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POWER CALCULATION whether the receiver is inductive or capacitive

Stanislaw elektryk12 8850 14
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Treść została przetłumaczona polish » english Zobacz oryginalną wersję tematu
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  • #1 16933940
    Stanislaw elektryk12
    Level 8  
    Posts: 10
    Hello, I am looking for some way to determine whether a given receiver is inductive or capacitive in the AC waveform, I need to determine what I am dealing with, whether it is a coil or a capacitor.
    Is it possible to calculate exactly with the calculated active, reactive and apparent powers?
    calculate the exact value of capacitive and inductive power?
    Please give me some guidance suggestions.
    Is following the power triangle and trigonometry the right way?
    RESULTS
    ACTIVE POWER -32,499 W
    REACTIVE POWER -11.75 VaR
    APPARATIVE POWER -34.56 VA
    SHIFT ANGLE approx. 20 DEGREES
    COSINUS fi -0.9403935185
    I am looking for this formula to be able to determine what these receivers are and apply reactive power compensation.
    PLEASE HELP
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    #2 16934211
    Anonymous
    Level 1  
  • #3 16946600
    Stanislaw elektryk12
    Level 8  
    Posts: 10
    Buddy, did you give a formula to calculate the capacitive power? Thanks in advance,
  • #4 16946655
    Anonymous
    Level 1  
  • #5 16946853
    bodzio507
    Level 30  
    Posts: 1785
    Help: 22
    Rate: 579
    Did you take these measurements?
    Is it a three-phase system?
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  • #6 16947805
    mateusz25109
    Level 11  
    Posts: 34
    Help: 3
    Rate: 5
    It would be best if you upload a schematic of this circuit. Then it will be easier to help you :)
  • #7 16947813
    Anonymous
    Level 1  
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  • #8 16947913
    bodzio507
    Level 30  
    Posts: 1785
    Help: 22
    Rate: 579
    If we have a measurement system, then after deflecting the meter we can determine the nature of the receiver.

    Kraniec_Internetów wrote:
    It seems to me that this is an existing installation with dozens if not hundreds of receivers.

    It is in such systems that the power factor changes every now and then and we have special regulating systems for this.
    Do you think hundreds of receivers use 32 W of power and 11 var of reactive power?
    I don't normally compensate for that,
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  • #9 16947933
    Anonymous
    Level 1  
  • #10 16947956
    bodzio507
    Level 30  
    Posts: 1785
    Help: 22
    Rate: 579
    Kraniec_Internetów wrote:
    Depends on what meter.


    I mean an analog wattmeter. But the digital wattmeter should also show positive or negative power.

    Kraniec_Internetów wrote:
    Not necessarily. The devices can work continuously with the same power. For example, fans.

    This is of course possible. But much more often these are systems in which the receivers are turned on irregularly.

    Here we are to compensate the power in a given case.
    I think that you can choose a capacitor and if we turn it on and the current increases, it means that the receiver is capacitive, and when the current decreases, it is inductive.

    In my opinion, we have too little data to define it.
  • Helpful post
    #11 16947983
    Anonymous
    Level 1  
  • #12 16948065
    bodzio507
    Level 30  
    Posts: 1785
    Help: 22
    Rate: 579
    Kraniec_Internetów wrote:
    How do you want to measure reactive power with a wattmeter (active power meter)?

    This is possible in a three-phase system. You must turn on the wattmeter properly.
  • #13 16948182
    Anonymous
    Level 1  
  • #14 16951876
    Stanislaw elektryk12
    Level 8  
    Posts: 10
    It is a single-phase system.
  • #15 16951918
    bodzio507
    Level 30  
    Posts: 1785
    Help: 22
    Rate: 579
    Then you have the answer above.

    And by the way, what are you compensating for?
    The current limitation will be small. You will spend more on a capacitor than you will save on electricity.

    Usually the reactive power is compensated to cos (phi) 0.95
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Topic summary

✨ The discussion revolves around determining whether a given AC receiver is inductive or capacitive based on power calculations. Key indicators include the signs of reactive power: negative for capacitive loads and positive for inductive loads. The participants suggest that ordinary multimeter measurements are insufficient, and specialized meters are recommended for accurate readings. Various formulas for calculating reactive power are mentioned, including S^2 = P^2 + Q^2. The conversation also touches on the implications of reactive power compensation, emphasizing that active power remains unchanged while reactive power can be adjusted using capacitors. The system in question is identified as single-phase, and considerations regarding cost-effectiveness of compensation are discussed.
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FAQ

TL;DR: To tell if a load is inductive or capacitive, check Q: your −11.75 var means capacitive; “reactive power for the capacitive load is negative.” [Elektroda, Anonymous, post #16934211]

Why it matters: Knowing Q’s sign lets you size compensation and avoid higher currents or penalties. This FAQ is for electricians and techs asking how to identify load type and pick capacitors from basic P–Q–S data.

