Extended heat loss maths sums - help needed please

[EliteHeat]

I need a boffin to help me with the following dead hard sums that I would like to incorporate into my quotation system. I must admit that it is beyond the powers of my brain – which is starting to hurt a bit.

I want to be able to give an accurate indication of how much it costs in terms of kilowatt-hours to maintain a temperature of x degrees, given an outside temperature of y degrees.

Because I am the conscientious type, I have already worked out the heat loss using the whole of house method – so that is the only bit I am sure of.

So, the idea is that I would create a table of some sort plotting outside temperature against desired inside temperature and factor in a price per kilowatt-hour.

I know this is all a bit university challenge but hey, it’s Christmas and I’ve got nothing better to do.

Thanks in advance for any brain-boxes who would care to have a stab at this.

 

[ChrisR]

Weeeel, it depends where you're starting.

If you have a heatloss of HL killowatts, say to give you a boiler size, that'll be for something like a 22 degree difference between inside and out. SO HL/22 gives you a figure for the heat loss per degree.

Look at your gas bill for a price of gas per KwHour, call it P

Pick a number for the number of hours the heating would be on in your day

Then guess how efficient the boiler/heating is - depending on boiler, etc. Say 80%.

So the price per degree per day is

(HL/22) x P x hours /0.80

Is that the sort of thing you wanted?

 

[Hazelb]

from memory (!) the heatloss isn't a linnear relationship.

I'm not famililar with 'the whole house' method' for calulating heat loss, but I would guess that somewhere in it you have a variable ( value) for outside temp in deg C and inside temp in deg C.
Run the calcs a few times for each external temp ( with different internal house temps ) You should then be able to plot a graph of how heat loss varies for different internal temps ( If you can get an excel spreadsheet to do it for you it will even plot a graph ).

That might help abit. ( I won't mention that wind speed effects heat loss)!

 

[EliteHeat]

To take ChrisR's example of say a heat loss of 12kW and assume that I pay 3p per kilowatt hour, outside temp is -1 and I want to achieve 21 degrees, boiler runs at 80% efficiency and heating is on for 10 hours per day:-

12/22 = 0.545
0.545 * 0.03 = 0.0163
0.0163 * 10 = 0.163
0.163 * 0.8 = 0.130

Or 13 pence a day - which would be nice if true.

As I mentioned, I am no brain-box, but it would seem to me that the formula would have to include the following:-

Total output of all radiators, plus 10% for pipe loss, adjusted for efficiency of boiler - assuming that the boiler modulates and only generates the exact amount of heat required (?). This would then have to be adjusted to account for the heat loss every hour though the fabric of the building. If the temperature differential could be ignored then this would be quite simple but obviously it is crucial to the calculation.

I do not understand the relationship between temperature differential and heat loss / heat gain.

Thanks so far, any ideas anyone?

 

[EliteHeat]

from memory (!) the heatloss isn't a linnear relationship.

I'm not famililar with 'the whole house' method' for calulating heat loss, but I would guess that somewhere in it you have a variable ( value) for outside temp in deg C and inside temp in deg C.
Run the calcs a few times for each external temp ( with different internal house temps ) You should then be able to plot a graph of how heat loss varies for different internal temps ( If you can get an excel spreadsheet to do it for you it will even plot a graph ).

That might help abit. ( I won't mention that wind speed effects heat loss)!

Whole of house method is very simple and probably extremely inaccurate - but hey ho it's John Prescots idea. All you get at the end of it is a figure in kilowatts that is lost to the atmosphere every hour. There is no account taken of temperature differential but interestingly it does take into account the area of the country - so maybe this has something to do with it. If this is the case then I need something more accurate.

 

[Bigburn]

Why dont you just bring a laptop along to your customer and show them the sedbuk ratings which show annual heatings costs or alternatively just arrange a thermal imaging survey.
Very professional dont you think.


http://www.infraredvision.co.uk/pages/content.php?subID=48&catID=8

 

[Bigburn]

OOh and dont forget this -

http://www.sap-online.co.uk/

 

[TicklyT]

I do not understand the relationship between temperature differential and heat loss / heat gain.

