On Sunday, July 2, 2023 at 8:49:20 PM UTC-4, Ricky wrote:> 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.So am I.> Why would the contacts on the heat control need to be shielded? What do you think people bake? Heck, every wall switch throws sparks!Those control really throw sparks. Is your wall switch connected to a 2 kW load? And on top of that being switched on/off continuously? There are two types of control. One places the bimetal heater in series with the load, the other puts the bimetal heater in parallel with the mains voltage. The second is rarer to save wiring. The first kind will compress its range of adjustment with an oversized load, and likely could burn up the bimetal heater. BTW the "knob" , which pushes onto the stem of the control, is just a convenient means of manually manipulating the control, it doesn't do anything electrically.> > -- > > Rick C. > > -+-- Get 1,000 miles of free Supercharging > -+-- Tesla referral code - https://ts.la/richard11209
half wave rectifier
Started by ●June 29, 2023
Reply by ●July 3, 20232023-07-03
Reply by ●July 3, 20232023-07-03
On Monday, July 3, 2023 at 10:41:39 AM UTC-4, Fred Bloggs wrote:> On Sunday, July 2, 2023 at 8:49:20 PM UTC-4, Ricky wrote: > > 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. > So am I. > > Why would the contacts on the heat control need to be shielded? What do you think people bake? Heck, every wall switch throws sparks! > Those control really throw sparks. Is your wall switch connected to a 2 kW load? And on top of that being switched on/off continuously?1.44 kW, yes. The space heaters that draw the maximum contain exactly the bimetal "thermostat" you are talking about. There are zero issues with operation of such contacts. My water heater has a similar type of thermostat with similar contacts, that draws 4.5 kW. I know, because I've had it apart before. So, when you talk about "throwing sparks", you are talking about the normal operation of the contacts. I should have known. What you don't know fills volumes. -- Rick C. -+-+ Get 1,000 miles of free Supercharging -+-+ Tesla referral code - https://ts.la/richard11209
Reply by ●July 3, 20232023-07-03
On Monday, July 3, 2023 at 11:52:35 AM UTC-4, Ricky wrote:> On Monday, July 3, 2023 at 10:41:39 AM UTC-4, Fred Bloggs wrote: > > On Sunday, July 2, 2023 at 8:49:20 PM UTC-4, Ricky wrote: > > > 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. > > So am I. > > > Why would the contacts on the heat control need to be shielded? What do you think people bake? Heck, every wall switch throws sparks! > > Those control really throw sparks. Is your wall switch connected to a 2 kW load? And on top of that being switched on/off continuously? > 1.44 kW, yes. The space heaters that draw the maximum contain exactly the bimetal "thermostat" you are talking about. There are zero issues with operation of such contacts. My water heater has a similar type of thermostat with similar contacts, that draws 4.5 kW. I know, because I've had it apart before. >Those products are totally shielded against throwing sparks outside the enclosure or they couldn't be sold. The water heater controls are beneath a sealed cover that's screwed down, that's what's called enclosed. Surface burner controls are intended to mount behind a closed up panel, and are not suited for operation otherwise.> So, when you talk about "throwing sparks", you are talking about the normal operation of the contacts. I should have known. What you don't know fills volumes.Can't say the same about you, a vacuum doesn't fill anything.> > -- > > Rick C. > > -+-+ Get 1,000 miles of free Supercharging > -+-+ Tesla referral code - https://ts.la/richard11209
Reply by ●July 3, 20232023-07-03
