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ECE 2207: 2020 Semester Final: Explanation Style Answers

RUET · ECE Dept · 2nd Year Odd Semester 2020

Deep tutorial-style explanations focusing on first-principles physics, "why over what", step-by-step logic, and practical engineering intuition. Core cross-cutting theory topics are detailed in 2018_2024_answer.md.

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SECTION - A (Transformers: Q1 to Q4)

Question 1

Q1(a): Why Are Transformer Cores Laminated?

The Physics of Eddy Current Loss:

The iron core sits in an alternating magnetic field. By Faraday's Law, any closed conducting path enclosing alternating flux has an EMF induced in it. Because electrical steel is a metal with finite electrical conductivity, circulating eddy currents swirl throughout the core perpendicular to the magnetic flux lines, generating parasitic $I^{2}R$ heat loss.

The power dissipated by eddy currents in a solid core is: $$P_e=k_e{f}^2{B}_m^2{t}^2{V}_{\text{core}}$$ where $t$ is the thickness of the conducting sheet perpendicular to the flux. Notice that eddy loss scales with the square of thickness ($t^2$)!

How Lamination Eliminates the Loss:

If a solid core of thickness $T$ is sliced into $n$ thin sheets (laminations) of thickness $t=T/n$, each insulated from the next by a thin layer of varnish or chemical oxide: $$P_{e,\text{laminated}}=n\times k_e{f}^2{B}_m^2{\left(\frac{T}{n}\right)}^2=\frac{k_e{f}^2{B}_m^2{T}^2}{n}=\frac{P_{e,\text{solid}}}{n}$$ By dividing a 35 mm solid core block into 100 laminations of $0.35\text{ mm}$ thickness ($n=100$), eddy current power loss plummets by a factor of 100!

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Q1(b): Equivalent Circuit Derivation

See full step-by-step first-principles derivation with diagrams below in Q3(c): Step-by-Step Equivalent Circuit of a Transformer Referred to Primary Side.

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Q1(c): Working of a Transformer Under No-Load

When the secondary winding is open-circuited ($I_2=0$), the primary acts as an iron-cored inductor:

1. Primary draws a small no-load current $I_0$ (2% to 6% of rated current).
2. $I_0$ splits into two orthogonal physical components:
   • Magnetizing Component ($I_m$): Quadrature component in phase with core flux ${\Phi }_m$ (90° lagging applied voltage $V_1$). Provides the MMF ($N_1{I}_m$) required to magnetize the ferromagnetic core.
   • Core-Loss Component ($I_c$): In-phase active component that draws real power to supply hysteresis and eddy-current dissipation in the laminations ($P_0=V_1{I}_c$).
3. The alternating core flux $\Phi (t)$ induces primary counter-EMF $E_1\approx V_1$ and secondary output voltage $V_2=E_2=4.44f{N}_2{\Phi }_m$.
4. No-load power factor is very low: $\cos {\varphi }_0=\frac{I_c}{I_0}\approx 0.1\text{ to }0.25$.

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Q1(d): Primary Current and Power Factor Calculation Under Load

Given Data:

• Turns ratio $a=\frac{N_1}{N_2}=4$ (step-down)
• No-load excitation: $I_0=10\text{ A}$ at $\cos {\varphi }_0=0.2\text{ lagging}⟹{\varphi }_0=78.46°$
• Secondary load: $I_2=200\text{ A}$ at $\cos {\varphi }_2=0.85\text{ lagging}⟹{\varphi }_2=31.79°$

Step-by-Step Phasor Addition (Taking $V_1$ as reference along $+Y$ or horizontal $+X$):

Resolving into active (in-phase with $V_1$) and reactive (quadrature, lagging $V_1$ by 90°) components:

