Convert Flow and Temperature Difference to kWh

[kristen]

I want to convert measurements from my TDC3 to kWh.

I have Flow and Return temperatures, and flow in L/min.

I'll have a go, but I'd be grateful if you could tell me if I've got it right.

My system is logging values every 10 seconds.

I have started with (1°C * 1 Litre * 4200 J ) in J / 1,000,000 [convert to MJ] / 3.6 = 0.00117 kWh

My pump is running at about 6.95 L/min, so in an hour will pump 6.95 * 60 = 417 L

Lets assume that the dT is 10°C, over the course of an hour I would get:

(10°C * 417 Litres * 4200 J ) in J / 1,000,000 [convert to MJ] / 3.6 =  4.865 kWh (I'm actually seeing around 7kW so this look wrong )

So for the 10 second timeslice that was measured I got

4.865 kWh / 60 minutes / 6 [a 10-second slice] = 0.0135138888 kWh

So for a 10 second time slice the kWh generated is:

(10°C [delta-T] * 6.95 [pump speed in Litres / minute] * 60 [litres per hour] * 4200 J ) / 60 [minutes in an hour] / 6 [10 second timeslice] in MJ / 3.6

Miraculously that gives me the same 0.0135138888 answer

The bad news is that is giving me figures around 1,700 for the 10 second timeslice, and the TDC is telling me I'm getting about 7kW - 8kW and I can't marry the two

Thanks.

[ericw]

Kristen

Does this version of the formula help ?

The relationship between flowrate, temperature rise and power input is :-
 flow rate(l/min) = 14*power input (kW)/temp rise (degree C).

This gives 5kW

[kristen]

That's a lot more simple Eric, thanks.

At about 10°C, and 6.95 L/min flow rate, I'm seeing the TDC report 7kW to 8kW.

Unless I have misunderstood it feels like something is "adrift" there? What do you think?

[rob26440]

I reckon my TDC3 is reporting nearly double reality.  It's difficult to prove accurately but with 5 sensors up the side of the cylinder so I can get a reasonable picture of the stratification/average temp - and not using any hot water for a number of hours on a good, sunny day - that's the conclusion I came to.

[dhaslam]

Presumably the controller does the same calculation so the error has to be in reading the flow rate or the temperature difference.   Temperatures are displayed in even degrees but it may use more decimal places in the calculation, that wouldn't explain 50% in a ten degree differential but the "about 10" part may explain a bit as well.  

I set the flow rate on the TDC3 down until the output seemed about right for the day but  I didn't allow enough for heat losses in the system.  The only thing I noticed was that low output seems a bit understated and the reverse for high output.  This could be explained by higher cooling rate when the cylinder is hotter.     

[kristen]

"the "about 10" part may explain a bit as well"

Sorry, I didn't mean the differential was vague, rather that my reading was when the dT was around 10, rather than when it was around 20

I have found the total kWh per day ties reasonably well with my own calculations (my figures are in another thread), so I'm happy to trust it, I just can't get the exported Ethernet data to tally with the controller's instantaneous nor daily elapsed values

[kristen]

"The relationship between flowrate, temperature rise and power input is :-
 flow rate(l/min) = 14*power input (kW)/temp rise (degree C).

This gives 5kW"

Right, I'll have another go

 flow rate(l/min) = 14*power input (kW)/temp rise (degree C).

therefore:

 power input (kW) =  flow rate(l/min) * temp rise (degree C) / 14

I've just had 20 minutes of pump run:

Flow 68C Return 53C = 15C. Pump at 6.95 L/min.   Power = 7.3 kW

As the dT falls e.g. Flow 65C Return 53C = 12C it drops to 5.2kW

That's what I've been seeing on the TDC screen, so "feels right"

OK ... so then I have taken each 10 second slot and calculated the kWh

10 seconds is 1/360th of an hour, so 7.3kW for 10 seconds is 7.3 / 360 = 0.020277 kWh

Done that, totalled up all the ones when the pump was on ... and it came to 6.29 kWh

TDC reports 21kWh for the day, my calculation based on tank temperature rise, estimate standing losses, and metered hot water used is 23.76kWh



I am stuck now

[Brandon]

am I correct in thinking that the formula is:

kW = (flowrate[l/min]*deltaT[^oC])/14

  ??

[hiccup]

Hi

Strangely enough I've been wondering the same thing.

My TDC4 gives me the flow and return temps and a flow rate.

By the way, the number from the TDC4 for flow appears to be 60 times the actual flow rate in l/min displayed on the lcd so its l/hr. (Dumb mistake corrected)

When I put some numbers into this formula

power input (kW) =  flow rate(l/min) * temp rise (degree C) / 14

I get nothing like the displayed kW values.

e.g.

Flow 75C
Return 66.2C
Flow rate 3.68l/min
Displayed Power 2118W

Calculated Power 2.313kW

The calculated value always comes out higher than the displayed value.

So does anyone know how the TDC4 calculates the power?

Hic!

[KLD]

Brandon's formula is a useful rule-of-thumb approximation. And for that, 2313W and 2118W are close enough. The real thing is:

Power = Energy per time = Energy per mass and per temperature difference * mass per volume * change of volume per time * temperature difference

Energy per mass and per dT is the specific heat of your solar fluid, for water that's about 4200 J/kg K, with rising glycol concentrations this value drops.
Mass per Volume is density, for water that 1000 kg/m³. again, the glycol concentration changes this.
Change of volume per time is your flow rate, 1 m³/s = 1000L/(1/60)min =60000 L/min
temperature difference is the dT you measure, for instance between flow and return pipe.

So, for pure water you'd have

4200(J/kg K)*1000(kg/L)*3.68(L/min)/60000{(m³/s) / (L/min)}*8.8K  =  2267W

the remaining difference is due to the glycol.

Klaus

[hiccup]

Wow

Super response - I'd have googled for hours to find that. Many many thanks Klaus.

So, I just need to record a few more sets of readings to work out the effective fudge factor.

I see from your explanation the "14" works out as 14.286 for pure water, more (EDIT: not less) with glycol so that tallies quite well. 14.2857*1.034 = 14.77 @40%glycol

Hic!

[StBarnabas]

am I correct in thinking that the formula is:

kW = (flowrate[l/min]*deltaT[^oC])/14

  ??

Hi Brandon I get a constant which is slightly larger for pure water assuming a specific heat capacity of 4.2 J/[^oC]/cc.

1/(60e3*4.2/3.6e6)

ans =

   14.2857
Most additives will drop this. Sea water for example has a capacity of  3.8 J/[^oC]/cc.  I will probably get a reprimand for not using pure SI!

Edit Klaus has answered this a bit more thoroughly. Either the Navitron server or my connection is not good tonight - it seems very slow.

[EccentricAnomaly]

So, do you tell the TDC the proportion of glycol you have in your loop?

[KLD]

Google found these here, http://www.engineeringtoolbox.com/propylene-glycol-d_363.html
That should help you determine the glycol fudge factor.

Klaus

[KLD]

EA,
Yes, you do, in the heat metering section.

Klaus