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Dimension Analysis and buckingham pi

  • Thread starter princejan7
  • Start date
  • #1
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Homework Statement



can someone explain why we are interested in forming dimensionless products
and why only n-j of them should be formed from the problem's variables?

Homework Equations


Step 1: List the variables in the problem
Step 2: Express each of the variables in terms of basic dimensions
Step 3: Determine the number of Π parameters. Buckingham Pi Theorem says we have k=n-j terms
Step 4: Select j repeating variables from the n variables
Step 5: Use the three repeating non-repeating variables to form k=n-j Π terms
Step 6: Express the final form: Π1 = f(Π2,Π3)

The Attempt at a Solution


the only thing i know so far is that physical laws are independent of the system of units used
 

Answers and Replies

  • #2
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It has to do with carrying out experiments on a physical system and also on doing small scale experiments so that you can scale up to a full sized system.

If you do experiments on a system, and there are many variables that you can control, you want to save money by doing as small a number of experiments as you can. You don't want to have to vary every individual parameter over a range of values. By working with dimensionless groups, you can characterize the system behavior with fewer experiments, and convey the results in a more concise form.

If you are designing a full sized system, you don't want to spend a lot of money buying several versions of the full sized equipment. You would like to do your experiments with much smaller pieces of experimental equipment, but then have the results translate directly into the design of the full sized equipment (so that you only need to buy it once).

Chet
 

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