Go to Post Just remember that you can get just as much inspiration from making a simple system's details work really well as you can from just making a complex system work... - Rob [more]
Home
Go Back   Chief Delphi > Technical > Motors
CD-Media   CD-Spy  
portal register members calendar search Today's Posts Mark Forums Read FAQ rules

 
Closed Thread
Thread Tools Rate Thread Display Modes
  #1   Spotlight this post!  
Unread 14-11-2010, 19:05
Ether's Avatar
Ether Ether is offline
systems engineer (retired)
no team
 
Join Date: Nov 2009
Rookie Year: 1969
Location: US
Posts: 8,125
Ether has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond repute
CIM quiz

A CIM motor is attached to a small dynamometer which exerts a fixed constant load torque of 100 oz-in on the motor shaft. The CIM is directly attached to a 12VDC source (assume no internal resistance) with heavy gauge wire and connectors (assume no resistance). Assume the CIM's specs are exactly as specified in the attached screenshot.

Questions:

1) What is the CIM's speed?

2) What is the CIM's current?

3) How much power (watts) is being input to the CIM?

4) How much power (watts) is the CIM delivering into the load (dynamometer)?

5) How do you account for the difference between #4 and #3 above?



Attached Thumbnails
Click image for larger version

Name:	CIM specs.png
Views:	212
Size:	47.1 KB
ID:	9423  
  #2   Spotlight this post!  
Unread 14-11-2010, 20:08
PAR_WIG1350's Avatar
PAR_WIG1350 PAR_WIG1350 is offline
Registered User
AKA: Alan Wells
FRC #1350 (Rambots)
Team Role: Alumni
 
Join Date: Dec 2009
Rookie Year: 2009
Location: Rhode Island
Posts: 1,190
PAR_WIG1350 has a reputation beyond reputePAR_WIG1350 has a reputation beyond reputePAR_WIG1350 has a reputation beyond reputePAR_WIG1350 has a reputation beyond reputePAR_WIG1350 has a reputation beyond reputePAR_WIG1350 has a reputation beyond reputePAR_WIG1350 has a reputation beyond reputePAR_WIG1350 has a reputation beyond reputePAR_WIG1350 has a reputation beyond reputePAR_WIG1350 has a reputation beyond reputePAR_WIG1350 has a reputation beyond repute
Re: CIM quiz

Spoiler for Maybe:
Spoiler for ???:

~3763.7 RPM
~40.64 Amperes
487.68 Watts in
~256 Watts out
The inefficiencies of the motor (mostly friction) prevent power from being transfered without loss
__________________
  #3   Spotlight this post!  
Unread 14-11-2010, 20:45
Ether's Avatar
Ether Ether is offline
systems engineer (retired)
no team
 
Join Date: Nov 2009
Rookie Year: 1969
Location: US
Posts: 8,125
Ether has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond repute
Re: CIM quiz

Quote:
Originally Posted by PAR_WIG1350 View Post
.
Re-check your output power calculation.

Ponder your answer to #5 a bit more.



  #4   Spotlight this post!  
Unread 14-11-2010, 23:47
sanddrag sanddrag is offline
On to my 16th year in FRC
FRC #0696 (Circuit Breakers)
Team Role: Teacher
 
Join Date: Jul 2002
Rookie Year: 2002
Location: Glendale, CA
Posts: 8,519
sanddrag has a reputation beyond reputesanddrag has a reputation beyond reputesanddrag has a reputation beyond reputesanddrag has a reputation beyond reputesanddrag has a reputation beyond reputesanddrag has a reputation beyond reputesanddrag has a reputation beyond reputesanddrag has a reputation beyond reputesanddrag has a reputation beyond reputesanddrag has a reputation beyond reputesanddrag has a reputation beyond repute
Re: CIM quiz

Quote:
Originally Posted by Ether View Post
3) How much power (watts) is being input to the CIM?

