My bathroom extract fan's wax actuator has failed. This is used to
slowly open the shutters, when the fan is triggered. This one..

https://www.ebay.co.uk/itm/223481092415?ul_noapp=true

It looks as if it has over-heated in the area, before it actually
failed. As it is marked 120/240v I'm thinking to install a dropper to
reduce the voltage across it to around 120v, but I cannot find a
specification for the wattage or current to determine a suitable
dropper.

Anyone know please?

It happens that Harry Bloomfield formulated :
My bathroom extract fan's wax actuator has failed. This is used to slowly
open the shutters, when the fan is triggered. This one..

https://www.ebay.co.uk/itm/223481092415?ul_noapp=true

It looks as if it has over-heated in the area, before it actually failed. As
it is marked 120/240v I'm thinking to install a dropper to reduce the voltage
across it to around 120v, but I cannot find a specification for the wattage
or current to determine a suitable dropper.

Anyone know please?

I've just had another idea - a 400v diode will reduce the consumption
to around 50%, without the extra heat dissipation of a resistor.

Just trying to extend the working life of the actuator.

On 05/06/2019 09:55, Harry Bloomfield wrote:
It happens that Harry Bloomfield formulated :
My bathroom extract fan's wax actuator has failed. This is used to
slowly open the shutters, when the fan is triggered. This one..

https://www.ebay.co.uk/itm/223481092415?ul_noapp=true

It looks as if it has over-heated in the area, before it actually
failed. As it is marked 120/240v I'm thinking to install a dropper to
reduce the voltage across it to around 120v, but I cannot find a
specification for the wattage or current to determine a suitable dropper.

Anyone know please?

I've just had another idea - a 400v diode will reduce the consumption to
around 50%, without the extra heat dissipation of a resistor.

Just trying to extend the working life of the actuator.

Good idea, but 400V doesn't give much margin on 240V RMS - go for a
1N4007 1kV diode. This used to be a trick for keeping a
non-thermostatic soldering iron on standby, you put a diode in a torpedo
switch.

(Although the motor does say .02A which would make it ~5W @ 240V.)

Cheers
--
Clive

Clive Arthur submitted this idea :
Good idea, but 400V doesn't give much margin on 240V RMS - go for a 1N4007
1kV diode.

Good point, I have plenty of 4004's around here.
This used to be a trick for keeping a non-thermostatic soldering
iron on standby, you put a diode in a torpedo switch.

Yes, no you mention it, I have used that trick myself for irons.



(Although the motor does say .02A which would make it ~5W @ 240V.)

Yes, I spotted the .02A but I wasn't clear that it meant amps and it
doesn't indicate .02A at which voltage.

As it was, it was quite quick to respond to open, it will open a bit
slower now with the diode in circuit - once the replacement actuator
arrives.

Harry Bloomfield explained on 05/06/2019 :
Good point, I have plenty of 4004's around here.

I meant 4007's

I doubt if what you want to do is going to make much difference. These
things are designed to not get hot, if they do then they will be faulty. All
a dropper will do is make a lot of heat wherever you site it, what is the
point of that. I don't know how these devices work but it might well be that
the actual mechanism has become stiff and hence caused the failure. Back in
the old days Greenhouses had a purely mechanical device that opened and
closed the vents.
Brian

--
----- --
This newsgroup posting comes to you directly from...
The Sofa of Brian Gaff...

Blind user, so no pictures please
Note this Signature is meaningless.!
"Harry Bloomfield" wrote in message
...
My bathroom extract fan's wax actuator has failed. This is used to slowly
open the shutters, when the fan is triggered. This one..

https://www.ebay.co.uk/itm/223481092415?ul_noapp=true

It looks as if it has over-heated in the area, before it actually failed.
As it is marked 120/240v I'm thinking to install a dropper to reduce the
voltage across it to around 120v, but I cannot find a specification for
the wattage or current to determine a suitable dropper.

Anyone know please?

Brian Gaff pretended :
These things are designed to not get hot, if they do then they will be
faulty. All a dropper will do is make a lot of heat wherever you site it,
what is the point of that. I don't know how these devices work but it might
well be that the actual mechanism has become stiff and hence caused the
failure.

Quite the contrary, they are designed to get hot. The heat melts a high
expansion wax, which pushes out a plunger. The work in a similar way to
the wax thermostat used in car engine cooling systems - heat in the
coolant, opens the valve. In this case a tiny heating element heat the
wax, causing a plunger to extend, this then opens the fan cowl up.

