Omni-Jet

[omnicron]

Ok, another thread...I know I know but this should have it's own post as its well really cool!
Being Octa-jet has already been coined I had to come up with a different name...so Omni-Jet lol.

So I have to give littlenugget1 a lot of credit. He's helping out with cutting and is going to tig everything together once I get the holes for the inductor's cut.

As you can see, it's going to have 8 - 1/2"ID tubes, the compression chamber is 5x10" for a 6" jet.

It's going to be a dual inlet jet with the inlets per Hosers spec's as viewed in this sketch in the red.





Still have a lot to do but I think it's looking wicked!

Hope it works good.

 

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[principedeleon]

Ok, another thread...I know I know but this should have it's own post as its well really cool! Being Octa-jet has already been coined I had to come up with a different name...so Omni-Jet lol. So I have to give littlenugget1 a lot of credit. He's helping out with cutting and is going to tig everything together once I get the holes for the inductor's cut. As you can see, it's going to have 8 - 1/2"ID tubes, the compression chamber is 5x10" for a 6" jet. It's going to be a dual inlet jet with the inlets per Hosers spec's as viewed in this sketch in the red. <img src="http://www.treasurenet.com/forums/attachment.php?attachmentid=947881"/> <img src="http://www.treasurenet.com/forums/attachment.php?attachmentid=947884"/> Still have a lot to do but I think it's looking wicked! Hope it works good.

How big is the hoes going to be ? Isnt it suppose to be around 1.5 inches of water for a 6"?

 

[omnicron]

2 - P180's

2 - 2" lines as in the drawing marked in the red.

 

[omnicron]

It's designed like this.
Each 2" line is going to get a 1" reduction That's going to build pressure.
If you do the math:

3.14 x 1sqrd = 3.14

3.14 x sqr(.5) = .25 x 4 = 3.14

It should hopefully work good.
I base this on my experience of my 4" log I built. It had a 1" reduction and 1" inductor.
With the 8 inductor pipes, it should be close to the efficiency of a infinity.

 

[principedeleon]

So each pipe will have a 1/4 inch hoe right ? I know your trying to make a quick jet. But you should have tried to eliminate the retriction lost that is caused by those small tubes. Would be better if water flared in like in a log jet even some tri jets and other ones i seen water flaring in.

 

[omnicron]

So each pipe will have a 1/4 inch hoe right ? I know your trying to make a quick jet. But you should have tried to eliminate the retriction lost that is caused by those small tubes. Would be better if water flared in like in a log jet even some tri jets and other ones i seen water flaring in.

Um no, 1/2 pipe = 1/2 hole err well 1/2x 1 3/4" elongated.

 

[omnicron]

P, There is only 2 designs that will have no loss. Jet logs and well I'm not going to give away this other design yet...There may be others but for the most part youor going to have loss. It's like with a sluice. you could spends thousands of dollars trying to capture every single fleck of gold, it's just not going to happen. Your always going to lose some. You have to come to a point when you realize this and do what you can to minimize this. I call it picking your battles. This jet cost me about $35 to build...I did what I could to minimize pressure loss. To do the tubes in a cone shape is a losing battle. That's custom work that I cant do at home.

I will admit, I have no idea how this is going to work, it may work awesome of be a complete failure. I ran my numbers by a few people and they said it should work. I liken this to Timbers Octajet.

 

[Hoser John]

CONVALUTION IS NOT THE SOLUTION. What you are looking for is a flow that fully engages the inlet flow. That's why I stopped at 3 inlets as 4-5-6 did NOT help for squat and cut'm up for parts. You reach the point of diminishing return. Pressure head and supply is already built into the pressure chamber that feeds the induction tubes so just more convalution and duplication of science already applied. Make sure to weld as written as much easier to get the first flange straight with a T-double weld both inside and out as done in this fashion as induction is always the point of galvanic action. Also you are using a P-180 which is 2 1/2" out x 2' pumps so you have 4" total from 2. Now you propose to run 1/2 x 8=4" so WRONG as no restriction/backpressure within pressure chamber just 4 " and 4" total out. Simplify is the answer. I have a pic of a 8" hydra 12" made for alaska with 8 inductors and it was GARBAGE setup as your are proposing. Slow down and think,what has happened every time you complicate something to do with mining-YOU LOSE YOUR ARSE-KISS works everytime BUT have a ball man-.John PI(3.14 )x radius squared is area in squared inches of round induction-1" pipe= .5 x .5 x 3.14=.785 inches.

 

[omnicron]

Scratch this project!
LOL
John what size should my inductors be with my pump config?

I'm confused now. After your post, I discovered that I was all screwed up. Like my math on the 1/2 pipes for one.

3.14 x .25(.25) = .196
1/2 of 2" pipe is 1
3.14 x 1(1) = 3.14
so there is 2- 2" feeds so 6.28
8 - 1/2 pipes so
.196 x 8 = 1.57
This project is scraped before I get to far.

