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