TMdrive -70 Product Application Guide - Tmeic.com

Inverter ExampleWhen specifying an inverter, start from the process requirements and work through the motor to the inverter.The following example illustrates this process.1Define processrequirements.2Select motor based on processrequirements and computerequired inverter kVA.• 6500 kW (8700 hp)• 500 rpm, 3100 V• Efficiency = 0.965• Power factor = 1.00• Service factor = 1.0• Synchronous3Compute continuouscurrent requirements forthe inverter based onthe selected motor.4Select inverter based oncontinuous current andoverload requirements.Scan the 150% entries in theinverter tables for a frame wherethe continuous current ratingexceeds 1234 amps.The 8000 frame meets thiscriterion (1360 amps) and isappropriate for this application.kW Shaft = 6500 kW (8700 hp)500 rpmThe motor delivers constant torque fromzero to base speed of 500 rpm and 6500kW (8700 hp).Duty cycle requires 150% for 10 sec. but hasrms duty cycle of 6500 kW (8700 hp).I ac Inverter =kW Shaft x 1000 x SF Mtr3 x V Motor rated voltage x Eff Mtr x PF Mtr= 6500 x 1000 x 1.03 x 3150 V x 0.965 x 1.0= 1234 ampsCurrentA ac136011661020907816AllowableOverload %150175200225250Regenerative Converter (TMdrive-70) ExampleWhen specifying a converter, start from the process requirements and work through the motor to the inverter, and then theassociated converter. The following example illustrates this process (continuation of inverter application example from above):1 Compute kW requirementsinto the inverter. It isassumed that the converteris dedicated to the inverterspecified in the applicationexample above.It is also assumed that theconverter is controlled to unitypower factor.kW ac2I ac ConverterCompute continuous ac currentrequirement of the converter based on itspower requirements.3kW= ac x 10003 x V Converter line-to-line voltage x Eff drive= 6736 kW x 10003 x 3550 V x 0.985= 1112 amps= kW ShaftEff MtrNote: For sizing systems with peak powers in regenerative mode, adifferent equation is used to compute power requirements.= 6500 kW0.965kW ac = kW Shaft x Eff Mtr= 6736 kWScan the regenerativeconverter table forentries that exceedyour overload (150%), time (60sec) and continuous currentrequirements (1112 amps). In thiscase the 8000 frame TMdrive-P70meets the requirement and isappropriate for this application.CurrentA ac136011661020907816Overload –Time150% – 60s175% – 60s200% – 60s225% – 60s250% – 60s13