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Pitch factor, distribution factor and EMF equation

Short answer

Pitch factor kp = cos(α/2) accounts for emf reduction when coils are short-pitched, which suppresses harmonics. Distribution factor kd = sin(mβ/2)/(m sin(β/2)) arises from distributed coils, giving a more sinusoidal wave. RMS emf per phase: Eph = 4.44 kp kd f Φ T V, where T is turns per phase.

Define the pitch factor and distribution factor and their significance in synchronous machine. Derive the e.m.f equation of an alternator.

Pitch factor (coil span factor), kpk_p

The pitch factor is the ratio of the emf induced in a short-pitched (chorded) coil to the emf that would be induced in a full-pitched coil.

If the coil is short of full pitch (180° electrical) by an angle α\alpha, the emfs in the two coil sides are not in phase and add as phasors:

kp=phasor sumarithmetic sum=cos⁡α2k_p = \frac{\text{phasor sum}}{\text{arithmetic sum}} = \cos\frac{\alpha}{2}

Significance: short pitching slightly reduces the emf (kp<1k_p < 1) but it eliminates or reduces harmonics (e.g. α=36∘\alpha = 36^\circ removes the 5th harmonic, since kp5=cos⁡(5α/2)=0k_{p5} = \cos(5\alpha/2) = 0), saves copper in end connections and reduces leakage reactance.

Distribution (breadth) factor, kdk_d

The distribution factor is the ratio of the phasor sum to the arithmetic sum of the emfs of coils distributed in several slots of a phase group.

With mm slots per pole per phase and slot angle β=180∘slots per pole\beta = \dfrac{180^\circ}{\text{slots per pole}}:

kd=sin⁡(mβ/2)msin⁡(β/2)k_d = \frac{\sin(m\beta/2)}{m\sin(\beta/2)}

Significance: distributing the winding gives a more sinusoidal emf, better use of the core and better cooling; kd<1k_d < 1 only slightly reduces the emf.

EMF equation of an alternator

Let

  • PP = number of poles, NsN_s = speed in rpm,
  • Φ\Phi = flux per pole (Wb), ZZ = conductors in series per phase, T=Z/2T = Z/2 turns per phase.

In one revolution, each conductor cuts flux PΦP\Phi in time 60/Ns60/N_s seconds.

Average emf per conductor=dΦdt=PΦ60/Ns=PΦNs60Since f=PNs120,PNs60=2fAverage emf per conductor=2fΦ VAverage emf per turn=2×2fΦ=4fΦ VAverage emf per phase=4fΦTRMS emf per phase=1.11×4fΦT=4.44 fΦT V\begin{aligned} \text{Average emf per conductor} &= \frac{d\Phi}{dt} = \frac{P\Phi}{60/N_s} = \frac{P\Phi N_s}{60} \\ \text{Since } f = \frac{PN_s}{120},\quad \frac{PN_s}{60} &= 2f \\ \text{Average emf per conductor} &= 2f\Phi\ \text{V} \\ \text{Average emf per turn} &= 2 \times 2f\Phi = 4f\Phi\ \text{V} \\ \text{Average emf per phase} &= 4f\Phi T \\ \text{RMS emf per phase} &= 1.11 \times 4f\Phi T = 4.44\, f\Phi T\ \text{V} \end{aligned}

(form factor of a sine wave = 1.11). Including the winding factors for a short-pitched, distributed winding:

Eph=4.44 kp kd f Φ T voltsE_{ph} = 4.44\, k_p\, k_d\, f\, \Phi\, T\ \text{volts}

For a star-connected machine, line voltage EL=3 EphE_L = \sqrt{3}\,E_{ph}; for delta, EL=EphE_L = E_{ph}. The product kw=kpkdk_w = k_p k_d is called the winding factor.

✓ Answer checked by Claude Sonnet 5.5

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