Blauberg KWK 50x30-3
- Designation key
- Overall dimensions
- Pressure losses
- Calculation diagram
Supply air cooling for ventilation systems in various premises. Suitable for installation into supply ventilation or into air handling units to provide air cooling.DESIGN
Galvanized steel casing. The cooling elements are made of copper tubes and aluminum plates. Available in three-coil modifications and rated for maximum operating pressure 1.5 MPa (15 bar). Polypropylene droplet separator and drain pan for condensate drainage and removal included. Droplet separator is efficient at an air flow not exceeding 4 m/s.
Only horizontal mounting by means of flanged connection. Air evacuation and condensate drainage must be provided. Air filter installation upstream of the cooling unit to prevent the unit soiling. Installation position must ensure uniform air flow distribution in the section. Mounting upstream or downstream of the supply fan. The minimum air duct length downstream of the fan must be 1 m to ensure air flow stabilization. The maximum cooling capacity is attained if the cooling unit is connected on counter-flow basis. The attached charts are valid for counter-flow connection. If water is used as a cooling agent, the cooling unit is suitable for indoor use only with the ambient temperature not below 0 °C. If antifreezing solution, for example, ethylene glycol solution, is used as a cooling agent, the cooling unit is suitable for outdoor use as well. While mounting the cooling unit provide condensate drainage through the U-trap. The U-trap height must be selected with respect to the total fan pressure, refer to the table and diagram below.
For a proper and safe operation of the cooling unit it should be connected to a control system for integral control and automatic cooling capacity regulation.
|Parameter||KWK 50х30-3||Measurement unit|
|Casing material||galvanized steel||-|
|Air duct||for rectangular air ducts||-|
|Series||Flange size (WxH) [cm]||Number of water (glycol) coil rows|
|KWK||40x20; 50x25; 50x30; 60x30; 60x35; 70x40; 80x50; 90x50; 100x50||–||3|
|KWK 50х30-3||500||540||570||300||395||252||56||G 3/4’’|
|The air flow is 2000 m³/h and the air speed in the cooling unit is 3.75 m/s ①.|
|To calculate the coldest air temperature find the intersection point of the air flow line ① with the rated outer summer temperature shown in blue line (e.g., +32 °С) and draw the line ② to the left until it crosses the outdoor air humidity curve (e.g. 50 %). From this point draw a vertical line to the supply air temperature downstream of the cooling unit (+20.6 °С) ③. To calculate the power of the cooling unit find the intersection point of the air flow ① with the rated summer temperature (e.g. +32 °С) and draw the line ④ to the right until it crosses the air humidity curve (e.g. 50 %). From this point draw a vertical line to the cooling unit power axis (13.6 kW) ⑤.||To calculate the required water flow in the cooling unit prolong this line ⑥ downwards to the water flow axis (0.54 l/s). To calculate the water pressure drop in the cooling unit find the intersection point of the line ⑥ with the pressure loss curve and prolong the line ⑦ to the right on the water pressure axis (27.0 kPa).|