For balancing an alternator motor by the field-balancing technique, the experimental results obtained are listed in Table 12.44. Determine the correct balance masses that should be placed in two planes for complete dynamic balancing of the motor

Question

Table 12.44
Plane B		Plane A		Trial mass (kg)	Trial number
Phase angle deg	Amplitude cm	Phase angle deg	Amplitude cm	Trial mass (kg)	Trial number
65	[latex]4.2 imes 10^{-4}[/latex]	15	[latex]3.5 imes 10^{-4}[/latex]	0	1
81	[latex]3.1 imes 10^{-4}[/latex]	75	[latex]4.5 imes 10^{-4}[/latex]	2 at plane A	2
155	[latex]2.2 imes 10^{-4}[/latex]	35	[latex]4.0 imes 10^{-4}[/latex]	2 at plane B	3

Accepted Answer

[latex] ext { Fig.12.63(a) shows the vectors } A_{1}, A_{2} ext { and } A_{3} ext { to some scale and Fig.12.63(b) shows } B_{1}, B_{2} ext { and } B_{3} ext { drawn to the same scale.}[/latex]

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

[latex] A_{2}-A_{1}=4.1 imes 10^{-4} e^{-i\left(123^{\circ} ight)}, A_{3}-A_{1}=1.4 imes 10^{-4} e^{-} [/latex].

[latex] B_{2}-B_{1}=1.5 imes 10^{-4} e^{-i\left(211^{\circ} ight)}, B_{3}-B_{1}=4.75 imes 10^{-4} e^{-i\left(218^{\circ} ight)} [/latex].

From the given data

[latex] A_{1}=3.5 imes 10^{-4} e^{-i\left(15^{\circ} ight)}, B_{1}=4.2 imes 10^{-4} e^{-i\left(65^{\circ} ight)} [/latex].

[latex] a=\frac{4.2 e^{-i\left(65^{\circ} ight)} imes 1.4 e^{-i\left(94.5^{\circ} ight)}-3.5 e^{-i\left(15^{\circ} ight)} imes 4.75 e^{-i\left(218^{\circ} ight)}}{4.1 e^{-i\left(123^{\circ} ight)} imes 4.75 e^{-i\left(218^{\circ} ight)}-1.4 e^{-i\left(94.5^{\circ} ight)} imes 1.5 e^{-i\left(211^{\circ} ight)}} [/latex].

[latex] =\frac{5.88 e^{-i\left(159.5^{\circ} ight)}-16.625 e^{-i\left(233^{\circ} ight)}}{19.475 e^{-i\left(341^{\circ} ight)}-2.1 e^{-i\left(305.5^{\circ} ight)}} [/latex].

[latex] \beta=\frac{3.5 e^{-i\left(15^{\circ} ight)} imes 1.5 e^{-i\left(211^{\circ} ight)}-4.2 e^{-i\left(65^{\circ} ight)} imes 4.1 e^{-i\left(123^{\circ} ight)}}{19.475 e^{-i\left(341^{\circ} ight)} imes 2.1 e^{-i\left(305.5^{\circ} ight)}} [/latex].

[latex] =\frac{5.25 e^{-i\left(226^{\circ} ight)}-17.22 e^{-i\left(188^{\circ} ight)}}{19.475 e^{-i\left(341^{\circ} ight)}-2.1 e^{-i\left(138^{\circ} ight)}} [/latex].

[latex] ext { The valves of } \alpha ext { and } \beta ext { are obtained graphically as explained in Fig.12.63(c). } [/latex]

[latex] \alpha=\frac{1.59 e^{-i\left(74^{\circ} ight)}}{1.77 e^{-i\left(345^{\circ} ight)}}=0.902 e^{-i\left(-271^{\circ} ight)} [/latex].

[latex] \beta=\frac{1.36 e^{-i\left(350^{\circ} ight)}}{1.77 e^{-i\left(345^{\circ} ight)}}=0.768 e^{-i\left(5^{\circ} ight)} [/latex].

[latex] ext { Balance mass at plane } A=0.902 imes 2=1.804 kg ext { at } 271^{\circ} cw ext { or } 89^{\circ} ccw [/latex].

[latex] ext { Balance mass at plane } B=0.768 imes 2=1.536 kg ext { at } 5^{\circ} cc w ext { from the position of trial mess }[/latex].

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

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