Heat loss calculations - advanced sums - clever dick needed!

[ArtfulBodger]

while my van is being serviced I'm in the process of redesigning my standard quotation system to utilise the "energy efficiency best practise whole of house sizing method" to justify proposals made regarding the killowattage of the boiler.

Whilst this is easy and quick to do I would like to add a bit of value to the potential customer and, ahem, show them just how hard I've worked on their quotation.

For instance, it doesn't take much extra effort to quantify, in terms of money, how much they would save by increasing their roof insulation etc.

But, there is one thing that baffles me completely. Does anyone know a way of calculating the cost of, or amount of generated heat required to, effect a rise in temperature by x degrees for a given volume, taking into account the heat loss previously calculated.

I know that by using the whole of house method makes it inherently rather inaccurate, but I would like to know anyway.

Thanks in advance to any boffins out there!

 

[ArtfulBodger]

Any interest on this one gentlemen?

 

[oilman]

It's not so simple. First most heating systems seem to be fitted with 22mm pipes from what I've seen, so trying to increase the boiler power wont help. Next getting the temperature up assumes the outside temperature is not dropping, next if the sky is clear the and the sun isn't shining (night) the losses will be greater, unless the air is very still, when they wont. If the wind is blowing and its drizzling and it's just above freezing, the losses will be MUCH greater. Now how do we quantify all this, and what's the point anyway when people don't understand what a thermostatic radiator valve does?

 

[ArtfulBodger]

It's not so simple. First most heating systems seem to be fitted with 22mm pipes from what I've seen, so trying to increase the boiler power wont help. Next getting the temperature up assumes the outside temperature is not dropping, next if the sky is clear the and the sun isn't shining (night) the losses will be greater, unless the air is very still, when they wont. If the wind is blowing and its drizzling and it's just above freezing, the losses will be MUCH greater. Now how do we quantify all this, and what's the point anyway when people don't understand what a thermostatic radiator valve does?

Cheers oilman, I know it's a bit of a poser. But I was sort of hoping that there was a standard set of figures or assumptions that could be made based on the known U values for a property.

 

[PEDANTICVINDICTIVEMAN]

Plus with people buying boilers and rads in cheapshite packs from sheds and guessing everything whats the point in bothering with any of it or having a effaeces heating certificate.

 

[chrishutt]

I'm not sure if I've understood your requirements, but if I have the answer is quite straightforward.

Any heat loss calculation will be based on a design temperature for the outside, typically -1 C, and for the house / room, say 21 C, giving a temperature differential of 22 C in this case.

The resulting heat loss figure is proportional to the temperature differential, so that a whole house heat loss of say 10 kW at 20 C differential would be 11 kW at 22 C and 12 kW at 24 C (or 0.5 kW per degree C).

Converting this to financial cost is also simple. For example a load of 10 kW supplied by a boiler with an efficiency of 90% would require an input of about 11 kW which is then multiplied by the unit (1kWh) price of gas (say 3p) for each hour that the heating is on, making 33p per hour.

Remember that the above figure only applies when the outside temp. is -1 C. When its warmer outside, the heat loss is reduced, so in the example if it's 10 C outside, the heat loss will 5 kW, boiler input 5.5 kW, cost per hour 16.5p. As oilman pointed out, there are lots of other variables, so these figures are a theoretical guideline.

To continue the example, if the heating runs for 8 hours per day in the winter months, one could say to the customer that each 1 degree C increase in the internal temperature above the external temperature will cost them 0.6 (kW) x 8 (hours) x 3 pence per day = 14.4 per day.

I hope that makes sense.

 

[ArtfulBodger]

I'm not sure if I've understood your requirements, but if I have the answer is quite straightforward.

Any heat loss calculation will be based on a design temperature for the outside, typically -1 C, and for the house / room, say 21 C, giving a temperature differential of 22 C in this case.

The resulting heat loss figure is proportional to the temperature differential, so that a whole house heat loss of say 10 kW at 20 C differential would be 11 kW at 22 C and 12 kW at 24 C (or 0.5 kW per degree C).

