Tables, Formulas, Diagrams
Fundamentals of Vacuum Technology
D00.156
LEYBOLD VACUUM PRODUCTS AND REFERENCE BOOK 2001/2002
Fig. 9.7: Nomogram for determination of pump-down time tp of a vessel in the rough vacuum pressure range
p
END
– p
end, p
mbar
p
START
< 1013 mbar
p
START
= 1013 mbar
p
START
– p
end, P
p
END
– p
end, P
R =
Column ➀: Vessel volume V in liters
Column ➁: Maximum effective pumping speed S
eff,max
at the vessel in (left) liters per second or
(right) cubic meters per hour.
Column ➂: Pump-down time t
p
in (top right) seconds
or (center left) minutes or (bottom right)
hours.
Column ➃: Right:
Pressure p
END
in millibar at the END of the
pump-down time if the atmospheric
pressure p
START
( p
n
= 1013 prevailed at the
START of the pump-down time. The desired
pressure p
END
is to be reduced by the
ultimate pressure of the pump p
ult,p
and the
differential value is to be used in the co-
lumns. If there is inflow q
pV,in
, the value
p
end
– p
ult,p
– q
pV,in
/ S
eff,max
is to be used
in the columns.
Left:
Pressure reduction ratio R = (p
START
– p
ult,p
– q
pV,in
/ S
eff,max
)/(p
end
– p
ult,p
– q
pV,in
/
S
eff,max
), if the pressure p
START
prevails at
the beginning of the pumping operation
and the pressure is to be lowered to p
END
by pumping down.
The pressure dependence of the pumping
speed is taken into account in the nomo-
gram and is expressed in column ➄ by
p
ult,p
. If the pump pressure p
ult,p
is small in
relation to the pressure p
end
which is
desired at the end of the pump-down
operation, this corresponds to a constant
pumping speed S or S
eff
during the entire
pumping process.
Example 1 with regard to nomogram 9.7:
A vessel with the volume V = 2000 l is to be pumped
down from a pressure of p
START
= 1000 mbar
(atmospheric pressure) to a pressure of p
END
= 10
-2
mbar by means of a rotary plunger pump with an
effective pumping speed at the vessel of
S
eff,max
= 60 m
3
/h = 16.7 l · s
-1
. The pump-down time
can be obtained from the nomogram in two steps:
1) Determination of τ: A straight line is drawn
through V = 2000 l (column ➀) and S
eff
= 60 m
3
/h
-1
= 16.7 l · s
-1
(column ➁) and the value t = 120 s
= 2 min is read off at the intersection of these straight
lines with column ➂ (note that the uncertainty of this
procedure is around ∆τ = ± 10 s so that the relative
uncertainty is about 10 %).
2) Determination of t
p
: The ultimate pressure of the
rotary pump is p
ult,p
= 3 · 10
-2
mbar, the apparatus is
clean and leakage negligible (set q
pV,in
= 0); this is
p
START
– p
ult,p
= 10
-1
mbar – 3 · 10
-2
mbar = 7 · 10
-2
mbar. Now a straight line is drawn through the point
found under 1) t = 120 s (column ➂) and the point
p
END
– p
ult,p
= 7 · 10
-2
mbar (column ➄) and the
intersection of these straight lines with column ➃
t
p
= 1100 s = 18.5 min is read off. (Again the relative
uncertainty of the procedure is around 10 % so that
the relative uncertainty of tp is about 15 %.) Taking
into account an additional safety factor of 20 %,
one can assume a pump-down time of
t
p
= 18.5 min · (1 + 15 % + 20 %)
= 18.5 min · 1.35 = 25 min.
Example 2 with regard to nomogram 9.7:
A clean and dry vacuum system (q
pV,in
= 0) with
V = 2000 l (as in example 1) is to be pumped down
to a pressure of p
END
= 10
-2
mbar. Since this pres-
sure is smaller than the ultimate pressure of the
rotary piston pump (S
eff,max
= 60 m
3
/h = 16.7 l
( s
-1
= 3 · 10
-2
mbar), a Roots pump must be used
in connection with a rotary piston pump. The former
has a starting pressure of p
1
= 20 mbar, a pumping
speed of S
eff,max
= 200 m
3
/h – 55 l · s
-1
as well as
p
ult,p
– 4 · 10
-3
mbar. From p
start
= 1000 mbar to p =
20 mbar one works with the rotary piston pump and
then connects the Roots pump from p
1
= 20 mbar to
p
END
= 10
-2
mbar, where the rotary piston pump acts
as a backing pump. For the first pumping step one
obtains the time constant τ = 120 s = 2 min from the
nomogram as in example 1 (straight line through
V = 2000 l, S
eff
= 16.7 l · s
-1
). If this point in
column ➂ is connected with the point
p
1
- p
ult,p
= 20 mbar – 3 · 10
-2
mbar = 20 mbar (p
ult,p
is ignored here, i.e. the rotary piston pump has a
constant pumping speed over the entire range from
1000 mbar to 20 mbar) in column 5, one obtains t
p,1
= 7.7 min. The Roots pump must reduce the
pressure from p
1
= 20 mbar to p
END
= 10
-2
mbar, i.e.
the pressure reduction ratio R = (20 mbar – 4 · 10
-3
mbar) / (10
-2
mbar-4 · 10
-3
) = 20/6 · 10
-3
mbar =
3300.
The time constant is obtained (straight line V = 2000
l in column ➀, S
eff
= 55 l · s
–1
in column ➁) at = 37
s (in column ➂). If this point in column ➂ is
connected to R = 3300 in column ➄, then one
obtains in column ➃ t
p, 2
= 290 s = 4.8 min. If one
takes into account t
u
= 1 minfor the changeover
time, this results in a pump-down time of
t
p
= t
p1
+ t
u
+ t
p2
= 7.7 min + 1 min + 4.8 min = 13.5 min.
D00 E 19.06.2001 21:41 Uhr Seite 156