Ok, so we know a little more, that helps. Most likely a WFI pump of that size is going to be centrifugal, I cannot imagine a PD pump like a gear pump or a progressive cavity pump being made to be able to stay clean enough for pharma Water For Injection processes. I've never seen one anyway.
So assuming it is centrifugal, it is entirely possible that the 350HP rating is so high specifically BECAUSE they intended it to be used at high pump shaft speeds, regardless of how that was attained. So knowing the torque you WANT at the high speed, you work backward to get a HP at a base speed that you need to purchase. I mention this just to get beyond the initial issue of possible mis-sizing. It may be intentional for all we know.
So now addressing only the higher current issue, there are other possibilities here. As you increase above motor base speed, like I said, the loss of torque can result in an increase in slip, which means more current draw. When the speed increase was done with a gearbox, the motor would have been at a fixed speed and rated slip, then the gearbox was increasing the shaft speed at the pump. One thing we dont know yet for sure (do we?) is the base motor speed of the original moror. My suspicion is that maybe what really happened here is that the original motor was 4 pole, to which they added a gearbox to speed it up to run the pump faster. Someone may have looked at that and said, "Why don't we just use a 2 pole motor and start off at a higher speed?" Seems reasonable to me. The problem would be that although you would be eliminating mechanical losses in the gearbox, i.e. friction and heat, you may at the same time be just shifting the looses to the motor now. If I determined I needed 350HP from a 4 pole motor to have enough shaft torque at the pump at a higher speed, I would need to replace that 350HP 4 pole motor with a 700HP 2 pole motor to get the same shaft torque at the pump! This is what's behind LionelHutz comment by the way, it's an old trick, although I've never seen it done at this power level. Without increasing the HP then, the faster motor has LESS shaft torque, therefore more slip and it draws more current. It might have been a LOT more were it not for the elimination of the losses in the gearbox, so all you are seeing is the current delta in the overall SYSTEM EFFICIENCY differences.
If however this WAS all still done correctly, and let's say the original design was a 4 pole 200HP motor that they replaced with a 2 pole 350HP motor (because of removing the gearbox losses), then the current difference may still be explained in other ways:
1) With the added mass of the gearbox, the 4 pole motor was running at closer to full load so it was at a better power factor. Now the 2 pole motor has less load on it, so the PF is technically worse. You will see higher CURRENT as a result, but the actual kW would more likely be LOWER. I find trying to explain PF issues to wrench turners to be less than a satisfying experience, I usually have to finish the conversations with someting like "Because that's the way it is in the electrical world". Good luck with that if this is the case. The infamous "Beer Mug" analog often helps, if for no other reason than to find something else to talk about with them...
2) All other issues are right with the world and all my above hypothesizing is completely in left field. But... They now added in the issue of running over base frequency. The current, if measured from the INPUT side of the VFD, is now slightly higher because you have added losse. Inside of the VFD, running at a higher output frequency adds switching losses that you did not have before. Its not much, but there would also be added heat in the motor, which is also a total throughput efficieny loss, which would also show up as a current increase.
"Will work for (the memory of) salami"
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