Swimming pool heating by 16 car radiators on hot loft

[Sten Larsen]

During summer my swimming pool is 23° C and my loft air is 53° C. (73° F / 127° F)

The sun shines on the metal roof with 1x10^5 to 1x10^6 W and I hope to move 10.000 - 20.000 W to the pool.

To move the heat down by an external water circuit to the water/water heat exchanger by the pool pump, I bought sixteen medium sized VW car radiators for the loft including pipes, pump etc..

Assuming each car radiator collect 1.300 Watts, they would heat 2 tons water 10° K (by using an estimated 10 T of air). And the 50 m3 pools temperature should raise 4° K per day if the sun shines ten hours/day.

Question
I won’t be able to assemble the system before April 2015, but I can't wait to figure out the effect per car radiator given a temperature difference of 30° K. I feel confident in every other aspect of the systems efficiency.

1.300 W is fantastic, 800 W is great, 400 W is all right and everything below is disappointing compared to my costs and the time I spend designing everything.

The only arguments I know are that:
- a car radiator should dissipate say 100.000 W of heat at 100 km/h because, that would be a certain fraction of the engines nominal effect in e.g. Horse Power.
- a hunch that a 53° C car radiator would heat an air flow as much as a 1000 Watt electrical heater on a given electrical heater-fans air flow.
- experimentally I could add 53° C water to a car radiator and then use the fan above to measure air temperature or the change in water temperature per second.

This is largely an engineering question, but I know that physicists are great at applying good assumptions on scattered information and think backwards.

This is the car radiator:
http://www.alibaba.com/product-detail/auto-radiator-OEM-No-191121253D-191121253K_299595219.html

Here are a few articles that I found too advanced, too general or too specific:
http://iosrjournals.org/iosr-jmce/papers/RDME-Volume2/RDME-11.pdf
http://www.academia.edu/4400629/Automotive_Radiator_-_Design_and_Experimental_Validation
http://www.pan-ol.lublin.pl/wydawnictwa/TMot10a/Kulikov.pdf
http://www.carcraft.com/howto/ccrp_0707_high_performance_cooling_system/
http://www.google.dk/url?sa=t&rct=j...=TQ9Gf4piXNSnbRJ0A37Wpw&bvm=bv.74649129,d.bGQ
http://www.saldanaracingproducts.com/Cooling%20System%20Principles.pdf
http://deepblue.lib.umich.edu/bitstream/handle/2027.42/57958/me450f07projec?sequence=1

More info:
This is the latitude and longitude of my house: 55°46'32.84"N 12°23'14.94"E
I am not going to use glycol, but maybe some aluminum corrosion inhibitors.
Excess water will leave the system if it raises more than 1½ meter above the normal loft floor, so the system cannot pressurize.
The system will use mostly 40 mm PVC pipes and hold 75-100 L water.
When the pump (Grundfos UPS 25-40 K180) and the four fans are running they use 100 - 250 W.
A fan being an X, and each radiator a |, each fan and four radiators will look like this (seen from above): "||X||" with this order of water entrance "13X42". Each fan and four radiators get the water directly from a four-way manifold on the loft.
The air is sucked down from the ridge through IKEA children’s play hoses.

All answers are appreciated, except uncontrolled laughter.
Thanks
Sten Larsen, software developer and former physics student from Copenhagen

 

[billy_joule]

It's a roundabout way of heating a pool that's for sure.


It'll be difficult to predict the steady state temperatures of all the relevant bodies. You may find that after running the system for an hour or two the loft has cooled because the heat transfer through your roof is too low to keep up with the outgoing heat.

The sun shines on the metal roof with 1x10^5 to 1x10^6 W

Only a small fraction of that makes it through to the loft air.



I'd suggest a simpler, more direct, tried and tested heating method like this:

 

[elegysix]

I agree with billy. Here's a rough way to check if what he said will happen.

Monitor the loft temperature over the course of a hot/sunny day (circulation is good here).
Estimate the maximum rate of change of the loft temperature.
Estimate the volume of air in the loft.

Calculate how much power it would take to change that volume's temperature at that maximum rate.
(if memory serves me right, use P = c*V*ρ*dT/dt, approximate values for c and ρ can be found here)

See how that compares with the power you want to send to the pool.

