We don’t like to have the heat on all the time, because we’re cheap, and I like it cold at night while I’m sleeping, and fortunately it’s quite mild here (Maryland) in the Winter.
We would like to know if it is more cost-effective to have the heat on a lower temperature (65-68 degrees F) continually, or to have it turned off, and turn it on for a few hours a few times in the day.
It’s always cheaper not to use it.
So, as long as you aren’t turning the thermostat up during the times you use the heat, it will be cheaper to turn it off for several hours, then on again.
Right. To use a fluid analogy, say we’re trying to keep a bucket filled that has a hole in it. Our faucet has a constant flowrate (just like your furnace does; it’s either on or off, no in between). Will we use less water trying to keep that bucket filled up all the time, or only filling it up twice a day?
The closer your indoor temperature is to the outdoor temperature, the less heat will be lost. Less heat loss equals greater savings. It’s always cheaper to re-heat than continuously heat, but comfort must also be considered. My thermostat goes down to 17 C (63 F) at night and when we’re out, and goes back to 21 (70) when we’re home. Seems to work fine.
What do you mean by “lower temperature”? Lower than what? Lower than your “normal” set point? Lower than summertime temperatures?
Do you mean, (for example, using some numbers):
a) Is it more cost-effective to set the thermostat to 65 degrees F continually, or to turn the thermostat off, then turn it on at 65 degrees F (the same temp) for a few hours a few times during the day?
or
b) Is it more cost-effective to set the thermostat to 65 degrees F continually, or to turn the thermostat off, then turn it on at 70 degrees F (a higher temp) for a few hours a few times during the day?
If you mean (a), then other posters have already answered your question. If you mean (b), then the answer depends on the temperature swing in the house over time.
It’s not quite that simple. The *only *advantage to reducing the thermostat (or turning the furnace completely off) is to get the reduced heat loss rate by bringing the temperature of the house closer to the ambient temperature (see post #4). Without that benefit, the amount of savings you gain by turning it off is exactly offset by the additional energy needed to bring the temperature back up when you turn it back on.
Snap, I was wondering something similar. Is it more cost effective to have the heat on for half an hour, off for half an hour, then back on for half an hour, or for one hour solid?
Doesn’t matter, for the most part*.
Your furnace is either off or on. If it’s on, it’s using a certain amount of gas or oil or electricity every minute*. If your furnace is on for 60 minutes, it uses 60 minutes worth of fuel, whether that 60 minutes is straight through or in two 30-minute chunks. Now, the temperature in your house will be different between the two schemes, so the “value” to you (the amount of comfort you’re paying for v. the amount of money you pay) will be different, but the cost will be the same.
*Small caveat: there are likely some startup inefficiencies as dampers open and flames ignite and so forth, so I would expect the two-half-hours plan may have a bit more cost associated with it (since it has to start twice). But my guess is that any difference is negligible, on the order of inaccuracies in timing 30 minutes exactly.
The important thing to note is that the only savings with cyclical heating compared to constant heating is due to the different flow rates of heat at the higher and lower indoor temperatures. Heat loss is proportional to the temperature differential between outside and inside. So with cyclical heating, you save a certain amount of energy due to reduced heat loss during the lower temperature phase. For any given outdoor temperature, the saved energy is proportional to the difference between the high and low temperatures.
However, with cyclical heating you spend extra energy every time you heat up the house to the higher temperature. This energy expenditure is also proportional to the difference between the high and low temperatures. So you end up with two opposing effects. Unless I am mistaken, the two effects (heat loss reduction versus heating the air in the house) are independent and therefore cyclical heating may actually use MORE energy (it could vary from house to house). In any case the difference is most likely small and you are sacrificing your comfort for no good reason.
I’ve heard this explanation, but it doesn’t seem to add up - if there is no difference in the rate of heat loss during the warm and cold periods, why would there be *any *energy savings? You would just be throwing away heat at the beginning of the cold period, then having to use extra energy to increase the temperature for the warm period. Nothing would balance out, it’s not like the outdoor environment is going to store the heat for you and give it back when you need it. Please enlighten me if I’ve missed something here.
Eh, if anyone cares, here are some earlier thoughts. (But please ignore some of the attempted wisecracking in OP of the first link. The guy’s a blithering idiot, and obviously had some unfounded reason for thinking he could be witty.)
