Calculating Gear Train Efficiency & Input Power | Homework Solution

In summary, the questions involve determining the input power required at shaft 1, specifying the efficiency of the gear train, and deriving an equation for the efficiency of the gear train in terms of the load on shaft 2. The gear train consists of four gears lying on a line, with various specifications for each gear. The total power loss due to friction can be calculated using the conservation of energy principle. The resulting equation for the efficiency of the gear train is TL/(TL + 10.25).

Homework Statement

a) Determine the input power required at shaft 1.
b) Specify the efficiency of the gear train as a percentage.
c) Determine an equation for the efficiency of the gear train in terms of the load (torque) on shaft 2 (all other factors remaining constant).

Knows:
4 gear train lie on a line.
NA=80, rA=80mm, ωA=10π/3 rads-1 (input shaft)
ND=20, rD=20mm, ωD=-40π/3 rads-1 (output shaft)

All shafts have a friction resistance of Tfr=5Nm

The Attempt at a Solution

a)
Total torque TT=TL+2*Tfr=200+10=210Nm
P=TTD=210*(-40π/3)=-2800π=-8.7965kW

Required input power at shaft 1 due to load and friction is 8.7965kW.

b)
gear train efficiency η=(-power at output/power at input)*100%
TA={[ωD*(TL+Tfr)]/ωA}-Tfr=815Nm

input power=TAA=815*10π/3=2716.6667π=8534.6600W=8.5347kW

Something not right here as input should be greater then output?

C)
Will need some help with it, not sure where to start.

Total torque TT=TL+2*Tfr=200+10=210Nm

Can you explain how you get 2*Tfr ? All four shafts have friction and have different ratios between where that friction occurs and where the total Torque is calculated.

CWatters said:
Can you explain how you get 2*Tfr ? All four shafts have friction and have different ratios between where that friction occurs and where the total Torque is calculated.

I thought when it says shafts it means the input and output shafts, not all 4 gears. That is way I've 2*Tfr

I've got no example in my txt book on how to calculate torque when friction is present so I'm not entirely sure if it's correct.

It says all shafts and I'd assume that includes the shafts for the two middle gears.

You could work out the torque at the input that is attributable to friction in each gear but there is a slightly easier way. The question doesn't actually ask you to calculate the torque at the input just the power at the input. Have you considered applying conservation of energy to the gearbox? Have a think about that. If you can't work out what I mean let us know and I'll give another hint.

When do you have to submit this homework?

CWatters said:
It says all shafts and I'd assume that includes the shafts for the two middle gears.

You could work out the torque at the input that is attributable to friction in each gear but there is a slightly easier way. The question doesn't actually ask you to calculate the torque at the input just the power at the input. Have you considered applying conservation of energy to the gearbox? Have a think about that. If you can't work out what I mean let us know and I'll give another hint.

When do you have to submit this homework?

I have done some research and I might found a way, however it seams a bit simple, maybe to simple.

a)
Pin=Pout+Ploss

Pout=TLD=200*40π/3=8000π/3=8377.5804W

Power losses due to friction on each gear
Gear A
P=TfrA=5*10π/3=50π/3=52.3599W
Gear B & C (ωBC)
P=TfrB/C=5*16π/3=80π/3=83.7758W
Gear D
P=TfrD=5*40π/3=200π/3=209.4395W

Total power loss=2*83.7758+52.3599+209.4395=429.351W

Pin=8377.5804+429.351=8806.9314W=8.8069kW

b)
η=(Pout/Pin)*100%=(8377.5804/8806.9314)*100%=09512*100%=95.12%

Is a) & b) correct now?

I don't have a deadline for the homework, however I've been on it for some time now and I need to move on.

My attempt for c)

η=output power/input power

where input power=output power + power loss

therefore

η=(TLD)/(TLD+power loss)=(40π/3*TL)/(40π/3*TL+429.351)=(41.8879*TL)/41.8879*TL+429.351)

How's that look like??

Post #5 looks good to me. Exactly the approach I was hinting at.

As for c). Yes the approach looks right but try and avoid substituting the value for the friction power loss. Then pi will probably cancel.

CWatters said:
Post #5 looks good to me. Exactly the approach I was hinting at.

As for c). Yes the approach looks right but try and avoid substituting the value for the friction power loss. Then pi will probably cancel.

Thanks for the help on that. As of c) - I was thinking about simplifying the term too, however as there is sum in the denominator I don't think it's possible.

η=(TL*ωD)/(TL*ωD + power loss)

Should simplify because all terms contain angular velocity and all are specified in terms of Pi.

Edit: Actually all contain Pi/3

Last edited:
For what it's worth the simplest I could get was..

Power Loss = 410Pi/3

η=(TL*ωD)/(TL*ωD + power loss)

= (TL*40Pi/3) / ((TL*40Pi/3 + 410Pi/3)

Pi/3 cancels

= (TL*40) / (TL*40 + 410)

= TL/(TL + 10.25)

CWatters said:
For what it's worth the simplest I could get was..

Power Loss = 410Pi/3

η=(TL*ωD)/(TL*ωD + power loss)

= (TL*40Pi/3) / ((TL*40Pi/3 + 410Pi/3)

Pi/3 cancels

= (TL*40) / (TL*40 + 410)

= TL/(TL + 10.25)

Forgot that I can show the power loss in terms of π. You learn something new everyday. Thanks for your help.

1. What is gear train efficiency?

Gear train efficiency refers to the ratio of output power to input power in a gear system. It measures how much of the input power is actually transferred to the output and how much is lost due to friction and other inefficiencies.

2. How is gear train efficiency calculated?

Gear train efficiency can be calculated by dividing the output power by the input power and multiplying by 100 to get a percentage. This is known as the efficiency ratio.

3. What factors affect gear train efficiency?

There are several factors that can affect gear train efficiency, including the type and quality of gears used, the lubrication of the gears, and the alignment and spacing of the gears. Additionally, the speed and load placed on the gears can also impact their efficiency.

4. What is the purpose of calculating gear train efficiency?

Calculating gear train efficiency is important because it helps engineers and scientists understand the performance of a gear system and identify areas for improvement. It also allows for more accurate predictions and calculations in designing and using gear systems for various applications.

5. How can gear train efficiency be improved?

Gear train efficiency can be improved by using high-quality gears with proper lubrication, ensuring proper alignment and spacing of the gears, and reducing the speed and load placed on the gears. Regular maintenance and monitoring can also help improve efficiency. Additionally, using alternate gear configurations, such as helical or bevel gears, may also improve efficiency in certain applications.

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