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Chapter 8

Q. 8.9

Torque in a Planetary Geartrain

In part (a) of the previous example, the carrier and sun gear are driven by engines producing 2 hp and 5 hp, respectively (Figure 8.42). Determine the torque applied to the shaft of the output ring gear.

Approach
To calculate the torque applied to the output shaft, we recognize that, to balance the power supplied to the geartrain, the output shaft must transfer a total of 7 hp to a mechanical load. To apply Equation (8.5)

P = T_{ω} (torque)                      (8.5)

in dimensionally consistent units, first convert the units for power from horsepower to (ft ·lb)/s with the conversion factor in Table 7.2.

Table 7.2
Conversion Factors Between Various Units for Power in the USCS and SI

(ft . lb)/s W hp
1 1.356 1.818×10^{-3}
0.7376 1 1.341×10^{-3}
550 745.7 1

Step-by-Step

Verified Solution

In dimensionally consistent units, the speed of the ring gear’s output shaft is

\omega _{r} = (4800 rpm)\left(0.1047 \ \frac{rad/s}{rpm} \right)

= 502.6 (\cancel {rpm}) \left(\frac{rad/s}{\cancel {rpm}} \right)

= 502.6 \frac{rad}{s}

The total power supplied the geartrain is P = 7 hp, and, in dimensionally consistent units, this becomes

P = (7 \ hp)\left(550 \frac{(ft.lb)/s}{hp} \right)

= 3850(\cancel{hp})\left( \frac{(ft.lb)/s}{\cancel{hp}} \right)

= 3850 \frac{ft .lb}{s}

The output torque is therefore

T =\frac{3850 \ (ft.lb)/s}{502.6 \ rad/s}              \longleftarrow \left[P=T\omega \right]

= 7.659 \left(\frac{ft.lb}{\cancel {s}} \right)\left(\frac{\cancel {s}}{\cancel{rad}} \right)

= 7.659 \ ft.lb

Discussion
As before, we can directly cancel the radian unit when calculating the torque because the radian is a dimensionless measure of an angle. For this geartrain as a whole, the input power is equivalent to the power transferred to the mechanical load to the extent that friction within the geartrain can be neglected.

T = 7.659 ft · lb