On a sunny day (Sun, 2 Jul 2023 10:39:42 +0100) it happened Martin Brown <'''newspam'''@nonad.co.uk> wrote in <u7rgkv$39fkn$1@dont-email.me>:>You run up against the maximum permitted ripple current PDQ. Of Rapid's >stock today 3x 470uF electrolytics in parallel would get you ~160uF and >7A ripple. Obviously need 6 in total and a couple of low drop diodes.I presume you mean in series... Or is this in an other Universe string theory?
half wave rectifier
Started by ●June 29, 2023
Reply by ●July 2, 20232023-07-02
Reply by ●July 2, 20232023-07-02
On 02/07/2023 11:03, Phil Allison wrote:> Martin Brown wrote: > ----------------------------- >> Phil Allison wrote: >>> Don wrote: >> >>>> >>>> This (so far unanswered) question was previously asked by me: >>>> >>>> Does the -3 dB half power point play a role with kW (eg not kWh)? >>>> >>> >>> ** I suggested adding a series capacitor so the load voltage falls to half power. >>> The capacitive reactance then needs to be the same as the load resistance. >>> Same calculation as the -3dB point. >>> C= 1/2.pi.f.R >> >> >> A non-polar 160uF capacitor rated for peak voltage 450v 50Hz and 10A >> continuous ripple current isn't going to be cheap and cheerful. >> > > ** Why not a bank of 4 x 40uF popypropylenes rated at 250VAC ( continuous ) . > In use, each would have 165VAC applied and carry a tad over 2 amps - so no problem. > Likely cost $80 or so. > Lot safer than the live diode idea.That still isn't very cheap though. I take your point that it is safer. -- Martin Brown
Reply by ●July 2, 20232023-07-02
Ricky wrote:> Don wrote: >> Ricky wrote: >> > TTman wrote: >> >> Ed wrote: >> >> > >> >> > The communication hasn't worked so far, perhaps it's been a bit >> >> > unclear. I'll try a bit differently. The key point is that doubling >> >> > the resistance does NOT halve the power. >> >> > >> >> > Take the resistance of your heater (HR), which is about 20.8 ohms, >> >> > and figure what voltage needs to be applied to it to run it at 1500 >> >> > watts. Here's the math: >> >> > Solve for heater resistance (HR): >> >> > P=E^2/HR so HR=E^2/P so HR=250*250/3000 so HR=62500/3000 = 20.83 >> >> > >> >> > Now what voltage must be applied to HR to equal 1500 watts? >> >> > 1500 = E*E/20.83 so 20.83 * 1500 = E^E so 31250 = E*E so >> >> > E = sqroot 31250 so E = 176.77volts >> >> > >> >> > Ok, once you know that voltage you can solve for the rest. >> >> > We need to add a series resistance (AR) to drop the voltage >> >> > across HR from 250 to 176.77volts >> >> > >> >> > Compute the current through the series string of HR plus AR >> >> > >> >> > Solve for I through HR at 1500 watts: >> >> > P=E*I so 1500 = 176.77 * I so I = 1500/176.77 = 8.485amps >> >> > >> >> > The total voltage (250) minus VHR (176.77) equals VAR, the >> >> > voltage across the (AddedResistance) AR >> >> > >> >> > 250-176.77 = 73.223volts >> >> > >> >> > Solve for power dissipated in AR >> >> > >> >> > P=I*E= 8.486*73.223=621.32 watts in AR >> >> > >> >> > I hope this makes it a bit clearer. >> >> Too complicated...HR= 20 Ohms. Add another resistor in series = 20 Ohms >> >> making Rt - 40 Ohms. >> >> If I apply 250 V across 40 Ohms, I get 125V across each resistor. >> > >> > and... you get half the current, so each resistor will dissipate a quarter >> > of the original power. The total power will be half of the original power, >> > but if I understand your limitations on the design, the series resistor can >> > not be used to heat the pot. >> > >> > I'm not going to continue to beat you over the heat to get you to pay >> > attentn to the equations. But here are equations for power, one more time. >> > >> > P = E * I >> > P = E^2 / R >> > P = I^2 * R >> > >> > Apply them to the appropriate devices, rather than just saying, "twice the >> > resistance, half the power". Also, try solving for the actual current and >> > the voltage on each device. This isn't hard, but sometimes it takes a bit >> > of work to get used to it. >> Thanks guys. You helped me put it all together. If we keep 250VAC >> constant then V^2 (eg 250VAC^2) is a constant. Initially, my mind >> misrepresents the V^2 constant as a slope in a linear equation. >> But V^2 is not a slope because P=V^2/R is not linear. The >> multiplicative inverse term, 