Hydronic heating system partial load behavior

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chakil

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Hi,
I have a question :

-with a room temperature set to 70F and with a baseboard
-the boiler output is 60,000 BTU/hr with 6gpm and delt t 20f
-minimum boiler temperature is 160F and limit is 180F with 10F differential
- Let's ignore the thermal mass effect

Let's say the load is 30000 btu/h, half design load. Which mean the room needs only 500 btu/min
1) Which one is going to cycle the boiler or the room thermostat?
2) How long would be the cycle length? frequency?

Thanks
 

John Gayewski

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Hi,
I have a question :

-with a room temperature set to 70F and with a baseboard
-the boiler output is 60,000 BTU/hr with 6gpm and delt t 20f
-minimum boiler temperature is 160F and limit is 180F with 10F differential
- Let's ignore the thermal mass effect

Let's say the load is 30000 btu/h, half design load. Which mean the room needs only 500 btu/min
1) Which one is going to cycle the boiler or the room thermostat?
2) How long would be the cycle length? frequency?

Thanks
The boiler will cycle more with less of a load. The thermostat is what sets the boiler off and tells it to fire. Not sure what your after here.
 

wwhitney

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You can't answer that question while ignoring thermal mass effects. It will be determined by the thermal mass of the appropriate system component.

For example, say you have 60,000 btu/hr of emitters (to match the boiler output of 60,000 btu/hr) in your location with a 30,000 btu/hr heat loss. The boiler and emitters should be happily run forever dumping 60,000 btu/hr into the location. The system behavior will be governed by a thermostat.

If the thermostat is set for a particular temperature band, then the thermal mass of the location will dictate the cycle length and frequency. E.g. if it's set to 70F, and will turn on at 69F and turn off at 71F, then just after the system shuts off, the location starts cooling down. You say it's losing 30,000 btu/hr, but say the thermal mass of the location is 30,000 btu/F. That means it will take 2 hours to cool down from 71 F to 69F, causing the thermostat to turn on the heating. Then the heater will be net adding 30,000 btu/hr (since the location is still losing 30,000 btu/hr) and will run 2 hours to get the temperature back to 71F.

The exact times in the example depend on that thermal mass of 30,000 btu/F (which I expect is ridiculously high, but I wanted to make the math easy). Halve that that and the on and off cycles will 1 hour each, etc. Likewise it depends on the thermostat behavior.

If instead you have only 30,000 btu/hr of emitters on your 60,000 btu/hr boiler, then whenever the system is running the hydronic loop temp is going to go to be rising, from the extra 30,000 btu/hr of boiler output. Now the system behavior will be governed by the loop temperature high limit and the thermal mass of the loop. Otherwise similar to the first example, with the loop temp controls playing the role of the thermostat.

At least, the above is my answer from first principles; I don't have much direct experience with hydronic systems (currently learning on my first system).

Cheers, Wayne
 

chakil

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For example, say you have 60,000 btu/hr of emitters (to match the boiler output of 60,000 btu/hr) in your location with a 30,000 btu/hr heat loss. The boiler and emitters should be happily run forever dumping 60,000 btu/hr into the location. The system behavior will be governed by a thermostat.
Why would the system run forever? the room needs only 30 000 btuh the thermostat should turn off the boiler
My understanding is the boiler would raise water temperature from 160F to 170F then stopped by thermostat since: 30 000btu = 500 *6*10
emitters heat output is linear with water supply temperature - room temperature



If the thermostat is set for a particular temperature band, then the thermal mass of the location will dictate the cycle length and frequency. E.g. if it's set to 70F, and will turn on at 69F and turn off at 71F, then just after the system shuts off, the location starts cooling down. You say it's losing 30,000 btu/hr, but say the thermal mass of the location is 30,000 btu/F. That means it will take 2 hours to cool down from 71 F to 69F, causing the thermostat to turn on the heating. Then the heater will be net adding 30,000 btu/hr (since the location is still losing 30,000 btu/hr) and will run 2 hours to get the temperature back to 71F.
why would the heater need to run 2hours? that is 120 000 btu
my understanding is the termostat from 69F to 71F would need 60000 btu plus 30000btu for the location thermal mass

At least, the above is my answer from first principles; I don't have much direct experience with hydronic systems (currently learning on my first system).

