Pump Operating Torque is higher than Output Power of the Gearbox

[bmw318be]

I have some doubt is pump initial Sizing application

50 Hz Motor with 150 HP VFD Motor 4 Poles (1450RPM) with a gearbox ratio of 5.3 with Safety Factor 2.3 is having Gearbox Shaft Torque Maximum of 3500 Nm


I am getting a motor torque from a motor and multiple by the Gearbox ratio, correct me if I am wrong because I am getting very low output torque here.

I am operating a PD internal gear Pump for various service condition:

My issue is that I could not accommodate some of service condition as it has higher running torque than the Output Gearbox Torque. Although the BHP of the pump is still within the Motor Limit of 150 HP

Some cases is running at multiple viscosity of 1500 CST, Q = 30 m3/hr at 110 Degree C , SG of 0.9
My BKW is 18 HP
Pump rated speed of 35 RPM

My calculated Operating torque or Power Required to drive at this operating condition is 3660 Nm
However, looking at VFD speed curves, I am running at approximately at 8 Hz at 35 RPM and the Output Motor torque after Gearbox is 75% of Maximum Torque which is 2625 Nm


I need to check if the pump operating Torque has to be lesser than output Gearbox Torque in order to run smoothly. I am having a VFD motor characteristic as followed as speed goes down the torque goes down too from 50 Hz to lower Hz. However, I am still operating at a speed between 5 Hz to 50 Hz.

Is there anyway to increase the Motor Torque with any coupling as my understanding Motor Torque is fixed and I am not able to upsize the motor due to range limitation. It is clear that I am having higher torque required because of lower rated speed of the pump and higher Break Horse Power .


Some terminology of Internal Gear pump that it is constant Torque.

However it seems, when we operate at various flow rate, pressure or viscosity, the BHP (Break Horse Power) and Rated RPM would various and My understanding is that we could not reduce the torque by lowering the flow, lowering the flow would increase the torque as speed is inversely proportional to the Torque and BHP or Break Horse Power is directly proportional to the Torque.

Torque (Lb ft) = ( BHP * 5252) / RPM




Lastly, I have seen that the Power for BHP of the pump can be calculated by

Pump HP = Flow (GPM) * Pressure (PSI) / (1714 * efficiency of pump )

How this formula is valid for BHP calculation for Internal Gear Pump as some liquid properties such as (Viscosity,Specific Gravity & Temperature are not part of the formula here.

Anyway, my BHP above case based on the Pump manufacturer calculation, however it far off the calculated using this formula.

 

[bmw318be]

Pumpsonly, thanls some of your point is relevant, yes I met a motor supplier who told me the same, the derating factor of the torque at lower speed is used to prevent overheating .I need to understand more materials on Exproof motor design.


However, using 6 poles vs 4 poles at full speed running it at the same rated speed after gearbox. It shows the torque Of 6 poles motor is half of 4 poles so even after gearbox, it is not comParatively higher.
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I observed that I am having an issue of insufficient torque running the pump at below 10 Hz because the % Torque on motor is below 80 % from 50 Hz Torque.

Is it advisible to still run it ?


Another question on inverter that you recommend to use the same brand, however my motor supplier do not have VFD so I would to know if the inverter for constant torque what does it implies ?


Is the constant torque VFD would follow my motor % torque characteristic or it would deviate lower ? If lower i would need to ensure I got sufficient margin or it could be a big problem.

Another VFD supplier explained that its constant torque VFD would try to match the torque to my pump's torque demand which I still doubts that it could do so