On Monday, July 3, 2023 at 12:05:13 PM UTC-4, Fred Bloggs wrote:> On Monday, July 3, 2023 at 11:52:35 AM UTC-4, Ricky wrote: > > On Monday, July 3, 2023 at 10:41:39 AM UTC-4, Fred Bloggs wrote: > > > On Sunday, July 2, 2023 at 8:49:20 PM UTC-4, Ricky wrote: > > > > 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. > > > So am I. > > > > Why would the contacts on the heat control need to be shielded? What do you think people bake? Heck, every wall switch throws sparks! > > > Those control really throw sparks. Is your wall switch connected to a 2 kW load? And on top of that being switched on/off continuously? > > 1.44 kW, yes. The space heaters that draw the maximum contain exactly the bimetal "thermostat" you are talking about. There are zero issues with operation of such contacts. My water heater has a similar type of thermostat with similar contacts, that draws 4.5 kW. I know, because I've had it apart before. > > > Those products are totally shielded against throwing sparks outside the enclosure or they couldn't be sold. The water heater controls are beneath a sealed cover that's screwed down, that's what's called enclosed. Surface burner controls are intended to mount behind a closed up panel, and are not suited for operation otherwise. > > So, when you talk about "throwing sparks", you are talking about the normal operation of the contacts. I should have known. What you don't know fills volumes. > Can't say the same about you, a vacuum doesn't fill anything.So, you agree that stove and oven controls are in enclosures and so, can be used for more than 2,800 W? Glad we got that settled. -- Rick C. -++- Get 1,000 miles of free Supercharging -++- Tesla referral code - https://ts.la/richard11209
Reply by ●July 3, 20232023-07-03
On Monday, July 3, 2023 at 2:50:09 PM UTC-4, Ricky wrote:> On Monday, July 3, 2023 at 12:05:13 PM UTC-4, Fred Bloggs wrote: > > On Monday, July 3, 2023 at 11:52:35 AM UTC-4, Ricky wrote: > > > On Monday, July 3, 2023 at 10:41:39 AM UTC-4, Fred Bloggs wrote: > > > > On Sunday, July 2, 2023 at 8:49:20 PM UTC-4, Ricky wrote: > > > > > 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. > > > > So am I. > > > > > Why would the contacts on the heat control need to be shielded? What do you think people bake? Heck, every wall switch throws sparks! > > > > Those control really throw sparks. Is your wall switch connected to a 2 kW load? And on top of that being switched on/off continuously? > > > 1.44 kW, yes. The space heaters that draw the maximum contain exactly the bimetal "thermostat" you are talking about. There are zero issues with operation of such contacts. My water heater has a similar type of thermostat with similar contacts, that draws 4.5 kW. I know, because I've had it apart before. > > > > > Those products are totally shielded against throwing sparks outside the enclosure or they couldn't be sold. The water heater controls are beneath a sealed cover that's screwed down, that's what's called enclosed. Surface burner controls are intended to mount behind a closed up panel, and are not suited for operation otherwise. > > > So, when you talk about "throwing sparks", you are talking about the normal operation of the contacts. I should have known. What you don't know fills volumes. > > Can't say the same about you, a vacuum doesn't fill anything. > So, you agree that stove and oven controls are in enclosures and so, can be used for more than 2,800 W? Glad we got that settled.None of them are in enclosures outside the appliance. You wouldn't want to use an oven control as they're on the pricey side. They don't use a bimetal. They use a diaphragm activated switch with adjustable bias for temperature control. The diaphragm is activated by a calibrated gas pressure in a capillary tube in the oven chamber. The electronic control ovens use an RTD sensor enclosed in copper tube projecting into the oven chamber. It will be tricky adapting either of them to deliver half power to an immersion heater. It's a waste of time to even discuss it.> > -- > > Rick C. > > -++- Get 1,000 miles of free Supercharging > -++- Tesla referral code - https://ts.la/richard11209
Reply by ●July 3, 20232023-07-03
On Monday, July 3, 2023 at 3:35:44 PM UTC-4, Fred Bloggs wrote:> On Monday, July 3, 2023 at 2:50:09 PM UTC-4, Ricky wrote: > > On Monday, July 3, 2023 at 12:05:13 PM UTC-4, Fred Bloggs wrote: > > > On Monday, July 3, 2023 at 11:52:35 AM UTC-4, Ricky wrote: > > > > On Monday, July 3, 2023 at 10:41:39 AM UTC-4, Fred Bloggs wrote: > > > > > On Sunday, July 2, 2023 at 8:49:20 PM UTC-4, Ricky wrote: > > > > > > 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. > > > > > So am I. > > > > > > Why would the contacts on the heat control need to be shielded? What do you think people bake? Heck, every wall switch throws sparks! > > > > > Those control really throw sparks. Is your wall switch connected to a 2 kW load? And on top of that being switched on/off continuously? > > > > 