4) How much power (watts) is the CIM delivering into the load (dynamometer)?
4. 278.4 Watts
__________________
Teacher/Engineer/Machinist - Team 696 Circuit Breakers, 2011 - Present
Mentor/Engineer/Machinist, Team 968 RAWC, 2007-2010
Technical Mentor, Team 696 Circuit Breakers, 2005-2007
Student Mechanical Leader and Driver, Team 696 Circuit Breakers, 2002-2004
  #5   Spotlight this post!  
Unread 15-11-2010, 00:23
davidthefat davidthefat is offline
Alumni
AKA: David Yoon
FRC #0589 (Falkons)
Team Role: Alumni
 
Join Date: Jan 2011
Rookie Year: 2010
Location: California
Posts: 792
davidthefat has much to be proud ofdavidthefat has much to be proud ofdavidthefat has much to be proud ofdavidthefat has much to be proud ofdavidthefat has much to be proud ofdavidthefat has much to be proud ofdavidthefat has much to be proud ofdavidthefat has much to be proud ofdavidthefat has much to be proud of
Re: CIM quiz

Quote:
Originally Posted by Ether View Post
Re-check your output power calculation.

Ponder your answer to #5 a bit more.


Heat
__________________
Do not say what can or cannot be done, but, instead, say what must be done for the task at hand must be accomplished.
  #6   Spotlight this post!  
Unread 15-11-2010, 00:54
AustinSchuh AustinSchuh is offline
Registered User
FRC #0971 (Spartan Robotics) #254 (The Cheesy Poofs)
Team Role: Engineer
 
Join Date: Feb 2005
Rookie Year: 1999
Location: Los Altos, CA
Posts: 803
AustinSchuh has a reputation beyond reputeAustinSchuh has a reputation beyond reputeAustinSchuh has a reputation beyond reputeAustinSchuh has a reputation beyond reputeAustinSchuh has a reputation beyond reputeAustinSchuh has a reputation beyond reputeAustinSchuh has a reputation beyond reputeAustinSchuh has a reputation beyond reputeAustinSchuh has a reputation beyond reputeAustinSchuh has a reputation beyond reputeAustinSchuh has a reputation beyond repute
Re: CIM quiz

Quote:
Originally Posted by davidthefat View Post
Heat
Elaborate a bit more. Heat due to what?
  #7   Spotlight this post!  
Unread 15-11-2010, 08:55
Unsung FIRST Hero
Al Skierkiewicz Al Skierkiewicz is offline
Broadcast Eng/Chief Robot Inspector
AKA: Big Al WFFA 2005
FRC #0111 (WildStang)
Team Role: Engineer
 
Join Date: Jun 2001
Rookie Year: 1996
Location: Wheeling, IL
Posts: 10,798
Al Skierkiewicz has a reputation beyond reputeAl Skierkiewicz has a reputation beyond reputeAl Skierkiewicz has a reputation beyond reputeAl Skierkiewicz has a reputation beyond reputeAl Skierkiewicz has a reputation beyond reputeAl Skierkiewicz has a reputation beyond reputeAl Skierkiewicz has a reputation beyond reputeAl Skierkiewicz has a reputation beyond reputeAl Skierkiewicz has a reputation beyond reputeAl Skierkiewicz has a reputation beyond reputeAl Skierkiewicz has a reputation beyond repute
Re: CIM quiz

1) What is the CIM's speed?
If free speed is 5310 RPM and stall torque is 343.4 and the calculated speed is the free speed - reduction in speed due to load then speed = free speed - (rated torque * (free speed/max torque)) or
5310 - (100 oz.in. * (5310/343.4)) = 3763.69 RPM

2) What is the CIM's current?
If stall torque is 343.4 oz. in. and stall current is 133, then the current constant is .3873A/oz.in. therefore current at 3763.69 RPM is the constant * RPM + I min
I= (3763.69RPM * .3873A/oz.in.) + 2.7A = 38.73 + 2.7A = 41.43A

3) How much power (watts) is being input to the CIM?
P IN = I * V = 41.43A * 12V = 497.16 W