It is designed to work with 120/240v and has heat destroyed it, my idea
is to fit a replacement but halve the 240v here, to average 120v, using
a 1N4007 diode. It will no doubt open up little slower, but no problem.

On Wednesday, 5 June 2019 21:57:20 UTC+1, Harry Bloomfield wrote:
Brian Gaff pretended :

These things are designed to not get hot, if they do then they will be
faulty. All a dropper will do is make a lot of heat wherever you site it,
what is the point of that. I don't know how these devices work but it might
well be that the actual mechanism has become stiff and hence caused the
failure.

Quite the contrary, they are designed to get hot. The heat melts a high
expansion wax, which pushes out a plunger. The work in a similar way to
the wax thermostat used in car engine cooling systems - heat in the
coolant, opens the valve. In this case a tiny heating element heat the
wax, causing a plunger to extend, this then opens the fan cowl up.

It is designed to work with 120/240v and has heat destroyed it, my idea
is to fit a replacement but halve the 240v here, to average 120v, using
a 1N4007 diode. It will no doubt open up little slower, but no problem.

All Brian's points are incorrect except that they do often get stiff due to limescale or corrosion.
A diode will give 240v x 0.707 rms volts, which is more than 120.


NT

On 05/06/2019 22:21, wrote:
On Wednesday, 5 June 2019 21:57:20 UTC+1, Harry Bloomfield wrote:
Brian Gaff pretended :

These things are designed to not get hot, if they do then they will be
faulty. All a dropper will do is make a lot of heat wherever you site it,
what is the point of that. I don't know how these devices work but it might
well be that the actual mechanism has become stiff and hence caused the
failure.

Quite the contrary, they are designed to get hot. The heat melts a high
expansion wax, which pushes out a plunger. The work in a similar way to
the wax thermostat used in car engine cooling systems - heat in the
coolant, opens the valve. In this case a tiny heating element heat the
wax, causing a plunger to extend, this then opens the fan cowl up.

It is designed to work with 120/240v and has heat destroyed it, my idea
is to fit a replacement but halve the 240v here, to average 120v, using
a 1N4007 diode. It will no doubt open up little slower, but no problem.

All Brian's points are incorrect except that they do often get stiff due to limescale or corrosion.
A diode will give 240v x 0.707 rms volts, which is more than 120.

Correct, the unit with a diode at 240V will consume half the power at
240V without.

I suspect a diode is the cheapest and most reliable method but this is
an ideal candidate for using a capacitive dropper.

On 05/06/2019 09:52, Harry Bloomfield wrote:
My bathroom extract fan's wax actuator has failed. This is used to
slowly open the shutters, when the fan is triggered. This one..

https://www.ebay.co.uk/itm/223481092415?ul_noapp=true

It looks as if it has over-heated in the area, before it actually
failed. As it is marked 120/240v I'm thinking to install a dropper to
reduce the voltage across it to around 120v, but I cannot find a
specification for the wattage or current to determine a suitable dropper.

Anyone know please?


It says on it 240V, 0.02A

Has it overheated because the connections to the spade terminals were
not perfect?

Was it actually overheating or just discoloured due to it being used in
a damp environment and dust in the atmosphere has just stuck to the
casing over the years that it has been in use.

I would just replace like with like without modifications.

--
mailto : news {at} admac {dot} myzen {dot} co {dot} uk

On 05/06/2019 21:57, Harry Bloomfield wrote:


It is designed to work with 120/240v and has heat destroyed it, my idea
is to fit a replacement but halve the 240v here, to average 120v, using
a 1N4007 diode. It will no doubt open up little slower, but no problem.

Will it not get to the same temperature but just take longer? Putting
2.5W or 5W into a device that cannot dissipate that power to heatsink or
air flow will cause the temperature to rise above that required to push
out the plunger to its maximum travel.

If the shutters take longer to open will the actuator be in a static or
a lower airflow for longer?




--
mailto : news {at} admac {dot} myzen {dot} co {dot} uk

On 06/06/2019 10:25, alan_m wrote:
On 05/06/2019 21:57, Harry Bloomfield wrote:


It is designed to work with 120/240v and has heat destroyed it, my
idea is to fit a replacement but halve the 240v here, to average 120v,
using a 1N4007 diode. It will no doubt open up little slower, but no
problem.

Will it not get to the same temperature but just take longer?* Putting
2.5W or 5W into a device that cannot dissipate that power to heatsink or
air flow will cause the temperature to rise above that required to push
out the plunger to its maximum travel.

