Calculation of thermal output???

[Dogsbody]

I've just had an air to water heat pump installed at my house, and I wanted to get a rough idea of what it's performance is. Can anyone help with a reasonably simple way to calculate the real world COP as a cross-check against the manufacturers BS?

I appreciate that COP is a crude measure, that it's only a snapshot, and that the average across the whole season is more useful etc etc, but indulge me for the sake of my curiosity.

The heat pump control tells me the inlet and outlet water temperatures, the Wilo circulation pump shows the water flow rate, and I have the electricity consumption metered separately. What would the formula be, and do I need any other measurements? Anyone?

 

[doitmyself]

The heat pump control tells me the inlet and outlet water temperatures, the Wilo circulation pump shows the water flow rate, and I have the electricity consumption metered separately. What would the formula be, and do I need any other measurements? Anyone?

The heat output is calculated using the formula: kW = flow x ΔT x 4.18

Flow is measured in litres/sec
ΔT is the difference between inlet and outlet temperature.

e.g

flow =0.5 l/sec
ΔT = 10°C

kW = 0.5 x 10 x 4.18 = 20.9kW

If the system produces this for one hour then 20.9 kWh will have been produced.

As well as knowing the electricity consumption you also need the time over which it was consumed. Only the time when the system is running is relevant.

 

[Dogsbody]

doitmyself: Thanks for taking the time to help. So, the 4.18 is the specific heat of water I think? More importantly, can you expand a bit on your last point? How does the time get plugged into the equation? Come to that, exactly what is the equation? And am I right in thinking that COP is a snapshot taken at a point in time?

 

[Nostrum]

Which Wilo pump is it? Are you sure it's displaying flow and not power consumption or metres head?

Also, which heat pumps have you had installed?

 

[Dogsbody]

It's the Stratos Pico circulation pump, and yes, it definitely shows flow rate. I'd like to keep to myself the heat pump make for the time being as I don't want the debate to get sidetracked by pros and cons of different brands. I don't think it's relevant to my original question is it?

 

[JonG1]

Might be easier to get a heat meter installed along with an electricity meter to calculate seasonal COP that would be more useful for you.

 

[Dogsbody]

Might be easier to get a heat meter installed along with an electricity meter to calculate seasonal COP that would be more useful for you.

Cheers. Yeah, I may do that eventually, but at the moment I just want a quick fix as a starting point.

 

[doitmyself]

So, the 4.18 is the specific heat of water I think? More importantly, can you expand a bit on your last point? How does the time get plugged into the equation? Come to that, exactly what is the equation?

You are right about 4.18.

As for time, remember that one watt is actually one Joule per second. So one kWh is therefore 3.6 Megajoules (1000 x 60 x 60).

If the system in my previous example runs for one minute it will produce 20.9 x 60 = 1254 Joules of energy. Let's say the electricity consumption over the minute is 0.1 kWh = 0.1 X 60 x 60 = 360 Joules. Then the COP is 1254/360 = 3.48.

I have no idea about the COP part of your question, but a snapshot seems the most likely method. It would be similar to the way a gas boiler is rated by measuring the gas consumption over two minutes. Measuring over a long time would be difficult as the temperatures would not be constant.

 

[doitmyself]

If the system in my previous example runs for one minute it will produce 20.9 x 60 = 1254 Joules of energy. Let's say the electricity consumption over the minute is 0.1 kWh = 0.1 X 60 x 60 = 360 Joules. Then the COP is 1254/360 = 3.48.

Sorry, but that's not correct. I forgot the kilo prefix. The heat produced is 1254 kJ and electricity used is 360 kJ. The COP doesn't change.

 

[Dogsbody]

......As for time, remember that one watt is actually one Joule per second. So one kWh is therefore 3.6 Megajoules (1000 x 60 x 60).

If the system in my previous example runs for one minute it will produce 20.9 x 60 = 1254 Joules of energy. Let's say the electricity consumption over the minute is 0.1 kWh = 0.1 X 60 x 60 = 360 Joules. Then the COP is 1254/360 = 3.48.

