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an exponential moment 8 aug 1997 arithmetic can help solar house heating, eg ohm's law with units of heat, btu/hr = (t1-t2)/r: temperature replaces voltage, heatflow in btu/hr or watts replaces amps, and r is a thermal resistance, eg a us r-value divided by an area in ft^2. insulation can be a single resistor, and thermal mass a single capacitor. ee prof rick perry might see a 1' cube with 10 pounds of water surrounded by r6 insulation like this: r=1 (=r6/6ft^2) ta---www------tc | --- c=10 btu/f --- | --- - with a time constant rc = 10 hours, if the water inside the box has an initial temp of say, tc = 100 f, and we put the box in a room with a constant temp of, say, ta = 50 f, after t hours, tc = ta + (ta-tc)exp(-t/(rc)) = 50 + (100-50)exp(-t/10), where "exp" is the exponential function, the inverse of the natural logarithm, "e-to-the-x" on a $20 scientific calculator. after 10 hours, the initial temp diff (100-50) is squashed about 1/3 by our friendly decaying exponential, exp(-10/10), ie tc = 50 + (100-50)exp(-1) = 50 + 18.4 = 68.4 f. after a long long time, exp(-oo) is 0, and the water acquires the room temp... what's the water temperature after 24 hours? tc = ta + (ta-tc)exp(-t/(rc)) = 50 + (100-50)exp(-24/10) = 54.5 f. how would that change with an 8' cube full of water with r20 insulation? water weighs about 64 pounds per cubic foot. ignoring the insulation thickness, rc = r20/(6x64)x64x8^3 = 1707 hours or 71 days. after 1 day, tc = ta + (ta-tc)exp(-t/(rc)) = 50 + (100-50)exp(-1/71) = 99.3 f. heating is similar: putting the box with 40 f water in a 70 f room, with an initial temp difference of (70-40), we need a formula with initial and final temps and a decaying exponential to squash the initial temp difference, eg tc = 70+(40-70)exp(-t/(rc))... how warm will the water be after 5 hours? tc = ta + (tc-ta)exp(-t/(rc)) = 70 + (40-70)exp(-5/10) = 51.8 f. 10 hours? tc = ta + (tc-ta)exp(-t/(rc)) = 70 + (40-70)exp(-10/10) = 58.9 f. 50 hours? tc = ta + (tc-ta)exp(-t/(rc)) = 70 + (40-70)exp(-50/10) = 69.8 f. now then (warning: calculus ahead), if we put 2 pounds of water inside a 1 liter black-painted soda bottle in full sun, in 70 f air with no wind, with a thermometer sticking out of the top of the bottle, how will the temperature change with time? if the sun is a current source, providing about 300 btu/h per square foot of peak heatflow, ie 75 btu/h into the side of this bottle, and the bottle wall has a large thermal conductance compared to the still air film conductance of 1.5 btu/h-f-ft^2, and the whole bottle has about 1 ft^2 of surface, then r = 1/1.5 = 2/3 in this case, and ignoring reradiation, we have: i=75 tc ---- | r=2/3 |--| -> |----+----www------ta ---- | ir--> --- ic --- c=2 btu/f | | --- - the water gains 75 btu/h of heat, while losing (tc-ta)1ft^2/r2/3 = 1.5(tc-ta) to the air. every time the water gains 2 btu of heat from the heatflow ic down into the water, its temperature rises 1 f. it will eventually warm up enough so the incoming solar heatflow just balances the heat lost to the air... assuming ta = 0 for now, we have: tc = ir r = 1/c integral ic dt = (i-ic) r (1). so, multiplying by c, integral ic dt = (i-ic) rc, and differentiating both sides, ic dt = -rc dic. separating variables, dt dic --- = ---. integrating, -rc ic -t --- + a = ln ic (2), where a is an integration constant, rc and ln is the natural logarithm. since ic = i at t = 0, because the water and air have the same temperature, -0 --- + a = ln i, rc so a = ln i. plugging this back into (2), -t --- + ln i = ln ic, and e-to-the-x-ing both sides, rc ic = exp(ln i) exp(-t/(rc)) = i exp(-t/(rc)). plugging this back into (1), we have tc = ir(1-exp(-t/(rc))), or if the air temp is 70 f, tc = 70 + ir(1-exp(-t/(rc))). intuitively, there's an initial value, 70 f, a final value, 70 + ir, and a decaying exponential squashing the difference to start with. rearranging this, it looks like the heating equation, tc = 120 + (70-120)exp(-t/(rc))), with an final value of 120 f, and an initial negative temperature difference being gradually squashed by the our friendly decaying exponential. for the 1 liter bottle, tc = 70 + 75x2/3(1-exp(-t/(2/3x2))) = 70 + 50(1-exp(-t/1.33h) = 120 - 50exp-t/1.33h). with rc in minutes, tc = 120 - 50exp(-t/80m). so after 10 minutes, tc = 120 - 50exp(-10/80) = 75.9 f. after an hour, tc = 120 - 50exp(-60/80) = 96.4 f, and after 8 hours, tc = 120 - 50exp(-480/80) = 119.9 f. now suppose we cut the bottom out of a 2 liter bottle and slip it over another blackened 1 liter bottle with another thermometer sticking out of the top, and race the bare bottle against the one under the clear solar cozy... c stays the same, with the cozy, and i stays about the same, but r increases to about 5/3 vs 2/3, so tc = 70 + 75x5/3(1-exp(-t/(5/3x2))) = 70 + 125(1-exp(-t/3.33h) = 195 - 125exp(-t/200m). so after 10 minutes, tc = 195 - 125exp(-10/200) = 76.1 vs 75.9 f. after an hour, tc = 195 - 125exp(-60/200) = 102.4 vs 96.4 f, and after 8 hours, tc = 195 - 125exp(-480/200) = 183.7 vs 119.9 f. the bare bottle poops out while the one under the cozy is still pulling ahead. looks like the cozy will win the day's race, and it will be warmer at dawn... a cozy with foil on the back might make steam. nick |