With that sort of actuator, even once it has opened, the power stays on,
adding heat energy. It will also be losing heat to the environment. The
temperature will stabilise when energy inflow equals energy outflow.

If you halve the inflow, you halve the outflow at equilibrium and as the
rate of energy loss is proportional to temperature difference (body to
air), that dictates that equilibrium will be established at a lower
actuator body temperature - so it *WILL* run cooler.

If the shutters take longer to open will the actuator be in a static or
a lower airflow for longer?

The actuator will still be fully extended by the time it reaches its
normal fully open temperature. It will continue to rise to its
equilibrium temperature, but the vent is already fully open by then.

SteveW

On Thursday, 6 June 2019 10:25:39 UTC+1, alan_m wrote:
On 05/06/2019 21:57, Harry Bloomfield wrote:


It is designed to work with 120/240v and has heat destroyed it, my idea
is to fit a replacement but halve the 240v here, to average 120v, using
a 1N4007 diode. It will no doubt open up little slower, but no problem.

Will it not get to the same temperature but just take longer?

no

Putting
2.5W or 5W into a device that cannot dissipate that power to heatsink or
air flow

obviously it does dissipate that

will cause the temperature to rise above that required to push
out the plunger to its maximum travel.

that's how they work

If the shutters take longer to open will the actuator be in a static or
a lower airflow for longer?

alan_m wrote on 06/06/2019 :
Has it overheated because the connections to the spade terminals were not
perfect?

The wire is soldered onto the terminals with no sign of overheating
there. The fan casing had become discoloured and brittle, the end of
the actuator opposite the piston/plunger had actually burst away. It
has quite a strong spring built into the actuator, to return it once it
has cooled.


Was it actually overheating or just discoloured due to it being used in a
damp environment and dust in the atmosphere has just stuck to the casing over
the years that it has been in use.

on 05/06/2019, supposed :
All Brian's points are incorrect except that they do often get stiff due to
limescale or corrosion.
A diode will give 240v x 0.707 rms volts, which is more than 120.

No, it will be 50% of the AC waveform, less a bit for the fraction of a
volt lost across the junction.

0.707 x the peak AC is the RMS voltage.

If you know the RMS, then multiplying by 1.41 will give you the peak
voltage. If you charged a cap via a bridge without any loading, that
would achieve 240v x 1.41 = 338.4v peak, less junction losses.

On Thursday, 6 June 2019 15:12:49 UTC+1, Harry Bloomfield wrote:
on 05/06/2019, tabbypurr supposed :

All Brian's points are incorrect except that they do often get stiff due to
limescale or corrosion.
A diode will give 240v x 0.707 rms volts, which is more than 120.

No, it will be 50% of the AC waveform, less a bit for the fraction of a
volt lost across the junction.

do you ever get anything right?

0.707 x the peak AC is the RMS voltage.

If you know the RMS, then multiplying by 1.41 will give you the peak
voltage. If you charged a cap via a bridge without any loading, that
would achieve 240v x 1.41 = 338.4v peak, less junction losses.

On 06/06/2019 15:12, Harry Bloomfield wrote:
on 05/06/2019, supposed :
All Brian's points are incorrect except that they do often get stiff
due to limescale or corrosion.
A diode will give 240v x 0.707 rms volts, which is more than 120.

No, it will be 50% of the AC waveform, less a bit for the fraction of a
volt lost across the junction.

Quite, so half the power? Same voltage but half the duration?

If you halved the voltage, what would happen to the power?

0.707 x the peak AC is the RMS voltage.

And 0.707 of the rms voltage gives you half the power.

snip

On 06/06/2019 15:12, Harry Bloomfield wrote:
on 05/06/2019, supposed :
All Brian's points are incorrect except that they do often get stiff
due to limescale or corrosion.
A diode will give 240v x 0.707 rms volts, which is more than 120.

No, it will be 50% of the AC waveform, less a bit for the fraction of a
volt lost across the junction.

0.707 x the peak AC is the RMS voltage.

Of a sinewave, not a half-wave rectified sine wave.

Sinewave: 339V / sqrt(2) = 240V RMS

Half-wave rectified sinewave: 339V / 2 = 169.5V RMS

SteveW

Steve Walker expressed precisely :
Of a sinewave, not a half-wave rectified sine wave.

