seventofour
Mechanical
Not really sure if this is a gearbox question or a motor question or if I am even on the right path.
We have a pipe fabrication turntable used to rotate the pipe while a submerged arc welding machine welds the joint. The total length of the pipe is 25ft long and is supported by roller stands. The pipe itself is 24 inch OD (ID = 23.625) 304 stainless steel and weighs 10,000lbs, being welded onto this pipe is a 24 inch and has a valve stem and controller 78" from center of pipe to the top of the valve. The Valve weighs 12,000lbs. For a total of 22,000lbs.
The motor is a 5HP motor that the user can control the output rpm with a vfd. The output shaft of the motor goes to a 7" center distance gearbox with a 50:1 ratio, so when the motor sees all 60Hz the gearbox input is around 1750rpm and the output is 35rpm and 13011 in-lbs of torque. The output of the gearbox goes through another reducer with a 6:1 ratio and gives an output around 5.8rpm.
I tried to calculate the moment of inertia by chopping up the valve into 3 pieces (the valve body, stem, and actuator)and then the pipe into one. Added them all together.
Most of the dimensions for the valve are guesses since the piece in question is no longer in our shop.
The valve I=mass(OD^2 +ID^2)/2: with mass in slugs
body,mass 9600lbs = 298.37slugs
I=298.37 ( 26^2+24^2)/2
I=186780lb-sec2/in
Stem, mass 175lbs = 5.44slugs
I=5.44(8^2+6^2)/2
I=272lb-sec2/in
Valve Actuator mass 2225lbs =69.2slugs
I=69.2(20^2+2.5^2)/2
I=14056lb-sec2/in
Valve Total 201108lb-sec2/in
Pipe is 92.6lb per foot
2315lbs = 71.9 slugs
I=71.9(24^2+23.625^2)/2
I=40772lb-sec2/in
Weldment total = 241881lb-sec2/in
Alpha = (Final velocity*2pi/60)/Time
Alpha = (2.5*2pi/60)/3.5seconds
Alpha = 0.0748 rad/sec2
Torque =I*Alpha
T=241881 * 0.0748
T=18092.6 in-lbs
But wouldn't this starting torque value (if I did this correctly) differ depending on where the end of the valve is located such as 12 o-clock would need less torque compared to 3 o-clock?
We have a pipe fabrication turntable used to rotate the pipe while a submerged arc welding machine welds the joint. The total length of the pipe is 25ft long and is supported by roller stands. The pipe itself is 24 inch OD (ID = 23.625) 304 stainless steel and weighs 10,000lbs, being welded onto this pipe is a 24 inch and has a valve stem and controller 78" from center of pipe to the top of the valve. The Valve weighs 12,000lbs. For a total of 22,000lbs.
The motor is a 5HP motor that the user can control the output rpm with a vfd. The output shaft of the motor goes to a 7" center distance gearbox with a 50:1 ratio, so when the motor sees all 60Hz the gearbox input is around 1750rpm and the output is 35rpm and 13011 in-lbs of torque. The output of the gearbox goes through another reducer with a 6:1 ratio and gives an output around 5.8rpm.
I tried to calculate the moment of inertia by chopping up the valve into 3 pieces (the valve body, stem, and actuator)and then the pipe into one. Added them all together.
Most of the dimensions for the valve are guesses since the piece in question is no longer in our shop.
The valve I=mass(OD^2 +ID^2)/2: with mass in slugs
body,mass 9600lbs = 298.37slugs
I=298.37 ( 26^2+24^2)/2
I=186780lb-sec2/in
Stem, mass 175lbs = 5.44slugs
I=5.44(8^2+6^2)/2
I=272lb-sec2/in
Valve Actuator mass 2225lbs =69.2slugs
I=69.2(20^2+2.5^2)/2
I=14056lb-sec2/in
Valve Total 201108lb-sec2/in
Pipe is 92.6lb per foot
2315lbs = 71.9 slugs
I=71.9(24^2+23.625^2)/2
I=40772lb-sec2/in
Weldment total = 241881lb-sec2/in
Alpha = (Final velocity*2pi/60)/Time
Alpha = (2.5*2pi/60)/3.5seconds
Alpha = 0.0748 rad/sec2
Torque =I*Alpha
T=241881 * 0.0748
T=18092.6 in-lbs
But wouldn't this starting torque value (if I did this correctly) differ depending on where the end of the valve is located such as 12 o-clock would need less torque compared to 3 o-clock?
