Mike’s solar pool - How best to measure the gain from our tubes?

[Mike N.]

We are now at the stage where we are collecting impressive amounts of heat from our tubes (less than 2 hours of electrical heating required for the pool in the last 18 days since I installed the tubes).

I would like to be able to quantify what we are saving and would value the advice of those who understand such things.

My initial thoughts have included doing sums involving the difference in temperature of water in and out of the panels or heat exchanger or more simply perhaps just using the rise in temperature of the pool each day.

Data available on a minute by minute basis include the following:

Temperature of water in and out of panels
Temperature of solar water in and out of heat exchanger
Temperature of pool water in and out of heat exchanger
Temperature of pool water
Volume of pool water
Volume of water circulating in the solar circuit.
Flow rate of water circulating in the solar circuit.
Duration of solar pump running on a daily basis (though this doesn’t yet equate with times it’s actually pumping heat. We are still experimenting with different triggers for pump on and off involving absolute and relative temperatures of panels and pool)

Ideas from maffs boffs please.

Mike

[dhaslam]

The only measurement you can make is the temperature rise multiplied by the volume of water circulating  because the pool gains solar heat and also looses heat.  In bright sunshine there would be quite a bit of heat gain direct to water.    Our local boiler maker used to use two buckets, a stop watch, a scales and a thermometer to measure the output of back  boilers.   

[ericw]

Mike,

I would suggest that to make things much easier you rig the pump to run continually.

Then as you know the flow rate, you can calculate the instantaneous energy flow at various points of the solar loop by using the temperature differences across the components.

If you integrate this you will get a total value which you could compare with that estimated from the temperature rise of the pool, as a sanity check.
Although this doesn't equate to 'real life' if the losses when the energy is not being added to the system by the sun are small it should be close.

[Mike N.]

The only measurement you can make is the temperature rise multiplied by the volume of water circulating  because the pool gains solar heat and also looses heat.  In bright sunshine there would be quite a bit of heat gain direct to water.   

As the pool is indoors and fairly well insulated, I wouldn't expect too much in the way of direct heating. Certainly this would be an easy calculation to make. Would it not be the temperature rise multiplied by the volume of the pool rather than the volume circulating?

Mike

[Mike N.]

I would suggest that to make things much easier you rig the pump to run continually.

Then as you know the flow rate, you can calculate the instantaneous energy flow at various points of the solar loop by using the temperature differences across the components.

If you integrate this you will get a total value which you could compare with that estimated from the temperature rise of the pool, as a sanity check.
Although this doesn't equate to 'real life' if the losses when the energy is not being added to the system by the sun are small it should be close.

Hi Eric

I'm concerned about running the pump continually. I did this for the first week (told it to turn off when the tubes dropped below 30, which they never did as the pool continually warmed them!). I would be using the pool water to keep the manifold on the roof warm all winter, though I guess it is well insulated.

I like the idea of being able to measure the energy gained at different points but am still having a problem getting my head round the sums. I can see that I only need the flow rate and temp in and out of the item in question, with perhaps the magic number 1164 thrown in somewhere. Can anybody string these figures together in the right order and throw in a few maths symbols? I assume I'd have to divide the day into bite sized chunks and add them all together too?

Mike

[dhaslam]

If the pool is close to ambient temperature then the measurement will be accurate.   You need to plot the temperature at regular intervals.   The heat gain in kilowatts in an hour is the volume in litres multiplied by heat gain divided by 830.  One cubic metre of water is 1000 litres.

[ericw]

Mike,
I wasn't suggesting that you had the pump continually running as a permanent thing but just while you were in a measurement phase.
(Although I'm sure an argument could be made for controlling the pump from the level of solar radiation rather than the delta T.)
 
This will reduce the number of variables present and enable you to get a handle on the input and the various losses so you could then hopefully decide that you could ignore some of them, if you try and analyse the normal operation.

You could filter out measurement 'slots' where the panels had a negative temp rise (due to the losses exceeding the solar gain) as these wouldn't exist with a delta T controlled environment.

I think the other formula you need is    dT(*C) = 14.3 * kW / flow(l/min)

Eric

[Mike N.]

