### Re: [Dorset] OT: Keeping costs down

On 29/08/2022 14:16, Peter Merchant via dorset wrote:
I will now try and investigate why my gas boiler keeps on firing up
when it kicks in, even though I have used solar to heat my tank. It
has been suggested that the solar heats the top part of the tank via
the immersion heater, and the gas heats the bottom part of the tank
via the heating coil.

That's right.  The thermostat for the gas-heated water is near the
bottom of the tank (mine's about 1/4 of the way up).   The thermostat
for the immersion heater is usually in the body of the device, at the
top.  The boiler therefore runs until nearly all of the tank has reached
the required temperature and the immersion cuts out when the top half of
the tank is hot.

--
Terry Coles

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### Re: [Dorset] OT: Keeping costs down

using those figures and the cost of electricity from my latest bill (27.10
p/Kwh ) using the kettle costs 1.49p which isn't a lot more than using the hob
(1.125 p).

It's not really a game changer, so thanks everyone for contributing.

I will now try and investigate why my gas boiler keeps on firing up when it
kicks in, even though I have used solar to heat my tank. It has been suggested
that the solar heats the top part of the tank via the immersion heater, and the
gas heats the bottom part of the tank via the heating coil.

Peter.

On 29/08/2022 13:24, Ian Morris via dorset wrote:

So I just boiled my kettle (500ml, fresh from tap). Mains here = 242 volts,
Kettle takes 12.3 A (pretty much constant) Power taken = 0.055kWh (so pretty
close to my calculation!) I allowed the kettle to trip using the steam
generated by the then boiling water (so addressing both the loss and latent
heat aspects of the question.)

On 29/08/2022 12:28, Peter Merchant via dorset wrote:

I had considered that, but it's so long ago in my ancient past that I studied
thermodynamics, I didn't look into calculating it. My concern about it is the
cost of raising the temperature to 100 deg, then the latent heat of
vaporisation to actually boil it.

Peter M.   [ Applies to any form of boiling water of course. ]

n 29/08/2022 11:31, Ian Morris via dorset wrote:

500ml of water = 500g water.  let's assume you want to go from 20-100 C =>
increase of 80C
Specific heat capacity of water = 4.184 J/g-K
=> 167,360 J = 0.0465 kWh Electricity @27p/kWh = 1.2552p

So rather similar (OK, I don't know efficiency of electric kettle ... but
heating elements are 100% efficient... sure there will be some losses to both
the kettle and the wider environment, but i doubt somehow that they are
huge) Of course this is about cost for the consumer ... I'm totally
ignoring the ineffeciency of turning gas into electricity and transmitting it
around the country in the first place.

On 29/08/2022 09:33, Peter Merchant wrote:
I was curious, so I took the water from the kettle and put it in a pot on the
stove. It was very close to 500ml.

Watching the gas meter while it boiled it used 0.014 m(cubed) of gas, which
converted to 0.15788Kwh and at my current rate of 7.123p/Kwh cost me 1.1.25p to
boil up.

Unfortunately I can't see my electricity meter easily to see what it costs me
to boil that amount in the kettle or by Microwave. Does anybody have that
facility?  And also there is always other electricity being used.

Cheers,

Peter

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### Re: [Dorset] OT: Keeping costs down

So I just boiled my kettle (500ml, fresh from tap). Mains here = 242
volts, Kettle takes 12.3 A (pretty much constant) Power taken = 0.055kWh
(so pretty close to my calculation!) I allowed the kettle to trip using
the steam generated by the then boiling water (so addressing both the
loss and latent heat aspects of the question.)

On 29/08/2022 12:28, Peter Merchant via dorset wrote:
I had considered that, but it's so long ago in my ancient past that I
studied thermodynamics, I didn't look into calculating it. My concern
about it is the cost of raising the temperature to 100 deg, then the
latent heat of vaporisation to actually boil it.

Peter M.   [ Applies to any form of boiling water of course. ]

n 29/08/2022 11:31, Ian Morris via dorset wrote:
500ml of water = 500g water.  let's assume you want to go from 20-100
C => increase of 80C

Specific heat capacity of water = 4.184 J/g-K
=> 167,360 J = 0.0465 kWh Electricity @27p/kWh = 1.2552p

So rather similar (OK, I don't know efficiency of electric kettle
... but heating elements are 100% efficient... sure there will be
some losses to both the kettle and the wider environment, but i doubt
somehow that they are huge) Of course this is about cost for the
consumer ... I'm totally ignoring the ineffeciency of turning gas
into electricity and transmitting it around the country in the first
place.

