Calculating Gear Train Efficiency & Input Power | Homework Solution

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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.
N_A=80, r_A=80mm, ω_A=10π/3 rads^-1 (input shaft)
N_B=50, r_B=50mm, ω_B=-16π/3 rads^-1
N_C=50, r_C=50mm, ω_C=16π/3 rads^-1
N_D=20, r_D=20mm, ωD=-40π/3 rads^-1 (output shaft)

All shafts have a friction resistance of T_fr=5Nm
Load on shaft 2 T_L=200Nm

Homework Equations

The Attempt at a Solution

a)
Total torque T_T=T_L+2*T_fr=200+10=210Nm
P=T_T*ω_D=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%
T_A={[ω_D*(T_L+T_fr)]/ω_A}-T_fr=815Nm

input power=T_A*ω_A=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.

 

[CWatters]

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.

 

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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*T_fr

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.

 

[CWatters]

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?

 

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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)
P_in=P_out+P_loss

P_out=T_L*ω_D=200*40π/3=8000π/3=8377.5804W

Power losses due to friction on each gear
Gear A
P=T_fr*ω_A=5*10π/3=50π/3=52.3599W
Gear B & C (ω_B=ω_C)
P=T_fr*ω_B/C=5*16π/3=80π/3=83.7758W
Gear D
P=T_fr*ω_D=5*40π/3=200π/3=209.4395W

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

P_in=8377.5804+429.351=8806.9314W=8.8069kW

b)
η=(P_out/P_in)*100%=(8377.5804/8806.9314)*100%=09512*100%=95.12%

Is a) & b) correct now?

I don't know where to start with c).

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.

 

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My attempt for c)

η=output power/input power

where input power=output power + power loss

therefore

η=(T_L*ω_D)/(T_L*ω_D+power loss)=(40π/3*T_L)/(40π/3*T_L+429.351)=(41.8879*T_L)/41.8879*T_L+429.351)

How's that look like??

 

[CWatters]

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.

 

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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.

 

[CWatters]

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

 

[CWatters]

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)

 

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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.