# How Long Will It Take A Tank Of Water To Warm to 70 Degrees

## Main Question or Discussion Point

Hopefully someone can help me with finding the answer to the 2 related questions/problems below:
QUESTION/PROBLEM
1. how long will it take a 200 gallon tank of water to warm up to 70 degrees Fahrenheit, in a room with an ambient temperature = 80 degrees Fahrenheit with 50% relative humidity?
2. how long will it take a 300 gallon tank of water to warm up to 70 degrees Fahrenheit, in a room with an ambient temperature = 80 degrees Fahrenheit with 50% relative humidity?

VARIABLES
• Water Starting Temp = 45 degrees Fahrenheit
• Tank Material = HDPE (High-density polyethylene)
• Tank Wall Thickness = ~1/4 inch (0.25)
• Tank Lid = 1-inch rigid insulation panel; Insulation R-Value = R3.85
• Tank Location = indoors - no exposure to sun; very minimal exposure to light
• Ambient Room Temp = 80 degrees Fahrenheit
• Room Relative Humidity (RH) = 50%

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berkeman
Mentor
Welcome to the PF. What is the application? (I'm assuming this is not a homework question?)

Is there any mixing in the tanks? Or is the water just real still inside each one? Do you have any experimental data already?

Welcome to the PF. What is the application? (I'm assuming this is not a homework question?)

Is there any mixing in the tanks? Or is the water just real still inside each one? Do you have any experimental data already?
Thanks for the reply berkeman - correct, this is not for homework. The water tank will feed an automated drip watering system. The tank will have an air pump submersed in the water (http://elemental-solutions.com/product/Details/EOCP284), to aerate it

We do not have any experimental data

OmCheeto
Gold Member
I think we'll need to know a bit more about the tank: Shape and dimensions.

berkeman
Mentor
The tank will have an air pump submersed in the water (http://elemental-solutions.com/product/Details/EOCP284), to aerate it
That may be the main way to warm the water. Is there any way you can heat the air that you are blowing through the water, or will it just be the ambient 80F air from the room?

That may be the main way to warm the water. Is there any way you can heat the air that you are blowing through the water, or will it just be the ambient 80F air from the room?
Actually we don't want to heat the water - we want to know how long it will take to reach 70 degrees under the room conditions (80 degree ambient temp and 50% RH), if the water has a starting temp = 45 degrees.

If all 200 or 300 gallons reach 70 degrees within a certain time threshold, we may have to install a chiller - we want the water to remain under 70 degrees until it is used each day

OmCheeto
Gold Member
Actually we don't want to heat the water...
In that case, I would ixnay the air pump, as it will only increase the heat transfer rate.

What is the purpose of the pump? Mixing fertilizer?

jim hardy
Gold Member
2019 Award
Dearly Missed
Analytically, it's a two part solution

80F and 50% RH is dewpoint of 59.7F
(per https://www.calculator.net/dew-poin...dewpointunit=fahrenheit&humidity=50&x=68&y=20 )

so you'll have condensation
and on account of 'heat of vaporization' of that water you're removing from the room air, faster warming until the tank wall reaches that temperature.

Afterward plain old convection takes over.

Being not a heat transfer guy, i know my limits...
So I think i'd wrap it with one of those water heater insulating blankets , install a thermometer and see what happens. old jim

#### Attachments

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jrmichler
Start by setting the tank on a 2" thick piece of foam insulation. Then a simplified heat transfer calculation is as follows:

R-value of top = 3.85
R-value of sides = 1 (rough estimate)
R-value of bottom = 10

Area of top and bottom = 24 ft^2
Area of sides = 44 ft^2

Heat loss at start:
Heat loss top = 24 ft^2 / 3.85 hr-ft^2 - deg F / BTU * (80-45) deg F = 218 BTU/hr
Heat loss sides = 44 / 1 * 35 = 1540 BTU/hr
Heat loss bottom = 24 / 10 * 35 = 84 BTU/hr

Total heat loss = 1842 BTU/hr
Initial rate of temperature rise = 1842 BTU/hr / 300 gallons / 8.34 lb/gallon / 1 BTU/lb-deg F = 0.7 deg F / hour.

That will decrease as the water temperature gets closer to ambient temperature. It will increase if the dew point is greater than the water temperature as @jim hardy said. The increase due to condensation will be less if the lid fits closely so that ambient air does not circulate freely. Make the lid 2" thick and put a couple of rocks on it.

Tank insulation is far cheaper than a chiller. Tank insulation should have a vapor barrier on the outside to prevent condensation on the tank wall. The best tank insulation would be a closed cell spray foam because that is its own vapor barrier. Or just let the water warm to the dew point temperature quickly.

• hmmm27, berkeman and jim hardy
Start by setting the tank on a 2" thick piece of foam insulation. Then a simplified heat transfer calculation is as follows:

R-value of top = 3.85
R-value of sides = 1 (rough estimate)
R-value of bottom = 10

Area of top and bottom = 24 ft^2
Area of sides = 44 ft^2

Heat loss at start:
Heat loss top = 24 ft^2 / 3.85 hr-ft^2 - deg F / BTU * (80-45) deg F = 218 BTU/hr
Heat loss sides = 44 / 1 * 35 = 1540 BTU/hr
Heat loss bottom = 24 / 10 * 35 = 84 BTU/hr

Total heat loss = 1842 BTU/hr
Initial rate of temperature rise = 1842 BTU/hr / 300 gallons / 8.34 lb/gallon / 1 BTU/lb-deg F = 0.7 deg F / hour.

