Infer radiator spec and system volume using the MyHeatpump app

I’ve added an experimental feature to the MyHeatpump app to try and infer radiator spec and system volume.

This can be found at the bottom of the power view, DETAIL section:

How to use it

Indoor temperature sensor & feed required: These calculations require indoor temperature data to work.

1. Select a period with continuous stable output, flow and return temps need to be stable and not increasing during this period e.g 17:20 to 17:30 in the example below:

selected section:

Emitter spec result: 15.6 kW @ DT50 (My heat loss calculation suggests it should be 15.0 kW, so not bad! considering there’s probably a few hundred watts emitted from pipes etc). Not expecting super accuracy here, this is very much ballpark figure. The accuracy is impaired by needing to know an accurate picture of average indoor temperature, a single sensor in the living room is not necessarily representative of the whole house!

DT50 here refers to the standard radiator datasheet output at 70C mean water temperature and 20C room temperature, a difference in temperature of 50K. I know that does not relate to heat pump temperatures but it’s the input required for the radiator output equation https://docs.openenergymonitor.org/heatpumps/basics.html#radiator-output.

1. Click on ‘Use for volume calc’ to save this emitter output value. Zoom out and find a heat pump cycle where the system is warming up (a portion of the heat is going to heating the system volume):

Zoom in to select a part of the cycle where DT is fairly uniform:

During this example period the average flow temperature was 30.3C and return temp 23.9C a mean water temperature of 27.1C. The average heat output according to the heat meter was 4.9KW.

My radiator system cannot output anywhere near 4.9 kW at 27.1C, the actual heat output to the room should be 15.6×((27.1-20)÷50)^1.3 = 1.23 kW. This means that 3.7 kW is going to heating the water, copper and steel in the system.

The 3.7 kW increases the system water temperature by 4.3K over ~10mins

``````system volume = heat input (Joules) / (4150 J/K.L x 4.3K)
(3700×60×10.1)÷(4150×4.3) = ~125L
``````

We can use this value to work out the effective system volume, which in this case the calculation suggests is about 125 Litres… Now my own calculation suggests that my system volume should be closer to 73L, not including the volume in the heat pump heat exchanger itself or the heat capacity of the copper pipe and steel radiators.

Given the wide spread of system volume values that come out of this (for mine 100-134L is fairly typical), it’s not super accurate, it’s more of a rough indication.

2 Likes

Here’s an example result from Adam’s (HeatGeek) system which has much more system volume than mine (large cast iron radiators):

and system volume:

2 Likes

Excellent feature!

“Effective volume” might be a better label - you’re inferring how much water is circulating NOT how much water there is in the system

2 Likes

Thanks @marko yes that makes sense.

Very interesting, Trystan. I’ve been playing with this all week and it seems to match my system quite well.

I calculate I have a total radiator output of 16,655 W, and the app gives me figures around there.
Some of my radiators have fans fitted to them, and I estimate they increase output by 25% on those radiators, taking the total output to 18,000 W.

This morning started with 2 of the radiator fans not running (17,000W), and switched on at 6am. You can maybe see the effect they had on this heating cycle, as the flow temperatures drop in the second half when the fans turn on.

In the flat bit before 6am, 16.8 kW @ DT50. After the fans turn on the reported size is 18 kW.

The indoor temp I’m been using is pretty coarse though; I have an alternate feed I might try that gives average house temp, see if that give better numbers.

The estimated volume seems to be about 130 litres. I have no idea if that’s right.

1 Like

Thanks Tim, that’s great to hear!

Yes definitely take the system volume figure with a pinch of salt for now. I think it may need very specific conditions to be semi accurate. E.g if there is very little temperature rise later in a cycle it can suggest wildly large volumes.

I did measurements again using the average house temperature, which is about 1 C higher, and got figures 18.2 kW and 19.8 kW for the same heating period.

1 Like

It would be interesting to see how well the standard radiator output equation matches the data:

``````Delta_T = Mean water temperature (MWT) - Room temperature
Heat_output = Rated_Heat_Output x (Delta_T / Rated_Delta_T) ^ 1.3
``````

I’d like to build a test setup in the office here at some point with a radiator, an inline variable output electrical heating element, say 200-1000W (to give an accurate known heat input). A heat meter for heat meter accuracy validation. Pressure sensors, variable speed pump. option to add pressure drop, option to add a volumiser. Option to turn fans on on the radiator and so on.

Would make for an easier way to validate some of these things!

2 Likes

Tried this from this morning data, after completing the balancing/opening up of rads the other day.

Inferred emitter spec: 18.5kW @ DT50
Inferred system volume: 98 Litres

That’s amazing, as I calculated rads and pipework to come to approx 95 Litres!!

Rad sizes I can be sure of, but the pipework isn’t always going to be to the exact metre.

1 Like

I’ve been working the other way: adjusting the heating parameters to match the equations:

I wish it was that accurate on yours I was just trying to get reasonable values earlier and was getting numbers all over the shop

Where’s the fun in making it took easy.

It doesn’t matter what the theory is. You have the actual “effective volume” and “effective emitter” outputs. Discrepancy is mostly imbalance.