# Heat Transfer solar collector system Problem

Fish4Fun
I am planning a solar collector system for domestic hot water and heating. I am attempting to calculate the heat loss in a storage tank. If we assume:

Tank Volume = 1m^3
Tank Surface Area = 6m^2
Tank Contents = 1000L water
Initial Water Temp = 90C (194F)
The Temp Outside the Tank = 10C (50F)
Tank Insulation Thickness = 5.08cm (2in)
Insulation Material = Closed Cell Urethane Foam
Insulation k = 0.02 W/mC

Our initial energy in the tank =

1000L * (90 + 273)K * 4184J = 1,518,792,000J

Calculating the heat loss we have:

k * A * dT * d^-1 ==> .02 * 6 * 80 / .0508 = 189W = 189J/s

So, 1,518,792,000J - 189J = 1,518,791,811J is the energy remaining after 1 second. The new temperature is then:

T = (1,518,791,811 /(4184 * 1000)) - 273 = 89.9999548C

If we then repeat the process 60 times our water should be
89.997245C with total energy being 1,518,780,662J.

After 3600 Iterations (1 hour) our water should be:
89.837566C with the total energy being 1,518,112,376J

After 86,400 Iterations (24 Hours) our water should be:
86.191268C with the total energy being 1,502,856,267J

After 2,592,000 Iterations (30 Days) our water should be:
28.515970C with the total energy being 1,261,542,820J

But I just do not believe these numbers. Even if I change k from .02 to .2 After 86,400 Iterations (24 Hours) our water should be:
69.412071C with the total energy being 1,432,652,104J

(I do understand that the "useful" energy is considerably less than the total energy, with 1000L of water @ 90C having roughly 61.6kWh of "useful energy" if we define "useful energy" is any water temp > 37C. That is Useful Energy = Total Energy - Energy @ 37C).

Anyway, I have had a fair amount of life experience with attempting to keep things cold and hot, and I simply do not believe 1000L of 90C water will only "cool down" to 86C in 24 hours with a mere 2in of closed cell foam surrounding it. What am I doing wrong? Obviously I have something wrong, I just cannot figure out what it is.

Fish

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Gordianus
Assuming a lumped parameter model, the time constant is given by:

(c*m*d)/(k*A)

where c is the specific heat of water (4184 J/Kg C), m is the mass (1000 Kg), d is the insulation thickness (0.05 m) and A is the exposed area (6 m2).

Upon replacing the numbers, I obtained a time constant of about 15 days. No wonder the temperatures lowers so slowly

Gold Member
One thing that seems off is that the area is only 6m^2, but the volume is 1m^3. This is actually impossible. The shape that has the least surface area containing a given volume is a sphere. If we assume your container is a sphere, we would have:
$$Volume=V=\frac{4}{3}\pi r^3 \rightarrow r = (\frac{3}{4\pi}V)^{\frac{1}{3}}$$

$$Surface Area = A = 4\pi r^2 = 4\pi (\frac{3}{4\pi}V)^{\frac{2}{3}} \approx 6.97m^2$$
This is for a perfect sphere. The real surface area needs to be higher than this.

Fish4Fun
LeonhardEuler,

Hrmm, a 1m cube is what I was thinking, and each side being 1m x 1m with 6 sides brings us to 6m^2, for the inside surface area of the tank. Obviously I am ignoring the 0.0508m added to all of the edges to achieve the slightly larger external surface area of, but in reality only 6m^2 of the internal area is coupled with the external environment regardless of how much insulation is added to the outside.

And your math is a bit off on the Sphere:

V = 1 = 4/3*Pi*r^3 => r = 0.620m
A = 4 * Pi * 0.620^2 = 4.836m^2

(BTW, If you plug "1" into your above formula, you get the same 4.836m^2. I assume you simply hit a wrong button somewhere along the line.)

I guess the best thing to do is build a small scale model and test it, lol.

Fish

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RocketSci5KN
Is your temp outside the tank 10C or 32F ?

Fish4Fun
It seems the answer lies in economy of scale. For a 10 liter cube the numbers seem somewhat more reasonable to me, dropping from 90C to 72.69C in 24 hours and within 2 degrees of equilibrium in 15 days. With the 1000L tank the time to 2 degrees from equilibrium is ~38 days.