1/R, makes it non-linear. >> Here's a plot: <https://crcomp.net/misc/pr.png> [1]. As expected, it >> looks similar to a 1/x plot. > > Of course it does, that's the way you've constructed it. Your graph is not > particularly relevant to adding a series resistor. The pot heater is the > only resistance which provides useful heat. The added series resistance is > just to lower the power in the pot heater. To figure the power in the pot > heater, the reduced voltage on the pot heater needs to be taken into account. > > >> This (so far unanswered) question was previously asked by me: >> >> Does the -3 dB half power point play a role with kW (eg not kWh)? > > Not sure what role you are referring to. -3 dB, is a way of expressing > half power. What more are you thinking of? > >> Ordinarily the half power point pertains to bandwidth. > > It has nothing to do with bandwidth, other than it is the point chosen > for specifying an attenuation. > >> It's a little >> eccentric to apply it to a constant frequency. > > Why? Because you've not see this before? > >> But it can be done. By >> inspection, the half power point of 3000 W yields 40 Ohms. > > You are solving an irrelevant problem. No one cares about reducing the > total power to half. That's the mistake that TTman is making. I > originally said he could add an equivalent heater in series with the one > he's already using and the total power would be half. He said that won't > work because only the pot heater is producing useful heat. > > If the pot heater (at 18 or 20 ohms, pick one) is dissipating 3,000 watts, > how much resistance do you need to add in series to make the pot heater > dissipate 1,500 watts?By inspection, from my plot, you need to add 0 Ohms in series in order to dissipate 3000 W from 250 VAC. On the other hand, when you add 20 additional Ohms, in series to the original 20 Ohms, you obtain 1500 W. Again, by inspection. As stated above, 250 VAC^2 is a constant. Let's call it V'. When you double the original resistance, Ro, you obtain: P = V'(1/(2*Ro)) = V'(1/2)(1/Ro) = Po / 2 where Po represents the original power, 3000 W. My post pertains to plot theory. It taught me how a slope only appears in a linear equation, where a first order independent variable is simply multiplied by a coefficient - a multiplicative inverse doesn't qualify. My newly acquired analytic insight enables me to know exactly what to expect from the 1/R term in a power equation. Don't let my plot's implication of doubling the resistance to obtain half the power spook you. Danke, -- Don, KB7RPU, https://www.qsl.net/kb7rpu There was a young lady named Bright Whose speed was far faster than light; She set out one day In a relative way And returned on the previous night.
Reply by ●July 2, 20232023-07-02
On Sunday, July 2, 2023 at 12:07:24 AM UTC-7, Ricky wrote:> On Sunday, July 2, 2023 at 1:32:34 AM UTC-4, whit3rd wrote: > > On Saturday, July 1, 2023 at 11:20:23 AM UTC-7, Ricky wrote: > > > On Saturday, July 1, 2023 at 1:45:10 PM UTC-4, whit3rd wrote: > > > > On Saturday, July 1, 2023 at 5:33:42 AM UTC-7, TTman wrote: > > > > > > > > > It's absolutely clear to me, and no doubt to many others on here that > > > > > you have no idea. Which part of 3kW heater running off 250V don't you > > > > > understand ? I want to halve the power in the 3kW heater. Show me the > > > > > maths... > > > > If you want to halve the power in a heater, use an electric oven control > > > > knob. Those are available as repair parts, and they cycle the heating > > > > element they control, to give proportional heat control. > > > > > > > > Every broken down cooktop has four of those, and three of 'em will work. > > > > > I've never seen an oven or stove control that gave proportional control. Every one I've seen was an on/off switch with some hysteresis, around 10-15 degrees. What am I missing? > > If you want to 'halve the power in a heater', that's the way an electric resistance stovetop > > acts with the knob set on 'medium'; it's a time-proportioning switch, with an internal heater> What you are calling "Time proportioning" is called Bang-Bang. It's not proportional, it's on or off.And, a half-wave rectifier is 'not proportional', it's also on or off. Is that relevant? Either achieves the half-power goal.