Sinewave: 339V / sqrt(2) = 240V RMS

Half-wave rectified sinewave: 339V / 2 = 169.5V RMS

SteveW

169.5v x 0.707= 119.83v RMS..

expressed precisely :
On Thursday, 6 June 2019 15:12:49 UTC+1, Harry Bloomfield wrote:
on 05/06/2019, tabbypurr supposed :

All Brian's points are incorrect except that they do often get stiff due
to
limescale or corrosion.
A diode will give 240v x 0.707 rms volts, which is more than 120.

No, it will be 50% of the AC waveform, less a bit for the fraction of a
volt lost across the junction.

do you ever get anything right?

Yes, quite often actually and I am right this time too.

On 08/06/2019 09:47, Harry Bloomfield wrote:
expressed precisely :
On Thursday, 6 June 2019 15:12:49 UTC+1, Harry BloomfieldÂ* wrote:
on 05/06/2019, tabbypurr supposed :

All Brian's points are incorrect except that they do often get stiff
due to limescale or corrosion.
A diode will give 240v x 0.707 rms volts, which is more than 120.

No, it will be 50% of the AC waveform, less a bit for the fraction of
a volt lost across the junction.

do you ever get anything right?

Yes, quite often actually and I am right this time too.

Dont see tabbypurr as he is plonked, but yes, he is talking cock and for
a resistive load, a diode will indeed halve the power.

Its the same watts for half the time so to speak.

'RMS volts' on a clipped waveform is almost not worth even thinking
about. Its a seriously damaged concept by then.

--
Renewable energy: Expensive solutions that don't work to a problem that
doesn't exist instituted by self legalising protection rackets that
don't protect, masquerading as public servants who don't serve the public.

The RMS voltage is halved - the voltage profile stays the same for one half-cycle (hence same RMS for this portion) and is zero for the next half-cycle, so the average over a full cycle is halved (ignoring diode voltage drop).. Nonetheless, the confusion on this point accidentally gets the correct answer that the half-cycle version of 240v is effectively equivalent to 170v in terms of power, but thats because for a resistive loud the power consumption goes as v^2 (ignoring any temperature dependence of heater resistance)

On 08/06/2019 09:45, Harry Bloomfield wrote:
Steve Walker expressed precisely :
Of a sinewave, not a half-wave rectified sine wave.

Sinewave: 339V / sqrt(2) = 240V RMS

Half-wave rectified sinewave: 339V / 2 = 169.5V RMS

SteveW

169.5v x 0.707= 119.83v RMS..

Peak voltage is 339V for both full and half-wave rectified sinewaves.
The equations for RMS voltage differ in each case. A quick google will
show you them and they are as I quoted above. You have taken the RMS
(not peak) voltage of a half-wave rectified sinewave and multiplied it
by the factor for converting a full-wave sinewave's peak voltage into an
RMS one!

SteveW

The Natural Philosopher wrote:

On 08/06/2019 09:47, Harry Bloomfield wrote:
expressed precisely :
On Thursday, 6 June 2019 15:12:49 UTC+1, Harry Bloomfield wrote:
on 05/06/2019, tabbypurr supposed :

All Brian's points are incorrect except that they do often get stiff
due to limescale or corrosion.
A diode will give 240v x 0.707 rms volts, which is more than 120.

No, it will be 50% of the AC waveform, less a bit for the fraction of
a volt lost across the junction.

do you ever get anything right?

Yes, quite often actually and I am right this time too.

Dont see tabbypurr as he is plonked, but yes, he is talking cock and for
a resistive load, a diode will indeed halve the power.

Its the same watts for half the time so to speak.

'RMS volts' on a clipped waveform is almost not worth even thinking
about. Its a seriously damaged concept by then.

Unlike RMS power, which is a seriously confused concept, RMS voltage of
any repetitive waveform, regardless of shape, remains a precisely
defined quantity and, importantly, continues to goven the average power
delivered to a resistive load over that repetitive waveform. The
relationship to peak voltage is, of course, arbitrary in the general
case. But a rectified sine wave still allows simple, accurate
calculation. But not by me, at least at this time in the morning.



--

Roger Hayter

On 08/06/2019 10:21, wrote:
The RMS voltage is halved - the voltage profile stays the same for one half-cycle (hence same RMS for this portion) and is zero for the next half-cycle, so the average over a full cycle is halved (ignoring diode voltage drop). Nonetheless, the confusion on this point accidentally gets the correct answer that the half-cycle version of 240v is effectively equivalent to 170v in terms of power, but thats because for a resistive loud the power consumption goes as v^2 (ignoring any temperature dependence of heater resistance)

Its all a matter of definition. RMS volts is, strictly I guess, heating
power, but in a weird waveform it doesnt make much sense to think in
those terms.


