Conquering Alternator Fundamentals: Solved Problems for Smooth Sailing

Demonstrative Video

Problem-1

Calculate the pitch factor for the following cases:

  1. 36 stator slots, 4 poles, coil-span 1 to 8

  2. 72 stator slots, 6 poles, coil-span 1 to 10

  3. 96 stator slots, 6 poles, coil-span 1 to 12

  1. 36 stator slots, 4 poles, coil-span 1 to 8

  2. 72 stator slots, 6 poles, coil-span 1 to 10

  3. 96 stator slots, 6 poles, coil-span 1 to 12

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\[\begin{aligned} & \text { (a) Coil span falls short by}~(2 / 9) \times 180^{\circ}=40^{\circ}\\ & \alpha=40^{\circ} ~\Rightarrow ~ k_c=\cos 40^{\circ} / 2=\cos 20^{\circ}=\mathbf{0 . 9 4} \\ & \text { (b)} \alpha=(3 / 12) \times 180^{\circ}=45^{\circ} \quad \Rightarrow k_c=\cos 45^{\circ} / 2=\cos 22.5^{\circ}=\mathbf{0 . 9 2 4} \\ & \text { (c) } \alpha=(5 / 16) \times 180^{\circ}=56^{\circ} 16^{\prime} \quad \Rightarrow k_c=\cos 28^{\circ} 8^{\prime}=\mathbf{0 . 8 8 2} \\ & \end{aligned}\]

Problem-2

Determine \(k_d\) for an alternator with 9 slots per pole for the following cases :

  1. One winding in all the slots

  2. one winding using only the first \(2 / 3\) of the slots/pole

  3. three equal windings placed sequentially in \(60^{\circ}\) group .

  1. One winding in all the slots

  2. one winding using only the first \(2 / 3\) of the slots/pole

  3. three equal windings placed sequentially in \(60^{\circ}\) group .

\(\beta=180^{\circ} / 9=20^{\circ}\) and number of slots in a group (\(m\)) are 9, 6 and 3 respectively.

\[\boxed{k_d = \dfrac{\sin(m\beta/2)}{m\sin(\beta/2)}}\] \[\begin{aligned} \text{(i)}~m&=9, \quad \beta=20^{\circ}, \quad k_d=\frac{\sin 9 \times 20^{\circ} / 2}{9 \sin 20^{\circ} / 2}=0.64~\left[\frac{\sin \pi / 2}{\pi / 2}=0.637\right] \\ \text{(ii)}~m&=6, \quad \beta=20^{\circ}, \quad k_d=\frac{\sin 6 \times 20^{\circ} / 2}{6 \sin 20^{\circ} / 2}=0.83~\left[\frac{\sin \pi / 3}{\pi / 3}=0.827\right]\\ \text{(iii)}~m&=3, \quad \beta=20^{\circ}, \quad k_d=\frac{\sin 3 \times 20^{\circ} / 2}{3 \sin 20^{\circ} / 2}=0.96~\left[\frac{\sin \pi / 6}{\pi / 6}=0.955\right] \end{aligned}\]

Problem-3

A part of an alternator winding consists of six coils in series, each coil having an e.m.f. of \(10 \mathrm{~V}\) r.m.s. induced in it. The coils are placed in successive slots and between each slot and the next, there is an electrical phase displacement of \(30^{\circ}\).

\[\begin{aligned} \beta& =30^{\circ} \quad m =6 \\ k_d&=\frac{\sin m \beta / 2}{m \sin \beta / 2}=\frac{\sin 90^{\circ}}{6 \times \sin 15^{\circ}}=\frac{1}{6 \times 0.2588} \\ & \\ & \text{Arithmetic sum of voltage induced in 6 coils} =6 \times 10=60 \mathrm{~V} \\ & \text{Vector sum} = k_d \times \text{arithmetic sum} =60 \times 1 / 6 \times 0.2588=\mathbf{3 8 . 6 4 V} \end{aligned}\]

Problem-4

A \(3\phi\), 50 Hz alternator is running at 600 rpm has a 2-layer winding, 12 turns/coil, 4 slots/pole/phase, and coil-pitch of 10 slots. Determine the induced EMF per phase if the flux/pole is 0.035 Wb. \[\begin{aligned} \text{No. of poles}~p& = \dfrac{120f}{n} = 4 \\ \text{Total slots} ~ S & = 4\times 3\times 10 = 120 \\ \beta & = \dfrac{180^{\circ}\times \text{coil-pitch}}{\text{Total slots}} \\ & =\dfrac{180^{\circ}\times 10}{120}=15^{\circ} \\ k_d & = \dfrac{\sin\dfrac{4\times 15^{\circ}}{2}}{4\times \sin\dfrac{15^{\circ}}{2}} = 0.958 \end{aligned}\]

\[\begin{aligned} \text{Pole-pitch} & = \dfrac{\text{Total slots}}{\text{coil-pitch}} \\ &= \dfrac{120}{10}=12~\text{slots} \\ \alpha & = (12-10)\times 15^{\circ} = 30^{\circ} \\ k_p & = \cos\dfrac{30}{2} = 0.966 \\ E_{rms} & = 4.44 \cdot f \cdot T \cdot k_p \cdot k_d \\ & = 4.44 \times 0.035 \times 50\times 480\times 0.958\times 0.966\\ & = 3451~\mathrm{V} \end{aligned}\]

