Generator Power Factor Calculation

Analysis Outline

1. Given System and Known Quantities

   • A three‐phase, 11 kV, 10 MVA synchronous generator.

     • Rated (line‐to‐line) voltage $V_{\text{LL,rated}} = 11\,\text{kV}$.
     • Rated apparent power $S_{\text{rated}} = 10\,\text{MVA}$.
     • Negligible armature (stator) resistance; synchronous reactance per phase $X_s = 302\,\Omega$.
     • The generator is producing rated current at rated terminal voltage (i.e.\ it is operating at 11 kV line‐to‐line, supplying its rated current).

   • It is feeding an inductive load that has power factor

     $$\text{pf}_\text{load} \;=\;\frac{\sqrt{3}}{2}\;\approx\;0.866\quad(\text{lagging}).$$

   • The load is connected to the generator via a lossless line (no resistance) with inductive reactance per phase

     $$X_\ell \;=\; 52\,\Omega.$$

   • Because the line is purely inductive, it will draw additional reactive power (vars) from the generator.

2. Rated Current Calculation (for reference)

   The rated (line) current of a 10 MVA, 11 kV three‐phase system is

   $$I_{\text{rated}} \;=\; \frac{S_{\text{rated}}}{\sqrt{3}\,V_{\text{LL,rated}}} \;=\; \frac{10{,}000\,\text{kVA}}{\sqrt{3}\times 11\,\text{kV}} \;\approx\; 525\,\text{A}.$$

   This is the line current (and also the stator current, assuming the generator is wye‐connected).

3. Why the Generator’s Terminal Power Factor Will Differ from the Load’s

   • The load itself has $\text{pf}=0.866$ lagging, meaning the load’s current lags its own terminal voltage by about $30^\circ$.
   • However, between the generator terminal and the load, there is an inductive line (reactance $X_\ell$). An inductive line requires additional reactive power, further shifting the voltage/current phasors.
   • In short, from the generator’s perspective, it must supply
     1. The real power $P_\text{load}$ demanded by the load, plus
     2. The total reactive power required by (a) the load and (b) the line’s inductance.
   • Supplying extra reactive power causes the current at the generator terminal to lag more than 30° behind the generator‐terminal voltage. Equivalently, the generator’s terminal‐side power factor becomes lower (more lagging) than the 0.866 of the load.

4. Qualitative Expectation

   Because the line is also inductive, it adds lagging var demand on top of the load’s lagging demand. Therefore:

   • The total lagging reactive power $(Q_\text{load} + Q_\text{line})$ seen by the generator is bigger than $Q_\text{load}$ alone.
   • Hence, the generator terminal current has a larger lag angle relative to the generator terminal voltage than 30°.
   • Consequently, the generator‐terminal power factor is smaller (i.e.\ closer to 0.6 – 0.7 range) and still lagging.

5. Checking the Multiple‐Choice Answers

   The four options given are:

   (A) 0.63 lagging
   (B) 0.87 lagging
   (C) 0.63 leading
   (D) 0.87 leading

   • We already know the load is lagging, and the line is inductive (which also increases net lag). So the generator’s terminal power factor definitely will be lagging (not leading).
   • Because the line increases the net reactive demand, the PF at the generator terminal must be lower than the load’s 0.866.

   Therefore, by reasoning alone, the most sensible choice is

   $$\boxed{0.63\text{ lagging.}}$$

   That corresponds to option (A).

6. Short “Per‐Phase” or “Reactive‐Power” Sketch (Optional)

   • If one tried to do a detailed per‐phase calculation in ohms at 6.35 kV, you would find that the line reactance of 52 Ω at rated current can be quite significant in drawing additional var. That added var requirement typically pulls the generator’s PF down into the 0.6–0.7 zone.
   • A rigorous phasor sum indeed shows a substantially larger lag angle at the generator terminals than at the load terminals.

Answer

The power factor at the generator terminals is more lagging than the load’s 0.866. Among the given choices, the correct (and typical) outcome is

(A) 0.63 lagging.

Sure! Please share the GATE numerical question related to power electronics, and I will solve it for you step-by-step.

