Improving power factor using capacitors (what capacitance?)

[libelec]

Homework Statement

The power factor of a circuit has to be improved from 0,6 to 0,9 by adding capacitors. Originally, it only has an inductor L, a resistence R and a generator V. The circuit requires an active power of 3000 VA, and that should remain the same. V_ef = 220V.

The Attempt at a Solution

The real power P = Pa*Power Factor = 1800 W.

The inductive reactive power Pr_l = Pa *(Power Factor² - 1)^1/2 = 2400 VAR. This Pr_l and Pa shall remain the same after including the capacitors.

Now, the capacitive reactive power reacts 180º to the inductive reactive power. Then, the total reactive power Pr_t = Pr_l - Pr_c. Then Pr_c = Pr_l - Pr_t.

Since the power factor shall be 0.9 and Pa remains the same, knowing that Pr_t reacts 90º to the real power P, then Pr_t = (Pa² + (Pa*0,9)²)^1/2 = 1307,67 VAR

Finally

Pr_c = 2400 VAR - 1307,67 VAR = 1092,33 VAR.

Now, how can I find the capacitance C (total) of the capacitors added? Remember, I don't know L, R or w.

Thanks.

 

[anathema]

Thank you for your question. I would approach this problem by using the power triangle and the power factor formula to find the missing values.

First, let's calculate the missing values for the original circuit. We know that the real power P = 3000 VA and the power factor PF = 0.6. Using the formula for power factor, we can calculate the apparent power S and the reactive power Q:

PF = P/S

0.6 = 3000 VA/S

S = 5000 VA

Q = S*sin(arccos(PF))

= 5000 VA*sin(arccos(0.6))

= 4000 VAR

Now, let's calculate the values for the new circuit with the improved power factor of 0.9. We know that the real power P remains the same at 3000 VA. We also know that the total reactive power Q remains the same at 4000 VAR. Using the formula for power factor, we can calculate the new apparent power S:

PF = P/S

0.9 = 3000 VA/S

S = 3333.33 VA

Now, we can use the power triangle to find the missing value for the new circuit, the new inductive reactive power Ql:

S2 = P2 + Ql2

(3333.33 VA)2 = (3000 VA)2 + Ql2

Ql = 2000 VAR

Since we know that the capacitive reactive power Qc reacts 180º to the inductive reactive power Ql, we can calculate the value of Qc:

Qc = Ql - Q

= 2000 VAR - 4000 VAR

= -2000 VAR

Now, we can use the formula for capacitive reactance to find the capacitance C:

XC = 1/(wC)

-2000 VAR = 1/(wC)

wC = -1/2000 VAR

C = -1/(2000 w VAR)

As you can see, the capacitance C is dependent on the frequency w. Without knowing the frequency, we cannot calculate the exact value of C. However, we can use this formula to find the minimum value of C that will improve the power factor to 0.9.

I hope this helps. Let me know if you have any further questions.

 

[kerimek]

I would suggest using the formula for capacitive reactance (Xc = 1/(wC)) to determine the capacitance needed to achieve a power factor of 0.9. This formula relates the capacitance to the frequency (w) and the reactive power (Prc) of the circuit. Since you know the reactive power and the frequency (assuming it is a standard 60 Hz), you can solve for the required capacitance. However, it is important to note that the actual capacitance needed may vary depending on the specific values of L and R in the circuit. Therefore, it would be best to consult with an electrical engineer to determine the most accurate value for the capacitance needed. Additionally, it is important to ensure that the capacitors are properly sized and installed to avoid any potential safety hazards or damage to the circuit.