 

[davenn]

Got to agree with billy_joule on this one

There are much more efficient ways of heating water ... as shown is one of them

Dave

 

[Sten Larsen]

Thanks. I have been worried about the temperature drop of the loft air, but i feel confident that the 100 m2 / 1000 ft2 thin metal sheets (Decra) slooping towards the sun should transfer the heat well enough.

Assume that at equilibrium the difference in temperature is 20 K and not 30 K, and the fans blows 10 tons of air per hour through the 16 radiators that flows 2 tons of water per hour.

I would love to hear from anybody that has or can find details on car radiator physics. I am not able to use heavy integrals, so please translate it into something linear.

P.S.: I discarded the idea of adding pipes and similar things on top of the roof because of aestetics, costs and complexity of stormproofing it. I have been told that the necessary area needed for having an effect by using solar panels is 2/3 of the pool surface area.

 

[CWatters]

Thanks. I have been worried about the temperature drop of the loft air...

The stable 53C temperature you currently measure is somewhat irrelevant. At 53C the amount of energy entering and leaving the loft sums to zero. It has to be or the temperature would rise/fall further. If you plan on extracting more energy from the loft the temperature must fall. You can't easily predict how much it will fall unless you have a lot more data.

 

[Sten Larsen]

Car radiator physics and loft physics

Thanks for all your answers about the 16 car radiators in my loft heating the swimming pool.

My question is still about car radiator physics, and less about loft physics, but I would like to comment on your suspicion that the loft temperature will drop too much when the system is turned on.

When the sun is shining and equilibrium is reached then the temperatures could be named:
- T_off, when the radiators are off, and
- T_on, when the radiators are on.

The roof is gaining solar energy by up to 1000 W/m2 (http://en.wikipedia.org/wiki/Sunlight) and the surface area towards the sun is 100 m2 adding up to a potential of 100.000 W. The roof is almost black.

The roof's 200 m2 (sunny side + shadow side) is losing energy by:
1. radiation,
2. conduction by contact with air on the outside. Some air may also escape through the ridge.

At T_off the 200 m2 or 2150 ft2 dark roof is about 30 K warmer than the outside environment. So I believe that the roof at T_off is a system that receives and looses energy at a very high rate.

If 10% of the suns potential effect is transferred by the loft before escaping, the air of the loft should have a throughput of 10.000 W.

This rate increases if the temperature drops to T_on, because less energy escapes from the roof surface as radiation and more enters the loft by conduction (mostly).

So I believe that there should be plenty of energy available in the loft even if the radiators are removing 5000W and equilibrium T_on is reached.

Anyway! The hardware is bought, and my family, friends and colleagues expect nothing but success or failure. So the most important thing is to declare the project a complete success, when it is up and running.

I expect it to be very easy to get equilibrium in the pool at +1 K. And +2 K should be four times less easy, and so forth. I hope for +4 K, and guess that +8 K is next to impossible.

Please share more good ideas on how to estimate the effect of the radiators.

Thanks for your qualified concern
Sten

P.S. Later I might experiment and try to enhance the lofts convection or chimney-effect by adding a plastic barrier below the rafters.

 

[mfb]

There are several points that have to work with this small temperature difference, and I'm a bit skeptical about two of them. Note that those temperature drops are all additive, and just the last one is used to heat the pool.

Heating the walls with the sun: If you don't have an insulation there (and the high temperature suggests this), that's probably fine. The temperature drop can be small here.

Heating the air with the walls:
200m^2 roof area probably correspond to something like 400m^3 of air, which has a weight of about 500kg, so you get 500kJ/K. Extracting 5kW of power would cool the room by 6K in 10 minutes, which is enough time for the walls to deliver more heat - but it will come with some temperature drop.

Heating the radiators with the air:
That's a serious issue. I don't understand the description of your setup, but you'll need some good method (read: fans) to generate a strong air flow through the radiators. If you want to keep the temperature drop at 6K, you'll have to cycle the whole air volume of the loft through the radiators in 10 minutes.

