Resonance and high frequency

[Classic]

This is a very nice collection from wikipedia or similar. To be honest I have expected something else .. to dive a bit deeper in this phenomena.

I am well aware of the practice of power grid suppliers and their unscrupulous greed. All these explanations do not hold the water as to say. And my aim wasn't necessarily against such behaviour but to explore real options outside of such "conservative" approach.

I do appreciate the effort to reply and thank you for taking time to reply.

Mainly if anyone is looking for "free energy" first of all must evaluate capabilities of device to have one part of the system which is able to isolate (not draining) the input ... or capabilities to harvest additional energy available in local environment, or capabilities to conservate/recycle the input.

For example: how much reactive power can be recovered and converted back into real power ? What type of reactive power will a certain system produce: capacitive or inductive ? Voltage lagging or current lagging ?
Would make sense to design a system able to devour 10kw of electric power with a power factor of 0.3 if I have a capacitive/inductive solution to recover 98% of reactive power. Yes, I am saying to recover, not to improve power factor ... what would mean this ? I can run continuous the system with a single charge and have an excess.

If you improve power factor doesn't mean you pay less bills to your supplier, it means you pay the right amount for usage and make an effort to absorb all energy drawn from source. Someone may say it is fair and square as you improved efficiency ... yes, you improve efficiency of consumption if you think like that, but is not how the nature works and in this way you need to constant hunt for every single particle of energy available to drain it with your efficient system.

[kampen]

Great questions! Let's break them down logically.
1. How much reactive power can be recovered and converted back into real power?

• Reactive power (VARs) itself cannot be directly converted into real power (Watts) because it represents energy oscillating between the source and load rather than being consumed.
• However, power factor correction techniques (like capacitor banks, synchronous condensers, or active power factor correction systems) can reduce the reactive power drawn from the grid.
• If your system has a way to store and reuse reactive power efficiently, it can improve real power delivery by reducing losses, improving voltage stability, and reducing stress on the power system.

2. What type of reactive power will a certain system produce?

• Inductive Loads (lagging power factor): Motors, transformers, inductors, and most industrial machinery consume reactive power, causing the current to lag behind the voltage.
• Capacitive Loads (leading power factor): Capacitor banks, underground cables, and some power electronics generate reactive power, causing the voltage to lag behind the current.
• Your system type determines whether you need power factor correction using capacitors (for inductive loads) or inductors (for capacitive loads).

3. Voltage lagging vs. current lagging?

• Voltage lagging (leading power factor): Happens when the system is capacitive (current leads voltage).
• Current lagging (lagging power factor): Happens when the system is inductive (voltage leads current).
• In most real-world scenarios, inductive loads dominate, meaning the current lags the voltage.

4. Would it make sense to design a system consuming 10 kW at PF = 0.3 if 98% of reactive power is recovered?

• Apparent Power (S) = Real Power (P) / Power Factor (PF) S=10kW0.3=33.3kVAS = \frac{10kW}{0.3} = 33.3 kVA
  
  • This means your system is drawing 33.3 kVA, out of which only 10 kW is real power and the rest is reactive power.
• Reactive Power (Q) Calculation Q=S×sin�(cos�−1(0.3))Q = S \times \sin(\cos^{-1}(0.3)) Q≈32.1kVARQ \approx 32.1 kVAR
  
  • If you recover 98% of the reactive power, that means you're effectively reducing the reactive burden to: 32.1×(1−0.98)=0.64kVAR32.1 \times (1 - 0.98) = 0.64 kVAR
• System Feasibility:
  
  • If your system can truly recover 98% of reactive power, your new power factor will improve significantly, reducing the apparent power drawn from the grid.
  • This can reduce transmission losses, improve efficiency, and lower electricity costs due to power factor penalties.
  • However, running at PF = 0.3 initially is extremely poor, meaning power delivery infrastructure would be heavily loaded even if reactive power is mostly recovered.
  • The design could make sense if the power factor correction is built into the system and dynamically adjusted.

[kampen]

Practical Implementation of a System That Recovers 98% of Reactive Power
If you're designing a system consuming 10 kW at a poor power factor of 0.3, but you want to recover 98% of reactive power, you need an advanced power factor correction (PFC) system.
1. Key Components for Reactive Power Recovery
To achieve high-efficiency recovery, you can use:
(a) Passive Power Factor Correction (PFC) - Capacitor Banks / Inductors

• If your system is inductive (most likely), capacitor banks can supply the reactive power locally.
• If your system is capacitive, inductive reactors (or chokes) can help balance it.
• Passive PFC works well for fixed loads but lacks dynamic adaptability.

(b) Active Power Factor Correction (APFC) - Power Electronics

• Uses IGBT-based power converters (like active rectifiers, STATCOMs, or PWM inverters) to dynamically adjust reactive power.
• Pros: Can handle rapid load changes and compensate for nonlinear loads (e.g., motors, power supplies).
• Cons: More expensive and requires control electronics.

