Comfortable indoor climate
The human body can sense a temperature difference of about 1.5 °C.
The maximum temperature difference between in- and outdoor, for which the comfort level has to be sustained, is set at 30 °C. The required ventilation rate of fresh air for health is 150 m³h-¹. Heating by the fresh air (an efficient DC) fan is about 0.5 °C. This results in a maximum allowable temperature difference between the fresh air leaving the recuperator and the room air of 2 °C.
The temperature effectiveness of the recuperator therefore has to satisfy:
ε= realized temperature difference / maximum temperature difference =
(30 – 1.5 – 0.5) / 30 = 93 [%] (1)
The heat capacitance flow (product of the air mass flow and the specific heat capacity) is:
C = airflow [m³s-¹] * specific density [kgm-³] * specific heat capacity [Jkg-¹] =
150 / 3600 * 1.23 * 1002 = 51 [WK-¹] (2)
The temperature effectiveness of a recuperator under mass balanced flow is:
ε= kA / (kA + C) (3)
In which kA, the heat exchanging capacitance, is the product of the heat transfer coefficient (k [Wm-²K-¹]) and the heat-exchanging surface of the recuperator (A [m²]).
From (1), (2) and (3) it follows:
kA = ε/ (1 - ε) * C = 650 [WK-¹] (4)
The quality of the heat transfer is the ratio (also called the Number of Transfer Units) of the heat exchanging capacitance with the heat capacity flow:
kA / C = NTU = 13 (5)
The relative humidity has to stay between 40 and 50% for a good comfort level at 21 °C. Therefore a maximum of 5 g water per kg air is needed to humidify the indoor air, during times of heating the ambient air. At the required 150 m³h-¹ the moisture ml , that has to be recuperated, is:
ml = 150 * 1.23 * 5 /1000 = 0.92 [kgh-1] (6)
Part of this moisture is produced by the inhabitants. At 0.06 kgh-1 per person this adds up to 0.15 kgh-1 for the average 2.5 persons. The effectiveness of the moisture to be recuperated is:
εl = (0.92 – 0.06) / 0.92 = 85 [%] (7)
For dehumidification a maximum outdoor condition is taken at 35 °C and 85 % RH. Therefore 25 g of water has to be removed per kg air, resulting in:
ml = 150*1.23 *25 / 100 + 0.15 = 4.8 [kgh-¹] (8)
With 100 % effectiveness of the recuperator only 3.1 kgh-¹ can be removed, so the enthalpy (both sensible and latent heat) recuperation has to be as high as possible. For a further reduction, cooling is needed. As the production of moisture inside can vary, a control system for the moisture recuperation is needed.