10.2: Operation- Energy Balance

Now we are ready to see how the induction machine actually works. Assume for the moment that Figure 2 represents one phase of a polyphase system and that the machine is operated under balanced conditions and that speed is constant or varying only slowly. “Balanced conditions” means that each phase has the same terminal voltage magnitude and that the phase difference between phases is a uniform. Under those conditions, we may analyze each phase separately (as if it were a single phase system). Assume an RMS voltage magnitude of \(\ V_{t}\) across each phase.

The “gap impedance”, or the impedance looking to the right from the right-most terminal of \(\ X_{1}\) is:

\(\ Z_{g}=j X_{m} \|\left(j X_{2}+\frac{R_{2}}{s}\right)\label{46}\)

A total, or terminal impedance is then

\(\ Z_{t}=j X_{1}+R_{a}+Z_{g}\label{47}\)

and terminal current is

\(\ I_{t}=\frac{V_{t}}{Z_{t}}\label{48}\)

Rotor current is found by using a current divider:

\(\ I_{2}=I_{t} \frac{j X_{m}}{j X_{2}+\frac{R_{2}}{s}}\label{49}\)

“Air-gap” power is then calculated (assuming a three-phase machine):

\(\ P_{a g}=3\left|I_{2}\right|^{2} \frac{R_{2}}{s}\label{50}\)

This is real (time-average) power crossing the air-gap of the machine. Positive slip implies rotor speed less than synchronous and positive air-gap power (motor operation). Negative slip means rotor speed is higher than synchronous, negative air-gap power (from the rotor to the stator) and generator operation.

Now, note that this equivalent circuit represents a real physical structure, so it should be possible to calculate power dissipated in the physical rotor resistance, and that is:

\(\ P_{s}=P_{a g} s\)

(Note that, since both \(\ P_{a g}\) and \(\ s\) will always have the same sign, dissipated power is positive.) The rest of this discussion is framed in terms of motor operation, but the conversion to generator operation is simple. The difference between power crossing the air-gap and power dissipated in the rotor resistance must be converted from mechanical form:

\(\ P_{m}=P_{a g}-P_{s}\label{52}\)

and electrical input power is:

\(\ P_{i n}=P_{a g}+P_{a}\label{53}\)

where armature dissipation is:

\(\ P_{a}=3\left|I_{t}\right|^{2} R_{a}\label{54}\)

Output (mechanical) power is

\(\ P_{\mathrm{Out}}=P_{a g}-P_{w}\label{55}\)

Where \(\ P_{w}\) describes friction, windage and certain stray losses which we will discuss later.

And, finally, efficiency and power factor are:

\(\ \eta=\frac{P_{\mathrm{Out}}}{P_{i n}}\label{56}\)

\(\ \cos \psi=\frac{P_{\mathrm{in}}}{3 V_{t} I_{t}}\label{57}\)

% -----------------------------------------------------­

% Torque-Speed Curve for an Induction Motor

% Assumes the classical model

% This is a single-circuit model

% Required parameters are R1, X1, X2, R2, Xm, Vt, Ns

% Assumed is a three-phase motor

% This thing does a motoring, full speed range curve

% Copyright 1994 James L. Kirtley Jr.

% ------------------------------------------------------­

s = .002:.002:1; \(\ \quad\) % vector of slip

N = Ns .* (1 - s); \(\ \quad\) % Speed, in RPM

oms = 2*pi*Ns/60; \(\ \quad\) % Synchronous speed

Rr = R2 ./ s; \(\ \quad\) % Rotor resistance

Zr = j*X2 + Rr; \(\ \quad\) % Total rotor impedance

Za = par(j*Xm, Zr); \(\ \quad\) % Air-gap impedance

Zt = R1 + j*X1 +Za; \(\ \quad\) % Terminal impedance

Ia = Vt ./ Zt; \(\ \quad\) % Terminal Current

I2 = Ia .* cdiv (Zr, j*Xm); \(\ \quad\) % Rotor Current

Pag = 3 .* abs(I2) .^2 .* Rr; \(\ \quad\) % Air-Gap Power

Pm = Pag .* (1 - s); \(\ \quad\) % Converted Power

Trq = Pag ./ oms; \(\ \quad\) % Developed Torque

subplot(2, 1, 1)

plot(N, Trq)

title(’Induction Motor’);

ylabel(’N-m’);

subplot(2,1,2)

plot(N, Pm);

ylabel(’Watts’);

xlabel(’RPM’);

Example of Operation

The following MATLAB script generates a torque-speed and power-speed curve for the simple induction motor model described above. Note that, while the analysis does not require that any of the parameters, such as rotor resistance, be independent of rotor speed, this simple script does assume that all parameters are constant.

Example

That MATLAB script has been run for a standard motor with parameters given in Table 1.

Torque vs. speed and power vs. speed are plotted for this motor in Figure 3. These curves were generated by the MATLAB script shown above.

Table 1: Example, Standard Motor

Rating	300	kw
Voltage	440	VRMS, l-l
	254	VRMS, l-n
Stator Resistance R1	.0073	Ω
Rotor Resistance R2	.0064	Ω
Stator Reactance X1	.06	Ω
Rotor Reactance X2	.06	Ω
Magnetizing Reactance Xm	2.5	Ω
Synchronous Speed Ns	1200	RPM