Determining power requirements for an AC Motor

  • Thread starter Whatamiat
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I am trying to determine the power requirements for an AC Motor for a Plunger type wave maker.

Attached is the layout of the plunger and tank.

The formulas that I think might be involved in solving this problem are below:
P.E = mgh
W= F x d
Power= W/time
Pressure= F/ area

What I know(guesstimates to try help me understand the problem) :
Volume of plunger =.125m3 => amount of water displaced by plunger = 125kgs of water
Distance = 50cm = plunger moves from start position 50cm downwards.
Time= 1 sec for plunger to move downward 50cm.
10000kg of water in the tank.
Horizontal force of water in tank= 375kg force per unit area = 3678.75 N

What I don't know:
How is Force calculated in this situation
Which of the above formula(if any) can be used in figuring this out
Does it make a difference that the external force used to move the water is acting vertically rather than horizontally?

Any idea anyone?
 

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  • #3
jrmichler
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There is no easy way to hand calculate the force required. There is, however, a way to estimate the force. The estimate will be crude, so some safety factors will be built into the calculation. Some assumptions are needed to start solving this problem:

1) This is a one time move, followed by a slow return. Continuous cyclic motion introduces additional complexity from waves bouncing back against the plunger.
2) Motion will be driven by a servomotor. A normal induction motor will not easily make a controlled one second move.
3) The motion profile will accelerate for 0.5 second, then decelerate for 0.5 second.
4) The plunger with associated moving parts has mass 125 kg. Why 125 kg? It seems like a reasonable number.
5) Friction is negligible. This assumption assumes that friction has been calculated and designed for.
6) I'll assume that 500 kg of water is directly moved by the plunger. The water immediately adjacent to the plunger is directly moved by the plunger, while water farther away is moved by the wave, so this amount is purely a WAG. Improving this number can be done by studying existing wave making machines or by CFD. For those not familiar with engineering terms, search wag swag definition.
7) The tank is 1 meter wide, based on the numbers in your diagram.
8) Assume that the peak mass is subject to peak velocity and acceleration. This is conservative, but conservative is the way to go in a system with this many unknowns.
9) Static forces are ignored because they reduce the force of the downward move. The buoyancy force on the plunger is also ignored.
10) The motor and plunger drive mechanism are designed so that the peak motor RPM is appropriate for the motor and drive combination. The peak motor RPM will be close to, but less than, the maximum allowable motor RPM. The peak motor torque must be less than the peak allowable motor torque, but is allowed to be larger than the continuous motor torque rating (based on Assumption #1).

Now for the calculations:
1) Total moving mass = 125 kg of plunger plus 500 kg of water equals 625 kg.
2) Acceleration required, from distance = 0.5 * acceleration * time^2 = 0.25m * 2 / 0.5 sec^2 = 2.0 m/sec^2.
3) Peak velocity = 2.0 m/sec^2 * 0.5 second = 1.0 m/sec.
4) Force = 2.0 m/sec^2 * 625 kg = 1250 N.
5) Power = 1250 N * 1.0 m/sec = 1250 watts.
6) Calculate the total drive gear ratio based on Assumption #10.
7) Alternative method #1: Find a similar wave machine, and copy their drive. Scale as needed.
8) Alternative method #2: Start with CFD. Make sure that you properly verify the CFD pressure and force results before using them to size the drive system.
9) There will be a static force from the water piling up against the plunger. This will mostly be against the bearings supporting the plunger. I think it will be small compared to the forces estimated above because of the relatively low acceleration and peak velocity. But I could be wrong.
 
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