If the pool is close to ambient temperature then the measurement will be accurate.   You need to plot the temperature at regular intervals.   The heat gain in kilowatts in an hour is the volume in litres multiplied by heat gain divided by 830.  One cubic metre of water is 1000 litres.

Thanks for that.  Where does the magic number of 830 come from?

Can you check my sums please Sir?

Looking at results for the most consistent part of the day -  the one hour period from 15:28 to 16:27 -
Average temp of water returning to the tubes was 34.3 and leaving was 41.6, so heat gained was 7.3 degrees C over that one hour period.
In line water meter over a 10 hour period gave a total flow of 2.436 cu metres so I'm assuming that in the one hour period we are discussing this amounts to 243.6 litres passing through the manifold

Using your calculation I make this 243.6 x 7.3 / 830 which gives a heat gain of 2.14kW

As a reality check, Ivan's tables suggest that an average May day could give 33.8kWh a day.
Assuming a 10 hour day, this could be 3.38 kW in an hour
So our 2.14 kW suggests that today was 2.14 / 3.38 which was a 2/3 average May day - which I think it probably felt like .
QED - Looks about right I think.

Looking at the pool water calcs is a bit more dubious as it only rose 0.1 over that hour, and with the sensors only reporting to the nearest 0.1, the true rise could of course be anything from 0 to 0.2, but let's assume 0.1
Pool volume is 12.2 cu m so sums are 12200 x 0.1 / 830 which makes 1.47kW.
Allowing for the potential errors in the sensors reporting over such a narrow interval, this figure could actually be anything from 0 to 2.9, so the 2.14 we calculated for the tubes output does at least fall in the right ballpark - very satisfying.

Before I get Geraint to write some clever sums software, can you confirm that these are the correct sums to be doing please?

Mike

[Mike N.]

Mike,
I wasn't suggesting that you had the pump continually running as a permanent thing but just while you were in a measurement phase.
(Although I'm sure an argument could be made for controlling the pump from the level of solar radiation rather than the delta T.)
 
This will reduce the number of variables present and enable you to get a handle on the input and the various losses so you could then hopefully decide that you could ignore some of them, if you try and analyse the normal operation.

You could filter out measurement 'slots' where the panels had a negative temp rise (due to the losses exceeding the solar gain) as these wouldn't exist with a delta T controlled environment.

I think the other formula you need is    dT(*C) = 14.3 * kW / flow(l/min)

Eric

Hi Eric

I wasn't ignoring your post - it came in whilst I was off doing sums and didn't notice - sorry

Can you explain what you mean by "other formula"  . I'm a bit confused by your formula - which bits are in brackets due to being part of the calc and which are by way of explanation? Is dT the difference between the in and out temp of the tubes, or is that C?

On the pump on/off bit we did get a bit of cycling on and off as the sun came out with the cooled water turning it off again, so we then told it to wait until the flow from the tubes was 3 degrees hotter than the pool which seems to have sorted it for the moment

Mike

[dhaslam]

The conversion figure should be 860 rather than 830. Sorry, my bad memory.  The conversion comes about because the calorie is based on heat rise per gram per second and the kilowatt is based on joules or  units of work done and is related to newtons and metres.  The calorie is equivalent to 4.184 joules but when you take an hours worth of calories, 3600 and divide by 4.184 you get 860.

[ericw]

Mike,

To clarify:-   Temp rise (in degrees C)  = 14.3 x power (in Kw) / flow (in l/min)

While this is most useful for working out how much power is being absorbed by the tubes, it can also be used to calculate the losses in the pipes.
Using your figures
   Flowrate - 4.06 l/min
   Temp rise  7.3 degrees
   Power input = 2.07 kW

Which is roughly the same as you got.

I think to be able to reduce the potential errors in the temperature rise measurement of the pool you are going to have to do it over a whole day.
As the sun is unlikely to shine for a whole day you will have to use minute by minute measurments and add up the total energy input for the day.
If you kept the pump running you would eliminate the stop start uncertainty but probably have some negative gain periods, but the overall total for the day should stack up with the result from the pool temperature rise calculation.

Eric