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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?
On Friday, June 30, 2023 at 10:25:41â¯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.
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.
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.
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.
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.
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 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,
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.
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 ?
It\'s absolutely clear to me, and no doubt to many others on here thatOn Friday, June 30, 2023 at 5:31:50â¯PM UTC-4, TTman wrote:
On 30/06/2023 16:39, Ricky wrote:
On Friday, June 30, 2023 at 7:54:26â¯AM UTC-4, TTman wrote:
P = E^2 / R
If you string two heaters in series, you have the same voltage, but twice the resistance, so half the power.
No, you don\'t understand this english stuff. Twin heaters are rare in the UK
Not sure what you are talking about. You want to rig a diode to create half wave AC to drive a heater. That\'s not exactly \"standard\" practice anywhere.
Correct. Not standard. Read my original post that explains why.
If you have mechanical limitations, that\'s what you have. Electrically, this makes sense.
No mechanical limitations.
You could add a resistance in series. It would need to be about 30% of the resistance of the heater. The main element would dissipate half of the original heat. The heat in this added resistance would be about 22% of the heat dissipated in the original heating element. One nice thing about this is that you can tailor the heat in the main element to other than half the original heat level.
No chance. 3kW = ~18 Ohms. Where do I get 18 Ohms 1.5kW rating?
You don\'t design much electronics, do you? You need about 600 ohms at 700 watts. Six 1 ohm, 150 watt resistors will do the job.
I used to. Back in the day, 250V @3kW = 12Amps. Also R = V/I so 250/12 =
20 Ohms ish, give or take a bit. So, back in the day, to halve the
power, one had to double the resistance. Still with me ? So double 20 =
40. Correct? So what I need is a 20 ohm resistor capable of dissipating
half the power.i.e. 1500 watts in series with the 3kW heater. 1500 watts
will go in heat into the resistor, the other 1500 watts in the 3kW
heater element will go to heat the water.
Have things changed so much in modern technology that the basic
equations have been repaced with garbage ? P=ZxI and V=I/R back in the day.
So what are these new equations that tell me I need 600 Ohms at 700 watts ?
https://www.digikey.com/en/products/filter/chassis-mount-resistors/54?s=N4IgjCBcoExaBjKAzAhgGwM4FMA0IB7KAbXAAYyA2ATgHYR8wywbqHyYYBWNxqsgMwAOdmDGUKosABZaYXuDBCYlSqOllqVdWKFd1QiXEZcytLmpO0Bp0ZRjTK%2BxvaHUIL6RePhly57600oaibrQqotQCDmSRQtLSAuwwzPLSyWDCIfgwYOHUPpxcwrE5Ata0liDclAJiyfZc0oVO1NIe1dYCssnmeSI55gJmvaryozRVMJXhAdOUfb2VCUuqc7SV1FMbQnnJhpTj%2BMNkQkKlIMPDWuxXPFVXThfDjpQDl2SOdOxv7T6sGh%2B1Gou3Y4VSYJSXHo%2BHC3XSsPBAQ2MHKYNmMJAGwKPlo8TAPiEAlUCJAei4TXYwLomJkzAp6g2R3AFj8oRSegydS4HVRqmJIAAuvgAA4AFygIAAymKAE4ASwAdgBzEAAXxyp300BASEgaCweEIJHAAAIAPIACwAtpghaKJZAQABVRXysXm5AAWWwqEwAFdZdh1RqQGodfKACaSnziyUyhUq9higCeIuDTr9SDVaqAA
The equations are all the same. But you have to apply them correctly. Maybe I don\'t understand your setup, but my mental image is a resistance heater built into a pot of some sort. You want to heat the pot at half power. Your calculations are figuring the power in the entire load. Only the power in the pot heater is useful. The power in the added resistors is waste heat, no?
My calculations give you half power in the pot heater. Please remember that the added resistor will cut both the voltage and the current to the pot heater. If you are trying to do something else, I have no understanding of it at this point.
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
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...
On 30/06/2023 23:08, Ricky wrote:
On Friday, June 30, 2023 at 5:31:50â¯PM UTC-4, TTman wrote:
On 30/06/2023 16:39, Ricky wrote:
On Friday, June 30, 2023 at 7:54:26â¯AM UTC-4, TTman wrote:
P = E^2 / R
If you string two heaters in series, you have the same voltage, but twice the resistance, so half the power.
No, you don\'t understand this english stuff. Twin heaters are rare in the UK
Not sure what you are talking about. You want to rig a diode to create half wave AC to drive a heater. That\'s not exactly \"standard\" practice anywhere.
Correct. Not standard. Read my original post that explains why.
If you have mechanical limitations, that\'s what you have. Electrically, this makes sense.
No mechanical limitations.
