Engineers Edge
Torque,power calculation and Motor selection Don
Post Reply
 
Mechanical Engineering Forum
Posted by: enahs

09/06/2010, 07:03:14

Author Profile
eMail author
Edit

Dear All,
It will be a great help if anyone could help me solving this problem. I am a beginner in this and not able to conclude certain things. I need to select a geared motor for the below system.
System Details
It has a hollow shaft and a Drum
Drum is of ID 3 m, OD 7m and width 5m
The drum is attached to the shaft through steel rods as shown in the figure.
Weight of the drum is 235 ton (density = 1500 kg/m3)

Shaft is 10 m in length, ID of 450mm and OD 600mm and is simply supported by bearings at both the ends.( I have used combined bending and torsion equations to arrive at the diameter the shaft and designed for fatigue.)
The shaft material density is 7850 kg/m3

The weight of the system is calculated as 250 tons
In short
A 250 ton weighing system has to rotate in 1000 rpm using a geared motor.

The calculation I used are as follows
To find power, I need to know the torque
2 pi NT /60

To find torque (T), I used equations like
Method - 1 T = I a ( a=accleration)
a = 0.17 (to reach 1000rpm in 60 min)
Method - 2 T = F R
Where F = m vr2(v=angular velocity)
Is this the correct approach and how do I chose a correct motor for this system


System_pic.JPG (36.3 KB)  






Post Reply
Tell a Friend (must be logged in)
Alert Admin About Post
View All   Previous | Next |

 
Share |

Replies to this message

: Torque,power calculation and Motor selection
: Torque,power calculation and Motor selection -- enahs
Post Reply
Top of thread
Mechanical Engineering Forum
Posted by: enahs

09/07/2010, 04:56:19

Author Profile
eMail author
Edit

I will try to expalin in detail.Please correct me
if I wrong.
I have calculated
T = I a
where a = angular acceleration
I = mass moment of inertia
a = v/t
where v= angular velocity
t = time in seconds
v= 2 pi N /60 where N = rpm

Here N= 1000rpm
therefore v = 104.66 rad/sec

Now a= v/t
= 104.66 /(60*60) For60 minutes
= 0.029 rad/sec^2
I = m (R^2 - r^2)/2
Here m = 180 000 kg
R = 600mm
r = 400mm

Mass moment of inertia = 4.68 * 10^9 Kgm^2

T = I a
= 1357.2 * 10^4 N-m

This torque seems to be very huge.The torque increases as we
reduce the time required to reach 1000rpm ie
when we decrease time, acceleration increases and there by
tourque increases.So we are asuming that we can take the machine to 1000rpm in 60 minuteswhich gives lesser torque.

To select a motor with this torque seems to be hard when we
calculate the power required for this torque
from the equation
(2 pi N T) / 60
Is this the correct method ?


please give your valuble comments.








Post Reply
Tell a Friend (must be logged in)
Alert Admin About Post
Where am I? Original Top of thread Previous |   |


© Copyright 2000 - 2019, by Engineers Edge, LLC All rights reserved.  Disclaimer