hot tub heat loss

[chaimech]

hello, i was wondering if anyone would like to offer their input on hot tub heater control. assuming the hot tub is used every day and heating with a resistance heater would it be more cost effective to leave the heater on all the time so the users can hop in whenever and have the water ready to go? or to have the heater on only when it is used - just preheat it then hop in?
and if you can direct me to proper formulas i would greatly appreciate it.

thanks

 

[IRstuff]

"proper formulas" ???

Besides conduction, convection, evaporation, radiation, and Joule heating?

Note that the answer depends heavily on factors you have not presented, i.e., insulation, temperature, duty cycle, etc.



TTFN

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[chaimech]

thanks, sorry for leaving out a lot. i am assuming 102 tub temp, 65 ambient; tub is 30" height x 120" length x 120" width; walls and cover with R-20 insulation.

Thanks

 

[IRstuff]

You can start with a downloadable heat transfer text:

http://web.mit.edu/lienhard/

http://www/ahtt.html

Just bear in mind that Lienhard is a bit abstruse. Wikipedia articles on heat transfer can get you a long ways.

TTFN

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[coldmaster]

Also you may have to consider how far from the hot tub to locate the cooler, so as not to unduly heat the beer...

 

[Compositepro]

Heat losses are proportional to time and water temperature. The longer the water is at an elevated temperature and the hotter the water, the greater your heat loss. Turning-off the heater will always save energy. The only other consideration is convenience of not having to plan or wait for heat-up. If it takes 24 hours for heat-up you may not have any option but to keep the heater on.

 

[IRstuff]

So, why is letting it cool better?

Say the heat loss is L for 102°F and 65°F ambient. Then every day of heating incurs the same loss L. However, if you allow the tub to cool, each additional day results in a lower temperature, and hence, a lower L. So, for say, 20 days, with heat you incur 20L of heat input. but for 20 days of cooling, the heat loss is <<20L, which is what you have to make up to heat back to temperature.

So, that's why shutting off the heat is more economical.

TTFN

FAQ731-376
Chinese prisoner wins Nobel Peace Prize

 

[chaimech]

thank you all for the input. i am trying to figure out how long it would take to heat the tub with a 4 kW heater. to make it simple, if i just convert the kW to BTU, subtract the heat loss through the walls, the remaining BTU would be going directly to the water, correct?

thanks

 

[IRstuff]

volume * density * specific heat * delta temperature

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FAQ731-376
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[IRstuff]

4 kW??

Wow... That's smaller than the main burner on my stovetop.

TTFN

FAQ731-376
Chinese prisoner wins Nobel Peace Prize

 

[willard3]

You are trying to oversimplify this. Try reading what people post before you restate the problem.

There is convection, conduction, radiation, evaporation and air velocity to consider.

 

[KiwiMace]

I feel this stuff is best left to field experimentation. The days will never be exactly 65 all day so what is the point?

You can mess around with some maths that you clearly have no prior experience with, making random assumptions for parameters that you have no feel for and getting results you can't verify...

...or you can casually observe the steady state heater cycle. If the heater on-time per day (leaving it on) exceeds the heat up and use time (when turned off), then you have your solution.

WIRT*, is the first time you go away for a week on holiday and leave it on will be the last time you leave it on overnight.

* "What I really think" - you heard it here first.