Atom Jet with applied EVG

[PyroJoe]

In this topic I will attempt to apply EVG to a Atom Jet. The Eckstein Valve Glide origins can be found here:

dynamic modeling of a strip valve
by WebPilot
viewtopic.php?f=2&t=4984&st=0&sk=t&sd=a


page 5:
WebPilot wrote:

"I call this phenomenum 'valve gliding'.

At the end of the -tive portion of the cycle, the valve just glides in for a soft landing at a slow, ever decreasing velocity. Not only is the tip velocity zero, but the whole valve just touches the seat with a velocity of zero."


(Plans for the Atom Jet can be found under the sticky in the valved section named Looking for plans? Look here.)

[PyroJoe]

This is a walk through of previous work, although any foul ups, mis-steps or fast and loose play with terms and numbers are my own.

The key to valve glide lies in utilizing a dfr of 0.43

Ok, so what is dfr? dfr is the "driving frequency ratio"

Why is it important?
As posted above, the valve glides in slowly for a soft landing on the surface of the valve head, as opposed to a constant detrimental hammering effect of less than optimal conditions.


How do we find the driving frequency ratio? dfr=f/fn

We will need 2 frequencies to begin:

1st-frequency created by the body of the jet (f)

2nd-the resonant frequency of the valves (fn)

[PyroJoe]

--Finding f --

One tool useful for finding f is Pulse Jet Design Aid by Forrest Eckstein is provided here:

viewtopic.php?f=2&t=5584&start=0

Lets plug in the values of the Atom Jet and see what the frequency is.
(Note: to convert mm to inches divide the mm value by 25.40)

-first it is asking to pick a volume shape: 1. Cylinder 2. Frustrum of cone
Starting with the CC so lets start cylinder 1 and input:
diameter 2.126
length 1.929

the program gives a volume of 6.847in^3, and do you wish to add another volume?

input: y

now we will do the frustrum 2 and input:
diameter 1 2.126
diameter 2 1.102
length 2.953

program gives volume of 6.244in^3 and do you wish to add another volume?

input: y

now we will do the tail pipe cylinder 1 and input:
diameter 1.102
length 11.024

program gives volume of 10.515 and do you wish to add another volume?

input: n

program gives total volume 23.606, total length 15.906

Cycle frequency is 276.397 cps [Hz]

slenderness ratio is 11.571



So if all the inputs are good, f should be approx. 276.397 cps [Hz]

[PyroJoe]

--finding f -- (part 2)

Another tool that can be used is Eric Becks Pulsejet Engine Calculator 1.4 found on the main page here:

http://www.pulse-jets.com/

Using 3.815 pounds for the thrust figure gives a tailpipe diameter of 1.102"
and a estimated frequency on the low side of 283.500 close to ball park.

(the diameters and lengths are not the same, but that is to be expected, as I basically backed the tailpipe diameter into the calculator.)

[PyroJoe]

Next up is to find fn

To determine the fn of the Atom Jet petal valve, will be using the method found under the topic:
the hunt for an easier way ...
by Webpilot

viewtopic.php?f=2&t=5514


using the script file Forrest created, but utilizing BASIC found under the topic:

Back to Basic
viewtopic.php?f=2&t=5570&start=0

(apologize for all the links, but just trying to sew this stuff together)

[PyroJoe]

--Finding fn --
One of the first steps is to draw the valve similar to what was done with the Dyna Jet valve. Geez, the Atom Jet valve looks a little stubby next to the Dyna Jet valve.

[WebPilot]

 I'm just letting you know I am quietly paying attention to this, Joe. You reminded me of something I need to check and see if I finished it or not. I got sidetracked over the holidays. Nice work.

[Mark]

Interesting to compare the reeds. My Tigerjet reed is ~1.5 inches in diameter. Each petal is about 8mm at the widest, roughly measuring it just now. I wonder what's the ideal number of reeds for a flower petal pulsejet of a given size? I noticed the Atom has 8.

[PyroJoe]

Thanks for posting Forrest, and thanks for doing the heavy lifting! I think I see why you started into modeling the valve retainer now. I'll post some interesting stuff soon.

Mark, are those petals of the same thickness? It looks like simply a scale change between the two.

Joe

[PyroJoe]

Ok, have drawn the petal with the 5 main parameters needed for the BASIC program.

input1: .5811 length of rectangle 1
input2: .0823 width of rectangle 1
input3: .008 thickness of the valve
input4: .3562 distance to center of mass of side lobe
input5: .0294 area of side lobe

[PyroJoe]

the program returns:
what is the length of CAD rectangle 1?.5811
what is the width of CAD rectangle 1?.0823
what is the thickness of petal valve?.008

'Continuous beam', partial system frequency
Young's modulus, E: 30000000 lbf/in^2
density: 0.283 lb m/in^3
thickness of beam strip: 0.008 in
width of beam strip: 0.1646 in
length of beam strip: 0.5811 in
1st moment of inertia: 0.70229333e-8 in^4
mass per unit length: 0.3726544e-3 lbm/in
wl = 4866.1225 rad/sec
fl = 774.467449 Hz

What is the longitudinal distance to the CM of a side lobe?.3562
What is the area of one side lobe?.0294

Lumped mass, partial system frequency
w2 = 6371.34819 rad/sec
f2 = 1014.03156 Hz

System resonant frequency
ws = 3867.224 rad/sec
fs = 615.487815 Hz


So if all things are good. the program has found fn to be 615.48 Hz

(I think there is a way to double check this, if one has the finished valves on the valve head and retainer in place, simply blow air through the intake and record the sound. If I'm not mistaken the recorded frequency should match the above)

[PyroJoe]

So now have found f and fn.

to find dfr simply take f/fn...in this case.... 276.397/615.48= 0.449

fairly close to 0.43 so what can be done to bring it to 0.43?

Lets try altering the retainer flat contact area.

[PyroJoe]

After some plug and play, lets try increasing the retainer contact diameter from .6299" to .6530"

[PyroJoe]

input the new figures into BASIC:

input1: .5696 length of rectangle 1
input2: .0823 width of rectangle 1
input3: .008 thickness of the valve
input4: .3447 distance to center of mass of side lobe
input5: .0294 area of side lobe

and the results:

what is the length of CAD rectangle 1?.5696
what is the width of CAD rectangle 1?.0823
what is the thickness of petal valve?.008

'Continuous beam', partial system frequency
Young's modulus, E: 30000000 lbf/in^2
density: 0.283 lb m/in^3
thickness of beam strip: 0.008 in
width of beam strip: 0.1646 in
length of beam strip: 0.5696 in
1st moment of inertia: 0.70229333e-8 in^4
mass per unit length: 0.3726544e-3 lbm/in
wl = 5064.59623 rad/sec
fl = 806.055524 Hz

What is the longitudinal distance to the CM of a side lobe?.3447
What is the area of one side lobe?.0294

Lumped mass, partial system frequency
w2 = 6692.8376 rad/sec
f2 = 1065.19819 Hz

System resonant frequency
ws = 4038.61508 rad/sec
fs = 642.765553 Hz

Now find dfr again f/fn....... 276.397/642.765553=0.430

From a mere .0231" (.5867mm) increase in the diameter of the contact area!

[PyroJoe]

Maybe the radius should be machined by .02" (near the flat) to allow the contact area to stand off, and be easily measured?
(With some deductive reasoning, it becomes obvious how important it is to accurately center the valves to the retainer)