How Does Reactive Power Compensation Affect Power Factor and System Efficiency?

[snoggerT]

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.

 

[Valkarie]

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

 

[Mene]

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