Thanks so far, any ideas anyone?

I believe the heat loss is proportional to the square of the temperature differential, but at the moment I haven't got any figures to confirm this.

 

[ChrisR]

Heat loss through the fabric - eg brick, is roughly proportional to the temperature difference each side. Near enough . Heat loss from the outer surface of the brick does depend on the airflow, whether it's natural convection or wind. (Or rain!). That could vary a lot. The total certainly isn't a square law, because the limiting factor is the material of the building.
The U values we use do take some account of the external environment, but it's obviously a bit of a compromise.

Elite - no you don't understand, mate!
To take ChrisR's example of say a heat loss of 12kW and assume that I pay 3p per kilowatt hour, outside temp is -1 and I want to achieve 21 degrees, boiler runs at 80% efficiency and heating is on for 10 hours per day:-

12/22 = 0.545 WRONG - You're dividing by the temp difference - that gives you a cost PER DEGREE
0.545 * 0.03 = 0.0163
0.0163 * 10 = 0.163
0.163 * 0.8 = 0.130 You should have DIVIDED by 0.8
Then the figure you'd have would be
((12 / 22) * 0.03 * 10) / 0.8 = 0.204545455

Or 13 pence a day - which would be nice if true. It isn't

20p is per degree increase over outside per day, so at the moment about 3 quid.

As I mentioned, I am no brain-box, but it would seem to me that the formula would have to include the following:- .....

It doesn't - you've already calculated that the heatloss for the building is 12kW with a 22º differential. HOW it's lost isn't important for the cost.

 

[Agile]

I wonder why hazelb "thinks" its a non linear relationship?

Surely heat loss is created by thermal transfer which is proportion to the temp difference and the area involved and inversely proportional to the thickness?

Wind effects will probably be non linear at first as with no wind the outside air will act as an additional insulator.

All these effects are pretty irrelevant as occupancy aspects are more important. I know school teachers who wear corderoy trousers and thick pullovers and live at 15-16 °C and I know West Africans who want lagos temperatures of 30 °C.

Tony

 

[the plumbing miracle]

sad very sad have u lot got nothing better to do

 

[EliteHeat]

Heat loss through the fabric - eg brick, is roughly proportional to the temperature difference each side. Near enough . Heat loss from the outer surface of the brick does depend on the airflow, whether it's natural convection or wind. (Or rain!). That could vary a lot. The total certainly isn't a square law, because the limiting factor is the material of the building.
The U values we use do take some account of the external environment, but it's obviously a bit of a compromise.

Elite - no you don't understand, mate!
To take ChrisR's example of say a heat loss of 12kW and assume that I pay 3p per kilowatt hour, outside temp is -1 and I want to achieve 21 degrees, boiler runs at 80% efficiency and heating is on for 10 hours per day:-

12/22 = 0.545 WRONG - You're dividing by the temp difference - that gives you a cost PER DEGREE
0.545 * 0.03 = 0.0163
0.0163 * 10 = 0.163
0.163 * 0.8 = 0.130 You should have DIVIDED by 0.8
Then the figure you'd have would be
((12 / 22) * 0.03 * 10) / 0.8 = 0.204545455

Or 13 pence a day - which would be nice if true. It isn't

20p is per degree increase over outside per day, so at the moment about 3 quid.

As I mentioned, I am no brain-box, but it would seem to me that the formula would have to include the following:- .....

It doesn't - you've already calculated that the heatloss for the building is 12kW with a 22º differential. HOW it's lost isn't important for the cost.

Thanks Chris - that makes more sense

 

[Agile]

A long while ago when CH was starting to be the in thing to have installed in your home some of the installers were advertising "guaranteed temperatures".

After a few years they stopped. I would presume that was because they got into difficulties when the temperatures could not be reached with the rad sizes they had fitted.

It may not be a very good idea to suggest that any particular temperature might be reached. Particularly when the condensing boiler flow is better set at 70 ° C rather than the 80 ° which the system might have been designed for.

Tony

 

[Bigburn]

Fitting one of these on the clients premises before the new boiler goes in will tell you a lot more.

http://www.picotech.com/data.html

 

[ChrisR]

How?