Pump Down Times - Vacuum Pump, Equation & Examples

Often we need to predict the time it's going to take for a vacuum pump to evacuate a chamber or process from atmospheric pressure down to the operating vacuum. We call that the pump down time and here is the standard equation that we can use to calculate it.
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⦿ 00:01 What is pump down time?
⦿ 00:15 Equation to calculate pump down time.
⦿ 01:21 Example
⦿ 03:13 Pump Curves
The time taken is the volume of the chamber divided by the pumping speed multiplied by the logarithm of the first pressure divided by the end pressure - T= V/S x ln (P1/P2).
Looking at the equation it's clear that bigger volumes will take a longer time or bigger pumps will reduce the time required. Looking at the second term we can see that it's the ratio of absolute pressures which is important, it's not the δP. So that tells us pumping with a constant flow rate from 1,000 mbar down to 100 mbar would take exactly the same time as pumping from 500 mbar to 50 mbar or indeed from 10 down to 1 mbar. This, in turn, illustrates how much harder it can be to pump gas at low absolute pressures.
So creating nine millibar pressure difference can take the same time and effort as 900 millibar pressure difference in a more rough pressure regime. The natural logarithm relates the continuous change in pressure back to the time taken.
You can think of it in a similar way to compound interest so if I have $100 with a yearly interest payment of 12% then after the first year I have a hundred and twelve dollars but if I had twelve-monthly interest payments of one percent then I would end up with more money and that's because in the first month of the year I will learn $1 in the second month I own 1% of 101 dollars and in the third 1% of the new total and so on until in the final month I end up with nearly a hundred and twelve dollars and seventy cents. A whole seventy cents earned because of the compound growth rate.
Let's try to bring that back two pumps. So imagine a 30 meter cubed in our vacuum pump working in a single stroke each hour on a 60 m³ chamber held at 1,000 mbar. In the first hour it will remove in a stroke half of the gas in the vessel and the pressure will be reduced to 50% or 500 mbar. Now imagine a similar pump but instead of one single 30 m³ stroke it has two strokes of 15 m³ every half hour. The first stroke will remove 15 m³ of gas and reduce the pressure by 25% to 750 millibar then the second stroke will now remove 25% of only the remaining gas leaving the pressure at 75 percent of 750 which is 563 millibar.
The natural log function in the equation accounts for the difference between instantaneous work and gradual incremental work. It's worth noting here that the 'S' in this equation relates to the capacity rating of the vacuum pump and that's important because it's assuming a single capacity figure. But we know by looking at pump curves that they never have a totally flat volumetric performance across every pressure. So what we need to do to compensate for that is split the calculation into sections or pressure bands and we assume an average figure for 'S' in each band then we calculate the pump downtime for each section and the total pump downtime is the sum of all the sections. Lucky for us, we now have calculators and programs which can read performance curves of vacuum equipment and take the heavy lifting out of the equations. But it's important to have a feel for what that program is calculating when it provides pump down figures for us.
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[@laurasofiaguerra42]

What about this formula t=-(V/S)*(P2-P1)? i used the one you said for higher pressures but the result its crazy and no way posible. instead for a range of 1*e-3 Torr- 7.6e-6 Torr its more accurate

[@damonwu1216]

easy to understand

[@goldendragon776]

Thank you for your video. But, how is your neck?

[@PinguExpert]

🤣

[@asfpd]

Question: If we take the a vac pump of say 1200m³/h therefore S = 20m³/min and V = 1m³ closed vessel and want to draw a vacuum from say P1 = 1000mbar(g) to P2 = 10mbar(g) then, using your equation.......... T = 1/20 x log(1000/10) = 1/20 x 2 min = 2/20 min = 1/10 min = 6s. Is this a realistic expectation because actual draw down time I'm given for a larger capacity GHS1600VSD is nearer to 18s. What am I missing here, please?

[@amatyakotamraju3838]

The log should be natural log not to the base 10

[@PinguExpert]

Good video, but why are you giving us your back?