Question 12.46: For balancing an alternator motor by the field-balancing tec...

From Figs. 12.63(a) and (b), we get

$A_{2}-A_{1}=4.1 \times 10^{-4} e^{-i\left(123^{\circ}\right)}, A_{3}-A_{1}=1.4 \times 10^{-4} e^{-}$.

$B_{2}-B_{1}=1.5 \times 10^{-4} e^{-i\left(211^{\circ}\right)}, B_{3}-B_{1}=4.75 \times 10^{-4} e^{-i\left(218^{\circ}\right)}$.

From the given data

$A_{1}=3.5 \times 10^{-4} e^{-i\left(15^{\circ}\right)}, B_{1}=4.2 \times 10^{-4} e^{-i\left(65^{\circ}\right)}$.

$a=\frac{4.2 e^{-i\left(65^{\circ}\right)} \times 1.4 e^{-i\left(94.5^{\circ}\right)}-3.5 e^{-i\left(15^{\circ}\right)} \times 4.75 e^{-i\left(218^{\circ}\right)}}{4.1 e^{-i\left(123^{\circ}\right)} \times 4.75 e^{-i\left(218^{\circ}\right)}-1.4 e^{-i\left(94.5^{\circ}\right)} \times 1.5 e^{-i\left(211^{\circ}\right)}}$.

$=\frac{5.88 e^{-i\left(159.5^{\circ}\right)}-16.625 e^{-i\left(233^{\circ}\right)}}{19.475 e^{-i\left(341^{\circ}\right)}-2.1 e^{-i\left(305.5^{\circ}\right)}}$.

$\beta=\frac{3.5 e^{-i\left(15^{\circ}\right)} \times 1.5 e^{-i\left(211^{\circ}\right)}-4.2 e^{-i\left(65^{\circ}\right)} \times 4.1 e^{-i\left(123^{\circ}\right)}}{19.475 e^{-i\left(341^{\circ}\right)} \times 2.1 e^{-i\left(305.5^{\circ}\right)}}$.

$=\frac{5.25 e^{-i\left(226^{\circ}\right)}-17.22 e^{-i\left(188^{\circ}\right)}}{19.475 e^{-i\left(341^{\circ}\right)}-2.1 e^{-i\left(138^{\circ}\right)}}$.

$\alpha=\frac{1.59 e^{-i\left(74^{\circ}\right)}}{1.77 e^{-i\left(345^{\circ}\right)}}=0.902 e^{-i\left(-271^{\circ}\right)}$.

$\beta=\frac{1.36 e^{-i\left(350^{\circ}\right)}}{1.77 e^{-i\left(345^{\circ}\right)}}=0.768 e^{-i\left(5^{\circ}\right)}$.

$\text { Balance mass at plane } A=0.902 \times 2=1.804 kg \text { at } 271^{\circ} cw \text { or } 89^{\circ} ccw$.

$\text { Balance mass at plane } B=0.768 \times 2=1.536 kg \text { at } 5^{\circ} cc w \text { from the position of trial mess }$.