Reply by ●July 2, 20232023-07-02
On Sunday, July 2, 2023 at 11:10:52 AM UTC-4, whit3rd wrote:> On Sunday, July 2, 2023 at 12:07:24 AM UTC-7, Ricky wrote: > > On Sunday, July 2, 2023 at 1:32:34 AM UTC-4, whit3rd wrote: > > > On Saturday, July 1, 2023 at 11:20:23 AM UTC-7, Ricky wrote: > > > > On Saturday, July 1, 2023 at 1:45:10 PM UTC-4, whit3rd wrote: > > > > > On Saturday, July 1, 2023 at 5:33:42 AM UTC-7, TTman wrote: > > > > > > > > > > > It's absolutely clear to me, and no doubt to many others on here that > > > > > > you have no idea. Which part of 3kW heater running off 250V don't you > > > > > > understand ? I want to halve the power in the 3kW heater. Show me the > > > > > > maths... > > > > > If you want to halve the power in a heater, use an electric oven control > > > > > knob. Those are available as repair parts, and they cycle the heating > > > > > element they control, to give proportional heat control. > > > > > > > > > > Every broken down cooktop has four of those, and three of 'em will work. > > > > > > > I've never seen an oven or stove control that gave proportional control. Every one I've seen was an on/off switch with some hysteresis, around 10-15 degrees. What am I missing? > > > If you want to 'halve the power in a heater', that's the way an electric resistance stovetop > > > acts with the knob set on 'medium'; it's a time-proportioning switch, with an internal heater > > > What you are calling "Time proportioning" is called Bang-Bang. It's not proportional, it's on or off. > And, a half-wave rectifier is 'not proportional', it's also on or off. Is that relevant? > Either achieves the half-power goal.I was simply pointing out a misuse of the term "proportional control". In control circuits, "proportional" has a specific meaning of the control being able to set the thing being controlled over a range, rather than just on or off. -- Rick C. --++ Get 1,000 miles of free Supercharging --++ Tesla referral code - https://ts.la/richard11209
Reply by ●July 2, 20232023-07-02
On 7/2/2023 9:24 AM, Don wrote:> Ricky wrote: >> Don wrote: >>> Ricky wrote: >>>> TTman wrote: >>>>> Ed wrote: >>>>>> >>>>>> The communication hasn't worked so far, perhaps it's been a bit >>>>>> unclear. I'll try a bit differently. The key point is that doubling >>>>>> the resistance does NOT halve the power. >>>>>> >>>>>> Take the resistance of your heater (HR), which is about 20.8 ohms, >>>>>> and figure what voltage needs to be applied to it to run it at 1500 >>>>>> watts. Here's the math: >>>>>> Solve for heater resistance (HR): >>>>>> P=E^2/HR so HR=E^2/P so HR=250*250/3000 so HR=62500/3000 = 20.83 >>>>>> >>>>>> Now what voltage must be applied to HR to equal 1500 watts? >>>>>> 1500 = E*E/20.83 so 20.83 * 1500 = E^E so 31250 = E*E so >>>>>> E = sqroot 31250 so E = 176.77volts >>>>>> >>>>>> Ok, once you know that voltage you can solve for the rest. >>>>>> We need to add a series resistance (AR) to drop the voltage >>>>>> across HR from 250 to 176.77volts >>>>>> >>>>>> Compute the current through the series string of HR plus AR >>>>>> >>>>>> Solve for I through HR at 1500 watts: >>>>>> P=E*I so 1500 = 176.77 * I so I = 1500/176.77 = 8.485amps >>>>>> >>>>>> The total voltage (250) minus VHR (176.77) equals VAR, the >>>>>> voltage across the (AddedResistance) AR >>>>>> >>>>>> 250-176.77 = 73.223volts >>>>>> >>>>>> Solve for power dissipated in AR >>>>>> >>>>>> P=I*E= 8.486*73.223=621.32 watts in AR >>>>>> >>>>>> I hope this makes it a bit clearer. >>>>> Too complicated...HR= 20 Ohms. Add another resistor in series = 20 Ohms >>>>> making Rt - 40 Ohms. >>>>> If I apply 250 V across 40 Ohms, I get 125V across each resistor. >>>> >>>> and... you get half the current, so each resistor will dissipate a quarter >>>> of the