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When I was having all of the rads replaced I contacted Stelrad tech support to help me choose. They said with the figures I gave them they could not do any calcs. I ended up installing regular 1100 x 600 double type 22 rads from screwfix. Home was very warm through that harsh winter we had this year. Now I've spent the summer adding another 200mm insulation to underfloor areas, I'm hoping to reduce heat losses even further.
I suppose they couldn't do the heat loss calcs so then couldn't recommend the correct output rad.
If you are wondering how the correction factors that Siricosm quoted then they are (obviously) based on a T50 rad or sometimes called a 50 deg rad, this is based on the rad flow and return temps and the required room temp which is often assumed at 20C, the "deg rad" is then the mean rad temp-the required room temp, if you had flow/return temps of 75/65 and a required room temp of 20 then you would have a ((75+65)/2)-20, a
50 deg (or T50) rad, you can see from the table below that a 40 deg rad will give ~ 75% output of a 50 deg rad so the correction factor is 1/0.748 or 1.34 and so on. the output is the ("deg rad"/50)^1.3. You probably know this anyhow but no harm to show that rads have to be very oversized if using very low flow temps like with Heat Pumps.
If you are wondering how the correction factors that Siricosm quoted then they are (obviously) based on a T50 rad or sometimes called a 50 deg rad, this is based on the rad flow and return temps and the required room temp which is often assumed at 20C, the "deg rad" is then the mean rad temp-the required room temp, if you had flow/return temps of 75/65 and a required room temp of 20 then you would have a ((75+65)/2)-20, a
50 deg (or T50) rad, you can see from the table below that a 40 deg rad will give ~ 75% output of a 50 deg rad so the correction factor is 1/0.748 or 1.34 and so on. the output is the ("deg rad"/50)^1.3. You probably know this anyhow but no harm to show that rads have to be very oversized if using very low flow temps like with Heat Pumps.
Flow temp | 75 | 65 | 55 | 45 |
Ret temp | 65 | 55 | 45 | 35 |
Mean temp | 70 | 60 | 50 | 40 |
"deg rad" | 50 | 40 | 30 | 20 |
Output | 100.0% | 74.8% | 51.5% | 30.4% |
X Factor | 1.00 | 1.34 | 1.94 | 3.29 |
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Heating a room is simply replacing the heat lost from the room. Raising the temperature of air or water is a linear scale so for every degree drop you have a set figure to replace it. So in a room with high insulation and controlled circulation, once heated then keeping on top is the balance. Leaving the heating on 24/7 has created a constant house temperature (I don't think rooms) @ 18 degrees (the personal optimum for my home). Gas consumption is down 64% when compared to the old combi that was replaced. BTW, this is not a heat pump, its a combi
I'm not so sure about that, if your house was at the same temperature as the outside air then, depending on insulation it will take a certain amount of heat to increase that temperature by say 5C but to increase it by the next 5C will take more energy as the heat loss at ambient+10C will be greater that at ambient+5C? or to think of it in another way, If you switched off your boiler at your desired room temp of 18C it will take a fixed amount of time to fall by 5C to 13C but if you had had your room temp at say 22C then it will IMO fall by 5C to 17C in a shorter period as heat loss is greater. If the temperature rise/fall was linear then by definition once any room is at its desired temperature then it only takes the same amount of heat to maintain it at its desired temperature, my house certainly requires less heat to keep it at 16C which is my set back temperature when we go out than to maintain it at 22C which is our normal desired temperature.
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It takes less than 2 seconds. With no one using hot water overnight, I find the plate exchanger is warm, so I guess it must thermosiphon.
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so are you saying that the specific heat energy is not linear? Pretty sure there have been no changes to the basic laws of physics in the last few decades
The heat loss varies with the difference between inside temperature and the outside. So Hometech and John.g are looking through opposite ends of a lens but effectively seeing the same thing (what kind of mixed metaphor is that?).
I've worked it out this way and feel free to correct me if I've made a mistake somewhere... To take John's example, if external is 13°C, the room at 22°C is 9 degrees above it, whereas the room at 18° is only 5 degrees above it. Heat losses for the warmer room will be 80% greater i.e. if it takes 1000W to maintain 18°C, it will take 1800W to maintain 22°C. There is a linear relationship between heat lost and the difference in temperature. In this example, 200W per degree above the external ambient.
This is consistent with heat loss calculations (radiator output is heat lost from the emitter into the room) which are calculated based on building elements having a heat loss expressed in W/m2K (K=inside/outside difference).
