Gear train torque / power and load?

[LDC1972]

Homework Statement

This is a final hurdle of a long question designing a simple gear train. That is done but now am not sure of my calculations for power in / out and torque in and out, so hopefully someone can help?

OK - I have 4 spur gears inline, Gear A = 100 rpm 80 Teeth (input motor - unknown power?)
Gear B = 160 rpm 50 Teeth (idler)
Gear C = 160 rpm 50 Teeth (idler)
Gear D = 400 rpm 20 Teeth (output against a load of 200 Nm)
And ALL shafts carry a frictional resistance of 5 Nm

1/ What is the input power at shaft 1 (gear A)?
2/ What is the efficiency percentage?

Homework Equations

P = TW
T = P/W
T1 x W1 = P input

The Attempt at a Solution

I have done the following:

(TA x WA) - (TB x WB) - (TC - WC) - (TD - WD) = ZERO
And TA x WA = Power input

Convert the rpm to rad s:
Gear A = 100 rpm = 10.47197551 rad s
Gears B and C = 160 rpm = 16.75516082 rad s
Gear D = 400 rpm = 41.88790205 rad s

Input Power (Pi) = (TB x WB) + (TC x WC) + (TD + WD)
= (5 Nm x 16.75516082 rad s) + (5 Nm x 16.75516082 rad s) + (5 Nm + 200
Nm x 41.88790205 rad s)
Input Power = 8.755 Kw

Torque in = Input power / WA = 836 Nm
Torque out = 205 Nm
Power out = TW = 205 Nm x 41.88790205 rad s) = 8.587 Kw

Efficiency = Output power / Input power = 98 %


Does this look correctly tackled? Any advice very welcome!

I like the 205 Nm out as the gear ratio is 1 : 4 and the output is 1/4 input, which makes sense?

 

[CWatters]

As I see it the power lost in each gear shaft will be it's angular velocity * frictional torque. So for each shaft the losses are..

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

Add it up and Total lost in the gearbox = 430W approx.

Power in must = Power Out + losses in gearbox

Power out = 41.89 * 200 = 8378W
So
Power in = 8378 + 430 = 8808W

Efficiency (%) = 100 * Power Out/Power In = 100* 8378/8808 = 95%

Not really my field but I can't see anything I've missed.

 

[LDC1972]

Thank you so much!

A concise and helpful reply - I've 'thanked' you.

I think 95% sounds much more realistic, so I'll take a look and do it your way.

Thanks again!

 

[Big Jock]

LDC1972 following this answer and was it correct? I found the efficiency to be different to yours...

 

[Big Jock]

Any help going folks?