You could add a resistance in series. It would need to be about 30% of the resistance of the heater. The main element would dissipate half of the original heat. The heat in this added resistance would be about 22% of the heat dissipated in the original heating element. One nice thing about this is that you can tailor the heat in the main element to other than half the original heat level.
No chance. 3kW = ~18 Ohms. Where do I get 18 Ohms 1.5kW rating?
You don\'t design much electronics, do you? You need about 600 ohms at 700 watts. Six 1 ohm, 150 watt resistors will do the job.
I used to. Back in the day, 250V @3kW = 12Amps. Also R = V/I so 250/12 =
20 Ohms ish, give or take a bit. So, back in the day, to halve the
power, one had to double the resistance. Still with me ? So double 20 =
40. Correct? So what I need is a 20 ohm resistor capable of dissipating
half the power.i.e. 1500 watts in series with the 3kW heater. 1500 watts
will go in heat into the resistor, the other 1500 watts in the 3kW
heater element will go to heat the water.
Have things changed so much in modern technology that the basic
equations have been repaced with garbage ? P=ZxI and V=I/R back in the day.
So what are these new equations that tell me I need 600 Ohms at 700 watts ?
https://www.digikey.com/en/products/filter/chassis-mount-resistors/54?s=N4IgjCBcoExaBjKAzAhgGwM4FMA0IB7KAbXAAYyA2ATgHYR8wywbqHyYYBWNxqsgMwAOdmDGUKosABZaYXuDBCYlSqOllqVdWKFd1QiXEZcytLmpO0Bp0ZRjTK%2BxvaHUIL6RePhly57600oaibrQqotQCDmSRQtLSAuwwzPLSyWDCIfgwYOHUPpxcwrE5Ata0liDclAJiyfZc0oVO1NIe1dYCssnmeSI55gJmvaryozRVMJXhAdOUfb2VCUuqc7SV1FMbQnnJhpTj%2BMNkQkKlIMPDWuxXPFVXThfDjpQDl2SOdOxv7T6sGh%2B1Gou3Y4VSYJSXHo%2BHC3XSsPBAQ2MHKYNmMJAGwKPlo8TAPiEAlUCJAei4TXYwLomJkzAp6g2R3AFj8oRSegydS4HVRqmJIAAuvgAA4AFygIAAymKAE4ASwAdgBzEAAXxyp300BASEgaCweEIJHAAAIAPIACwAtpghaKJZAQABVRXysXm5AAWWwqEwAFdZdh1RqQGodfKACaSnziyUyhUq9higCeIuDTr9SDVaqAA
The equations are all the same. But you have to apply them correctly. Maybe I don\'t understand your setup, but my mental image is a resistance heater built into a pot of some sort. You want to heat the pot at half power. Your calculations are figuring the power in the entire load. Only the power in the pot heater is useful. The power in the added resistors is waste heat, no?
My calculations give you half power in the pot heater. Please remember that the added resistor will cut both the voltage and the current to the pot heater. If you are trying to do something else, I have no understanding of it at this point.
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...
On Saturday, July 1, 2023 at 5:33:42â¯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.
On 30/06/2023 23:08, Ricky wrote:
On Friday, June 30, 2023 at 5:31:50â¯PM UTC-4, TTman wrote:
On 30/06/2023 16:39, Ricky wrote:
On Friday, June 30, 2023 at 7:54:26â¯AM UTC-4, TTman wrote:
P = E^2 / R
If you string two heaters in series, you have the same voltage,
but twice the resistance, so half the power.
No, you don\'t understand this english stuff. Twin heaters are
rare in the UK
Not sure what you are talking about. You want to rig a diode to
create half wave AC to drive a heater. That\'s not exactly
\"standard\" practice anywhere.
Correct. Not standard. Read my original post that explains why.
If you have mechanical limitations, that\'s what you have.
Electrically, this makes sense.
No mechanical limitations.
You could add a resistance in series. It would need to be about
30% of the resistance of the heater. The main element would
dissipate half of the original heat. The heat in this added
resistance would be about 22% of the heat dissipated in the
original heating element. One nice thing about this is that you
can tailor the heat in the main element to other than half the
original heat level.
No chance. 3kW = ~18 Ohms. Where do I get 18 Ohms 1.5kW rating?
You don\'t design much electronics, do you? You need about 600 ohms
at 700 watts. Six 1 ohm, 150 watt resistors will do the job.
I used to. Back in the day, 250V @3kW = 12Amps. Also R = V/I so 250/12 =
20 Ohms ish, give or take a bit. So, back in the day, to halve the
power, one had to double the resistance. Still with me ? So double 20 =
40. Correct? So what I need is a 20 ohm resistor capable of dissipating
half the power.i.e. 1500 watts in series with the 3kW heater. 1500 watts
will go in heat into the resistor, the other 1500 watts in the 3kW
heater element will go to heat the water.
Have things changed so much in modern technology that the basic
equations have been repaced with garbage ? P=ZxI and V=I/R back in
the day.
