Practical values of power factor: Difference between revisions
(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 = …') |
(modified to proper format (paragraphs, images ...)) |
||
Line 1: | Line 1: | ||
{{Menu_Power_factor_correction_and_harmonic_filtering}} | {{Menu_Power_factor_correction_and_harmonic_filtering}} | ||
__NOTOC__ | |||
The calculations for the three-phase example above are as follows: | 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''') | |||
{| style="width: 774px; height: 418px" cellspacing="1" cellpadding="1" width="774" border="1" | {| style="width: 774px; height: 418px" cellspacing="1" cellpadding="1" width="774" border="1" | ||
Line 111: | Line 129: | ||
|} | |} | ||
'''''Fig. L6:'''''<i> Values of cos<span class="texhtml">φ</span> and tan<span class="texhtml">φ</span> for commonly-used equipment</i><br> | '''''Fig. L6:'''''<i> 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]
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