Hi,
I wondered if anyone would like to help me check my calculations?

We have an ASHP that runs 24/7 at 45 litres per second. The average yearly outdoor temp for where we live is 8 degrees C. We like to have a room temperature of 21 degrees C.

When in my (swedish) college, to calculate the energy requirement to heat intake air, we used this formula

P [kW] = q [m3/s] x R [kg/m3] x delta T [degrees C]

where:

R = 1.2
delta T = indoor temp - average outdoor temp

We'd then multiply the result by 8766 hours to get the (approximate) yearly requirement in kWh.

So for my house, that would give:

P = 0.045 x 1.2 x (21 - 8)
= 0.702 kW

0.702 kW x 8766 hours = 6154 kWh.

Sounds reasonable?

Thanks Ed :)

I see that once again I've misremembered the English acronyms - we have an EAHP, not a ASHP, although I guess that exhaust air is an air source ... Anyway, yes, we have heat recovery.

Ah yes, 1 kJ/(kg·K) that was in the original lecture notes, but the teachers had removed it from the revision sheets. Presumably as we were using 1 kJ ...

The 45 l/s comes from the fan setting on the heatpump - we can choose between 20 l/s up to around 70 or 80 l/s. Our house is 115 m2, but 45 l/s seems to work for the four of us, - it's a compromise, as we cannot vary the flow - 45 l/s clears the bathroom reasonably quickly, and we get ok heat recovery; at higher flows we get rather "dried out" eyes, noses, etc.

I'm doing the calculations as our house (mid 1980s) originally had MVHR (and presumably an electric heater for heating and hot water); I'm contemplating reinstalling MVHR, so this helps with dimensioning (?) any unit we may buy. I agree that 0.45 AC/h is high, so would hopefully get something that senses the bathroom and kitchen humidity (maybe even VOCs)

Need to do heat loss through the building now ... :)