Heat loss in watts, stored heat in joules - how to compare?
 

Hi all, i'm new :)
We are designing a passive solar house, and we have worked out how much heat we can store in its thermal mass, and how much solar energy passes through the glazing, both in joules. The heat leaking to the outside is all expressed in watts. How can i calculate how many joules it takes to maintain the same temperature inside the house?

I have a book that explains the whole business in Btu's (Kachadorian - The Passive Solar House), and i'm going to do the calculation that way, and then convert it, but it would be a lot easier if i could stick to metric - we live in metric-land, and most of the materials we are checking out are listed with metric specs. Can anyone help?

1 MBtu = 1,000 BTU
1 Therm = 100,000 BTU
1 BTU = 0.293071 Wh
1 watt = 3.41214 BTU/h (or 1 wh is about 3.41214BTU)
1 BTU = 1 055.05585 joules (most people round this off and I've seen it vary a little depending on what source you are looking at)

I think this is what you are looking for. Let us know if you want any clarification.

I've gotten some conversions like this, but my confusion is about joules and watts in this context.
Here's the thing - the thermal mass as currently designed would have to shed 429,000 kJ to lower 1 C in temperature. Over the course of a model extra-harsh winter in its intended location (Hamilton, Ontario, Canada), the current design of the house would lose 21,359 kW of heat to the exterior through the walls, roof, and foundation. But what does that mean in terms of kilojoules? A watt is one joule per second. Does that mean that in this case, the two things are essentially interchangeable?

(Air exchange heat loss i calculated without problems, i could do it all in joules.)

Watts measure power, and Joules measure energy. Watts are a rate, while Joules are a total amount.

Consider a pot of water on the stove. Joules would measure how much energy it would take to make the water boil. Watts measure how quickly you're putting energy into the water.

Power = (Energy) / (elapsed time). If your house is losing 21000kW through the walls, you're losing 21,000kJ per second. How long would it take to lose 429000kJ? Elapsed time = energy / power = 429000kJ/21000kW = 20s. I'd say one of your figures is not correct.

The house loses a total of 21,359 kW over the course of the entire winter. I took the total r values of all the surfaces, over their whole area, at the given temperature difference between exterior and interior, over the time in question - i'm using November 1 to March 31 - and got that figure. Kachadorian expresses it in Btu's, working in W/m2K as the insulative value, and m2 and K (otherwise known as degrees Celsius) the result comes out in watts.

Using the specific heat, in J/kgK (joules per kilogram per degree Celsius), i worked out that the thermal mass socks away 429,000 kJ for every degree Celsius it increases in temperature. Needless to say, it isn't a typical structure. Actually, it's a greenhouse, without going into details. If the design process yields a viable plan, then i can say more.

(If anyone is curious what the thermal mass is, it's mud. Courtesy of an article from Mother Earth in the 70's, which came up on one of those kick-*** Google Books results. I'll see if i can find the link.)

Here it is...
JOSEPH ORR' S FABULOUS " " MUD HEAT-STORAGE" " SOLAR GREENHOUSE

Nope. 21MW is the output of a small power plant. Also, power is a rate, so losing 21MW over the course of the entire winter is the same as losing 21MW each hour. I would wager that your math is incorrect, or that your model is not an adequate representation of the actual greenhouse.

Heat transfer is an interesting thing. The R-value of a 3mm thick glass pane is approximately zero, which would lead you to conclude that a very huge amount of energy is being transferred. That would be true, except there are other barriers to heat transfer involved. Heat must be transferred first within the mud, then from mud to air, then air must move, then heat must flow from air to glass, then glass to ambient air. Not that you can't set up a model of this yourself. Just that conductive heat transfer may not be the most significant mode. Good luck!

I think you are mixing up your units. kW or kilowatts is a figure of power, like my refrigerator, for example, uses 150 watts while its running. If it runs for 11.62 hours out of a 24 hour day like it did last summer when my air conditioner was out of service and my kitchen was over 80 degrees during the day, then it used 1743 watthours or 1.743 kwh per day.

