Calculation of time to replace a volume of extracted air by heating it...

[js.b1]

Can someone check my calculation...

Time to replace 15m^3 of air extracted at 20oC, with air heated by a

650 Watt heater from 0oC to 20oC.

Air-Volume = 15 m^3 Temp-Change = 20oC less 0oC = 20oC Heater-Watts = 650W where 1W = 1J/Sec Air SPH = 716 J/Kg K Air Density = 1.3kg / m^3

Time (secs) = Temp-Change * Air-Density * Air SPH / Heater-Watts * Air- Volume Time (secs) = 20 * 1.3 * 716 / 650 * 15 = 430 secs or just over 7 mins

Repeated 3x a day is 0.37hr @ 0.65kW @ 11p/kWhr = 2.65p or bugger all. That is running a fan 3x a day for 10mins assuming it removes 100% of free flow (which is not likely, so the running time could be as much as 20min to achieve 15m^3 of air). Whatever it costs very little indeed yet enforces a near 40% air change 3x a day.

Is my calculation correct?

[Java Jive]

Yes (assuming a typo), but you've made things unnecessarily difficult. All that matter are:

The total mass of air to be heated = # of times x vol x density The specific heat of air The raise in temperature The conversion ratio between Joules and kWHr = 1/3.6M The cost of 1kWHr

Thus: 3 x 15 x 1.3 x 716 x 20 x 0.11/ 3600000 = £0.0256 or 2.56p

This of course assumes that no heat is be>

[js.b1]

:-)

Noted.

It is indeed true. The fan heater is 650-1300-2000W unit so would be used on 2000W, a fan heater given very rapid warm up of the bulk body of air also.

#15:

Will do, thanks.