rvim002 Posted April 6, 2008 Report Share Posted April 6, 2008 Hi All, The quality of the content I get from this forum is just fantastic. Thanks for all the info. I am in a process of specifying the transformer capacity for starting three 1250HP motors. I looked into a few websites for the transformer sizing guidelines and this is what I have come up with. I assume that it is going to be a 5.7% impedance transformer. I would require the two motors running and the third motor to be started, with a maximum of 10 % voltage drop at the motor terminals. Motor nameplate details: 1250 HP 3300V Locked rotor current: 1254 A Full load current: 198 A Full load power factor: 86% Code Letter: G Calculation of motor starting kVA: For Code G motor, starting kVA = 6.29 times Horsepower So, Motor starting kVA = 6.29 x 1250 = 7.862 MVA Transformer sizing: Total transformer kVA size required = ( Running kVA rating of two 1250 HP motors) + ( Transformer rating required to start the 3rd motor with 10 % Volt drop) Running kVA Rating of 1250 HP motors = 1.732 * ( Rated Voltage) * ( Rated FL Current) / Assumed average power factor of 85% =1.732 * 3300 * 198 / 0.85 = 1332 kVA So, Running kVA of two 1250 HP Motors = 2 * 1332 kVA = 2.664 MVA Generally, Volt drop due to motor starting in % = Motor starting kVA / ( Motor starting kVA + Transformer short circuit kVA) So, 10% = (7.862MVA) / ( 7.862 MVA + Transformer short circuit capacity) So, calcualted transformer short circuit capacity = 70.76 MVA Transformer Short circuit capacity = Transformer kVA rating / (Transformer impedance) So, transformer kVA Rating = Transformer short circuit capacity * Transformer impedance So, the transformer rating required to start the third motor with 10 % volt drop = 70.76 MVA * .057 transformer rating required to start the third motor with 10 % volt drop = 4.033 MVA Recollect Total transformer kVA size required = ( Running kVA rating of two 1250 HP motors) + ( Transformer rating required to start the 3rd motor with 10 % Volt drop) = 2.664 MVA + 4.033 MVA = 6.697 MVA Does this approach seem right? Or am I going wrong somewhere? I would actually be using softstarters in this case, but I need to prove that 5MVA transformer is not sufficient for DOL starting of third 1250 HP motor, with two other 1250 HP motors running. Thanks in advance Cheers Link to comment Share on other sites More sharing options...

rvim002 Posted April 21, 2008 Author Report Share Posted April 21, 2008 Bump Bump Link to comment Share on other sites More sharing options...

marke Posted May 1, 2008 Report Share Posted May 1, 2008 Hello rvim002 Essentially, you are on the right lines. Iwould probably use a different calculation method but the concept is similar. The calculated running current is probably a little high. The power rating is actually the shaft power, so the electrical power is the shaft power divided by the efficiency. A motor of this size could reasonable have an efficiency of 96% and a power factor of 0.95 (your figure of 0.85 power factor is probably on the low side but does include a safety margin. By my calculations: 1250HP = 933KW = 933/0.96/0.95 = 1023KVA full load per motor. Start KVA would be in the order of 7 - 8MVA. Allow for 8MVA to be sure. The load with two motors runing and one starting would be in the order of 10MVA. If we assume a 5.7% transfomer impedance and a voltage drop of 10%, then the transformer rating would be 5.7/10 x 10MVA so would be in the order of 5.7MVA. My calculations yeild a reduced transformer size, but this is primarily due to the reduced runing KVA compared to your estimation. You would need to get a good indication of the full load power factor and efficiency to firm up on these figures. Best regards, Mark Empson | administratorSkype Contact = markempson | phone +64 274 363 067LMPForum | Power Factor | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | Pressure Transducers | Smart Relay | GSM Control | Mark Empson Website | AuCom | Soft Starters Link to comment Share on other sites More sharing options...

rvim002 Posted May 4, 2008 Author Report Share Posted May 4, 2008 Thanks Marke The manufacturer datasheets suggests that the full load power factor is 86% and the efficiency is 96%. Is there an industry standard governing the volt drop during the motor startup? I see that the Nema Fire pump standards NFPA 20 suggests 15% volt drop during startup is OK. But it is only a guideline and not a standard. Have you come across any industry standard on volt drop during startup? Thanks again. Hello rvim002 Essentially, you are on the right lines. Iwould probably use a different calculation method but the concept is similar. The calculated running current is probably a little high. The power rating is actually the shaft power, so the electrical power is the shaft power divided by the efficiency. A motor of this size could reasonable have an efficiency of 96% and a power factor of 0.95 (your figure of 0.85 power factor is probably on the low side but does include a safety margin. By my calculations: 1250HP = 933KW = 933/0.96/0.95 = 1023KVA full load per motor. Start KVA would be in the order of 7 - 8MVA. Allow for 8MVA to be sure. The load with two motors runing and one starting would be in the order of 10MVA. If we assume a 5.7% transfomer impedance and a voltage drop of 10%, then the transformer rating would be 5.7/10 x 10MVA so would be in the order of 5.7MVA. My calculations yeild a reduced transformer size, but this is primarily due to the reduced runing KVA compared to your estimation. You would need to get a good indication of the full load power factor and efficiency to firm up on these figures. Best regards, Link to comment Share on other sites More sharing options...

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