5.3 Calculate the terminal EMF on open circuit.

Here are the calculations for the alternator and induction motor:

5.3 Calculate the terminal EMF on open circuit.

Given:

• Poles $P = 4$
• Frequency $f = 50 \text{ Hz}$
• Total slots $S = 38$
• Conductors per slot $Z_{slot} = 18$
• Flux per pole $\Phi = 0.04 \text{ Wb}$
• Connection: Mesh-connected (Delta)

Step 1: Calculate the number of turns per phase ($T_{ph}$).
Total conductors $Z_{total} = S \times Z_{slot} = 38 \times 18 = 684$.
Total turns $T_{total} = \frac{Z_{total}}{2} = \frac{684}{2} = 342$.
For a three-phase winding, turns per phase $T_{ph} = \frac{T_{total}}{3} = \frac{342}{3} = 114 \text{ turns}$.

Step 2: Calculate the distribution factor ($k_d$).
Slots per pole $N_{s/p} = \frac{S}{P} = \frac{38}{4} = 9.5$.
Slots per pole per phase $m = \frac{N_{s/p}}{3} = \frac{9.5}{3} = \frac{19}{6} \approx 3.167$.
Electrical angle between adjacent slots $\beta = \frac{180^\circ}{N_{s/p}} = \frac{180^\circ}{9.5} \approx 18.947^\circ$.
The distribution factor is given by:

$$k_d = \frac{\sin\left(\frac{m\beta}{2}\right)}{m\sin\left(\frac{\beta}{2}\right)}$$ $$k_d = \frac{\sin\left(\frac{\frac{19}{6} \times 18.947^\circ}{2}\right)}{\frac{19}{6}\sin\left(\frac{18.947^\circ}{2}\right)}$$ $$k_d = \frac{\sin(30^\circ)}{\frac{19}{6}\sin(9.4735^\circ)}$$ $$k_d = \frac{0.5}{3.167 \times 0.1645} \approx 0.960$$

Assume pitch factor $k_p = 1$ (full pitch).

Step 3: Calculate the EMF per phase ($E_{ph}$).
The EMF equation is:

$$E_{ph} = 4.44 k_p k_d f \Phi T_{ph}$$ $$E_{ph} = 4.44 \times 1 \times 0.960 \times 50 \text{ Hz} \times 0.04 \text{ Wb} \times 114 \text{ turns}$$ $$E_{ph} \approx 972.8 \text{ V}$$

Step 4: Determine the terminal EMF ($V_L$).
For a mesh (delta) connected alternator, the line voltage is equal to the phase voltage.

$$V_L = E_{ph} = \boxed{\text{972.8 V}}$$

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5.4 A three-phase, 50 Hz induction motor has four poles and runs at a speed of 1 440 rev/min when the total torque developed by the rotor is 80 Nm.

Given:

• Poles $P = 4$
• Frequency $f = 50 \text{ Hz}$
• Rotor speed $N_r = 1440 \text{ rev/min}$
• Torque developed $T = 80 \text{ Nm}$

5.4.1 The total input (in kilowatts) to the rotor.

Step 1: Calculate the synchronous speed ($N_s$).

$$N_s = \frac{120f}{P} = \frac{120 \times 50 \text{ Hz}}{4} = 1500 \text{ rev/min}$$

Step 2: Calculate the synchronous angular speed ($\omega_s$).

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