Cheers, Wayne
Thank you for the answer
 

wwhitney

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Why would the system run forever? the room needs only 30 000 btuh the thermostat should turn off the boiler
My comment was that the boiler and the emitters would happily run forever dumping 60,000 Btu/hr into the location. If eg there were no thermostat, or if the location needed 60,000 Btu/hr. I.e. the behavior is thermostat controlled, not driven by the loop temperature controls, unlike the second example.

My understanding is the boiler would raise water temperature from 160F to 170F then stopped by thermostat
Sounds like you know more about boiler loop temperature controls than I do, I don't understand the exact meaning of "minimum boiler temperature is 160F and limit is 180F with 10F differential"

why would the heater need to run 2hours? that is 120 000 btu
my understanding is the termostat from 69F to 71F would need 60000 btu plus 30000btu for the location thermal mass
The location thermal mass was 30,000 Btu/F, so to raise the temp 2F it will need to net add 60,000 btus. The location is still loosing 30,000 btus/hr, so while the boiler is emitting 60,000 btus/hr, the location is only gaining 30,000 btus/hr. So it takes 2 hours to net add the 60,000 btus to raise the temperature 2F.

Cheers, Wayne
 

chakil

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Sounds like you know more about boiler loop temperature controls than I do, I don't understand the exact meaning of "minimum boiler temperature is 160F and limit is 180F with 10F differential"
The aquastat controlls the boiler temperature, minimum is set to 160F with 10F differential so the burner turns on at 150F and off at 160F, high limit is set to 180F with 10F differential, burner turns off at 180F and turns on at 170F

The location thermal mass was 30,000 Btu/F, so to raise the temp 2F it will need to net add 60,000 btus. The location is still loosing 30,000 btus/hr, so while the boiler is emitting 60,000 btus/hr, the location is only gaining 30,000 btus/hr. So it takes 2 hours to net add the 60,000 btus to raise the temperature 2F.
I am not sure what you mean by "location thermal mass "
 

wwhitney

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The aquastat controlls the boiler temperature, minimum is set to 160F with 10F differential so the burner turns on at 150F and off at 160F, high limit is set to 180F with 10F differential, burner turns off at 180F and turns on at 170F
That doesn't quite make sense to me. I agree that a high limit of 180F means that the boiler turns off at 180F. And a target loop temp of 160F makes sense. But doesn't the 10F differential have something to do with incoming temperature vs outgoing temperature? In which case if it's part of the boiler behavior control, either the boiler firing rate, or the flow rate, must be variable and under control of the algorithm

I am not sure what you mean by "location thermal mass "
I mean the thermal mass of the location with the emitters. 30,000 btu/hr is a large load, maybe too large for a single room, more like a house. In which case it would be the thermal mass of the house.

The computations assumed that heat emitted from the system goes into the location thermal mass immediately and evenly, and that the thermostat measures that temperature. In reality there would be another response loop between the emitters and the location thermal mass and wherever the thermostat is measuring, so the behavior would be more complicated.

Cheers, Wayne
 

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Hi limit is used only as a safety would want a minimum differential of 25-30° and a set point of 190. Operating and outdoor reset should be in series. Cast boiler minimum water 140°temp minimum 15° differential for a longer burn time. Without a blower door test doesn't matter what insulation u have it would be only a guess to what the infiltration is. The test should be before the insulation is in as to seal leaks that smoke stick would find. Energy recovery system to bring fresh air in at a set amount.
 

chakil

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That doesn't quite make sense to me. I agree that a high limit of 180F means that the boiler turns off at 180F. And a target loop temp of 160F makes sense. But doesn't the 10F differential have something to do with incoming temperature vs outgoing temperature? In which case if it's part of the boiler behavior control, either the boiler firing rate, or the flow rate, must be variable and under control of the algorithm


I mean the thermal mass of the location with the emitters. 30,000 btu/hr is a large load, maybe too large for a single room, more like a house. In which case it would be the thermal mass of the house.