Heating the pool with a water cycle:
Assuming you get 5K temperature difference between the flow directions, pumping 5kW will need ~250ml/s. That looks fine.Car radiators are much more effective in cars because they have a huge air flow and a much better temperature difference to work with.

 

[elegysix]

I really think 10,000 W is far too high power for heating the loft air, unless you have a substantial volume you're talking about. Let me explain -

You can work that equation I posted backwards. If you suppose there's 10,000 W of power heating the 53 C loft air, then you're looking at roughly 10kW/(1.09*V) = dT/dt

Supposing V is 1,000 m³, and 10,000 W went into heating your loft, your loft would be heating at a rate of 10kW/(1.09*1,000) =~0.009K/s, at a temp of ~53C.

Just guessing here, but that heating rate seems to be high, maybe by a factor of 10? It also depends strongly on the volume. If your volume is half that, the rate would be twice as much.
I would expect a heating rate to be 'in the neighborhood' of 30K /6 hours = 5K/hr = 5K/3600s ~.0014 K/s

This is why I would record the temperature of the loft throughout the day, and figure out dT/dt, to solve for the power input, P, rather than assuming it's 10,000 W.

 

[mfb]

elegysix: Somewhere in your calculation is a factor of 1000 missing (air has about 1000kJ/(m^3*K), probably there).

In addition, most of the heat capacitance is in the walls, just taking the air underestimates the energy flow significantly.

 

[billy_joule]

elegysix: Somewhere in your calculation is a factor of 1000 missing (air has about 1000kJ/(m^3*K), probably there).

roughly:
Density of air = 1 kg/m^3
Cp air = 1 kJ/Kg.K

So
=> Cp,v = 1 kJ/m^3.K

Just guessing here, but that heating rate seems to be several orders of magnitude off. I would expect a heating rate to be 'in the neighborhood' of 30K /6 hours = 5K/hr = 5K/3600s ~.0014 K/s

Having been in >100 roof cavities/attics/lofts as an electrician & heat recovery system installer I can say that is a very good guess.

 

[elegysix]

Thanks guys. I edited it. That's what I get for doing back of the napkin calculations and sharing them with the world lol.

 

[Sten Larsen]

Thanks. I will look at your ideas tonight.

Heres a few quick numbers:

The roof has about 250 m3 or 300 kg of air.

The air flow is estimated to 10 tons per hour because the four box fans (≈ 1 m2) could have an airspeed of 2.5 m/s. With 3600 seconds/hour and 1.2 kg/m3 it roughly equals 10 tons of air. I think I checked this number with nominal flows from hardware webshops and high school physics papers.

The dimensions of the piping and the pump is fine, and the pipes are insulated.

The water/water heat exchanger at the pool pump has nominal current of 8 m3/hour. It will have 1-2 m3 on the loft side and up to 8 m3 on the pool side. I hope this suggest that the water returning to the loft would be cooled significantly.

The throughput of 10.000 for the loft air and 5.000 W for the radiators are the numbers that need to be looked into. But what does the 200 m2 and 30 K difference at T_off tell you about throughput (energy lost to the environment)?

I will add drawings or pictures later this week and the final system will have thermometers and a flowmeter to calculate the power.

/Sten

 

[Sten Larsen]

Thanks for all the good advice so far.

You are welcome to browse the pictures and drawings. They may reveal more details.

This discussion has led me to a small design change.

Originally I put the pump and the expansion tank in the loft next to the manifold to ease the water filling procedure in the spring. Now I will place the pump down by the pool pump behind the garage, so I have the future option to use the pressure to distribute to more heat sources if deemed necessary after the summer season 2015. Also any leaks during the night will let air into the system and let the water drop into the 120 L open expansion tank.

Notes for the Pictures:

Picture1: (two blue boxes) I bought two 316 steal heat exchangers second hand and I will use them in parallel to lower resistance. Arrows are 40 mm pvc flex hoses. It will also have 4 thermometers and a flow meter. The manifold at the bottom of the picture distributes from 40 mm pvc pipes to 4 x 25mm pvc hoses for the car radiators. A long the pipes the purple Ts with yellow studs are for letting air and water out, they represent the highest point over the garage and the lowest point by the drive way. The green square is the expansion tank and it will change position together with the pump to right before the water/water heat exchangers.

