Simplifying the Buckingham equation

  • Thread starter stevemilw
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In summary, the apparent answer leaves me with a negative square root and no divider. I tried re-arranging the formula to make V2 the subject, but my textbook gave me an answer that I do not understand.
  • #1
stevemilw
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Hello, i have a question which is too hard for me to type in here, so I've scanned in the image.
At the top, as explained in the image, i have the answer I've found in my lecture notes, but then I've worked it out for myself and i do not get the same answer.
Can someone please tell me where I am going wrong?
Im making V2 the subject

All help is appretiated, don't be shy.
Thanks!

heres the link: http://www4.picturepush.com/photo/a/6324532/640/6324532.jpg

p.s i originally posted this in yahoo answers but got no response.
 
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  • #2
hello stevemilw! :smile:

i don't understand what you mean by "the apparent answer leaves me with a negative square root and no divider" :confused:

(and your z1 + z2 should be 2g(z1 + z2))
 
  • #3
Hello
What i meant by that is my answer (at the very bottom) gave me the top part with the pressures devided by the bottom part.
Also I am not sure what you meant by it should be 2G(Z1+Z2) because as i x2g on both sides, shouldn't i only times by one of the sides either side of the plus? writing this now, that seems stupid. You must be right.

What i mean by the whole thing, is you have the buckingham equation at the very top.
Second equation is the answer that my textbook says i should get when i re-arrange the formula to make V2 the subject.
The third equation annd the next and so on are my attempts. And i do not the the answer my textbook gives as you can see at the bottom.

Thanks
 
  • #4
stevemilw said:
Second equation is the answer that my textbook says i should get when i re-arrange the formula to make V2 the subject.
The third equation annd the next and so on are my attempts. And i do not the the answer my textbook gives as you can see at the bottom.

they look the same to me :confused:

(with X-1/2 instead of 1/√X )
 
  • #5
ooooh, i feel really stupid now.
I didnt realize that x^-1/2 was 1/sqrt x

God, you wouldn't believe the amount of hassel that's given me!

I know -1 was one over, to the negetive half is 1 over the sqrt.

Thanks, this is something I am not going to forget in a while! lol
 

Related to Simplifying the Buckingham equation

What is the Buckingham Pi Theorem?

The Buckingham Pi Theorem, also known as the Pi Theorem or the Buckingham Pi Method, is a mathematical principle used to reduce the number of variables in a physical problem while maintaining the underlying physical relationships. It is commonly used in dimensional analysis to simplify complex equations.

How is the Buckingham Pi Theorem applied?

The Buckingham Pi Theorem is applied by selecting a set of dimensionally independent variables, known as Pi terms, and grouping all other variables in the problem into these terms. This results in a dimensionless equation, which can then be solved for the desired variable.

What are some advantages of using the Buckingham Pi Theorem?

One advantage of using the Buckingham Pi Theorem is that it simplifies complex equations, making them easier to solve and interpret. Additionally, it can help identify relationships between variables and identify any missing or redundant variables in a problem.

What are the limitations of the Buckingham Pi Theorem?

The Buckingham Pi Theorem may not be applicable to all physical problems, as it relies on the variables being dimensionally homogeneous. It also assumes that the variables have linear relationships, which may not always be the case. Furthermore, it does not take into account any non-dimensional constants that may be present in the original equation.

Are there any practical applications of the Buckingham Pi Theorem?

Yes, the Buckingham Pi Theorem has many practical applications in various fields such as physics, engineering, and chemistry. It is commonly used in fluid dynamics, heat transfer, and chemical kinetics to simplify and analyze complex systems. It is also used in experimental design to reduce the number of experiments needed to determine the relationships between variables.

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