IPFS

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Institute for Plasma Focus Studies

 

Internet Workshop on Numerical Plasma Focus Experiments

 

Module3; You may also wish to refer to the supplementary notes part2supplementary.htm.

 

Summary:

 

This module is a  consolidation of Modules 1 & 2. Module 3 is divided into two parts..

 

For the first part we look at a commonly encountered situation when Lo is given only as a nominal or very approximate value and ro is not even mentioned. Then there are 6 fitting parameters; and the process becomes more involved. Nevertheless we have found that, despite that, it is still possible to get a reasonable fit. In these sessions participants will be taken through that experience which will enhance our ability and confidence to fit.

 

In the second part is an exercise in fitting the DPF78, with bank, tube and operating parameters all provided; but with Lo nominal and ro not given. The participant will fit the current curves. The properties of the DPF78 will then be placed on the comparison Excel Sheet PFcomparison.xls which you have saved from last week’s work.

 

Steps:   Part 1:

            (a) To configure the code for the PF1000 using nominal Lo, trial ro and trial model parameters.

             (b) To place a published PF1000 current waveform on Sheet 2.

            (c) To place the computed current waveform on Sheet 2 in the same figure

            (d) To vary Lo, ro and the model parameters until the two waveforms achieve the best match.

 

Part 2:  Exercise 4: Given measured current waveform data for the DPF78; given bank (with nominal Lo and no ro) , tube and operating parameters; participant will fit computed to measured current waveform. Then tabulate DPF78 computed properties into the PFcomparison.xls file which was saved from Week 2; or use attached PFcomparisonpf1000pf400.xls provided for your convenience.

 

 

The material:

 

You need RADPF5.15dd.xls for the following work. You should have a clean copy in your Reference Folder. Copy and Paste a clean copy on your Desktop. You should have RADPF5.15dd.xls on your Desktop before the next step.

 

You need the file PF1000dataNom.xls and DPF78dataNom.xls. You also need the file PFcomparison.xls, saved from last week’s work.[or the one attached for your convenience PFcomparisonpf1000pf400.xls]

 

(a) Configure the code for PF1000

 

Double click on RADPF5.15dd.xls (Excel logo RADPF5.15dd.xls on your Desktop).

Click on enable macros

The worksheet opens.

[Type in cell B3: PF1000; for identification purposes.]

 

The PF1000, at 40kV, 1.2 MJ full capacity, is one of the biggest plasma focus in the world. It is the flagship machine of the International Centre for Dense Magnetised Plasmas. On their website, inductance was quoted as 9nH for short circuit.

 

We searched through PF1000 publications and found figures for Lo of ‘around 20nH’ mentioned. For this work we assume we are looking at PF1000 for the first time and all we got for Lo is the figure Lo=20nH. There is no mention of  ro. So we use a starting value of  ro=0.4mW; this being 0.1 of the bank impedance (Lo/Co)^0.5 taking Lo as 20nH.

 

We use the following bank, tube parameters and operating conditions.

Bank:               Lo=20 nH (nominal), Co=1332 mF, ro=0.4mW (guess value).

Tube:                b=16 cm, a=11.55 cm, zo=60 cm

Operation:        Vo=27kV, Po=3.5 Torr, MW=4, A=1, At-Mol=2

 

We assume that we are starting to look at PF1000 for the first time; and  that  we do not know the model parameters. We will use the trial model parameters recommended in the code (See cells P9-V9)

 

Model Parameters:         massf=0.073, currf=0.7, massfr=0.16, currfr=0.7;  first try.

 

Configuring:   Key in the following: (e.g. in cell A5 key in 33.5 [for 33.5nH], in cell B5 key in 1332 [for 1332mF] etc)

                        A5       B5        C5       D5       E5        F5

                        20        1332    16        11.55   60        0.4

 

            Then     A9       B9        C9       D9       E9

                        27        3.5       4          1          2

 

            Then     A7       B7        C7       D7

                        0.063  0.7        0.16     0.7  for first try

 

Fire the PF1000 with these parameters.

 

(b) Place the published PF1000 current waveform on Sheet 2

 

We repeat the procedure to place the published PF1000 current waveform on sheet 2.

With RADPF5.15dd.xls (fired as PF1000 with first try parameters) open; open PF1000data.xls; click the Edit Tab; scroll down and click 'Move or Copy file'. A window pops out. In the 'To book: choose RADPF5.15dd.xls’; then choose ‘move to end’; click ‘OK’. Rename ‘Sheet1(2)’ as Sheet2.

The measured current waveform is now displayed in the chart in Sheet2 of RADPF5.15dd.xls.

 

(c) Place the computed current waveform on Sheet 2 in the same figure

 

Place the computed current waveform on the same chart following the same procedure we did in Part 2; using the strings: “=sheet1!$a$20:$a$6000” [without the quotation marks] and “=sheet1!$b$20:$b$6000” [without the quotation marks].

 

The pink trace is the computed current trace transferred from Sheet 1.

