On Friday, June 30, 2023 at 8:26:53 AM UTC-4, TTman wrote:> On 30/06/2023 11:17, Jan Panteltje wrote: > > On a sunny day (Fri, 30 Jun 2023 10:55:12 +0100) it happened TTman > > <kraken...@gmail.com> wrote in <u7m8q1$2gqfg$4...@dont-email.me>: > > > >> On 30/06/2023 00:04, John Larkin wrote: > >>> On Thu, 29 Jun 2023 23:37:22 +0100, TTman <kraken...@gmail.com> > >>> wrote: > >>> > >>>> As part of an experiment with solar, I want to drive my 220V 3kW > >>>> immersion heater with a suitable diode so that the effective power is > >>>> ~1.5kW and can be driven by a smart switch so that it doesn't suck too > >>>> much out of the 9.5kWh battery or solar panels.. What diode? This is UK > >>>> spec. > >>> > >>> Is the source the AC line, what you folks call "mains" ? > >>> > >>> If not, the DC component could be a problem. > >>> > >>> I have conjectured that a lot of DC current will slow down an electric > >>> meter, disk or electronic. > >>> > >> Yes, AC line, 220/250V 50Hz.Mains :) > > > > I'd prefer a triac over a diode to limit any DC (saturation inductors etc) > > Google has many examples, here is one: > > https://www.homemade-circuits.com/how-to-make-25-amp-1500-watts-heater/ > > > I bought this... It works, but I thought a diode would be a simpler > solution... > https://tinyurl.com/4ndkfddw > AC 220V 4000W SCR Thyristor Digital Control Electronic Voltage RegulatorThat is probably best because it eliminates the DC bias. Looks like the input is entered as a percentage between 0-100. Enter 70% for half power. I doubt it's any kind of real regulator, but it doesn't need to be for your use. It's going to put harmonics and RFI on the line, but your use is so brief, it shouldn't matter. You have the same problem with the diode. You can wire 220V receptacle to discrete wire breakout that is permanently wired into the lugs of the regulator. Packaging is not good. You really can't enclose it anything, and it doesn't look very spill proof.> -- > This email has been checked for viruses by Avast antivirus software. > www.avast.com
half wave rectifier
Started by ●June 29, 2023
Reply by ●June 30, 20232023-06-30
Reply by ●June 30, 20232023-06-30
On Friday, June 30, 2023 at 10:25:41 PM UTC+10, Clive Arthur wrote:> On 29/06/2023 23:37, TTman wrote: > > As part of an experiment with solar, I want to drive my 220V 3kW > > immersion heater with a suitable diode so that the effective power is > > ~1.5kW and can be driven by a smart switch so that it doesn't suck too > > much out of the 9.5kWh battery or solar panels.. What diode? This is UK > > spec. > > > A big fuck-off Variac is what you need, if you can find/borrow one. >** Why not a series capacitor bank with the needed value? C = 1/ ( 2.pi.50.18) = 175uF No heat losses and clean current waveform. .... Phil
Reply by ●June 30, 20232023-06-30
On Friday, June 30, 2023 at 9:53:05 PM UTC-4, Phil Allison wrote:> On Friday, June 30, 2023 at 10:25:41 PM UTC+10, Clive Arthur wrote: > > On 29/06/2023 23:37, TTman wrote: > > > As part of an experiment with solar, I want to drive my 220V 3kW > > > immersion heater with a suitable diode so that the effective power is > > > ~1.5kW and can be driven by a smart switch so that it doesn't suck too > > > much out of the 9.5kWh battery or solar panels.. What diode? This is UK > > > spec. > > > > > A big fuck-off Variac is what you need, if you can find/borrow one. > > > ** Why not a series capacitor bank with the needed value? > > C = 1/ ( 2.pi.50.18) = 175uF > > No heat losses and clean current waveform.Whacked out power factor. -- Rick C. --+ Get 1,000 miles of free Supercharging --+ Tesla referral code - https://ts.la/richard11209
Reply by ●June 30, 20232023-06-30
Ricky wrote: --------------------> > > > > > > > A big fuck-off Variac is what you need, if you can find/borrow one. > > > > > ** Why not a series capacitor bank with the needed value? > > > > C = 1/ ( 2.pi.50.18) = 175uF > > > > No heat losses and clean current waveform. > Whacked out power factor. >** LOL - what complete nonsense. ...... Phil
Reply by ●July 1, 20232023-07-01
