Dave Dickey Forum Moderator 292 Posts
Re: Agitator force20 December 2006 at 1:29pmTo calculate the force on the cold bar, you must first calculate the torque delivered by the mixer:
torque [in-lbs] = 63025 (horsepower [hp]) / (rotational speed [rpm])
= 63025 (150) / (84) = 112545 [in-lbs]
Assuming 100% of the load being transferred to the cold bar (actually some of the torque will be transferred to the walls of the container, so the 100% will be conservative). The cold bar must withstand the force associated with all of the torque from the mixer. In other words, the load on the bar will be the torque divided by the distance from the axis of rotation of the mixer to the centerline location of the cold bar:
force [lbs] = (112545 [in-lbs]) / (distance from axis of rotation to bar [in])
The force will be distributed over the length of the bar exposed to the dough. The design should be based on a cantilevered bar. To design the cold bar (stainless steel), use the following allowable stress limits: 6,000 psi shear stress and 10,000 psi tensile stress. Those limits are good for fatigue requirements in typical mixer applications.
I have an agitator which is used to mix the dough. The agitator is running at 150 hp motor at 84 rpm. Taking 1005 efficiency, how do I calculate how much force will the dough exert on a fixed cold bar? The idea is that the dough gets squeezed between the agitator and the fixed cold bar.
Have an insight or suggestion?
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