Question

A three-phase Y-connected four-pole winding is installed in 24 slots on a stator. There are 40 turns of wire in each slot of the windings. All coils in each phase are connected in series. The flux per pole in the machine is 0.060 Wb, and the speed of rotation of the magnetic field is 1800 r/min. (a)What is the frequency of the voltage produced in this winding?(b)What are the resulting phase and terminal voltages of this stator?

Answer 1

Given Information:

Number of slots = 24

Number of turns = 40

Flux per pole = 0.060 Wb

Speed of rotation of magnetic field = 1800 rev/min

Required Information:

a) Frequency = ?

b) Phase and Terminal voltage = ?

Answer:

a) Frequency = 60 Hz

b) Phase voltage = 2560 V

Terminal voltage = 4434 V

Step-by-step explanation:

(a)What is the frequency of the voltage produced in this winding?

The relation between frequency of the voltage produced and the speed of rotation of magnetic field is given by

`f = (nP)/(120)`

Where n is the speed of rotation of magnetic field and P is the number of poles in the winding.

`f = (1800* 4)/(120)\\\\f = (7200)/(120)\\\\f = 60 \: Hz`

(b)What are the resulting phase and terminal voltages of this stator?

The phase voltage is given by

`E = √(2) \cdot \pi\cdot N\cdot \phi\cdot f`

Where N is the number of turns, Ф is the flux per pole and f is the frequency calculated in part a.

`E = √(2) \cdot \pi\cdot 40\cdot \ 0.060\cdot 60\\\\E = 639.77\\\\E = 640 \:V\\\`

There are total 24 slots so each phase has 24/3 = 8 slots

We know that the number of poles are 4 so that means each phase has 4 sets of coils.

So the voltage in each phase is

Vp = 4*640

Vp = 2560 V

Since it is a Y-connected machine, The terminal voltage is will be

`V_T = √(3) V_p\\\\V_T = √(3) \cdot2560\\\\V_T = 4434 \: V\\\\`