- #1

- 18

- 0

## Homework Statement

I'm on the very last section of a question for the design of a simple gear train. I've managed to design the gear train with all criteria fulfilled and I am confident that it is correct. I just need clarification that my calculations thus far are correct and also a fresh pair of eyes on the final section, as I'm struggling to interpret it properly. Thanks in advance!

So I have four spur gears in a straight line:

Gear A = 100 rpm 80 Teeth (Unknown Input Power)

Gear B = 160 rpm 50 Teeth (Idler Gear)

Gear C = 160 rpm 50 Teeth (Idler Gear)

Gear D = 400 rpm 20 Teeth (Output against a load of 200 Nm)

All shafts carry a frictional resistance of 5 Nm

I need to find:

1) Input Power at Shaft 1.

2) Efficiency of the gear train as a percentage.

3) Determine an equation for the efficiency of the gear train in terms of the load (torque) on shaft 2 (all other factors remaining constant).

**2. The attempt at a solution**

**1) I**'ve started by converting all RPM to rad/s -

Gear A = 100 rpm = 10.47 rad/s

Gears B & C = 160 rpm = 16.76 rad/s

Gear D = 400 rpm = 41.89 rad/s

Then calculated power lost in each shaft:

Gear A = 100 rpm = 10.47 rads/s then 10.47 * 5Nm = 52.4W

Gear B = 160 rpm = 16.76 rads/s then 16.76 * 5NM = 83.8W

Gear C = 160 rpm = 16.76 rads/s then 16.76 * 5Nm = 83.8W

Gear D = 400 rpm = 41.89 rads/s then 41.89 * 5Nm = 209.5W

Total Power Loss in Gearbox = 430W

Calculated Power Output:

P = W * T

P = 41.89*200

P = 8378W

Calculated Power Input by adding on the total losses in the gearbox:

8378 + 430= 8808W

2) For the efficiency it's simply:

Efficiency = Power Out/Power In * 100

Efficiency = 8378/8808 * 100

Efficiency = 95%

3) Now this is the question I am not getting at all. The way I'm starting to interpret it is I need to calculate the efficiency using only the 200Nm load on shaft 2. Surely I've already done this with the above efficiency calculation?

Any help will be majorly appreciated, no matter how small. Just need that lightbulb moment.