 

 

Comparison of traces: Note that there is very poor matching of the traces; using nominal Lo, guessed ro and the first try model parameters.

 

(d) Varying model parameters and Lo and ro to obtain better matching of computed current to measured current traces

 

To vary model parameters:

 

(i)                              Note: that the computed current dip comes much too early;

             that the computed current rise slope much too high;

                   that the computed current maximum is much too large.

 

Suppose we do not know that Lo is not a correct value.

 

Try varying axial model parameters, which as we know control the current trace up to nearly the start of the roll-over  region of the current trace. To make the dip come earlier try increasing fm; which will slow down the axial speed (but as we know now, that will also reduce the circuit loading, leading to an even larger current; we got to try something anyway). The deviation is very large, so take a large step; say put fm=0.8 [note: max

allowed value of fm is 1]. That improves the time position of the dip, but as we expected the current got even bigger. Next try increasing fc, which will increase the dynamic loading effect of the dynamics on the circuit. Put fc to its max allowed value of 1

 

The time position of the dip is now good and the peak current has improved, but is still way too large. There is not much else we can do with fm and fc. (you could try reducing them, but you know by now that you are not going to see any improvement). Perhaps we could increase ro; which will lower the whole current profile. Again large difference, need large change. Try ro=2mW. Improvement, but not enough. Try ro=10mW. Possible improvement, but looks like we have gone beyond. Next try 7mW.

 

The topping profile  deviation has now improved, even touching the measured current profile at one place. But the top is too droopy; and the decreased current has pushed the dip too late. At the same time the current rise rate is still too high. Try reducing fm to 0.4.

 

There are now points of agreement; but the current rise slope is still too steep and the topping profile is still too droopy.

 

It is now clear that in all the things we have tried, the rising slope of the current profile is still too steep. How do we reduce the slope? From capacitor discharge behaviour, we know that increasing Lo would do it. (So would increasing Co; but in this case we are fairly sure that the given value of Co is more reliable than the nominal value of Lo.) So let’s try Lo=25nH; at last we see the slope beginning to match. Next try 30nH; even better. We can see now that at last we are getting onto a better track. It is therefore better to go back to more normal values of fm and fc ( rather than the unusual values we tried in our desperation) Go back to fm=0.15 and fc=0.7. The matching is improving, but there is still that extra slight droop at the top. Try reducing ro to 6mW.

 

It looks like we are getting there, but the rising slope could on average be improved by a larger Lo, which would also lower the top. Try Lo=33nH. The slope match is now pretty good on average, top still too high. Making small changes to Lo and ro, one comes to a final best fit for these two bank parameters which will not be too far away from 33nH and 6mW. The rising slope profile and the topping profile up to the rollover region of the current trace are now fairly well fitted.

 

Next make adjustments to fm and fc until the final best fit is obtained for the axial phase up to the region of rollover from the current top to the dip.

 

However we note that the radial phase is yet to be fitted and currently has fmr=0.16 and fcr=0.7. [We have already done this part of the fitting in S3S4 when we fitted the same curve for PF1000, except that then we were given the correct value of Lo, which in that case made the fitting of the axial phase much more simple. The fitting of the radial phase as suggested below should sound familiar]

 

Note that the computed current dip is too steep, and dips to too low a value. This suggests the computed radial phase has too high a speed. Try increasing the radial mass factor, say to 0.2. Observe the improvement (dip slope becomes less steep) as the computed current dip moves towards the measured. Continue making increments to massfr. When you have reached the massfr value of 0.4; it is becoming obvious that further increase will not improve the matching; the computed dip slope has already gone from too steep to too shallow, whilst the depth of the dip is still excessive. To decrease the depth of the dip try reducing fcr to say 0.68. Notice a reduction in the dip. By the time we go in this direction until fcr  is 0.65, it becomes obvious that the dip slope is getting too shallow; and the computed dip comes too late.

 

One possibility is to decrease massfr. Try 0.35

The fit is quite good now except the current dip could be steepened slightly and brought slightly earlier in time. Try decreasing massfr, say to 0.35.

The fit has improved, and is now quite good, except that the dip still goes too low.

 

However we can check the position of the end of radial phase which is at time=9.12 us. Putting the cursor on the pink curve at the point t=9.12, we note that the agreement of the computed curve with the measured curve up to this point is fair.

 

The best fit? Anyway, a good working fit!

 

 

 

So after finding the correct values of Lo and ro and fitting the model parameters, we should have gained more confidence in the ability of this method of finding a good fit. We repeat that  after this fit we have confidence that the gross features of the PF1000 including axial and radial trajectories, axial and radial speeds, gross dimensions, densities and plasma temperatures, and neutron yields up to the end of the radial phase may be compared well with measured values.

 

Moreover the code has been tested for neutron and SXR yields against a whole range of machines and once the computed total current curve is fitted to the measured total current curve, we have confidence that the neutron and SXR yields are also comparable with what would be actually measured.

