Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Homework Help
Engineering and Comp Sci Homework Help
How Do You Calculate Gear Train Efficiency Based on Load Torque?
Reply to thread
Message
[QUOTE="JohnnyS, post: 6073577, member: 652392"] 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: [U]Gear 2 [/U] Gear 1 / Gear 2 = 80 / 50 = 1.6 = 1:1.6 100 x 1.6 = 160 revs min-1 [U]Gear 3 [/U] Gear 2 / Gear 3 = 50 / 50 = 1 = 160 revs min-1 [U] Gear 4[/U] Gear 3 / Gear 4 = 50 / 20 = 2.5 = 1:2.5 = 400 revs min-1 [U]Power loss at each gear[/U] 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[U]Power Output[/U] P = ωT ω = 400revs min-1 = (400 x 2π) / 60 = 41.89 rad s-1 = 41.89 x 200 T = 200Nm P = 8378W [U]Power Input[/U] Power Input = Power Output + Power Loss Pin = 8378 + 429.4 = 8807.4W [B][U]Power Input = 8.81kW[/U][/B] Efficiency = Power Out / Power In x 100 Efficiency = 8378 / 8807.4 x 100 Efficiency = 0.9515 x 100 [B][U]Efficiency = 95.15%[/U][/B] 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? [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Engineering and Comp Sci Homework Help
How Do You Calculate Gear Train Efficiency Based on Load Torque?
Back
Top