That will decrease as the water temperature gets closer to ambient temperature. It will increase if the dew point is greater than the water temperature as @jim hardy said. The increase due to condensation will be less if the lid fits closely so that ambient air does not circulate freely. Make the lid 2" thick and put a couple of rocks on it.

Tank insulation is far cheaper than a chiller. Tank insulation should have a vapor barrier on the outside to prevent condensation on the tank wall. The best tank insulation would be a closed cell spray foam because that is its own vapor barrier. Or just let the water warm to the dew point temperature quickly.
Wow! Thanks everyone for taking the time to help. I'll respond in the order I received the replies:
1) OmCheeto: the purpose of the air pump is to aerate the water, but for purposes of the calculation on how long it would take for the tank water to warm to 70 degrees, you can forget about the pump
2) Old Jim: thanks for the "bare-knuckle" solution - I like it!
3) RMichler: so just ballparking it (not trying to calculate the faster increase in temp until the water reaches the dew point), using your result of the water temp rising 0.7 degrees F/hour, it would take approx. 35.71 hours for the water temp to rise from 45 degrees F to 70 degrees F (25 degrees/0.7) - if this is correct, does the result apply to both the 300 gallons and 200 gallons scenarios?

Chestermiller
Mentor
What about the convective heat transfer resistance at the outside surface of the tank.? This is likely to be high (i.e., low heat transfer coefficient).

jrmichler
A single pane glass window has an R-value about 1. That includes the thermal conductivity of the glass and boundary layers on both sides. A plastic tank with water will have higher thermal conductivity at the inside boundary layer, lower thermal conductivity through the plastic, and the same thermal conductivity on the outside. I think that an R-value of 1 for the tank walls is fairly close or slightly conservative.

This type of simplified calculation is not intended to be an exact calculation. It does show that the water will stay below 70 deg for a few hours, but not for a few days. It shows that the weak point is the tank sides. It shows that, if the OP wants to set the tank on a warm floor, the heat loss will increase significantly. And, as @jim hardy pointed out, dew point needs to be considered. Heat loss from condensation can be much larger than heat loss due to convection.

The heat gain from air bubbled through needs to be added to the calculation. That can be calculated by measuring the air temperature out of the air pump and assuming that the air bubble out of the water at water temperature and 100% RH. Get enthalpies from a psychrometric chart, flow rate by bubbling into an inverted jug, and calculate BTUs. The air flow needs to be measured because it is affected by the back pressure. I do not trust an air pump rated in GPH. Air moving devices are normally rated in CFM or CFH at a specified back pressure, and normally come with a curve of flow rate vs back pressure.

Easiest solution: Set it on a foam slab and put in a thermometer. If it gets warm too fast, add insulation to the tank walls.

• jim hardy
Thanks again everyone - I think I have the info I need - the simplified calc is enough - since the water will be drained daily, knowing that it will remain under 70 degrees for several hours is enough info...but if someone can confirm my math was correct using a water temp rise of 0.7 degrees F/hour, that'd be awesome.

I subtracted 45 (the starting water temp) from 70 (the threshold temp) = 25 degrees; and then divided 25 degrees by 0.7 degrees/hour to get ~35.71 hours for the water to rise from 45 degrees to 70 degrees...even is the margin of error is 25 - 30%, the result is still useful

berkeman
Mentor
I subtracted 45 (the starting water temp) from 70 (the threshold temp) = 25 degrees; and then divided 25 degrees by 0.7 degrees/hour to get ~35.71 hours for the water to rise from 45 degrees to 70 degrees...even is the margin of error is 25 - 30%, the result is still useful
I'm no expert in heat transfer, but since the water is approaching the temperature of the heat source (the ambient air), you can't really assume a constant delta-T as it heats up. It will be closer to an exponential curve, flattening out as the water temperature gets closer to the ambient air temperature (like the red curve below). #### Attachments

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russ_watters
Mentor
if this is correct, does the result apply to both the 300 gallons and 200 gallons scenarios?
Assuming the same shape a larger tank warms slower because the volume goes up faster than the surface area.

jrmichler
Think in terms of +/- 50% error for the simplified calculation. But it's good enough to give an idea whether you have a problem, and it shows you what to do if you need to make it better.

Don't forget to add in the heat from the air pump. To calculate that heat, assume that 100% of the electricity to the pump is heat added to the water, plus the air enters at room temperature and humidity and leaves at water temperature and 100% RH. Get the BTU per pound for air at a humidity from a psychrometric chart.

• berkeman
Think in terms of +/- 50% error for the simplified calculation. But it's good enough to give an idea whether you have a problem, and it shows you what to do if you need to make it better.

Don't forget to add in the heat from the air pump. To calculate that heat, assume that 100% of the electricity to the pump is heat added to the water, plus the air enters at room temperature and humidity and leaves at water temperature and 100% RH. Get the BTU per pound for air at a humidity from a psychrometric chart.
Thanks - I'll work on that and if I run into problems or issues, or have any questions - I'll post them - this has all been very helpful!

jim hardy
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