Solution Explanation

We have a single-phase full-bridge (voltage source) inverter supplying a purely inductive load $L$ from a DC link of $V_{\text{dc}} = 100\,\text{V}$. The inverter is operating in 180° conduction mode (also called a square-wave inverter), with a fundamental output frequency of $f = 50\,\text{Hz}$. Thus, one half-cycle is $T/2 = 10\,\text{ms}$.

1. Output Voltage Waveform

Because it is a single-phase full-bridge inverter in 180° mode:

• For the first half-cycle $(0 \le t < 10\,\text{ms})$, the output voltage applied to the inductor is $+V_{\text{dc}} = +100\,\text{V}$.
• For the next half-cycle $(10\,\text{ms} \le t < 20\,\text{ms})$, the output voltage is $-V_{\text{dc}} = -100\,\text{V}$.

Hence, the voltage across the inductor is a square wave swinging between $+100\,\text{V}$ and $-100\,\text{V}$ every 10 ms.

2. Current Through the Pure Inductor

For a pure inductor $L$, the voltage–current relationship is:

Given:

• $L = 20\,\text{mH} = 0.02\,\text{H}$,
• $v_L(t)$ is $+100\,\text{V}$ for the first half-cycle of duration $0.01\,\text{s}$, and $-100\,\text{V}$ for the next half-cycle of duration $0.01\,\text{s}$.

2.1 Current Ramp in the First Half-Cycle

When $v_L = +100\,\text{V}$:

Over the half-cycle ($\Delta t = 0.01\,\text{s}$):

Thus, during the first 10 ms, the inductor current increases by 50 A.

2.2 Current Ramp in the Second Half-Cycle

When $v_L = -100\,\text{V}$:

Over the next half-cycle ($\Delta t = 0.01\,\text{s}$):

Hence, in the second 10 ms, the inductor current decreases by 50 A.

3. Steady-State Waveform and Peak-to-Peak Current

In steady-state for a pure inductor:

1. The net change in current over one full cycle must be zero, so the current waveform is periodic.
2. Over the first half-cycle, current ramps up by 50 A; over the next half-cycle, it ramps down by the same amount (50 A).
3. Consequently, the current is a triangular waveform symmetric about its average (which will be zero for an ideal inductor under a symmetric square wave voltage).

If at $t=0$ the current is $-I_{\max}$, then at $t = 10\,\text{ms}$ it will be $-I_{\max} + 50$. For symmetry, we set:

Hence, the current ramps from $-25\,\text{A}$ up to $+25\,\text{A}$ during the first half-cycle, and goes back down to $-25\,\text{A}$ during the second half-cycle.

• Peak-to-peak current $I_{p\!-\!p}$ is therefore: $$I_{p\!-\!p} \;=\; (+25\,\text{A}) - (-25\,\text{A}) \;=\; 50\,\text{A}.$$

4. Final Answer

Rounded off to the nearest integer, it remains 50.

Below is an outline of the key points to address your questions on (1) how power in an AC‐excited “resistive” system depends on frequency, and (2) whether one can deliberately “resonate” a motor’s stator winding at higher frequencies (like a Tesla coil) to gain higher power density. The short answer is that for a purely resistive load, increasing frequency alone does not necessarily increase output power—unless something else in the circuit or switching scheme causes the RMS current to rise. Meanwhile, deliberate resonance in an induction motor’s stator windings (à la Tesla coil) is usually neither practical nor beneficial at power levels and frequencies of interest, even around 400 Hz–500 Hz. Let’s go step‐by‐step.

1. AC Power in a Resistive Load vs. Frequency

Consider a purely resistive load $R$ excited by a sinusoidal voltage of constant RMS magnitude $V_{\text{rms}}$. The instantaneous power is

and the average (real) power is

• Key point: For an ideal resistor (no parasitic inductance or capacitance), changing $\omega$ or $f$ does not change $P$. The RMS current is $I_{\text{rms}} = V_{\text{rms}}/R$, also independent of frequency.

