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Gear train efficiency help please

  • Thread starter JohnnyS
  • Start date
  • #1
14
2
Moved from a technical forum, no template.
I have been set the following question in my assignment and I have answered all parts with the exception of:-
(d) Determine an equation for the efficiency of the gear train in terms of
the load (torque) on shaft 2 (all other factors remaining constant).
I am struggling here and would appreciate any help with regards to a starting point. Thanks in advance.

2. A gear train is to be designed to transmit power from shaft 1 to shaft 2
with the following condition:

(i) Shaft 1 (driven by an electric motor) rotates at 100 revs min–1.
(ii) Shaft 2 is to rotate at 400 revs min–1 in the opposite direction to shaft
1 against a load of 200 Nm.
(iii) The centre of shaft 1 is 300 mm from shaft 2.
(iv) The minimum number of teeth on any gear is 15, all gears must have
a multiple of 5 teeth and have a module of 2 mm.
(v) The maximum number of gears permissible is 4 gears and the
diameter of the maximum gear must be minimised.
(vi) The centres of the gears should lie on a line which is as close to
straight as possible.
(vii) All shafts have a frictional resistance of 5 Nm.

Carry out the following:
(a) Design a gear train which satisfies the above criteria. Use a sketch to
illustrate your design, label the gears A, B, C, etc. from the driver to
the driven gear, and state the number of teeth on each gear.
(b) Determine the input power required at shaft 1.
(c) Specify the efficiency of the gear train as a percentage.
(d) Determine an equation for the efficiency of the gear train in terms of
the load (torque) on shaft 2 (all other factors remaining constant).
 

Answers and Replies

  • #2
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I moved this thread to an appropriate homework forum. We require that you show us your attempt at the solution before our helpers are allowed to help.
 
  • #3
14
2
Ok so I have answered parts a to c.

2a)
The design states that the shafts must rotate in opposite directions, this means that there must be an even number of gears. If I use 2 gears they will rotate in opposite directions but each gear will have to be large in diameter to maintain the stipulated distance between centers of 300mm. As the design requires the maximum gear diameter to be kept to a minimum the amount of gears used will have to be increased in order to reduce their individual diameters. The design criteria also stipulates that the maximum amount of gears that can be used is 4. This leaves using 4 gears as the only option to fulfill both of these design criteria.

Shaft 1 rotates at 100 revs min-1 and shaft 2 rotates at 400 revs min-1 which gives a ratio of 400:100 or G = 4:1. Therefore shaft 2 must be 4 times smaller than shaft 1 to achieve the desired rotational speeds.

To calculate the smallest diameter any gear can be when the minimum number of teeth is 15 and the module is set at 2mm the following equation can be used:-

m = D / N m = 2mm

D = mN N =15

D = 2 x 15

D = 30mm

Shaft 1/Gear 1

Shaft 2 is 4 times smaller than shaft 1 so this would make shaft 1 4 x 30 = 120mm. With shaft / gear 1 at 120mm diameter and shaft 2 / gear 4 at 30mm diameter this leaves the gap in between 225mm to be filled by gears 2 and 3.

Gears.PNG


As 225mm is to be split between 2 gears each of which are required by design to have at least 15 teeth with the number of teeth a multiple of 5 and a module of 2 it is not possible to meet these requirements with the remaining gap of 225mm whilst keeping the gear centers in a straight line. Although all the design criteria can be met with the above configuration the gears would have to be moved of center in order to keep the distance between the shafts at 300mm. I have opted to keep all the gear centers in line by increasing gear 4 (shaft 2) to 40mm. Due to the ratio established earlier of 4:1 this will increase the diameter of gear1 (shaft 1) to 160mm leaving a gap of 200mm to be filled by gears 2 and 3.

Gear Train.PNG



To calculate the number of teeth for each gear the following equation will be used:-

m = D / N

N = D / m

Gear 1 (Shaft 1)

N = 160 / 2 m = 2

N = 80 teeth D = 160mm

Gear 2

N = 80 / 2 m = 2

N = 40 teeth D = 80mm

Gear 3

N = 120 / 2 m = 2

N = 60 teeth D = 120mm

Gear 4 (Shaft 2)

N = 40 / 2 m = 2

N = 20 teeth D = 40mm

To establish the order of the 2 idler gears the revs min need to be calculated for each gear. They are given for gear 1 (shaft 1) as 100 revs min-1 and gear 4 (shaft 2) must be 400 revs min-1 as set by the design criteria. The following calculations show that gear 2 must be 80mm diameter and gear 3 120mm in order to achieve 400 revs min-1 at gear 4 (shaft 2).


Gear 2

Gear 2 / Gear 1 = 40 / 80 = 1 / 2 = 1:2

100 x 2 = 200 revs min-1

Gear 3

Gear 3 / Gear 2 = 60 / 40 = 1.5 / 1 = 1.5:1

200 / 1.5 = 133.333 revs min-1

Gear 4

Gear 4 / Gear 3 = 20 / 60 = 1 / 3 = 1:3

133.333 x 3 = 400 revs min-1

2b) The input required at shaft 1 =

Input Power = Output Power + Power Loss

Power loss at each gear

Gear 1 = ω x Fr ω = 100 revs min-1 = (100 x 2π) / 60 = 10.47 rad s-1

= 10.47 x 5 Fr = 5Nm

= 52.35W

Gear 2 = ω x Fr ω = 200 revs min-1 = (200 x 2π) / 60 = 20.94 rad s-1

= 20.94 x 5 Fr = 5Nm

= 104.7W

Gear 3 = ω x Fr ω = 133.333 revs min-1 = (133.333 x 2π) / 60 = 13.96 rad s-1

= 13.96 x 5 Fr = 5Nm

= 69.8W

Gear 4 = ω x Fr ω = 400revs min-1 = (400 x 2π) / 60 = 41.89 rad s-1

= 41.89 x 5 Fr = 5Nm

= 209.45W

Total Power loss = 52.35 + 104.7 + 69.8 + 209.45

= 436.3W

Power Output

P = ωT ω = 400revs min-1 = (400 x 2π) / 60 = 41.89 rad s-1

= 41.89 x 200 T = 200Nm

P = 8378W

Power Input

Power Input = Power Output + Power Loss

Pin = 8378 + 436.3

= 8814.3W

Power Input = 8.81kW

2c)

