How to calculate heat loss?

[geno]

I've been looking for a way to calculate the change in tank temp (at night) assuming the tank water temp = 80 and the night time ambient temp = 65.

Let's assume no heaters, no lights, no additional heat from pumps, i.e. keep the computation simple.

Anyone know the formula for this?

thanks,

 

[stickx911]

how does the temp reach 80? Lighting I assume. How long are the lights out? do you have sump...this really does not seem like it could be simple.

There are some more intelligent physicists on the boards, but I would think that it would be difficult to ignore flow, volume and surface area to get a good number. You could just get a glass of saltwater at w/c time and set it asside on your off day and see how long it takes to loose temp.

 

[geno]

I picked 80 degrees as a nominal number -- it could be 75, 77, 78 -- it doesn't matter -- I'm more interested in how the ambient temp affects the water in the tank.

There must be a formula for this. I found one, but not sure it's what I should use.

<p style="text-align:left;"><span style="font-family: Palatino Linotype;"><span style="color: #0000ff;">Q/t = kA(T_hot - T_cold)/d</span></span>
<span style="font-family: Palatino Linotype;"><span style="color: #0000ff;"></span></span>
<span style="font-family: Palatino Linotype;"><span style="color: #0000ff;">where:</span></span>
<span style="font-family: Palatino Linotype;"><span style="color: #0000ff;">Q/t = heat transferred in time t</span></span>
<span style="font-family: Palatino Linotype;"><span style="color: #0000ff;">k = thermal conductivity of the barrier (in this case glass)</span></span>
<span style="font-family: Palatino Linotype;"><span style="color: #0000ff;">A = area (here I'm not sure what to use -- the entire tank or just the exposed surface -- I used the entire tank)</span></span>
<span style="font-family: Palatino Linotype;"><span style="color: #0000ff;">T_hot = highest temp, i.e. 80-degree</span></span>
<span style="font-family: Palatino Linotype;"><span style="color: #0000ff;">T_cold = nighttime ambient temp of 65-degrees</span></span>
<span style="font-family: Palatino Linotype;"><span style="color: #0000ff;">d = thickness of the barrier (glass)</span></span>

 

[george]

It's a fairly simple calculation, but it's also meaningless since the exposed surface area that looses the most heat is presumably constantly changing (the top).

I'll see if I can find the simple version, though.

 

[geno]

Thanks, George!

 

[enderg60]

Q=M*C*dT

Q = your heat in or out(in BTU)
M = your mass of water(8.556 * # of gallons)
C = specific heat(0.94 for salt water)
dT = your change in temp in F

 

[george]

The calculation is similar to the heat loss calculation for a house. You have area time the temperature gradient divided by the thermal resistance of the wall (glass, in this case).

From http://hyperphysics.phy-astr.gsu.edu/hbase/tables/rvalue.html#c1">this site</a> and the calculations:

loss in BTU/hr == (36.5 ft^2 area) * (80 - 65) / (.89 per 1/8" of glass)

So roughly, 36.5 * 15 / (.89 * 4) if your 125 has 1/2" glass or about 154 BTU/hr or 45 watts.

However the far greater heat transfer is directly water to air on the surface which is also in motion.

Edit: I just saw all the replies. I agree that there are so many more factors that affect the heat loss (most of which excede the loss through glass) that it's not very useful.

 

[enderg60]

That is exactly the forumla for heat loss of a house.

you can use that to find out how big a heater you need to keep your temps stable at night(just convert that to watts) or how big a chiller you need to keep temps down(12000BTU = 1hp)

how much heat you loose through the glass and such is pretty pointless to know.

 

[geno]

Thanks, guys, for the formulas.

Ares, my point was one of knowledge and curiosity. I was curious regarding the heat loss of the my tank when the night time thermostat allows the room temperature to drop to say 65 degrees. I picked 80 as the temp in the tank even though it's probably 78 or so. Of course, I have heaters, but some on this board claim that heaters may not be needed for a 125g tank with 30g sump while in a heated space. So, I became interested in the heat loss of the water.

If this doesn't interest you -- you need not comment further.

 

[enderg60]

that all depends on the kind of pumps you use, how open your stand is, if you use glass tops, what lights you use....blah blah blah blah

anyone who tells you not to run a heater should be ignored, even if they are not needed all the time its there to keep the temp stable should something change, much like a chiller in most cases.

 

[geno]

I appreciate the wise advice to always use heaters. It was never my intent not to -- in fact I use two.

This whole thing came up as a result of my love for the hobby and my desire to learn things that I've never been exposed to before, i.e. water chemistry, plumbing, lighting spectrum, heat loss, etc.

Sorry if I wasted anyone's time.

 

[stickx911]

don't apologize, lol. It's just normal banter around here to make arguments. No one really means anything by it most of the time.

 

[enderg60]

learning is never a waste of time