Problem-5

A 3-phase, 16-pole alternator has a star-connected winding with 144 slots and 10 conductors per slot. The flux per pole is \(0.03 \mathrm{~Wb}\), Sinusoidally distributed and the speed is 375 r.p.m. Assume full-pitched coil.

\[\begin{aligned} f & =P N / 120=16 \times 375 / 120=50 \mathbf{H z}\\ k_c & = 1 \quad (\text{not given})\\ n & =144 / 16=9 \\ \beta&=180^{\circ} / 9=20^{\circ} \\ m& =144 / 16 \times 3=3 \\ k_d & =\sin 3 \times\left(20^{\circ} / 2\right) / 3 \sin \left(20^{\circ} / 2\right)=0.96 \\ Z & =144 \times 10 / 3=480 \\ T&=480 / 2=240 / \text { phase } \\ E_{p h} & =4.44 \times 1 \times 0.96 \times 50 \times 0.03 \times 240=\mathbf{1 5 . 3 4} \mathbf{V} \\ E_L & =\sqrt{3} E_{p h}=\sqrt{3} \times 1534=\mathbf{2 6 5 8} \mathbf{V} \end{aligned}\]

Problem-6

Find the no-load phase and line voltage of a star-connected 3-phase, 6-pole alternator which runs at \(1200~ \mathrm{rpm}\), having flux per pole of \(0.1~ \mathrm{~Wb}\) sinusoidally distributed. Its stator has 54 slots having double layer winding. Each coil has 8 turns and the coil is chorded by 1 slot.

\[\begin{aligned} & \text{Winding is short-pitched by}~ 1 / 9 \Rightarrow 180^{\circ} / 9=20^{\circ} \\ k_c & =\cos 20^{\circ} / 2=0.98 \\ f& =6 \times 1200 / 120=60 \mathrm{~Hz} \\ n & =54 / 6=9 \\ \beta& =180^{\circ} / 9=20^{\circ}\\ m&=54 / 6 \times 3=3 \\ k_d & =\sin 3 \times\left(20^{\circ} / 2\right) / 3 \sin \left(20^{\circ} / 2\right)=0.96 \\ Z & =54 \times 8 / 3=144 \\ T&=144 / 2=72\\ E_{p h} & =4.44 \times 0.98 \times 0.96 \times 60 \times 0.1 \times 72=1805 \mathrm{~V} \\ E_L & =\sqrt{3} \times 1805=\mathbf{3 1 2 5} \mathrm{V} . \end{aligned}\]

Problem-7

Fundamental E.M.F \[\begin{aligned} \alpha & =\left(180^{\circ}-150^{\circ}\right)=30^{\circ} \\ k_{c 1}& =\cos \alpha / 2=\cos 15^{\circ}=0.966 \\ m &= 2 \\ \text { No. of slots } / \text { pole }& =6 \\ \beta& =180^{\circ} / 6=30^{\circ} \\ k_{d 1}& =\frac{\sin m \beta / 2}{m \sin \beta / 2}=\frac{\sin 2 \times 30^{\circ} / 2}{2 \sin 30^{\circ} / 2}=0.966 \\ Z& =10 \times 2 \times 4=80 \\ \text { turn } / \text { phase, } T&=80 / 2=40 \\ \text {E.M.F./phase }&=4.44 k_c k_d f \Phi T \\ E_1&=4.44 \times 0.966 \times 0.966 \times 50 \times 0.12 \times 40\\ &=995 \mathrm{~V} \end{aligned}\]

Harmonic E.M.F. \[\begin{aligned} K_{c 3}&=\cos 3 \alpha / 2=\cos 3 \times 30^{\circ} / 2=\cos 45^{\circ}=0.707 \\ k_{d 3}&=\frac{\sin m n \beta / 2}{m \sin n \beta / 2} \qquad n=3 \\ &=\frac{\sin 2 \times 3 \times 30^{\circ} / 2}{2 \sin 3 \times 30^{\circ} / 2}=\frac{\sin 90^{\circ}}{2 \sin 45^{\circ}}=0.707\\ f_2& =50 \times 3=150 \mathrm{~Hz} \\ \Phi_3&=(1 / 3) \times 20 \% \text { of fundamental flux }\\ &=(1 / 3) \times 0.02 \times 0.12=0.008 \mathrm{~Wb} \\ E_3&=4.44 \times 0.707 \times 0.707 \times 150 \times 0.008 \times 40=106 \mathrm{~V} \\ E \text { per phase }&=\sqrt{E_1^2+E_3^2}=\sqrt{995^2+106^2}=1000 \mathrm{~V} \end{aligned}\] Note. Since phase e.m.fs. induced by the 3rd, 9th and 15 th harmonics etc. are eliminated from the line voltages, the line voltage for a \(Y\)-connection would be \(=995 \times \sqrt{3}\) volt.