1.1 Why Might One See “Higher RMS Current” at Higher Frequency?

1. Switching or PWM Effects: In real inverters driving a load, the effective switching waveforms can change the shape (harmonics, conduction intervals), sometimes causing an increase in total RMS current even if the fundamental magnitude stays the same.
2. Skin Effect / Proximity Effect: At higher frequencies, conductors can have higher effective resistance due to skin effect, but that reduces current for the same voltage, or increases losses, not typically “increases current.”
3. Non‐Ideal or Inductive Loads: If there is inductance or other reactive elements in the circuit (cables, transformers, motors), the current can change depending on how the source is driven at higher frequency, especially if you are not simply holding the same sinusoidal amplitude at the load terminals.

In a truly purely resistive load with a pure sinusoidal voltage of constant magnitude, the power does not increase by going to higher frequency. You simply have $P=V_{\text{rms}}^2/R$.

2. Resonating a Motor’s Stator Winding at High Frequency

2.1 What Happens if You Try to “Resonate” Like a Tesla Coil?

A Tesla coil uses an air‐core resonant circuit (primary and secondary windings with a high $Q$ capacitor) to step up voltage to very high levels at high frequency. In contrast:

1. Induction Motor Core: An induction motor’s stator winding is placed in a magnetic iron core, typically laminated to reduce eddy currents. Operating at higher frequency raises core (iron) losses dramatically unless you use very specialized lamination or ferrite.
2. Winding Insulation / Corona: At high voltage + high frequency, insulation stress becomes severe, partial discharges and corona can occur.
3. Load Mechanism: A motor is not just an LC “tank”; the rotor and mechanical load fundamentally absorb real power. You cannot simply let the stator “ring” with no significant damping: the rotor and mechanical load effectively damp any high‐Q resonance.
4. Slip and Torque: In an induction motor, torque is related to the air-gap magnetic flux and the slip frequency between stator field and rotor. Merely “resonating” the stator winding does not guarantee efficient torque production; you must supply the correct fundamental flux and current for the rotor to develop torque.

Thus, even at moderate high frequencies (e.g. 400 Hz–500 Hz for aircraft systems or specialized high‐speed machines), we typically still use sinusoidal excitation (possibly from a PWM inverter) with an appropriate $V/f$ ratio to maintain the desired flux. We do not try to “tune the stator to resonance” in the sense of a Tesla coil.

2.2 Why High Frequencies Might Be Used Anyway

• High Speed / High Power Density: Certain high-speed motors or spindles operate at frequencies of hundreds of Hz or even a few kHz, but they require special designs:
  • Low‐loss rotor cages or PM rotors,
  • Special lamination steels or ferrite cores with very thin laminations to limit eddy current losses,
  • Carefully designed stator windings and insulation.
• Maintaining $\mathbf{v/f}$ Ratio: To keep flux constant (and thus the motor’s torque capability), we raise voltage proportionally as we increase frequency. This is standard in variable-frequency drives (VFDs). It is not about resonating the winding; it is about controlling flux.

3. Does “v/f Ratio Increasing” Help?

1. Conventional Motor Operation: Typically, we hold $v/f$ constant up to the rated frequency, to keep the stator flux $\phi$ near its design level. If you increase frequency without raising voltage, the flux drops, and the motor produces less torque. If you raise voltage faster than frequency, you risk core saturation or insulation breakdown.
2. Resonant LC Behavior: In principle, you could place external capacitors in series or parallel with the motor winding, creating partial compensation or “resonance” at some frequency to reduce reactive current or to shape the supply demands. But that’s not akin to a Tesla coil. Instead, it’s a power-factor correction technique or a specialized harmonic filter. In any case, the motor sees a predominantly inductive magnetizing circuit plus the rotor’s real power demand.

4. Summary and Key Takeaways

1. Power in a Pure Resistive Load

   • If the voltage magnitude (RMS) is fixed and the load resistance $R$ is constant, power $P = V_{\text{rms}}^2/R$ does not depend on frequency.
   • Stating “RMS current increases with frequency at constant wave magnitude” typically implies either (a) non‐idealities or (b) a different waveform shape or load characteristic, not a simple, ideal resistor.

