Practical values of power factor: Difference between revisions

Revision as of 15:51, 23 August 2013

(see Fig. L6)

Equipment and appliances			cos φ	tan φ
Common induction motor	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
Incandescent lamps Fluorescent lamps (uncompensated) Fluorescent lamps (compensated) Discharge lamps			1.0 0.5 0.93 0.4 to 0.6	0 1.73 0.39 2.29 to 1.33
Ovens using resistance elements Induction heating ovens (compensated) Dielectric type heating ovens			1.0 0.85 0.85	0 0.62 0.62
Resistance-type soldering machines Fixed 1-phase arc-welding set Arc-welding motor-generating set Arc-welding transformer-rectifier set			0.8 to 0.9 0.5 0.7 to 0.9 0.7 to 0.8	0.75 to 0.48 1.73 1.02 to 0.48 1.02 to 0.75
Arc furnace			0.8	0.75

Fig. L6: Values of cos $φ$ and tan $φ$ for commonly-used equipment

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-{{Menu_Power_factor_correction_and_harmonic_filtering}}
+{{Menu_Power_Factor_Correction}}
-__NOTOC__
-The calculations for the three-phase example above are as follows:
-Pn = delivered shaft power = 51 kW<br>
-P = active power consumed
-<math>P=\frac {Pn}{\rho}=\frac{51}{0.91}=56\, kW</math>
-S = apparent power
-<math>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 <span class="texhtml">φ</span> corresponding to a cos <span class="texhtml">φ</span> of 0.86 is found to be 0.59
-Q = P tan <span class="texhtml">φ</span> = 56 x 0.59 = 33 kvar
-Alternatively:
-<math>Q=\sqrt{S^2 - P^2}=\sqrt{65^2 - 56^2}=33\, kvar</math>
-[[Image:FigL05.jpg|none]]
-'''''Fig. L5:''''' ''' Calculation power diagram'''
 == Average power factor values for the most commonly-used equipment and appliances ==
 (see '''Fig. L6''')
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 | valign="top" | 0.75
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 '''''Fig. L6:'''''<i>&nbsp;Values of cos <span class="texhtml">φ</span> and tan <span class="texhtml">φ</span> for commonly-used equipment</i><br>
 [[ru:Практические значения коэффициента мощности]]
 [[zh:功率因数的实际值]]