Going back to post 9, I don't therefore follow the example of a radiator at T30 having to be 2.4x the size. I make it 1.67x. If you divide the output of a rad calculated at T50 by 50 (the temperature difference between room temperature to be maintained and the mean emitter temperature), so the same rad at T30 would have:
T50 rated output/50x30 = 60% of the output
So if we want, say, 1000W output from a radiator run at T30, surely we need a radiator rated at 1667W (T50)? 60% of 1667W is 1000W.
The above logic conflicts with radiator manufacturers (e.g. Stelrad) who claim a radiator run at T30 will have an output 52% of the same radiator run at T50 and would therefore select a 1923W model rather than the 1667W one my logic would dictate. We cannot both be correct.
I've worked it out this way and feel free to correct me if I've made a mistake somewhere... To take John's example, if external is 13°C, the room at 22°C is 9 degrees above it, whereas the room at 18° is only 5 degrees above it. Heat losses for the warmer room will be 80% greater i.e. if it takes 1000W to maintain 18°C, it will take 1800W to maintain 22°C. There is a linear relationship between heat lost and the difference in temperature. In this example, 200W per degree above the external ambient.
This is consistent with heat loss calculations (radiator output is heat lost from the emitter into the room) which are calculated based on building elements having a heat loss expressed in W/m2K (K=inside/outside difference).
Going back to post 9, I don't therefore follow the example of a radiator at T30 having to be 2.4x the size. I make it 1.67x. If you divide the output of a rad calculated at T50 by 50 (the temperature difference between room temperature to be maintained and the mean emitter temperature), so the same rad at T30 would have:
T50 rated output/50x30 = 60% of the output
So if we want, say, 1000W output from a radiator run at T30, surely we need a radiator rated at 1667W (T50)? 60% of 1667W is 1000W.
The above logic conflicts with radiator manufacturers (e.g. Stelrad) who claim a radiator run at T30 will have an output 52% of the same radiator run at T50 and would therefore select a 1923W model rather than the 1667W one my logic would dictate. We cannot both be correct.
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While specific heat energy may be linear and the amount of energy required to raise the temperature by 1C is the same I would think that as the temp rises the heat losses increase so the total energy required is the specific heat energy+the increase in losses if one wants to keep the same rate of rise, don't know if this is still a linear relationship of energy vs temperature rise??.
Re rad output a 30 deg rad does not emit 30/50 or 60% of a 50 deg rad, it emits (30/50)^1.3 or "only" 51.5% of that of a 50 deg rad.
That's very interesting about the 200 watts/degC even though I still find it hard to grasp that if you keep adding exactly 200 watts/degC constantly that it can be described as linear as I think that the rate of temp rise would/will be slower the higher the room temperature even though, if the above is correct, its easy to see that the temperature will increase from 13C to 19C if the heat input is increased to 1800 watts immediately.
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Ah, right. So 30/50 or 60% but then hat 1.3. What does hat 1.3 mean? You're losing me 🙁 I've already stated that Stelrad uses 52% (rounded from a correction factor of 0.515), but I don't understand how they get that figure. Not saying they are wrong, but what I'm saying is I fail to understand how they calculate that figure as it conflicts with the logic of heat loss calculations used in U values. Perhaps U values are only a guideline are are known to be imprecise?
To be pedantic, W or Wh/h (flow of energy) is power, whereas Wh (quantity) is energy and I think this might be the source of the confusion. So obviously it will take more than twice the energy to heat a room/house by 10 degrees that it would take to heat it by 5 as it's losing heat the whole time. Obviously, if you could stick the house in a Thermos flask and heat it, 10° would only take a little over double as there would be almost no heat lost. But the energy required to maintain the higher temperature (Wh/h) is only double.
So, yes, I agree with you, basically.
Obviously we also overlook the fact that the heating system will struggle as the delta T of emitters will fall as the rooms get warmer and that since even an unheated house will be a few degrees above ambient due to solar gain and incidental gains from the body heat of the occupants and any electrical appliances in use, the first few degrees of warming are essentially already compensated for heat loss.
A long time ago I couldn't figure out why a 30 deg rad etc wasn't emitting 60% of that of a 50 deg rad and so on, so I found a table giving the deg rad vs output and put into a spreadsheet and by clicking on the trendline excel will show the formula that determines the trend and I saw that the calc is the deg rad/50 to the power (^) 1.3. Now I don't know if all manufacturers use this or not but I would think that its some number very close to 1.3. I will post a screen shot of the above shortly.
I have a cup of tea sitting outside in the winter. The tea is 60 degrees, the air temp is zero
My tea goes cold quicker than if it were 20 degrees outside...agree.
Now insulate the cup...does it take longer to cool down even in the same outside temperature? Yes.
Now add a insulated lid to the cup..does it extend even further the time it takes to cool down...Yes.
