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Let's say you have a house with natural gas atmospheric heater water radiator system. Does it matter in terms of natural gas expenditure, if the house is kept at higher or lower temperature, assuming the outside temperature is always lower than inside?

Let's say you have the thermostat which regulates the heating system to keep more or less constant air temperature. You are leaving the house for a number of days. The common wisdom is to turn down the heating system to lower temperature so that it somehow saves energy.

However, unless you can completely turn the system off for the duration of your absence, does it matter, if the thermostat is kept at 15°C or at the 21°C you use when you are in?

Quite possibly I am wrong here, my instinct was that the house loses heat at a more or less constant rate, which is independent of the inside temperature of the house.

At the start of your absence, the inside of house is broadly at 21°C. If you turn the thermostat to 15°C upon leaving, there will be no gas spent during the time house cools down. Then the gas spent to maintain 15°C will be about the same amount as to maintain 21°C. Then when you return, you have to heat the house up and you expend the energy you saved during cooling, to heat the house back up.

Is it right or am I missing some basic physics here? Or perhaps it is dependant on actual temperature differences between inside and outside?

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    $\begingroup$ If energy loss didn't depend on the temperature difference, then your body would be just as comfortable in a -100C environment as in a 20C environment. $\endgroup$ – James Nov 20 '19 at 13:03
  • $\begingroup$ @James quite right, beats me how I didn't think of that. $\endgroup$ – Gnudiff Nov 20 '19 at 20:36
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Is it right or am I missing some basic physics here? Or perhaps it is dependant on actual temperature differences between inside and outside?

I like Gonzonator's answer but want to add a bit of math to it.

The heat loss of the house can be described by Newton's Cooling Law:

$$\frac{\text{d}Q}{\text{d}t}=-uA(T_i-T_o)$$

where:

  • $\frac{\text{d}Q}{\text{d}t}$ is the heat flux. Quite literally it's the power (energy per unit of time) needed to maintain the house at a constant inside temperature of $T_i$. Note that here it is negative because it represents a loss of heat (energy).
  • $u$ is the overall heat transfer coefficient. The lower it is, the better. Low values are obtained with good double (or triple) glazing, cavity insulation, insulating wall paper, loft insulation, etc.
  • $T_i$ and $T_o$ are the inside and outside temperatures respectively. Clearly high $T_i$ and/or low $T_o$ increases heat loss.
  • $A$ is the total surface area of the house exposed, to the elements.

The heat required to bring a house back to the target temperature $T_i$ from a hibernation temperature $T_h$ ($<T_i$) can also be modeled:

$$\Delta Q=mC(T_h-T_i)$$

where:

  • $\Delta Q$ is the amount of heat energy needed.
  • $m$ is the total mass of the house.
  • $C$ is the specific heat energy of the house.

At the start of your absence, the inside of house is broadly at 21°C. If you turn the thermostat to 15°C upon leaving, there will be no gas spent during the time house cools down. Then the gas spent to maintain 15°C will be about the same amount as to maintain 21°C. Then when you return, you have to heat the house up and you expend the energy you saved during cooling, to heat the house back up.

Here, for an amount of time $\Delta t$, we can compare two quantities:

$$\Delta Q_1=-uA(T_i-T_o)\Delta t$$

and with 'hibernation':

$$\Delta Q_2=-uA(T_h-T_o)\Delta t+mC(T_h-T_i)$$

Whichever of the two quantities is the least negative is the most cost-effective strategy.

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  • $\begingroup$ Thank you. Both answers are relevant, however, this one shows under which conditions which variables are more important. $\endgroup$ – Gnudiff Nov 20 '19 at 18:12
  • $\begingroup$ Thank you for the upvote. $\endgroup$ – Gert Nov 20 '19 at 19:40
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I'm afraid your instinct is wrong, thermal transfer rate is very much dependent on the thermal gradient. In a given environment, hotter things cool at a faster rate.

It's quite easy to test this. Make two cups of coffee. Add cold milk to one immediately, and the same volume of cold milk to the other one after two minutes. Measure their temperatures. You will find that the second cup has cooled considerably faster than the first, as the coffee was hotter.

Equivalently, you could do the reverse. Maintain the temperature of your house at a steady 21°C. Then measure your energy usage to maintain that temperature at different outside temperatures. You will find (instinctively) that the colder it is outside, the more energy is required to maintain the internal temperature.

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    $\begingroup$ This is true, but if the energy saved by leaving the house at a lower temperature while you're gone is less than the extra energy spent to bring it back up to the normal temperature when you come back, this isn't economical. Generally, the longer you're gone, the more economical this becomes. $\endgroup$ – probably_someone Nov 20 '19 at 10:56
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    $\begingroup$ probably_someone This makes no sense. Let's assume that your house reaches exactly 15°C on your return. This is the optimal case. At every point since you left, and until the house warms up, the house has been cooler than 21°C, and the heat loss from your house has been less than it would have been had your house stayed at 21°C. The only way it could use more energy is if the boiler was somehow less efficient at high utilisation than at low utilisation, or if you neglect to use a thermostat and overshoot. $\endgroup$ – Gonzonator Nov 20 '19 at 12:33

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