--
No Apple devices were knowingly used in the preparation of this post.

On 08/06/2019 10:39, Roger Hayter wrote:
The Natural Philosopher wrote:

On 08/06/2019 09:47, Harry Bloomfield wrote:
expressed precisely :
On Thursday, 6 June 2019 15:12:49 UTC+1, Harry Bloomfield wrote:
on 05/06/2019, tabbypurr supposed :

All Brian's points are incorrect except that they do often get stiff
due to limescale or corrosion.
A diode will give 240v x 0.707 rms volts, which is more than 120.

No, it will be 50% of the AC waveform, less a bit for the fraction of
a volt lost across the junction.

do you ever get anything right?

Yes, quite often actually and I am right this time too.

Dont see tabbypurr as he is plonked, but yes, he is talking cock and for
a resistive load, a diode will indeed halve the power.

Its the same watts for half the time so to speak.

'RMS volts' on a clipped waveform is almost not worth even thinking
about. Its a seriously damaged concept by then.

Unlike RMS power, which is a seriously confused concept, RMS voltage of
any repetitive waveform, regardless of shape, remains a precisely
defined quantity and, importantly, continues to goven the average power
delivered to a resistive load over that repetitive waveform. The
relationship to peak voltage is, of course, arbitrary in the general
case. But a rectified sine wave still allows simple, accurate
calculation. But not by me, at least at this time in the morning.



Easiest is to simply halve the power of a full wave as it only is there
half the time and take the root.



--
€œBut what a weak barrier is truth when it stands in the way of an
hypothesis!€�

Mary Wollstonecraft

Steve Walker wrote:

On 08/06/2019 09:45, Harry Bloomfield wrote:
Steve Walker expressed precisely :
Of a sinewave, not a half-wave rectified sine wave.

Sinewave: 339V / sqrt(2) = 240V RMS

Half-wave rectified sinewave: 339V / 2 = 169.5V RMS

SteveW

169.5v x 0.707= 119.83v RMS..

Peak voltage is 339V for both full and half-wave rectified sinewaves.
The equations for RMS voltage differ in each case. A quick google will
show you them and they are as I quoted above. You have taken the RMS
(not peak) voltage of a half-wave rectified sinewave and multiplied it
by the factor for converting a full-wave sinewave's peak voltage into an
RMS one!

SteveW

You can confirm your result by simple inspection. The mean square
voltage of a half wave rectified sine wave must be half that of a full
wave sine wave because half the readings are replaced by 0^2. But for
RMS we take the square root of mean square voltage. Square root of a
half is 1/1.414 which is 0.707. So half wave rectification halves the
power into a resistive load. Any other result would be surprising
because it is applying the same waveform for half the time, ignoring
polarity which we can for a resistive load.

--

Roger Hayter

wrote:

The RMS voltage is halved - the voltage profile stays the same for one
half-cycle (hence same RMS for this portion) and is zero for the next
half-cycle, so the average over a full cycle is halved (ignoring diode
voltage drop). Nonetheless, the confusion on this point accidentally
gets the correct answer that the half-cycle version of 240v is
effectively equivalent to 170v in terms of power, but that's because for
a resistive loud the power consumption goes as v^2 (ignoring any
temperature dependence of heater resistance)

Sorry (a passive aggressive sorry) but if you halved the RMS voltage
then you'd get a quarter of the power. This is intuitively nonsense
because the load is getting the same power for half the time. So it is
not surprising that the actual calculation shows that the RMS voltage of
a half wave rectified sine wave is actually 0.707 of the RMS voltage of
a full sine wave. The RMS voltage is not linearly related to the mean
voltage.

--

Roger Hayter

The Natural Philosopher wrote:

On 08/06/2019 10:39, Roger Hayter wrote:
The Natural Philosopher wrote:

On 08/06/2019 09:47, Harry Bloomfield wrote:
expressed precisely :
On Thursday, 6 June 2019 15:12:49 UTC+1, Harry Bloomfield wrote:
on 05/06/2019, tabbypurr supposed :

All Brian's points are incorrect except that they do often get stiff
due to limescale or corrosion.
A diode will give 240v x 0.707 rms volts, which is more than 120.

