Heat loss calculation check

[SBMS]

We’ve had some heat loss calcs back from the ASHP supplier. I just wanted to check the maths. Given the following for the heat loss through a 1.0 Uw window:

 

the window area is 32.9m2 and the resultant heat loss is 881w. Am I right in thinking therefore that this is calculating for an outside temperature of about -5.8 degrees (delta t of 26.7 degrees)? Not sure I understand the maths of the above regarding the heat loss through the wall/window and ceiling…

[Alan Ambrose]

You could double check by putting similar assumptions into Jeremy’s spreadsheet?

[SBMS]

7 minutes ago, Alan Ambrose said:

You could double check by putting similar assumptions into Jeremy’s spreadsheet?

I’ve done this already. The MCS calcs are miles out because they can’t account for MVHR. However I’m unclear what formula they are applying. I can make the heat loss work for the window but for the wall or floor I don’t arrive at the same value for heat loss… 

[Gordo]

32.9m2 x 1w/m2/k x 27k = 888wh sounds right

[SteamyTea]

7 hours ago, SBMS said:

Not sure I understand the maths of the above regarding the heat loss through the wall/window and ceiling

You do understand it.  It is the U-Value [W.m^-2.K^-1 or W/m².K (not W/m²/K as above)] multiples by the surface area [m²] multiplied by the temperature difference [ΔK].

 

So with a U-Value of 1 [W.m^-2.K^-1], a surface area of 32.9 [m²], the unknown ΔK, but a solution of 881 W.

 

881 [W] / 1 [W.m^-2.K^-1] x 32.9 [m²] = 26.8 [ΔK]

 

To work out the assumed outside air temperature [OAT] you just subtract the external temperature from the internal temperature [IAT] i.e. 20°C.  20 [°C] - 26.8 [°C] = -6.8 [°C or K].

 

Now that is the power (the W) not the energy [Wh] losses over a set time period when the OAT is -6.8°C i.e. a cold winter day.

To calculate the energy lost, you have to multiply by the time in hours [h] to get to Wh, which is how we purchase energy (we purchase in kWh, so need to divide by 1000).

 

Now we do not have fixed temperature differences, they vary over the year i.e. large in winter, small in spring and autumn, irrelevant in summer (they they can be positive as well, we call that solar gain). Temperature probabilities, based on a typical metrological year [TMY] are used to calculate the amount of time spent at different OATs.

 

Now you may have noticed that I have mixed descriptors for temperature, sometimes I have used celsius [°C] and other time kelvin [K].  This is considered very bad practice but is often used to make it easier to read.

There is a reason that it is bad practice and that is if you are adding or subtraction temperatures then the scale name makes no difference as the size of the graduation is the same.  But if you are multiplying or dividing, then you must use the kelvin scale or you get odd results.  This is very important when working out radiative losses and the efficiency of CoP of heat pumps.

To convert celsius to kelvin, you add 273.15.  So -6.8°C is 266.35 [K].

(We should really use the joule [J] for energy and not the Wh, but that is another lecture)

 

 

[MikeSharp01]

9 hours ago, SBMS said:

about -5.8 degrees (delta t of 26.7 degrees)?

-5.8 feels like a low number, where are you installing? 

[joth]

Agree -5°C seems low for a design parameter, but also I can't see how it's consistent with other energy loss rates from the same table. 

E.g. for the external walls we have 

99÷(0.15×20.5×2.3) = 14 K delta T. Which yields is 7°C OAT

 

What gives?

[SBMS]

2 hours ago, joth said:

Agree -5°C seems low for a design parameter, but also I can't see how it's consistent with other energy loss rates from the same table. 

E.g. for the external walls we have 

99÷(0.15×20.5×2.3) = 14 K delta T. Which yields is 7°C OAT

 

What gives?

Completely agree - this is what I'm trying to find out from the supplier as it doesn't seem to work.

[JohnMo]

5 hours ago, SBMS said:

Completely agree

There are advantages for you if the supplier uses a lower design temp of -5 or lower.

 

First if they set the flow temp at 45 for -5, in reality you have a system that will most likely work well at 40(ish) at -3, so get a better CoP. When the heat pump is running at -3 you will be in defrost mode so loosing approx 10% of output, so don't fret too much.

