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What size of capacitor can bring the power factor to 0.90 lagging, then what will be the real power?

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In summary, in order to bring the power factor of the system with a balanced 3 phase load of 100kW at 0.75 power factor and a system voltage of 12500V line to line, grounded wye to 0.90 lagging, a capacitor with a size of 0.416mF per phase is needed. This will result in a real power of 520kW and a total reactive power before of 321kVAR and after of 151.62kVAR.

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What size of capacitor can bring the power factor to 0.90 lagging, then what will be the real power?

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- #2

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Would anyone please verify my approach is correct?

Initial 3 phase load of Ptotal =520kW, Pf1=0.85, and Vline=31540volts

3*Pf=Ptotal

Pf=520kW/3=173.33kW

For grounded wye,

Vf=Vline/SQRT3=18209volts

Pf1=0.85,

Pf1=Sf1 * (cosf1), where Sf1= Pf1/(cosf1)

Sf1=173.33W/0.85=203.92kVA

Qf1=SQRT(Sf12-Pf12)=107kVAR

Pf2=0.96,

Sf2=Pf2/(cosf2) = 173.33kW/0.96=180.55kVA

Qf2=SQRT(Sf22-Pf22)=50.54kVAR

Xf(reactive) = Vf2/Qf=182092/Qf=331.567E6/ Qf

Xf(before) = 331.567E6/107k=3086Ω

Qfcap= Qf1-Qf1=107k-50.54k=56460VAR

Xfcap= Vfcap2/Qfcap= 331.567E6/56460= 5872 Ω

a) The size of the capacitor per phase C is defined as

C=1/(2*p*f* Xfcap)=1/(376.99*5872)=0.416mF since it is assumed f=60Hz

b) Real power should remain same since we are not changing anything about the resistance part of the impedance so

Ptotal = 3* Pf =520kW

Pf=173.33kW per phase

C) Qtotal = 3*Qf

So Qtotal before = 3*107kVAR=321kVAR and Qtotal after = 3*50.54kVAR=151.62kVAR

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I would first like to clarify that power factor is a measure of the efficiency of an electrical system and is defined as the ratio of real power (measured in watts) to apparent power (measured in volt-amperes). A power factor of 1 indicates a perfectly efficient system, while a power factor less than 1 indicates a less efficient system.

Given the information provided, we can calculate the apparent power of the 3-phase AC system using the equation: Apparent Power = (Voltage * Current * √3), where voltage is the line-to-line voltage of 12500V and current is the total current drawn by the load which can be calculated as (100kW/0.75)/(√3) = 115.47 amps.

Therefore, the apparent power of the system is (12500V * 115.47A * √3) = 199.89 kVA.

To bring the power factor to 0.90 lagging, we need to introduce a capacitive load in the system. The size of the capacitor needed can be calculated using the equation: Capacitance = (Apparent Power * tan θ)/(2π * Frequency * Voltage^2), where θ is the angle between the voltage and current in the system.

Assuming a frequency of 60 Hz, the required capacitance would be (199.89kVA * tan 25.84°)/(2π * 60Hz * 12500V^2) = 64.82 microfarads.

Finally, to determine the real power in the system with a power factor of 0.90, we can use the equation: Real Power = Apparent Power * power factor = 199.89kVA * 0.90 = 179.90 kW.

In conclusion, in order to bring the power factor of the 3-phase AC system to 0.90 lagging, a capacitor with a capacitance of 64.82 microfarads would be needed. This would result in a real power of 179.90 kW in the system.

Power factor is a measure of the efficiency of an electrical system in converting electrical energy into useful work. It is important because a low power factor can result in increased energy costs and inefficiencies in the system.

To calculate power factor for a 3-phase AC system, you need to measure the real power (kW) and apparent power (kVA) using a wattmeter and voltmeter. Then, divide the real power by the apparent power to get the power factor.

The 100kW refers to the amount of real power, or the amount of power actually being used by the electrical system. The 12500V refers to the voltage of the system, which can affect the power factor by increasing or decreasing the amount of reactive power.

A low power factor results in increased current flow, which can cause overheating and inefficiencies in the system. This can lead to increased energy costs and potential damage to electrical equipment.

One way to improve power factor is to use power factor correction devices, such as capacitors, to reduce the amount of reactive power in the system. Another way is to balance the loads in the system to reduce the overall demand for reactive power. Regular maintenance and upgrades to equipment can also help improve power factor.

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