"Backwards" heat-loss calculation?

Thor

Is there a simple formula for working backwards from measured fuel consumption, burner efficiency, degree-days, and minimum exterior temperature to heat-loss?



My apartment complex recently had a state-mandated energy efficiency report written by some consultants whom I guess I'd characterize as "well-intentioned". They were clearly plugging numbers into some kind of complex modeling software they didn't understand, and charging us a fee for the results. It didn't cost much, but what to do with those results is a bit of a puzzle since we don't trust them.



If I knew the structure's heat loss, it would be easy to check a lot of their estimated savings against reality. But calculating it in the "usual" way looks like a major engineering exercise (15 separate buildings of 4-6 stories, partially unknown construction methods, many basic variables like infiltration basically unmeasurable).



However I know how much fuel we burned every month last year, I have the climate data, and the measured burner efficiency. I think I ought to be able to get from there to an approximation of the heat loss but I'm hoping someone else here already knows how, because it looks like quite an algebra exercise starting from the "forwards" computations I have examples of!



Anyone?

Jamie Hall

It does look formidable, doesn't it?

But it isn't really, provided two things: first, that you will be happy with approximate results (good to, at most, two significant digits) and second, that you take it step by step.



Start off with your known fuel burn and convert that into BTU input.  Multiply that by your known efficiency, and you have gross BTU output to the building complex.  Set that aside for the moment (sounds like a cake recipe).  You don't really need the minimum temperature -- what you need is the total degree days for the same time period as the fuel burn.  You have that.



What you are looking for is BTU per hour per degree (the structure heat loss).  OK, well a degree days is basically the product of temperature difference between some standard and time -- days.  So a bit of units conversion here; multiply degree days (total) by 24 -- hours per day -- and you get degree hours. 



Now divide your gross BTU output to the building by the total degree hours, and you will wind up with BTU per hour per degree. 



Should work.  But I've only just gotten up, and haven't had my second cup of coffee yet...

nicholas bonham-carter

[fuel use]/[degree day]

when we put in our new boiler, i was able to get a printout from the gas company, for the previous 24 mos, including the degree days for colder months.

this information enabled us to calculate that our new boiler, at the end of the first winter had burnt a third less gas. this will tell you at the end of the winter if any improvements made before the winter actually saved fuel.

i suppose you could compare a heat loss study with the actual fuel used, however the "unknown construction methods" make it difficult to fill in the blanks. this would be a perfect use for an  IR camera.

unfortunately these consultants may not know much about hydronics. does the present heating system seem to function well?--nbc 

Gordy

Variables to weed out

Is domestic hot water heated by the same fuel? May be even with the same heat source example boiler with indirect water heater. Gas stoves



Gas bills can be deceiving month to month. If the company estimates one month, and reads actual the next then what ever they were off the estimated month is made up on the actual reading month. So neither month is truly accurate information for what was burned in that month most of the time.



Knowing this you would have to use the years total gas usage hopefully billing starts with an actual read, and ends with an actual reading for the year. Then you need to deduct any fuel burning appliances consumption not used to actually heat the buildings



  One way is an hour meter on the gas valve. You know the efficiency of the burner. You need to know how long the burner fires in a 24 hour period. This burn time frame should co inside with the same time frame the degree days are calculated. Then you can calculate your BTU per Hour per DEGREE DAY.



  Even this method is not truly accurate because during a heat call the burner is not always firing. Its going from high limit to low limit then firing again but the circulator is still moving hot water to distribute btus between burns until the thermostat is satisfied.



You could rig the hour meter to the circulator, or most new thermostats have built in timers for how long ther is a call for heat in a 24 hour period and yyou can go bacck to the previous days reading, but there is the time the circulator starts before the burner fires to actually start making btus. This is usually the time it takes the boiler to go through all the proving of saftey components before the gas valve opens to fire. Depending on how many heat calls in a 24 hour period you get this could add up to skew things a bit.



  There are also btu meters on the market that hook into the piping. A little pricey to buy, and install but more accurate as to the btus the building is using.



It all depends on how anal you want to get.



Me I just use the thermostat timer, and my boiler output calculate my btu per degree day per square foot  good enough for me.  I believe 5 btus an hour per square foot per degree day is a really good number for an efficient super insulated home. Most homes fall in the 10 to 20  range. Higher than that time to tighten the envelope.



Gordy

Thor

Average, or maximum heat loss?

Jamie -- I think the formula you give will yield average heat loss per hour over the heating season (or over one fuel bill), rather than maximum heat loss on the coldest day.



Given that I have a data point for each fuel bill, and the daily temperature information, I could further estimate maximum loss from average loss if I knew the shape of the curve for structural heat loss as a function of temperature. Which I don't. Do you?

Jamie Hall

In a sense

it is an average.  It will give you the BTU per hour per degree, on average.  Multiplying that by the temperature difference on your maximum day will give you the BTU per hour for that maximum day -- under otherwise average conditions.  However.  There are, as Gordy pointed out, other variables than just temperature difference.  While all else being equal, the heat flow rate is a linear function of temperature difference between the two surfaces, wind can throw it way off, as it transfers heat much faster from the outer surface (this isn't the same as wind chill, though).  Also infiltration can make a huge difference.



To illustrate, for the museum I supervise the number comes out to about 2400 BTU per hour per degree.  On a still winter day, with a temperature difference of 70 degrees inside to outside, that gives me a heat loss rate of around 170,000 BTU per hour, which the fuel burn on such a day also gives.  However, add a 30 mph wind into that, and the same building burns through almost 300,000 BTU per hour -- which is just about what my boiler can manage!

Mike Kusiak_2

Heat loss vs. delta T

Conductive heat loss is a linear function of temperature differential. It is directly proportional. If you double the temperature difference between inside and outside, you double the heat loss.



But as Jamie has said, there can be other factors that can throw it off from the theoretical.

Thor

I should read more carefully!

BTU *per degree* per hour. D'oh. You'd already answered my question. Thanks for answering it again!



It looks like my Audel book has a table of corrections for wind speeds, ignoring infiltration. This building's so old, and the location so windy, that infiltration is a big issue. But I should be able to at least sanity-check the figures in the original report, which is what I wanted; I suspect if I want better results for the peak values we will have to arrange to measure them.