- #1
Rocket795
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Hello,
I'm stuck on what should be a fairly simple problem. Trying to calculate the size of motor required to rotate steel pipe at 0.33RPM around its axis, on 2 sets of rollers, 4 wheels total (hard rubber).
Pipe
48" OD Mild Steel pipe 0.5WT
24m in Length
Rollers
4 x 12" Rubber wheels (40" Center to Center) on bearings
This is where is stand.
Torque required to accelerate the system (0.00 RPM to 0.33 RPM in 1 sec)
Ta=Iα
I=(∏/32)ρL(D1^4-D2^4)
1.219 D1 – External diameter of the disc [m]
1.194 D2 – Internal diameter of the disc [m]
7,850.000 ρ – Density of the load [kg/m3]
9,069.050 m – Weight of the load [kg]
24.000 L – Length of the load [m]
I=3,300.689
α=(w1-w0)/t
0.000 ω0 – Initial velocity of the motor [rad/s]
0.035 ω1 – Final velocity of the motor [rad/s] or 0.33 RPM
1.000 t – Time for velocity change
α=0.035
Ta=114.064 N*m
Now the next part i am not too sure about,
Torque required to overcome rolling resistance :
Rolling resistance coefficient - f = 0.0077m Hard Rubber on Steel
http://www.roymech.co.uk/Useful_Tables/Tribology/co_of_frict.htm
F = f x W/R
Fr=(0.0077m*9069.050kg*9.81 kg/s^2)/0.610m
Trr= 1123.031 N *.610m = 685.049 N*m
Total Torque required to rotate the pipe at 0.33rpms in 1 sec.
T=Ta+Trr=799 N*m
or
589 Lb*ft
This seems low to me and this is only the torque at the pipe, once geared that torque will be significantly lower. will I know I'm missing the bearing friction torque required but that can't be that much.
What else should I be calculating?, if anything.
I have attached a drawing of the system, once I find the torque required for rotate the pipe I should have the required input torque dividing that Torque by the gear ratios.
One more thing, Bonus question. Some of these types of pipe have a seam welded down the middle along the length of the pipe, it projects from the pipe a few mm sort of like a speed bump on the pipe. Is there a way to calculate the torque required to over come the rollers once the seams comes in contact with the roller wheels?
Thanks in advance.
I'm stuck on what should be a fairly simple problem. Trying to calculate the size of motor required to rotate steel pipe at 0.33RPM around its axis, on 2 sets of rollers, 4 wheels total (hard rubber).
Pipe
48" OD Mild Steel pipe 0.5WT
24m in Length
Rollers
4 x 12" Rubber wheels (40" Center to Center) on bearings
This is where is stand.
Torque required to accelerate the system (0.00 RPM to 0.33 RPM in 1 sec)
Ta=Iα
I=(∏/32)ρL(D1^4-D2^4)
1.219 D1 – External diameter of the disc [m]
1.194 D2 – Internal diameter of the disc [m]
7,850.000 ρ – Density of the load [kg/m3]
9,069.050 m – Weight of the load [kg]
24.000 L – Length of the load [m]
I=3,300.689
α=(w1-w0)/t
0.000 ω0 – Initial velocity of the motor [rad/s]
0.035 ω1 – Final velocity of the motor [rad/s] or 0.33 RPM
1.000 t – Time for velocity change
α=0.035
Ta=114.064 N*m
Now the next part i am not too sure about,
Torque required to overcome rolling resistance :
Rolling resistance coefficient - f = 0.0077m Hard Rubber on Steel
http://www.roymech.co.uk/Useful_Tables/Tribology/co_of_frict.htm
F = f x W/R
Fr=(0.0077m*9069.050kg*9.81 kg/s^2)/0.610m
Trr= 1123.031 N *.610m = 685.049 N*m
Total Torque required to rotate the pipe at 0.33rpms in 1 sec.
T=Ta+Trr=799 N*m
or
589 Lb*ft
This seems low to me and this is only the torque at the pipe, once geared that torque will be significantly lower. will I know I'm missing the bearing friction torque required but that can't be that much.
What else should I be calculating?, if anything.
I have attached a drawing of the system, once I find the torque required for rotate the pipe I should have the required input torque dividing that Torque by the gear ratios.
One more thing, Bonus question. Some of these types of pipe have a seam welded down the middle along the length of the pipe, it projects from the pipe a few mm sort of like a speed bump on the pipe. Is there a way to calculate the torque required to over come the rollers once the seams comes in contact with the roller wheels?
Thanks in advance.