I let the tea cool to 15 degrees and need it reheated to 60. It takes the same energy input to raise the tea temperature at whatever the outside temperature it is heated in. Slow down the rate and sources of heat loss and you have totally different needs in a radiator. Like a combi and the ratio it can modulate at. Higher modulation, greater efficiency. A number of principles of physics are being rolled into one here...
My tea goes cold quicker than if it were 20 degrees outside...agree.
Now insulate the cup...does it take longer to cool down even in the same outside temperature? Yes.
Now add a insulated lid to the cup..does it extend even further the time it takes to cool down...Yes.
I let the tea cool to 15 degrees and need it reheated to 60. It takes the same energy input to raise the tea temperature at whatever the outside temperature it is heated in. Slow down the rate and sources of heat loss and you have totally different needs in a radiator. Like a combi and the ratio it can modulate at. Higher modulation, greater efficiency. A number of principles of physics are being rolled into one here...
I'm afraid I can't raise my cup to that, it will only take the same energy to heat it from 15C to 60c if its fully insulated, remove the insulation and it will take more energy as the cup is losing heat to the outside but must still be heated to 60C, exactly as you say in reverse, the more the insulation the longer it takes to cool down so less energy needed to heat up, if it was 100% insulated it wouldn't take any energy to heat it up as it would never cool down in the first place.
I can only see it your way if you fully insulated the cup and then heat it from 15C to 60C, remove the insulation, let it cooldown, replace the insulation, same exact energy required, do not replace the insulation more energy required because you are heating up the cup + making up for the losses due to no insulation, maybe we are just going around in circles here.
I can only see it your way if you fully insulated the cup and then heat it from 15C to 60C, remove the insulation, let it cooldown, replace the insulation, same exact energy required, do not replace the insulation more energy required because you are heating up the cup + making up for the losses due to no insulation, maybe we are just going around in circles here.
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One last go at it, if you had a 1 kg cup of water at 15C and you heated it in a fully insulated cup to 60C it will require 1X(60-15)/860, 0.0523kwh or 52.3 wh. If you heated it in uninskulated cup it will require 52.3 wh to heat the water plus whatever wh are lost to outside, if you put only 52.3 wh into the uninsulated cup then it will not reach 60C.
I turned the tank down to 55C and it seems to make hot water fine, but that is 55C at the bottom of the tank, so I am guessing the top is warmer. It is really a shame those tanks don't have pockets for thermometers.
I'll have to see if it needs to be turned up when the weather gets colder, and the incoming water temperature drops.
I wonder if anyone makes a weather compensating cylinder stat.
I'll have to see if it needs to be turned up when the weather gets colder, and the incoming water temperature drops.
I wonder if anyone makes a weather compensating cylinder stat.
What is the boiler setpoint temperature?
You have a temperature gauges on the hot water inlet (from boiler) to the tank so that will tell you the temperature at the tank top (and to the Hx)., depending on the boiler output and circ pump flow rate and boiler set point the water to the tank top could be 20C higher than the temperature at the stat, the water is being heated from the top down (vs from the bottom up with a immersion heater or tank coil) so the positioning of the stat is very important, but if the boiler setpoint is set to say 5C above the stat temperature setpoint then the top of the tank can't be more than (in your case), 60C). IMO the boiler setpoint should always be linked to the tank stat setpoint, say tankstat SP+5C?. So if you want a tank temp of 55C, set the tankstat to 50C and the boiler SP to 55C.
You have a temperature gauges on the hot water inlet (from boiler) to the tank so that will tell you the temperature at the tank top (and to the Hx)., depending on the boiler output and circ pump flow rate and boiler set point the water to the tank top could be 20C higher than the temperature at the stat, the water is being heated from the top down (vs from the bottom up with a immersion heater or tank coil) so the positioning of the stat is very important, but if the boiler setpoint is set to say 5C above the stat temperature setpoint then the top of the tank can't be more than (in your case), 60C). IMO the boiler setpoint should always be linked to the tank stat setpoint, say tankstat SP+5C?. So if you want a tank temp of 55C, set the tankstat to 50C and the boiler SP to 55C.
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I have an old Ideal FF360, which has no gauges. The manual says the max temperature is 82C when it is turned all the way up. With a thermal store I assumed that you would always just run the boiler on the maximum setting. I don't think there is any advantage to turning it down, is there?
I have some thermometers in the pipework, but I get the feeling they report a lower temperature than reality, they say the water coming from the boiler never gets much over 70C or so, and the return temperature reads about 5C lower than the thermostat setting when it cuts out. I am going to insulate the pipework soon, so maybe that will change.
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