On Friday, June 30, 2023 at 10:16:45 PM UTC-4, Phil Allison wrote:> Ricky wrote: > -------------------- > > > > > > > > > > > A big fuck-off Variac is what you need, if you can find/borrow one. > > > > > > > ** Why not a series capacitor bank with the needed value? > > > > > > C = 1/ ( 2.pi.50.18) = 175uF > > > > > > No heat losses and clean current waveform. > > Whacked out power factor. > > > ** LOL - what complete nonsense.So, another topic we've found that Phil A. knows nothing about. It is not uncommon to add capacitance to mitigate inductive loads, restoring the power factor to close to 1.0. I've never heard of adding inductors to mitigate capacitive power factors. I guess that's because there are so few large capacitive loads, and they would likely be countered by other inductive loads. The reason a power factor not close to 1.0 is undesired, is that it means the current is flowing out of phase with the voltage, resulting in larger current flow than is needed to transmit the "real" power, with excess power losses in the wiring. The power company doesn't like losing power in their wires. They only bill for the power delivered to your meter, and only the real power at that. -- Rick C. -+- Get 1,000 miles of free Supercharging -+- Tesla referral code - https://ts.la/richard11209
Reply by ●July 1, 20232023-07-01
Mad Ricky wrote: --------------------------- Phil Allison wrote:> > > > > Clive Arthur wrote: > > > > > > > > > > > A big fuck-off Variac is what you need, if you can find/borrow one. > > > > > > > > > ** Why not a series capacitor bank with the needed value? > > > > > > > > C = 1/ ( 2.pi.50.18) = 175uF > > > > > > > > No heat losses and clean current waveform. > > > > Whacked out power factor. > > > > > ** LOL - what complete nonsense. > > So, another topic we've found that Phil A. knows nothing about.** FFS you do LOVE making idiotic assumptions. Using a cap in series to get 1/2 power produces a PF of 0.7, which is not "wacked out" - whatever that means. The 3 posted ways to half the power (using a diode, triac or cap) have the same PF, making nonsense of your claim. PF = true power / VA where V and A are rms values. The true power, is 1500W in each case. The rms current value is also the same since the load is a fixed resistance. The current wave is the same in the load and the supply each time. ....... Phil
Reply by ●July 1, 20232023-07-01
On Saturday, July 1, 2023 at 3:38:59 AM UTC-4, Phil Allison wrote:> Mad Ricky wrote: > --------------------------- > Phil Allison wrote: > > > > > > > Clive Arthur wrote: > > > > > > > > > > > > > A big fuck-off Variac is what you need, if you can find/borrow one. > > > > > > > > > > > ** Why not a series capacitor bank with the needed value? > > > > > > > > > > C = 1/ ( 2.pi.50.18) = 175uF > > > > > > > > > > No heat losses and clean current waveform. > > > > > > Whacked out power factor. > > > > > > > ** LOL - what complete nonsense. > > > > So, another topic we've found that Phil A. knows nothing about. > ** FFS you do LOVE making idiotic assumptions. > > Using a cap in series to get 1/2 power produces a PF of 0.7, which is not "wacked out" - whatever that means. > The 3 posted ways to half the power (using a diode, triac or cap) have the same PF, making nonsense of your claim. > > PF = true power / VA where V and A are rms values. > > The true power, is 1500W in each case. > The rms current value is also the same since the load is a fixed resistance. > The current wave is the same in the load and the supply each time.You ignored using a series resistor, which has zero impact on the power factor. The utility does not care about YOUR real power. They care about THEIR imaginary power which wastes power in the power line. With a power factor of 0.7, the power company would see significantly higher line losses. "For example, if the load power factor were as low as 0.7, the apparent power would be 1.4 times the real power used by the load." Current would be 1.4 times higher, with the line losses doubled. No, the power company does not like that! "Utilities typically charge additional costs to commercial customers who have a power factor below some limit, which is typically 0.9 to 0.95." Both quotes from Wikipedia. https://en.wikipedia.org/wiki/Power_factor#Importance_in_distribution_systems
Reply by ●July 1, 20232023-07-01
