Guest Posted May 18, 2005 Report Share Posted May 18, 2005 Hi All, I hope that I am putting this in the right spot. I need to explain to new students in electrical theory this: In a single phase motor as you add load to the shaft, the current goes up, the watts goes up, and the power factor increases. WHY?Here is how I am tackling this now:As you add load to the motor, the motor slows, thus the Counter EMF in the rotor is smaller. Therefore more current is drawn by the rotor. This increased current is flowing more in the resistive component of the motor than the reactive component. Therefore the PF angle decreases, the Watts and vars go up but the Watts more so. Is this a valid answer to the question? Link to comment Share on other sites More sharing options...
marke Posted May 18, 2005 Report Share Posted May 18, 2005 HiIf you register, you can get email notification of replies. - It is a free service so there are no hidden costs! I prefer to look at the equivilent circuit of the motor and show a parallel inductive current path which is the magnetising current, and the resistive load current. You can show that the magnetising current is load independent and them use the two prime current vectors to show the changing resultant with the constant magnetising current and changing load current (resistive). There are many ways of looking at this, but the illustration of the constant magnetising current etc eases the understanding of power factor correction. 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...
Guest Posted May 18, 2005 Report Share Posted May 18, 2005 That is great. I am a big user of phasor diagrams with my students. So if I get this correctly, the magnetizing current phasor is at 90 degrees leading the resistive current phasor. And when load is added only the resistive current changes (as the magnetising current is not load dependant) and the vector sum is the total new current when load is added. THerefore a smaller PF angle and a larger total current phasor. Cool, I thinks I got it. I will certainly register, thank you. Link to comment Share on other sites More sharing options...
GerryB Posted May 18, 2005 Report Share Posted May 18, 2005 THere we go, all registered up.... Link to comment Share on other sites More sharing options...
marke Posted May 19, 2005 Report Share Posted May 19, 2005 Hi GerryB Welcome to the forum. You will find that there are a number of very knowlegable and helpful people active on here. The way that you describe is is correct, except if you study the equivilent circuit oof the motor, you will see that there a small series inductive and resistive element (leakage reactance and copper loss that do have some small influence on changing loads, but these are small and so can be largely ignored in explaining the effect of load on current and power factor. 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...
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