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Technical System Catalogue Climate Control

Cooling units Project planning Calculate your required cooling output: Practical tips . . In all situations where optimum operating temperatures are QE = QV – k A ΔT tures, a Rittal enclosure cooling unit can provide the rightrequired inside an enclosure, even at high external tempera- solution. It is even possible to cool the interior temperature of the enclosure to well below the ambient temperature. Condensation and dehumidification of enclosure air The favourable aerodynamic arrangement of the air inlet and when using cooling units outlet openings in the internal and external circuits ensures One unavoidable side-effect of using cooling units is the optimum air circulation inside the enclosure. This sample dehumidification of the enclosure’s interior air. As it cools calculation will show you a quick, time-saving method of down, part of the humidity contained in the air condenses selecting a cooling unit. on the evaporator coil. This condensate must be reliably dis- charged from the enclosure. The amount of condensate Example: occurring depends on relative humidity, the air temperature A cooling unit with a cooling output of 1500 watts begins inside the enclosure and on the evaporator coil, and the air operation with a temperature setting of Ti = 35 °C. volume present in the enclosure. The Mollier h-x diagram The relative ambient air humidity is 70%. If air at 35 °C is shows the water content of air depending on its temperature passed over the evaporator coil, the surface temperature of and relative air humidity. the evaporator coil (evaporation temperature of the refriger- ant) is approximately 18 °C. At the boundary layer adhering to the surface of the evapo- rator coil, water is deposited at the dew point. The difference Mollier h-x diagram Δx = x1 – x2 indicates how much condenstion is produced per for calculating the water content of air. kg of air with complete dehumidification. The leak-tightness of the enclosure has a decisive effect on the quantity of con- x densation. 0 5 10 15 20 25 30 35 40 x2 x1 50 45 10% 40 20% The quantity of condensation is calculated from the following 35 30% equation:➀ 40% 50% 30 60% 70% 100% W = V ρ Δx 25 90% 80% 20 W = Water quantity in g V = Volume of the enclosure in m3 T 15 r = Density of air kg/m3 10 Dx = Difference of water content in g/kg dry air (from the Mollier h-x diagram) 5 Enclosure door closed: 0 Only the enclosure volume is dehumidified. –5 V = W H D = 0.6 m 2 m 0.5 m V = 0.6 m3 –10 W = V r Dx –15 = 0.6 m3 1.2 kg/m3 11 g/kg –20 W = 7.92 g 8 ml 0 5 10 15 20 25 30 35 40 45 50 55 60 Pd Poorly sealed cable entries, damaged door seals and the Pd = Water vapour partial pressure (mbar) fitting of display media to enclosure surfaces lead to T = Air temperature (°C) increased rates of leakage in the enclosure. Hence, with x = Water content (g/kg dry air) a leakage rate of, say, 5 m3/h, a continuous condensate ➀ = Relative humidity volume of up to 80 ml/h may occur. Summary: Enclosure cooling units should only operate with the door closed. Seal the enclosure on all sides. Use a door limit switch. Use TÜV-tested equipment. Only set the enclosure internal temperature as low as is actually needed. 3 - 17 Technical System Catalogue/Climate control


Technical System Catalogue Climate Control
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