OK, I'm struggling a bit with the maths.... My figures are as follows:

Flow rate: 66.66 litres per minute
DeltaT: 14 C
Energy produced = 66.66 x 14 x 4.18 = 3900 Joules
Electricity consumed: .067 kW per minute x 60 x 60 = 241 Joules
COP = 3900/241 = 16.2

A COP of 16.2 seems a little unlikely. Can you see what I am doing wrong? I'm sure it's really basic. I think all the 60s are confusing me. (Didn't see your second post, but as you say, it doesn't change the COP)

 

[joni os]

Joules confuse me, but 66.66 X 14 X 4 X 60 equals 224,000 BTU/hr Approx 67 Kw. This seems high. I would seek confirmation of flow rate or temp. diff.

 

[Nostrum]

Your DT is way to high which suggests a flow rate issue. The reason I asked about which heat pump you have is to understand its capabilities and whether or not it was an inverter driven compressor etc, not because I wanted to slag it of or suggest a different model.

When did you take the DT and under what conditions? You need to allow the heat pump to run for long enough for it to adjust to whatever load you're placing it under.

 

[Dogsbody]

Joules confuse me, but 66.66 X 14 X 4 X 60 equals 224,000 BTU/hr Approx 67 Kw. This seems high. I would seek confirmation of flow rate or temp. diff.

Now I'm worried. The temp readings are on the heat pump control panel, reading off sensors inside. The numbers they give seem to make sense when compared with the operation of the thermostat on the buffer tank, and with the return temperature off the UFH as given by the manifold thermometers. Could be wrong, but they seem to stack up. The flow rate is off the Wilo circulation pump. This has an LCD screen that gives electricity consumption and flow rate amongst other things. I guess I'd assumed this would be reasonable reliable, given Wilo's reputation. Probably naive?

 

[Dogsbody]

Your DT is way to high which suggests a flow rate issue. The reason I asked about which heat pump you have is to understand its capabilities and whether or not it was an inverter driven compressor etc, not because I wanted to slag it of or suggest a different model.

When did you take the DT and under what conditions? You need to allow the heat pump to run for long enough for it to adjust to whatever load you're placing it under.

It's not inverter driven. I wasn't thinking you were about to slag it off, but trust me that if I do mention the brand, there will be a whole chorus of jeering, telling me it's rubbish. I'm basing this on posts in other threads. Don't think it matters for now, I'll share later if people are interested.

The DT was even higher till I cranked up the flow rate!! The Wilo Stratos Pico has adjustable flow rate, and the screen readout tells me I'm at maximum at 4000 litres/hour.

I've been letting the heat pump run and stabilise, although to be honest, it makes little difference. Seems to start out at a DT of 14 and stay there with outlet temperature gradually rising from 43 to 48 when the thermostat cuts out. I mentioned in another post, that I'm running the heat pump into a buffer tank. Would that result in the stable DT? Why do you think it's high?

 

[joni os]

Unfamiliar with pump, but does the 4000 litre/hr. refer to theoretical flow against zero head resistance.

 

[Dogsbody]

Unfamiliar with pump, but does the 4000 litre/hr. refer to theoretical flow against zero head resistance.

Good question. No idea. Would have thought it was a bit odd to display theoretical flow rate on a changing display that can be adjusted by user? Dunno.

 

[joni os]

4000 seemed too round a figure for actual flow rate. Since posting I have goggled pump data and flow drops off considerably with head.

 

[Nostrum]

The heat pump does matter as the controller is what makes a lot of difference with a heat pump.

The basic principles aren't all that different for most heat pumps, maybe different refrigerant etc

How the heat pump output is controlled is another matter. I'm guessing it's a Dream or similar and this is a self fit / non RHI type install?

What size and type of pipework is used from the outdoor unit to the buffer and over what distance?

How is the buffer connected to the heat pump and system, is it in series/parallel, 2 pipe, 4 pipe etc etc

I'm afraid if you want help or answers you'll have to start disclosing more information.

 

[doitmyself]

OK, I'm struggling a bit with the maths.... My figures are as follows:

Flow rate: 66.66 litres per minute

If, as you say in a later post, the flow is 4m³/hr that is 66.66 litres/min. However, my formula needs the flow in litres/sec so the output is 66.66/60 x 14 x 4.18 = 65kW or 65 J/sec. The output over one minute is therefore 3900 Joules. (I know it's the same as you got but, as one of my maths teachers used to say, "right answer, wrong method!")

