We have a 375 gallon hot tub. On a refill , how long do you expect it to take for yours to heat up?
How long does it take your water to heat up?
- Thread starter Sjde
- Start date
This is a tough question to know without knowing the starting water temp (or maybe even the capacity of the heater), so I'm simply throwing this out for comparison.
Our 435-ish gallon tub went from 41 to 102 on Friday in about 10 hours (not sure when it got to temp, but we were in the tub 10 hours after it was filled). I got impatient initially because it was going up slow, so I after about an hour of heating, I dumped some water and emptied the 105 gallon hot water heater into the tub to speed the process.
Did an awesome treatment on Monday evening, and when I refilled at 1am, I used both regular water and whatever the hot water heater had left, starting me at about 60 degrees. That refill water was at 100 at 7am the next morning...
Our 435-ish gallon tub went from 41 to 102 on Friday in about 10 hours (not sure when it got to temp, but we were in the tub 10 hours after it was filled). I got impatient initially because it was going up slow, so I after about an hour of heating, I dumped some water and emptied the 105 gallon hot water heater into the tub to speed the process.
Did an awesome treatment on Monday evening, and when I refilled at 1am, I used both regular water and whatever the hot water heater had left, starting me at about 60 degrees. That refill water was at 100 at 7am the next morning...
Thanks. I think our tap water is close to 50 degrees year round. It took 18 hours this time to warm to 102 degrees!
My spa will go for 65 to 90 in about 2 hours. Mine is about 6’ in diameter..
I should mention I have a 400K BTU pool heater so when I move to just the spa it's fairly fast
I should mention I have a 400K BTU pool heater so when I move to just the spa it's fairly fast
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Takes me about 8 1/2 hours to go from fill temp to 100 degrees. 500 gal.
I shouldn’t have taken the salesman’s word for it . He said 8 hours. When I called the manufacturer she said 3-5 degrees / hour, so that’s 10-18 hours, and it seems we’re on the longer end.
I refilled yesterday. Here's some more quantifiable data for you - I refilled with some hot water, so my starting temp was 57 degrees. This is with the spa covered. Ambient air temp is listed in the table. Tub is 435 gallons, and has a circulation pump running a 5.5kW heater. Tub was set to 100 degrees, and I got 6 degrees / hour:
It may not, but since I was gathering data anyway, figured I'd include it.
calinb
0
Just skip to "The Bottom Line", below, if you're only interested in making warm-up time estimates (the topic at hand of this thread).
I just made several repairs on an old outdoor (gazebo covered) spa that came with the house we bought 1-1/2 years ago. I powered it up yesterday. After correcting several leaks, failed switches and controller wiring errors (I still need to try to disassemble and clean the buzzing heater contactors), I was pleased to see it function correctly after the first post-rewiring power-up!
It's running on 120 V / 20 A service and it has a "1500 W" heater. (I don't know why a 240 V / 20 A heating system wasn't installed, because there are two electrical service lines out there in the gazebo.) I measured the heater resistance at 10 ohms and I also measured the heater contactor supplied voltage of 120 V at the heater leads (a slight drop from the open line voltage of 126 V).
Ohms law says my heater thusly makes V^2 / R = 120^2 / 10 = 1440 W.
1 W = 860 calories / hour so my heater generates 860 * 1440 = 1,238,400 cal / hour of power.
1 calorie of energy will raise the temperature 1ml of water by 1 degree C so my heater can raise the temperature of 1,238,400 ml. of water by 1C in one hour, if there are no heat losses (but of course there are losses).
According to the seller's sales sheet, my hot tub holds 1,324,894 ml. (350 gal.) of water.
Therefore, the time it takes to increase the spa's water temperature by 1C is 1,324,894 / 1,238,400 = 1.07 hour. (Again, not accounting for any heat losses or heater element resistance increase with temperature, etc..)
1 C / 1.07 hour = 0.93 degrees C / hour.
Multiplying 0.93 C / hour by 9/5 converts to 1.68 degrees F / hour.
I started my hot tub 20 hours ago and the water temp was 53F. Now it's up to 88F.
So the temperature rise over the 20 hours was 88-53/20 = 1.75 degrees F / hour.
The ambient air temperature has been lower than the water temperature the entire time (the weather has been cold and rainy here in N. Central Idaho) so the ambient air certainly didn't help to heat the water. How did the hot tub heater beat the above 1.68 F /hour estimate, which didn't even include losses or heater element temperature rise sensitivity? My guess is the 350 gallon tub size specified by the seller is either wrong or it's the tub capacity when filled to its very brim (and there's a lot of volume in the top few inches)! Nonetheless, my tub is very well insulated with what the sales sheet calls the "Alaskan" insulation package, and the above measurements and calculations show that the heat loss is obviously small compared to the heat energy input.