4) How much power (watts) is the CIM delivering into the load (dynamometer)?
The conversion factor for oz. in and speed is 0.00074 and Pout = Torque * RPM * conversion factor,
Therefore P out = 100 oz. in. * 3763.69 * 0.00074 = 278.51W

5) How do you account for the difference between #4 and #3 above?
As the lost electrical power is the current squared * motor winding resistance, and R motor is 12V/133A (stall current) = .0902 ohms, then electrical power loss is 41.43A^2 * .0902ohms = 154.87W.
If the input power is 497.16W and electrical loss is 154.87 and power out is 278.51W then all other losses amount to 497.16-154.87-278.51=63.78W
These losses are attributable to bearing friction, brush friction, back EMF, friction of armature with the air, etc.
__________________
Good Luck All. Learn something new, everyday!
Al
WB9UVJ
www.wildstang.org
________________________
Storming the Tower since 1996.
  #8   Spotlight this post!  
Unread 15-11-2010, 11:37
Ether's Avatar
Ether Ether is offline
systems engineer (retired)
no team
 
Join Date: Nov 2009
Rookie Year: 1969
Location: US
Posts: 8,125
Ether has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond repute
Re: CIM quiz

Quote:
A CIM motor is attached to a small dynamometer which exerts a fixed constant load torque of 100 oz-in on the motor shaft. The CIM is directly attached to a 12VDC source (assume no internal resistance). Assume the CIM's specs are exactly as specified here.

Questions:

Quote:
1) What is the CIM's speed?
Solution: Assume that the speed vs torque curve is linear, and use the given load torque (100 oz-in) to interpolate the rpm:

rpm = 5310*(1-100/343.4) = 3763.7 rpm


Quote:
2) What is the CIM's current?
Solution: Assume that the current vs torque curve is linear, and use the given load torque (100 oz-in) to interpolate the current:

amps = 2.7 + (100/343.4)*(133-2.7) = 40.64 amps


Quote:
3) How much power (watts) is being input to the CIM?
Solution: watts = volts*amps = 12*40.64 = 487.7 watts


Quote:
4) How much power (watts) is the CIM delivering into the load (dynamometer)?
Solution: watts = oz_in*rpm/1352.3 = 100*3763.7/1352.3 = 278.3 watts


Quote:
5) How do you account for the difference between #4 and #3 above?
Solution: The total power loss is 487.7-278.3 = 209 watts. Most of this loss is due to I^2R heat loss in the motor windings. The motor winding resistance is 12/133 = 0.09023 ohms, so the I^2R loss is (40.64)^2*(0.09023) = 149 watts. The remaining loss (60 watts) is due to a combination of motor bearing friction, windage, and motor core losses (hysteresis losses and induced eddy currents).



Now, suppose I open the circuit and insert a 0.1 ohm resistor in series with the motor. Assume that the load torque remains the same, at 100 oz-in, and that the resistor is large enough that it does not overheat.

Questions:

1) What is the CIM's speed?

2) What is the CIM's current?

3) How much power (watts) is being input to the CIM?

4) How much power (watts) is the CIM delivering into the load (dynamometer)?

5) Explain any assumptions you had to make to get the answers.



  #9   Spotlight this post!  
Unread 15-11-2010, 17:00
PAR_WIG1350's Avatar
PAR_WIG1350 PAR_WIG1350 is offline
Registered User
AKA: Alan Wells
FRC #1350 (Rambots)
Team Role: Alumni
 
Join Date: Dec 2009
Rookie Year: 2009
Location: Rhode Island
Posts: 1,190
PAR_WIG1350 has a reputation beyond reputePAR_WIG1350 has a reputation beyond reputePAR_WIG1350 has a reputation beyond reputePAR_WIG1350 has a reputation beyond reputePAR_WIG1350 has a reputation beyond reputePAR_WIG1350 has a reputation beyond reputePAR_WIG1350 has a reputation beyond reputePAR_WIG1350 has a reputation beyond reputePAR_WIG1350 has a reputation beyond reputePAR_WIG1350 has a reputation beyond reputePAR_WIG1350 has a reputation beyond repute
Re: CIM quiz