No, it will be 50% of the AC waveform, less a bit for the fraction of
a volt lost across the junction.

do you ever get anything right?

Yes, quite often actually and I am right this time too.

Dont see tabbypurr as he is plonked, but yes, he is talking cock and for
a resistive load, a diode will indeed halve the power.

Its the same watts for half the time so to speak.

'RMS volts' on a clipped waveform is almost not worth even thinking
about. Its a seriously damaged concept by then.

Unlike RMS power, which is a seriously confused concept, RMS voltage of
any repetitive waveform, regardless of shape, remains a precisely
defined quantity and, importantly, continues to goven the average power
delivered to a resistive load over that repetitive waveform. The
relationship to peak voltage is, of course, arbitrary in the general
case. But a rectified sine wave still allows simple, accurate
calculation. But not by me, at least at this time in the morning.



Easiest is to simply halve the power of a full wave as it only is there
half the time and take the root.

Agreed!

--

Roger Hayter

On Saturday, 8 June 2019 09:45:19 UTC+1, Harry Bloomfield wrote:
Steve Walker expressed precisely :

Of a sinewave, not a half-wave rectified sine wave.

Sinewave: 339V / sqrt(2) = 240V RMS

Half-wave rectified sinewave: 339V / 2 = 169.5V RMS

SteveW

169.5v x 0.707= 119.83v RMS..

Lol.

On Saturday, 8 June 2019 09:59:16 UTC+1, The Natural Philosopher wrote:
On 08/06/2019 09:47, Harry Bloomfield wrote:
tabbypurr expressed precisely :
On Thursday, 6 June 2019 15:12:49 UTC+1, Harry BloomfieldÂ* wrote:
on 05/06/2019, tabbypurr supposed :

All Brian's points are incorrect except that they do often get stiff
due to limescale or corrosion.
A diode will give 240v x 0.707 rms volts, which is more than 120.

No, it will be 50% of the AC waveform, less a bit for the fraction of
a volt lost across the junction.

do you ever get anything right?

Yes, quite often actually and I am right this time too.

Dont see tabbypurr as he is plonked, but yes, he is talking cock and for
a resistive load, a diode will indeed halve the power.

Half power yes, which equates to 240v x 0.707 rms volts, as I said. Wakey wakey.


NT

On Saturday, 8 June 2019 10:34:29 UTC+1, Steve Walker wrote:
On 08/06/2019 09:45, Harry Bloomfield wrote:
Steve Walker expressed precisely :

Of a sinewave, not a half-wave rectified sine wave.

Sinewave: 339V / sqrt(2) = 240V RMS

Half-wave rectified sinewave: 339V / 2 = 169.5V RMS

SteveW

169.5v x 0.707= 119.83v RMS..

Peak voltage is 339V for both full and half-wave rectified sinewaves.
The equations for RMS voltage differ in each case. A quick google will
show you them and they are as I quoted above. You have taken the RMS
(not peak) voltage of a half-wave rectified sinewave and multiplied it
by the factor for converting a full-wave sinewave's peak voltage into an
RMS one!

SteveW

I don't think Harry has a clue what he's done.

On 08/06/2019 11:16, Roger Hayter wrote:
Steve Walker wrote:

On 08/06/2019 09:45, Harry Bloomfield wrote:
Steve Walker expressed precisely :
Of a sinewave, not a half-wave rectified sine wave.

Sinewave: 339V / sqrt(2) = 240V RMS

Half-wave rectified sinewave: 339V / 2 = 169.5V RMS

SteveW

169.5v x 0.707= 119.83v RMS..

Peak voltage is 339V for both full and half-wave rectified sinewaves.
The equations for RMS voltage differ in each case. A quick google will
show you them and they are as I quoted above. You have taken the RMS
(not peak) voltage of a half-wave rectified sinewave and multiplied it
by the factor for converting a full-wave sinewave's peak voltage into an
RMS one!

SteveW

You can confirm your result by simple inspection. The mean square
voltage of a half wave rectified sine wave must be half that of a full
wave sine wave because half the readings are replaced by 0^2. But for
RMS we take the square root of mean square voltage. Square root of a
half is 1/1.414 which is 0.707. So half wave rectification halves the
power into a resistive load. Any other result would be surprising
because it is applying the same waveform for half the time, ignoring
polarity which we can for a resistive load.

Different ways of getting to the same result, unlike Harry's oddball
guess. Yours of course does have the advantage of showing how it gets to
the result, whereas mine just applies the different equations.

SteveW