This is an archival version of the original KnowledgePoint website.

Interactive features have been disabled and some pages and links have been removed.

Visit the new KnowledgePoint website at https://next.knowledgepoint.org.

 

Revision history [back]

click to hide/show revision1
initial version
Daniel Fayolle gravatar image
RedR

There are an important rule to observe with this kind of problem: always start the largest motor first and also always make sure the full load rating of the generator is never exceeded, especially upon starting a new motor load. Remember that squirel cage induction electric motors take up to 6 times their full load current upon starting. This is why when these pump motors are fed from a "limited capacity" generator they are always started by means of assisted device like an autotransformer or a star-delta motor starter.

Another rule: never run the generator on full load permanently. Aim at 80% FL if the load is permanent. Never above 80% of generator capacity.

The type of starter is not mentionned. But I would guess they are star-delta starters, quite suitable for centrifugal pumps motors. With this kind of starter the starting current of the motor is reduced to 2 times full load current.

The method for appraising the generator size is as follows:

Starting with largest motor first: 1) 195 kVA horizontal pump, 296A FL, load on generator is 592A 2) 195 kVa pump, 296A FL, load on generator: 296 + 592A (2X296) = 888 A 3) 165 kVA horizontal pump, 250A FL, load on generator: 592 + 500A = 1092A 4) 110 kVa pump, 162A FL, load on generator: 296 + 296 + 250 + 324 = 1166 A 5) 110 kVA pump, 162 A FL, load on generator: 296+296+250+162+ 324= 1328A 6) 110 kVA pump, 162A FL, load on generator = 296+296+250+162+162+ 324 = 1490A 7) 110 kVA , 162A FL, load on generator: 296+296+250+162+162+162+324= 1652A 8) 40 kVA , 62A FL, load on generator: 296+296+250+162+162+162+162+124 (2X62)= 1614 A upon starting of last generator. Thus generator minimum size required is : 1.732X380X1614= 1062 KVA minimum.

So you would need another generator of 1062 kVA size. Preferably 1200 kVA. Or a second one of circa 500 kVA to operate in paralell with the existing one.

This calculation above is based on the assumption that all electric motors are star-delta started and that all pumps and motors run on full load.

Daniel FAYOLLE RedR Expert

click to hide/show revision2
No.2 Revision
KnowledgePointAdmin gravatar image
RedR CCDRR

There are an important rule to observe with this kind of problem: always start the largest motor first and also always make sure the full load rating of the generator is never exceeded, especially upon starting a new motor load. Remember that squirel cage induction electric motors take up to 6 times their full load current upon starting. This is why when these pump motors are fed from a "limited capacity" generator they are always started by means of assisted device like an autotransformer or a star-delta motor starter.

Another rule: never run the generator on full load permanently. Aim at 80% FL if the load is permanent. Never above 80% of generator capacity.

The type of starter is not mentionned. But I would guess they are star-delta starters, quite suitable for centrifugal pumps motors. With this kind of starter the starting current of the motor is reduced to 2 times full load current.

The method for appraising the generator size is as follows:

Starting with largest motor first:

1) 195 kVA horizontal pump, 296A FL, load on generator is 592A 592A

2) 195 kVa pump, 296A FL, load on generator: 296 + 592A (2X296) = 888 A A

3) 165 kVA horizontal pump, 250A FL, load on generator: 592 + 500A = 1092A 1092A

4) 110 kVa pump, 162A FL, load on generator: 296 + 296 + 250 + 324 = 1166 A A

5) 110 kVA pump, 162 A FL, load on generator: 296+296+250+162+ 324= 1328A 1328A

6) 110 kVA pump, 162A FL, load on generator = 296+296+250+162+162+ 324 = 1490A 1490A

7) 110 kVA , 162A FL, load on generator: 296+296+250+162+162+162+324= 1652A 1652A

8) 40 kVA , 62A FL, load on generator: 296+296+250+162+162+162+162+124 (2X62)= 1614 A upon starting of last generator.

Thus generator minimum size required is : 1.732X380X1614= 1062 KVA minimum.

So you would need another generator of 1062 kVA size. Preferably 1200 kVA. Or a second one of circa 500 kVA to operate in paralell with the existing one.

This calculation above is based on the assumption that all electric motors are star-delta started and that all pumps and motors run on full load.

Daniel FAYOLLE FAYOLLE

RedR Expert

click to hide/show revision3
No.3 Revision
KnowledgePointAdmin gravatar image
RedR CCDRR

There are is an important rule to observe with this kind of problem: always start the largest motor first and also always make sure the full load rating of the generator is never exceeded, especially upon starting a new motor load. Remember that squirel cage induction electric motors take up to 6 times their full load current upon starting. This is why when these pump motors are fed from a "limited capacity" generator they are always started by means of assisted device like an autotransformer or a star-delta motor starter.

Another rule: never run the generator on full load permanently. Aim at 80% FL if the load is permanent. Never above 80% of generator capacity.

The type of starter is not mentionned. But I would guess they are star-delta starters, quite suitable for centrifugal pumps motors. With this kind of starter the starting current of the motor is reduced to 2 times full load current.

The method for appraising the generator size is as follows:

Starting with largest motor first:

1) 195 kVA horizontal pump, 296A FL, load on generator is 592A

2) 195 kVa pump, 296A FL, load on generator: 296 + 592A (2X296) = 888 A

3) 165 kVA horizontal pump, 250A FL, load on generator: 592 + 500A = 1092A

4) 110 kVa pump, 162A FL, load on generator: 296 + 296 + 250 + 324 = 1166 A

5) 110 kVA pump, 162 A FL, load on generator: 296+296+250+162+ 324= 1328A

6) 110 kVA pump, 162A FL, load on generator = 296+296+250+162+162+ 324 = 1490A

7) 110 kVA , 162A FL, load on generator: 296+296+250+162+162+162+324= 1652A

8) 40 kVA , 62A FL, load on generator: 296+296+250+162+162+162+162+124 (2X62)= 1614 A upon starting of last generator.

Thus generator minimum size required is : 1.732X380X1614= 1062 KVA minimum.

So you would need another generator of 1062 kVA size. Preferably 1200 kVA. Or a second one of circa 500 kVA to operate in paralell with the existing one.

This calculation above is based on the assumption that all electric motors are star-delta started and that all pumps and motors run on full load.

Daniel FAYOLLE

RedR Expert