Inductive vs non-inductive load testing

[obruno]

I have just been assigned to review our in-house 15year old testing procedures. I ran across a procedure that has never been performed other than during its initial conception and no one that performed the procedure is still around and very limited details as what was used. It calls for testing a switch at its rated 240VAC and 150% rated current while connected to an inductive load with a 70% power factor for “x” amount of cylces/minute. The question is, can I use a “non-inductive” load actually measured about 98%pf and modify the load to get 105% which is 70% of 150% since the only thing being tested is the switch? The goal here is to avoid purchasing the inductive loads, since there’s no trace of an inductive load to connect. If there is no other choice also advice as to best place to purchase. Thanks!

 

[davidbeach]

The more inductive the load the harder it is for the switch to interrupt the current, so a resistive load does not test the switch the same way an inductive load would.

 

[rbulsara]

Your statement

The question is, can I use a "non-inductive" load actually measured about 98%pf and modify the load to get 105% which is 70% of 150% since the only thing being tested is the switch?

also does not make sense.

70% pf is 70% pf or 0.7. If the switch is say rated 10 amps, you need to find a load that draws 15 amps (150% of 10A) at 0.7 pf at 240V, to meet your spec.

Assuming a single phase switch, that equates to 240*15*0.7= 2420 watts and 240*15*.71=2556 var of reactive load connected in parallel. You can get combination of resistive and reactive load bank from any reputed load bank supplier. Do a web search.

(due to math when active and reactive components are equal, the power factor will be 0.707)

Rafiq Bulsara

http://www.srengineersct.com

 

[electricpete]

I agree. Inductive load is much harder to interrupt than resistive load of same magnitude. I assume the purpose of the test is to test switching function (not just the continuous current carrying function) of the switch.

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[obruno]

Thanks for pointing out the badly worded statement:

"The question is, can I use a "non-inductive" load actually measured about 98%pf and modify the load to get 105% which is 70% of 150% since the only thing being tested is the switch?"

I was thinking current and power factor percentage at the same time, sorry about that. My question was answered by mentioning "Inductive load is much harder to interrupt than resistive load of same magnitude." And yes the purpose is to test the switching function so inductive it shall be. Thanks to all.