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G. g. — G1 = G8 . ( H1 / H8 )0,525 = 330 . /m v = 0,47 m/s — Pressure loss: ∆P1-0 = 5,0 . g. g. 740 l/h Note: The emitters on the lower floors are supplied at excessive flow rates. See chapter entitled “CHARACTERISTICS OF SIMPLE CIRCUITS”. Chosen diameter ø = 1 1/4” Σξ = 2,0 (no variation in diameter) v3 = 0,71 m/s Chosen diameter ø = 1 1/2” Σξ = 3,5 (variation in diameter) v2 = 0,75 m/s Chosen diameter ø = 1 1/2” Σξ = 2,0 (no variation in diameter) v1 = 0,80 m/s Chosen diameter ø = 2” Σξ = 3,5 (variation in diameter) 17 18 PRACTICAL CALCULATION OF SIMPLE CIRCUITS There are no completely reliable general ‘rule of thumb’ methods for the practical design of simple circuits are there are too many variables to be taken into account: Only experience can provide knowledge of short cuts which can be used to reduce the complexities of the calculations.
G. g. g. g. g. G6 = G7 . F = 366 . 1,12 = 410 l/h G5 = G6 . F = 410 . 1,10 = 451 l/h G4 = G5 . F = 451 . 1,08 = 487 l/h G3 = G4 . F = 487 . 1,07 = 521 l/h G2 = G3 . F = 521 . 1,06 = 552 l/h G1 = G2 . F = 552 . /m. e. g. 702 l/h 696 l/h 3/4” ø = 3/4” ø= 1” 2” 23 24 Note: With the following summary table the data obtained using the basic calculations and the data determined using the less complex method can be compared. g. g. From this comparison it will be seen that the differences between the two methods fall within the acceptable limits of design calculations for air conditioning systems (See 1st Manual under TOTAL HEAD LOSSES).
1,37 = 452 l/h 412 . 1,37 = 466 . 1,37 = 564 l/h 677 l/h New flow rate of the fan coil unit 3: G3 = 494 . 1,37 = 529 . 1,37 = New flow rate of the fan coil unit 2: G2 = 562 . 1,37 = 770 l/h 598 . 740 . 123 l/h Note: This example shows that the new differential pressure assumed at the base of the pipework provides a significant increase in the flow rates across the fan coil units. Consequently the difference between the required nominal flow rates and those which can be affectively supplied is increased to an even greater extent.