 

All depends where you live for design temp - MCS are wanting the system designed for 99.6% likely lowest temp.  For us that would be -9.

[MikeSharp01]

1 hour ago, JohnMo said:

For us that would be -9.

Yes but you live in the frozen North! 

 

The whole issue here is that the numbers don't add up / work out makes you wonder how good the spreadsheet they are using is.

[SteamyTea]

I didn't look at the walls this morning, so here is my take on it.

 

Total wall area is 20.5 [m] x 2.3 [m] = 47.15 m²

Minus the window area of 32.9 [m²] gives 14.25 [m²]

 

Wall U-Value is 0.15 [W.m^-2.K^-1]

Power loss though wall is 99W

 

0.15 [W.m^-2.K^-1] x 14.25 [m²] =  2.1 [W.K^-1]

99 [W] / 2.1 [W.K^-1] = 47 [ΔT]

 

Something is wrong.  Is it really only 14.25 m² of wall?

[SBMS]

37 minutes ago, SteamyTea said:

I didn't look at the walls this morning, so here is my take on it.

 

Total wall area is 20.5 [m] x 2.3 [m] = 47.15 m²

Minus the window area of 32.9 [m²] gives 14.25 [m²]

 

Wall U-Value is 0.15 [W.m^-2.K^-1]

Power loss though wall is 99W

 

0.15 [W.m^-2.K^-1] x 14.25 [m²] =  2.1 [W.K^-1]

99 [W] / 2.1 [W.K^-1] = 47 [ΔT]

 

Something is wrong.  Is it really only 14.25 m² of wall?

It is a small section in my example because two of the walls are internal (it’s a corner room). 
 

The maths still doesn’t make sense. I’ve asked the question but think the supplier has finished for Xmas so won’t get an answer before. 
 

I wonder if there is some sort of magic number for thermal bridging that increases the heat loss?

 

Also - Does the ventilation heat loss look correct? The room is 181m3 and they’ve said the ACH to 0.5. I did mention that we are having MVHR but I don’t think the ventilation heat loss of 698w takes any heat recovery into account? How would I calculate the ventilation heat loss at 0.5ACH (without MVHR and then with MVHR at 96% efficiency - zehnder q600)?

[SBMS]

4 minutes ago, SBMS said:

It is a small section in my example because two of the walls are internal (it’s a corner room). 
 

The maths still doesn’t make sense. I’ve asked the question but think the supplier has finished for Xmas so won’t get an answer before. 
 

I wonder if there is some sort of magic number for thermal bridging that increases the heat loss?

 

Also - Does the ventilation heat loss look correct? The room is 181m3 and they’ve said the ACH to 0.5. I did mention that we are having MVHR but I don’t think the ventilation heat loss of 698w takes any heat recovery into account? How would I calculate the ventilation heat loss at 0.5ACH (without MVHR and then with MVHR at 96% efficiency - zehnder q600)?

I asked ChatGPT and it stated 695W so sounds like that part of the heat loss is correct?  Am I right in assuming this drops to something like 28W with MVHR?

Edited Friday at 17:41 by SBMS

[SteamyTea]

15 minutes ago, SBMS said:

I wonder if there is some sort of magic number for thermal bridging that increases the heat loss

There is the PSI number for linear metre of exposed wall, plus the air film numbers for the wall area. Off the top of my head I don't know them, but think there is something in the Building Regs about it.

 

MVHR is additional loss to normal ventilation losses. Think of it as a bathroom with the window open (normal ventilation losses) then turn on the extractor fan, it does not make the window close. So even with heat recovery, it still increases the overall losses.

Why it is vitally important to get the airtightness detail right.

 

I am out now, but to calculate the air losses at 0.5 ACH, you multiply the volume by the air density, 1.25 kg.m^-3, then multiply by the ACH, that will give you about 115 kg of air, then multiply by the specific heat capacity if air, 1 kJ.kg^-1.K^-1 and finally by the ∆T, should bring you to about 3 MJ.h^-1.