Efficiency = Power Out / Power In x 100

Efficiency = 8378 / 8814.3 x 100

Efficiency = 0.9505 x 100

Efficiency = 95.05%
 

Attachments

  • #4
Baluncore
Science Advisor
2019 Award
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To establish the order of the 2 idler gears the revs min need to be calculated for each gear. They are given for gear 1 (shaft 1) as 100 revs min-1 and gear 4 (shaft 2) must be 400 revs min-1 as set by the design criteria. The following calculations show that gear 2 must be 80mm diameter and gear 3 120mm in order to achieve 400 revs min-1 at gear 4 (shaft 2).
The idler gear ratio is not important. No matter what size an idler gear is, the same number of teeth will pass on each side per minute. The only requirement is to fill the 200 mm gap with 100 teeth. So why do you select idlers of 40 and 60 teeth?

Is efficiency improved by using two 50 tooth gears, with diameters of 100mm each, because of the 5 Nm loss per idler shaft?
What about 15 and 85 tooth idlers? or 85 and 15? What about the orderly series of 80 to 60 then 40 to 20.
 
  • #5
14
2
Honestly I used 60 and 40 as it was the first combination I tried through trial and error. I can see now that as long as the input spins at 100 revs min with a ratio of 1:4 then the final gear will always be 400 revs min-1.
If I used idler gears both with 50 teeth, diameters of 100mm then from what I understand this will effect the efficiency as below:

Gear 2

Gear 1 / Gear 2 = 80 / 50 = 1.6 = 1:1.6

100 x 1.6 = 160 revs min-1

Gear 3

Gear 2 / Gear 3 = 50 / 50 = 1

= 160 revs min-1

Gear 4


Gear 3 / Gear 4 = 50 / 20 = 2.5 = 1:2.5

= 400 revs min-1

Power loss at each gear

Gear 1 = ω x Fr ω = 100 revs min-1 = (100 x 2π) / 60 = 10.47 rad s-1

= 10.47 x 5 Fr = 5Nm

= 52.35W

Gear 2 = ω x Fr ω = 160 revs min-1 = (160 x 2π) / 60 = 16.76 rad s-1

= 16.76 x 5 Fr = 5Nm

= 83.8W

Gear 3 = ω x Fr ω = 160 revs min-1 = (160 x 2π) / 60 = 16.76 rad s-1

= 16.76 x 5 Fr = 5Nm

= 83.8W

Gear 4 = ω x Fr ω = 400revs min-1 = (400 x 2π) / 60 = 41.89 rad s-1

= 41.89 x 5 Fr = 5Nm

= 209.45W

Total Power loss = 52.35 + 83.8 + 83.8 + 209.45

= 429.4W


Power Output

P = ωT ω = 400revs min-1 = (400 x 2π) / 60 = 41.89 rad s-1

= 41.89 x 200 T = 200Nm

P = 8378W

Power Input

Power Input = Power Output + Power Loss

Pin = 8378 + 429.4

= 8807.4W

Power Input = 8.81kW

Efficiency = Power Out / Power In x 100

Efficiency = 8378 / 8807.4 x 100

Efficiency = 0.9515 x 100

Efficiency = 95.15%

So yes if I were to use 2 idler gears with 50 teeth on each I would improve efficiency. I will go ahead and try to see if any of the other combinations give me a better efficiency. Thanks for the help on that.

Can you provide guidance on d) please?
 
  • #6
14
2
Correction to the above calculation;

Efficiency = 0.9513 x 100

Efficiency = 95.13%
 
  • #7
14
2
For d) I have got as far as this:
efficiency = Power out / Power in x 100
Power out = ωT
Power in = Power out + Power loss
Power loss = ωFr

Efficiency = ωT / (ωT) + (ωFr)
I am pretty sure this is not the final answer.
 
  • #8
14
2
Sorry forgot the x 100

Efficiency = (ωT / (ωT) + (ωFr)) x 100
 
  • #9
JohnnyS did you ever get to an answer for d?

I'm on the same question and have pretty much the same answers for a-c but I'm stuck on d. Not really sure what the question is asking for.
 
  • #10
14
2
I got my results earlier today and got a distinction. I got the first part of the gear train question wrong but the rest right based on my train. The gear train they are after is not in a straight line and some of the gears are slightly off centre.
With regards to question d I will have a look through it tomorrow when I am back at my computer and post something to help you out.
 
  • #11
14
2
This is what I got for 2d) hope it helps.

Efficiency = Power Out / Power In

Power Out = ωT

Power In = Power Out / Power Loss

Power Loss = 429.4W = 410π / 3

Gear 4 ω = 400revs min-1 = (400 x 2π) / 60 = 41.89 rad s-1 = 40π / 3

Efficiency = ((ω x T) / (ω X T + power loss))

=( (40π / 3 x T) / (40π / 3 x T)+ (410π / 3))

= 40T / (40T + 410)

Efficiency = T / (T + 10.25)
 
  • #12
You're a star.
Cheers pal
 

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