Practical values of power factor: Difference between revisions
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Revision as of 11:32, 28 January 2013
The calculations for the three-phase example above are as follows:
Pn = delivered shaft power = 51 kW
P = active power consumed
[math]\displaystyle{ P=\frac {Pn}{\rho}=\frac{51}{0.91}=56\, kW }[/math]
S = apparent power
[math]\displaystyle{ S=\frac{P}{cos \phi}=\frac {56}{0.86}= 65\, kVA }[/math]
So that, on referring to diagram Figure L5 or using a pocket calculator, the value of tan φ corresponding to a cos φ of 0.86 is found to be 0.59
Q = P tan φ = 56 x 0.59 = 33 kvar
Alternatively:
[math]\displaystyle{ Q=\sqrt{S^2 - P^2}=\sqrt{65^2 - 56^2}=33\, kvar }[/math]
Fig. L5: Calculation power diagram
Average power factor values for the most commonly-used equipment and appliances
(see Fig. L6)
Equipment and appliances | cos φ | tan φ | ||
|
loaded at |
0% 25% 50% 75% 100% |
0.17 0.55 0.73 0.80 0.85 |
5.80 1.52 0.94 0.75 0.62 |
|
1.0 |
0 | ||
|
1.0 |
0 | ||
|
0.8 to 0.9 |
0.75 to 0.48 | ||
|
0.8 | 0.75 |
Fig. L6: Values of cos φ and tan φ for commonly-used equipment