1. No-Load Current Components:
   • In-phase: $I_{0w}=I_0\cos {\varphi }_0=10\times 0.2=2.00\text{ A}$
   • Quadrature: $I_{0m}=I_0\sin {\varphi }_0=10\times \sin (78.46°)=10\times 0.9798=9.80\text{ A}$
2. Reflected Load Current Components ($I_2^{\prime }=I_2/a=200/4=50\text{ A}$):
   • In-phase: $I_{2w}^{\prime }=I_2^{\prime }\cos {\varphi }_2=50\times 0.85=42.50\text{ A}$
   • Quadrature: $I_{2m}^{\prime }=I_2^{\prime }\sin {\varphi }_2=50\times \sin (31.79°)=50\times 0.5268=26.34\text{ A}$
3. Total Primary Current Components:
   • Total in-phase component: $I_{1w}=2.00+42.50=44.50\text{ A}$
   • Total quadrature component: $I_{1m}=9.80+26.34=36.14\text{ A}$
4. Primary Current Magnitude and Operating Power Factor: $$I_1=\sqrt{(I_{1w}{)}^2+(I_{1m}{)}^2}=\sqrt{(44.50{)}^2+(36.14{)}^2}=\sqrt{1980.25+1306.10}=\sqrt{3286.35}=57.33\text{ A}$$ $$\cos {\varphi }_1=\frac{I_{1w}}{I_1}=\frac{44.50}{57.33}=0.776\text{ lagging}$$

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Question 2

Q2(a): Comparison Between Two-Winding Transformers and Autotransformers

Feature	Two-Winding Transformer	Autotransformer
Winding Architecture	Two electrically isolated windings	Single continuous tapped winding
Galvanic Isolation	Full electrical isolation between primary and secondary	No galvanic isolation (conductive connection)
Power Transfer Mode	100% inductive (via magnetic core flux)	Dual mode: partly conductive, partly inductive
Copper Requirement	High (volume $\propto 2S$)	Low (saving $=k\times \text{reference copper}$)
Efficiency	High (95–98%)	Even higher (less copper and lower core loss)
Leakage Reactance & VR	Higher leakage reactance, larger VR	Very low leakage reactance, superior VR
Short-Circuit Hazard	Lower fault current due to higher impedance	Very high fault currents; requires robust protection
Optimal Use Case	Large voltage transformations, grid isolation	Close voltage ratios ($k\to 1$), variacs, motor starters

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Q2(b): Copper Saving Proof in an Autotransformer

In any electromagnetic coil, the weight of copper required is proportional to the product of turns and rated current (total ampere-turns): $$\text{Weight of copper }W\propto NI$$

1. Two-Winding Transformer: $$W_{\text{two}}\propto N_1{I}_1+N_2{I}_2$$ Since $N_1{I}_1\approx N_2{I}_2$: $$W_{\text{two}}\propto 2N_1{I}_1$$
2. Autotransformer (Step-Down, $k=N_2/N_1<1$):
   • Series section has $(N_1-N_2)$ turns and carries current $I_1$.
   • Common section has $N_2$ turns and carries the difference current $(I_2-I_1)$. $$W_{\text{auto}}\propto (N_1-N_2)I_1+N_2(I_2-I_1)$$ $$W_{\text{auto}}\propto N_1{I}_1-N_2{I}_1+N_2{I}_2-N_2{I}_1=N_1{I}_1+N_2{I}_2-2N_2{I}_1$$ Substitute $N_2{I}_2=N_1{I}_1$ and $N_2=k{N}_1$: $$W_{\text{auto}}\propto 2N_1{I}_1-2k{N}_1{I}_1=2N_1{I}_1(1-k)$$
3. Ratio of Copper Weight: $$\frac{W_{\text{auto}}}{W_{\text{two}}}=\frac{2N_1{I}_1(1-k)}{2N_1{I}_1}=1-\mathbf{k}$$ $$\boxed{\text{Copper Saved}=W_{\text{two}}-W_{\text{auto}}=k\cdot W_{\text{two}}}$$

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Q2(c): Circuit Diagrams and Physical Setup for OC and SC Tests