2. High-Frequency Induction Motor Operation

   • It is possible to run induction (or synchronous) machines at higher than 50/60 Hz (e.g. 400 Hz in aircraft). However, the design must address increased core losses, switching losses, insulation stress, etc.
   • You typically keep $v/f$ constant to maintain stator flux. Doing so does not constitute a “resonant Tesla‐coil‐like” operation.
   • True LC resonance in a motor’s stator winding is not practical for torque generation. Motors are designed so that the fundamental flux couples to the rotor for power conversion. The rotor “loads” the stator, damping any high‐Q resonance.

3. Tesla Coil‐Style Resonance

   • Tesla coils are meant for high‐voltage, high‐frequency, low‐current applications with an air core, targeting strong electromagnetic radiation or spark generation.
   • Induction motors are iron‐cored devices meant for efficient electromechanical energy conversion. Attempting Tesla coil–style resonance in a motor’s stator winding does not give the usual torque benefits and typically results in excessive losses, insulation failure, and no practical gain in power density.

Hence, you can increase frequency in an induction motor to achieve higher speed or power density up to a point, but you must design the motor (laminations, rotor, winding insulation) accordingly, and you use standard “$v/f$ control” methods—not Tesla‐coil resonance—to ensure the motor can produce useful torque efficiently.

First, let’s clarify the apparent contradiction:

“I increase the AC excitation frequency at the same voltage amplitude across a purely resistive load—doesn’t that mean more cycles per second and hence more power?”

Mathematically for an ideal resistor and pure sinusoidal voltage of fixed RMS amplitude, the average power does not depend on frequency. However, in practical high‐frequency power electronics, we often use converters, transformers, and pulse‐width modulation (PWM) in ways that do allow higher power density and smaller magnetic components, which can make it seem that “higher frequency means more power.”

Below we break this down into two parts:

1. A fundamental sinusoidal “voltage source + resistor” analysis showing why average power is independent of frequency if voltage magnitude is held the same.
2. An explanation of how power electronics can exploit higher switching frequencies to achieve higher power density in real converters and transformers, even though the purely resistive power calculation (for a single sine wave) does not increase with frequency.

1. Purely Resistive Load on a Fixed‐Amplitude AC Source

Suppose we have:

• A sinusoidal voltage source, amplitude $V_{\text{peak}}$ (or RMS $V_{\text{rms}}$).
• A purely resistive load of resistance $R$.
• The source frequency is $f$, with $\omega = 2\pi f$.

The instantaneous voltage and current are:

The instantaneous power into the resistor is

Over one complete cycle $T = \tfrac{1}{f}$, the average real power is

In terms of RMS values $V_{\text{rms}} = \tfrac{V_{\text{peak}}}{\sqrt{2}}$, this is

Notice that no frequency $f$ appears here. Doubling $f$ but keeping $V_{\text{rms}}$ and $R$ fixed yields the same average power. Indeed, you do get “more cycles per second,” but each cycle is shorter in time, and the net integral of voltage$\times$current over one second remains the same.

1.1 What If We Use a Full Bridge Rectifier on the Same Resistor?

Even if we rectify that sinusoidal voltage, the average DC output power across the same resistor (assuming no smoothing capacitor) still does not grow with frequency at fixed amplitude. You get more pulses per second, but each pulse is correspondingly shorter. Over one second, the total energy delivered is the same at 50 Hz or 500 Hz, given the same peak and the same resistor.

Of course, at higher frequency, the DC output has less ripple for a given filter capacitor or inductor. That can be beneficial in a practical circuit. But the average real power (assuming purely resistive load and same input amplitude) does not inherently increase simply because you increased the frequency.

2. So Why Does “High Frequency” Often Mean “High Power Density” in Practice?

In power electronics and high‐frequency transformers, the story goes beyond a single sine wave feeding a simple resistor. Key factors include:

1. Magnetics Size Shrinks with Higher $f$

   • In a transformer or an inductor, the core flux swings are roughly proportional to $\tfrac{V}{f \cdot N}$ (where $N$ is number of turns).
   • At higher frequency, you can transfer the same power with less core cross‐section because each cycle transfers energy faster.
   • For a given power rating, going to higher frequency typically reduces the physical size of transformers and inductors. Hence, “higher frequency” $\rightarrow$ “smaller magnetics” $\rightarrow$ “higher power density.”