original power. The total power will be half of the original power, >>>> but if I understand your limitations on the design, the series resistor can >>>> not be used to heat the pot. >>>> >>>> I'm not going to continue to beat you over the heat to get you to pay >>>> attentn to the equations. But here are equations for power, one more time. >>>> >>>> P = E * I >>>> P = E^2 / R >>>> P = I^2 * R >>>> >>>> Apply them to the appropriate devices, rather than just saying, "twice the >>>> resistance, half the power". Also, try solving for the actual current and >>>> the voltage on each device. This isn't hard, but sometimes it takes a bit >>>> of work to get used to it. >>> Thanks guys. You helped me put it all together. If we keep 250VAC >>> constant then V^2 (eg 250VAC^2) is a constant. Initially, my mind >>> misrepresents the V^2 constant as a slope in a linear equation. >>> But V^2 is not a slope because P=V^2/R is not linear. The >>> multiplicative inverse term, 1/R, makes it non-linear. >>> Here's a plot: <https://crcomp.net/misc/pr.png> [1]. As expected, it >>> looks similar to a 1/x plot. >> >> Of course it does, that's the way you've constructed it. Your graph is not >> particularly relevant to adding a series resistor. The pot heater is the >> only resistance which provides useful heat. The added series resistance is >> just to lower the power in the pot heater. To figure the power in the pot >> heater, the reduced voltage on the pot heater needs to be taken into account. >> >> >>> This (so far unanswered) question was previously asked by me: >>> >>> Does the -3 dB half power point play a role with kW (eg not kWh)? >> >> Not sure what role you are referring to. -3 dB, is a way of expressing >> half power. What more are you thinking of? >> >>> Ordinarily the half power point pertains to bandwidth. >> >> It has nothing to do with bandwidth, other than it is the point chosen >> for specifying an attenuation. >> >>> It's a little >>> eccentric to apply it to a constant frequency. >> >> Why? Because you've not see this before? >> >>> But it can be done. By >>> inspection, the half power point of 3000 W yields 40 Ohms. >> >> You are solving an irrelevant problem. No one cares about reducing the >> total power to half. That's the mistake that TTman is making. I >> originally said he could add an equivalent heater in series with the one >> he's already using and the total power would be half. He said that won't >> work because only the pot heater is producing useful heat. >> >> If the pot heater (at 18 or 20 ohms, pick one) is dissipating 3,000 watts, >> how much resistance do you need to add in series to make the pot heater >> dissipate 1,500 watts? > > By inspection, from my plot, you need to add 0 Ohms in series in order > to dissipate 3000 W from 250 VAC. On the other hand, when you add 20 > additional Ohms, in series to the original 20 Ohms, you obtain 1500 W. > Again, by inspection. > As stated above, 250 VAC^2 is a constant.No. You make a resistive voltage divider when you add a series resistor. The voltage across the original resistor is reduced (except if the added R is 0 ohms). Whatever leads one to think the voltage is constant is erroneous or not properly understood. Ed Let's call it V'. When you> double the original resistance, Ro, you obtain: > > P = V'(1/(2*Ro)) = V'(1/2)(1/Ro) = Po / 2 > > where Po represents the original power, 3000 W. > > My post pertains to plot theory. It taught me how a slope only appears > in a linear equation, where a first order independent variable is simply > multiplied by a coefficient - a multiplicative inverse doesn't qualify. > My newly acquired analytic insight enables me to know exactly what > to expect from the 1/R term in a power equation. > > Don't let my plot's implication of doubling the resistance to obtain > half the power spook you. > > Danke, >