1. Open-Circuit (No-Load) Test:

• LV Side Excited at Rated Voltage, HV Side Open:
  • Safety: Eliminates dangerous high voltages on test benches.
  • Current Metering: No-load current $I_0$ is large enough on LV side for precision measurement.
  • Instrument: Low-power-factor (LPF) wattmeter is mandatory because $\cos {\varphi }_0\approx 0.1\text{–}0.2$.
• Why Wattmeter Equals Core Loss Only: $I_0$ is tiny (~2–5%), so primary copper loss is $I_0^2{R}_1\approx 0$. Secondary current is zero. Wattmeter reading $W_0$ measures purely core hysteresis and eddy-current dissipation ($P_{Fe}=W_0$).

2. Short-Circuit Test:

• HV Side Excited with Reduced Voltage (~5–10%), LV Solidly Shorted:
  • Safety & Current Handling: Rated current is smaller on HV side, matching standard lab instruments.
  • Precision Variac Control: 5–10% of HV rating allows fine adjustment of test current to exact rated value.
• Why Wattmeter Equals Copper Loss Only: Because applied voltage $V_{sc}$ is small, core flux $\Phi \propto V_{sc}$ is tiny. Core loss scales as $V_{sc}^2\approx (0.05{)}^2\approx 0.25\%$ of rated core loss (negligible). All power measured by wattmeter $W_{sc}$ represents total full-load copper loss ($P_{Cu,FL}=W_{sc}$).

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Q2(d): No-Load Test Parameter Extraction

Given Data:

• Primary rating $V_1=220\text{ V}$, Secondary $V_2=110\text{ V}$
• No-load test readings: $V_0=220\text{ V},I_0=0.5\text{ A},W_0=30\text{ W}$

Step-by-Step Parameter Extraction:

1. Total Core Loss (Iron Loss): $$P_{Fe}=W_0=30\text{ W}$$
2. Loss (Core-Loss) Current Component ($I_c$): $$I_c=\frac{W_0}{V_0}=\frac{30}{220}=0.1364\text{ A}$$
3. Magnetizing Current Component ($I_m$): $$I_m=\sqrt{I_0^2-I_c^2}=\sqrt{(0.5{)}^2-(0.1364{)}^2}=\sqrt{0.25-0.0186}=\sqrt{0.2314}=0.4810\text{ A}$$

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Question 3

Q3(a): Why Are Transformers Rated in kVA Instead of kW?

The output power rating of electrical equipment is limited strictly by internal heating and temperature rise, which degrades insulation life. In a transformer:

1. Iron Loss ($P_{Fe}$) depends exclusively on core flux density, which is determined solely by voltage ($V$).
2. Copper Loss ($P_{Cu}$) depends exclusively on $I^{2}R$ heating, which is determined solely by current ($I$).
3. Neither iron loss nor copper loss depends on the phase angle (power factor $\cos \varphi$) of the connected load!
   • A transformer carrying rated 100 A at 1000 V dissipates the exact same internal heat whether powering a resistive heater ($\cos \varphi =1.0$, $P=100\text{ kW}$) or an unloaded inductor ($\cos \varphi =0$, $P=0\text{ kW}$).
4. Since the manufacturer cannot predict the power factor of the consumer's load, the machine is rated by its maximum allowable volt-ampere product: kVA.

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Q3(b): Copper Saving in Autotransformer (Detailed Proof)

(See derivation in Q2(b) above: $\text{Copper Saved}=k\cdot W_{\text{two}}$).

