5.2 Horizontal air flow
Design data
Example
■
Geometry of the room
■
Fabric heat losses
■
Room temperature
■
Extract air temperature
(Because of the low height, room temperature = extract air temperature is used in this example.)
■
Distance from ceiling
Geometry ..................................15 × 20 × 3 m
Fabric heat losses .....................38 kW
Room temperature ....................20 °C
Extract air temperature..............20 °C
Distance from ceiling .................0.6 m
Mounting height
■
Calculate the actual mounting height (= distance between the floor and the bottom edge of
the units).
H = Hall height – distance from ceiling – unit height
■
Compare the actual mounting height with the minimum and maximum mounting height
GV-3/30
→
OK
GV-5/50
→
OK
Minimum number of units
■
Determine the minimum number of units for each unit type. Take into account the following criteria:
–
Heat output
–
Unit clearances and reach
Type
Required number
Minimum number
of units
Heat output
Unit clearances
and reach
GV-3/30
2
2
2
GV-5/50
1
2
2
■
Choose the final solution from the remaining possibilities, depending on the geometry of the hall and the costs.
When positioning the units consider the following:
■
Do not direct the air current directly at persons.
■
Do not install the units at too great a distance from the
ceiling, in order to avoid the formation of warm air pockets.
■
The units can also be arranged opposite each other or
opposite and offset.
Fig. D3: Offset arrangement of the units on opposing walls
D
41
TopVent
®
GV
Design example
Design example
