ᐅ Heat Pump Calculation Formula Based on Heating Load

The definition of the U-value assumes a steady state condition (!), meaning that temperatures on both "sides" are constant. In reality, of course, wind makes a difference. However, for all building components that are reasonably insulated (double or triple-glazed windows, walls with a U-value around 0.3 or better), the external heat transfer coefficient has such a small impact that the variation in total heat transfer between calm conditions and strong wind is only within a low percentage range. This is generally sufficient for a rough estimate of whether the heating system is operating properly.

Convection (I meant: ventilation system) is quite easy to include, for example, when calculating total energy use simply by multiplying airflow rate * temperature difference * specific heat capacity of air.

Saruss schrieb:
The definition of the U-value requires a steady state (!), meaning constant temperatures on both "sides".

I am not entirely confident with the strict U-value definition in building services engineering. Generally, there are three mechanisms: heat transfer from the room to the wall surface, heat conduction through the wall (simplest case: a solid wall), and heat transfer from the outside wall surface to the surrounding air. The last one strongly depends on whether the air is still or moving.

Saruss schrieb:
Convection (I said: ventilation system) is quite easy to include, for example, for total energy use simply airflow rate * temperature difference * heat capacity of air

You just need to know the airflow rate ;-).

Oh, and don’t forget internal heat gains (people, electrical devices).

The airflow rate in modern controlled residential ventilation systems is very easy to determine. In the best case, it has even been individually adjusted for each room to suit the house.

In theory, you are correct that heat transfer depends on the airflow, but in the case of very poor thermal conductivity, which applies here to the walls, windows, and roof, this is actually irrelevant. So little energy passes through the material that the surface effect makes no difference; the exterior surface essentially has the outside temperature regardless of how much wind is blowing.

Only with poor insulation would it happen that the exterior side of the "wall" heats up due to heat conduction and then, naturally, more energy is carried away with higher airflow or wind than with less wind.

To calculate the U-value,
difference = Outside -12°C (10°F) – Inside 20°C (68°F) = 32

Walls = envelope area 389 * 0.2 (aerated concrete 42.5) + surcharge 0.05 * 32 (difference) = 3111.2
Ceiling = 90 * 0.14 + 0.04 * 32 = 547.2
Floor = same as ceiling = 547.2
Windows = 26.6 * 0.6 + 0.05 * 32 = 553.28
Total 4760 W = 4.76 kW

Now I would need to know how often the heating system needs to turn on and then divide by the COP, right?

PS.
The problem was found: the heating system was not set up correctly. Every time the utility lockout occurred, the outdoor unit was switched off, and the heating system thought it was frozen. As a result, the frost protection activated twice a day for 1.5 hours each.
Now the consumption is back to about 15 kWh.

Saruss schrieb:
Well, the airflow rate in modern controlled residential ventilation systems is quite easy to determine.

But people do open a door or a window sometimes. And a bit of uncontrolled air infiltration exists as well (blower door test).

@Bitknight: The values already look reasonable for a house like this. Ventilation losses in my case (for a larger house, so more volume) are about 1800W at -14°C (7°F) during normal operation of mechanical ventilation with heat recovery (MVHR), but this is a pessimistic estimate assuming low heat exchanger efficiency. You might be a bit below that.
That would translate to roughly 6kW heating power at -12°C (10°F) for your house. It won’t be -12°C (10°F) all day long (it’s best to calculate with the average daily temperature difference; I monitor and calculate this automatically with sensors and have found over the last 4 years that the heating demand is exactly proportional to this value!). If it were constant, that would mean about 24 × 6 = 144 kWh heating energy needed per day. This amount, divided by an average coefficient of performance, is the electrical energy consumption at that temperature.
If the average difference (daytime/partly warmer than -12°C (10°F), maybe around -6°C (21°F) on average?) is smaller, for example instead of your 32 it’s only 26, then it would be about 117 kWh (at 0°C (32°F) around 92 kWh). This shows that the actual consumption reflects both the pessimistic nature of our calculation and values (I have no idea how your MVHR system or ventilation strategy performs, since that share is not insignificant) and also a decent coefficient of performance.

@Bieber0815
Compared to the 150 m³ (about 5300 ft³) or more that a mechanical ventilation with heat recovery system moves every hour (continuously!), occasionally going through the door is not much. And windows? I don’t know who your “people” are, but personally, in recent years I have only opened windows for a few minutes in cases of strong odors (which can happen with kids, pets, or cooking...), otherwise they stay closed with no forced ventilation needed. Actually, this is no problem even in the bedrooms.
By the way, infiltration air is also not a big issue, because with today’s required values for the blower door test, the exchange rate due to infiltration is quite low (also consider the differential pressure at which the test is performed and valid—I was there during mine, and you really notice it clearly in your ears when ramping up and down during the tests). Moreover, this value can also be detected by the mechanical ventilation with heat recovery system.