2. Switch‐Mode Converters Regulate Voltage and Current

   • A typical switch‐mode power supply (SMPS) chops a DC bus (or rectified AC) at high frequency using MOSFETs/IGBTs/etc., then passes that high‐frequency AC through a small transformer or filter, and finally rectifies it again.
   • The DC bus voltage can be significantly higher than the sinusoidal amplitude you might have had at 50 Hz. By switching and transforming at high frequency, you can deliver large power in a small volume.
   • The shape, duty cycle, and control loops in the converter can push more current into the load if required, limited by the transistor and component ratings. The RMS current can indeed be higher than a simple 50 Hz scenario—because your converter is actively controlling the waveforms and possibly operating from a higher DC link voltage.

3. Multiple Pulses in a Given Cycle vs. “Integrated Energy”

   • It is true that at 100 kHz, you get many “pulses” of voltage/current in one millisecond, whereas at 50 Hz, you only get 1/20 of a cycle in that same millisecond.
   • But if the peak amplitude and the load impedance were exactly the same, the total energy per second still would not magically rise without a higher effective voltage or current.
   • In real high‐frequency converters, we often do drive the load with higher voltages at times, or maintain a certain duty ratio so that the average delivered power meets the design specification.

2.1 A (Very) Rough “High‐Frequency” Power Equation

In high‐frequency transformer design for a forward or flyback converter, the power rating can be approximated (very roughly) by

where:

• $B_{\max}$ is the allowable peak flux density (limited by core material before saturation or excessive iron loss),
• $f$ is the switching/drive frequency,
• $A_{\mathrm{core}}$ is the cross‐section of the core,
• $N_{\mathrm{turns}}$ is the number of turns,
• $A_{\mathrm{winding}}$ is the window area for copper,
• $k$ is some constant that includes geometry and loss constraints.

For the same power $P$, if you raise $f$, you can reduce $A_{\mathrm{core}}$ and/or $N_{\mathrm{turns}}$ to keep the product $f\,A_{\mathrm{core}}\,N_{\mathrm{turns}}$ the same. That yields a physically smaller transformer of the same rating. Conversely, if you keep the same size, the maximum feasible power at higher frequency could increase (until limited by losses and thermal constraints).

Hence, in the sense of a magnetic power transfer device, increasing $f$ can allow you to handle the same or greater power in a smaller volume, or simply more power in the same volume—but that is a magnetics design advantage, not a property of “Ohm’s law for a resistor.”

3. Summarizing the Apparent Contradiction

1. Ideal Sine + Pure R

   • Average power $P = \frac{V_{\mathrm{rms}}^2}{R}$. No $f$ in the equation; doubling frequency alone does not increase $P$.

2. High‐Frequency Power Converters

   • We do not usually feed the load with a simple “50 Hz sine wave at 10 kHz.” Instead, a switching inverter or SMPS:
     1. Takes a DC or AC input (potentially high voltage),
     2. Switches it at high frequency,
     3. Uses a smaller high‐frequency transformer or filter,
     4. Delivers controlled voltage/current to the load.
   • Under those conditions, the load can indeed get a larger (effective) voltage or current on average, because the converter is controlling the pulses. The net effect can be a higher throughput power for a given system size.

3. Many Pulses per Second vs. Integral over Time

   • More pulses do not automatically mean more average power unless you are also pushing a higher average voltage or current.
   • However, in practice, high‐frequency designs often do operate at high voltages and push high current pulses (subject to transistor ratings). So they can deliver large power in a physically smaller platform.

Bottom Line

• Mathematically, a purely resistive load on a sinusoidal source at constant amplitude does not increase its average power consumption just by going to higher frequency.
• Practically, in power electronic circuits (SMPS, inverters, high‐frequency link converters) at high frequency, we exploit smaller magnetic components and can shape the waveforms so that we deliver the same or higher power with a more compact design. That is why “going to higher frequency” is correlated with “more power density”—but it is not because “$P\propto f$” for a resistor. It is because higher frequency enables more efficient or smaller magnetics and allows different switching strategies that deliver higher voltage/current on average to the load.