Here's the Bottom Line that's pertinent to this thread (I think):
You can do some arithmetic similar to the above or probably reasonably figure about 0.4 F degree rise per hour for every watt/gallon of your spa, if the tub is thickly covered and well-insulated.
As an example of using the 0.4 multiplier, if I were to run the second power line to my "350 gallon" spa, buy a slightly higher resistance heating element and rewire the controller for a 240 V heater (new GFCI...more $$$), I could speed things up with the roughly 4000 watts of heating power that would be available (along with the 1/15 HP circ pump and other even smaller loads). Using the quick and dirty 0.4 water temperature rise rate multiplier:
0.4 * 4000W / 350 gal = 4.57 F degrees per hour instead of only 1.68 F degree per hour. Wow! With 4000 W of heat power, I could probably "afford" to turn the heater off during periods of non-use and not worry about pre-planning each spa session 24 hours in advance or so!
Now if the weather would just warm up a wee-bit more, I could open my pool. Next week, sez the wx-man!
I just made several repairs on an old outdoor (gazebo covered) spa that came with the house we bought 1-1/2 years ago. I powered it up yesterday. After correcting several leaks, failed switches and controller wiring errors (I still need to try to disassemble and clean the buzzing heater contactors), I was pleased to see it function correctly after the first post-rewiring power-up!
It's running on 120 V / 20 A service and it has a "1500 W" heater. (I don't know why a 240 V / 20 A heating system wasn't installed, because there are two electrical service lines out there in the gazebo.) I measured the heater resistance at 10 ohms and I also measured the heater contactor supplied voltage of 120 V at the heater leads (a slight drop from the open line voltage of 126 V).
Ohms law says my heater thusly makes V^2 / R = 120^2 / 10 = 1440 W.
1 W = 860 calories / hour so my heater generates 860 * 1440 = 1,238,400 cal / hour of power.
1 calorie of energy will raise the temperature 1ml of water by 1 degree C so my heater can raise the temperature of 1,238,400 ml. of water by 1C in one hour, if there are no heat losses (but of course there are losses).
According to the seller's sales sheet, my hot tub holds 1,324,894 ml. (350 gal.) of water.
Therefore, the time it takes to increase the spa's water temperature by 1C is 1,324,894 / 1,238,400 = 1.07 hour. (Again, not accounting for any heat losses or heater element resistance increase with temperature, etc..)
1 C / 1.07 hour = 0.93 degrees C / hour.
Multiplying 0.93 C / hour by 9/5 converts to 1.68 degrees F / hour.
I started my hot tub 20 hours ago and the water temp was 53F. Now it's up to 88F.
So the temperature rise over the 20 hours was 88-53/20 = 1.75 degrees F / hour.
The ambient air temperature has been lower than the water temperature the entire time (the weather has been cold and rainy here in N. Central Idaho) so the ambient air certainly didn't help to heat the water. How did the hot tub heater beat the above 1.68 F /hour estimate, which didn't even include losses or heater element temperature rise sensitivity? My guess is the 350 gallon tub size specified by the seller is either wrong or it's the tub capacity when filled to its very brim (and there's a lot of volume in the top few inches)! Nonetheless, my tub is very well insulated with what the sales sheet calls the "Alaskan" insulation package, and the above measurements and calculations show that the heat loss is obviously small compared to the heat energy input.
Here's the Bottom Line that's pertinent to this thread (I think):
You can do some arithmetic similar to the above or probably reasonably figure about 0.4 F degree rise per hour for every watt/gallon of your spa, if the tub is thickly covered and well-insulated.
As an example of using the 0.4 multiplier, if I were to run the second power line to my "350 gallon" spa, buy a slightly higher resistance heating element and rewire the controller for a 240 V heater (new GFCI...more $$$), I could speed things up with the roughly 4000 watts of heating power that would be available (along with the 1/15 HP circ pump and other even smaller loads). Using the quick and dirty 0.4 water temperature rise rate multiplier:
0.4 * 4000W / 350 gal = 4.57 F degrees per hour instead of only 1.68 F degree per hour. Wow! With 4000 W of heat power, I could probably "afford" to turn the heater off during periods of non-use and not worry about pre-planning each spa session 24 hours in advance or so!
Now if the weather would just warm up a wee-bit more, I could open my pool. Next week, sez the wx-man!
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calinb
0
It turns out that even the heat from my recirc pump 1/15 HP can keep the water from dropping lower than 99F when the night air isn't too cold (near 70) so that's how I "beat" the physics of the heater only output! My spa is well insulated, but most of insulation is in the walls and floor of the box that holds it rather than on the tube itself. The tub itself has a relatively thin coating of foam (no plumbing is "buried" in the foam) so heat transfers from the warm inside of the box to the water in the tube fairly well, I'm sure.
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