Air Leakage in Vacuum Vessels
The chart, Fig. 1, is calculated from the
formula:
∆p .
MLA = 00.071*
,071 * ×.
×V
t
Where:
MA = Air leakage in kg/h
t
= Corresponding time in min
V
= Plant volume in m³
Air leakage in kg/hr
∆p
p = Change of pressure in mbar
How airtight is a vacuum plant?
Is the suction capacity of the vacuum
pump large enough?
Why does it take so long for the plant
to reach the vacuum?
Must the vacuum pump extract leak
air as well as gases from the
product?
Evacuate the vessel to a vacuum under
500 mbar, e. g. 60 mbar. Isolate the vacuum pump from the vessel and completely
seal off the vessel. Measure the pressure
increase in the vessel and determine the
corresponding time. The pressure increase in mbar divided by the time in minutes gives the vacuum loss in
mbar/minute. With this value and the volume of the vessel under vacuum the air
leakage rate in kg/hr can be found in
chart fig.1.
* the exact value is 0.071289977 based
on:
J
universal gas constant = 8.31441
,
mol K
absolute Temperature 293.15 K and
mol mass for air 28.96 kg/kmol
Vacuum loss in mbar/min
V= Plant volume
Fig 1
Example: A vessel of 20 m³ volume is
evacuated to 60 mbar and isolated. Within 10 minutes the vacuum drops to 120
mbar = ∆p 60 mbar. The pressure change
is 60 : 10 = 6 mbar/minute, therefore the
vacuum loss is 6 mbar/minute. From the
chart fig. 1 an air leakage rate of 8.5 kg/h
is found.
In the high vacuum range the air leakage
rate or the quantities of gases and
vapours are measured in mbar · litre/s.
1 mbar · litre/s 0.0043 kg/hr air at 20 °C
Admissible Velocity of Flow in Vacuum Pipelines
The admissible velocity of flow in a vacuum pipeline depends on how high the
pressure loss of this pipeline is allowed to
be. Higher pressure loss means increased
energy requirements for the vacuum
pump. A pressure loss of 10 % of the total pressure can generally be accepted.
This is shown in graph fig. 1 and valid for
air at 20 °C. It is calculated according to
the formula:
0.2
R ×. TT
0,2 ×. R
l
1 + 40 ×.
d
wadm = admissible velocity of flow in m/s
wadm
W
zul =
R
= gas constant in J/kg K
R=
For reasons of simplication, the calculation is based on an average pipe friction
coefficient of λ = 0.04 (this is ~ λ max.)
and on a free-of-loss acceleration from 0
to w m/s, i. e. with a well rounded
pipeline inlet.
Range: 2 mbar ≤ p ≤ 1000 mbar.
r
m
r = 8314.3 J/kmol K= universal
gas constant
m = Mol mass in kg/kmol
T
= Temperature in K
I
= Length of pipeline in m
d
= Diameter of pipeline in mm
A 15
Furthermore the graph contains lines for
constant volume flow V in m³/hr. The
graph is meant for rapid, rough dimensioning of a vacuum line. An exact pressure loss calculation can be made with
the help of sheet 2000 abl 9.
2000
abl 8
∆p
at p = 0.1 or 10%
Admissible velocity of flow in m/s
Admissible Velocity of Flow in Vacuum Pipelines
Nominal diameter of pipeline in mm
l= lenght of pipeline, V= volume flow
Fig 2
Example: calculation of mass flow
840 3
m / kg
p
vA = specific volume of air at 20°C
vvAL =
Given:
Pipeline ND 100
Length of pipeline I = 10 m
Flow medium: Air at 20 °C
Total pressure p = 10 mbar
p = pressure in mbar:
1650 ×. 10
kg
≈ 20kg
M=
840
Wanted:
1) Admissible velocity wadm in m/s
2) Volume flow V in m³/hr
3) Mass flow M in kg/hr
From the graph fig. 1 can be found:
1) wadm u 58m/s
2) V
u 1650 m³/hr
f=
Calculation:
3)
M=
Conversion to other gases and
vapours
If instead of air at 20 °C another gas or
vapour at the temperature in °C is to be
conveyed, the values from the graph have
to be multiplied by the factor:
273 + ϑ
m ×. 10
m = Mol mass of gas or vapours
V
kg/hr
kg
/h
v
Equivalent length of pipeline of pipe
bends and gate valves are installed
in the pipeline
Example
Ie
= equivalent length of pipeline in m
Given are:
Pipeline DN 600
Length of pipeline l = 100 m
Flowing medium: air at 20°C
Total pressure p = 10 mbar
4 pipe bends 90°; 1 gate valve
I
= length of pipeline in m
Wanted:
d
= diameter of pipeline in mm
1) Equivalent length of pipeline Ie in m
ζ
= resistance coefficients:
IÄe = I +
d
×. Σζ
40
pipe bend D/d = 3; 90°
ζ = 0.16
gate valve with restriction ζ = 1.0
If in the above example values for water
vapour with a temperature ϑ = 7 °C, p =
10 mbar and m = 18 kg/kmol are inserted,
the following formula results:
w’adm = f · wadm =
273 + 7 .
× 58 ≈ 72 m / s
18 ×. 10
from diagram fig. 2:
Volume flow V’ u 2050 m³/hr
by calculation:
Mass flow
M' =
V'
kg/hr
kg
/h
v'
from Water vapour temperature table zu
ϑ = 7°C; v’ = 129.1 m³/kg
therefore:
2050
M' =
≈ 16 kg
/h
kg/hr
129.1
129,1
By calculation we find:
1)
600 . .
IeÄ = 100 +
× (4 × 0,16 + 1)
40
= 124.6 m
From graph fig. 2 we find:
2) wadm u 44 m/s
2) Admissible velocity of flow wadm in
m/s
3) Volume flow V in m³/hr
4) Mass flow in M kg/hr
A 16
3) V
Hence:
4)
M=
u 45000 m³/hr
45000 .
535
≈ 10
535u kg
/ h kg/hr
840