On Saturday, July 1, 2023 at 3:38:59 AM UTC-4, Phil Allison wrote:> Mad Ricky wrote: > --------------------------- > Phil Allison wrote: > > > > > > > Clive Arthur wrote: > > > > > > > > > > > > > A big fuck-off Variac is what you need, if you can find/borrow one. > > > > > > > > > > > ** Why not a series capacitor bank with the needed value? > > > > > > > > > > C = 1/ ( 2.pi.50.18) = 175uF > > > > > > > > > > No heat losses and clean current waveform. > > > > > > Whacked out power factor. > > > > > > > ** LOL - what complete nonsense. > > > > So, another topic we've found that Phil A. knows nothing about. > ** FFS you do LOVE making idiotic assumptions. > > Using a cap in series to get 1/2 power produces a PF of 0.7, which is not "wacked out" - whatever that means. > The 3 posted ways to half the power (using a diode, triac or cap) have the same PF, making nonsense of your claim. > > PF = true power / VA where V and A are rms values. > > The true power, is 1500W in each case. > The rms current value is also the same since the load is a fixed resistance. > The current wave is the same in the load and the supply each time.I should also point out that your analysis is faulty, but it will be harder to explain it to you, since it involves complex math, which I'm guessing you don't understand. While the current in the cap and resistor are the same, also the same as the current in the line, the voltage from the line is not on the resistor. The resistor voltage is E = I * R. However, the voltage on the cap is 90° out of phase with the current. The complex sum of the capacitor voltage and resistor voltage are equal to the line voltage. This is equivalent to the three sides of a right triangle, where the hypotenuse is the line voltage, the cap and resistor voltages are the other two sides with the right angle between them. The current is in phase with the voltage on the resistor, and so, out of phase with the line voltage. This means the product of the RMS of the current and the RMS of the line voltage are higher than the power in the resistor. The excess power is flowing in and out of the capacitor, out of phase with the real power. This extra power causes line loses significantly larger than if a real load were being powered. If this doesn't make sense to you, try reading some source on power factor, or maybe you need to learn more about complex math? Good luck, -- Rick C. -++ Get 1,000 miles of free Supercharging -++ Tesla referral code - https://ts.la/richard11209
Reply by ●July 1, 20232023-07-01
Mad Ricky wrote: -------------------------------> You ignored using a series resistor,** You should too - fuckwit. ... Phil
Reply by ●July 1, 20232023-07-01
Mad Ricky wrote: ---------------------------> > Phil Allison wrote: > > > > > > > > > Clive Arthur wrote: > > > > > > > > > > > > > > > A big fuck-off Variac is what you need, if you can find/borrow one. > > > > > > > > > > > > > ** Why not a series capacitor bank with the needed value? > > > > > > > > > > > > C = 1/ ( 2.pi.50.18) = 175uF > > > > > > > > > > > > No heat losses and clean current waveform. > > > > > > > > Whacked out power factor. > > > > > > > > > ** LOL - what complete nonsense. > > > > > > So, another topic we've found that Phil A. knows nothing about. > > ** FFS you do LOVE making idiotic assumptions. > > > > Using a cap in series to get 1/2 power produces a PF of 0.7, which is not "wacked out" - whatever that means. > > The 3 posted ways to half the power (using a diode, triac or cap) have the same PF, making nonsense of your claim. > > > > PF = true power / VA where V and A are rms values. > > > > The true power, is 1500W in each case. > > The rms current value is also the same since the load is a fixed resistance. > > The current wave is the same in the load and the supply each time. > I should also point out that your analysis is faulty,** No it fucking is not.> > The current is in phase with the voltage on the resistor, and so, out of phase with the line voltage. > This means the product of the RMS of the current and the RMS of the line voltage are higher than the power in the resistor. > The excess power is flowing in and out of the capacitor, out of phase with the real power. > This extra power causes line loses significantly larger than if a real load were being powered. >** FFS you just over explained how the VA becomes greater than the true watts. Take you long to Google that ? ... Phil