Reply by ●July 2, 20232023-07-02
On 7/1/2023 6:22 PM, TTman wrote:> >> >> The communication hasn't worked so far, perhaps it's been a bit >> unclear. I'll try a bit differently. The key point is that doubling >> the resistance does NOT halve the power. >> >> Take the resistance of your heater (HR), which is about 20.8 ohms, >> and figure what voltage needs to be applied to it to run it at 1500 >> watts. Here's the math: >> Solve for heater resistance (HR): >> P=E^2/HR so HR=E^2/P so HR=250*250/3000 so HR=62500/3000 = 20.83 >> >> Now what voltage must be applied to HR to equal 1500 watts? >> 1500 = E*E/20.83 so 20.83 * 1500 = E^E so 31250 = E*E so >> E = sqroot 31250 so E = 176.77volts >> >> Ok, once you know that voltage you can solve for the rest. >> We need to add a series resistance (AR) to drop the voltage >> across HR from 250 to 176.77volts >> >> Compute the current through the series string of HR plus AR >> >> Solve for I through HR at 1500 watts: >> P=E*I so 1500 = 176.77 * I so I = 1500/176.77 = 8.485amps >> >> The total voltage (250) minus VHR (176.77) equals VAR, the >> voltage across the (AddedResistance) AR >> >> 250-176.77 = 73.223volts >> >> Solve for power dissipated in AR >> >> P=I*E= 8.486*73.223=621.32 watts in AR >> >> I hope this makes it a bit clearer. >> Ed > Too complicatedNo. ...HR= 20 Ohms. Add another resistor in series = 20 Ohms> making Rt - 40 Ohms. > If I apply 250 V across 40 Ohms, I get 125V across each resistor. >Correct. Now for goodness sake do the next step in the math. 125 volts across 20 ohms produces 781.25 watts in the pot heater and 781.25 watts in the added resistor. But you said you want 1500 watts in the pot heater. Ed
Reply by ●July 2, 20232023-07-02
On Saturday, July 1, 2023 at 2:20:23 PM UTC-4, Ricky wrote:> On Saturday, July 1, 2023 at 1:45:10 PM UTC-4, whit3rd wrote: > > On Saturday, July 1, 2023 at 5:33:42 AM UTC-7, TTman wrote: > > > > > It's absolutely clear to me, and no doubt to many others on here that > > > you have no idea. Which part of 3kW heater running off 250V don't you > > > understand ? I want to halve the power in the 3kW heater. Show me the > > > maths... > > If you want to halve the power in a heater, use an electric oven control > > knob. Those are available as repair parts, and they cycle the heating > > element they control, to give proportional heat control. > > > > Every broken down cooktop has four of those, and three of 'em will work. > I've never seen an oven or stove control that gave proportional control. Every one I've seen was an on/off switch with some hysteresis, around 10-15 degrees. What am I missing?Most of those controls can only handle 2800 W, some less. They throw sparks and will need to be enclosed.> > -- > > Rick C. > > ++- Get 1,000 miles of free Supercharging > ++- Tesla referral code - https://ts.la/richard11209
Reply by ●July 2, 20232023-07-02
Ed wrote:> Don wrote: >> Ricky wrote: >>> Don wrote: >>>> Ricky wrote: >>>>> TTman wrote: >>>>>> Ed wrote: >>>>>>> >>>>>>> The communication hasn't worked so far, perhaps it's been a bit >>>>>>> unclear. I'll try a bit differently. The key point is that doubling >>>>>>> the resistance does NOT halve the power. >>>>>>> >>>>>>> Take the resistance of your heater (HR), which is about 20.8 ohms, >>>>>>> and figure what voltage needs to be applied to it to run it at 1500 >>>>>>> watts. Here's the math: >>>>>>> Solve for heater resistance (HR): >>>>>>> P=E^2/HR so HR=E^2/P so HR=250*250/3000 so HR=62500/3000 = 20.83 >>>>>>> >>>>>>> Now what voltage must be applied to HR to equal 1500 watts? >>>>>>> 1500 = E*E/20.83 so 20.83 * 1500 = E^E so 31250 = E*E so >>>>>>> E = sqroot 31250 so E = 176.77volts >>>>>>> >>>>>>> Ok, once you know that voltage you can solve for the rest. >>>>>>> We need to add a series resistance (AR) to drop the voltage >>>>>>> across HR from 250 to 176.77volts >>>>>>> >>>>>>> Compute the current through the series string of HR plus AR >>>>>>> >>>>>>> Solve for I through HR at 1500 watts: >>>>>>> P=E*I so 1500 = 176.77 * I so I = 1500/176.77 = 8.485amps >>>>>>> >>>>>>> The total voltage (250) minus VHR (176.77) equals VAR, the >>>>>>> voltage across the (AddedResistance) AR >>>>>>> >>>>>>> 250-176.77 = 73.223volts >>>>>>> >>>>>>> Solve for power dissipated in AR >>>>>>> >>>>>>> P=I*E= 8.486*73.223=621.32 watts in AR >>>>>>> >>>>>>> I hope this makes it a bit clearer. >>>>>> Too complicated...HR= 20 Ohms. Add another resistor in series = 20 Ohms >>>>>> making Rt - 40 Ohms. >>>>>> If I apply 250 V across 40 Ohms, I get 125V across each resistor. >>>>> >>>>> and... you get half the current, so each resistor will dissipate a quarte >>>>>>> I hope this makes it a bit clearer. >>>>>> Too complicated...HR= 20 Ohms. Add another resistor in series = 20 Ohms >>>>>> making Rt - 40 Ohms. >>>>>> If I apply 250 V across 40 Ohms, I get 125V across each resistor. >>>>> >>>>> and... you get half the current, so each resistor will dissipate a quarter >>>>> of the original power. The total power will be half of the original power, >>>>> but if I understand your limitations on the design, the series resistor can >>>>> not be used to heat the pot. >>>>> >>>>> I'm not going to continue to beat you over the heat to get you to pay >>>>> attentn to the equations. But here are equations for power, one more time. >>>>> >>>>> P = E * I >>>>> P = E^2 / R >>>>> P = I^2 * R >>>>> >>>>> Apply them to the appropriate devices, rather than just saying, "twice the >>>>> resistance, half the power". Also, try solving for the actual current and >>>>> the voltage on each device. This isn't hard, but sometimes it takes a bit >>>>> of work to get used to it. >>>> >>>> Thanks guys. You helped me put it all together. If we keep 250VAC >>>> constant then V^2 (eg 250VAC^2) is a constant. Initially, my mind >>>> misrepresents the V^2 constant as a slope in a linear equation. >>>> But V^2 is not a slope because P=V^2/R is not linear. The >>>> multiplicative inverse term, 1/R, makes it non-linear. >>>> Here's a plot: <https://crcomp.net/misc/pr.png> [1]. As expected, it >>>> looks similar to a 1/x plot. >>>> >>>> This (so far unanswered) question was previously asked by me: >>>> >>>> Does the -3 dB half power point play a role with kW (eg not kWh)? >>>> >>>> Ordinarily the half power point pertains to bandwidth. It's a little >>>> eccentric to apply it to a constant frequency. But it can be done. By >>>> inspection, the half power point of 3000 W yields 40 Ohms. >>>> >>>> Note. >>>> >>>> [1] octave code: >>>> >>>> R=linspace(10,50,40); >>>> P=250^2 ./ R; >>>> plot(R, P); >>>> xlabel('R (Ohms)'); >>>> ylabel('P (Watts)');<snipped>>>> You are solving an irrelevant problem. No one cares about reducing the >>> total power to half. That's the mistake that TTman is making. I >>> originally said he could add an equivalent heater in series with the one >>> he's already using and the total power would be half. He said that won't >>> work because only the pot heater is producing useful heat. >>> >>> If the pot heater (at 18 or 20 ohms, pick one) is dissipating 3,000 watts, >>> how much resistance do you need to add in series to make the pot heater >>> dissipate 1,500 watts? >> >> By inspection, from my plot, you need to add 0 Ohms in series in order >> to dissipate 3000 W from 250 VAC. On the other hand, when you add 20 >> additional Ohms, in series to the original 20 Ohms, you obtain 1500 W. >> Again, by inspection. >> As stated above, 