u values, heating costs

[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?

[The Natural Philosopher]

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]

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.

[John Rumm]

Looks promising. You may need to add a bit for air changes unless the place is totally air tight.

See

details.

Well lets say you are maintaining the temperature at 5 degrees above ambient, then you already know the heat loss for that. The fact that the heater will cycle on and off does not really matter since its stat will keep the room averaging around the target temp. So the nett rate heat input needs to match the nett loss for equilibrium to be maintained.

In reality its a bit better than that since there will only be certain times when the outside temp is actually that cold. Even most winter days will reach 5 (in the south anyway). If there are any windows, you might get some solar gain as well.

[Andy Wade]

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,

or 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.

[Mike Buckley]

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.