On 29/08/2022 09:33, Peter Merchant wrote:
I was curious, so I took the water from the kettle and put it in a
pot on the stove. It was very close to 500ml.

Watching the gas meter while it boiled it used 0.014 m(cubed) of gas,
which converted to 0.15788Kwh and at my current rate of 7.123p/Kwh
cost me 1.1.25p to boil up.

Unfortunately I can't see my electricity meter easily to see what it
costs me to boil that amount in the kettle or by Microwave. Does
anybody have that facility?  And also there is always other
electricity being used.

Cheers,

Peter

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### Re: [Dorset] OT: Keeping costs down

My 1965 'Basic Tables in Electrical Engineering' gives the Heat of Vaporization
of Water as 972 Btu/Lb   so there is quite a bit of conversion required!

P.
On 29/08/2022 12:28, Peter Merchant via dorset wrote:

I had considered that, but it's so long ago in my ancient past that I studied
thermodynamics, I didn't look into calculating it. My concern about it is the
cost of raising the temperature to 100 deg, then the latent heat of
vaporisation to actually boil it.

Peter M.   [ Applies to any form of boiling water of course. ]

n 29/08/2022 11:31, Ian Morris via dorset wrote:

500ml of water = 500g water.  let's assume you want to go from 20-100 C =>
increase of 80C
Specific heat capacity of water = 4.184 J/g-K
=> 167,360 J = 0.0465 kWh Electricity @27p/kWh = 1.2552p

So rather similar (OK, I don't know efficiency of electric kettle ... but
heating elements are 100% efficient... sure there will be some losses to both
the kettle and the wider environment, but i doubt somehow that they are
huge) Of course this is about cost for the consumer ... I'm totally
ignoring the ineffeciency of turning gas into electricity and transmitting it
around the country in the first place.

On 29/08/2022 09:33, Peter Merchant wrote:
I was curious, so I took the water from the kettle and put it in a pot on the
stove. It was very close to 500ml.

Watching the gas meter while it boiled it used 0.014 m(cubed) of gas, which
converted to 0.15788Kwh and at my current rate of 7.123p/Kwh cost me 1.1.25p to
boil up.

Unfortunately I can't see my electricity meter easily to see what it costs me
to boil that amount in the kettle or by Microwave. Does anybody have that
facility?  And also there is always other electricity being used.

Cheers,

Peter

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### Re: [Dorset] OT: Keeping costs down

I had considered that, but it's so long ago in my ancient past that I studied
thermodynamics, I didn't look into calculating it. My concern about it is the
cost of raising the temperature to 100 deg, then the latent heat of
vaporisation to actually boil it.

Peter M.   [ Applies to any form of boiling water of course. ]

n 29/08/2022 11:31, Ian Morris via dorset wrote:

500ml of water = 500g water.  let's assume you want to go from 20-100 C =>
increase of 80C
Specific heat capacity of water = 4.184 J/g-K
=> 167,360 J = 0.0465 kWh Electricity @27p/kWh = 1.2552p

So rather similar (OK, I don't know efficiency of electric kettle ... but
heating elements are 100% efficient... sure there will be some losses to both
the kettle and the wider environment, but i doubt somehow that they are
huge) Of course this is about cost for the consumer ... I'm totally
ignoring the ineffeciency of turning gas into electricity and transmitting it
around the country in the first place.

On 29/08/2022 09:33, Peter Merchant wrote:
I was curious, so I took the water from the kettle and put it in a pot on the
stove. It was very close to 500ml.

Watching the gas meter while it boiled it used 0.014 m(cubed) of gas, which
converted to 0.15788Kwh and at my current rate of 7.123p/Kwh cost me 1.1.25p to
boil up.

Unfortunately I can't see my electricity meter easily to see what it costs me
to boil that amount in the kettle or by Microwave. Does anybody have that
facility?  And also there is always other electricity being used.

Cheers,

Peter

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### Re: [Dorset] OT: Keeping costs down

500ml of water = 500g water.  let's assume you want to go from 20-100 C
=> increase of 80C

Specific heat capacity of water = 4.184 J/g-K
=> 167,360 J = 0.0465 kWh Electricity @27p/kWh = 1.2552p

So rather similar (OK, I don't know efficiency of electric kettle
... but heating elements are 100% efficient... sure there will be some
losses to both the kettle and the wider environment, but i doubt somehow
that they are huge) Of course this is about cost for the consumer
... I'm totally ignoring the ineffeciency of turning gas into
electricity and transmitting it around the country in the first place.

On 29/08/2022 09:33, Peter Merchant wrote:
I was curious, so I took the water from the kettle and put it in a pot
on the stove. It was very close to 500ml.