Picture2: Mockup. The system will have 4 fans and 16 car radiators distributed in 4 IKEA bed drawers.

Picture3: Hot air from the top of the loft is pushed through the radiators. The studs point upward to let more air pass automatically.

Picture4: The loft with no insulation in the roof.

Picture5: A cheap hose - perfect for this project. I would have liked if none of the 300 components for this system had cost me more than a dollar. Pipes, hoses, valves and advanced fittings are expensive and add up quickly. Another trick has been to solder a 10$ heat controller kit and let its 5 V control an old USB controlled 230 V extension cord.

This system has cost me 50% or maybe only 30% percent of a recommended solar panel system or a professional heat pump. It is hidden, it is supposed to be noiseless, it may cool the house a bit, and it uses only 100 - 250 W when running.

I will definitely look closely at everybody’s arguments again and try to improve the design. (Right now I am just communicating.)

More ideas of how to estimate the throughput of the loft and the radiators are welcome.

Thanks
Sten

 

[mfb]

Where are the pictures?

 

[billy_joule]

Picture one: too small, can't read text.

Picture two/three: I would recommend having all the radiators in parallel. Overall mass airflow will be higher as the CSA will be doubled (air fans don't like constrictions) and the temperature differential at each radiator will be greater. Make sure there are no air gaps, expanding foam/duct tape etc will do.

Picture four; seal the air gaps at the eaves (can see light coming through on the left). I don't know about your house specifically but in my experience significant drafts appear on hot days due to the buoyancy of the hot air in the roof space. You may have noticed it when you open the manhole on a hot (relatively still) day?

Good luck!

 

[billy_joule]

The throughput of 10.000 for the loft air and 5.000 W for the radiators are the numbers that need to be looked into. But what does the 200 m2 and 30 K difference at T_off tell you about throughput (energy lost to the environment)?

It doesn't tell us anything about the size of the input and output. The equilibrium point is where the input is equal to the output. Hopefully your losses are large (eg drafts through air gaps) which would mean your input is large.

EDIT: I've thought about this some more and figured the losses are actually quite small (therefore the input is small). I've lived in three houses with ventilation systems that blow warm/hot air from the roof cavity down to the living quarters. The roof cavity temp is displayed on a control panel in the hallway. All houses had similar gable size and the same roof material as yours.
Like yours, the cavity often reaches >50 Celsius and it's not uncommon to still be >40 C as late as 10pm, hours after the sun (input) is gone i.e. the losses are small which means the input is small.

 

[Sten Larsen]

...
Like yours, the cavity often reaches >50 Celsius and it's not uncommon to still be >40 C as late as 10pm, hours after the sun (input) is gone ...

I hear. Maybe the 40 C is from the heat capacity of the loft floor/1.floor ceiling. There may be 10 tons of material. The energy would stay by the 2. or 4. root of time, and tend to go up.

Now look at this...

I know it is a loose argument, but how many 2000 W electrical fans/ovens would it take to heat the loft to equilibrium of 20° C during a -10° C cloudy winter day with no wind. More than one - maybe 4 or 10?

This optimistic approach makes me look like a happy camper
/Sten

 

[Sten Larsen]

I just calculated that the net radiation from my roof to the outside environment is 37.000 W.

To me it indicates that the throughput in the loft air is very high, and the 37.000 W does not include conduction and convection.

I used Georgia State university's online Radiation Calculator .

Please see the attached screen dump.



The calculation was done with a temperature difference that I meassured daily through May, June and July. August was rainy. Below are the assumptions I used to calculate the result:

1.
The difference in air temperature outside and inside the loft is about 30 K / 50 F on a sunny day.

2.
Most of the energy radiated from the sunny side is directly from the sun, but then again the roof on the sunny side is also much warmer. To make things easy I'd say that the surface of the 200 m2 / 2150 ft2 roof - sunny side or not - is 53° C / 127 F - like the loft air temperature I measure, when it is 23° outside.

3.
The blue sky above probably absorbs more radiation than it gives back. But I will assume that the environment surrounding the house is a body with a temperature of 23 C°.

