip alone. Since \(P_{cu2} = sP_g\) and \(P_m = (1-s)P_g\), dividing one by the other removes \(P_g\):

No resistance, no current and no voltage were needed — the slip carries all the information.

Add it back to get the air-gap power (the rotor input):

Check: \(P_m/P_g = 101.2/103.055 = 0.982 = 1-s\), as it must be.

Cross the air gap backwards by restoring the two stator losses:

The line current now comes from the stator input, remembering the \(\sqrt3\) of a three-phase measurement:

The star connection does not enter here: \(\sqrt3 V_LI_L\cos\phi\) is the total three-phase power whatever the connection.

The efficiency is the ratio of the two ends of the chain:

The completed ledger, which is worth writing out because it is the shape of every problem in this set:

Total loss 8.995 kW on 108.995 kW in — and the rotor copper loss, at 1.855 kW, is only the fourth largest of the five.

Two torques, one machine. The developed torque uses the air-gap power and synchronous speed; the shaft torque uses the output and the actual speed:

The 23 N·m difference is exactly the friction and windage torque, 1200 W at 491 rpm.

The air-gap power is what survives the stator. Both stator losses are taken before the gap:

Now read the slip straight off the loss ratio. Rotor copper loss is defined as the slip fraction of the air-gap power, so the slip is a measured quantity here rather than a speed measurement:

This is the practical value of the \(1 : s : (1-s)\) split: it turns a wattmeter reading into a speed without a tachometer.

The speed and rotor frequency follow at once:

The other two terms of the split:

Check the split adds up: \(1.12 + 26.88 = 28\ \text{kW} = P_g\).

Torque and efficiency. The developed torque uses synchronous speed:

Where the 3.6 kW of loss went, as a check that nothing has been counted twice:

And \(30 - 3.60 = 26.40\ \text{kW}\), the shaft output already found.

Stator input from the line quantities:

The delta connection now matters. The 0.4 Ω is the resistance of one winding, and each winding carries only \(1/\sqrt3\) of the line current:

Using the line current instead would have given \(3\times42^2\times0.4 = 2117\ \text{W}\) — three times too much, and enough to wreck the efficiency figure at the end.

Cross the air gap by removing both stator losses:

The slip comes from the speed, and with it the rotor copper loss:

Mechanical power and shaft output:

Shaft torque and efficiency:

For comparison, the developed torque is \(P_g/\omega_s = 23\,519/157.08 = 149.7\ \text{N·m}\); the 2.3 N·m difference is the friction and windage.

Slip from the two speeds:

Restore the friction and windage to reach the mechanical power the rotor actually develops:

This step must come first. The \(1:s:(1-s)\) split relates \(P_g\) to \(P_m\), not to \(P_{out}\) — the friction and windage sits outside the rotor circuit entirely.

The rotor copper loss from the ratio form of the split:

The rotor input is the sum, and can be checked two ways:

Equivalently the rotor is 96% efficient in itself: \(\eta_{rotor} = 1-s = 0.96\), independent of size, voltage or load.

The two torques:

Overall efficiency, once the stator losses are added at the front of the chain:

Input power:

Stator copper loss. The stator is in star, so the phase current is the line current:

Air-gap power and slip:

The rotor split and the shaft output:

Shaft torque and efficiency:

The developed torque is \(16\,188/157.08 = 103.1\ \text{N·m}\), so 2.7 N·m of the torque produced is spent turning the machine itself.

Where the 2408 W of loss sits:

Fixed losses account for 1100 W and variable losses for 1309 W — close enough to equality that this machine is near its maximum-efficiency load, which is the maximum-efficiency condition of Set 20 for the transformer and holds here for the same reason.

Synchronous speed and slip. A full-load speed of 1425 rpm at 50 Hz can only sit just below 1500 rpm, so the machine has four poles:

The standstill rotor e.m.f. The stator is in star, so each stator phase sees \(440/\sqrt3\), and the turns ratio scales that across to the rotor:

Full-load torque from the standard expression, with \(n_s = 25\) rev/s:

Rotor current, from the same impedance triangle:

Rotor copper loss, three phases of it:

Note the rotor actual resistance and the rotor actual current are used together — both are true rotor quantities, so no referral is needed.

Mechanical power and output. The cleanest route is through the air-gap power, since \(P_{cu2} = sP_g\):

Check against the torque: \(P_m = \omega_mT_{fl} = (2\pi\times1425/60)\times78.87 = 149.2\times78.87 = 11\,770\ \text{W}\) ✓ — the two routes agree exactly, as they must.

Maximum torque occurs where the rotor resistance equals the rotor reactance at that slip:

The slip speed at pull-out is \(s_bN_s = 150\) rpm, and the second form of \(T_{max}\) shows that its value does not contain \(R_2\) at all — only the slip at which it appears does.

Starting current and starting torque, both at \(s = 1\), where the full standstill e.m.f. drives the full standstill reactance:

The picture the six answers paint, which is the whole argument for a starting device:

At standstill the machine draws more than twice its full-load rotor current and produces a quarter of its full-load torque — because at \(s=1\) the rotor power factor is only \(0.4/4.02 = 0.10\), so almost none of that large current is doing useful work.