Heat Pump Sizing?

FreeBear

michaels said: I know that although we have a 12kw heat pump our rads can not emit more than about 7.5-8kw of heat even with a 55C (max) flow temp and it is taking a while for the temperature to recover even though that output is fine to maintain the temp even with -5 outside temp as we have now.

Is that 7.5-8kW based on a ∆T50°C (pretty standard when looking at radiator specs) ?

A ∆T50°C equates to a flow temperature of 70°C, and when running at 55°C. your radiator output would drop to ~6kW. At 45°C, the output would only be 4kW..

michaels

FreeBear said:

michaels said: I know that although we have a 12kw heat pump our rads can not emit more than about 7.5-8kw of heat even with a 55C (max) flow temp and it is taking a while for the temperature to recover even though that output is fine to maintain the temp even with -5 outside temp as we have now.

Is that 7.5-8kW based on a ∆T50°C (pretty standard when looking at radiator specs) ?

A ∆T50°C equates to a flow temperature of 70°C, and when running at 55°C. your radiator output would drop to ~6kW. At 45°C, the output would only be 4kW..

No that is at deltaT30 ie a flow of about 50 and room temp of 20.
TBH the pipe size also means that I would struggle to transfer more than 7 to 8 kw at a reasonable flow rate. Plan is to upsize 9 of 16 rads to give the same heat transfer at 40C

michaels

QrizB said:

michaels said:

Random question, but how much more energy does it need to raise the temperature as opposed to maintain the temperature? 
...
So it would appear that increasing the temp takes a fair bit more power than maintaining the temp and maybe this needs to be factored in alongside the heat loss calcs?

It depends on the effective heat capacity of your house.

https://en.wikipedia.org/wiki/Heat_capacity

It might be easier to imagine a hot water storage tank (the maths is easier). A tank will have a rated heat loss. Let's assume we've got a 100-litre tank with a rated heat loss of 1.2kWh/day when filled with water at 80C. 1.2kWh/day is equivalent to a continuous power of 50 watts. So it'll only need 50 watts of heat to keep the tank at 80C once it's hot.

Now imagine we've used all the hot water and it's been refilled from the mains, and that cold mains water is only at 10C. We need to heat the water up from cold to 80C, that's an increase of 70C.

Heating a litre of water by 1C takes 4200 joules (ie. watt-seconds), or 1.167 watt-hours. We've got 100 litres to heat by 70C, so that's 70 x 100 x 4200J which is 29.4 million joules or 8.167 kWh. you need to add that much energy in addition to the continuous 50 watts (arguably you could get there with 50 watts but it would take weeks).

So you'll bump your 50 watts up by an extra kilowatt and have a new taankful of hot water after eight hours or so.

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It's not quite the same with houses; their heat loss is higher (mine's around 5kW right now, per my Octopus Mini I'm using less than 6kWh of gas an hour) and the change in temperature is smaller, but also their heat capacity is much higher. Bricks are around 800J per kg C, and a house might need 40 tonnes of bricks. That's 32MJ (almost 9kWh) per degree C, and that's just for the bricks.

Does that help at all?

Thanks. Really helpful. I wonder what the typical ratio of maintenance power to raise by 1c power is or alternatively typically how much extra power you would typically need to raise temp by saying 2c per hour 

FreeBear

michaels said: I wonder what the typical ratio of maintenance power to raise by 1c power is or alternatively typically how much extra power you would typically need to raise temp by saying 2c per hour 

I'm currently using around 4kW to heat at the rate of 1°C per hour with a flow temperature of 45°C and -2°C outside. This is in a 3 bed late 1920s semi with about 100m² floor space. But every house is different.

Reed_Richards

QrizB said:

 Let's assume we've got a 100-litre tank with a rated heat loss of 1.2kWh/day when filled with water at 80C. 1.2kWh/day is equivalent to a continuous power of 50 watts. So it'll only need 50 watts of heat to keep the tank at 80C once it's hot.

Now imagine we've used all the hot water and it's been refilled from the mains, and that cold mains water is only at 10C. We need to heat the water up from cold to 80C, that's an increase of 70C.

