As for the if it's linear or logarithmic, I believe it's linear - water needs a certain amount of energy to raise 1 degree regardless of temp.True, but the difference between water temp and ambient air temp has a huge impact on the heat loss out the top during use. It's intuitive that if there is a 50 degree difference between water and air temp, the heat loss (cover open) will be quite a bit more and quite a bit faster than if there is only a few degrees of difference. And - That differential is effectively increased by causing the water to move around and the surface to be turbulant, via jets or blower systems.
Chas and Vinny,Ah. OK, with those parameters set, then my answer becomes much easier to arrive at: I don't know.
Based upon your responses, I didn't make the scenario very clear though Vinny addressed more of what I was looking. Taking the same spa with the same usage trends under the same conditions (understanding that a poorly insulated tub/cover would amplify any energy needs), what would the difference in energy consumption be if operated at an outside temp of 70 and a tub temp of 97 and again at 104? Starting at 97 as a baseline, would the difference at 104 be 7 times greater than the difference at 98 or would it be greater or less than 7 times?
Saturday morning I too was having an early morning soak in 103-degree water, feeling the fall chill in the air, when a bug of some kind landed in the spa. As I watched closely, he struggled between life and death in the hot water and I wondered if he was more frightened by the hot water or the fact that he was drowning.... Suddenly I had what most of you would call an "epiphany" and I knew clearly what I had to do........
STOP DRINKING BEFORE NOON! ;D 8)
Ah. OK, with those parameters set, then my answer becomes much easier to arrive at: I don't know.
But I will look into it. I may even ask one of my brainy kids - and get back to you. This is a very good question, and very apropos to this forum!
I didn't do well with word problems in school. Shouldn't one of these scenarios involve a train going east?? ;)
Windsurfdog,
If you want an exact answer then it's going to take some time to research. Those books are long gone from my possession - replaced by hot tub brochures.
ASSUMING ALL THINGS ARE EQUAL IN THE TUB
For heat loss > it's generally an equation of temp difference and insulation factor (I believe that someone posted the formula on this site a while back). All things being equal the closer the two temps are the slower the heat loss.
For heating water > it's generally an equation of (temp desired - actual temp ) x kw/ degree (F or C). Example: if the water temp is 50 F and you want 97 F and ASSUME each degree of temp rise takes 13kw then (97-50) = 47 x 13kw = 611kw, using the same info except the desired water temp is now 104 (104 - 50) = 54 x 13 = 702 kw.
Now once the water gets to the desired temp, the other equation takes place to keep the water warm. Actually the heat loss calculation comes in while heating the water and as the water's temp moves further away from the outside air temp the tub loses more heat (I believe Chas mentioned that before)
These are over simplifications and the actual calculations are probably more complex (I forget what they are).
Windsurfdog, if you really want the calculations, I will try to come up with them but and this is a BIG BUT - YOU will start turning into a Engineering Techno GEEK and won't be happy with the simpler pleasures in life. We are an elite group of people that will ask questions like "How do you really know that your pump is putting out the correct HP" and "how many lumens does that 7 LED light put out vs a halogen bulb vs the 22 LED bulb". You will buy a UV light meter so that you know how much output your ozonator bulb is producing. Friends might shun you but you'll have a wealth of useless knowledge! ??? ;D
Please take my advice ponder on more important things in life - Beer or Wine in the tub!
Anyway, thanks for the good word to the spatopian goddess!
I'm still working on a simple equation to show the rate of heat loss off still water with a different delta t. But in the meantime, a thought struck me. Don't worry, it happens several times a year and I wasn't hurt this time....
A practical way of looking at this would be to do a simple calc of the cost of running the heater. For the sake of this flimsy example, let's use a one-hour soak.
If you have a 5000 watt heater, it will cost you a little over 5KWH (kilowatthours) to run for one hour. If you are paying $.25 per KWH, then your heater will cost $1.25 to run for an hour. I think a Quarter is a high rate, most non-California residents pay less.
Just for fun, imagine a spa with very low temp difference between water and ambient air. The heater may run for as little as ten minutes out of the hour.
That would be 1.25 / 6 or $.21 to run the heater for that one-hour soak.
So, the higher the temp differential between water and air, the longer the heater will run. Keep in mind that in extremely low air temp situations, the heater may run for more than an hour for a one-hour soak - taking a few minutes AFTER the lid is closed to get back to the heat setting - but I think that would only happen in extremely cold climates, and nobody with any sense would live there. ;)
At $.25/KW hour USD, you'd have to be crazy to live where you do! I pay $.085 CDN :D 8)
Saturday morning I too was having an early morning soak in 103-degree water, feeling the fall chill in the air, when a bug of some kind landed in the spa. As I watched closely, he struggled between life and death in the hot water and I wondered if he was more frightened by the hot water or the fact that he was drowning.... Suddenly I had what most of you would call an "epiphany" and I knew clearly what I had to do........
STOP DRINKING BEFORE NOON! ;D 8)
I was halfway through this topic and my head started hurting so much I had to retract that vow! ;D
8)
Ok so at any point does the body temperature of the soakers enter as a variable? For example: if you have 6 soakers with body temps of 98.6 and we are trying to keep the water at 102, do the soakers actually have a downward influence on temp? Conversely, If you put 6 slightly feverish people with a temp of 100, and you trying to keep the water temp 95, would they raise the temp... and.... at what ambient air temp could you drop to where the 6 bodies would maintain the temp. of the water at 95?
Well, I check at home in the morning as I am checking my regular mail ( and trying to decide whether or not to order viagra or some "extensions" ) then I check at night when I get home ( assuming I can get computer time from my 14 y/o son.
Windsurfdog, you see how being a Techno Geek is; a pondering thought gets us to explain all the mysteries of life - See I told you a wealth of useless information!
I wouldn't say useless at all, Vinman.....
Thanks to all that contributed both scientifically and humorously......I certainly agree with rocket, nhbeacon, stuart, salesdvl and especially Chas.....this forum is both informational AND enjoyable. (Now if I could just understand how Wisoki's panties got so wadded.....)
8)
Heat Loss - calculating the rate of heat loss due to conductance
Q = k A (Tb- Ta) / d, where
Q is the heat lost due to conduction
k is the thermal conductivity of the tub's insulation layer
A is the surface area through which the heat is being lost
Tb is the temperature of the tub
Ta is the temperature of the surrounding air
d is the thickness of the tub's insulating layer
WAAAY to tired to try and plug numbers in this morning. Maybe after I get out in the fresh air on my first delivery.
Heat Loss - calculating the rate of heat loss due to conductance
Q = k A (Tb- Ta) / d, where
Q is the heat lost due to conduction
k is the thermal conductivity of the tub's insulation layer
A is the surface area through which the heat is being lost
Tb is the temperature of the tub
Ta is the temperature of the surrounding air
d is the thickness of the tub's insulating layer
WAAAY to tired to try and plug numbers in this morning. Maybe after I get out in the fresh air on my first delivery.
I regularly warm my tub water by having 8 friends with 103 degree fevers get into the tub and pee in the water.
Temp goes from 98 to 99 instantaneously. This makes it a HTP tub, I guess.
Then that would make it a linear relationship??
Not only is this a fun and enjoyable forum but I actually am learning a few things too. I hope that the otential spa customers enjoy this as much as we do.
Here's the answer to the bodies in the water.
Ya gotta wonder how many times a statement like this has been made by guys with names like Vinny...? 8) Fugetaboutit!
You have the right to remain silent....