Question 9.8: Determining the Efficiency of an Epicyclic Gear Train.* Find...
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.
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.
R=\pm \frac{\text { product of number of teeth on driver gears }}{\text { product of number of teeth on driven gears }}=\frac{\omega_{L}-\omega_{\text {arm }}}{\omega_{F}-\omega_{\text {arm }}} (9.14)
\text { if }|R| \geq 1 \text {, then } \rho=R \text { else } \rho=1 / R (9.15)
\rho=\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 (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:
\eta=\frac{E_{0}(\rho-1)}{\rho-E_{0}}=\frac{0.9928(1.000567-1)}{1.000567-0.9928}=0.073=7.3 \% (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 \text { Train }^{[4]} |
| \begin{array}{ccccccccc}\text { Case } & \rho & \begin{array}{l}\text { Fixed } \\\text { Shaft }\end{array} & \begin{array}{l}\text { Input } \\\text { Shaft }\end{array} & \begin{array}{c}\text { Train } \\\text { Ratio }\end{array} & T _{1} & T _{2} & T _{\text {arm }} & \text { Efficiency (η) } \\\hline 1 & >+1 & 2 & 1 & 1-\rho & -\frac{T_{\text {arm }}}{1-\rho E_{0}} & \frac{\rho E_{0} T_{\text {arm }}}{1-\rho E_{0}} & T_{\text {arm }} & \frac{\rho E_{0}-1}{\rho-1} \\2 & >+1 & 2 & \text { arm } & \frac{1}{1-\rho} & T_{1} & -\rho \frac{T_{1}}{E_{0}} & \left(\frac{\rho-E_{0}}{E_{0}}\right) T_{1} & \frac{E_{0}(\rho-1)}{\rho-E_{0}} \\3 & >+1 & 1 & 2 & \frac{\rho-1}{\rho} & \frac{T_{\text {arm }}}{\rho E_{0}-1} & -\frac{\rho E_{0} T_{\text {arm }}}{\rho E_{0}-1} & T_{\text {arm }} & \frac{\rho E_{0}-1}{E_{0}(\rho-1)} \\4 & >+1 & 1 & \text { arm } & \frac{\rho}{\rho-1} & -\frac{E_{0}}{\rho} T_{2} & T_{2} & -\left(\frac{\rho-E_{0}}{\rho}\right) T_{2} & \frac{\rho-1}{\rho-E_{0}} \\5 & \leq-1 & 2 & 1 & 1-\rho & -\frac{T_{\text {arm }}}{1-\rho E_{0}} & \frac{\rho E_{0} T_{\text {arm }}}{1-\rho E_{0}} & T_{\text {arm }} & \frac{\rho E_{0}-1}{\rho-1} \\6 & \leq-1 & 2 & \text { arm } & \frac{1}{1-\rho} & T_{1} & -\rho \frac{T_{1}}{E_{0}} & \left(\frac{\rho-E_{0}}{E_{0}}\right) T_{1} & \frac{E_{0}(\rho-1)}{\rho-E_{0}} \\7 & \leq-1 & 1 & 2 & \frac{\rho-1}{\rho} & \frac{E_{0} T_{\text {arm }}}{\rho-E_{0}} & -\frac{\rho T_{\text {arm }}}{\rho-E_{0}} & T_{\text {arm }} & \frac{\rho-E_{0}}{\rho-1} \\ 8 & \leq-1 & 1 & \text { arm } & \frac{\rho}{\rho-1} & -\frac{T_{2}}{\rho E_{0}} & T_{2} & -\left(\frac{\rho E_{0}-1}{\rho E_{0}}\right) T_{2} & \frac{E_{0}(\rho-1)}{\rho E_{0}-1}\end{array} |