250 VAC^2 is a constant. > > No. You make a resistive voltage divider when you add a series > resistor. The voltage across the original resistor is reduced > (except if the added R is 0 ohms). Whatever leads one to > think the voltage is constant is erroneous or not properly > understood."If we keep 250VAC constant then V^2 (eg 250VAC^2) is a constant" (as stated above) is my premise. This 250VAC is treated as Vmains by me. My mind mechanically treats Vmains as a constant - because it is. Most electric outlets in my world source about 120 VAC regardless of the load connected. When Vmains is held constant, a doubling of Rmains halves Pmains. This happens because if the voltage in P=V^2/R is held constant then the power must proportionally change with resistance in order to keep the equation balanced. Allow me to share some graphical interpretations. Here's a plot of Pmains versus Rmains: <https://crcomp.net/misc/pr.png> [1] Use conductance to linearize the resistive power plot: <https://crcomp.net/misc/pg.png> [2] The key insight you offer above is to drop voltage Vmains down to Vheater in order to lower Pheater to 1500 W. In graphical form, lower Vheater to 176.77 VAC to bring Pheater down to 1500 W: <https://crcomp.net/misc/pd.png> [3] where the dotted vertical line shows conductance Gheater (1 / Rheater) equal to 0.048 siemens. Note. [1] octave code: R=linspace(20,70); P=250^2 ./ R; plot(R, P); xlabel('Rmains (Ohms)'); ylabel('Pmains (Watts)'); ht = text(40, 2000, '250 VAC'); set(ht, "color", "blue"); grid; [2] octave code: R=linspace(20,70); G= 1 ./ R; P=250^2 .* G; plot(G, P); xlabel('Gmains (Siemens)'); ylabel('Pmains (Watts)'); ht = text(0.03, 1500, '250 VAC'); set(ht, "color", "blue"); grid; [3] octave code: R=linspace(20,70); G= 1 ./ R; P=250^2 .* G; Ph=176.77^2 .* G; plot(G, P, '', G, Ph, '', [0.048, 0.048], [0, 4000], 'k:'); xlabel('G (Siemens)'); ylabel('P (Watts)'); mt = text(0.015, 1500, '250 VAC'); set(mt, "color", "blue"); ht = text(0.035, 1000, '176.77 VAC'); set(ht, "color", "red"); hc = text(0.038, 500, '0.048 siemens ->'); set(hc, "color", "black"); grid; Danke, -- Don, KB7RPU, https://www.qsl.net/kb7rpu There was a young lady named Bright Whose speed was far faster than light; She set out one day In a relative way And returned on the previous night.
Reply by ●July 2, 20232023-07-02
On Sunday, July 2, 2023 at 6:27:57 PM UTC-4, Fred Bloggs wrote:> On Saturday, July 1, 2023 at 2:20:23 PM UTC-4, Ricky wrote: > > On Saturday, July 1, 2023 at 1:45:10 PM UTC-4, whit3rd wrote: > > > On Saturday, July 1, 2023 at 5:33:42 AM UTC-7, TTman wrote: > > > > > > > It's absolutely clear to me, and no doubt to many others on here that > > > > you have no idea. Which part of 3kW heater running off 250V don't you > > > > understand ? I want to halve the power in the 3kW heater. Show me the > > > > maths... > > > If you want to halve the power in a heater, use an electric oven control > > > knob. Those are available as repair parts, and they cycle the heating > > > element they control, to give proportional heat control. > > > > > > Every broken down cooktop has four of those, and three of 'em will work. > > I've never seen an oven or stove control that gave proportional control. Every one I've seen was an on/off switch with some hysteresis, around 10-15 degrees. What am I missing? > Most of those controls can only handle 2800 W, some less. They throw sparks and will need to be enclosed.Hmmm... I am talking about the sort of burners or ovens in the home used for cooking. Why would the contacts on the heat control need to be shielded? What do you think people bake? Heck, every wall switch throws sparks! -- Rick C. -+-- Get 1,000 miles of free Supercharging -+-- Tesla referral code - https://ts.la/richard11209
