Chain Turbine Mechanics- Discussions with Duy

Chain Turbine Mechanics- Discussions with Duy!!!

----- Original Message -----

From: Nguyen Minh Duy

To: cfay@frenchriverland.com

Sent: Tuesday, June 05, 2007 10:59 PM

Subject: Chain turbine

 Dear William K. Fay,

Do you have this paper?

"Schneider, D. J., et al., 1979, Schneider Lift Translator System, Low Head Hydro Feasibility Study Final Report, Schneider Lift Translator Co., NTIS No. DOE/RA/01693-T1. "

Can you send for me a copy of this paper?

I cannot buy it, I need this paper to do my final report!

I don't have any experience about  chain turbine. So It is very difficult.

Thank you so much!

Duy

 Dear Duy:

     I will try to get a copy of the paper.

 My son and I have been discussing the design parameters of your turbine. We considered the following:

1) It is a gravity machine. It is also a water motor. The power output is still KW=( efficiency*Q*deltaH)/11.81

 2) The speed of the machine can be determined by the average velocity that one bucket volume of water reaches, as it passes from the top of the machine, from where it is filled, to the bottom of the machine, were it empties. Of course this would translate directly into the average velocity of the chain to which the buckets are attached. This is independent of the mass of the object per Sir Isaac Newton.

 3) The upper limit of the power of the machine is dictated by P=eff*Q*H. This will be true, if the buckets are large enough to catch all the water passing into the machine. If the buckets are too small, a volumetric inefficiency will result.

 4) The gross head across the machine will have to be adjusted. The performance of this type of machine is dramatically effected by the buckets impinging the tailwater (this is called "wading"). In order to not  have the buckets drag in the tailwater, you need to have them discharge at a distance above the tailwater. This distance results in a loss of gross head and is an inefficiency of the machine.

 5) The velocity of the water entering the buckets should be minimized. It would be best if the water entered from the headrace through a sluiceway or weir. It would be difficult, with the shape of the proposed buckets, to recover much of the intake velocity head.

 6) Another inefficiency is the water lost between the filling of each bucket. If the buckets are not closely spaced on the chain, water will be lost between bucket charging. This will probably happen to some extent even with closely spaced bucket and some type of shrouding. This is another type of volumetric inefficiency.

 As memory serves me:

 v= dx/dt, a= dv/dt or a= v* dv/dx integrating from vo to v and from 0 to t we have:

v= vo+at

also vdv=adx, integrating we obtain:

v^2=v0^2+2*a*(x-x0)

 now for your gravity machine, a=g= acceleration of gravity. When you first fill the bucket the machine is at rest so v0=0.  The height the buckets drop is deltaX= X-X0= delta H= H

 so the equation reduces to Vterminal= square root(2*g*H) This is the maximum speed of rotation. The average speed of descent is Vavg= 1/2 Vt

 In all turbo machines, you can apply a dynamometer to their shaft and measure the torque and rpm. Torque times rpm is power. If you plot the graph of horsepower versus rpm, you will obtain a parabola, that goes to zero power, on either end. This makes sense. If you stall the machine, you have maximum torque but zero rpm and

 zero rpm*max torque=0

 On the other end, the speed of the machine is being countered by the friction in the bearings, which is very small. You have maximum speed but negligible torque and again Maximum speed * minimal torque=0

 Now, it is true that this parabola has a maximum and it is in the middle of the parabola. This is the point of maximum horsepower and also the point of maximum machine efficiency. So to obtain your operational speed, to size your pulleys for your motor, take V terminal and divide by 2 again.

 So,  Voperational= 1/4*(2*g*H)^0.5

 The RPM of the sprocket would be :

 revolution/second=Voperational/(phi*d) where V is in meters/second and d=diameter is in meters.

     I am sure you have already considered a similar analysis. This is a very preliminary and simplified one. You could check these equations on your operational machine.

 I hope this has been of some help.

 Sincerely,

 Bill Fay

Thorndike, MA

 ---- Original Message -----

Dear Duy:

     In addition, to determine the number of buckets and their volume, one would need to consider the velocity of the chain and the volumetric flowrate entering the machine. If the water is entering at one cubic meter per second and the chain was descending at one meterpersecond and the head was ten meters you would need ten, one cubic meter buckets, on the descending chain, to catch all the water. In order to prevent spillage, I would make the volume of the buckets 20 percent larger.

 Best regards and have a nice day.

 Sincerely,

 Bill Fay

----- Original Message -----

----- Original Message -----

From: Nguyen Minh Duy

To: Celeste Fay

Sent: Wednesday, June 06, 2007 9:37 PM

Subject: Re: Chain turbine

 Dear Bill Fay:

Thank you very much!

I try to solve it. After that I will send the result for you!

Best Regards,

Duy

----- Original Message -----

From: Nguyen Minh Duy

To: Celeste Fay

Sent: Friday, June 08, 2007 9:51 AM

Subject: Re: Chain turbine

Dear Mr. Bill Fay,

How are you?

I receive both of your email. They are very useful. Thank you!

I calculate the chain turbine with flow rate (Q=1 cubic meter /sec) and Head (H = 20 m). The capacity of this plant about 120kW (I assume the efficiency of this system about 0.6).