An interesting note, a cylinder of Volume 1m^3 with a Diameter = Height = 1.083m has a surface area of 5.35m^2 which leads to a 42.8% increase in "useful kWh" stored after 10 days (3.67kWh / 6.42kWh). While a storage sphere would further reduce losses, the complexities of DIYing a sphere are considerably higher than building a cylinder. But a cylinder is not really any more or less difficult to construct than a cube.

If anyone can find anything wrong with my process/math/conclusions I sure would love to hear about it now rather than find out after I start building!

Fish

Fish4Fun
RocketSci5KN,

I have run it both ways, (and many others), for the numbers I posted I used 10C (which is actually lower than anything I expect, the tank will be buried and ground temp should not drop much below 20C).

Thanks,

Fish

Fish4Fun
This is just an unashamed "bump". I do not want to re-post in the engineering forums, but I would love a bit of confirmation or refutation of the basic math/iteration process I am using to predict the temperature of the water in my tank. I think the rather long time to equilibrium that seems to defy my "real-world-experience" has more to do with the size of the system than the basic principals of thermodynamics, but I would love some reassurance.

Fish

Gordianus
The only "odd" thing if find is that, instead of solving the differential equation of cooling for a lumped paremeter system, you resorted to a discretization in steps of about 1 sec, which is a very reasonable choice since the temperature changes slowly.
Disregarding the error that 1000 l of water can't be hold in a container whose area is 6 m^2, the other part is O.K.
I think you don't believe in your results because, I guess, in a real world situation heat losses may be larger, for example trough conduction in input/output pipes

Homework Helper
Your calculation says you are losing about 200W of heat form the tank.

OK, so how long does it take an electric kettle to heat 1 liter of water from 10C to 90C?

Answer: heat energy required = M Cp delta T = 1 x 4184 x 80 = 335 kJ
Suppose it is a 2 kW kettle, then the time = 335/2 = 167 sec. That seems reasonable to me.

So a 200W heater would take 10 times as long, which is about 0.5 hours.

So to heat 1000 liters with a 200W heater would take 1000 times as long, i.e. 500 hours or about 21 days.

I don't think you are doing anything wrong, except your intuition isn't calibrated properly for a 2 inch thickness of VERY good insulating material.

In fact the thermal gradient through the insulation will decrease as the water cools down, so the cooling will take longer than your estimate.

Fish4Fun
Gordianus,

I used 1 second steps because the passive heat losses are only a small part of the overall heat gain/loss calculations I want to run (I want to model "usage" and "Solar Input" to appropriately size my tank and collectors.)

Disregarding the error that 1000 l of water can't be hold in a container whose area is 6 m^2,

As stated above, for a cube:

1m x 1m x 1m = 1000L
Surface Area = 6 * 1m = 6m^2

For a Cylinder:
Code:
Cylinder    Cylinde r   Cylinder     Cylinder    Cylinder
1.08385     0.541926	1.083852	1	5.53581044

For a Sphere:
V = 1 = 4/3*Pi*r^3 => r = 0.620m
A = 4 * Pi * 0.620^2 = 4.836m^2

How is my estimate of 6m^2 surface area unreasonable for 1000L of water? Granted, I will leave the tank less than 100% full, but even if the tank is 1.1m^3, for a cylinder the surface area can be minimized to 5.9m^2.

Anyway, thank you for your response!

AlephZero,

Thanks for the response. I couldn't find anything wrong with my math, it just seemed too good to be true! My only issue now is figuring out what to do with all the excess heat in the summer, LOL.

Fish

Gold Member
2021 Award
fish4fun....
been following the thread with interest, not sure why so many were having a difficult time understanding the volume of a m^3 container, and its surface area.... pretty basic huh :)

But I am more interested to see what your experiments reveil and how much, maybe what, different types of insulation material have on heatloss of the system.
Please update us all in the future :)