For very short periods, this is a good approximation But, since the heat loss is proportional to the indoor-outdoor temperature differential, what I said is correct. Think about the bucket analogy Santo Rugger presented above.
no cite here, but I vaguely remember reading a study online where the percieved temperature depended on the temperature of the walls and floor … if they were cold, even if the air was 70, if the walls were close to outside ambient in the winter you felt like 5 or more degrees cooler … but this was last year sometimes.
I do know that when we have the walls warm from the wood stove, the thermometer may tell us the air is cold, but we dont feel as cold. If the walls are ice cold from us being away and not burning the stove, it is a long time until the room feels warm.
As long as the bucket does not empty before being refilled, there will be no difference (assuming the hole is in the bottom and the flow rate out of the hole is constant).
I agree that as the cycle length increases, savings due to reduced heat flow increase. Since the cost of re-heating is fixed this improves the advantage of cyclical heating. However this does not prove that any particular cycle length will necessarily be better than static heating - please see my previous post and let me know if you find a flaw.
See above.
If you have ceiling or under floor radiant heat, you’ll have considerable savings by limiting its use to when you are conscious or present. The radiant heat will immediately affect the degree to which you radiate heat providing instant relief.
This is a link to another of Una’s threads. It’s about using a programmable thermostat.
I set our thermostat to get colder at night and start to warm up about 5:00 am(everyone gets up at 5:30-6:00).
Then it turns down during the day and starts to warm up again about a 1/2 hour before we get home.
The thermostat was very easy to install myself.
Una Persson - I saw your post from a few years back where you did some sample calculations, however it seems possible that your result was dependent on the various assumptions you made. Here is my computationally simpler but more rigorous general analysis, please let me know if there are mistakes.
Let:
H = total heat needed (from furnace) during one cycle
t2= comfortable indoor temperature (eg. 72F)
t1= lower indoor temperature (eg. 66F)
t0 = outdoor temperature
k = heat loss per degree per unit time (due to difference between outdoor and indoor temperature)
C = heat required to heat house by 1 degree
T = cycle period (length of 1 complete cycle)
Assumptions:
k is independent of absolute temperatures (indoor and outdoor)
C is independent of absolute temperatures
t1, t2 are always greater than t0
House is at t1 50% of the time and t2 50% of the time
(this one is not necessary but it avoids clutter with too many variables)
Case 1 - Cyclical Heating: H1 = 0.5Tk(t1-t0) + 0.5Tk(t2-t0) + C*(t2-t1)
Case 2 - Constant Temp: H2 = Tk(t2-t0)
“Savings” of cyclical heating (after a little algebra)
H2-H1 = (0.5T*k-C)(t2-t1)
So if k>2C/T there is an advantage to cyclical heating.
If k<2C/T there is a disadvantage to cyclical heating.
k depends on surface area of the house, insulation quality, number of windows and doors, etc.
C is based on the amount and composition of material making up the house (including the air).
I don’t know enough about materials and insulation to know if k would normally be greater than or less than 2C/T.
Much like the bucket (where flow rate out the hole is dependent on the water pressure, and water pressure decreases as the water level lowers), A house’s heat loss rate is not constant, and is dependent on the delta between the indoor and outdoor temperatures.
Ah, but the flow rate out of the hole isn’t constant: It depends on the pressure, which depends on the depth of the water.
I know this of course, I thought he was trying to make the point that even in a heavily oversimplified case cyclical heating/filling would have an advantage.
However, my analysis above should apply equally well to to the water in the bucket problem. In this case t represents height of water in the bucket. k (flow rate out of the hole) depends on the shape of the bucket (tall and skinny versus short and fat), volume of the bucket, size of the hole and viscosity of water (which is known). H (amount of water needed to fill the bucket up per unit of height) is just the cross-sectional area of the bucket.
There still seems to be two possible answers (advantageous or disadvantageous) depending on the specifics of the situation. Intuitively we are more likely to think there is an advantage because we picture the level slowly decreasing at a constant rate, rather than quickly plummeting and having to turn the faucet on again very shortly.
Note: I just realized the difference - in the house case I assumed that when the temperature changes from high to low the heat was being quickly dumped through a mechanism other than normal heat loss due to temperature gradient. However if heat loss is relatively high (the house cools from t2 to t1 in a time much smaller than T) then this would not significantly affect the answer.