1.44 kW, yes. The space heaters that draw the maximum contain exactly the bimetal "thermostat" you are talking about. There are zero issues with operation of such contacts. My water heater has a similar type of thermostat with similar contacts, that draws 4.5 kW. I know, because I've had it apart before. > > > > > > > Those products are totally shielded against throwing sparks outside the enclosure or they couldn't be sold. The water heater controls are beneath a sealed cover that's screwed down, that's what's called enclosed. Surface burner controls are intended to mount behind a closed up panel, and are not suited for operation otherwise. > > > > So, when you talk about "throwing sparks", you are talking about the normal operation of the contacts. I should have known. What you don't know fills volumes. > > > Can't say the same about you, a vacuum doesn't fill anything. > > So, you agree that stove and oven controls are in enclosures and so, can be used for more than 2,800 W? Glad we got that settled. > None of them are in enclosures outside the appliance. You wouldn't want to use an oven control as they're on the pricey side. They don't use a bimetal. They use a diaphragm activated switch with adjustable bias for temperature control. The diaphragm is activated by a calibrated gas pressure in a capillary tube in the oven chamber. The electronic control ovens use an RTD sensor enclosed in copper tube projecting into the oven chamber. It will be tricky adapting either of them to deliver half power to an immersion heater. It's a waste of time to even discuss it.LOL! I've never seen you to not be happy wasting everyone's time. I never said this was a good idea. I said it is not proportional. whit3rd referred to the on/off thermostat control as "proportional" which I corrected. You then jumped in blabbing about limited power. You continue to backpedal until we are finally in agreement. But now you are still talking about using this control. You should bring that up with someone else. It's not a point I was ever discussing. -- Rick C. -+++ Get 1,000 miles of free Supercharging -+++ Tesla referral code - https://ts.la/richard11209
Reply by ●July 3, 20232023-07-03
> No. > ...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. > > EdThat's a surprise... -- This email has been checked for viruses by Avast antivirus software. www.avast.com
Reply by ●July 3, 20232023-07-03
On 7/2/2023 7:35 PM, Don wrote:> 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.Correct. He'll get half power out of the combination of HR (the heater resistance) plus AR (the added resistance). The original power was 3000 watts. Half power is 1500 watts. So he'll get 750 watts out of AR and 750 watts out of HR. BUT THE OP WANTS 1500 WATTS OUT OF HR. Your math solves the wrong problem. From the op: "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..." Ed <snip>
Reply by ●July 3, 20232023-07-03
On Monday, July 3, 2023 at 5:40:51 PM UTC-4, TTman wrote:> > No. > > ...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 > That's a surprise...That's the point. You don't understand this and seem to be unwilling to learn anything new. You say you don't understand what others post, then repeat the wrongthink that you have in your head. Maybe now you will pay a bit more attention to what others have posted and learn something about the problem. -- Rick C. +--- Get 1,000 miles of free Supercharging +--- Tesla referral code - https://ts.la/richard11209
Reply by ●July 3, 20232023-07-03
On Monday, July 3, 2023 at 6:54:37 PM UTC-4, ehsjr wrote:> On 7/2/2023 7:35 PM, Don wrote: > > 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. > Correct. He'll get half power out of the combination of HR (the heater > resistance) plus AR (the added resistance). The original power was 3000 > watts. Half power is 1500 watts. So he'll get 750 watts out of AR and > 750 watts out of HR. > > BUT THE OP WANTS 1500 WATTS OUT OF HR. > > Your math solves the wrong problem. > From the op: "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..."The OP's problem has been solved many times in this thread, in different ways. But the OP doesn't understand any of them. Now, people are just trying to educate the OP. -- Rick C. +--+ Get 1,000 miles of free Supercharging +--+ Tesla referral code - https://ts.la/richard11209
