TPI 19 VVKR

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Diffusers
Swirl Diffuser VVKR Calculation and Design Example
Given parameters:
Room dimensions:
Width: 18,0 m
Length: 24,0 m
Height: 3,65 m
Air exchange in the room: i = 12 times
Room temperature: T = 24 °C
Temperature differential between the room air temperature and the supply air temperature ΔT
0
= -8 K
Average air ow velocity between two diffusers at distance H
1
, v
H1
< 0,2 m•s
-1
Requested sound power level L
WA
< 30 dB(A)
Requested parameters:
1. Q
V
- supply air ow volume into the room
2. q
V
- supply air ow volume into one VVKR
3. VVKR type
Solution:
1. We determine the supply air volume for the room 24 m long, 18 m wide and 3.65 m high with a plasterboard
ceiling. Q
V
= 24 × 18 × 3,65 × 1 = 24 m × 18 m × 3,65 m × 12 = 18922 m
3
• h
-1
2. Architect’s denition for this particular case is that the minimim distance from a vertical wall is 3 m (Y = 3 m).
Then the remaining area for VVKR distribution is 12 × 18 m. If the axial distance between end diffusers
at a transverse direction is 12 m and the distance between rows is B = 3 m, then the number of rows is 5 and if the
axial distance between end diffusers at a longitudinal direction is 18 m and the distance between columns is A = 3
m, then the number of columns is 7.
We calculate the supply air ow volume into one VVKR
q
V
= Q
V
/ number of rows (B) × number of columns (A) = 18922 / (5 × 7) = 540 m
3
• h
-1
3. Based on Tab. 6 we determine a preliminary size design for q
V
= 540 m
3
• h
-1
, which is VVKR-A-S-600×40
Calculation of the amount of pressure drop, sound power level and effective velocity
for a swirl diffuser
Given parameters:
VVKR-A-S-600×40
Air supply into box - horizontally
Measurable air ow volume q
V
= 540 m
3
• h
-1
Free area at dimensions 600 × 40 A
V
= 0,0409 m
2
(from Tab. 4)
1. diffuser pressure drop: Δp
t
= 22 Pa
2. sound power level: L
WA
= 29 dB(A)
Requested parameters:
1. L
WA
2. Δp
t
Solution:
Based on page 13 (Diagram 1) for an open damper in VVKR-A-S-600×40, where q
V
= 540 m
3
• h
-1
we get: