Electrical Machines

A transformer is a static device that transfers electrical energy from one circuit to another at the same frequency by electromagnetic induction, usually changing the voltage level.

Principle (mutual induction). It has two windings, primary and secondary, wound on a laminated silicon-steel core. When an alternating voltage $V_1$ is applied to the primary, an alternating current flows and produces an alternating flux $\Phi$ in the core. This flux links the secondary winding and, by Faraday's law, induces an emf in it:

$$e=-N\frac{d\Phi}{dt}$$

The same flux links both windings, so the emf per turn is the same, giving

$$\frac{E_2}{E_1}=\frac{N_2}{N_1}=K$$

If $N_2>N_1$ it is a step-up transformer; if $N_2<N_1$ it is step-down. There is no electrical connection between windings, and ideally input power = output power ($V_1I_1=V_2I_2$).

Efficiency of a transformer at load fraction $x$ (of full-load current), power factor $\cos\phi$:

$$\eta=\frac{x\,S\cos\phi}{x\,S\cos\phi+P_i+x^2P_{cu}}$$

where $S$ is the rated VA, $P_i$ the iron loss (constant) and $P_{cu}$ the full-load copper loss.

For simplicity, write it with load current $I_2$ and secondary voltage $V_2$ (assumed constant):

$$\eta=\frac{V_2I_2\cos\phi}{V_2I_2\cos\phi+P_i+I_2^2R_{2e}}$$

where $R_{2e}$ is the equivalent resistance referred to the secondary.

Condition for maximum. Divide numerator and denominator by $I_2$:

$$\eta=\frac{V_2\cos\phi}{V_2\cos\phi+\dfrac{P_i}{I_2}+I_2R_{2e}}$$

$\eta$ is maximum when the denominator is minimum. Since $V_2\cos\phi$ is constant, differentiate $\left(\frac{P_i}{I_2}+I_2R_{2e}\right)$ with respect to $I_2$ and equate to zero:

$$-\frac{P_i}{I_2^2}+R_{2e}=0 \ \Rightarrow\ P_i=I_2^2R_{2e}$$

Result: the efficiency is maximum when the iron loss equals the copper loss (variable loss).

Load at maximum efficiency:

$$x=\sqrt{\frac{P_i}{P_{cu}}}, \qquad \text{kVA at } \eta_{max}=\text{kVA}_{rated}\times\sqrt{\frac{P_i}{P_{cu}}}$$

and

$$\eta_{max}=\frac{xS\cos\phi}{xS\cos\phi+2P_i}$$

Given: $V_L=400$ V, output $=80$ kW, $f=50$ Hz, $P=4$ poles, $s=0.05$, stator : rotor turns $=1.5$. The stator is taken as star-connected.

a) Synchronous speed:

$$N_s=\frac{120f}{P}=\frac{120\times50}{4}=1500\ \text{rpm}$$

b) Motor speed:

$$N=N_s(1-s)=1500\times0.95=1425\ \text{rpm}$$

c) Rotor current frequency:

$$f_r=sf=0.05\times50=2.5\ \text{Hz}$$

d) Rotor induced voltage at running speed. Stator phase voltage:

$$V_1=\frac{400}{\sqrt3}=230.9\ \text{V}$$

Standstill rotor emf per phase $E_2=\frac{V_1}{1.5}=153.9$ V. At slip $s$ the rotor emf is $sE_2$:

$$E_{2r}=0.05\times153.9=7.70\ \text{V per phase}$$

(Line value $= \sqrt3\times7.70 = 13.33$ V.)

e) Rated torque (shaft output 80 kW at 1425 rpm):

$$\omega=\frac{2\pi\times1425}{60}=149.23\ \text{rad/s}, \qquad T=\frac{P_{out}}{\omega}=\frac{80000}{149.23}=536.1\ \text{N·m}$$

Answer: a) 1500 rpm, b) 1425 rpm, c) 2.5 Hz, d) 7.70 V per phase, e) 536.1 N·m.

Back emf. When the armature of a DC motor rotates, its conductors cut the main flux, and by Faraday's law an emf is induced in them. By Lenz's law this emf opposes the applied voltage (the cause producing it), so it is called the back (counter) emf:

$$E_b=\frac{P\phi ZN}{60A}$$

and the armature current is

$$I_a=\frac{V-E_b}{R_a}$$

Significance:

1. Acts as a governor. If load increases, speed falls, $E_b$ falls, $I_a$ rises, and torque increases to meet the load. If the load falls, the opposite occurs. The motor thus adjusts the input current automatically.
2. Limits the armature current. At standstill $E_b=0$, so the starting current $V/R_a$ is very large (a starter is therefore needed). At running speed, $E_b$ is close to $V$ and current is small.
3. Indicates energy conversion. The electrical power converted to mechanical power is $E_bI_a$; without back emf there would be no conversion.
4. Determines speed: $N\propto\frac{E_b}{\phi}$.

A synchronous generator (alternator) has two main parts, the stator (armature) and the rotor (field).

   Stator frame (cast iron / steel)
  +---------------------------------+
  |  Stator core (laminated steel)  |
  |   +-------------------------+   |
  |   |  armature winding in    |   |
  |   |  slots (3-phase)        |   |
  |   |    +---------------+    |   |
  |   |    | Rotor (field) |    |   |
  |   |    |   N   [shaft] |    |   |
  |   |    |       S       |    |   |
  |   |    +---------------+    |   |
  |   +-------------------------+   |
  +---------------------------------+
    Slip rings + brushes -> DC supply
    (exciter) to field winding
    Prime mover coupled to the shaft

Components:

1. Stator frame: outer cast iron/steel body that supports the core and protects the machine.
2. Stator core: laminated silicon-steel stampings, with slots on the inner surface, that reduce eddy-current loss.
3. Armature winding: three-phase winding placed in the slots, where the emf is induced and the output taken.
4. Rotor: carries the DC field winding, either salient-pole (low speed, hydro) or cylindrical (high speed, turbo).
5. Field winding: supplied with DC through slip rings and carbon brushes, producing the main flux.
6. Exciter: a small DC source that supplies the field current.
7. Shaft and bearings: transmit the prime mover's torque to the rotor.
8. Cooling system and air gap: a small air gap separates rotor and stator.

Given: $S=300$ kVA, $P_i=1.8$ kW, full-load $P_{cu}=2.5$ kW.

(i) Full-load efficiency at 0.8 lagging pf.

$$\text{Output}=300\times0.8=240\ \text{kW}$$ $$\text{Total loss}=1.8+2.5=4.3\ \text{kW}$$ $$\eta=\frac{240}{240+4.3}\times100=98.24\%$$

(ii) kVA at maximum efficiency. Efficiency is maximum when copper loss = iron loss:

$$x=\sqrt{\frac{P_i}{P_{cu}}}=\sqrt{\frac{1.8}{2.5}}=0.8485$$ $$\text{kVA}=300\times0.8485=254.56\ \text{kVA}$$

(iii) Maximum efficiency at 0.8 pf. Output $=254.56\times0.8=203.65$ kW. At this load, the copper loss equals the iron loss $=1.8$ kW, so the total loss is $2\times1.8=3.6$ kW:

$$\eta_{max}=\frac{203.65}{203.65+3.6}\times100=98.26\%$$

Answer: (i) 98.24%, (ii) 254.56 kVA, (iii) 98.26%.

