Thanks. I shall read more..
K2
K2
Monotube Flash Boiler Design
- Thread starter Toymaker
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De Laval Nozzle...
Watched a U-tube presentation on it...
Sorry, the maths didn't "click"...
I need a dictionary to translate the language into something I comprehend.
e.g. "Isentropic"? - My non-existent Latin and Greek can't translate this into something...
But I learned that sub-sonic gases slow down in the nozzle, whereas supersonic gases accelerate in the nozzle. Those words had meaning for me.
I Just don't understand what is happening in the centrifugal filter now..
Does air injected at the equator accelerate when injected tangentially into the sphere as it moves to the poles? (supersonic flow?) - or decelerate? (sub-sonic flow?).
I suspect the end of the tangential feed-in pipe is square ended, so creates turbulence at the periphery of the jet, and maybe some laminar flow towards the middle of the jet of gas? I.E. it is not a De Laval nozzle, so does not work the same way?
And when the air - less oil droplets - leaves the flow (upwards) at the middle of the sphere, how does the velocity of flow convert into the pressure of flowing air in the pipe?
All I can imagine is that there are some frictional losses inside the sphere, so air entering is from a pipe at pressure Pi, and leaving at a lower pressure Pe...?
Oil droplet mass at Vin has a momentum and energy exchange to the walls of the sphere so Vin is reduced to zero... But how does this help the air velocity and Pe?
So I shall give up on that one.... Life is too short.
I did get slightly involved with gas centrifuges separating U238 from U235...., I.E. the same technology as this oil separator, but I cannot discuss that...
K2
Watched a U-tube presentation on it...
Sorry, the maths didn't "click"...
I need a dictionary to translate the language into something I comprehend.
e.g. "Isentropic"? - My non-existent Latin and Greek can't translate this into something...
But I learned that sub-sonic gases slow down in the nozzle, whereas supersonic gases accelerate in the nozzle. Those words had meaning for me.
I Just don't understand what is happening in the centrifugal filter now..
Does air injected at the equator accelerate when injected tangentially into the sphere as it moves to the poles? (supersonic flow?) - or decelerate? (sub-sonic flow?).
I suspect the end of the tangential feed-in pipe is square ended, so creates turbulence at the periphery of the jet, and maybe some laminar flow towards the middle of the jet of gas? I.E. it is not a De Laval nozzle, so does not work the same way?
And when the air - less oil droplets - leaves the flow (upwards) at the middle of the sphere, how does the velocity of flow convert into the pressure of flowing air in the pipe?
All I can imagine is that there are some frictional losses inside the sphere, so air entering is from a pipe at pressure Pi, and leaving at a lower pressure Pe...?
Oil droplet mass at Vin has a momentum and energy exchange to the walls of the sphere so Vin is reduced to zero... But how does this help the air velocity and Pe?
So I shall give up on that one.... Life is too short.
I did get slightly involved with gas centrifuges separating U238 from U235...., I.E. the same technology as this oil separator, but I cannot discuss that...
K2
NapierDeltic
Well-Known Member
My position - you can find tons of images/documents on Net to figure how complicate is this "simple" device. But the overall impact is not that significant in Toymaker's powerplant so I wouldn't stuck here .
Of course we can discuss - clarifications are always welcome. And ... it is working on -now- common house vacuum cleaners , not by magic
.NapierDeltic
Well-Known Member
As the cherry on the cake, in 80's I have read a book about Ranque-Hilsch Vortex Tube - RHVT and its current (or intended) applications.
https://www.airtx.net/airtx-vortex-tubes-review
Today you can struggle to find something about it....
But at those days, mathematical description/modeling of processes involved was not possible; though research was on, theoretical and practical.
That would be interesting to understand entirely
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If you start with accepting that the oil-laden air enters the sphere with relatively high kinetic energy, (ie, it enters with fairly high sub-sonic velocity) then it might help to think of the air as oil coated BBs. As the BBs rub against the sphere's inner wall, some of the oil rubs off onto the wall via direct contact, and as the BBs are forced to spiral around the inner wall, centrifugal force will pull more oil off the BBs and onto the sphere's wall. Gas expansion is not a part of how the oil is removed from the air, it's all done through momentum and centrifugal force.
As soon as air leaves the nozzle it begins to decelerate.
I posted this pic of the nozzle back in post #419. As is shown, the nozzle is just a plain drilled hole with no attempt to shape it into a De Laval configuration. So, no supersonic flow anywhere.
Pressure inside the sphere is only slightly lower than pressure from the compressor. As the air's velocity slows after it leaves the nozzle, the pressure rises,...which is just part of the conservation of energy,...as velocity lowers (ie, kinetic energy is lower) pressure energy (which is a form of potential energy) rises.
The Ignoble Troll
Active Member
Isentropic = same energy.
So the working fluid has the same energy before and after but expressed differently.
Like how 1 gallon of fluid moved at 1714psi is 1 hp and 1714 gallons moved at 1psi is 1 HP.
With the perfect pumps/fluid motors at magical 100% efficiency, you could transition between the two mass/pressure volumes. If you did, the process would be isentropic, no energy added or removed.
