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Topic: How do I calculate the forces on a baffle?

posed by  Last edited: 30 July 2012 At 3:15pm
I am trying to design a tank baffle. I do not know how to calculate the forces on it. The parameters are:
Tank height: 7.4m
Tank Diameter: 7.5m
Baffle width: .625m
Agitator power: 22kW
Material in tank: Slurry of finely ground rock
SG of slurry mixture: approximately 2.0

Please could you tell me how to determine the pressure on the baffleplate due to the rotation of the slurry?
  • Avatar Dave Dickey Forum Moderator 324 Posts

    Re: How do I calculate the forces on a baffle?

    Last edited: 30 July 2012 at 3:15pm
    Your question is a good one, but you have failed to provide essential information, along with some unnecessary information. The only information you need is the maximum torque of the mixer, which is power divided by speed and the tank diameter. You have provided the motor power, which is essential, but you failed to provide the rotational speed of the mixer which is necessary to calculate the mixer torque. The input torque from the mixer must be taken out by the baffles, so the torque load by the mixer becomes the force load on the baffles divided by the tank diameter. The tank height, baffle width, and slurry density will have no effect on the maximum load on the baffles. The only other piece of information of possible interest is the number of baffles. The logical assumption would be that the load on each baffle would be the same and a fraction of the total imposed torque. The answers by this expert are based on the best available interpretation of the information provided. The consequences of the application of this information are the responsibility of the user. If clarification is needed, please submit a further question.
  • Avatar Filip B Community Member 2 Posts

    Re: How do I calculate the forces on a baffle?

    Last edited: 31 October 2012 at 12:10pm
    I have thought over of you way of thinking - it is interesting. However when we mixing, nevermind what, but in tank without baffles the agitator is loading by torque. I suppose that main part of torque take over the liquid and some energy is taking by the wall and additional equipment the rest is losses. When we mixing in tank with baffles we have to reserve more power about 20% (it is depend on dimensions and number of baffles). Could we consider about it? I hope that you don't mind that I presented my point of view. Formula to calculate force acting on the baffle is in attachment - full torque is taking over by the baffles.