Condition for maximum torque

The torque of a three-phase induction motor is

$$T=\frac{3}{\omega_s}\cdot\frac{sE_2^2R_2}{R_2^2+(sX_2)^2}$$

where $E_2$, $R_2$, $X_2$ are rotor emf, resistance and reactance per phase at standstill, and $s$ is the slip. For constant supply, $E_2$ is constant. Differentiate with respect to $s$ and equate to zero (equivalently, maximise $\frac{sR_2}{R_2^2+s^2X_2^2}$):

$$\frac{d}{ds}\left[\frac{sR_2}{R_2^2+s^2X_2^2}\right]=0 \Rightarrow R_2^2+s^2X_2^2-2s^2X_2^2=0$$ $$R_2^2=s^2X_2^2 \Rightarrow s_m=\frac{R_2}{X_2}$$

Result: torque is maximum when the rotor resistance equals the rotor reactance at that slip ($R_2=s_mX_2$). Substituting,

$$T_{max}=\frac{3}{\omega_s}\cdot\frac{E_2^2}{2X_2}$$

$T_{max}$ is independent of the rotor resistance, but the slip at which it occurs ($s_m=R_2/X_2$) depends on it.

Torque-slip characteristic

 Torque
  |         Tmax
  |        .--*--.
  |      /         \
  | Tst /            \
  |   /                \
  | /                    \___
  +----------------------------> slip
  1.0 (start)  sm      0 (sync)

• At $s=1$ (standstill) the torque is the starting torque $T_{st}$, which is fairly low.
• From $s=1$ to $s_m$, torque rises to $T_{max}$ (unstable region, as $R_2\ll sX_2$ makes $T\propto 1/s$).
• Between $s_m$ and zero (the normal working region), $T\propto s$ and the curve is nearly straight and steep; a small slip (2-5%) gives full-load torque, so speed is nearly constant.
• At $s=0$ (synchronous speed) torque is zero.
• Increasing the rotor resistance shifts $s_m$ towards 1 and improves starting torque; when $R_2=X_2$, $T_{st}=T_{max}$.

A synchronous generator (alternator) works on Faraday's law of electromagnetic induction: an emf is induced in a conductor when there is relative motion between the conductor and a magnetic field.

Working:

1. The rotor carries a field winding supplied with DC through slip rings, which creates alternate N and S poles.
2. A prime mover (turbine or engine) rotates the rotor at constant synchronous speed $N_s$.
3. The rotating field flux cuts the stationary three-phase armature conductors in the stator slots, inducing an alternating emf in each phase.
4. The three phase windings are placed $120^\circ$ apart in space, so the emfs are $120^\circ$ apart in time, giving a three-phase output.
5. The frequency depends on the speed and the number of poles:

$$f=\frac{PN_s}{120}$$

For 50 Hz and 4 poles, the speed is 1500 rpm. The rms emf per phase is $E=4.44f\Phi T_{ph}K_w$. Because the rotor speed is locked to the frequency, the machine is called synchronous.

Given: $P=4$, lap winding so $A=P=4$, $Z=980$, $\phi=20$ mWb, $V=230$ V, line current $I_L=26$ A, $R_a=0.12\ \Omega$, $R_{sh}=130\ \Omega$.

Shunt field current:

$$I_{sh}=\frac{V}{R_{sh}}=\frac{230}{130}=1.769\ \text{A}$$

Armature current:

$$I_a=I_L-I_{sh}=26-1.769=24.23\ \text{A}$$

Armature torque (using $T_a=\frac{P\phi Z}{2\pi A}I_a$):

$$T_a=\frac{4\times0.02\times980}{2\pi\times4}\times24.23=\frac{78.4}{25.13}\times24.23=3.119\times24.23=75.59\ \text{N·m}$$

Check through back emf and speed:

$$E_b=V-I_aR_a=230-24.23\times0.12=227.09\ \text{V}$$ $$N=\frac{60AE_b}{P\phi Z}=\frac{60\times4\times227.09}{4\times0.02\times980}=695.2\ \text{rpm}$$ $$T_a=\frac{E_bI_a}{2\pi N/60}=\frac{227.09\times24.23}{72.80}=75.59\ \text{N·m}$$

Answer: Armature torque $T_a=75.59$ N·m (at about 695 rpm).

Definition. A transformer is a static electrical device that transfers electric power from one circuit to another at the same frequency through electromagnetic induction, normally raising or lowering the voltage.