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Q3(c): Step-by-Step Equivalent Circuit of a Transformer Referred to Primary Side

Physical Premise:

A real transformer has two separate electrical coils linked purely by magnetic flux. Standard electrical circuit analysis tools (KVL, KCL, mesh analysis, Thevenin theorem) require a single, continuous electrical circuit. To create this unified circuit model, we systematically account for the four physical departures from an ideal transformer:

1. Winding Resistances: Copper wires have finite conductivity, generating $I^{2}R$ heat losses ($R_1$ in primary, $R_2$ in secondary).
2. Leakage Fluxes: Magnetic flux lines (${\Phi }_{l1},{\Phi }_{l2}$) that fail to link both windings pass through air or insulation, creating reactive series voltage drops ($X_1,X_2$).
3. Core Excitation & Iron Losses: The ferromagnetic core has finite permeability requiring magnetizing current ($I_m$), and alternating flux creates hysteresis and eddy-current losses ($I_c$).
4. Turns Ratio & Electrical Isolation: The primary and secondary are galvanically isolated with turns ratio $a=N_1/N_2$.

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Step 1: The Ideal Transformer Core Model

An ideal transformer represents the lossless magnetic coupling core:

• Zero winding resistance ($R_1=R_2=0$)
• Zero leakage flux ($X_1=X_2=0$)
• Infinite core permeability (${\mu }_r\to \infty$, hence no-load current $I_0=0$)
• Zero core losses ($P_c=0$)

Voltages and currents transform purely according to turns ratio $a=N_1/N_2$: $$\frac{E_1}{E_2}=\frac{N_1}{N_2}=a,\frac{I_1}{I_2}=\frac{N_2}{N_1}=\frac{1}{a}$$

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Step 2: Incorporating Winding Resistances and Leakage Reactances

Practical conductors have finite resistance and leakage flux induces reactive back-EMFs:

• Primary winding: Series resistance $R_1$ and leakage reactance $X_1=2\pi f{L}_{l1}$.
• Secondary winding: Series resistance $R_2$ and leakage reactance $X_2=2\pi f{L}_{l2}$.

Applying Kirchhoff's Voltage Law (KVL) to both sides: $${\mathbf{V}}_1={\mathbf{E}}_1+{\mathbf{I}}_1(R_1+j{X}_1)$$ $${\mathbf{E}}_2={\mathbf{V}}_2+{\mathbf{I}}_2(R_2+j{X}_2)$$

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Step 3: Adding the Core Excitation Shunt Branch ($R_c\parallel j{X}_m$)

A real core draws exciting current ${\mathbf{I}}_0$ even under no-load conditions ($I_2=0$), connected across the primary induced EMF ${\mathbf{E}}_1$: $${\mathbf{I}}_0={\mathbf{I}}_c+{\mathbf{I}}_m$$

1. Core-loss resistance $R_c$: Models real iron losses (hysteresis + eddy current) dissipating active power: $$I_c=I_0\cos {\varphi }_0,R_c=\frac{E_1}{I_c}=\frac{E_1^2}{P_c}$$
2. Magnetizing reactance $X_m$: Models reactive VARs required to establish the alternating mutual core flux ${\Phi }_m$: $$I_m=I_0\sin {\varphi }_0,X_m=\frac{E_1}{I_m}$$

By KCL at the primary junction: $${\mathbf{I}}_1={\mathbf{I}}_0+{\mathbf{I}}_2^{\prime }$$ where ${\mathbf{I}}_2^{\prime }$ is the load component of primary current that counteracts secondary demagnetization.

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Step 4: Transferring Secondary Parameters to Primary Side (Exact Equivalent Circuit)

To eliminate the ideal transformer block and form a single continuous network, all secondary quantities are referred across the turns ratio $a=N_1/N_2$ while preserving total volt-ampere and power relationships:

1. Referred voltage: $V_2^{\prime }=a{V}_2,E_2^{\prime }=a{E}_2=E_1$
2. Referred current: $I_2^{\prime }=I_2/a$
3. Referred resistance: $I_2^2{R}_2=(I_2^{\prime }{)}^2{R}_2^{\prime }⟹R_2^{\prime }=a^2{R}_2$
4. Referred leakage reactance: $I_2^2{X}_2=(I_2^{\prime }{)}^2{X}_2^{\prime }⟹X_2^{\prime }=a^2{X}_2$
5. Referred load impedance: $Z_L^{\prime }=a^2{Z}_L$

The ideal transformer is eliminated, yielding the Complete Exact Equivalent Circuit:

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Step 5: Approximate Equivalent Circuit Referred to Primary

Why we move the shunt branch: In power transformers, the exciting current $I_0$ is only $2\%-6\%$ of full-load rated current $I_1$. The series voltage drop ${\mathbf{I}}_0(R_1+j{X}_1)$ across the primary winding is negligible ($<1\%$ of $V_1$), meaning ${\mathbf{E}}_1\approx {\mathbf{V}}_1$.