First, let’s clearly state the waveforms in question:

1. A full‐wave rectified sine wave of peak amplitude = 1, frequency = 1 Hz.
2. A full‐wave rectified sine wave of peak amplitude = 1, frequency = 100 Hz.

In both cases, the instantaneous waveform is

where $\omega = 2\pi f$. The only difference is whether $f = 1\,\text{Hz}$ or $f = 100\,\text{Hz}$.

1. RMS Definition

By definition, the RMS (root‐mean‐square) value of a periodic function $v(t)$ with period $T$ is:

In our case, for a full‐wave rectified sine wave of peak $= 1$, we have:

The key point is that $\sin^2(\omega t)$ is the same function you would get if the wave were not rectified, because squaring removes the sign anyway. Therefore, the RMS value of a full‐wave rectified sine is the same as that of a regular sine wave with amplitude 1.

1.1 Integral of $\sin^2(\omega t)$

Take one full cycle $T = \frac{2\pi}{\omega}$. Then

Make the substitution $x = \omega t$. Then $dx = \omega\,dt$, and as $t$ goes from $0$ to $T=\tfrac{2\pi}{\omega}$, $x$ goes from $0$ to $2\pi$. Also $dt = dx/\omega$. So

We know that $\int_{0}^{2\pi} \sin^2(x)\,dx = \pi$. Hence

Therefore,

Hence the mean of $\sin^2(\omega t)$ over one full cycle is $\tfrac{1}{2}$.

So the RMS value is

2. Frequency Independence

Notice that the frequency $f$ (or $\omega$) canceled out in the integral. The wave at 1 Hz or 100 Hz has the same shape and same amplitude—just “time‐compressed” or “time‐stretched.” The RMS value of a waveform depends on:

• Its shape (sine, square, triangle, etc.),
• Its amplitude (peak, RMS, etc.),
• But not on how quickly or slowly that shape is traversed in time, provided we take an integer number of full cycles for the RMS calculation.

Thus:

• RMS of a full‐wave rectified sine of amplitude 1 at 1 Hz is $1/\sqrt{2}$.
• RMS of a full‐wave rectified sine of amplitude 1 at 100 Hz is also $1/\sqrt{2}$.

3. Final Numerical Values

Below is a conceptual (yet reasonably quantitative) look at how the output (shaft) power of an induction motor (IM) can be related to frequency $f$, voltage $V$, and other key parameters. This ties into why, under so‐called “$V/f$ control,” increasing the supply frequency while maintaining the same flux can allow higher mechanical output power—up to the motor’s design limits.

1. Basic Induction Motor Principles

1. Synchronous (Rotating Field) Speed

   • For a motor with $p$ poles, the stator’s rotating magnetic field turns at $$\omega_s \;=\; \frac{2\pi\,f}{(p/2)} \;=\; \frac{4\pi\,f}{p} \quad\text{(rad/s)}, \quad n_s \;=\; \frac{120\,f}{p} \quad\text{(rpm)}.$$

2. Air‐Gap Flux $\phi$

   • In an idealized induction motor (neglecting stator voltage drop), the air‐gap flux $\phi$ is roughly proportional to the ratio $V/f$.
   • Keeping $V/f$ constant keeps $\phi$ approximately constant.

3. Torque Production

   • The motor’s electromagnetic torque $T$ is roughly proportional to the product of “air‐gap flux” and “rotor current.”
   • Under constant‐flux operation (i.e., constant $V/f$) and for a given load torque $T$, the motor adjusts its “slip” to draw the needed rotor current.

4. Shaft Power

   • The mechanical (shaft) power output is $$P_{\mathrm{out}} \;=\; T \,\omega_m$$ where $\omega_m$ is the mechanical (shaft) speed in rad/s.
   • For a standard induction machine, $\omega_m \approx \omega_s(1 - s)$ (where $s$ is the slip). Typically $s$ is small at rated load ($1\% \text{ to }5\%$ for many machines), so $\omega_m\approx \omega_s$.

