Commonly used in homes down south, a heat transfer kit works by taking heated air from one room and depositing it into another. This has resulted in many people confused to why they don't seem to work in our Queensland homes.

In theory they do work to an extent that the transfer enough heated air to accommodate the room size but that's neglecting that our homes are designed to expel as much heat as possible. Considering a very simple room that is 3m by 6m and 2.4m ceiling with a 3m² window the thermal losses are considerable high.

• Thermal loses can be calculated with Q = U x A x ∆T
• U : Thermal transmittance
• A : Area (m²)
• T : Temperature difference between inside and outside(K)

Thermal Transmittance

Assuming the intake from the vent in the adjacent room is 60^oC and no losses in the exchange kit then the output would also be 60^oC.

Hot Air Mass required to heat

t = (m1 c1 t1 + m2 c2 t2 )/(m1 c1 + m2 c2 ) = (m1 t1+m2 t2)/(m1+m2)

With the final temperature of the room being 24^oC and the initial room temperature at 12^oC, the is a required 120kgs of heated air needed to heat the room as the room has initially 90kgs of cold air.

Heat Input into the Room

The heat transferring from 60^oC to 12^oC in the room is calculated using Q = m x c x ∆T

• Q : Heat added (W)


• m : mass (kg) 120kg of air added to the room.


• c : Specific Heat (1.005kJ/kg).


• T : Temperature difference between output and room temperature (^oC).


• Q = 120 x 1.005 x (60-12) = 5.8kW

As this result is much lower than the heat lost through the room this means that the transfer kit is unable to heat the room.

Note: This is based on the design of a Queensland home.