Moving the shunt branch ($R_c\parallel j{X}_m$) directly across the input terminals introduces negligible error while allowing primary and referred secondary impedances to combine into single lumped parameters: $$R_{01}=R_1+R_2^{\prime }=R_1+a^2{R}_2$$ $$X_{01}=X_1+X_2^{\prime }=X_1+a^2{X}_2$$ $${\mathbf{Z}}_{01}=R_{01}+j{X}_{01}$$

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Step 6: Simplified Series Equivalent Circuit (Neglecting $I_0$)

For short-circuit fault analysis, heavy-load calculations, and voltage regulation determinations, $I_0\ll I_2^{\prime }$ can be neglected entirely: $${\mathbf{V}}_1={\mathbf{V}}_2^{\prime }+{\mathbf{I}}_2^{\prime }(R_{01}+j{X}_{01})$$

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Parameter Transformation Summary (Referred to Primary)

Parameter	Actual Secondary Value	Transformation Rule	Referred to Primary ($a=N_1/N_2$)
Voltage	$V_2$	Multiply by $a$	$V_2^{\prime }=a{V}_2$
Current	$I_2$	Divide by $a$	$I_2^{\prime }=I_2/a$
Resistance	$R_2$	Multiply by $a^2$	$R_2^{\prime }=a^2{R}_2$
Leakage Reactance	$X_2$	Multiply by $a^2$	$X_2^{\prime }=a^2{X}_2$
Impedance	$Z_L$	Multiply by $a^2$	$Z_L^{\prime }=a^2{Z}_L$
Total Equivalent Resistance	—	$R_1+R_2^{\prime }$	$R_{01}=R_1+a^2{R}_2$
Total Equivalent Reactance	—	$X_1+X_2^{\prime }$	$X_{01}=X_1+a^2{X}_2$

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Question 4

Q4(a): Practical Applications of Autotransformers

1. Induction Motor Starting: Autotransformer starters reduce voltage to 50%, 65%, or 80% to limit starting inrush current.
2. Laboratory Variacs: Adjustable continuous AC power supplies.
3. Power Transmission Interties: Efficiently coupling transmission networks operating at close voltage levels (e.g., 400 kV to 220 kV or 132 kV to 66 kV).
4. Electric Railway Traction: Traction booster transformers along electrified lines (e.g., 50 kV / 25 kV systems).
5. Voltage Boosters / Stabilizers: Compensating for voltage drops on long transmission/distribution feeders.

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Q4(b): Service Continuity with One Burnt Transformer (Open-Delta 57.7% Proof)

See full proof and vector diagram in T-04: Open-Delta Connection: Why 57.7% and When to Use It.

• Closed delta bank of 3 single-phase units: $S_{\Delta }=3VI$.
• Open delta bank with 2 units: $S_V=\sqrt{3}VI$.
• Capacity ratio: $$\frac{S_V}{S_{\Delta }}=\frac{\sqrt{3}VI}{3VI}=\frac{1}{\sqrt{3}}=0.577=57.7\%$$

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Q4(c): Numerical: Two 25 kVA Transformers in Open-Delta

Given Data:

• Two single-phase units, each rated $S_1=25\text{ kVA}$.

