C - Constant according to speed of motor
n= 2800 1/Min C=0.23 mm
n= 1400 1/Min C=0.91 mm
n= 930 1/Min C=2.03 mm
Spring deflection S= 0.5 mm
Weight of system G= 250 kgf
Weight of motor(s) R= 26 kgf
Speed of motor n = 2800 1/Min
A system whose the total weight (incl. motor(s)) is 276 kgf will be vibrated with 0,5 mm of the spring deflection at 3000 rpm. Which motor must be use?
The chart below is used for selection of motor.
1. Find and mark the spring deflection on the left side of the chart at 3000 rpm.
2. Find and mark the total weight of system in the middle of the chart.
3. Join the two points with a straight line and lengthen this line to the right side of the chart.
4. Read the total centrifugal force on the right side of the chart.
5. Divide the total centrifugal force by the number of motor.
6. Find the centrifugal force on the table of the performance data.
7. "VEM 80-2-77" will provide the required centrifugal force at 3rd position.
|Frequency of motor at 3000 rpm is 50 t/sn,
|Frequency of motor at 1500 rpm is 25 t/sn,
|Frequency of motor at 1000 rpm is 16,7 t/sn.