4.
The roof has an emissivity of ≈90% (http://en.wikipedia.org/wiki/Emissivity) The metal roof is dark gray outside and a little lighter on the inside. The surfaces are not shiny at all.


I just now realized that if the temperature of the loft starts to drop, when the radiators are removing energy, then the net radiation from the sunny side of the roof to the lofts floor and other materials quickly grow big enough. And I am still not including conduction from the roof. The metals temperature may be as high as 65° C / 149° F. One day I meassured the rafters to be 60° C.)

At equilibrium at T_on the sunny 100 m2 roff metal might be 50° C, the loft air/loft floor might be 30 C, and that gives a net radiation into the loft of 12.500 W using the same calculator as before. (http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html#c3)




Elegysix made this nice calculation (thanks), and I think he is right and wrong.

I really think 10,000 W is far too high power for heating the loft air,
...
Supposing V is 1,000 m³, and 10,000 W went into heating your loft, your loft would be heating at a rate of 10kW/(1.09*1,000) =~0.009K/s, at a temp of ~53C.

...

Isn't this the rate you would expect in Space orbiting the Earth going from shadow and into the sun: 0.009K/s? Disaster and Science Fiction movies have many good examples of the sun appearing or suddon cold environments with little radiating heat. The Suns power is only about 50% greater in Space (locally) than on Earths surface.

But the rate of 0.009K/s is not un-natural.

The reason why we don't see this rate in anybodys roof on Earth is because the environment everywhere here is a great heat capacitor that absorbs/emits energy fast and thereby evens out any difference in temperature by radiation (not to mention conduction).

Energy is generally exchanged quickly, we just don't see it because usually the net transportation equals zero.


I might be missing something?

 

[billy_joule]

Forget radiation. Most heat will be lost from the hot iron by forced convection (or natural convection on totally still days)

 

[elegysix]

I don't know what you're talking about space and science fiction stuff. anyways.

But the rate of 0.009K/s is not un-natural.

You estimated the volume to be 250 m³ right? Then the rate estimate would be 0.036 K/s at 53 C. That estimate ignores a lot of things.

That aside, your approach with stefan-boltzmann is interesting. I like it.

At equilibrium at T_on the sunny 100 m2 roff metal might be 50° C, the loft air/loft floor might be 30 C, and that gives a net radiation into the loft of 12.500 W using the same calculator as before

But you know things are nowhere near 100% efficient. So if your max estimate is 12.5 kW... drawing 10kW is a steep goal. Don't you think all 200 m² are going to be ~50C? Your roof doesn't look too steep from the picture. Changing that will double your estimate to 25 kW, and that will make drawing 10kW start sounding possible.

 

[Sten Larsen]

Stefan-Boltzmann Law is great.

I don't know what you're talking about space and science fiction stuff. anyways.

You have a surface area of 2 m² and outer space has a temperature of -270° C (http://www.universetoday.com/77070/how-cold-is-space/)

In outer Space - let R₁ be the radiation from your body to Space and let R₂ be the radiation from Space to your body. This represents a net radiation of 1050 W.

On Earth - the net radiation is close to zero, but your body still radiate 1050 W to the environment, but luckily the environment also radiate 1050 W to you.

This shows me that a substantial fraction of the heat will radiate from the sunny roof to the loft.

If I find the time i will make the right temperature measurements and estimates and an illustrated spreadsheet model with equilibriums Toff+Ton, sunny and shadow roofs, air, loft materials, car radiators, air streams, and arrows of radiation, conduction, and convection.

Forget radiation. Most heat will be lost from the hot iron by forced convection (or natural convection on totally still days)

You can't neglect Stephan-Boltzmann. The roof surface is one hundred times bigger than the human body in Space and hence the temperature that I have measured daily have showed me that at temperature T_off (Equilibrium with pool heating off) the roof radiates ≈1.100.000 W into the loft(net is still zero) as well as 37.000 W out of the loft (net loss to the environment).

When temperature T_on (Equilibrium with pool heating on) is reached the sunny roof temperature drops to maybe 50° C (the shadow roof may be much colder e.g. T_on). The point here is that now at T_on) the loss to the outside environment is much lower, and this difference is the heat that is going into the swimming pool.

