Respuesta :
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
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
[tex]f = \frac{nP}{120}[/tex]
Where n is the speed of rotation of magnetic field and P is the number of poles in the winding.
[tex]f = \frac{1800\times 4}{120}\\\\f = \frac{7200}{120}\\\\f = 60 \: Hz[/tex]
(b)What are the resulting phase and terminal voltages of this stator?
The phase voltage is given by
[tex]E = \sqrt{2} \cdot \pi\cdot N\cdot \phi\cdot f[/tex]
Where N is the number of turns, ะค is the flux per pole and f is the frequency calculated in part a.
[tex]E = \sqrt{2} \cdot \pi\cdot 40\cdot \ 0.060\cdot 60\\\\E = 639.77\\\\E = 640 \:V\\\[/tex]
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
[tex]V_T = \sqrt{3} V_p\\\\V_T = \sqrt{3} \cdot2560\\\\V_T = 4434 \: V\\\\[/tex]