Operating principle. It works on mutual induction. An alternating voltage applied to the primary winding drives an alternating flux in the laminated iron core. This flux links the secondary winding and induces an emf in it (Faraday's law). The emf in each winding is proportional to its number of turns, so $\frac{E_2}{E_1}=\frac{N_2}{N_1}$.

emf equation. Let $N_1$ = primary turns, $N_2$ = secondary turns, $\Phi_m$ = maximum flux in the core (Wb), $f$ = frequency (Hz). The flux varies sinusoidally:

$$\Phi=\Phi_m\sin\omega t$$

The average rate of change of flux in a quarter cycle is

$$\frac{d\Phi}{dt}\Big|_{av}=\frac{\Phi_m}{T/4}=4f\Phi_m$$

So the average emf per turn is $4f\Phi_m$ volts. For a sine wave, the form factor is 1.11, so rms emf per turn $=1.11\times4f\Phi_m=4.44f\Phi_m$. Hence:

$$E_1=4.44\,f\,N_1\,\Phi_m \qquad E_2=4.44\,f\,N_2\,\Phi_m$$

Dividing, the transformation ratio is

$$\frac{E_2}{E_1}=\frac{N_2}{N_1}=K$$

Given: $S=25$ kVA, $N_1=500$, $N_2=40$, $V_1=3000$ V, $f=50$ Hz. The transformer is ideal.

Turns ratio: $K=\frac{N_2}{N_1}=\frac{40}{500}=0.08$.

(i) Full-load currents.

$$I_1=\frac{S}{V_1}=\frac{25000}{3000}=8.33\ \text{A}$$ $$I_2=\frac{I_1}{K}=\frac{8.33}{0.08}=104.17\ \text{A}$$

(ii) Secondary emf.

$$E_2=V_1\times\frac{N_2}{N_1}=3000\times0.08=240\ \text{V}$$

(Check: $25000/240=104.17$ A.)

(iii) Maximum core flux. From the emf equation $E_1=4.44fN_1\Phi_m$ (for an ideal transformer, $E_1=V_1$):

$$\Phi_m=\frac{3000}{4.44\times50\times500}=0.02703\ \text{Wb}=27.03\ \text{mWb}$$

Answer: (i) $I_1=8.33$ A, $I_2=104.17$ A; (ii) $E_2=240$ V; (iii) $\Phi_m=27.03$ mWb.

A DC machine is an electromechanical energy converter. As a generator it converts mechanical energy into DC electrical energy; as a motor it converts DC electrical energy into mechanical energy. Construction is the same for both.

   Yoke
  +------------------------------+
  |  Pole core + pole shoe  (N)  |
  |   field winding              |
  |      +---------------+       |
  |      | Armature core |<-shaft|
  |      | + winding     |       |
  |      | Commutator    |-brush |
  |      +---------------+       |
  |  Pole core + pole shoe  (S)  |
  +------------------------------+

Main parts:

1. Yoke (frame): the outer cast-iron or cast-steel ring. It supports the poles, carries the return flux and protects the machine.
2. Poles and pole shoes: laminated steel poles bolted to the yoke carry the field winding. The shoe spreads the flux evenly over the air gap and supports the coils.
3. Field winding: coils of insulated copper on the poles; excited with DC to produce the main flux.
4. Armature core: a cylinder of laminated silicon steel with slots, mounted on the shaft. Lamination reduces eddy-current loss.
5. Armature winding: copper conductors in the slots, connected as lap or wave winding, where the emf is induced.
6. Commutator: copper segments insulated by mica, joined to the armature coils. It converts the alternating emf in the armature to unidirectional output (generator) or reverses the current for continuous torque (motor).
7. Brushes and brush gear: carbon or graphite blocks held by springs against the commutator, to collect or supply current.
8. Shaft, bearings and end covers: carry the rotating parts and keep the air gap uniform.
9. Interpoles (optional) improve commutation, and a fan cools the machine.