What happens inside the roof is not important, as long as I know that the heat in there is moving around. Here radiation prooved simpler to estimate than forced convection. What really interests me is the difference in the roofs loss to the environment between T_off and T_on.

The difference in roof surface temperature indicates with Stefan-Boltzmann that energy is going into the pool.

Most of you agreed that the loft temperature would drop quickly and I argued that this is translated into a lower roof temperature, much lower loss of energy to the outside environment and a substantial amount of heat going into the pool instead.

Thanks for the input about the loft. Until a week ago I had nothing but a gut feeling.I am still interested in hints about car radiator physics.

 

[elegysix]

the net power radiated into the loft is P=εσA(T_roof⁴-T_loft⁴)

if T_roof= 50 C, T_loft=30 C, ε=.9, A=200 m²,
then P = 25 kW.

In other words, to keep the loft's temperature at 30 C, the sum of power lost and power sent to the pool must be 25 kW.

 

[billy_joule]

You can't neglect Stephan-Boltzmann.

In many situation you can, it is a classic example you'll find in all heat transfer textbooks; compare heat loss from a warm surface via radiation and convection. It's to show students that for many cases radiation losses can be ignored as they are small.

That's a large part of why most cooling systems use convection & conduction (such as your car 'radiators') because heat transfer via radiation is far too slow for most situations and temperatures we encounter.
It's also why vacuum flasks (thermos') are so effective at keeping drinks hot or cold.


A simple experiment; heat a brick or something up to around 50 C in your oven. Take it out, hold your hand about above the brick, then hold the brick above your hand the same distance.

In the first case you will feel the heat transfer to your hand by convection in the second you'll be lucky if you can feel anything as the radiation is too low. If you get close enough to the bottom you'll feel heat, but it'll mostly be from crossing the boundary layer.
http://en.wikipedia.org/wiki/Boundary_layer#Heat_and_mass_transfer

 

[Sten Larsen]

If Stefan-Boltzmann Law can be neglected in my loft because conduction or convection have say five times the effect, then I have an additional five times the effect directly from conduction/convection.

That makes the challenge of heating the swimming pool by car radiators much easier.

I used Stefan-Boltzmann to show that:
- enough energy is moving around inside the loft, and that
- the radiators are drawing enough heat effect (10.000 W) from the loft system just by radiation when the loft temperature drops by 5 or 10° C, because this translates into a lower loss to the environment. (e=0.9: dT=46-53° C, 200 m2)

To show the same thing with convection I would probably have needed a detailed computational model with time steps and 3D grids.

It's also why vacuum flasks (thermos') are so effective at keeping drinks hot or cold.

Thermo flasks use both vacuum and a silver-like core bottle with an emissivity close to zero.

 

[Sten Larsen]

Actual results:

The car radiators ended up giving 13.000 Watts of heat to the swimming pool and today the pool temperature was 32 C / 90 F. (The last ten days the air has been 15 C / 59 F during nights and 25 C / 77 F during the day and sunny)

When I use 1 Joule of energy I get 30 Joules back; k=30. A normal heat pump only has k=4 or 5. (The eight fans and the two pumps use 436 watts.)

I build the system with 24 car radiators and they produce 540 Watts each - and 540 Watts is the answer to my original question in this thread.

Dimensions of water pipes, pumps and flow:
50 meters of 40 mm pvc pipes distributed to 4 x 10 meters of 25 mm pvc tubes run by 100 Watts of pump power and that sends 1660 Liters of water back and forth every hour with a temperature increase of 5 or 6 degrees Celcius.

People may ask, "why didn't you just use regular solar water heating collector panels". Well my wife is a designer, and she said panels would look bad on the roof - end of discussion. Also i didn't like the idea of installing a heat pump, that would be expensive and noisy to run.

I think the system is a success, it was fun to build and it is fun to have, but its 400 components did cost 4000 dollars to buy and it will take 200 dollars to operate per year. Alternatively a heat pump would cost 2000-3000 dollars to buy and 1000 dollars to operate per year. (Danish prices).

I love watching free energy going into my pool.

 

[russ_watters]

Congrats! Great project! You collect more energy than I would have guessed.