half wave rectifier...

[Don]

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.

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)\');

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.

 

[Phil Allison]

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

....... Phil

 

[whit3rd]

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
run by a rheostat; \'high\' is always on, because heat pushes the bimetallic strip together.
The hysteresis is why the contacts don\'t make/break on a fast cycle (which would
wear out the high-current contacts). It\'s electrically on/off, but time proportioning.

Only a few fancy stovetops have thermostats, though ovens often do. Those haven\'t
half-heat-like settings.

 

[Ricky]

On Sunday, July 2, 2023 at 12:31:18 AM UTC-4, 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?

--

Rick C.

---+ Get 1,000 miles of free Supercharging
---+ Tesla referral code - https://ts.la/richard11209

 

[Ricky]

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.

--

Rick C.

--+- Get 1,000 miles of free Supercharging
--+- Tesla referral code - https://ts.la/richard11209

 

[Martin Brown]

On 01/07/2023 23:09, TTman wrote:

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, √ 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.

Whilst I don\'t like the heavy handed obfuscated way he has done it or
that fact that his solution sets power in the heater load to be 1.5kW
and leaves about 630W dissipated as waste heat in the dropper resistor.

So the total mains load is 2.1kW and it is one possible interpretation
of what you said you wanted to do at the outset.

1500W into Rh
and 1500W into Rs.

No. If you double the resistance you halve the current flowing in each
resistor and you also halve the voltage drop across each of them.

Power is V^2/R = 3kW in the original configuration
and V^2/2R = 1.5kW in the series configuration 750W in each.

The total dissipation it is then 1500W in *total* with 750W delivered in
each resistor. IOW you waste 50% of the energy supplied in the dropper.

The only way to use all the heat would be to have your additional series
resistor as two immersion heaters in series in the hot water tank.

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.

His objective was to reduce the load presented to his mains supply from
3kW to 1.5kW. An older >3kW industrial filament lighting or infrared
heater controller might be one option for the OP. Something like this
from Amazon depends a bit on whether it is really a thyristor (which
would be no better than his original power diode) or a triac. eg.

www.amazon.co.uk/dp/B0BY96W111/0TmFtZT1zcF9kZXRhaWwy

Spec says thyristor but it is anybody\'s guess if it is or not. Heaven
knows how much RFI a cheap and nasty Chinese one might throw out.

Better quality ones might be available scrap from old theatres that are
replacing their incandescent lights with much lower power LED systems.
(although most have already done so)

--
Martin Brown

 

[Martin Brown]

On 02/07/2023 06:17, 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

...... Phil

A non-polar 160uF capacitor rated for peak voltage 450v 50Hz and 10A
continuous ripple current isn\'t going to be cheap and cheerful.

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.

Or a pair of non-polar 80uF 450v motor start capacitors in parallel -
can\'t figure out their ripple current rating (Italian datasheet).

Phase shifting is OK for a 12W filament bulb nightlight but not very
realistic or cost effective for a 3kW load!

--
Martin Brown

 

[TTman]

On 02/07/2023 01:11, Ricky wrote:

On Saturday, July 1, 2023 at 6:09:30 PM UTC-4, TTman wrote:


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, √ 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.

Yes, that\'s what I figured.


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.

I\'m really sorry you were not willing to try to understand the math. Unfortunately, such simplistic talking about the problem, is not remotely the same as math. Once you learn how to do the math, come back and I\'ll be happy to discuss this with you. Meanwhile, you will find others agree with me, and not you. Try to recognize when there is something you need to learn.

Why do I need to consider all the RMS stuff? How is that relevant ? P= V x I

--
This email has been checked for viruses by Avast antivirus software.
www.avast.com

 

[Phil Allison]

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.



..... Phil

 

[Jan Panteltje]

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?

 

[Martin Brown]

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

 

[Don]

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.

 

[whit3rd]

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.

 

[Ricky]

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

 

[ehsjr]

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,

 

[ehsjr]

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 complicated

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

 

[Fred Bloggs]

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

 

[Don]

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.

 

[Ricky]

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

 

[Fred Bloggs]

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