Skip to main content

Frequency and voltage of 16-pole generator

Short answer

Frequency f = PN/120 = 16×375/120 = 50 Hz. Slots per pole per phase m=3, slot angle β=20°, distribution factor kd=0.960. Turns per phase T=240. Phase voltage E_ph = 4.44×1×0.96×50×0.03×240 = 1534 V; line voltage E_L = √3×1534 = 2657 V.

A 3 phase, 16 pole synchronous generator has star connected winding with 144 slots and 10 conductor per slot. The flux per pole is 0.03 Wb, sinusoidally distributed and speed is 375 rpm. Find the frequency, phase and line voltage. Assume full pitched coil.

Given: P=16P = 16, 144 slots, 10 conductors/slot, Φ=0.03\Phi = 0.03 Wb, N=375N = 375 rpm, star, full-pitched (kp=1k_p = 1).

Frequency

f=PN120=16×375120=50 Hzf = \frac{PN}{120} = \frac{16\times375}{120} = 50\ \text{Hz}

Distribution factor

m=slotspoles×phases=14416×3=3β=180∘×Pslots=180×16144=20∘kd=sin⁡(mβ/2)msin⁡(β/2)=sin⁡30∘3sin⁡10∘=0.50.5209=0.960\begin{aligned} m &= \frac{\text{slots}}{\text{poles}\times\text{phases}} = \frac{144}{16\times3} = 3 \\ \beta &= \frac{180^\circ\times P}{\text{slots}} = \frac{180\times16}{144} = 20^\circ \\ k_d &= \frac{\sin(m\beta/2)}{m\sin(\beta/2)} = \frac{\sin30^\circ}{3\sin10^\circ} = \frac{0.5}{0.5209} = 0.960 \end{aligned}

Turns per phase

Z=144×10=1440 conductors,Zph=14403=480T=4802=240 turns per phase\begin{aligned} Z &= 144\times10 = 1440\ \text{conductors},\quad Z_{ph} = \frac{1440}{3} = 480 \\ T &= \frac{480}{2} = 240\ \text{turns per phase} \end{aligned}

Phase and line voltage

Eph=4.44 kpkd f Φ T=4.44×1×0.960×50×0.03×240=1534.1 VEL=3×1534.1=2657.2 V\begin{aligned} E_{ph} &= 4.44\,k_pk_d\,f\,\Phi\,T = 4.44\times1\times0.960\times50\times0.03\times240 \\ &= 1534.1\ \text{V} \\ E_L &= \sqrt3\times1534.1 = 2657.2\ \text{V} \end{aligned}

Answer: f = 50 Hz; phase voltage = 1534 V; line voltage = 2657 V (k_d = 0.960).

✓ Answer checked by Claude Sonnet 5.5

Related questions

All Electric Machines II chapter 1 questions →