1. Maximum 3-Phase Load Served Without Overloading (Open-Delta): In open-delta, maximum safe load is: $$S_{V-V}=\sqrt{3}\times S_1=\sqrt{3}\times 25\text{ kVA}=43.30\text{ kVA}$$ (Explanation: Although the two transformers have a nominal combined rating of $50\text{ kVA}$, their 86.6% utilization factor limits safe delivery to $43.3\text{ kVA}$ to avoid thermal overload).
2. Total Load Served When Closed by a Third 25 kVA Unit: With three identical units in closed delta ($\Delta$-$\Delta$): $$S_{\Delta -\Delta }=3\times S_1=3\times 25\text{ kVA}=75.00\text{ kVA}$$ Adding one transformer increases system capacity by: $$75-43.3=31.7\text{ kVA}(+73.2\%)$$

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SECTION - B (Induction Motors: Q5 to Q8)

Question 5

Q5(a): AC Motor Classification Tree

AC Motors are broadly classified into two grand families:

1. Synchronous Motors: Run strictly at synchronous speed ($N=N_s=\frac{120f}{P}$). Require DC rotor excitation (or permanent magnets) and damper windings. Constant speed, power-factor adjustable.
2. Asynchronous (Induction) Motors: Must run at a speed strictly less than synchronous speed ($N<N_s$, slip $s>0$).
   • Squirrel-Cage Induction Motor: Rotor conductors are heavy copper/aluminum bars shorted by end rings. Extremely rugged, simple, low cost.
   • Wound-Rotor (Slip-Ring) Induction Motor: Rotor carries distributed 3-phase winding brought out to external slip rings. Allows inserting external rotor resistance for high starting torque and limited speed control.
   • Single-Phase Induction Motors: Split-phase, capacitor-start, capacitor-run, shaded-pole.

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Q5(b): Why Asynchronous Motor is Treated as a Rotating Transformer

• Primary: The stator winding acts as the primary, drawing electrical energy from the AC line.
• Secondary: The rotor cage acts as a short-circuited secondary winding, receiving power across the air gap purely by electromagnetic induction.
• Standstill Analogy ($s=1.0$): With the shaft locked, rotor frequency equals stator frequency ($f_r=f$). The machine is mathematically and physically identical to a short-circuited static transformer.
• Running Operation ($s<1.0$): As the rotor accelerates, the induced rotor frequency is scaled down by slip ($f_r=sf$), and electrical energy transferred across the gap is divided into internal rotor copper loss ($s{P}_g$) and mechanical shaft drive ($(1-s)P_g$).

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Q5(c): Equivalent Circuit Model of an Induction Motor (Step-by-Step)

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Question 6

Q6(a): Reversing the Direction of Rotation of a 3-Phase Induction Motor

To reverse the direction of rotation of a 3-phase induction motor, interchange any two of the three stator supply line leads (e.g., swap lines $A$ and $B$, or $R$ and $Y$):

• Physical Reason: Swapping two leads reverses the phase sequence of the stator currents from $A-B-C$ to $B-A-C$.
• Reversing the phase sequence reverses the direction of the rotating magnetic field ($\omega \to -\omega$).
• By Lenz's Law, the rotor immediately follows the new direction of field rotation, reversing its shaft motion.

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Q6(b): Induction Motor Power Equations and Impact of Voltage Sags

• Air gap power: $P_g=3I_2^2\frac{R_2}{s}$
• Rotor copper loss: $P_{r,Cu}=3I_2^2{R}_2=s{P}_g$
• Developed mechanical power: $P_m=(1-s)P_g$
• Power ratio: $P_g:P_{r,Cu}:P_m=1:s:(1-s)$
• Effect of 10% Voltage Drop on Constant-Torque Load: Operating slip increases by 23.46% ($s_2=s_1/(0.9{)}^2=1.2346s_1$). Rotor copper loss increases directly by 23.46%, causing severe overheating.

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Q6(c): Why Synchronous Motors Are Not Self-Starting

In a synchronous motor, the stator RMF rotates at synchronous speed $N_s$ (e.g., 3000 rpm for 2-pole 50 Hz) the instant power is switched on.

