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- Thread starter T C
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- #36

- #37

jrmichler

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Now for some calculations:

A 15" diameter duct flowing at 218 ft/sec = 16,000 CFM.

The velocity pressure at that speed is 10.6" w.c.

The duct has area 1.23 ft^2.

The annulus has area 0.36 ft^2.

The air velocity through the annulus will be 218 ft/sec X 1.23 / 0.36 = 743 ft/sec.

The velocity pressure in the annulus will be (neglecting compressibility) 123 in w.c. = 4.5 PSI.

If we assume that the turbine blades are oriented 30 degrees to the airflow at the outer radius, the tangential velocity of the turbine will be 372 ft/sec, or 5700 RPM.

Conclusions and things to be aware of:

1) Installing this turbine in that duct will create a very large pressure drop due to the reduced flow area caused by the center area blockage. A carefully designed downstream section can reduce, but will not eliminate, that pressure drop.

2) As mentioned above, the entrance needs a streamlined nose cone. Lack of a properly designed nose cone will further increase pressure loss.

3) A plastic rotor of this size spinning almost 6000 RPM will explode due to centrifugal force.

4) Any power drawn from a turbine will cause pressure drop. If the system can tolerate that pressure drop, you would be better off to reduce the system pressure. The power saved from reducing system pressure will be greater than the power generated by the turbine.

- #38

T C

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How you have calculated the rpm?If we assume that the turbine blades are oriented 30 degrees to the airflow at the outer radius, the tangential velocity of the turbine will be 372 ft/sec, or 5700 RPM.

How you have calculated the annulus?The annulus has area 0.36 ft^2.

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- #39

Baluncore

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“The total air flow will be through the 12.6" ID, 15" OD annulus”.How you have calculated the annulus?

Di = 12.6”; Ri = 6.3”; Do = 15”; Ro = 7.5”.

Area = Pi * ( Ro^2 – Ri^2 ) = 52.025 sq inches.

"The airspeed in the open duct is 66.5 m/s".How you have calculated the rpm?

The duct has an area of Pi * 7.5 * 7.5 = 176.7 sq inch.

But the middle is blocked = Pi * 6.3 * 6.3 = 124.69 sq inch.

Leaving 176.7 - 124.69 = 52.0 for the airflow.

The airspeed must increase through the constriction of the turbine throat.

To; 66.5 * 176.7 / 52.0 = 226.0 m/s.

A blade at 30° to the airflow will move sideways at; 226.0 * Tan( 30° ) = 130.5 m/s.

The crude average diameter of the duct is about ( 12.6 + 15 ) / 2 = 13.8”

The mid-blade circumference of the turbine is Pi * 13.8=43.354” =1.1 m.

It will do 130.5 / 1.1 = 118.62 revs per second; = 7117. RPM.

- #40

russ_watters

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The velocity really needs to be clarified by the OP because the way I first interpreted it, the 66.5 m/s was through the turbine, not in the duct ahead of it. If your interpretation is correct, that's a big problem.If I understand correctly, you want to place a turbine in a duct. The air in the duct will drive the turbine. The air in the duct is moving 66.5 m/sec = 218 ft/sec. The turbine will be 19 cm outside radius = 15 inches diameter. The turbine will be 16 cm inside radius = 12.6" inside diameter. The total air flow will be through the 12.6" ID, 15" OD annulus.

Now for some calculations:

A 15" diameter duct flowing at 218 ft/sec = 16,000 CFM.

The velocity pressure at that speed is 10.6" w.c.

The duct has area 1.23 ft^2.

The annulus has area 0.36 ft^2.

The air velocity through the annulus will be 218 ft/sec X 1.23 / 0.36 = 743 ft/sec.

The velocity pressure in the annulus will be (neglecting compressibility) 123 in w.c. = 4.5 PSI.

That airflow at that velocity is an airflow power of 200 kW for the kinetic energy alone (not including compressibility or the static pressure losses through the ductwork). An industrial fan for this purpose (if you can even find one) will cost many tens of thousands of dollars and the system requires real engineering design, not just guesswork and back-of-the envelope calculations.

This project is not achievable on the level of a home-made, non-expert/tinkerer experiment.

- #41

T C

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- #42

jrmichler

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This is in fact a turbine that will be placed in airlfow of 65-70 m/s velocity.

Based on these two quotes, my assumptions stand because standard practice is to report average duct velocity, not peak velocity around obstructions. Even at the lower velocity through the turbine, the rotor RPM at the 35.5 degree blade angle will be over 2500 RPM. I have not calculated the stresses, but it is unlikely that a 3D printable plastic rotor of your dimensions will survive that. Don't forget to calculate deflection and creep.The velocity is around 66.5 m/s.

As side note, small airplane propellers run at about 2500 RPM. The C172 I flew yesterday cruises nicely at 2400 RPM. Those propellers are made from aluminum, wood, or carbon fiber. Not thermoplastic.

As others have said, you need the nose cone to get smooth air flow into the turbine. Smooth air flow is necessary for proper operation.

And Point #4 in Post #37 still stands, with one addition. The total pressure drop is proportional to the power generated (divided by efficiency) plus the center section obstruction drag plus additional drag due to the turbine mounting system.

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cjl

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- #45

cjl

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That airflow at that velocity is an airflow power of 200 kW for the kinetic energy alone

Are you sure? I get 1/10 that, though I could be making a mistake here (it is rather late and I did the math quickly).

- #46

russ_watters

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The wording in the OP isn't exactly clear and we haven't seen a drawing, which would make it crystal clear. Essentially, the annulus/nose cone could be considered part of the duct or part of the device, but I agree with @jrmichler that it would be more common to see it as part of the device. Heck, without a complete picture or diagram we're still basically just guessing that there's a nose cone on it and that air isn't flowing through the middle ring.

Mixed-units are a pain here, but here's what I did (and I'm too lazy to write-out all the unit conversions and calcs, but let me know if you can't identify one):Are you sure? I get 1/10 that, though I could be making a mistake here (it is rather late and I did the math quickly).

16,000 CFM / (3.28

743 ft/s = 226.5 m/s

KE/s = .5mV

But that's not the OP's intent anyway. He's after this:

66.5 m/s through 52 in

KE/s = 6.0 kW

I think I did that right...

Again, this doesn't include static pressure losses in the ductwork system or, of course, the power output of the turbine (which also manifests as a static pressure loss) and efficiency of the fan. If we double or triple that, we're starting to run up against the limit of residential electrical supply as well.

You're fortunate to have several professional engineers helping you through this, and while it is your choice, keeping us guessing doesn't help your chances. Near as I can tell, you haven't started to address what the turbine is spinning and what is moving the air through it, but I suspect you have some ideas you haven't yet shared. Those choices/ideas will have a huge impact on whether this project will "work".It's my choice. Hope it will work. Want to know others opinion.

Simply put, the more real/rigorous effort you put into this, the better your odds of success will be.

- #47

T C

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- #48

cjl

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- #49

T C

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Can you explain why?

- #50

Baluncore

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The blades slice the air in a helical pattern. There is no advantage in slicing air that has already been sliced by another blade.Can you explain why?

The surface drag or whetted area of a wider airfoil is greater.

https://en.wikipedia.org/wiki/Blade_solidity

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