# I am dumb and need smart people to help me

• Requiem
In summary, the conversation discusses the need for an estimation of the energy required to maintain a large plastic tank filled with water at a constant temperature. The tank is heated in the morning and maintained at a specific temperature for a 16-hour day, while the outside air temperature varies. The water in the tank is heated by a circulating hose, and the energy consumption is affected by various factors such as the heat escaping from the open top, the material of the tank, and the load on the circulator. Suggestions are made to measure the power consumption and calculate the energy required using specific heat equations. Insulation is also mentioned as a way to reduce energy loss.

#### Requiem

Hello. Let me preface this by acknowledging that I am not a particularly bright person, and am therefore pleading for help from those who's IQs are not comparable with their belt size.

I need an estimation of the energy needed to maintain a tank of water at a constant temperature. The tank in question is a large plastic cylinder, 54" in diameter. It is filled to a height of 48" with water from the municipal supply, which for the sake of estimation is 10degC (50degF). The top of the tank is open to the air, the side wall is about .25" thick. I'm pretty sure the material is HDPE.

The tank is heated in the morning to 71.1degC (160degF) and maintained at that temperature for a 16hour day. Outside air temperature varies according to the season, but for the sake of estimation call it 20degC (68degF).
The water in the tank itself is not flowing. There is a closed loop hose that comes from a heating pump, passes into the tank and then back to the circulator. The heat from the water passing through the hose heats the tank water. The circulator runs at about 30liters per minute.

If the circulator's only job was to heat the tank I would just check it's power consumption, unfortunately it is used in several systems throughout the plant. Keeping the bath hot is just one of it's jobs. The reason I need this info is to get a rough costing on how much we spend each month keeping that huge open tub of water hot. I've played around with a bunch of specific heat equations, but I keep thinking there's something I'm missing.

If anyone can help me out with this I'd really appreciate it, I'd even go so far as to build a little statue in your honor and mount in somewhere near the tank.
Gracias

Greetings Requiem !

Unfotunetly, I can't tell you the equations
you're looking for. I hope someone who

I will try to suggest a few things, if I may, though:
(It's probably stuff you already considered but
I'll say'em anyway - it's just that I don't know
why but your post is kin'na touching and funny -
no offense I hope. )
1. At the top of this forum is a locked thread
with multiple links through which you can
go to many other multiple links which might help
you find what you're looking for.
2. In the General Discussion forum of this site
also just search for other forums and post your
question in each to maximize your chances at
3. If it's your plant - could you just measure
the power consumption by running the thing
when the rest is shut off outside of normal
working hours(if you don't have 24hr production).
4. If you can trace the specific heater's power
cables then you could theoreticly attach something
as simple as a "fluke" to them and record the
power consumption periodicly. (Or ask an
electrician to do it for you - as long as
it's not gon'na cost much...)
5. You could try and see if you can estimate
the average power consumption of everything else
but the tank.

Hope this helps.

Live long and prosper.

The biggest problem I run into is the number of variables involved.
- the big ass tank of water takes a certain amount of energy to heat from 10degC to 71degC
- The top is open, so heat will escape quickly from contact with the outside air
- The sides and bottom are plastic, so heat will escape slower through there than the top
- The water in the tank is not being directly heated by the circulator, it is being heated by the hose running in a loop from the circulator
- The energy to maintain the 71degC is going to fluctuate according to the outside air temperature, as well as the load on the circulator from other systems in the plant

and there are probably a few others I haven't thought of. As far as running the bath while nothing else is running, that unfortunately is not an option. The best I can hope for is that we'll get a slow week and as few of the auxilliary systems will be required as possible.

If you record the time to reach your set point you can compute the energy required to heat the tank up.
(do every thing metric to save headaches
Use
&Delta;Q = mc &Delta;T

to get the work done, divide by the time to get the number of Watts.

Once it is at temp record the temperatures of the circulating water at the input and output.

Now use the same equation

Where &Delta;T is the change in temp of the circulating water and m is the mass flow for some fixed time.

To get watts divide by the fixed time.

This should get you pretty close to your number.

You could also shut off the water flow and record the drop in temp for a set time.

Last edited:
To heat the water initially takes about 500,070 btu. 436.6 gallons = 3636lbs at 110 degF delta temp at 80% efficiency (you're probably below this figure unless you are using waste heat to heat the tank).
The calculation is btu = Gal*SH*Lbs*DT/Eff
SH=1
LBS = gallons x 8.33
DT = 110= Temp 160 - In temp 50
Eff = .8

Using a calculation for losses through uninsulated pipe gives a rough estimate that you are probably losing about 260,000 btu/hr not counting the losses due to water being replaced as water in the tank is used, evaporating, and losses to the air through the open top.

Based on 260,000 btuh and the time of 15 hours (figuring the first hour as heat up time) I get 3,900,000 btu + 500,000 btu = 4,400,000 btuh, which is about 30.5 gallons of #2 fuel oil at 144,000 btu/gal.

Insulating this tank could save you quite a bit. Just 1" thick of .25k insulation (and an insulated lid) would reduce your losses to about 1,300 btuh x 15 = 19,500 +500,000 (you still have to heat the water initially) = 519,500 btuh. Or about 3.6 gallons of #2 oil per day. Quite a savings.

Again this is an estimation and does not account for water that is replaced from the 50 deg F source after the initial startup. To be more precise, use Integral's suggestion for determining the losses once the tank is heated.

You guys are great. Pina Colodas for all!

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