 

[principedeleon]

Okay john explained that water issue i wanted to let you know.

try smaller hoes you could create more pressure from the pump to actually make it work better .

You are basically almost done anyways.
i think it would be dumb just throwing the project in a corner some where..

 

[omnicron]

Okay john explained that water issue i wanted to let you know.

try smaller hoes you could create more pressure from the pump to actually make it work better .

You are basically almost done anyways.
i think it would be dumb just throwing the project in a corner some where..

But I only have 1.57 square inches for my tubes.
I have 6.28 square inches of feed

If I understand what John was getting at, the compression chamber is creating the pressure. If I have 2 - 2" lines (4") coming in, I need 4" going out for my tubes. I'm 3 kinds of F'd up right now with confusion.

I don't want to waste this piece of pipe or I'll have to buy more.

 

[principedeleon]

But I only have 1.57 square inches for my tubes. I have 6.28 square inches of feed If I understand what John was getting at, the compression chamber is creating the pressure. If I have 2 - 2" lines (4") coming in, I need 4" going out for my tubes. I'm 3 kinds of F'd up right now with confusion. I don't want to waste this piece of pipe or I'll have to buy more.

A 3" normally haves 3/4" of water pressure, 4" haves 1" of water . ..

So a 6" should be around 1.5" of water.
But if you have a bigger pump than normal you could get away with 2" .

 

[omnicron]

On a standard log. If you read what John posted I gather this, the compression chamber builds the pressure, if you have 4" of water in you need 4" out...unless I'm not understanding what I am reading.

 

[principedeleon]

On a standard log. If you read what John posted I gather this, the compression chamber builds the pressure, if you have 4" of water in you need 4" out...unless I'm not understanding what I am reading.

Then thats what you have .. Right? 8x .5"=4"

 

[omnicron]

No 1.57 or somthing

 

[littlenugget1]

omni you need to have more water coming into the chamber or less going out.you can not have the same in as you do out. thats what i was trying to tell you. you have 4" coming in you need no more than 2 1/2" going out for it to build up pressure and do what you need.

 

[mxer47]

Idk about all this. I have a standard jet log. I will say as a t-net student and observer you have the water right. I think what John is saying is K.I.S.S. 4" in 4-1" out. I do agree about finishing. It looks to me you are 3/4 of the way there, besides it is nice looking you might be able to sell it as art on eBay if it doesn't work well.

 

[principedeleon]

No 1.57 or somthing

I dont know how are you calculating it .. I cant see how you with 8 tubes with lets say 3/8 of inch holes going to put out 1.5 " of water

 

[omnicron]

calculate area of a circle is pie 3.14 x r(r) r is radias. That number is 1/2 of the diameter. 2" is 1" .5" is .25

so I take one .5" tube. 3.14 x .25(.25) = 0.19625 then multiply that by 8 tubes.
0.19625 x 8 = 1.57"....so I have 1.57" area of tubes.

Two 2" water feeds...Take one...3.14 x 1(1) = 3.14 for one 2" line. I have 2 so 3.14 x 2 = 6.28" of area
6.28 - 1.57 = 4.71....
So I have 6.28" water coming in, I'm using 1.57" through the tubes and have 4.71" left over (pressure).

IDK It may build pressure, but is it going to have enough flow? My though is a infinity jet gains it's effiency from having more surface area to push with ie" using 2 hands to push a car verse using only 1 hand.
To me having 8 hands pushing would be better then 3 hands.
With that said, I don't know to figure out how much water is needed...

 

[principedeleon]

calculate area of a circle is pie 3.14 x r(r) r is radias. That number is 1/2 of the diameter. 2" is 1" .5" is .25 so I take one .5" tube. 3.14 x .25(.25) = 0.19625 then multiply that by 8 tubes. 0.19625 x 8 = 1.57"....so I have 1.57" area of tubes. Two 2" water feeds...Take one...3.14 x 1(1) = 3.14 for one 2" line. I have 2 so 3.14 x 2 = 6.28" of area 6.28 - 1.57 = 4.71.... So I have 6.28" water coming in, I'm using 1.57" through the tubes and have 4.71" left over (pressure). IDK It may build pressure, but is it going to have enough flow? My though is a infinity jet gains it's effiency from having more surface area to push with ie" using 2 hands to push a car verse using only 1 hand. To me having 8 hands pushing would be better then 3 hands. With that said, I don't know to figure out how much water is needed...

Im getting mixed up with this whole calculation thing.. I see a little what your saying but just cant figure out how you put 8 tubes of .5" in a 1.5" tube

Edit : i used a program and it cleared to me exactlly what your saying . So it should be right.

So even if i have 2 half inch water entering the jet doesnt equal up to the area of a 1 one inch pipe .
Thats maybe why i also generated so much pressure in my 4".