EcoRenovator - View Single Post - Heat loss in watts, stored heat in joules

briligg · 03-21-12, 01:16 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.)

03-21-12, 01:16 AM	#10
briligg Lurking Renovator Join Date: Feb 2012 Location: Mexico Posts: 14 Thanks: 2 Thanked 0 Times in 0 Posts	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.) Last edited by briligg; 03-21-12 at 06:24 PM..