How much energy is used in pumping down to high vacuum?
This seems to be an extremely simple question and I was quite surprised when I couldn't find an answer on the Internet - since the time required to pump down to a given vacuum is well-known, and addressed in all elementary texts on vacuum technology.
I had the feeling that the power consumed should increase as the vacuum level goes further down from the bar to the millibar to the microbar range... but it turns out not to be so...
Since I couldn't find an answer I asked my colleague Dr.Shiv Kumar who promptly sent me a thread on researchgate started by a researcher Xuezhong He of the Norway Institute of Science and Technology who asked the same question, along with some responses - to which I added my two cent's worth, as reproduced below:
"If we want to use a vacuum pump to suck a gas stream from 1 bar to 100mbar, how to estimate the power consumption for this vacuum pump? Is it identical to use a compressor from 100mbar to 1bar?
How can I estimate vacuum pump power consumption?"
Christian Binek (Univ. of Nebraska at Lincoln) responded:
This seems to be an extremely simple question and I was quite surprised when I couldn't find an answer on the Internet - since the time required to pump down to a given vacuum is well-known, and addressed in all elementary texts on vacuum technology.
I had the feeling that the power consumed should increase as the vacuum level goes further down from the bar to the millibar to the microbar range... but it turns out not to be so...
Since I couldn't find an answer I asked my colleague Dr.Shiv Kumar who promptly sent me a thread on researchgate started by a researcher Xuezhong He of the Norway Institute of Science and Technology who asked the same question, along with some responses - to which I added my two cent's worth, as reproduced below:
"If we want to use a vacuum pump to suck a gas stream from 1 bar to 100mbar, how to estimate the power consumption for this vacuum pump? Is it identical to use a compressor from 100mbar to 1bar?
How can I estimate vacuum pump power consumption?"
Christian Binek (Univ. of Nebraska at Lincoln) responded:
I would argue that the work Integral[PdV] is just a minuscule
and most likely negligible fraction of the internal power consumption of the
pump, e.g., due to friction losses and acceleration of mechanical parts.
Pavel Soucek of Masaryk University suggested that we just use a wattmeter, and Bhupendra Desai of IIT Roorkee suggested a clip-on wattmeter.
Pavel Soucek of Masaryk University suggested that we just use a wattmeter, and Bhupendra Desai of IIT Roorkee suggested a clip-on wattmeter.
Prashant Thankey of the Institute of Plasma Research added:
"The actual power consumed by a vacuum pump in its pumping action
is actually the throughput (Q = PS) of the vacuum system. As pumping continues,
Q generally decreases, and so the power consumption decreases. However, the
total power consumption of the pump is the sum of the power consumed in the
pumping action plus power consumed in overcoming the friction, heat losses etc."
However, let's just try to do the PdV thing:
Q = SP
Where s = speed of
pump (litre/sec), P = pressure in torr and flow rate Q is in torr-litre/sec.
The standard equation
is:
P
= P0 exp(- St/V)
Since PV has units of
energy, the total energy is:
E
= P0 òexp(- St/V) dt between P1 and P2.
If S is independent of
pressure and time:
E
=P0V[exp(-St2/V) - exp(-St1/V)]
But P = P0
at t = 0, so:
E
=P0V [1 – exp(-St/V)]
This means that the
rate of energy consumption is not constant with time because:
dE/dt
= P0S exp(-St/V)
But the actual
pump-down time from P0 to Pf is given as:
t
= (V/S) ln(P0/Pf)
Numerically, if V= 1 m3
and S = 0.1 m3/sec, P0 = 1 bar, and Pf = 10 millibar:
t = 46 sec.
The energy consumed in
this process is:
E = (1.013x105)(1)[1 – exp(-4.6)] = 1.003x105
J
Suppose pf
=1 mbar, then t = 69 secs.
As long as speed s
remains constant, each decade reduction in pf would only add 23
secs.
The power consumption
on going from 1 bar to 1 millibar is = 1x105/69 = 1450 W.
And there is literally
no extra energy consumed in this process… because the maximum energy consumed can
only be 1.013x105 J!
So this is just what Prashant Thankey also said...with equations, and some numbers.
However, the reason
extra energy is consumed is because any pump consumes power by being on,
overcoming internal friction etc.
Also, pump speed s will not
remain constant. As you approach the ultimate pressure, the speed will go to
zero – and each vacuum pump, has an ultimate pressure, whose
value depends upon the
operating principle. For example, cryopumps do not have mechanical moving
parts. But they do have an ultimate vacuum limit. In addition, one would have
to incorporate the effects of outgassing, diffusion and permeation that act to
limit the highest attainable vacuum level.
The power consumption
of most diffusion pumps is also of the order of 1500 W.
Thus, the power
consumed by pump-down is not a large fraction of the total power consumed – but
it is not ‘minuscule’ as Christian Binek would suggest.
However, the pump-down power
will indeed become a small fraction if we are talking about ultrahigh vacuum
since it takes hours (if not days!) to reach these vacuum levels.
Realistic calculations
would take into account actual geometry of a vacuum vessel including the vacuum
duct used and the effective speed at the mouth of the vessel.
However, that is a rather more detailed analysis than I plan to do!
As Soucek & Desai argued, it is best to just use a wattmeter and do the actual measurements. This is obviously the correct approach... but it is complicated a bit by the fact that that the total time required to reach a fairly high vacuum will depend upon the degree of outgassing, diffusion & permeation in the material of the vessel, assuming that there are no microleaks.
Endnote: Clearly, the power consumption is not going to increase as you go to higher vacuum levels (more rarefied), but you could argue that since the time required to attain ultrahigh vacuum is much more, it is the total energy consumption (in Joules) that will increase as the level of vacuum increases. However, the same problem occurs as in the previous para: it depends more on the details of the vessel (outgassing, diffusion, permeation etc), than on the pump...
However, that is a rather more detailed analysis than I plan to do!
As Soucek & Desai argued, it is best to just use a wattmeter and do the actual measurements. This is obviously the correct approach... but it is complicated a bit by the fact that that the total time required to reach a fairly high vacuum will depend upon the degree of outgassing, diffusion & permeation in the material of the vessel, assuming that there are no microleaks.
Endnote: Clearly, the power consumption is not going to increase as you go to higher vacuum levels (more rarefied), but you could argue that since the time required to attain ultrahigh vacuum is much more, it is the total energy consumption (in Joules) that will increase as the level of vacuum increases. However, the same problem occurs as in the previous para: it depends more on the details of the vessel (outgassing, diffusion, permeation etc), than on the pump...
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