I didn't calculate the output current, I drew a picture of the curve and guessed. That would be why I got it wrong. We haven't learned (much) about angular motion in physics yet
__________________
  #10   Spotlight this post!  
Unread 15-11-2010, 17:02
AdamHeard's Avatar
AdamHeard AdamHeard is offline
Lead Mentor
FRC #0973 (Greybots)
Team Role: Mentor
 
Join Date: Oct 2004
Rookie Year: 2004
Location: Atascadero
Posts: 5,526
AdamHeard has a reputation beyond reputeAdamHeard has a reputation beyond reputeAdamHeard has a reputation beyond reputeAdamHeard has a reputation beyond reputeAdamHeard has a reputation beyond reputeAdamHeard has a reputation beyond reputeAdamHeard has a reputation beyond reputeAdamHeard has a reputation beyond reputeAdamHeard has a reputation beyond reputeAdamHeard has a reputation beyond reputeAdamHeard has a reputation beyond repute
Send a message via AIM to AdamHeard
Re: CIM quiz

Quote:
Originally Posted by PAR_WIG1350 View Post
I didn't calculate the output current, I drew a picture of the curve and guessed. That would be why I got it wrong. We haven't learned (much) about angular motion in physics yet
Do you mean output power?

That can be calculated from torque x speed, give it a shot!
  #11   Spotlight this post!  
Unread 15-11-2010, 19:45
theprgramerdude theprgramerdude is offline
WPI Freshman
AKA: Alex
FRC #2503 (Warrior Robotics)
Team Role: Mentor
 
Join Date: Feb 2010
Rookie Year: 2008
Location: Brainerd, Minnesota
Posts: 347
theprgramerdude has much to be proud oftheprgramerdude has much to be proud oftheprgramerdude has much to be proud oftheprgramerdude has much to be proud oftheprgramerdude has much to be proud oftheprgramerdude has much to be proud oftheprgramerdude has much to be proud oftheprgramerdude has much to be proud oftheprgramerdude has much to be proud oftheprgramerdude has much to be proud of
Re: CIM quiz

Well, lets see.... Assuming the motor is mostly linear in performance, The No-Load Current is 2.7 amps, the constant (lets use this) D determining the ratio of current to torque is (64-0)/(27-2.7)= 2.633 oz-in per amp, the no-load 12v speed is 5310 rpm, with it losing (5310-4320)/(64) rpm per oz-in, or 5310-4320/27-2.7 per amp.
Oh, and 5310 RPM per 12 volts equates to 442.5 RPM's per volt.
Based on this:
5310 RPM no load
2.7 amps no load
2.633 oz-in per amp of torque
-15.46 rpm per oz-in torque, or 40.74 rpm per amp
442.5 RPM per volt, no load.

So, lets answer number two first. In order for there to be 100 oz-in of torque, the current applied (which in turn causes the force) must be the same, and I'm hoping this math was done correct before, 40.64 amps.


Which in turn brings us back to #1. The voltage loss across a .1 ohm resistor at 40.64 amps would be equal to V=IR, or .1*40.64= 4.064 volts (can't believe I used my TI-89 on that.)
12-4.064 volts = 7.936 volts left for the CIM motor.
A no load speed would be 3511.68 rpm. Subtracting the torque losses would result in an actual speed of -15.46 * 100 + 3511.68 = 1965.68 RPM.


Number 3: Power input to the CIM = VI= 7.936 V * 40.64 amps = 322.5 watts.

Numbah 4: Power output, mechanical: Speed * Force. 1966 RPM * 100 oz-in ~ using a random converter = 100 oz *.278N per oz* (32.76 revs * 2 pi inches * .0254m/in) per sec = OMG = 145.3 ? Watts.

Number 5: I believe I had to answer the second question first. You don't have to, but it eliminates a few extra steps of variable making.
  #12   Spotlight this post!  
Unread 15-11-2010, 21:00
Ether's Avatar
Ether Ether is offline
systems engineer (retired)
no team
 
Join Date: Nov 2009
Rookie Year: 1969
Location: US
Posts: 8,125
Ether has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond repute
Re: CIM quiz

Quote:
Originally Posted by theprgramerdude View Post
Well, lets see....
Nicely done


Now some food for further thought.