• The rotor has substantial physical inertia and is stationary at standstill.
• When an $N$-pole of the stator field sweeps past an $S$-pole of the rotor, it exerts an attractive torque forward.
• But within half a cycle ($1/100\text{ sec}$ at 50 Hz), the stator pole has moved 180° ahead, presenting an $N$-pole that exerts a repulsive torque backward!
• Due to rotor inertia, the shaft cannot accelerate in $0.01\text{ s}$. The net average starting torque over one AC cycle is strictly zero.
• Remedy: Damper winding bars (squirrel-cage bars) embedded in the rotor pole faces allow the motor to accelerate as an induction motor up to ~95% speed before DC field excitation is applied to lock into synchronism.

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Q6(d): Numerical: 400V, 50 Hz, 6-Pole Induction Motor Power Balance

Given Data:

• Poles $P=6,f=50\text{ Hz}⟹N_s=\frac{120\times 50}{6}=1000\text{ rpm}$
• Air-gap input power: $P_g=75\text{ kW}=75,000\text{ W}$
• Rotor EMF frequency: 100 alternations per minute. Since 1 cycle $=2$ alternations, rotor frequency is: $$f_r=\frac{100}{2\times 60}=\frac{50}{60}=0.8333\text{ Hz}\left(\text{or if 100 cycles/min: }{f}_r=\frac{100}{60}=1.667\text{ Hz}\right)$$ (Using the RUET convention where "alternations" represents half-cycles or cycles per minute: taking standard syllabus interpretation $f_r=\frac{100}{60}=1.667\text{ Hz}$):

1. Operating Slip: $$s=\frac{f_r}{f}=\frac{100/60}{50}=\frac{100}{3000}=0.0333(3.33\%\text{ slip})$$
2. Rotor Speed: $$N=N_s(1-s)=1000\times (1-0.0333)=966.7\text{ rpm}$$
3. Rotor Copper Loss (Per-Phase and Total):
   • Total 3-phase rotor Cu loss: $$P_{r,Cu}=s{P}_g=0.0333\times 75,000\text{ W}=2500\text{ W}=2.50\text{ kW}$$
   • Per-phase rotor Cu loss: $$P_{r,Cu,\varphi }=\frac{2500}{3}=833.3\text{ W/phase}$$
4. Gross Mechanical Power Developed: $$P_m=(1-s)P_g=(1-0.0333)\times 75,000=0.9667\times 75,000=72,500\text{ W}=72.50\text{ kW}$$

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Question 7

Q7(a): Star-Delta Starting of 3-Phase Induction Motors

• Starting Phase (Star Connection): Winding phase voltage is throttled down: $V_{\varphi }=\frac{V_L}{\sqrt{3}}$. Line starting current drops to: $I_{st,Y}=\frac{1}{3}{I}_{st,\Delta }$. Starting torque drops to: $T_{st,Y}=\frac{1}{3}{T}_{st,\Delta }$.
• Running Phase (Delta Connection): At ~80% synchronous speed, starter switches to Delta, restoring full line voltage and full motor torque capacity.

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Q7(b): Why Maximum Torque Varies Proportionally With $V^2$

The breakdown torque of an induction motor is: $$T_{\max }=\frac{k{E}_2^2}{2X_2}$$ Because rotor standstill EMF is induced by stator flux ($E_2\propto {\Phi }_m\propto V$): $$E_2=KV⟹E_2^2=K^2{V}^2$$ Therefore: $$\boxed{T_{\max }\propto V^2}$$ Physical Significance: A mere 10% drop in line voltage causes maximum torque to plunge to $(0.9{)}^2=0.81$ (an almost 20% loss in peak overload capacity), making induction motors extremely vulnerable to stalling during grid voltage sags.

