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A 600V, 60Hz, ∆-connected, 8-pole synchronous motor has a synchronous reactance of 0.15Ω and armature resistance of 0.02Ω. This motor has an output power rating of 46 hp with a 0.90 lagging power factor. Motor losses can be approximated at 1kW.

I don't want a solution.

I only have a question about the lagging power factor of 0.9, is it the angle of the phase current or the angle of the line current?

knowing that delta connected loads phase current relation to line current is:

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A 0.9 lagging power factor is :

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Which current has the angle of -25.8?

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  • \$\begingroup\$ How do you mean phase VS. line current? There is just one way you can measure the current, it is not like phase VS. line voltage. \$\endgroup\$ Jan 3 at 8:08

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Maybe study the phasor diagram for a delta load with a slightly lagging power factor (denoted by \$\Phi\$): -

enter image description here

Image from here.

If the power factor were 0°, \$I_{ab}\$ would be exactly in phase with \$V_{ab}\$. So, \$I_{ab}\$ can be called a phase current because it applies to the current that travels down the specific load phase directly connected between A and B. But, the line current (\$I_{a}\$) also involves \$I_{ca}\$; specifically \$I_{ac}\$ (the reverse of \$I_{ca}\$).

Thus, \$I_{a}=I_{ab}+I_{ac}\$.

I only have a question about the lagging power factor of 0.9, is it the angle of the phase current or the angle of the line current?

The PF angle is the angle of the phase current \$I_{ab}\$ relative to line voltage \$V_{ab}\$. Ditto for the other two pairs of phasors.

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