[Mike Buckley]

I'm trying to work out how much it would cost to heat my garage when
it's in more of a room like state, so there'll be good roof insulation
(already plasterboarded, just needs insulating), the walls are probably
going to be insulated plasterboard and the door will have some form of
insulation on it.

Using various online calcs and some finger in the air stuff I've come up
with some rough and ready values - are they realistic?

Sizes:

Roof - 36m2 with good insulation, u value estimated at .4
Walls - 72m2 with ok insulation (and crap on the door wall), u value .7
Floor - 36m2 no insulation, u value .7

Assuming a starting point of 0 degrees c, I'd like to work out how much
it would cost for frost protection (keep a steady 5 degrees) and also to
bump it up a bit further if I decide to put some of my more fragile
electronic stuff in there - so about 10 degrees.

So I figure I need to work out the heat loss, this seems to be total
size of area multiplied by the u value.

Roof - 36 x .4 28.8
Wall - 72 x .7 50.4
Floor - 36 x .7 25.2

Total - 104.4

To work out the watts needed, multiply the temp uplift by the heat loss,
so:

104.4 x 5 522w Frost Protection
104.4 x 10 1044w Ideal


So that gives me the size of heater I need - I was counting on something
like a 2Kw convector which we have in our conservatory and it really
does a great job, so this should easily be enough to heat the garage to
these levels, even allowing for underestimating the heat loss through
the door.

Assuming all that looks ok, and even if I'm quite a way off I've still
got another Kw+ before I need another heater, how do I calculate the
cost of the heater? I can easily do it per hour, but that assumes it's
on full blast all the time which it won't be - so how do you calculate
the cost based on partial use - is it even possible?


--
Mike Buckley
RD350LC2
XJ900S

[The Natural Philosopher[_2_]]

Mike Buckley wrote:
I'm trying to work out how much it would cost to heat my garage when
it's in more of a room like state, so there'll be good roof insulation
(already plasterboarded, just needs insulating), the walls are probably
going to be insulated plasterboard and the door will have some form of
insulation on it.

Using various online calcs and some finger in the air stuff I've come up
with some rough and ready values - are they realistic?

Sizes:

Roof - 36m2 with good insulation, u value estimated at .4
Walls - 72m2 with ok insulation (and crap on the door wall), u value .7
Floor - 36m2 no insulation, u value .7

Assuming a starting point of 0 degrees c, I'd like to work out how much
it would cost for frost protection (keep a steady 5 degrees) and also to
bump it up a bit further if I decide to put some of my more fragile
electronic stuff in there - so about 10 degrees.

So I figure I need to work out the heat loss, this seems to be total
size of area multiplied by the u value.

Roof - 36 x .4 28.8
Wall - 72 x .7 50.4
Floor - 36 x .7 25.2

Total - 104.4

To work out the watts needed, multiply the temp uplift by the heat loss,
so:

104.4 x 5 522w Frost Protection
104.4 x 10 1044w Ideal


So that gives me the size of heater I need - I was counting on something
like a 2Kw convector which we have in our conservatory and it really
does a great job, so this should easily be enough to heat the garage to
these levels, even allowing for underestimating the heat loss through
the door.

Assuming all that looks ok, and even if I'm quite a way off I've still
got another Kw+ before I need another heater, how do I calculate the
cost of the heater? I can easily do it per hour, but that assumes it's
on full blast all the time which it won't be - so how do you calculate
the cost based on partial use - is it even possible?


if it has a stat on it, you simply take the average temperature on say a
daily basis and multiply the thing as you have done already.

I think you can do better than you think on wall insulation.

but the figures seem not far out.

However, worry about the -15C and north wind blowing through every
cranny. That's the worst case.

[Ronald Raygun]

Mike Buckley wrote:

To work out the watts needed, multiply the temp uplift by the heat loss,
so:

104.4 x 5 522w Frost Protection
104.4 x 10 1044w Ideal

So that gives me the size of heater I need - I was counting on something
like a 2Kw convector which we have in our conservatory and it really
does a great job, so this should easily be enough to heat the garage to
these levels, even allowing for underestimating the heat loss through
the door.

Assuming all that looks ok, and even if I'm quite a way off I've still
got another Kw+ before I need another heater, how do I calculate the
cost of the heater? I can easily do it per hour, but that assumes it's
on full blast all the time which it won't be - so how do you calculate
the cost based on partial use - is it even possible?

You've already done it. The heater is presumably controlled by a thermostat
which you cat set to give the desired 5 or 10 degree inside temperature.
This thermostat will in effect run the heater at quarter blast (522W) or
half blast (1044W) by switching it onto full blast for a quarter or half
of the time. The actual on and off times will depend on the hysteresis of
your feedback loop, but as an example for 5 degrees it might be on for 2
minutes and then off for 6. In this case the cycle length would be 8 mins,
but for cost purposes it doesn't matter what the cycle length is, since if
the heater is on for a quarter of every 8 minute period, it is in effect
on for a quarter of any period, such as a day. So if on full blast it
would use 48kWh per day, on quarter blast it would use 12kWh per day.

[Andy Wade]

On 16/10/2010 17:08, Mike Buckley wrote:

[...]
Assuming all that looks ok, and even if I'm quite a way off I've still
got another Kw+ before I need another heater, how do I calculate the
cost of the heater? I can easily do it per hour, but that assumes it's
on full blast all the time which it won't be - so how do you calculate
the cost based on partial use - is it even possible?

You can get an idea of the upper limit (continuous heating) by using the
concept of degree-days. Degree-day data is readily available for the
UK, see, for example,
http://www.vesma.com/ddd/ or
http://www.degreedays.net/.

The latter site allows you to set your own base temperature. Setting a
5 deg. base would give you the consumption for frost protection.

Add up your individual U*A products and ventilation heat requirement to
get an overall total heat demand in watts per kelvin of temperature
difference. Then multiply by the degree-days for any period - this
gives the consumption for the period in watt-days. Multiply by 24/1000
to convert to the more useful unit of kWh.

--
Andy

[Mike Buckley]

In message , Andy Wade
writes
On 16/10/2010 17:08, Mike Buckley wrote:

[...]
Assuming all that looks ok, and even if I'm quite a way off I've still
got another Kw+ before I need another heater, how do I calculate the
cost of the heater? I can easily do it per hour, but that assumes it's
on full blast all the time which it won't be - so how do you calculate
the cost based on partial use - is it even possible?

You can get an idea of the upper limit (continuous heating) by using
the concept of degree-days. Degree-day data is readily available for
the UK, see, for example,
http://www.vesma.com/ddd/ or
http://www.degreedays.net/.

The latter site allows you to set your own base temperature. Setting a
5 deg. base would give you the consumption for frost protection.

Add up your individual U*A products and ventilation heat requirement to
get an overall total heat demand in watts per kelvin of temperature
difference. Then multiply by the degree-days for any period - this
gives the consumption for the period in watt-days. Multiply by 24/1000
to convert to the more useful unit of kWh.


Thanks for the help everybody, very useful.

That degree days thing looks more accurate and it's very interesting but
I don't need that level of detail (and my u values aren't accurate
enough either). Stashed the info for bed time reading though.

--
Mike Buckley
RD350LC2
XJ900S