Your calculations are correct, but a COP of over 16 seems unlikely. You mentioned another post about a buffer tank, but I can't find it. A link would be useful.

 

[Dogsbody]

4000 seemed too round a figure for actual flow rate. Since posting I have goggled pump data and flow drops off considerably with head.

OK, makes sense. I think there is a setting on the pump to tell it what head you are working with. Can't check now as weather is a bit dodgy. Will look tomorrow. I think it was set properly, but as you say, the 4000 sounds a bit nominal....

However, if I can get an accurate figure for flow, I still need confirmation that I am doing the sums right. Any help on that would be much appreciated.

For example, if I halve the flow rate, (seems extreme), I still get a COP of around 8, I think. Something still wrong?

 

[Nostrum]

It's not inverter driven. I wasn't thinking you were about to slag it off, but trust me that if I do mention the brand, there will be a whole chorus of jeering, telling me it's rubbish. I'm basing this on posts in other threads. Don't think it matters for now, I'll share later if people are interested.

The DT was even higher till I cranked up the flow rate!! The Wilo Stratos Pico has adjustable flow rate, and the screen readout tells me I'm at maximum at 4000 litres/hour.

I've been letting the heat pump run and stabilise, although to be honest, it makes little difference. Seems to start out at a DT of 14 and stay there with outlet temperature gradually rising from 43 to 48 when the thermostat cuts out. I mentioned in another post, that I'm running the heat pump into a buffer tank. Would that result in the stable DT? Why do you think it's high?

Do you have a pump valve you can throttle down to see if it effects the displayed flow rate on the display?

What size pump is it you've installed? Looking at the info, my guess would be a 1-6 which at max would give you a theoretical flow rate of about 4m3 p/hr, or 4000 litres...

 

[Dogsbody]

Do you have a pump valve you can throttle down to see if it effects the displayed flow rate on the display?

What size pump is it you've installed?

Well, as already mentioned, there is an adjustment on the pump to reduce flow rate. When I knock that down, the pump goes quieter and the readout shows a lower electricity consumption, so I guess it's working. The pump is the Wilo Stratos Pico 30/1-4.

But as already touched on, if I reduce flow rate, the DT goes even higher.

 

[joni os]

The maths that arrive at 65Kw is agreed, from data based on flow and temp.
Of the two, temp. is easily verified, even by touch. Determination of flow will resolve the issue.

 

[Nostrum]

Well, as already mentioned, there is an adjustment on the pump to reduce flow rate. When I knock that down, the pump goes quieter and the readout shows a lower electricity consumption, so I guess it's working. The pump is the Wilo Stratos Pico 30/1-4.

But as already touched on, if I reduce flow rate, the DT goes even higher.

If the flow rate was measured then external influences (such as reducing flow through a valve) would effect the displayed flow rate. Adjusting the pump on its setting won't.

According to the Wilo data, the pump you've got won't even perform at the given flow rate, so I'd take that with a pinch of salt.

I think it's also safe to say the heat pump isn't providing 65kw, or a COP of 16 for that matter.

Until you can give more info, you won't get the answers you need. I suspect you have flow issues. I'd also be very surprised if you can get anywhere near an accurate COP unless you calculate the real time flow rate, which again requires info from the manufacturer.

Good luck.

 

[Worcester]

Somethings wrong:

66 lpm is one heck of flow rate

Flow rate: 66.66 litres per minute
DeltaT: 14 C

= 65kW that is one heck of a heat load for a heat pump - in a domestic single phase environment the DNO wouldn't let you connect it...

I would expect and old leaky house of 600m2 for that load... That's one heck of sized house, and the radiators will be filling the walls

Two things sound wrong

1) Flow rate
2) delta T

We install Wilo pumps all the time, and they usually display either watts being consumed or m head.

If the system has been installed to proper standards, then installing a heat meter and electricity meter should be a very simple process. (installing them correctly requires some specific knowledge )

We do that quite often when the client really wants to know what's what.