Reactive power compensation

In summary, we first calculate the supply source impedance and then use it to find the Rs and Xs values. Using these values, we calculate the capacitance needed for load compensation to unity power factor. We then calculate the rms value of the needed distribution voltage before and after compensation. Finally, we calculate the active power loss at energy delivery before and after compensation, with no active power loss after compensation due to unity power factor being achieved.
  • #1
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Calculate the capacitance C needed for the load compensation to unity power factor. Calculate the rms value E of the needed distribution voltage e before and after compensation. Calculate the active power loss (delta)Ps at energy delivery before and after compensation.

U = 2.4kV (line to ground), load has active power P = 120 kW at power factor = 0.6. Short circuit power at the bus is Ssc = 2MVA, and Xs-to-Rs ratio to the supply source impedance qual = Xs/Rs = 3.


The Attempt at a Solution



- I'm not really sure where to start with this problem. I know how to find Rs and Xs using the formula for Zs, but I don't really know where to go from there. Please help me.
 
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  • #2
First, calculate the supply source impedance (Zs):Zs = U^2/Ssc = (2.4kV)^2 / 2MVA = 1.2 kΩThen, calculate the Rs and Xs values:Rs = Zs / (1 + Xs/Rs) = 1.2 kΩ / (1 + 3) = 0.4 kΩXs = Rs * Xs/Rs = 0.4 kΩ * 3 = 1.2 kΩNext, calculate the capacitance C needed for load compensation to unity power factor:C = P/(2πfU^2) = 120 kW / (2π * 50Hz * (2.4kV)^2) = 0.0056 FFinally, calculate the rms value E of the needed distribution voltage e before and after compensation:Before Compensation:E = U * √(P/Ssc) = 2.4kV * √(120kW/2MVA) = 1.67 kVAfter Compensation:E = U * √(1/Xs/Rs) = 2.4kV * √(1/3) = 1.81 kVCalculate the active power loss (delta)Ps at energy delivery before and after compensation:Before Compensation: (delta)Ps = P * (1-PF) = 120 kW * (1-0.6) = 48 kWAfter Compensation: (delta)Ps = 0 kW
 
  • #3


I can provide you with some guidance on how to approach this problem. Firstly, let's define reactive power compensation. It is a method used in power systems to improve the power factor and reduce the reactive power consumption in the system. This is achieved by adding capacitors in parallel with the load, which helps to balance out the reactive power and bring the power factor closer to unity (1).

To calculate the capacitance C needed for load compensation to unity power factor, we can use the following formula:

C = Q/(ωV^2)

Where Q is the reactive power, ω is the angular frequency, and V is the voltage. In this problem, we are given the active power P and the power factor, which means we can calculate the reactive power using the formula Q = P*tan(θ), where θ is the angle of the power factor. In this case, θ = arccos(0.6) = 53.13°. So, Q = 120kW*tan(53.13°) = 135.38 kVAr.

We are also given the voltage U = 2.4kV, so we can calculate the angular frequency ω using the formula ω = 2πf, where f is the frequency. Assuming a standard frequency of 60Hz, we get ω = 2π*60 = 377 rad/s.

Therefore, the capacitance needed for load compensation is:

C = 135.38kVAr/(377rad/s*2.4kV^2) = 149.7 μF

Next, let's calculate the rms value E of the needed distribution voltage e before and after compensation. Before compensation, the distribution voltage e is equal to the line voltage U, which is 2.4kV. After compensation, the distribution voltage e is given by the formula:

e = √(U^2 + Q^2/C^2)

Substituting the values, we get:

e = √(2.4kV^2 + (135.38kVAr)^2/(149.7μF)^2) = 2.58kV

Finally, let's calculate the active power loss (delta)Ps at energy delivery before and after compensation. The active power loss is given by the formula:

(delta)Ps = P*(1-cos(θ))

Before compensation
 

1. What is reactive power compensation?

Reactive power compensation is the process of correcting the power factor in an electrical system by adding or subtracting reactive power to balance the reactive and real power. This helps to improve the efficiency of the system and reduce energy costs.

2. What causes reactive power?

Reactive power is caused by inductive or capacitive loads in an electrical system. These loads create a phase difference between the voltage and current, which results in a reactive power component.

3. How is reactive power compensated?

Reactive power can be compensated by using devices such as capacitors and inductors to supply or absorb reactive power. These devices are connected in parallel with the load to balance the reactive power and improve the power factor.

4. Why is reactive power compensation important?

Reactive power compensation is important because it helps to reduce energy losses, improve the power factor, and increase the efficiency of an electrical system. It also helps to prevent voltage drops and maintain the stability of the system.

5. What are the benefits of reactive power compensation?

The benefits of reactive power compensation include improved power factor, reduced energy costs, increased efficiency and reliability of the system, and reduced stress on electrical equipment. It also helps to comply with utility requirements and regulations.

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