The computations assumed that heat emitted from the system goes into the location thermal mass immediately and evenly, and that the thermostat measures that temperature. In reality there would be another response loop between the emitters and the location thermal mass and wherever the thermostat is measuring, so the behavior would be more complicated.

Cheers, Wayne

By thermal mass i meant boiler mass and system water volume
By boiler i mean traditional cast iron boiler, it needs a minimum 140F to prevent condensation


I apologize for my poor english
 

chakil

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If
-the boiler output is 60,000 BTU/hr with 6gpm and delta t 20f
-system mass is 100 Gallons
-boiler temperature min 160F max 180F
-room temperature 70F

How long the boiler would take to supply the emitter with 180F water temperature?
 

wwhitney

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-the boiler output is 60,000 BTU/hr with 6gpm and delta t 20f
OK, that makes sense, where delta T is the outgoing water temp - incoming water temp. Since 8.3 lbs/gallon * 6 gpm * 20F * 60 mins/hr = 60,000 BTU/hr.

-system mass is 100 Gallons
OK, I'll take that to mean equivalent mass of water. I.e. it includes the cast iron and other system components, in the sense that for the whole system, it takes 8.3 lbs/gallon * 100 gallon * 1.0 BTU/lb-F = 830 BTUs to raise the system temperature 1 degree F. [Cast iron has a heat a capacity of 0.110 BTU/lb-F, only 1/9 as much as water. So 830 lbs equivalent of water could mean 730 lbs of water and 900 lbs of cast iron, e.g.]

-boiler temperature min 160F max 180F
Is boiler temperature = outgoing water temperature? Regardless, I'm getting the sense you want to compute a length of time for a water temperature at some point in the system to rise from 160F to 180F while the boiler is firing.
-room temperature 70F
Immaterial, if the emitter is still specified as emitting at 30,000 BTU/hr. The emitter behavior will actually depend on room temp and on incoming water temp, but if we are still to use the 30,000 BTU/hr figure specified in the OP, then that information has already gone into determining the 30,000 BTU/hr figure and we just need that result.
How long the boiler would take to supply the emitter with 180F water temperature?
OK, this question is well defined. If the emitters aren't emitting, and the boiler is dumping 60,000 BTUs/hr into the system, with no losses, then per the above it takes 830 BTUs to raise the system temp 1F. You want to raise it 20F, which takes 830 * 20 = 16,600 BTUs. 60,000 BTUs/hr = 1,000 BTUs/min, so the boiler would take 16.6 minutes to do that.

If the emitters are emitting at a constant 30,000 BTUs/hr, then the system is only heating up at a net rate of 30,000 BTUs/hr (60,000 BTUs/hr in from the boiler less 30,000 out from the emitters). In which case it would take twice as long for the system to see a 20F temperature rise, or 33.2 minutes.

Of course, that's an approximation, as the actual emitter behavior depends on the emitter incoming water temperature minus the room temperature. With an equation for the heat emission of the system as a function of system temperature, you could model the behavior of the system temperature over a firing cycle more accurately.

The upshot is that questions like this are a matter of energy accounting, tracking the BTU rates, and using the heat capacity to convert from BTUs to degrees F.

Cheers, Wayne
 
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chakil

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OK, this question is well defined. If the emitters aren't emitting, and the boiler is dumping 60,000 BTUs/hr into the system, with no losses, then per the above it takes 830 BTUs to raise the system temp 1F. You want to raise it 20F, which takes 830 * 20 = 16,600 BTUs. 60,000 BTUs/hr = 1,000 BTUs/min, so the boiler would take 16.6 minutes to do that.

If the emitters are emitting at a constant 30,000 BTUs/hr, then the system is only heating up at a net rate of 30,000 BTUs/hr (60,000 BTUs/hr in from the boiler less 30,000 out from the emitters). In which case it would take twice as long for the system to see a 20F temperature rise, or 33.2 minutes.

Of course, that's an approximation, as the actual emitter behavior depends on the emitter incoming water temperature minus the room temperature. With an equation for the heat emission of the system as a function of system temperature, you could model the behavior of the system temperature over a firing cycle more accurately.

