An epicyclic gear train, as shown in Fig.15.27, has a sun wheel S of 30 teeth and two planet wheels P of teeth 50 each. The planet wheels mesh with the teeth of internal gear A. The driving shaft carrying the sun wheel transmits 6 kW at 300 rpm. The driven shaft is connected to an arm which carries the planet wheels.
Determine the speed of the driven shaft and the torque transmitted, if the overall efficiency is 95%.

Question

Accepted Answer

[latex] ext { Given: } z_{s}=30, z_{n}=50 ext {, power transmitted }=6 kW , N=300 rpm ext {, efficiency }=0.95 [/latex].

Table 15.21 is used to find the speed of gears.

Table 15.21
Revolutions of				Operation
Annular A	Planet P, 50	Sungear S, 30	Arm	Operation
[latex] \frac{-z_{s}}{z_{a}} [/latex]	[latex] \frac{-z_{s}}{z_{p}} = \frac{-30}{50} [/latex] [latex] \frac{-3 }{5}[/latex]	+1	0	1. Arm fixed, +1 revolutions given to gear S, ccw
[latex] -x \frac{z_{s}}{z_{a}} [/latex]	[latex] \frac{-3 x}{5} [/latex]	+x	0	2. Multiply by x
[latex] y-x \frac{z_{s}}{z_{a}} [/latex]	[latex] y-\frac{3 x}{5} [/latex]	y+x	y	3. Add y

[latex] n_{s}=x+y=300 [/latex] (1).

Now [latex] d_{a}=2 d_{p}+d_{s} [/latex].

For same module, [latex] z_{a}=2 z_{p}+z_{s} [/latex].

=100 = 30 = 130.

For to be fixed, [latex] \frac{-x z_{s}}{z_{a}}+y=0 [/latex].

[latex] =\frac{-3 x}{13}+y=0 [/latex] (2).

From (1) and (2), we get

[latex] x=\frac{975}{4}, y=\frac{225}{4} [/latex].

Speed of driven shaft (arm) = 56.25 rpm.

[latex] T_{s}=6 imes 10^{3} imes \frac{60}{(2 \pi imes 300)}=190.986 Nm [/latex].

[latex] T_{a}=-T_{s} n_{s} \frac{\eta}{n_{a}}=190.986 imes 300 imes \frac{0.95}{56.25}=-967.7 N m [/latex].

Question 15.23: An epicyclic gear train, as shown in Fig.15.27, has a sun wh...

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