Datasheet

12
472
1
2
3
5
10
20
50
S
L
1 2 3 5 10 20
8
6
5
4
3
2
8
6
5
4
3
2
1.0
0.5
0.95
0.63
2.5
4.5
6.5
7.5
2.18
0.5
0.5
0.3
0.2
Subject to change
SI UNITS, SYMBOLS AND DIMENSIONING
Evacuation time of vacuum pads
How to calculate evacuation time for vacuum pads, and choosing ejector and tubing.
H
ere we show, using formulas and charts, how to
calculate how long it takes for a vacuum pad to
achieve the desired vacuum level.
Calculation of evacuation time
Average suction ow in ejector
Q
1
= 0.4 · Q
max
Tubing maximum ow
Q
2
= S · 11.1
Tubing volume between ejector and vacuum pad
V = 1 ÷ 1 000 · π ÷ 4 · D
2
· L
Evacuation time
T1 = V · 60 ÷ Q
T2 = 3 · T1
Example:
Ejector: ZH10BS-06-06
Max. vacuum (P
V
): –88 kPa
Max. suction ow (Q
max
): 24 l/min
Tubing length (L): 1 m
Tubing inner diameter (D): 6 mm
Vacuum pad diameter: 10 mm
Necessary vacuum: 63% of P
V
, no leakage
1. Calculate ejector’s average suction ow (Q
1
) by mul-
tiplying the maximum suction ow by 0.4.
Q
1
= 0.4 · 24 l/min = 9.6 l/min
2. Calculate maximum tubing ow (Q
2
) by nding the
tubing’s equivalent cross-sectional area (S) in chart 4
and multiplying this by 11.1.
Q
2
= 18 · 11.1 = 198 l/min
3. Calculate tubing volume between ejector and pad.
V = 1 ÷ 1 000 · π ÷ 4 · 6
2
· 1 = 0.028 l
4. Calculate evacuation time. Since Q
1
is lower than Q
2
this means Q = Q
1
i.e. 9.6 l/min. The time to reach
63 % of max. vacuum equals:
T1 = 0.028 · 60 ÷ 9.6 = 0.18 s
Chart 4 – tubing equivalent cross-sectional area Q
max
= Ejector’s maximum suction ow (l/min), see the
technical data
S = Tubing equivalent cross-sectional area (mm
2
),
see chart 4
V = Tubing volume (l) between ejector and vacuum
pad
T1 = Time (s) to reach 63% of maximum vacuum
level (P
V
)
T2 = Time (s) to reach 95% of maximum vacuum
level (P
V
)
Q = The lowest of Q
1
and Q
2
Pressure (vacuum)
Vacuum (P
V
)
Time
Tubing length (m)
Tubing inner diameter (mm)
Equiv. cross-section (mm
2
)
1
P
V
·
P
V
·
0
2 3 4
T1 T2