Hope to co-operation more.  The end of July, I finish my reports. I am going to co-operation with you to write a Vietnamese-English dictionary.

Thank you very much!

Duy

----- Original Message -----

From: Celeste Fay

To: Nguyen Minh Duy

Sent: Thursday, June 07, 2007 7:13 PM

Subject: Re: Chain turbine

 Dear Duy:

 Did you receive both of my recent e-mails regarding the machine? I just want to make sure. The first was about the poweroutput and speed of the chain. The second was about the number and sizing of the buckets. They were sent within minutes of each other. The first has more information and I wanted to make sure you received both e-mails. If not I will resend them.

 Best Regards,

 Bill Fay

 Duy:

 Xin Chao!!!

 Cam on    for your friendship!!

 Tam Biet!!!

 Bill Fay

 Original Message -----

From: Nguyen Minh Duy

To: Celeste Fay

Sent: Sunday, June 17, 2007 1:54 PM

Subject: Re: Chain turbine again

  Dear Mr. Bill Fay,

     Can I ask you again about chain turbine? I send you an my attached file. I solve the design parameters of chain turbine. Could you correct it for me?

    Last email, you sent me your idea about this turbine with the average speed of descent is Vavg=1/2Vt, but I query, this is only a velocity when water from the jet act on bucket. How about the velocity follow the force of buckets full of water? (have figure in attached file). Could you explain more?

       About my English, My English is still very poor! Hope you teach me more. I never go to American or countries using the English. Nowadays, Vietnamese can speak English very well. 

    Thank you very much!

    Best Regards

 Duy

Dear Mr. Bill Fay,

How are you?

Can you help me to check the calculation of my chain turbine (attached file).

With your last mail, please explain:

 

Now, it is true that this parabola has a maximum and it is in the middle of the parabola. This is the point of maximum horsepower and also the point of maximum machine efficiency. So to obtain your operational speed, to size your pulleys for your motor, take V terminal and divide by 2 again. (Why you divide Vt by 2?)

 So,  Voperational= 1/4*(2*g*H)^0.5

 

The RPM of the sprocket would be :

 

revolution/second=Voperational/(phi*d) where V is in meters/second and d=diameter is in meters. (The angular velocity the sprocket would be: Revolution/speed=Vop/R )

 

Let's me know if they are stupid questions.

Thank you very much.

Duy

 Dear Duy:

     We are confused about the head H. Is it approximately the height of the machine like an overshot waterwheel or is it a much higher head with a very large exit velocity for the jet/nozzle. Your analysis is correct and very similiar to a Pelton Wheel analysis if the available head is much more then the height of the machine. If the machine is approximately the height of the dam from headwater to tailwater then my analysis is more correct. If the head is much higher then the height of the machine, the water is going to splash out of the buckets, the energy regained will be from a momentum transfer and the useful work gained from the weight of the water will be minimal. In this case, a Pelton Wheel is much easier to design, build and has less moving parts then your chain drive machine. I need to know more about how you intend to apply/install these machines. I thought that if the distance between the axles was ten meters then you had 11 meters of head. Then the analysis is more like a gravity waterwheel. If the height of the machine is 10 meters and you are operating on 50 meters, then the machine needs to recover the velocity head. You need to turn the jet 180 degrees around in the bucket to recover the head.

     I need more information!

 Cam On!!!

 Tam Biet!!!!

 Dear Duy:

It is nice to hear from you. Did you give your presentation on the turbine design?

 I have spent the last five weeks in Sebec, Maine working on the Sebec Lake Hydro. When I come home on the weekends, Will and I have been replacing a 10' by 10' oak, flood gate at Winchendon. The stems weighed 1400 pounds apiece and we had to drag them across a tiny bridge over the tailrace with a rope, pulleys and my ford truck. The FERC inspector is coming next Monday for the every four year inspection so the kids are frantically cleaning and polishing!!!

    
---------- Forwarded message ----------
From: Nguyen Minh Duy < duyspar@gmail.com>
Date: Jul 16, 2007 11:26 PM
Subject: Emailing: final - Report Hydroelectric power plant
To: duyspvn@gmail.com, Nguyen Minh Duy <duyspar@gmail.com>, nmduy80@gmail.com

 The message is ready to be sent with the following file or link attachments:
 final - Report Hydroelectric power plant

Note: To protect against computer viruses, e-mail programs may prevent sending or receiving certain types of file attachments.  Check your e-mail security settings to determine how attachments are handled.

--
Nguyen Minh Duy

 ----- Original Message -----

From: Nguyen Minh Duy

To: Celeste Fay

Sent: Sunday, September 02, 2007 9:17 AM

Subject: Fwd: Emailing: final - Report Hydroelectric power plant

 Dear Mr. Celeste Fay:

 How are you?

I finished to report my presentation.

I send you my presentation.

Hope you like it!

Duy

 Dear Duy:

 Please see our website with your presentation available to the world!!!

Bill Fay

Chain Turbine Design- Link<<<Click here

This is a powerpoint presentation and takes a while to download. Please wait. It is worth it!!