Smart questions
Smart answers
Smart people
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Member Login




Remember Me
Forgot Password?
Join Us!

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips now!
  • Talk With Other Members
  • Be Notified Of Responses
    To Your Posts
  • Keyword Search
  • One-Click Access To Your
    Favorite Forums
  • Automated Signatures
    On Your Posts
  • Best Of All, It's Free!

Join Eng-Tips
*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

Donate Today!

Do you enjoy these
technical forums?
Donate Today! Click Here

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.
Jobs from Indeed

Link To This Forum!

Partner Button
Add Stickiness To Your Site By Linking To This Professionally Managed Technical Forum.
Just copy and paste the
code below into your site.

Apakrat (Civil/Environmental)
5 Aug 09 20:09

I have a working knowledge of geometry, algebra, and trigonometry but absolutely nothing about calculus.  

Usually, when using metrics, I convert to feet, gallons, etc to be comfortable because this "SI" is almost baffling if it were not for a couple of conversion tables found on the Internet.

I need to calculate the Theoretical Horse Power at the horizontal shaft axis of a waterwheel being propelled by the flowing force of river water against the paddles of the waterwheel.   

The animated image below is supposed to represent a Elevation view of a waterwheel in flowing water.

Quote (Formula):


F = MA2 ÷ 2
    F = power in ft-lb/sec
    M = moving volume in lb/ft 3 ÷ 32.2 lb/slug
    A = flow rate in ft/sec

F ft-lb/sec X 60 sec/min = F ft-lb/min

Theoretical Power (hp) = ( F ft-lb/min X  rpm ) ÷ Factor 5252
Is my understanding of the Formula logical?

Ignoring wind, water surface ripples, impure water, temperature, and mechanical efficiency, I calculated a 12 ft. diameter waterwheel with 3 ft wide paddles, in a river flowing at a rate of 6 ft per second, will develop a 68.5 Theoretical Horse Power at the waterwheel axis shaft rotating at 9.55 rpm.

At 74th year working on IR-One2 - - UHK PhD - - -

cpretty (Mechanical)
5 Aug 09 22:28
A first comment is that the wheel will not rotate at 9.55 rpm while the shaft is loaded.  To get any useful work out of the wheel, the wheel will travel slower than the water.

There will be a relationship between torque available and rpm range from high torque (not maximum) at 0, and no torque at 9.55rpm.

I will attempt to comment on the rest of the calc but am the opposite and need to convert to SI to even comprehend the problem.
Apakrat (Civil/Environmental)
7 Aug 09 12:06

Thank You "cpretty"!!

Have personally been concerned about the rpm thingy. I couldn't find a formula showing a ratio to the rpm.  Maybe it was in the calculus formulas which I couldn't comprehend.

Perhaps this is a better arrangement.

Quote (Altered Formula):

Theoretical Power (hp) = (  F ft-lb/min X 1/2 rpm  ) ÷ Factor 5252
I must apologize for failing to include metric equivalent of values used, with initial posting.    
        12 ft (3.6576 m) diameter waterwheel
         3 ft (0.9144 m) wide paddles
         6 ft (1.8288 m) per second flow
        68.5 Theoretical Horse Power (51 kw)

Am I getting closer???

At 74th year working on IR-One2 - - UHK PhD - - -

Apakrat (Civil/Environmental)
7 Aug 09 17:25

Thank You "IRstuff"

I went to the link suggested which has an article written by "Rudy Behrens".  

Played with the mathematics, along with the same numbers in formulas given at

All three have slightly different formula arrangements but  the final numerical results are almost identical.  

They all calculate about 10% more power than my initial formula as first posted above, which is based on Newton's law and torque power conversion to angular rotation.

Quote ("Rudy Behrens"):


"The most efficient energy transfer occurs when the wheel speed is between 67% and 90% of the water speed."

Curious!!        Curious!!        Curious!!

At 74th year working on IR-One2 - - UHK PhD - - -

FeX32 (Mechanical)
10 Aug 09 22:13
I don't see where you take into account the number of blades in the water at any one time. There are many assums.
Although, I don't see any major concern with the original equn.

BUT, personally, I would never trust a formula I haven't derived myself unless it was for a very good reason. I always feel better about a problem when I start from ground up....
For example, I may start by simplifying the problem so it has only one blade completely vertical and draw a conclusion using conservation of momentum. Then find out what may happen when we add more blades.  

peace
Fe

BigInch (Petroleum)
11 Aug 09 12:24
http://www.panoramio.com/photo/8214845

**********************
"Pumping accounts for 20% of the world's energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies) http://virtualpipeline.spaces.live.com/

GregLocock (Automotive)
12 Aug 09 3:53
Would your equation show a 33% benefit if the blades were curved? If not, it doesn't seem very useful, since that is a real test result.

  

Cheers

Greg Locock

SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.

Reply To This Thread

Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members!

Close Box

Join Eng-Tips® Today!

Join your peers on the Internet's largest technical engineering professional community.
It's easy to join and it's free.

Here's Why Members Love Eng-Tips Forums:

Register now while it's still free!

Already a member? Close this window and log in.

Join Us             Close