Power Stages and Power Flow | Electrical Machines

i Learning Objectives

By the end of this chapter you should be able to:

Trace the power flow and losses from stator input to shaft output.
Prove that \(P_2 : P_m : P_{cu2} = 1 : (1-s) : s\).
Calculate gross and shaft torque, and torque in synchronous watts.
Derive the torque equation in terms of \(E_2\), \(R_2\), \(X_2\) and \(s\).

Demonstrative Video

Section 63-1

The Power Stages

Power stages of an induction motor — Power flow from stator input to shaft output.

\[\begin{aligned} \text{Stator input } P_1 &= \text{stator output} + \text{stator losses (Cu + iron)}\\ \text{Rotor input } P_2 &= \text{stator output (air-gap power)}\\ \text{Rotor gross output } P_m &= P_2 - \text{rotor Cu loss}\\ \text{Shaft output } P_{out} &= P_m - \text{friction and windage} \end{aligned}\]

Stator iron loss (hysteresis and eddy current) depends on supply frequency and flux density. It is practically constant.
Rotor iron loss is negligible, because the rotor frequency \(sf\) is very small in normal running.
Total rotor copper loss is \(3I_2^2R_2\).

Section 63-2

The Key Ratios

The same gross torque \(T_g\) acts on the rotor. Viewed from the air gap, where the field moves at \(N_s\), and from the rotor, which moves at \(N\):

\[P_2 = T_g\times2\pi N_s, \qquad P_m = T_g\times2\pi N\]

\[\Rightarrow\ \text{rotor Cu loss} = P_2 - P_m = T_g\times2\pi(N_s - N), \qquad \frac{\text{rotor Cu loss}}{P_2} = \frac{N_s - N}{N_s} = s\]

Air-Gap Power Split

\(P_2 : P_m : P_{cu2} = 1 : (1-s) : s\). Rotor efficiency \(= P_m/P_2 = 1 - s = N/N_s\), and \(P_{cu2}/P_m = s/(1-s)\).

If there were no rotor copper loss, the rotor output would equal its input and the rotor would run at synchronous speed. The loss and the slip are inseparable.

Section 63-3

Gross Torque, Shaft Torque and Synchronous Watts

\[\begin{aligned} T_g &= \frac{P_2}{2\pi N_s/60} = 9.55\frac{P_2}{N_s} = 9.55\frac{P_m}{N}\\ T_{sh} &= 9.55\frac{P_{out}}{N} \qquad (N, N_s \text{ in rpm}) \end{aligned}\]

The synchronous watt is the torque that, at the machine's synchronous speed, would develop a power of 1 W. The torque in synchronous watts is therefore simply the rotor input:

\[T_{sw} = P_2 = \frac{\text{rotor Cu loss}}{s} = \frac{P_m}{1-s}\]

1 Worked Example — Power stages

Problem. A 4-pole, 50 Hz induction motor has a rotor input of 40 kW at 4 % slip. Friction and windage loss is 0.8 kW. Find the rotor copper loss, gross output, shaft output, gross torque and shaft torque.

\[P_{cu2} = 0.04\times40 = 1.6\ \text{kW}, \qquad P_m = 38.4\ \text{kW}, \qquad P_{out} = 37.6\ \text{kW}\]

\[N = 1500(1 - 0.04) = 1440\ \text{rpm}\]

\[T_g = 9.55\times\frac{40\,000}{1500} = 254.7\ \text{N·m}, \qquad T_{sh} = 9.55\times\frac{37\,600}{1440} = 249.4\ \text{N·m}\]

Section 63-4

The Torque Equation

Using \(P_2 = 3I_2^2R_2/s\) with \(I_2 = sE_2/\sqrt{R_2^2 + (sX_2)^2}\):

\[P_2 = \frac{3sE_2^2R_2}{R_2^2 + (sX_2)^2}, \qquad T_g = \frac{3}{2\pi N_s}\cdot\frac{sE_2^2R_2}{R_2^2 + (sX_2)^2}\]

With \(E_2 = KE_1\) (\(K\) = rotor/stator turns ratio) and \(N_s\) in rps:

Torque in Terms of Stator EMF

\(T_g = k\,\dfrac{sE_1^2R_2}{R_2^2 + (sX_2)^2}, \qquad k = \dfrac{3K^2}{2\pi N_s}\)

Section 63-5

Chapter Review

Practice Problems

P63.1 A motor's rotor copper loss is 500 W at 5 % slip. Find the air-gap power and the gross mechanical power.

Show answer
\(P_2 = 500/0.05 = 10\) kW, and \(P_m = 9.5\) kW.

Multiple-Choice Questions

MCQ 1. Rotor copper loss equals
(a) \(sP_2\) (b) \((1-s)P_2\) (c) \(P_2/s\) (d) \(P_m/s\)

Show answer
(a) \(sP_2\).
MCQ 2. Torque in synchronous watts equals
(a) shaft output (b) rotor input (c) stator input (d) gross output

Show answer
(b) rotor input.

Conceptual Questions

Why is rotor iron loss neglected in normal running?
Why is a high-slip motor necessarily inefficient?