This question is related to a 5 bedroom property, with a radiator in each room. Whereby heating is turned on/off in bursts of 2 hours (2 hours on, 2 hours off). 3 rooms are used constantly and 2 (smaller) are not. Assume the house is well insulated.

A DIY family member believes that heating the whole house is the better option, however I am under the impression that turning the radiators off in the 2 rooms makes more sense. DIY family member argues that in the 2 hours off period the house will cool down much slower if the whole house has been heated. I however argue the cost implications outweigh the probably negligible slower cool down?

Views on this?

As an additional question would leaving the doors of those 2 (smaller) rooms open/closed make an impact?

Hope that makes sense.

Many thanks.


4 Answers 4


Not heating a room will always be cheaper.

This doesn't matter if you're doing an on-off process or throttling heating to radiators according to heat needs. Heating a room costs money, not heating a room costs less.


Imagine you have two rooms with equal volume of air $V_r$ (m$^3$). They connect with each other through a door, and both rooms are insulated equally to the outside air at temperature $T_a$ (K). You have set points for the desired room temperatures as $T_1$ and $T_1$.

The heat $q$ (J) that you will need to bring a room to a desired set point temperature is

$$ q = V_r \rho_a \tilde{C}_p \left(T_N - T_{No}\right) $$

where $\rho_a$ is air density (kg/m$^3$), $\tilde{C}_p$ is air specific heat capacity (J/kg K), and $T_{No}$ is the initial temperature for the room.

All else being equal, you need less heat to reach a desired set point when the difference $(T_N - T_{No})$ is smaller.

The rooms loose heat to the outside. The rate of heat loss $\dot{q}$ (W) is controlled by the thermal resistance factor $R$ (W/m K) according to

$$\dot{q} = \left(T_N - T_a\right)/R $$

All else being equal, for any given increment in time $\Delta t$, the room with the smaller $T_N$ will loose less heat. This is why you turn down thermostats in the winter. The rate of heat loss from a colder room to the outside is smaller than that from a hotter room.

Suppose that you start with $T_1 > T_2$ as your desired set points. Suppose that the contact between the two rooms is perfectly insulated. No heat flows between them. In a given increment in time, room 2 will loose less heat to the outside than room 1 because $T_2 < T_1$ always. Therefore, room 2 will have a smaller change in its temperature than room 1 over the increment in time when heat is exiting the rooms. When the heat is to be given back, room 2 will need less heat to recover because $(T_2 - T_{2o}) < (T_1 - T_{1o})$. We confirm our heat savings by setting lower thermostat temperatures.

Now allow the wall between the two rooms to have some thermal conduction. Heat will flow from (hotter) room 1 to (colder) room 2 at all times. This will act to cause room 1 to cool faster. It will cause room 2 to have a higher set point temperature than in the case with a perfectly insulated wall. In the time when no heat is provided, room 2 will also cool faster to the outside air.

When you do not heat room 2 and the wall to room 1 has heat flow, you will need more heat to recover the set point $T_1$ in room 1 because some heat will flow to room 2. What about room 2? Since its set point is now hotter (with heat from room 1), its change in temperature over a period of time is larger. You will need more heat to recover the (higher) set point $T_2'$ in room 2.

Let's compare cases.

A) Actively heat rooms 1 and 2 (the entire house) to the same set point $T_A$. Turn off the heat. Over an increment in time $\Delta t$, the entire heat loss is $q_A$.

B) Actively heat room 1 to $T_A$ and allow room 2 to passively heat to $T_2' < T_A$ only by thermal conduction from room 2. Turn off the heat. Over the same increment in time $\Delta t$, the entire heat loss is $q_B$. It is less than $q_A$.

C) Actively heat room 1 to $T_A$ and allow room 2 to heat to $T_2''$ by thermal convection from room 2. In this case $T_2'' > T_2'$ because convection is more effective than just conduction to heat room 2. Turn off the heat. Over the same increment in time $\Delta t$, the entire heat loss is $q_C$. It is less than $q_A$ but more than $q_B$.

Your best improvement could likely be to thermostat each room and control the radiator flow values to each room separately by their thermostats. Also, keep the doors closed on the rooms that you want to set to hold to lower temperatures.


A simple and intuitive way to think about the problem is that each room is like a leaky bucket with heat leaking out through the walls, windows and ceiling on the upper floor.

Let's say that each bucket leaks 1 L/h. When keeping all five buckets topped up you would have to supply 5 L/h. If you turn off the supply to two of the rooms your supply requirements drop to 3 L/h.

The domestic situation will be complicated by the fact that having two cold rooms will increase the losses from the neighbouring rooms.


From a control standpoint, the cheapest way to achieve a given comfort level (a formally defined HVAC metric) is to have a steady temp and humidity, but that isn't easy to do when the load is most often only a small fraction of the installed capacity.

Given the interior walls have far less insulation than the exterior walls, the temperature of the occupied rooms will drop noticeably faster if you only heat 3 of the five rooms.

Control strategies for best comfort at medium/low run-time fractions for minimum cost are poorly researched. This is odd, because NREL suggests that the typical control strategy is about 40% more costly than an optimized strategy in modern, well insulated houses that require forced make-up air.

Well-insulated houses often require active dehumidification even during heating days. This problem gets substantially worse if you have unheated spaces. Condensation ruins stuff, and you can't always see it if it is behind the walls on the vapor barrier fabric.

For well insulated, tight homes with regulated make-up air exchange, two or three heat cycles per hour and central dehumidification to prevent condensation in all spaces is a reasonable rule of thumb for good economy.

Running a portable dehumidifier in the unoccupied rooms is a reasonable option also, but these things put out a lot of heat - the unoccupied rooms may end up at about the same temp anyway. This is my current strategy when I shut off part of the house in the winter.

Radiators were usually located on exterior walls and under windows to control condensation -and they worked well at that. Convection currents kept the air mixed. Household water pipes were usually routed along boiler pipe runs to prevent pipes from freezing. Your strategy potentially creates problems if the outdoor temps are getting down to freezing temps.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.