Dave

Fish4Fun
davenn,

Thank you for not asking how one could possibly get 1m^3 in 6m^2 LOL! As far as the REAL WORLD results, don't hold your breath, this is a long term project. We moved into the house in Oct 2008 with the notion of solar DHW & solar DH as primary concerns, but the down turn in the economy has suspended the reality. I installed a heat exchanger in out heat pump HVAC, and the DHW solar part is a no brainer, but the $15k worth of solar collectors are currently "going to college". When my son begins to support himself, I will have the money to "be green", until then, I can't afford it, LOL, so I just keep trying to refine the math. I have built a DIY solar collector, but the$/W are such that commercial collectors are more cost effective. Using evacuated tube technology, my proposed system will require ~10 * 30 tubes to get the ~60kWh/insolation day I would need to handle my Winter with 80% solar DHW/DH. The biggest issue is thermal storage of relatively "high-grade" heat (50C+). If you are truly interested in the project, please PM me. This forum is not terribly interested in this topic, and it would be better to keep track of this project in another venue. My interest here was to make certain my math was right.

Thanks for the interest!

Fish

dirtyhabanero
Hey guys, I'm new here but I think I've found what I've been looking for. I've only a small understanding of physics (velocity, acceleration etc), but I'd like to know more about heat transfer, namely between a liquid and its surroundings.

I've been using the equation above that Fish4Fun has been using (which I believe is heat conduction?), to try and calculate the amount of heat transferred (not sure if this is correct terminology) from my fish tank to its surroundings.

I've also been using Wiki to understand again all the conversions between units. As well as this site http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/heatra.html#c2"

I encounted two problems when I try and apply the equation:

1) What measurement or number do I use for 'insulation thickness' (d from the image above) if I am calculating Joules/sec for just the top surface area? for which it is just a water-to-air interface (no barrier)

If I leave d out of the equation, then it's the same as letting d=1. Note: I correct the dimension by cancelling out one of the [L] from above the vinculum

2) Carrying on from the example above, if there is no 'barrier', should I be using a k value or no? Currently I've been using the thermal conductivity of water, ~ 0.6 W/mK, so my calculations look a bit like this:

Joules/sec = (0.6W/mK * 2m^2 * 8K)/[L]^-1

Fish4Fun: Please add me on twitter if you are interested, I'm in the process of putting together a Compost Water Heater to keep my fish tank warm this winter (I'm in Australia), http://www.ecofilms.com.au/2009/12/27/compost-wheelie-bin-hot-water/" [Broken]
@dirtyhabanero

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Staff Emeritus
Homework Helper
The calculation of the initial heat content of the water in the tank is incorrect. Water at 90 C has an enthalpy of 376.96 kJ/kg and a density of 965.06 kg/m^3. Taking the volume of water as 1000 L (or 1 m^3), the heat content of the water in the tank is 363,789. kJ
You can't apply Q=mcDelta T over a temperature range which includes a phase change. For ice to change to liquid water, a certain quantity of heat called the heat of fusion must be added to the substance. In any event, the specific heat of ice is different from the specific heat of liquid water (which is also different from the specific heat of steam).

chompie
surely the kJ is lower by the temperature difference in and outside the store.

4.184*(90-10)*1000*1 = 334,720kJ = 79.964kWh (334720*0.0002389)
you have approximately oversized by 5 times the actual energy in the tank?

you had a heat loss of 189W/s which equates to 86.4kW/day(0.189*3600*24) I would expect around 1.89kW/day heat loss from a well insulated cylinder of 500l?

mpmitch

How can I calculate the electrical load of a MHRV system to passive house standards in the UK climate? Is there an equation I can apply to the specifications of the actual system I wish to install?
Any help would be much appreciated.

chompie
Why would you post that question in this topic, where it bears no relation and not as a new question or a topic better suited?

Homework Helper
Gold Member
I confirm your numerical result of 4C loss after 24 hours, assuming perfect temperature equalisation within the tank at all times.
if θ is the temperature difference from ambient then
dθ/dt = -θ.k.A/(D.s.V)
where k = conductivity of wall material, A = surface area, D = wall thickness, s = s.h. of water by volume, V = volume.
Integrating we get θ = α.exp(-kAt/DsV), where α is the initial temp difference.
With the values supplied, that gives 4.8% loss after 24 hours, i.e. temperature has dropped from 90 to
10 + (90-10)*0.952 = 86.2.

But note the assumption that temperature in the tank is always uniform. In practice, convection (inside and outside the tank) will arrange that the upper layers stay warmer. Not sure what that will do to the total heat loss.