Watching the gas meter while it boiled it used 0.014 m(cubed) of gas,
which converted to 0.15788Kwh and at my current rate of 7.123p/Kwh cost
me 1.1.25p to boil up.

Unfortunately I can't see my electricity meter easily to see what it
costs me to boil that amount in the kettle or by Microwave. Does anybody
have that facility?  And also there is always other electricity being used.

Cheers,

Peter

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### Re: [Dorset] OT: Keeping costs down

Terry, I only received one message, as you can see timestamped 9:51

I did wonder if a smart meter could give you a reading on that, but acknowledge
that electricity usage is continuously fluctuating, so unlikely.

I might have to borrow my neighbour's in-line device and test the kettle/
microwave.

Thanks John for that link.

Peter

On 29/08/2022 09:51, Terry Coles wrote:

On 29/08/2022 09:33, Peter Merchant wrote:

I was curious, so I took the water from the kettle and put it in a pot on the
stove. It was very close to 500ml.

Watching the gas meter while it boiled it used 0.014 m(cubed) of gas, which
converted to 0.15788Kwh and at my current rate of 7.123p/Kwh cost me 1.1.25p to
boil up.

Unfortunately I can't see my electricity meter easily to see what it costs me
to boil that amount in the kettle or by Microwave. Does anybody have that
facility?  And also there is always other electricity being used.

Peter,

I have a SMART Meter with associated In-Home Display (IHD) which is telling me
that my current electricity consumption is 85W, however, it is continuously
varying between ~25 W and ~220 W. Presumably, I could see how much the
consumption increases when I boil the kettle but I would then have to calculate
the total consumption against minutes and convert it to KWh and even then the
result would be a bit iffy.

You can get gadgets that measure the electricity consumption at the 'incomer'
(at the meters).  ( I have one which I use to measure the amount of juice being
generated by my solar Panels) and you can also get ones that do the same for a
single socket. If anyone has such a beast, then I think that the most accurate
result could be obtained that way.

Of course if we know the efficiency of the kettle we could calculate it!

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### Re: [Dorset] OT: Keeping costs down

On 29/08/2022 09:33, Peter Merchant wrote:
I was curious, so I took the water from the kettle and put it in a pot
on the stove. It was very close to 500ml.

Watching the gas meter while it boiled it used 0.014 m(cubed) of gas,
which converted to 0.15788Kwh and at my current rate of 7.123p/Kwh
cost me 1.1.25p to boil up.

Unfortunately I can't see my electricity meter easily to see what it
costs me to boil that amount in the kettle or by Microwave. Does
anybody have that facility?  And also there is always other
electricity being used.

Peter,

I have a SMART Meter with associated In-Home Display (IHD) which is
telling me that my current electricity consumption is 85W, however, it
is continuously varying between ~25 W and ~220 W. Presumably, I could
see how much the consumption increases when I boil the kettle but I
would then have to calculate the total consumption against minutes and
convert it to KWh and even then the result would be a bit iffy.

You can get gadgets that measure the electricity consumption at the
'incomer' (at the meters).  ( I have one which I use to measure the
amount of juice being generated by my solar Panels) and you can also get
ones that do the same for a single socket. If anyone has such a beast,
then I think that the most accurate result could be obtained that way.

Of course if we know the efficiency of the kettle we could calculate it!

--
Terry Coles

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### Re: [Dorset] OT: Keeping costs down

I heard this discussed on the radio recently. Using the kettle to boil
water is faster, but using gas is much cheaper.

Electricity, even before the current hikes, usually worked out more
expensive for a given amount of energy than other sources like gas, oil,
solid fuel etc. LPG like Calor is the most expensive after electricity.

Obviously prices move around for all of these but this has held true every
time I've looked into it.

Example analysis I found
https://nottenergy.com/resources/energy-cost-comparison/

On Mon, 29 Aug 2022, 09:33 Peter Merchant,
wrote:

> I was curious, so I took the water from the kettle and put it in a pot on
> the stove. It was very close to 500ml.
>
> Watching the gas meter while it boiled it used 0.014 m(cubed) of gas,
> which converted to 0.15788Kwh and at my current rate of 7.123p/Kwh cost me
> 1.1.25p to boil up.
>
> Unfortunately I can't see my electricity meter easily to see what it costs
> me to boil that amount in the kettle or by Microwave. Does anybody have
> that facility?  And also there is always other electricity being used.
>
> Cheers,
>
> Peter
>
>
> --
>   Next meeting: Online, Jitsi, Tuesday, 2022-09-06 20:00
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>
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