Solution:

Given data:
- Supply voltage, V = 220V
- Lap wound conductors, N = 540
- Supply current, I = 32A
- Output power, P = 5.5kW
- Field current, If = 1A
- Armature resistance, Ra = 0.09Ω
- Flux per pole, Φ = 30mWb

To find:
a. Speed, N
b. Torque developed, T

Formulae used:
- Output power, P = VIa
- Back EMF, Eb = ΦZN/60A
- Torque developed, T = (Eb - V)/Ra
- Speed, N = 60f/P

Calculation:
- Output power, P = 5.5kW = 5500W
- Total conductors, Z = 2N = 1080
- Flux per conductors, Φz = Φ/Z = 27.78μWb
- Back EMF, Eb = ΦZN/60A = 220V
- Armature current, Ia = (P/V) = 25A
- Torque developed, T = (Eb - V)/Ra = 369.7N-m
- Speed, N = 60f/P = 1080rpm

Answer:
a. Speed, N = 1080rpm
b. Torque developed, T = 369.7N-m

Explanation:
A shunt motor is a type of DC motor in which the field winding is connected in parallel with the armature winding. The shunt motor has a constant speed, which is proportional to the supply voltage and inversely proportional to the flux per pole. The torque developed by the motor is proportional to the armature current and the flux per pole.

In this problem, we are given the supply voltage, supply current, output power, field current, armature resistance, and flux per pole of the shunt motor. Using these values, we can calculate the back EMF, armature current, torque developed, and speed of the motor.

The back EMF is the voltage generated by the motor due to the rotation of the armature in the magnetic field. It is proportional to the flux per pole, speed, and number of conductors. The armature current is the current flowing through the armature winding of the motor. The torque developed is the force produced by the motor due to the interaction of the magnetic field and the armature current. The speed of the motor is the number of revolutions per minute of the motor shaft.

In conclusion, the speed and torque of a shunt motor can be calculated using the given parameters of the motor. These values are important in designing and analyzing the performance of the motor.