So what are these new equations that tell me I need 600 Ohms at 700
watts ?
https://www.digikey.com/en/products/filter/chassis-mount-resistors/54?s=N4IgjCBcoExaBjKAzAhgGwM4FMA0IB7KAbXAAYyA2ATgHYR8wywbqHyYYBWNxqsgMwAOdmDGUKosABZaYXuDBCYlSqOllqVdWKFd1QiXEZcytLmpO0Bp0ZRjTK%2BxvaHUIL6RePhly57600oaibrQqotQCDmSRQtLSAuwwzPLSyWDCIfgwYOHUPpxcwrE5Ata0liDclAJiyfZc0oVO1NIe1dYCssnmeSI55gJmvaryozRVMJXhAdOUfb2VCUuqc7SV1FMbQnnJhpTj%2BMNkQkKlIMPDWuxXPFVXThfDjpQDl2SOdOxv7T6sGh%2B1Gou3Y4VSYJSXHo%2BHC3XSsPBAQ2MHKYNmMJAGwKPlo8TAPiEAlUCJAei4TXYwLomJkzAp6g2R3AFj8oRSegydS4HVRqmJIAAuvgAA4AFygIAAymKAE4ASwAdgBzEAAXxyp300BASEgaCweEIJHAAAIAPIACwAtpghaKJZAQABVRXysXm5AAWWwqEwAFdZdh1RqQGodfKACaSnziyUyhUq9higCeIuDTr9SDVaqAA
The equations are all the same. But you have to apply them
correctly. Maybe I don\'t understand your setup, but my mental image
is a resistance heater built into a pot of some sort. You want to
heat the pot at half power. Your calculations are figuring the power
in the entire load. Only the power in the pot heater is useful. The
power in the added resistors is waste heat, no?
My calculations give you half power in the pot heater. Please
remember that the added resistor will cut both the voltage and the
current to the pot heater. If you are trying to do something else, I
have no understanding of it at this point.
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...
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...
I already did, but here it is again.
P = E^2 / Rh, you want to reduce the voltage on the heater by adding a resistor in series, i.e. replace Rh with Rh + Rs, where Rs is the added resistor. So, how large does Rs need to be for the power in Rh to be half the original value (1,500 W vs. 3,000 W)?
So, now we have Ph = Eh^2 / Rh. Since Ph is 1/2 * P from above, 2 * Eh^2 / Rh = E^2 / Rh.
Multiply by Rh / Eh^2 to get, â 2 = E / Eh = 1.414, Eh = E * 0.707.
So now, we know the voltage needed on the heater for it to dissipate half the original power.
Easy to get the required added resistance, Es = E * Rs / (Rs + Rh), E * 0.293 = E * Rs / (Rs + Rh), 0.293 = Rs / (Rs + Rh). 0.293 Rs + 0.293 Rh = Rs, 0.293 Rh = 0.707 Rs, Rs = Rh * 0.293 / 0.707 = Rh * 0.414.
Rh can be found from P = E^2 / Rh, Rh = E^2 / P = 16.133 ohms, so Rh = 6.6861 ohms.
Was there any part you didn\'t follow?
I would point out that anyone familiar with using dB to compare power levels has likely discovered that -3 dB is half power, but 0.707 of the voltage and current. Most designers simply know this off the top of their heads. I take it you don\'t use such calculations very often.
Too complicated...HR= 20 Ohms. Add another resistor in series = 20 OhmsThe 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
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...
I already did, but here it is again.
P = E^2 / Rh, you want to reduce the voltage on the heater by adding a resistor in series, i.e. replace Rh with Rh + Rs, where Rs is the added resistor. So, how large does Rs need to be for the power in Rh to be half the original value (1,500 W vs. 3,000 W)?
So, now we have Ph = Eh^2 / Rh. Since Ph is 1/2 * P from above, 2 * Eh^2 / Rh = E^2 / Rh.
Multiply by Rh / Eh^2 to get, â 2 = E / Eh = 1.414, Eh = E * 0.707.
So now, we know the voltage needed on the heater for it to dissipate half the original power.
Easy to get the required added resistance, Es = E * Rs / (Rs + Rh), E * 0.293 = E * Rs / (Rs + Rh), 0.293 = Rs / (Rs + Rh). 0.293 Rs + 0.293 Rh = Rs, 0.293 Rh = 0.707 Rs, Rs = Rh * 0.293 / 0.707 = Rh * 0.414..
Rh can be found from P = E^2 / Rh, Rh = E^2 / P = 16.133 ohms, so Rh = 6.6861 ohms.
Was there any part you didn\'t follow?
None of it.
You\'re over thinking it. Rh = Rs. Rh = ~18 Ohms. therefore
Rs = 18 ohms.Half the power = twice the resistance.Simple. 1500W into Rh
and 1500W into Rs.
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 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.