On air flow:
All nozzles: convergent, divergent, convergent-divergent must have a lower down stream pressure to operate. Pumping 10 psig into a nozzle does not allow you to pump into a static 10 psig filled receiver. The flow would stagnate.
Otherwise a loop of pipe with a restriction would form a perpetual motion machine.
Iirc( don't quote me) for supersonic flow at room temp you need something like double plus a bit PSIA to have supersonic flow. (A not G, atmospheric pressure joins the chat)
Ie. This is why a compressor air gun gets suddenly much louder as you increase flow to it, once you have achieved choked flow you get a sustained sonic boom which translates into a much louder HSSSSSSSS sound.
A centrifugal oil seperator works the same as a uranium centrifuge but you get to use Stokes equation and the air velocity to calculate size/droplet mass removal. But that's math I no longer remember how to apply in these circumstances...
Thanks. I wasn't involved in the centrifuge design as such. Just installation of power supplies to centrifuges. Can say no more.... signed Official Secrets paperwork. So I know NOTHING about the system.
Cheers! (Before they lock me away).
K2
Cheers! (Before they lock me away).
K2
Hilsch vortex tube is in common use, especially in cooling instrument cabinets and hazmat suits which need cooling where the cool air is placed usually at the neck or helmet area. But energy use is quite high when considering you need high compressed air at least 100psi.As the cherry on the cake, in 80's I have read a book about Ranque-Hilsch Vortex Tube - RHVT and its current (or intended) applications.
https://www.airtx.net/airtx-vortex-tubes-review
Today you can struggle to find something about it....
But at those days, mathematical description/modeling of processes involved was not possible; though research was on, theoretical and practical.
That would be interesting to understand entirely
A sphere is usually a poor shape for particulate separation and would not normally conform to separation physics. Has this geometry been tested for efficiency other then this one off use. Typically you inter with a high velocity and the walls pick up the particles and the exit is a much larger diameter with slower velocities as the entertainment is dependent on carrying velocity. Plus there is the need to remove the material on high load conditions.
I hope we can all agree that centrifugal force plays a major role in separating the oil particles from the air, so lets look at the math used to find centrifugal (or centripetal) force:
The air-oil mixture is injected near the equator of my sphere where the radius, r is greatest; the velocity, v of the air-oil mix will also be greatest at this point. As the air-oil mix circles the inside of the sphere, it's forced upwards towards the exit, and as the mixture moves upwards the radius, r becomes smaller and smaller, which, applying the formula, will cause the centrifugal force, F to increase. Clearly the mixture's velocity, v will slow and it's mass, m will also decrease as the oil is removed from the air, but the formula tells us that if the radius, r remained the same, as it would inside a straight tube, or if it became larger as it would in a conical shape, than the centrifugal force, F exerted on the air-oil mixture, would decrease compared to a sphere.
The proof of how well the sphere works is shown in the video in post #484.
If you Google: spherical separator you'll find they're often used where only a small space is available. However, I could not find any that use the same vortex principle I'm using.
Thanks Toymaker. I was envisioning the sphere as similar to a pair of cones - just non-linear - joined at their common large diameter. Air injected tangentially at the equator splits so half goes up and half down... both have reducing diameter vortices, depositing oil droplets on the inside walls as the swirl reduces diameter. The air going "generally upwards" turns at the top to pass downwards, and the air going generally down turns to go upwards, so I reckon the 2 streams of cleaned air would meet in the middle, and exchange velocity (kinetic energy) for pressure (very low momentum) so there is a local high pressure zone there, where air-streams meet? - Thus I would locate the exit orifice for clean air at the centre of the sphere.
Is this the right idea? - Or just too simple?
I successfully use simple oil separators on my steam engine exhausts: A simple piece of pipe, exhaust steam (with oil vapour and droplets) entering pointing downwards at one side of the closed cylinder, and exiting through a central pipe after a 180degree turn so the steam and water vapour goes upwards. That leaves me "clean air and water vapour" - free from oil droplets. Maybe I should now make a centrifugal separator - and see if this extracts more oil? - I have not detected oil in the exhaust after my simple separator, so probably unnecessary?
K2
Is this the right idea? - Or just too simple?
I successfully use simple oil separators on my steam engine exhausts: A simple piece of pipe, exhaust steam (with oil vapour and droplets) entering pointing downwards at one side of the closed cylinder, and exiting through a central pipe after a 180degree turn so the steam and water vapour goes upwards. That leaves me "clean air and water vapour" - free from oil droplets. Maybe I should now make a centrifugal separator - and see if this extracts more oil? - I have not detected oil in the exhaust after my simple separator, so probably unnecessary?
K2
You're almost right, except I only have the top half of the sphere to use as the centrifugal vortex separator because the bottom half is filled with oil, and the center section is where the pump sits.
Edit: added more detail to drawing.
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Well I googled spherical separator and they were all quite large operating as dead space removal with entrained filters to capture oil and liquid products. None used the vortex principle as that requires the establishment of a low pressure area in the center. However did find similar designs for dry sump oil systems. Oil was introduced tangent to the cylinder walls. In addition to be an injector the nozzle must be diverging converging venturi using a motive force. Of these systems I have had more than just a common knowledge base from low pressure steam recovery, oil savaging, condensate recovery and emergency pumps. But it appears you have built a dry sump system which will work quite well for your system.