21,359 kWh is likely what you are trying to represent, I see that number as 72,879,898 BTU which is a form of numbers that I'm used to. Oddly enough I've added less heat to my house over the course of the past two years in my house. To qualify in German passivhaus standards, I think you need less than 4,750 BTU per square foot every year. Is this house 15,343 square feet or 1425 square meters. Passive mansion? I'd like to tour one of those.

Robert -
Power is a rate, but watts can also be a quantity. A machine might use up 100 watts in one hour, or in one minute. Knowing how many watts are involved doesn't tell you how much time it took to use them up, just that 'x' watts is equivalent to 'x' joules for 'x' seconds. Your 21 megawatt power plant doesn't take all winter to generate 21 MW - it is pumping out 21 megawatts when it is running at full capacity. According to its strict literal definition, that would be 21 MJ/second, but that wouldn't make sense. What they mean is 21 MW per hour, which is enough for 20 or 30 thousand houses
The mud heat storage system mentioned in the article is not the same as what we are planning, but at any rate, thermal mass heating works by slow seepage, constantly, over a very large area, such that the bit of heat per square meter, over many square meters, adds up to be the equivalent of the highly concentrated heat that has to be distributed from the small area of a furnace running intermittently. It has to be designed carefully, especially in terms of heating incoming air (earth tubes do the trick) but it is very effective.

MN Renovator - No, the calculation i did doesn't yield kWh, it's kilowatts.
Kachadorian does the calculation this way: he multiplies the u-value, stated as Btu/hr x ft2 x deg. F, by the ft2 of each surface, to get the Btu/hr x deg. F. Then he multiplies that by the degree days charted out for American cities, which show the average number of degrees you will need to heat your home over the outside temperature each month, total. That is, maybe in September you will need to heat for 117 degree days, but that means 5 degrees one day, 3 degrees another, etc. The end result is a quantity in Btus.
Kilowatts are equivalent to Btu/hr - 3412 Btu/hr is 1 kW. They don't have an equivalent in Btu's.
U values in metric are stated as W/m2 x K. One of those is equivalent to 5.678 Btu/hr x ft2 x deg F. There is no time unit mentioned in the metric version.
But you need one to get a meaningful result. By running the numbers in Imperial and then converting to metric on a known quantity, i found that the typical u value listed in W/m2K is per hour, although that isn't stated. Following the same steps, you multiply the W/m2K by the m2 of all surfaces, and get 'x' W/K, which is really W/hr x K.
Multiply that by 24 to get W/day x K, and then by the average daily temperature in Celsius each month, by the number of days in the month, and you get a result in watts, because the other units cancel out.
That follows the same logic, but yields a number that can't be compared to Btu's. Or kWh. But it doesn't really matter, because what i need to compare it to is joules. My thermal mass energy storage has been tallied up in joules. That was straightforward. Specific heat x kg x degrees Celsius.
It makes sense to me, actually, that watts and joules would be equivalent in this situation. A watt is 1 joule for 1 second. Or, it could be a tenth of a joule for 10 seconds. Or, a joule that is spent in 10 seconds is a tenth of a watt for 10 seconds. The kilowatts that leak through the surfaces of the greenhouse are never more than 10 or 15 kW per day. Since it would take 429,000 kJ to lower the whole thermal mass 1 degree Celsius, it seems reasonable that 10 or 15 kJ would disperse from the thermal mass to the air of the greenhouse fast enough to compensate the heat loss through conduction to the exterior. The heat loss due to air exchanges, which are planned at a given rate, are harder to keep up with by diffusion from the thermal mass, but the earth tubes will handle that.
(Edit - actually, the structure loses 185 kWh on a typical Feb day if there is no sun, between heat conduction through the envelope and air exchange losses.)

