Low suction pressure cause overcurrent?

bornwild

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Join Date
May 2010
Location
Riyadh
Posts
429
Hello Guys,

We had problem with bent axial position pump run by vfd. Whenever the input suction pressure dropped , we would see over current faults on the vfd. I am curious to know why does it happen? Shouldn't the current drop when the suction line has low pressure?
 
Pumps do weird things!


I have one customer that figured out to monitor spray pressure they only have to use a current trip relay at a specific current setpoint.


If the pump is running as it is designed the current is low. If the pressure goes high the current goes up as expected, BUT if the pressure drops due to missing spray nozzles or an open fitting the current goes UP also, tripping the relay.



Common sense says if the pressure goes down the motor would be running easier and the current should also go down. The maintenance engineer there says that some pumps are designed for flow and some pumps are designed for pressure and if a pressure pump can't make the pressure it will burn itself up trying to.
 
Do you know if the pump is fixed or variable displacement?
Do you know if the suction pressure is still within the acceptable range for this pump?

Reduced suction pressure indicates higher flow (Bernoulli) or a low tank level. If you don’t hear cavitation noise then likely just higher flow.

Many systems are designed for high flow st low pressure or low flow at high pressure. It is up to the Engineeer to make sure the operating parameters don’t overtax the electric motor. The pressure is determined by conditions downstream from the pump so suction pressure is useless to determine what is happening after the pump.

The motor torque required is proportional to pressure times pump displacement. This is independent of rpm. You need to be looking at this.
 
Pumps do weird things!


I have one customer that figured out to monitor spray pressure they only have to use a current trip relay at a specific current setpoint.


If the pump is running as it is designed the current is low. If the pressure goes high the current goes up as expected, BUT if the pressure drops due to missing spray nozzles or an open fitting the current goes UP also, tripping the relay.



Common sense says if the pressure goes down the motor would be running easier and the current should also go down. The maintenance engineer there says that some pumps are designed for flow and some pumps are designed for pressure and if a pressure pump can't make the pressure it will burn itself up trying to.

But Pumps generate flow and not pressure. So if have a fixed displacement pump running at xxx rpm the oil delivery is also fixed . Pumps dont care about the pressure and they generate only flow. Correct me if I am mistaken
 
Do you know if the pump is fixed or variable displacement?
Do you know if the suction pressure is still within the acceptable range for this pump?

Reduced suction pressure indicates higher flow (Bernoulli) or a low tank level. If you don’t hear cavitation noise then likely just higher flow.

Many systems are designed for high flow st low pressure or low flow at high pressure. It is up to the Engineeer to make sure the operating parameters don’t overtax the electric motor. The pressure is determined by conditions downstream from the pump so suction pressure is useless to determine what is happening after the pump.

The motor torque required is proportional to pressure times pump displacement. This is independent of rpm. You need to be looking at this.

The pump is fixed displacement but the speed is controlled by the vfd so it delivers different volumes at different speed
 
But Pumps generate flow and not pressure. So if have a fixed displacement pump running at xxx rpm the oil delivery is also fixed . Pumps dont care about the pressure and they generate only flow. Correct me if I am mistaken




The OP is not mistaken per se, but what is written is perhaps an oversimplification.


A) Pumps perform work on a fluid, and the rate of that work over time is power: the power performed by a pump is proportional to the fluid flowrate times the pressure rise across the pump*.

B) Positive displacement pumps (e.g. the bent axial of the OP) generate an essentially fixed flow at a fixed RPM over a range of pressure rises, from zero to some maximum pressure rise.

C) The pressure rise is the pump outlet pressure minus the pump inlet pressure; so if the pump inlet pressure decreases, then the pressure rise increases**

A+B+C) If the flowrate is fixed and the the pressure rise increases (because the pump inlet pressure decreases), then the pump will do more work per fluid unit, and will output more power (fixed flow rate times an increased presssure rise) into the flowing fluid. [Power out] is [power in * efficiency], and [power in] is proportional to [voltage in * current in]***; since voltage and efficiency are constant, the overall corellation is that [power out] is proportional to current.


So reducing the inlet pressure requires more current to run.


I'm not at all an expert, but it sounds like the designer underestimated the maximum pressure rise for the system and as a result chose an undersized VFD.

It's amazing what can be done by simply multiplying by unity.



* To first order, and for an incompressible fluid.
** Assuming the pump outlet pressure is constant.
*** I think [power in] = voltage * current * [power factor], but that does not change the ultimate point, which is that, to first order, power is proportional to current
 
The pump is fixed displacement but the speed is controlled by the vfd so it delivers different volumes at different speed

Can you share the pump part number and motor and drive used?

Do you have a plot of the output pressure?

Does the pump draw directly from a reservoir or is there a charge pump involved?
 

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