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How do I determine the power necessary to accelerate a pump to meet a specific change in demand?

The design is a VFD driving a constant torque positive displacement pump with a varying capacity load. The VFD will control pump speed thus met the flow demand. The steady state power is easily addressed. I need to assure that I can accelerate the pump from 50% load to 75% load in one second.

What are the specification parameters that will address this? I have talked with the suppliers of each of the components and none have a formula or procedure to determine this.

After determining this necessary properties, I need to include the these parameters into a dynamic model of the plant in Matlab.

I also need to determine how to bring the pump down in speed from 100% to 0% as fast as practical and know how fast that is. The load rejection does not occur often so regenerative braking is not necessary, a resistor is expected to be used.

I have looked at this from a dimension analysis view but I can't seem to find what the appropriate relationships are.

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  • \$\begingroup\$ Basically, you are angularly accelerating a mass with a particular moment of inertia. So, you can use t=I*a (torque = moment of intertia * angular acceleration). When the pump is acting in steady state at speed S = S0, the torque applied by the pump opposing the motor is known (you should know what this is). Then you increase the torque to max, instantaneously. The NET torque is max motor torque minus torque at speed S0. This causes an acceleration of the rotational mass that you can calculate. Now you have a new pump torque which you can calculate based on the new speed. Etc. \$\endgroup\$ – mkeith Jan 15 at 6:03
  • \$\begingroup\$ It seems to me, though, that the amount of water moving in the pipe will also affect how quickly the flow rate changes (you can change the pressure as fast as you want, but the flow rate can only change so fast). You can't accelerate a body of moving water any faster than F=ma. So that may be another way to approach it. Via pressure and force and acceleration. \$\endgroup\$ – mkeith Jan 15 at 6:06
  • \$\begingroup\$ mkeith, Thank you. Yes, that is my challenge. I have a pipe full of fluid in motion, I need to accelerate that mass. The acceleration includes the mass itself and the flow resistance. The pump and motor rotational mass is small in comparison to the mass in the pipe. A further challenge is PD pumps are considered constant torque. When I apply the manufacturer's horse power calculation and then convert that to torque at a given speed/flow rate, the torque is always the same. The formula calculated power converts to torque at a given rpm and the torque is the same. OK, I repeated it. \$\endgroup\$ – rayj Jan 15 at 18:52
  • \$\begingroup\$ I didn't read your question carefully enough. Sorry. Obviously the pump is not constant torque when flow regime changes. I think torque and pressure are inextricably linked. Seems to me that motor output torque translates directly to linear force on the piston of the pump (even if it is not a piston pump, it ends up being the same because it is positive displacement). So to a first approximation, motor output torque will be linearly related to force applied to water which will be linearly related to pressure. Not my area of expertise. But I think you can maybe work with that. \$\endgroup\$ – mkeith Jan 15 at 19:19
  • \$\begingroup\$ What kind of pump is it though? Centrifugal or Positive Displacement? \$\endgroup\$ – J. Raefield Jan 16 at 0:26
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Increasing the load from 50% to 75% load is an indication of power = Torque * RPM = I*V/f *k for some constant k.

Thus a 50% increase in load , I would expect VI to increase 50%. But to accelerate to 50% increase in load in 1 second will demand full power as RPM increases then drop to the steady state V to achieve the 75% VI product at the RPM.

An estimate of this capacity is the step response to full speed at full load and this must be < 4 seconds to achieve a step load from 50 to 75% in 1 second using full voltage used for 100% RPM but at 50% RPM in order to accelerate at max rate then reduce V to 75% to 85% according to the the motor profile.

added

Since VFD's normally have V/F as a constant, it is unknown what the max demand/rating ratio is for this demand response time of 1 s. and how the load is related to RPM , V/F & I if torque is in fact constant. But I suspect the constant torque to to maintain constant water pressure, which has suddenly dropped or demand flow has increased x% .

The other thing is you don't want a sudden pressure change which could cause all sorts of vibrations in the network, unless there is some pressure regulation. Normally I might expect a tank ballast buffers the load variations. So the flow rate is limited by the pump speed response to move a certain mass of fluid in 1 second to a certain elevation change.

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  • \$\begingroup\$ Sunnyskyguy, Thank you. This looks like the right path. I need to determent the amount of mass and its acceleration to determine the power required. \$\endgroup\$ – rayj Jan 16 at 1:03

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