"Power is a rate, but watts can also be a quantity. A machine might use up 100 watts in one hour, or in one minute. Knowing how many watts are involved doesn't tell you how much time it took to use them up, just that 'x' watts is equivalent to 'x' joules for 'x' seconds. Your 21 megawatt power plant doesn't take all winter to generate 21 MW - it is pumping out 21 megawatts when it is running at full capacity"

No, if its using 100 watts, it's using 100 watts at a given point in time. If it used 100 watthours, then you could ask how long it was doing that for. You have the 21 megawatts part right though.

"MN Renovator - No, the calculation i did doesn't yield kWh, it's kilowatts."
I'm quite certain it didn't. I'll explain why your calculation didn't somewhere below this.
"Kilowatts are equivalent to Btu/hr - 3412 Btu/hr is 1 kW. They don't have an equivalent in Btu's."
Yes, they do. 1kWh is 3412 BTU. If you run a 1kW heater for an hour you have 3412
BTU of output. If you run the 1kW heater for two hours, it's 2kWh or 6824 BTU.

"U values in metric are stated as W/m2 x K. One of those is equivalent to 5.678 Btu/hr x ft2 x deg F. There is no time unit mentioned in the metric version."
Yes, there is! W/m2 * K for two hours at the same temperature will be whatever the result of W/m2 mulitiplied by 2 and the answer will be in Wh. When you take wattage and have time as a factor, it is watthours.

"That follows the same logic, but yields a number that can't be compared to Btu's. Or kWh. But it doesn't really matter, because what i need to compare it to is joules. My thermal mass energy storage has been tallied up in joules."
Yes, you can. 1 kWh is 3412 BTU. 1 kW is 3412 BTU/hr. If you can turn kWh into joules, then there you go. With the multiplier to convert that into joules, there you go. You can do the math directly.

21,359 kWh of heat is what it seems you have calculated for your winter heating needs, I'm sure a great quantity of that will be reduced from that heating needs through your glazing. If you already factored glass gains in, you've got a house that is colossal, not passive, or your math is fouled somewhere.

From what I'm seeing with my information, the metric definition of a Passive house is an annual heating/cooling requirement of under 15kWh/m2/year.

If you still think that your yearly heating needs are 21,359 kW all winter and lets say you theoretically used an electric furnace(bad idea, but it's an example) you'll be running enough electricity to where you would instantly pop your main circuit breaker. You would need to have a very large uninsulated house made of 30 gauge sheet metal with a -40 degree temperature outside, I'd imagine such a building would have to be the size of an office building to lose that much heat constantly over the course of a winter.

Please learn about the difference of the two units: watts and watthours.

MN Renovator - i edited my previous reply to be clearer and more complete just after you replied, so some things maybe don't match up anymore.
But i think the bulk of the confusion comes from the fact that we are in no way discussing anything electrical here, this purely deals with heat.
1 kWh is equal to 3412 Btu. But 1 kW is not. If it was 1 kW for 1 hr, it would be. Sure, there could be something wrong with my calculation, but it isn't so straightforward as that i need to tack on an h to the result. I tried to explain the method as clearly as i could, and how i've done my best to reproduce in metric the calculation as done in Imperial. The trip up comes when the metric unit doesn't mention a time-scale. Think about my point that a 21 MW power plant is actually one rated for 21 MW per hour. Or, i guess it's 21 MWh per hour. Watts are pinned to seconds - joules per second. 21 MJ per second. If it was pumping out 21 MJ in a tenth of a second, it would be a 210 MW power plant. Sometimes time-scales aren't stated, just implied. The 5.678 Btu/ hr x ft2 x deg F = 1 W/ m2 x K conversion rate works, if i assume that the rate is for one hour. I can illustrate the point with an example.... Tomorrow. G'night.

