Determining the Efficiency of an Epicyclic Gear Train.*
Find the overall efficiency of the epicyclic train shown in Figure 9-43. The basic efficiency E0 is 0.9928 and the gear tooth numbers are: $N_{ A }=82 t , N_{B}=84 t , N_{C}=86 t$ , $N_{D}=82 t , N_{E}=82 t , \text { and } N_{F}=84 t$ . Gear A (shaft 2) is fixed to the frame, providing a zero velocity input. The arm is driven as the second input.

Question

Determining the Efficiency of an Epicyclic Gear Train.*
Find the overall efficiency of the epicyclic train shown in Figure 9-43. The basic efficiency E0 is 0.9928 and the gear tooth numbers are: [latex] N_{ A }=82 t , N_{B}=84 t , N_{C}=86 t [/latex], [latex] N_{D}=82 t , N_{E}=82 t , ext { and } N_{F}=84 t [/latex]. Gear A (shaft 2) is fixed to the frame, providing a zero velocity input. The arm is driven as the second input.

Accepted Answer

1 Find the basic ratio ρ for the gear train using equations 9.14 and 9.15. Note that gears B and C have the same velocity as do gears D and E, so their ratios are 1 and thus are omitted.

[latex] R=\pm \frac{ ext { product of number of teeth on driver gears }}{ ext { product of number of teeth on driven gears }}=\frac{\omega_{L}-\omega_{ ext {arm }}}{\omega_{F}-\omega_{ ext {arm }}} [/latex] (9.14)

[latex] ext { if }|R| \geq 1 ext {, then } ho=R ext { else } ho=1 / R [/latex] (9.15)

[latex] ho=\frac{N_{F} N_{D} N_{B}}{N_{E} N_{C} N_{A}}=\frac{84(82)(84)}{82(86)(82)}=\frac{1764}{1763} \cong 1.000567 [/latex] (a)

2 The combination of ρ > 1, shaft 2 fixed and input to the arm corresponds to Case 2 in Table 9-12, giving an efficiency of:

[latex] \eta=\frac{E_{0}( ho-1)}{ ho-E_{0}}=\frac{0.9928(1.000567-1)}{1.000567-0.9928}=0.073=7.3 \% [/latex] (b)

3 This is a very low efficiency which makes this gearbox essentially useless. About 93% of the input power is being circulated within the gear train and wasted as heat.

TABLE 9-12 Torques and Efficiencies in an Epicyclic [latex] ext { Train }^{[4]} [/latex]

[latex]\begin{array}{ccccccccc} ext { Case } & ho & \begin{array}{l} ext { Fixed } \ ext { Shaft }\end{array} & \begin{array}{l} ext { Input } \ ext { Shaft }\end{array} & \begin{array}{c} ext { Train } \ ext { Ratio }\end{array} & T _{1} & T _{2} & T _{ ext {arm }} & ext { Efficiency (η) } \\hline 1 & >+1 & 2 & 1 & 1- ho & -\frac{T_{ ext {arm }}}{1- ho E_{0}} & \frac{ ho E_{0} T_{ ext {arm }}}{1- ho E_{0}} & T_{ ext {arm }} & \frac{ ho E_{0}-1}{ ho-1} \2 & >+1 & 2 & ext { arm } & \frac{1}{1- ho} & T_{1} & - ho \frac{T_{1}}{E_{0}} & \left(\frac{ ho-E_{0}}{E_{0}} ight) T_{1} & \frac{E_{0}( ho-1)}{ ho-E_{0}} \3 & >+1 & 1 & 2 & \frac{ ho-1}{ ho} & \frac{T_{ ext {arm }}}{ ho E_{0}-1} & -\frac{ ho E_{0} T_{ ext {arm }}}{ ho E_{0}-1} & T_{ ext {arm }} & \frac{ ho E_{0}-1}{E_{0}( ho-1)} \4 & >+1 & 1 & ext { arm } & \frac{ ho}{ ho-1} & -\frac{E_{0}}{ ho} T_{2} & T_{2} & -\left(\frac{ ho-E_{0}}{ ho} ight) T_{2} & \frac{ ho-1}{ ho-E_{0}} \5 & \leq-1 & 2 & 1 & 1- ho & -\frac{T_{ ext {arm }}}{1- ho E_{0}} & \frac{ ho E_{0} T_{ ext {arm }}}{1- ho E_{0}} & T_{ ext {arm }} & \frac{ ho E_{0}-1}{ ho-1} \6 & \leq-1 & 2 & ext { arm } & \frac{1}{1- ho} & T_{1} & - ho \frac{T_{1}}{E_{0}} & \left(\frac{ ho-E_{0}}{E_{0}} ight) T_{1} & \frac{E_{0}( ho-1)}{ ho-E_{0}} \7 & \leq-1 & 1 & 2 & \frac{ ho-1}{ ho} & \frac{E_{0} T_{ ext {arm }}}{ ho-E_{0}} & -\frac{ ho T_{ ext {arm }}}{ ho-E_{0}} & T_{ ext {arm }} & \frac{ ho-E_{0}}{ ho-1} \ 8 & \leq-1 & 1 & ext { arm } & \frac{ ho}{ ho-1} & -\frac{T_{2}}{ ho E_{0}} & T_{2} & -\left(\frac{ ho E_{0}-1}{ ho E_{0}} ight) T_{2} & \frac{E_{0}( ho-1)}{ ho E_{0}-1}\end{array}[/latex]

Question 9.8: Determining the Efficiency of an Epicyclic Gear Train.* Find...

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