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briligg 03-20-12 04:54 PM

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?

MN Renovator 03-20-12 05:20 PM

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.

briligg 03-20-12 05:37 PM

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.)

RobertSmalls 03-20-12 06:57 PM

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.

briligg 03-20-12 07:10 PM

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.

briligg 03-20-12 07:15 PM

(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.)

briligg 03-20-12 07:34 PM

Here it is...

RobertSmalls 03-20-12 09:49 PM


Originally Posted by briligg (Post 20804)
The house loses a total of 21,359 kW over the course of the entire winter.

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!

MN Renovator 03-21-12 12:24 AM

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.

briligg 03-21-12 02:16 AM

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.)

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