 

For example, the neutron yield computed in this shot of 8.6x10^10 is in agreement with the reported PF1000 experimental experiments; (range of 2-7x10^10 with best shots at 20x10^10).

 

 

(e) Exercise 4:

 

We are given the following parameters for the DPF78, operating at 60kV, 7.5 Torr D2. 

 

 Lo=44.5nH (nominal)  Co=17.2uF   b=5 cm,  a=2.5 cm  zo=13.7cm,

The DPF78 was a high voltage plasma focus operated at the IPF at Stuttgart. This current waveform (file DPF78dataNom.xls) was provided recently by H Schmidt.

 

Use our Universal Plasma Focus Laboratory code RADPF5.15dd.xls to configure the DPF78. Add the DPF78 data to Sheet 2. Then fit the computed current waveform to the measured.

 

[Hint 1: you need to assume a try value of ro in the same way we did for PF1000; ie try ro= 0.1*(Lo/Co)^0.5; which will print out in cell F13 RESF=0.1 where RESF=ro/(Lo/Co)^0.5.

Hint 2. the value of RESF very seldom goes below 0.05; so don’t put ro so small that RESF (F13) goes below 0.05.  Hint 3. The current rise slope is most controlled by value of Lo (also by Co, but in this case we are given a reliable value of Co).  Hint 4. Increasing fm  has the effect of reducing axial speed and increasing Ipeak; reducing fc produces similar effects. ]

 

After you are satisfied with the fit, add the DPF78 properties to the comparison tabulation that was saved from last week. PFcomparison.xls.    Or use the one provided for your convenience: PFcomparisonpf1000pf400.xls.

 

Fill in the following, copy and paste and e-mail to me by 26 April 2008.

Q1: My best fitted values for PF1000, 27kV 3.5 Torr Deuterium are:

fm=                     fc=                   fmr=                        fcr=

 

Q2: Insert an image of the discharge current comparison chart in Sheet 2 here.

 [Copy the Chart and paste onto a fresh Excel workbook (with just the chart on one worksheet). Save this workbook and then paste the workbook  here.

 

Q3. Add the newly computed properties of DPF78 to the file PFcomparison.xls saved from last week. Or use the provided PFcomparisonpf1000pf400.xls. That file already contains the properties of PF1000 and PF400. You may also calculate the ratio of PF1000/PF78 for each of the properties; as we did last week for PF1000/PF400. In other words we are using PF1000 as the reference; comparing PF400 as well as DPF78 with it.

 

 

Conclusion:

 

In these two sessions we experienced a common fitting situation when the given Lo is either nominal or wrong and ro is not given. Despite having to fit these two additional parameters we found that a reasonable fit could still be achieved. The participant then proceeded to fit a similar situation with the DPF78. The properties of the DPF78 obtained in the numerical experiment are then added to the comparative tabulation obtained earlier for the PF1000 and PF400. saving the file as PFcomparisonpf1000pf400dpf78.xls.

 

We note that the DPF78 was a high voltage plasma focus, obviously designed to test higher voltage, higher speed operations, resulting in an unusually high value of S; which is about a factor of 1.5 higher than the average value of S (close to 100) for most neutron-optimized plasma focus machines.

 

Study the comparative data in the light of the discussions last week, to strengthen and consolidate the main ideas**.

 

This table could be kept and added to from time to time with data from other plasma focus which you may be able to compute. Such comparative data could be useful for theses and publications.

 

[Suggestion: You are invited to fit your own plasma focus and add the data to PFcomparison.xls. I would appreciate a copy of all your fitted (and nominal) parameters, current trace comparison, and your PFcomparison.xls; to add to our database, which will be made available to all for downloading]

 

**Some notes (edited) kindly summarized by a participant:

 

As ‘a’ increases, rmin, zmax, and pinch duration increases;  approximately linear dependence; seen in these  numerical experiments as well as in agreement with general and theoretical observations.

 As ‘a’ increases, (pinch volume*pinch duration) increases; approximately to the 4th power of 'a'; ( 1 power from each dimension). Why is this factor important to think about?

 

S factor: additional note in comparing PF1000 to PF400:

The ratio of radial speed/axial speed depends on a factor of [(c^2-1)/lnc].

This factor  [(c^2-1)/lnc]~0.92/0.32~2.9 for PF1000;  and  ~5.8/0.96~6 for PF400; PF400 will have 2x radial speed as PF1000 (since axial speeds nearly the same] ; .and for supersonic plasmas:  Temp~speed^2  that is the main reason why PF400 has several times higher temperature than PF1000; although same speed factor.

In other same S means approx same axial speed; and also approx same radial speed; and also approx same temperature for cases where 'c' is the same. In this example, 'c' is not the same and favours higher radial speeds and T in PF400.

  

Yn scales with  Ipinch, because it is Ipinch that basically powers the pinching processes during which the neutrons are produced.
(You might wish to add other points.)

 

 

 

End of Part 3.