Quick Facts

How do I quickly tell if my AC load is inductive or capacitive?

Read the sign of reactive power Q. Negative Q means a capacitive load, positive Q means an inductive load. Your measured −11.75 var indicates a capacitive receiver. “Reactive power for the capacitive load is negative.” Confirm with a capable power meter. [Elektroda, Anonymous, post #16934211]

Can I use a standard multimeter to calculate Q and cos φ?

No. A typical multimeter measures RMS V and I, not phase angle or Q. Use a dedicated power meter that reports P, Q, S, cos φ, and Q’s sign. These instruments simplify classification and compensation work. [Elektroda, Anonymous, post #16946655]

What formula links apparent, active, and reactive power?

Use the power triangle: S² = P² + Q². Verify any two quantities and compute the third. This helps cross-check questionable readings before you decide on compensation. “S^2 = P^2 + Q^2.” [Elektroda, Anonymous, post #16946655]

My data shows P = −32.499 W and Q = −11.75 var. What does that mean?

The negative Q flags a capacitive nature. The negative P suggests your sign convention or wiring may be reversed. Focus first on Q’s sign for type identification, then confirm meter orientation and CT polarity to resolve P’s sign. [Elektroda, Anonymous, post #16934211]

How do I size a capacitor to correct 11.75 var at 230 VAC, single‑phase?

One worked example shows 11.75 var needs about 700 nF at 230 VAC to reach unity power factor. That corresponds to roughly 4.5 kΩ capacitive reactance at 50 Hz. Use this as a sizing check against your analyzer’s readings. [Elektroda, Anonymous, post #16947983]

What power factor should I target when compensating?

Aim for cos φ around 0.95 rather than unity. This avoids overcompensation and keeps currents reasonable across load changes. “Usually the reactive power is compensated to cos (phi) 0.95.” [Elektroda, bodzio507, post #16951918]

Is there a simple field test to confirm inductive vs capacitive?

Yes. Temporarily connect a known capacitor and watch current or apparent power. If current increases, the system was already capacitive. If current decreases, the system was inductive. Remove the test part after observing the change. [Elektroda, bodzio507, post #16947956]

Can a wattmeter measure reactive power directly?

A wattmeter reads active power. In three-phase setups, specific connections allow inferring Q. For single-phase, use a power analyzer that directly reports Q and cos φ to avoid wiring tricks. [Elektroda, bodzio507, post #16948065]

Is this problem about a single‑phase or three‑phase system?

The original case involved a single‑phase system. That affects instrument choice and any special wattmeter methods. Use a single‑phase power analyzer for clarity. [Elektroda, Stanislaw elektryk12, post #16951876]

What’s a practical 3‑step method to tune compensation on site?

  1. Measure P, Q, S, and cos φ with a power analyzer.
  2. Add a trial capacitor; observe Q trending toward zero and current change.
  3. Adjust to reach about cos φ = 0.95 under typical load. [Elektroda, bodzio507, post #16951918]

Will small loads benefit financially from compensation?

Savings may be negligible for very small P and Q. Hardware cost and labor can exceed any bill reduction. “You will spend more on a capacitor than you will save.” Prioritize larger offenders first. [Elektroda, bodzio507, post #16951918]

What if I have many receivers switching on and off?

Expect the power factor to change over time. Use automatic regulators or stepped capacitor banks to track the load. Fixed caps can mis-compensate when the mix shifts. [Elektroda, bodzio507, post #16947913]

How do constant-speed fans affect compensation planning?

Fans can run steadily, keeping power relatively constant. In such cases, a fixed compensation step may hold well. Verify stability with a data log before finalizing. [Elektroda, Anonymous, post #16947933]

What’s an edge case I should watch out for when testing with a capacitor?

If the load is already capacitive, adding capacitance will raise current instead of lowering it. That indicates overcompensation risk; remove or reduce capacitance immediately. [Elektroda, bodzio507, post #16947956]

What statistic from the thread confirms capacitive behavior?

The measured reactive power was −11.75 var with an apparent power of 34.56 VA and about −0.94 power factor. The negative Q establishes the capacitive character. [Elektroda, Stanislaw elektryk12, post #16933940]
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