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- How to calculate pump-down time

In practice, for instance, when estimating the cost of a planned vacuum plant, calculation of the pump-down time from the effective pumping speed S_{eff}, the required pressure p, and the chamber volume V by formulas presented would be too troublesome and time-consuming. Nomograms are very helpful here. By using the nomogram in Fig. 9.7, one can quickly estimate the pump-down time for vacuum plants evacuated with rotary pumps, if the pumping speed of the pump concerned is fairly constant through the pressure region involved. By studying the examples presented, one can easily understand the application of the nomogram.

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 pumpdown 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 columns. If there is inflow qpV,in, the value p_{end} – p_{ult},p – q_{p}V,in / S_{eff}, max is to be used in the columns.

Left:

Pressure reduction ratio R = (p_{START} – p_{ult},p – q_{p}V,in / S_{eff},max)/(p_{end} – p_{ult},p – q_{p}V,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 nomogram and is expressed in column ➄ _{ult,p}. If the pump pressure pult,p is small in relation to the pressure pend which is desired at the end of the pump-down operation, this corresponds to a constant pumping speed S or Seff 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 Δτ = ± 10 s so that the relative uncertainty is about 10 %).

2) Determination of tp: The ultimate pressure of the rotary pump is pult,p = 3 · 10-2 mbar, the apparatus is clean and leakage negligible (set qpV,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) τ = 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 pumpdown time of tp = 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 pressure 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 p1 = 20 mbar, a pumping speed of Seff,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, Seff = 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 (pult,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 ➄, one obtains t_{p,1} = 7.7 min. The Roots pump must reduce the pressure from p1 = 20 mbar to pEND = 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 min for 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.

The pump-down times of rotary vane and rotary piston pumps, insofar as the pumping speed of the pump concerned is constant down to the required pressure, can be determined by reference to example 1.

In general, Roots pumps do not have constant pumping speeds in the working region involved. For the evaluation of the pump-down time, it usually suffices to assume the mean pumping speed. Examples 2 and 3 of the nomogram show, in this context, that for Roots pumps, the compression ratio K refers not to the atmospheric pressure (1013 mbar), but to the pressure at which the Roots pump is switched on.

In the medium vacuum region, the gas evolution or the leak rate becomes significantly evident. From the nomogram 9.10, the corresponding calculations of the pump-down time in this vacuum region can be approximated.

The nomogram indicates the relationship between the nominal pumping speed of the pump, the chamber volume, size and nature of the inner surface as well as the time required to reduce the pressure from 10 mbar to 10^{-3} mbar.

**Example 1: **A given chamber has a volume of 70 m^{3} and an inner surface area of 100 m^{2}; a substantial gas evolution of 2 · 10^{-3} mbar · l · s^{-1} · m^{-2} is assumed. The first question is to decide whether a pump with a nominal pumping speed of 1300 m3/h is generally suitable in this case. The coordinates for the surface area concerned of 100 m^{2} and a gas evolution of 2 · 10^{-3} mbar · l · s^{-1} · m^{-2} result in an intersection point A, which is joined to point B by an upward sloping line and then connected via a vertical line to the curve that is based on the pumping speed of the pump of 1300 m^{3}/h (D). If the projection to the curve is within the marked curve area (F), the pumping speed of the pump is adequate for gas evolution. The relevant pump-down time (reduction of pressure from 10 mbar to 10^{-3} mbar) is then given as 30 min on the basis of the line connecting the point 1300 m^{3}/h on the pumping speed scale to the point 70 m^{3} (C) on the volume scale: the extension results in the intersection point at 30 min (E) on the time scale.

In **example 2 **one has to determine what pumping speed the

pump must have if the vessel (volume = approx. 3 m^{3}) with a

surface area of 16 m^{2 }and a low gas evolution of

8 · 10^{-5} mbar · l · s-1 · m^{-2} is to be evacuated from 10 mbar to

10^{-3} mbar within a time of 10 min. The nomogram shows that

in this case a pump with a nominal pumping speed of 150 m^{3}/h is appropriate.

In many applications it is expedient to relate the attainable pressures at any given time to the pump-down time. This is easily possible with reference to the nomogram 9.7.

As a first example, the pump-down characteristic – that is, the relationship pressure p (denoted as desired pressure p_{end}) versus pumping time t_{p} – is derived from the nomogram for evacuating a vessel of 5 m3 volume by the single-stage rotary plunger pump E 250 with an effective pumping speed of S_{eff }= 250 m^{3}/h and an ultimate pressure p_{end,p} = 3 · 10^{-1} mbar when operated with a gas ballast and at p_{end,p} = 3 · 10^{-2} mbar without a gas ballast. The time constant τ = V / S_{eff} (see equation 2.36) is the same in both cases and amounts as per nomogram 9.7 to about 70 s (column 3). For any given value of p_{end }> p_{end,p }the straight line connecting the “ 70 s point” on column 3 with the (pend – pend,p) value on the **right-hand** scale of column 5 gives the corresponding t_{p} value. The results of this procedure are shown as curves a and b in Fig. 2.77.

It is somewhat more tedious to determine the (p_{end},t_{p}) relationship for a combination of pumps. The second example discussed in the following deals with evacuating a vessel of 5 m^{3} volume by the pump combination Roots pump WA 1001 and the backing pump E 250 (as in the preceding example). Pumping starts with the E 250 pump operated without gas ballast alone, until the Roots pump is switched on at the pressure of 10 mbar. As the pumping speed characteristic of the combination WA 1001/ E 250 – in contrast to the characteristic of the E 250 – is no longer a horizontal straight line over the best part of the pressure range (compare this to the corresponding course of the characteristic for the combination WA 2001 / E 250 in Fig. 2.19), one introduces, as an approximation, average values of S_{eff}, related to defined pressure ranges. In the case of the WA 1001/ E 250 combination the following average figures apply:

S_{eff }= 800 m^{3}/h in the range 10 – 1 mbar,

S_{eff }= 900 m^{3}/h in the range 1 mbar to 5 · 10^{-2} mbar,

S_{eff} = 500 m^{3}/h in the range 5 · 10^{-2} to 5 · 10^{-3} mbar

The ultimate pressure of the combination WA 1001 / E 250 is: P_{end,p} = 3 · 10^{-3} mbar. From these figures the corresponding time constants in the nomogram can be determined; from there, the pump-down time tp can be found by calculating the pressure reduction R on the left side of column 5. The result is curve c in Fig. 2.77.

Of course, calculations for our industrial systems are performed by computer programs. These require high performance computers and are thus usually not available for simple initial calculations.

The preceding observations about the pump-down time are significantly altered if vapors and gases arise during the evacuation process. With bake-out processes particularly, large quantities of vapor can arise when the surfaces of the chamber are cleared of contamination. The resulting necessary pump-down time depends on very different parameters. Increased heating of the chamber walls is accompanied by increased desorption of gases and vapors from the walls. However, because the higher temperatures result in an accelerated escape of gases and vapors from the walls, the rate at which they can be removed from the chamber is also increased.

The magnitude of the allowable temperature for the bake-out process in question will, indeed, be determined essentially by the material in the chamber. Precise pump-down times can then be estimated by calculation only if the quantity of the evolving and pumped vapors is known. However, this is seldom the case except with drying processes.

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