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03-21-12, 01:50 AM | #11 |
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"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. |
The Following User Says Thank You to MN Renovator For This Useful Post: | briligg (03-21-12) |
03-21-12, 02:24 AM | #12 |
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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. |
03-21-12, 02:44 AM | #13 |
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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. |
The Following User Says Thank You to MN Renovator For This Useful Post: | briligg (03-21-12) |
03-21-12, 02:50 AM | #14 |
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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.
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03-21-12, 06:11 PM | #15 |
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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! |
03-21-12, 07:53 PM | #16 |
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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. |
03-22-12, 12:02 PM | #17 |
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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.) |
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conversion, heat loss, joules, thermal mass, watts |
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