House temperature rise rate?

[JohnD]

surely your calculated 7.6kW is calculated as the heat loss once the house is up to temp.

To bring it to that temp from cold will need additional heat, and the time will depend on the thermal mass of your home. e.g. thick stone walls take a long time to heat up, lightweight blocks with an insulated cavity much less.

I usually run my boiler at about 60C, but during the -5C nights we have had this week, I have turned it up to 75C which means the rads put out more heat.

I have a programmable stat so the house does not get very cold overnight.

 

[wipeoutdude]

surely your calculated 7.6kW is calculated as the heat loss once the house is up to temp.

To bring it to that temp from cold will need additional heat, and the time will depend on the thermal mass of your home. e.g. thick stone walls take a long time to heat up, lightweight blocks with an insulated cavity much less.

I usually run my boiler at about 60C, but during the -5C nights we have had this week, I have turned it up to 75C which means the rads put out more heat.

I have a programmable stat so the house does not get very cold overnight.

I think I am going to need to learn the system and adjust it for the different weather.. I have put it up to 75C today because its been so cold..

I have a programmable stat too.. I have it at 22C comfprt setting in the morning and evening and 19C overnight and through the day..

 

[D_Hailsham]

All the rads are getting hot which is what I thought the balancing was for.

Yes, they will all get hot. But if the system is not balanced properly, some will get hotter than others. Also TRVs will not work correctly.

So I will use the IR thermometer and then do I just keep closing the lock shield vales down? What return temperature am I aiming for? I know that to get a 20C drop the vales are just about closed.. Like 1/8 - 1/4 turn open.

It's better to start off with all LS valves about 1/3rd turn open.

As for what temperature you aim for, your boiler is designed to run with a differential of 20C (see page 8 of installation manual), so it's better to aim for that.

Balancing
1. Set water temp to 75C
2. Remove all TRV heads and set wheel valves fully open
3. Set all LS valves to 1/3 turn open
4. Set room stat to a high temperature so boiler keeps running
5. Allow system to warm up.
6. Measure differential for all rads and make a note of them.
7a. If all rads have too small a drop, close every LS valve by 1/12th turn
7b. If all rads have too large a drop, open every LS valve by 1/12th turn
Wait 5-10 minutes for system to settle down
Repeat 6 and 7 Until some with a larger drop than required and some with a smaller drop.
8. Adjust the one with the largest drop by opening the LS valve 1/12th turn or less.
9. Wait 5 minutes for system to settle down
10. Measure drop for all rads
11a. If balance has been achieved (within +/- 1C) STOP
otherwise
11b. Repeat from 8.
12. Turn room thermostat down!!
13. Replace all TRVs and set to required temperature.
END

That will provide 8kW, which should cater for 95% of the time. On the few days when it is extremely cold, just turn the flow temp up to 80C and you will get the full 9.4kW.

 

[D_Hailsham]

surely your calculated 7.6kW is calculated as the heat loss once the house is up to temp.

No. Heat loss calculators always give the heat required to bring the house from the base temperature (normally -1C) up to the desired temperature. They usually include 10-15% extra to allow for faster warm-up, e.g when the house is unoccupied for most of the day.

 

[JohnD]

so what would be the energy required to maintain it at that temperature, once reached?

 

[wipeoutdude]

Thanks for the info.. I will have a go at rebalancing this evening and see how things play out from there.. To get a 20C drop I think the rads will pretty much all be just on. Like 1/4 turn from closed.. The towel rails will just be cracked open.. Will be interesting to see what return temp I can reach..

 

[Dan Robinson]

Miniscule adjustments can make startling changes.

 

[D_Hailsham]

so what would be the energy required to maintain it at that temperature, once reached?

That all depends on how much you allow the temperature to drop before the heating comes back on.

 

[JohnD]

half a degree C. Lets suppose I have a digital thermostat which maintains an accurate temperature, and I just want to maintain the house at 20C all day.

In which case, the amount of energy I need to input is the same as what I would call the "heat loss"

BTW I am puzzled and surprised to read you say that the term "heat loss" is being used to describe something which is not heat loss (the rate at which energy, in the form of heat, transfers from a warmer to a cooler object; in this case, the inside of a house to the world outside), but "energy required to raise the temperature of a house" which is not at all the same thing

 

[D_Hailsham]

half a degree C. Lets suppose I have a digital thermostat which maintains an accurate temperature, and I just want to maintain the house at 20C all day.

Let's assume that the house needs 20kW to raise the temperature from 0C to 20C, i.e 1kW per degC. In that case, if the temperature drops by 0.5C, then 0.5kW needs to be put into the house to bring the temperature back to 20C.

In which case, the amount of energy I need to input is the same as what I would call the "heat loss"

Yes.

I'm not responsible for the terminology used by heating engineers.

 

[Agile]

Wouldn't energy be measured in kWh ?

 

[JohnD]

Not if we are talking about true heat loss, and the energy input needed to match it.

However, if we were talking about the amount of energy need to raise the temperature of a house of mass X by n degrees, yes, we would.

 

[D_Hailsham]

Energy is measured in Joules.

One Watt is the power of one Joule for one second.

One kilowatt is 1000 Joules per second and one kilowatt hour is 3,600,000 Joules = 3.6MJ.