You still don't understand. No, watts are NOT tied to seconds. Watts are not tied to time AT ALL!! If a wall is losing heat at a specific rate of 500 watts based on your calculation at a specific temperature and that temperature stays the same for an hour then you have 500 Wh of heat. The calculation is based on that specific point in time. My house loses heat at a rate between 285 and 300BTU/hr based on my calculations based on furnace runtime on different days when temperature was between -10 and -20f. I'll use 300BTU/hr for calculations sake which is 88 watts multiplied by the temperature difference(in F) between outside and inside. If its -10f outside and 70 inside that is 6.4 kilowatts. If I have no sun and it stays that temperature for 8 hours, that would be 51.2kWh.

If you still don't get it, please research the definition of watts and kilowatthours. The reason why the calculation doesn't have Wh in it is because the calculation doesn't factor the difference over a period of time. The calculation you are trying to make does factor in time though since you are putting it to a time frame, therefore the result will be in kWh. ...unless you took the total time of the winter, the heating months, or something similar and then divided it by that amount of time to come up with an average heat loss for that time period. It doesn't make for a very useful number for 21359kWh would be an average heat loss of 365 watts over the period of a year, that number isn't useful though. Figuring out that number over the period of a day, multiple days, weekly, or a monthly basis might be useful considering your thermal mass would reduce the size of any auxiliary heating device on very cold or cloudy days significantly.

Ahh, I just read back that you are building a greenhouse, which accounts for all of the heat loss. Glass is unfortunately very lossy and I don't know of any greenhouses that are heated in the winter in my climate. Any sort of growing is done indoors where the walls are insulated and there are grow lights used as you can't grow much with the short sun hours and cloudy days that come from being a northern area. Not to mention getting the snow off the roof grade glass. ..or plastic, whatever you are building with. I imagine your mud requirements will be tremendous.

MN Renovator
Thanks for hanging in there...
I see the process now. I maybe shouldn't have been responding so late at night. I got balled up with how the units cancel out and it seemed to me i was left with kW. So, with that, i can have everything come out in kWh, which makes life easy.

Southern Ontario has a fair number of commercial heated greenhouses that go year-round. What we want is just a personal garden space all year, where we can grow tropical trees and cacti and have grass between our toes in January. The roof will not need to be cleared of snow, large sections of it are not glazed, and most of what is, closes for the winter with reinforced insulation panels the snow can sit on. The open glazed part of the roof is angled sharply enough that the heat loss from within the greenhouse will cause any snow to slide off quickly. North, east, west walls are highly insulated, and so is the foundation. If the numbers i am now running aren't flawed (which is definitely not certain) the mud area doesn't need to be all that big.

(But watts are indeed tied to seconds - 1 joule for 1 second. It doesn't crop up with power calculations, but it is relevant to the power plant thing. A power plant is 21MW if it can process 21 MJ of its fuel stock in 1 second. If it can process that much in half a second, it is a 42 MW power plant.)
Cheers!

It sounds like you've thought the design through well enough to where it will be far superior to any greenhouse that is around here. The ones around here in commercial plant nurseries are all a combination of plastic and glass in a three season arrangement without any provision for being warm outside of reasonable weather conditions.

Southern Ontario makes it a little easier as it is likely south of my climate(at least Toronto is) which should make it a little more mild than where I am and more reasonable for what you are doing.

I think you might have the information you need for your unit conversion, do you have other questions? Sounds like an interesting plan and hopefully you'll share more details as planning and implementation progress. Good luck.

Hamilton is just slightly south of Toronto, on the west end of Lake Ontario. The Midwestern United States is a fair bit colder on average, but it also gets a lot more sun. Depending on your location in Minnesota, it is quite possible that the increased sunlight where you are would more than make up for the colder winter.

I do plan to say more about the plan once i'm confident the figures are correct. And i may well post those figures in a bit in a new thread because they strike me as too good to be true and i can't see the error. We are trying to come up with a design that is as cheap as possible, using newish materials like cellular concrete. My husband is an architect, but for the moment i've been placed in charge of research and preliminary calculations because he's busy. Besides, it was my idea. If it works well enough, building more will be part of a building business in Ontario. (Currently we are based in Mexico.)