Practical values of power factor: Difference between revisions

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(Created page with '{{Menu_Power_factor_correction_and_harmonic_filtering}} __TOC__ The calculations for the three-phase example above are as follows:<br>Pn = delivered shaft power = 51 kW<br>P = …')
 
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The calculations for the three-phase example above are as follows:<br>Pn = delivered shaft power = 51 kW<br>P = active power consumed<br><br><math>P=\frac {Pn}{\rho}=\frac{51}{0.91}=56\, kW</math>
The calculations for the three-phase example above are as follows:


S = apparent power <br><br><math>S=\frac{P}{cos \phi}=\frac {56}{0.86}= 65\, kVA</math> <br>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<br>Q = P tan <span class="texhtml">φ</span> = 56 x 0.59 = 33 kvar (see Figure L15).<br>Alternatively<br><br><math>Q=\sqrt{S^2 - P^2}=\sqrt{65^2 - 56^2}=33\, kvar</math>&nbsp;<br><br>'''Average power factor values for the most commonly-used equipment and appliances '''(see '''Fig. L6''')<br>
Pn = delivered shaft power = 51 kW<br>
P = active power consumed


----
<math>P=\frac {Pn}{\rho}=\frac{51}{0.91}=56\, kW</math>


<br>[[Image:FigL05.jpg|left]] <br><br><br><br><br><br><br><br><br><br>'''''Fig. L5:'''''<i>&nbsp;Calculation power diagram</i>
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''')


<br>


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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>
'''''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>

Revision as of 22:00, 4 January 2012


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]


FigL05.jpg

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
  • 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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