(c) Synchronous Condensers

• A synchronous motor running without a load can absorb or generate reactive power dynamically.
• Used in high-power industrial grids to stabilize voltage and improve efficiency.
• Pros: Can provide leading or lagging VARs as needed.
• Cons: Expensive and requires maintenance.

2. Step-by-Step System Design
Step 1: Calculate Reactive Power Needs

• From earlier calculations:
  
  • Real Power (P) = 10 kW
  • Power Factor (PF) = 0.3 → Apparent Power (S) ≈ 33.3 kVA
  • Reactive Power (Q) ≈ 32.1 kVAR
  • 98% Recovery → Remaining Q ≈ 0.64 kVAR (Negligible)

Step 2: Select Power Factor Correction Solution

• For an inductive system (current lagging):
  
  • Install a capacitor bank rated for ~32 kVAR.
  • Optionally, use an active PFC system for variable loads.
• For a capacitive system (current leading):
  
  • Use inductive reactors to absorb excess reactive power.
  • A synchronous condenser can dynamically adjust to load conditions.

Step 3: Implement Control and Monitoring

• For fixed loads: Use automatic capacitor banks that switch in stages.
• For variable loads: Use a real-time power factor controller with sensors to monitor voltage, current, and phase angle.
• For large industrial setups: Use STATCOM (Static Synchronous Compensator) to dynamically inject reactive power as needed.

3. Feasibility Analysis

• Will this reduce losses? Yes, by minimizing reactive power draw from the grid.
• Will this improve efficiency? Yes, by improving the power factor from 0.3 to nearly 1.0, reducing line losses and transformer loading.
• Does it make economic sense?
  
  • If you're paying power factor penalties, improving PF reduces costs.
  • An active PFC system costs more upfront but improves efficiency in dynamic loads.
  • A passive PFC (capacitor bank) is cheaper and easier but only works for steady loads.

4. Example Implementation - Industrial Case
Imagine an industrial motor running at 10 kW with a PF of 0.3. A 32 kVAR capacitor bank or STATCOM system could correct the power factor, ensuring that nearly all reactive power is supplied locally. This reduces strain on the grid, avoids PF penalties, and enhances energy efficiency.

[Classic]

It seems that I wasn't clear enough: if you design a device with a power factor of 0.3 or even less and you can convert 98% of reactive power, there would be enough energy to self sustain and have an excess.

You must run an extremely ineficient device (aka 0.3 power factor) in order to produce VAR which you recover and reuse. You can use a capacitor to bring curent in phase with voltage but capacity must meet the frequency.
If you can't refill capacitor at the rate of consumption you must increase frequency to build up the storage, or increase the voltage, or both. Always use cheap current ... I mean easy obtained higher voltage with very little current. Use quarter wave amplification.

Don't use AI for such questions, eventual use 2 or more AI and run parts of the project. I have discovered that AI uses all previous discussions on any new questions and even recommend to use systems described and analysed 1 year ago in different chats !
I am using this method to calculate parameters for expired patents considered non conventional due to over unity output, where I understand how they work and I can guide specific parts on AI to obtain comparable results for given parameters found using a different AI, eventually double check with a 3rd AI. Run your own experiments for POC and draw conclusions.

Can you almost double the current in a circuit without losing voltage ? Yes, you can. Take a supercapacitor 2.7 V +400 F and connect 2x1.5 V aa batteries ... monitor the voltage ... when voltage reach 2.5 V take a read of amps in cap, in battery and the system together, eventually to make sure you can measure amps in your batteries before charging the cap. No diodes, no switching or anything else, just battery and supercap. On top of this, voltage is constant all time.
This is really strange to see you haven't done anything but, the curent available is multiplied. Now of you want to take full advantage, use rechargeable batteries with supercapacitor banks, drain your capacitor bank while batteries are recharged.

If you need a shock absorber to limit inrush curent use incandescent light bulbs instead of resistors, compare the cost of resistor 25, 50, 100 or even 200 watts with the cost for the same watts for a resistor. Consider the light bulb with very little resistance when start and high resistance when fully lit.

Run an extension cable from your 150w power inverter powered from a 12 V battery, wrap a decent length of this extension cable with insulated multistrand 0.75-1.5 sqmm copper wire. Leave one end of wrapped wire on the cable and use the other end to power 50 w load, 100 w load, 200w load AC 50 Hz, instead of neutral use a separate earth connection and monitor battery with a battery meter. Do not touch un insulated wires or you'll be very sorry in best scenario. You can use magnet wire if you want to compare.
Eventually trow in equation a cheap SMPS 220/12 V and explore further whatever business you want through a battery or capacitor bank if results are within satisfactory parameters.