Converting this to financial cost is also simple. For example a load of 10 kW supplied by a boiler with an efficiency of 90% would require an input of about 11 kW which is then multiplied by the unit (1kWh) price of gas (say 3p) for each hour that the heating is on, making 33p per hour.

Remember that the above figure only applies when the outside temp. is -1 C. When its warmer outside, the heat loss is reduced, so in the example if it's 10 C outside, the heat loss will 5 kW, boiler input 5.5 kW, cost per hour 16.5p. As oilman pointed out, there are lots of other variables, so these figures are a theoretical guideline.

To continue the example, if the heating runs for 8 hours per day in the winter months, one could say to the customer that each 1 degree C increase in the internal temperature above the external temperature will cost them 0.6 (kW) x 8 (hours) x 3 pence per day = 14.4 per day.

I hope that makes sense.

Chris, fantastic - thanks very much

 

[ArtfulBodger]

I'm not sure if I've understood your requirements, but if I have the answer is quite straightforward.

Any heat loss calculation will be based on a design temperature for the outside, typically -1 C, and for the house / room, say 21 C, giving a temperature differential of 22 C in this case.

The resulting heat loss figure is proportional to the temperature differential, so that a whole house heat loss of say 10 kW at 20 C differential would be 11 kW at 22 C and 12 kW at 24 C (or 0.5 kW per degree C).

Converting this to financial cost is also simple. For example a load of 10 kW supplied by a boiler with an efficiency of 90% would require an input of about 11 kW which is then multiplied by the unit (1kWh) price of gas (say 3p) for each hour that the heating is on, making 33p per hour.

Remember that the above figure only applies when the outside temp. is -1 C. When its warmer outside, the heat loss is reduced, so in the example if it's 10 C outside, the heat loss will 5 kW, boiler input 5.5 kW, cost per hour 16.5p. As oilman pointed out, there are lots of other variables, so these figures are a theoretical guideline.

To continue the example, if the heating runs for 8 hours per day in the winter months, one could say to the customer that each 1 degree C increase in the internal temperature above the external temperature will cost them 0.6 (kW) x 8 (hours) x 3 pence per day = 14.4 per day.

I hope that makes sense.

Chris

I'm not sure that I understand this fully (now I'm sober!).

Temperature differential - no problem
Where does the figure of 0.5kW per hour per degree come from?
How does this relate to total heat loss already calculated? obviously volume is now out of the picture since heat loss was calculated using this

I know that it is all largely theoretical, but it would be nice to quantify, via and estimate, anticipated running costs for the householder for the proposed heating solution.

Many thanks for you patience

 

[chrishutt]

Where does the figure of 0.5kW per hour per degree come from?

The figure I gave was 0.5 kW per degree (no hour in my example - I'll come back to that later). This figure applies only to the example given.

In that case, if the heat loss is calculated on the basis of a temperature differential of 20 C and the result is a heat loss of 10 kW, then you divide the 10 by 20 to give 0.5 kW per degree.

The actual figure for heat loss per degree C temp. differential will be different for each house / room, but is calculated in the same way. You just divide the total heat loss (in kW) by the temp. differential that it's based on (degrees C) to get the heat loss per degree C temp. differential.

In fact you could do the original heat loss calculation based on losses per degree C differential. For example if a single glazed window is 1 metre high by 2 metres long (area 2 square metres) and has a U value of 5 Watts per square metre per degree C, then for that window the heat loss per degree C is 2 x 5 = 10.

Coming back to the point about kilowatts and hours, it's important to understand that a kW is a unit of power, or energy (heat) per unit of time, whereas a kWh (kilowatt hour) is unit of the quantity of energy (heat). I still find this difficult to grasp, so don't worry if you do too.

For example, a radiator with a rated output of 1 kW will emit 1 kWh in an hour but 8 kWh in 8 hours. Likewise an immersion heater rated at 3 kW will put 12 kWhs of heat into a cylinder in 4 hrs. The kWh is the Unit of heat for which we are billed by electricity and gas companies.

 

[ArtfulBodger]

Cheers Chris