The document explains how a capacitor bank improves power factor (p.f) and the calculations needed to determine the number of capacitors required for improvement. It details a case study where the p.f is initially 0.8 and shows how installing a capacitor bank raises it to 0.85, avoiding penalties from utility suppliers. Additionally, it discusses the relationship between active power, reactive power, and the power factor in inductive loads.

1. How Capacitor Bank Works
Mohamad Sabhi Hissam

2. Explain:
1. How does Capacitor Bank improve Power Factor (P.F).
2. How many capacitor needed to improve Power Factor (calculation).

3. Type of
Load
Resistive
Inductive
Capacitive

4. Type of
Load
Resistive
Inductive
Capacitive

5. Inductive
Load
Active
Power (W)
Reactive
Power (VAR)

7. •Power Factor (P.F)=
𝐴𝑐𝑡𝑖𝑣𝑒 𝑃𝑜𝑤𝑒𝑟 (𝑊)
𝐴𝑝𝑝𝑎𝑟𝑒𝑛𝑡 𝑃𝑜𝑤𝑒𝑟 (𝑉𝐴)
•VA = √(𝑊2
+ 𝑉𝐴𝑅2
)
•When VAR ↑, P.F ↓
•Large Motor Produce high VAR and result with
LOWER P.F
•When P.F < 0.85 Utility Supply will give Penalty
because they need to supply more reactive power
(VAR) → Higher Utility Bill

8. SO HOW TO IMPROVE POWER FACTOR TO
PREVENT PENALTY FROM UTILITY SUPPLIER ?????

9. CASE STUDY
Reactive Power
= 6 VAR
Active Power
= 8 W
𝑃. 𝐹 =
𝑊
√(𝑊2 + 𝑉𝐴𝑅2)
=
8
√(82 + 62)
P.F = 0.8

10. • P.F = 0.8
• We want to improve into 0.85 so there will be no penalty from utility
bill
ACTIVE POWER
REACTIVE POWER
(INDUCTIVE)
APPARENT POWER
 POWER FACTOR ANGLE
POWER FACTOR : cos =kW / KVA
(kVAR)
(kVA)
(kW)
Reverse Engineering
P.F = 0.85
𝑃. 𝐹 =
𝑊
√(𝑊2 + 𝑉𝐴𝑅2)
With Watt Constant
0.85 =
8
√(82 + 𝑉𝐴𝑅2)
VAR = 4.95

11. 8 W 8 W
6VAR
4.95VAR
P.F = 0.8 P.F = 0.85

12. 8 W
6 – 4.95= 1.05VAR (Supplied by Cap Bank)
8 W
6VAR
4.95VAR
P.F = 0.8 P.F = 0.85

13. Reactive Power
= 6 VAR
Active Power
= 8 W
BEFORE INSTALL CAP BANK
𝑃. 𝐹 =
𝑊
√(𝑊2 + 𝑉𝐴𝑅2)
=
8
√(82 + 62)
P.F = 0.8

14. AFTER INSTALL CAP BANK

15. Reactive Power
= 4.95 VAR
Active Power
= 8 W
Reactive Power
= 1.05 VAR
𝑃. 𝐹 =
8
√(82 + 4.952)
𝑃. 𝐹 = 𝟎. 𝟖𝟓 (𝑁𝑂 𝑃𝐸𝑁𝐴𝐿𝑇𝑌)

16. THANK YOU