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Q7(c): Equivalence Between Star-Delta Starter and Autotransformer of Ratio $1/\sqrt{3}$

• In Direct-On-Line (Delta) connection: Starting phase current: $I_{\varphi ,DOL}=\frac{V_L}{Z_s}$. Starting line current: $I_{L,DOL}=\sqrt{3}{I}_{\varphi ,DOL}=\frac{\sqrt{3}{V}_L}{Z_s}$.
• In Star connection: Starting phase current: $I_{\varphi ,Y}=\frac{V_L/\sqrt{3}}{Z_s}$. Starting line current: $I_{L,Y}=I_{\varphi ,Y}=\frac{V_L}{\sqrt{3}{Z}_s}$.
• Comparing line currents: $$\frac{I_{L,Y}}{I_{L,DOL}}=\frac{\frac{V_L}{\sqrt{3}{Z}_s}}{\frac{\sqrt{3}{V}_L}{Z_s}}=\frac{1}{3}$$
• In an autotransformer starter with tapping ratio $x=\frac{V_{\text{motor}}}{V_{\text{line}}}$: $$\frac{I_{\text{line,auto}}}{I_{\text{line,DOL}}}=x^2$$
• Equating the current reduction factors: $$x^2=\frac{1}{3}⟹\mathbf{x}=\frac{1}{\sqrt{3}}\approx 0.577=57.7\%$$ A Star-Delta starter is mathematically and electrically identical to an autotransformer starter tapped at 57.7% voltage.

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Question 8

Q8(a): Definitions of Plugging and Slip

1. Plugging: An emergency braking method achieved by swapping two stator power supply leads while running. This flips the direction of the rotating stator field, creating a large negative torque that rapidly brings the motor to rest.
2. Slip: The normalized fractional difference between synchronous field speed and rotor mechanical speed: $s=\frac{N_s-N}{N_s}$.

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Q8(b): Torque-Slip Characteristics and Stable Operating Zone

The torque-slip curve exhibits two distinct behavioral regions:

1. Stable Operating Zone ($0\le s\le s_{mT}$): In this low-slip range, $s^2{X}_2^2\ll R_2^2$, so torque is roughly linear with slip: $T\propto s$. If load increases, the motor slows down slightly (slip increases), which automatically increases motor torque to match the load demand, maintaining a stable equilibrium.
2. Unstable Operating Zone ($s_{mT}<s\le 1.0$): Rotor reactance dominates ($s^2{X}_2^2\gg R_2^2$). Torque decreases as slip increases ($T\propto 1/s$). If load exceeds breakdown torque, slowing down reduces torque further, causing the motor to rapidly stall.

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Q8(c): Double-Field Revolving Theory of 1-Phase IM

See full details in IM-03: Single-Phase Induction Motor: Double Revolving Field Theory.

• A pulsating flux $\Phi (t)={\Phi }_m\sin \omega t$ resolves into two counter-rotating fields: forward field ${\Phi }_f={\Phi }_m/2$ at $+N_s$ and backward field ${\Phi }_b={\Phi }_m/2$ at $-N_s$.
• At standstill ($s=1$), forward and backward torques cancel ($T_f=T_b⟹T_{\text{net}}=0$).

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Q8(d): V-Curves of a Synchronous Motor

A V-curve plots stator armature current $I_a$ versus rotor field excitation current $I_f$ at constant shaft load:

1. Normal Excitation (Unity Power Factor): At a specific field current, the motor operates at $\cos \varphi =1.0$. Armature current $I_a$ is strictly at its minimum (the bottom cusp of the "V").
2. Under-Excitation ($\cos \varphi$ Lagging): When field current is reduced ($I_f<I_{f,\text{normal}}$), the motor must draw lagging reactive magnetizing current from the AC supply to support its magnetic field. $I_a$ rises and lags terminal voltage.
3. Over-Excitation ($\cos \varphi$ Leading): When field current is increased ($I_f>I_{f,\text{normal}}$), the overexcited rotor supplies excess reactive VARs back into the AC supply. The motor draws leading current, acting as a synchronous capacitor/condenser to correct industrial factory power factor!

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Source: PrevYearQuestions/2020.md

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