Making Steam Again: I paced the burner assembly inside the boiler coils and made a little steam today. Below is a photo of my outdoor test stand; Lowest shelf shows the feed pump just to the left of the bright red Diesel fuel container. On the middle shelf, from left to right, is the air compressor assembly, center is the ECU (Engine Control Unit), far right is a power strip, 36 vdc power supply, and 1 liter fuel container. Top shelf, center is the boiler/burner assembly and the display screen.
I performed this test mostly to determine what numbers were needed for the feed pump at the burner's idle power setting, (with no pressure) so that I could plug that value into the software as this is how I want the boiler to start up. I initially set the feed pump for what I knew would be too much flow, and once wet steam began to exit, I slowly decreased the feed pump's RPM until I determined only dry steam was exiting the boiler tube. Once I had the feed pump RPM number, I turned off the burner and allowed the feed pump to continue pumping without changing the setting. After a few seconds of cooling down, the water from the boiler's steam outlet was flowing smoothly; placing a measuring cup to catch the water from the boiler, I timed the flow rate at 425 ml/minute. Since 1 ml of water = 1 gram, that meant my boiler was producing 425 grams of steam per minute. Converting to pounds; 425 grams/min = 0.937 lbs/min,...or 56.2 lbs/Hr.
Some folks consider Boiler Horsepower to be obsolete, but I've also noticed that many text still use it, and since I'm only looking for a ball-park value, seems appropriate to use here.
1 BHP = 34.5 lbs steam per Hr at atmospheric pressure.
Therefore, my boiler's output of 56.2 lbs/Hr divided by 34.5 = 1.63 BHP.
Since 1 BHP = 13.15 HP than 1.63 BHP x 13.15 HP/BHP = 21.4 HP,....at idle
I performed this test mostly to determine what numbers were needed for the feed pump at the burner's idle power setting, (with no pressure) so that I could plug that value into the software as this is how I want the boiler to start up. I initially set the feed pump for what I knew would be too much flow, and once wet steam began to exit, I slowly decreased the feed pump's RPM until I determined only dry steam was exiting the boiler tube. Once I had the feed pump RPM number, I turned off the burner and allowed the feed pump to continue pumping without changing the setting. After a few seconds of cooling down, the water from the boiler's steam outlet was flowing smoothly; placing a measuring cup to catch the water from the boiler, I timed the flow rate at 425 ml/minute. Since 1 ml of water = 1 gram, that meant my boiler was producing 425 grams of steam per minute. Converting to pounds; 425 grams/min = 0.937 lbs/min,...or 56.2 lbs/Hr.
Some folks consider Boiler Horsepower to be obsolete, but I've also noticed that many text still use it, and since I'm only looking for a ball-park value, seems appropriate to use here.
1 BHP = 34.5 lbs steam per Hr at atmospheric pressure.
Therefore, my boiler's output of 56.2 lbs/Hr divided by 34.5 = 1.63 BHP.
Since 1 BHP = 13.15 HP than 1.63 BHP x 13.15 HP/BHP = 21.4 HP,....at idle
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OK, extrapolating (have I understood this and got the maths right?) - 21.4HP = 21.4 x 0.746kW? = 16kW. in "Metric", if anyone is bothered about Metric!
A pretty good "idle" boiler! - Like a 3 1/2 in steam loco? or small traction engine?
From fuel flow you can probably extrapolate "max power", though not a real expectation as life is never that good.
K2
A pretty good "idle" boiler! - Like a 3 1/2 in steam loco? or small traction engine?
From fuel flow you can probably extrapolate "max power", though not a real expectation as life is never that good.
K2
Yes! You got the math right
BTW, I use the word "idle" to mean the lowest power output from the burner which consistently runs without the flame going out. I've never measured fuel consumption at the idle setting.
Thanks. I wasn't sure if fuel was metered, or just supply air (pressure?) to the jet? I think you have mentioned instrumentation and control system, somewhere...
If you have 16 kW of steam, it would not be unreasonable to guess something like 20 ~25 kW of fuel?
Which sounds like a largish burner to me....
What is your max fuel rate expected to be? I was thinking you could estimate max steam power by comparing fuel feed rates, from your idle, to give a potential max steam power?
K2
If you have 16 kW of steam, it would not be unreasonable to guess something like 20 ~25 kW of fuel?
Which sounds like a largish burner to me....
What is your max fuel rate expected to be? I was thinking you could estimate max steam power by comparing fuel feed rates, from your idle, to give a potential max steam power?
K2
Fuel flow through the venturi effect nozzle is controlled by air pressure from the air compressor, and compressor output, (RPM), is controlled by a PWM (Pulse Width Modulated) power supply inside the ECU. Measuring air pressure from the compressor hasn't been necessary, yet.
At 95% full power, the burner consumes 10 LPH (Liters Per Hour), which equates to 103 KWH or 138 HP-Hours (see post #405, page 21). I don't expect to see much increase from 95% to 100%.
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