Heating a litre of water by 1C takes 4200 joules (ie. watt-seconds), or 1.167 watt-hours. We've got 100 litres to heat by 70C, so that's 70 x 100 x 4200J which is 29.4 million joules or 8.167 kWh. you need to add that much energy in addition to the continuous 50 watts (arguably you could get there with 50 watts but it would take weeks).

So you'll bump your 50 watts up by an extra kilowatt and have a new taankful of hot water after eight hours or so.

I'm being very picky but if you have a continuous heat loss of 50 W when the tank is at 80 C then the tank won't be losing 50 W whilst you are heating the water in it up from cold.  It's the same with your house.  If it had cooled to 0 C inside then it would be barely losing any heat and the Delta T for your radiators would be 20 C more than their specified values so their outputs would be boosted.  As you raise the inside temperature than the heat loss from the house increases whilst the output from the radiators decreases.  

 If you want to heat your house as quickly as possible, turn off weather compensation and set the flow temperature to the highest your heat pump is capable of.  This will give you the least efficient operation of your heat pump but the fastest restoration of the normal house temperature.

QrizB

Reed_Richards said:

QrizB said:

 Let's assume we've got a 100-litre tank with a rated heat loss of 1.2kWh/day when filled with water at 80C. 1.2kWh/day is equivalent to a continuous power of 50 watts. So it'll only need 50 watts of heat to keep the tank at 80C once it's hot.

Now imagine we've used all the hot water and it's been refilled from the mains, and that cold mains water is only at 10C. We need to heat the water up from cold to 80C, that's an increase of 70C.

Heating a litre of water by 1C takes 4200 joules (ie. watt-seconds), or 1.167 watt-hours. We've got 100 litres to heat by 70C, so that's 70 x 100 x 4200J which is 29.4 million joules or 8.167 kWh. you need to add that much energy in addition to the continuous 50 watts (arguably you could get there with 50 watts but it would take weeks).

So you'll bump your 50 watts up by an extra kilowatt and have a new taankful of hot water after eight hours or so.

I'm being very picky but if you have a continuous heat loss of 50 W when the tank is at 80 C then the tank won't be losing 50 W whilst you are heating the water in it up from cold.

You're absolutely correct, I just didn't want to complicate it even further!

michaels

I'm wondering if the rate of cooling with the heating off might be used as a proxy for the loss of stored heat. Obviously a bit simplistic as potentially the air (measured temp) probably cools quicker than the fabric.

However to what extent is it fair to say that if the heating is off for an hour that would have needed 6kwh to maintain inside temp then you would need an extra 6kwh on top of the normal maintain temp input to return to the original temp?

Spies

Very informative post QrizB

Should I in that case have my heatpump offset to say +4c whilst the IOG hours are in play? 

QrizB

michaels said:

I'm wondering if the rate of cooling with the heating off might be used as a proxy for the loss of stored heat. Obviously a bit simplistic as potentially the air (measured temp) probably cools quicker than the fabric.

Unless you're living in a refrigerator, or have left a door open, or if it's summer and you're trying to keep cool, the air won't be any colder than the inner surface of your house. If it was cooler, the house would be trying to warm it up and heat would be flowing in the wrong direction.

michaels said:

However to what extent is it fair to say that if the heating is off for an hour that would have needed 6kwh to maintain inside temp then you would need an extra 6kwh on top of the normal maintain temp input to return to the original temp?

For a short outage (an hour or two) that's probably a reasonable approximation.

For a longer one, as the building cools down the delta-T gets smaller and the rate of heat loss will decrease (like R_R described, and the same phenomenon you see with your radiators). So the energy needed to warm back up will be less than simply multiplying 6kWh by the time it was off.

Spies said:

Very informative post QrizB
Should I in that case have my heatpump offset to say +4c whilst the IOG hours are in play? 

Maybe? There's probably a discussion to be had about using the building structure as a storage heater, but this might not be the right thread for it

matelodave

For those who wish to get into the weeds, this bloke has been writing some interesting stuff some of which is relevant to the discussions above. Check the articles in the menu on the right hand side

 https://protonsforbreakfast.wordpress.com/

He is Michael de Podesta, who worked for the National Physical Laboratory for 20 odd years and so has a pretty good grounding in data collection and analysis, He has been measuring and monitoring of his heatpump, solar and battery systems and trying to make sense of some of the data.