Download Dr. Sahar Abd El Moneim Moussa - Pharos University in Alexandria

Electrical Power & Machines
“Electrical Engineering Dept”
Prepared By:
Dr. Sahar Abd El Moneim Moussa
Dr. Sahar Abd El Moneim Moussa
1
Dr. Sahar Abd El Moneim Moussa
2
Power Factor Correction
 The power factor is the ratio between the active power of
the load(P in watt) to the total power of the load (S in VA)
P.F=
𝑃
𝑆
P
Ѳ
Q
S
Dr. Sahar Abd El Moneim Moussa
3
 Accordingly the power factor can be defined as
Cosine the angle between the voltage & the
current in the a.c. circuit
P.F= Cos θ
I
IC
IR
IL - IC
V
Ѳ
IL
Dr. Sahar Abd El Moneim Moussa
Itotal
4
 In inductive circuits  There is lagging P.F
 In capacitive circuits  there is leading P.F
 Most of loads are inductive in nature & have low lagging
P.F which is highly undesirable as it causes an increase in
current resulting in additional losses of active power in all
elements of the power system
 To improve the P.F, some devices taking leading power
should be connected in parallel to the load.
 This device draws a leading current & partly or completely
neutralizes the lagging reactive component of the load
current.
Dr. Sahar Abd El Moneim Moussa
5
Example 2:
A single phase 400 V, 50 Hz motor takes a supplying current 50 A
at a power factor 0.6 lagging. The motor p.f. has been improved to
0.9 lagging by connecting a capacitor in parallel. Calculate the
capacity of the capacitor.
Solution:
 The active current of the motor is
IL(active) = IL1 x Cos 1
= 50 x 0.6= 30 A
Dr. Sahar Abd El Moneim Moussa
6
Dr. Sahar Abd El Moneim Moussa
7
𝐼𝐿 (𝑎𝑐𝑡𝑖𝑣𝑒)
 Motor new current IL2=
𝐶𝑜𝑠 𝜃2
=
30
0.9
= 33.3 𝐴
 IL1 (Reactive) = IL1 x Sin 1=50 x 0.8 = 40 A
 IL2(Reactive) = IL2 x Sin 2=33.3 x 0.435 =14.53
 Current neutralized by the capacitor= IL1 - IL2 = 40-14.53 = 25.47
A
 Since the Capacitor is connected in parallel

𝑉
∴ 𝐼𝑐 =
𝑋𝑐
= V x 2  f C
𝐼𝑐
25.47
∴𝐶=
=
= 202 𝜇𝐹
𝑉 2 𝜋 𝑓 400 𝑋 2 𝑋 𝜋 𝑋 𝑓
Dr. Sahar Abd El Moneim Moussa
8