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Swimming pool heating by 16 car radiators on hot loft

  1. Sep 7, 2014 #1
    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.

    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:

    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 [Broken]
    http://www.saldanaracingproducts.com/Cooling System Principles.pdf

    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.
    Sten Larsen, software developer and former physics student from Copenhagen
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Sep 7, 2014 #2


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    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.

    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:

  4. Sep 7, 2014 #3
    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.
    Last edited: Sep 7, 2014
  5. Sep 7, 2014 #4


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    Got to agree with billy_joule on this one

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

  6. Sep 8, 2014 #5
    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.
  7. Sep 8, 2014 #6


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    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.
  8. Sep 8, 2014 #7
    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:
    - Toff, when the radiators are off, and
    - Ton, 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 Toff the 200 m2 or 2150 ft2 dark roof is about 30 K warmer than the outside environment. So I believe that the roof at Toff 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 Ton, 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 Ton 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

    P.S. Later I might experiment and try to enhance the lofts convection or chimney-effect by adding a plastic barrier below the rafters.
  9. Sep 8, 2014 #8


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    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.
  10. Sep 8, 2014 #9
    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.
    Last edited: Sep 8, 2014
  11. Sep 8, 2014 #10


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    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.
  12. Sep 8, 2014 #11


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

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

    Having been in >100 roof cavities/attics/lofts as an electrician & heat recovery system installer I can say that is a very good guess.
  13. Sep 8, 2014 #12
    Thanks guys. I edited it. That's what I get for doing back of the napkin calculations and sharing them with the world lol.
  14. Sep 9, 2014 #13
    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 Toff 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.

  15. Sep 9, 2014 #14
    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.


    Attached Files:

    Last edited: Sep 9, 2014
  16. Sep 9, 2014 #15


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    Where are the pictures?
  17. Sep 9, 2014 #16


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    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!
  18. Sep 9, 2014 #17


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    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.
    Last edited: Sep 9, 2014
  19. Sep 10, 2014 #18
    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 :smile:
    Last edited: Sep 10, 2014
  20. Sep 10, 2014 #19
    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:

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

    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.

    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°.

    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 Ton 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.

    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?

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  21. Sep 10, 2014 #20


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