That is around 850 W, which seems high, so I may have made a mistake somewhere (no pencil and paper to write the numbers down).

[SteamyTea]

Just now, MikeSharp01 said:

might bring it down to 70W

I am not sure if that is right. See my comment above.

 

I think the problem is that the air test, is a pressure test, so not the same as an ambient pressure test, but if, unpressured, a house replaced half the air every hour, then MVHR adds to the losses.

 

There are probably many old houses that loose that much air to 'natural' ventilation.

[MikeSharp01]

17 minutes ago, SteamyTea said:

I am not sure if that is right. See my comment above

No neither was I which is why I deleted my post - you must be having a very fun packed evening if you sample BH so often you picked it up in the 2 minutes it was on the site. Just enjoy your evening and we will just imagine the fun you are having.

[SteamyTea]

1 minute ago, MikeSharp01 said:

Just enjoy your evening and we will just imagine the fun you are having

I am going to see a play, can't even remember what it is called.

Been a very busy week at work with Christmas parties (85 people today, plus the walk ins), so unless I can park within 100m if the place, I will not bother to go, the local Am Dram gave my tenner anyway.

[SteamyTea]

Well I made it. Parked about 200m from the Newlyn tide height gauge. So nearly all of you are higher than me at the moment.

 

 

[DamonHD]

You could smoke something to fix that, or we could...

[joth]

1 hour ago, SteamyTea said:

That is around 850 W, which seems high, so I may have made a mistake somewhere (no pencil and paper to write the numbers down).

Your numbers look right. MCS uses an "Air Change Factor (W/m³K)" of 0.33 which is basically your 1.25 kg.m-3  *  1 kJ.kg-1.K-1  in watts.

Then multiplies that be (volume x delta-T x ACH).

 

The empty MCS calc can be downloaded here if you'd like to play more 😄

https://mcscertified.com/wp-content/uploads/2024/12/MCS-Heat-Pump-Calculator-Version-1.10-unlocked.xlsm

 

Basically, it's a very big room.

[JohnMo]

22 hours ago, SBMS said:

MCS calcs are miles out because they can’t account for MVHR

They can if you give them an air test certificate. They don't unless you provide a certificate and insist.

[SBMS]

2 hours ago, SteamyTea said:

then MVHR adds to the losses.

Surely only a tiny amount? If my house’s ACH50 is around 0.5, then the normalised infiltration rate I’ve seen used is 1/20 of this so 1/20 of the 698W is lost to ventilation. So the actual ACH is nowhere near 0.5 and the uncontrolled ventilation loss is around 35W.

 

I then add my MVHR which, let’s say I run at ACH 0.5 but it’s recovering 96% of that heat loss. So yes it adds to the ventilation losses but only about 4% of that 698W figure (28W).

 

so the actual losses to ventilation would be nearer 63W?

 

Does this sound right?

Edited Friday at 20:23 by SBMS

[SteamyTea]

10 minutes ago, SBMS said:

this sound right

Yes it does.

If you were loosing nearly 24 kWh a day, the heating bill would be horrible.

 

Half time on the play ATM.

[SBMS]

1 minute ago, SteamyTea said:

Yes it does.

If you were loosing nearly 24 kWh a day, the heating bill would be horrible.

 

Half time on the play ATM.

Thanks @SteamyTea. So is it fair to say the heat loss calculation done by the MSC supplier is horribly going to oversize the heatpump?

 

I imagine that I can adjust the ACH figures for natural ventilation once I have a blower test done pre ordering. We’ll target 1ACH50 - Obviously at 0.05 ACH I would die, hence I will use MVHR to make sure we don’t asphyxiate. I’ll probably run the MVHR at, say 0.3 to 0.5 which will push the ventilation rate back up to the MCS’ ventilation rate. 
 

My question then is whether MCS heat loss allows them to accommodate the heat recovery capability of the MVHR?

[SBMS]

49 minutes ago, JohnMo said:

They can if you give them an air test certificate. They don't unless you provide a certificate and insist.

Presumably this allows me to prove my natural/uncontrolled ventilation - but what about taking into account the MvHR controlled ventilation rate - and its heat recovery capability? Can MCS factor this in?