lindab | we simplify construction Facade system Fasadium Dimensioning For easy calculation please go to: Lindabs Quick Selection Tool on www.lindQST.com. Cooling capacity water P w Follow the instructions below to read off the effect from the diagram. 1. Calculate ∆ t rw . 2. Product length L minus 0.2 m, to obtain the active length L act . 3. Divide the primary air flow rate q a by the active length L act . Enter the result on the lower axis of diagram 2 and 3. 4. Follow the flow line to the right pressure, and then read off the specific cooling capacity P Lt per active metre. 5. Calculate the temperature difference in water circuit ∆ t w and find the capacity correction factor e ∆ tw in diagram 4. 6. Multiply the specific cooling capacity P Lt that was read off by e ∆ tw , Δ t rw and active length L act . Example 1 Cooling: What is the cooling capacity of Fasadium 1000 with 14 l/s and pressure of 150 Pa? The room's summer temperature is assumed to be 24.5ºC. The cooling water temperature in/out of Fasadium is 14/17ºC. Answer: Temperature difference: ∆ t rw = t r – (t wi + t wo ) / 2 ∆ t rw = 24.5 - (14+17) / 2 = 9 K Active length: L act = 1.0 m - 0.2 m = 0.8 m q a / L act = 14 l/s / 0.8 m = 17.5 l/(s m) Read off, from diagram 2: P Lt = 60.1 W/(m K) Diagram 4 shows a capacity correction factor e Δ tw = 0.968. This gives a cooling capacity: P w = 60.1 W/(m K) × 9 K × 0.8 x 0.968 = 419 W in the water circuit. NB! The capacity diagram applies for the nominal water flow of q wnom = 0.038 l/s. To obtain the right cooling capac- ity P w for other flows, read off the capacity correction fac- tor e qw from diagram 5, and then multiply the capacity, which is read off, by this factor. 10 Lindab reserves the right to make changes without prior notice 2021-10-18
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