As the thermostat calls for heat the water in the system is at room temperature 69F, so i think it should be some temperature mixing ?
The emitters are able to release 60 000 btu h in design conditions if supply water temperature is 180F, the emitters output is in linear relationship with supply water temperature minus the room temperature
Let's ignore the location thermal mass and assume the room has a heat loss of 30 000btu h
 

wwhitney

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As the thermostat calls for heat the water in the system is at room temperature 69F, so i think it should be some temperature mixing ?
The emitters are able to release 60 000 btu h in design conditions if supply water temperature is 180F, the emitters output is in linear relationship with supply water temperature minus the room temperature
Let's ignore the location thermal mass and assume the room has a heat loss of 30 000btu h
OK, now we have new parameters. With the emitters putting out 60,000 BTU/hr at 180F, and no model of room temp or discussion of thermostat controls, and the burner putting in 60,000 BTU/hr, the loop temp will never actually reach 180F, it will only approach it asymptotically.

The details:

The system temperature rises 1F for 830 BTUs (system thermal mass).
The burner fires at 60,000 BTUs/hr
The system temperature starts at 70F (round for simplicity).
The emitters emit 60,000 BTUs/hr * (System temp - 70F)/(180F - 70F)

The last equation assumes the room temp stays at 70F constantly (we can't model room temperature based on heat emitted, as we don't have a room thermal mass, so the best model possible is to ignore the heat loss/gain and assume a constant room temperature of 70F). And then that formula gives 0 emission at 70F system temp, and 60,000 BTUs/hr at 180F, with linear interpolation.

So let's call t the time in minutes since the burner started firing, T(t) the system temperature as a function of time, E(T) the emission rate, and B the burner rate. We have:

T(0) = 70
E(T) = 1,000 * (T-70)/110 for T>= 70
B = 1,000
dT/dt = (B - E)/830 = (1,000 * (180 - T)/110)/830

Let U = 180-T, then dU/dt = - dT/t, and we have

U(0) = 110
dU/dt = -1000/(110*830) * U = -0.011 U.

That simple differential equation has the exponential solution that U(t) = U(0) * exp(-0.011 t). That is T(t) = 70 + 110 * (1 - exp(-0.011 t)).

So in this model, the loop temp difference from 180F has a "halving time" 63 minutes. I.e. T(0) = 70F = 180 - 110; T(63) = 125F = 180 - 110/2 ; T(126) = 152.5 = 180 - 110/4, etc.

Cheers, Wayne
 

chakil

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OK, now we have new parameters. With the emitters putting out 60,000 BTU/hr at 180F, and no model of room temp or discussion of thermostat controls, and the burner putting in 60,000 BTU/hr, the loop temp will never actually reach 180F, it will only approach it asymptotically.

The details:

The system temperature rises 1F for 830 BTUs (system thermal mass).
The burner fires at 60,000 BTUs/hr
The system temperature starts at 70F (round for simplicity).
The emitters emit 60,000 BTUs/hr * (System temp - 70F)/(180F - 70F)

The last equation assumes the room temp stays at 70F constantly (we can't model room temperature based on heat emitted, as we don't have a room thermal mass, so the best model possible is to ignore the heat loss/gain and assume a constant room temperature of 70F). And then that formula gives 0 emission at 70F system temp, and 60,000 BTUs/hr at 180F, with linear interpolation.

So let's call t the time in minutes since the burner started firing, T(t) the system temperature as a function of time, E(T) the emission rate, and B the burner rate. We have:

T(0) = 70
E(T) = 1,000 * (T-70)/110 for T>= 70
B = 1,000
dT/dt = (B - E)/830 = (1,000 * (180 - T)/110)/830

Let U = 180-T, then dU/dt = - dT/t, and we have

U(0) = 110
dU/dt = -1000/(110*830) * U = -0.011 U.

That simple differential equation has the exponential solution that U(t) = U(0) * exp(-0.011 t). That is T(t) = 70 + 110 * (1 - exp(-0.011 t)).

So in this model, the loop temp difference from 180F has a "halving time" 63 minutes. I.e. T(0) = 70F = 180 - 110; T(63) = 125F = 180 - 110/2 ; T(126) = 152.5 = 180 - 110/4, etc.

Cheers, Wayne

Unfortunately that's too much math for me
Thanks anyway
 
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