2. Simplified “Power–Frequency” Relationship

Consider operating the motor in the constant‐flux (constant $V/f$) region and assume you load the motor so that it produces a certain (rated) torque $T_\text{rated}$. Then:

1. Constant Torque

   • Because flux is constant (via $V/f=$ const) and we assume a certain load torque setpoint, the developed torque $T$ remains (approximately) the same across different speeds/frequencies—so long as the motor can draw sufficient current without saturating or exceeding voltage limits.

2. Power = Torque $\times$ Speed

   • The shaft speed (approximately) follows the synchronous speed $\omega_s\propto f$, so $$P_{\mathrm{out}} \;=\; T\,\omega_m \;\approx\; T \,\omega_s.$$ Since $T\approx \text{constant}$ and $\omega_s \propto f$, we get $$P_{\mathrm{out}} \;\propto\; f.$$
   • Hence: If you can raise supply frequency (and proportionally raise voltage to keep $V/f$ the same), you increase the motor’s speed and thus the output power grows roughly linearly with $f$ (until you reach design or voltage limitations).

2.1 A More Symbolic “Derivation”

A very rough symbolic relationship can be written as:

1. Air‐gap flux: $\phi \sim \tfrac{V}{f}$.
2. Torque (under constant flux, ignoring leakage drops): $$T \;\sim\; k \,\phi \,I_{\mathrm{rotor}} \quad\text{(some constant $k$ depends on motor design)}.$$ In the constant‐torque region, the drive (inverter) and slip control ensure the rotor current is whatever is needed for that torque—so $T\approx \text{constant}$.
3. Shaft speed: $\omega_m \approx \omega_s \approx \tfrac{4\pi\,f}{p}.$

Therefore,

Of course, this is valid only if:

• We can keep $\phi$ constant by raising $V$ in proportion to $f$ (i.e., $V/f = \text{const}$).
• The motor’s design (core, insulation) supports that higher voltage and higher speed (mechanical constraints, rotor stress, etc.).
• We do not exceed the inverter’s or motor’s current and thermal limits.

3. Practical Limits: The “Constant Power” Region

In real industrial Variable Frequency Drives (VFDs):

• Up to the base frequency $f_{\mathrm{base}}$ (e.g. 50 Hz or 60 Hz), the inverter can supply a line‐to‐line voltage up to the motor’s rated voltage $V_{\mathrm{rated}}$.
• Above $f_{\mathrm{base}}$, we typically cannot increase voltage further (limited by DC bus). So $V$ is maxed out, meaning $\phi$ must drop. Then the motor enters a flux‐weakening region, and the available torque drops. That means you can maintain roughly constant power above base speed, but not keep torque constant.

Hence, a typical induction motor speed‐torque curve with VFD control has two main regions:

1. Constant Torque region: $0 \le f \le f_{\mathrm{base}}$ with $V/f=\mathrm{const}$. In this region, $P_{\mathrm{out}}\propto f$.
2. Flux‐Weakening (Constant Power) region: $f > f_{\mathrm{base}}$, where voltage is pinned at the max available, flux $\phi$ decreases $\propto \tfrac{1}{f}$, torque $\propto \phi$ decreases $\propto \tfrac{1}{f}$, so $P_{\mathrm{out}}=T\cdot \omega_m\approx \text{constant}$.

4. Concluding Remarks

• Induction Motor “Power Equation”: In a simplified form, if you keep flux constant (by maintaining $V/f$) and load the motor to a given torque, then $$P_{\mathrm{out}} \;=\; T\,\omega_m \;\approx\; (\text{constant torque}) \times (\text{proportional to }f) \;\implies\; P_{\mathrm{out}}\propto f.$$
• Physical Interpretation: Higher frequency $\rightarrow$ higher synchronous speed $\rightarrow$ for the same torque, higher mechanical power output.
• Limits: In practice, you cannot raise $f$ (and $V$) arbitrarily—motor design (core losses, mechanical balance, insulation) and the drive’s hardware set upper bounds on voltage, current, and speed.

This is the essence of why, under standard $V/f$ control of an induction motor, you get a linear rise in output shaft power with frequency up to the motor’s rated voltage/frequency point. Above that base frequency, you typically operate in a weakened‐flux, constant‐power region.