When calculating the unbalanced hydraulic force on an agitator impeller, one of the articles says that the calculation of force can be done by the formula:
Force = (2000*Output hp*12*Position factor)/(rpm*DIa. of turbine in inches). What does that position factor stand for, and what are the values for a top entry type agitator, with a 11kW motor? In another text it was written that the force can be computed by the formula F = Torque/(0.75*Radius of blade in mm). Which of the two gives an accurate result?
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Re: Calculating unbalanced hydraulic force
11 September 2006 at 1:29pmThe first formula is similar to the one I use, but the value of the constant is slightly different. The position factor may really be a load factor for the specific application, I will provide values later. However, impeller location is important. The hydraulic load is a lateral load, usually used to calculate a bending moment on the shaft. For the bending moment, the hydraulic load must be calculated for each impeller and then multiplied by the shaft length from the support bearing to the impeller location. The individual bending moments are added together to get a total bending load. Since units for the indicated calculation are not clear, let me provide one for which I know the units:Hydraulic Force [pounds force] = (19,000 * Motor Power [horsepower] * Service Factor)/(Rotational Speed [rpm] * Impeller Diameter [inches])
The Service Factor depends on the type of impeller and the operating conditions. Some service conditions will significantly increase the normal loads associated with operating submerged in liquid. Some typical values for the service factor are:
Significant time at liquid level: 2.0 - 3.5
Operating in boiling system: 1.5 - 3.0
Operating in gas sparged system: 2.0 - 3.5
Large volume of solid additions: 3.0 - 5.0
Impacting of large solids: 5.0 - 7.0
Start-up in settled solids: 5.0 - 7.0
Operating in a flow stream: 1.0 - 7.0I hope that this information answers the question and can be used to do shaft design.
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