In the first scenario, the motor is generating 100 oz-in of net torque (against the external load) plus torque losses at 3764 rpm (due to friction and windage).

In the second scenario, the motor is generating 100 oz-in of net torque (against the external load) plus torque losses at 1965 rpm (due to friction and windage).

In the second scenario the torque losses are less (due to the lower rpm). Therefore the total torque is less than the first scenario, and the current required should be slightly less. How much less ?



  #13   Spotlight this post!  
Unread 19-11-2010, 21:42
joek's Avatar
joek joek is offline
Team Welder and CAD tech
FRC #2052 (KnightKrawler)
Team Role: Mechanical
 
Join Date: Jan 2009
Rookie Year: 2009
Location: Irondale H.S.
Posts: 231
joek is an unknown quantity at this point
Talking Re: CIM quiz

the mechanical subteam's answer:

speed: go faster!
heat: it makes to much of this
torque: enough to work on accessories, and great for pulling or driving stuff if you put a 12.75:1 gearbox on
current: work with the battery, when run through the tan device
__________________


2012 Record (13-2-0)
lake superior regional finalists- thank you WAVE (2826) and Blue Twilight (2220)
  #14   Spotlight this post!  
Unread 19-11-2010, 21:57
AdamHeard's Avatar
AdamHeard AdamHeard is offline
Lead Mentor
FRC #0973 (Greybots)
Team Role: Mentor
 
Join Date: Oct 2004
Rookie Year: 2004
Location: Atascadero
Posts: 5,526
AdamHeard has a reputation beyond reputeAdamHeard has a reputation beyond reputeAdamHeard has a reputation beyond reputeAdamHeard has a reputation beyond reputeAdamHeard has a reputation beyond reputeAdamHeard has a reputation beyond reputeAdamHeard has a reputation beyond reputeAdamHeard has a reputation beyond reputeAdamHeard has a reputation beyond reputeAdamHeard has a reputation beyond reputeAdamHeard has a reputation beyond repute
Send a message via AIM to AdamHeard
Re: CIM quiz

Quote:
Originally Posted by joek View Post
the mechanical subteam's answer:

speed: go faster!
heat: it makes to much of this
torque: enough to work on accessories, and great for pulling or driving stuff if you put a 12.75:1 gearbox on
current: work with the battery, when run through the tan device
I would expect more out of a mechanical subteam.
  #15   Spotlight this post!  
Unread 21-11-2010, 16:01
joek's Avatar
joek joek is offline
Team Welder and CAD tech
FRC #2052 (KnightKrawler)
Team Role: Mechanical
 
Join Date: Jan 2009
Rookie Year: 2009
Location: Irondale H.S.
Posts: 231
joek is an unknown quantity at this point
Re: CIM quiz

me too
__________________


2012 Record (13-2-0)
lake superior regional finalists- thank you WAVE (2826) and Blue Twilight (2220)
Closed Thread


Thread Tools
Display Modes Rate This Thread
Rate This Thread:

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

vB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Forum Jump

Similar Threads
Thread Thread Starter Forum Replies Last Post
Bolt quiz Jay H 237 Chit-Chat 5 25-05-2009 10:00
Safety Quiz? morg102 Games/Trivia 1 20-01-2009 13:23
your thoughts on the FP CIM/2.5 CIM motor drive setup versus 2 2.5 CIM motors waialua359 Motors 11 14-12-2007 04:29
Gundam Quiz GAT-X105 STRIKE Games/Trivia 3 09-02-2005 17:11
Team Quiz MrsT General Forum 7 02-03-2002 18:54


All times are GMT -5. The time now is 15:06.

The Chief Delphi Forums are sponsored by Innovation First International, Inc.


Powered by vBulletin® Version 3.6.4
Copyright ©2000 - 2017, Jelsoft Enterprises Ltd.
Copyright © Chief Delphi