Dimensional Analysis: Combining 3 Variables

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Homework Help Overview

The discussion revolves around dimensional analysis involving four variables: Q, R, μ, and dp/dx. The original poster is attempting to determine if three of these variables (R, μ, dp/dx) can be combined into a dimensionless product.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster presents a method involving dimensional equations and a system of equations to argue that the selected variables cannot form a dimensionless parameter. Some participants question the validity of the dimensional combinations and suggest alternative considerations.

Discussion Status

Participants are engaging with the original poster's reasoning, with some affirming the method proposed. There is an exploration of the implications of dimensional analysis, and while some guidance has been offered, there is no explicit consensus on the original poster's conclusions.

Contextual Notes

There is an emphasis on the dimensions of the variables involved and the requirement for their power product to equal zero for dimensional analysis. The discussion reflects on the complexity of the dimensional relationships and the potential for multiple interpretations.

Saladsamurai
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Homework Statement



Hi. I have a function that contains 4 variables: Q, R, [itex]\mu[/itex], dp/dx

I wish to choose 3 of them, such that they cannot be combined into a dimensionless product.I have chosen (correctly) R, [itex]\mu[/itex], dp/dx and I would like to know if my method sounds correct:

If we know the dimensions: [R]=[L] [[itex]\mu[/itex]]=[ML-2T-2] and [dp/dx]=[ML-1T-1]and I know that in order for them to be dimensionless, their power product must equal zero:

[L]a[ML-1T-1]b[ML-2T-2]c=[MLT]0

or the system

a-b-2c=0
b+c=0
-b-2c=0

By inspection, this system can only be satisfied if a=0 but that does not make any sense since R is a physical quantity.

Hence I have reasoned that these 3 variable cannot form a non-dimensional parameter by themselves.Does this work? Thanks!
 
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Ignore length real quick. [tex][MT^{ - 1} ]^a [MT^{ - 2} ]^b[/tex] can NEVER be dimensionless (except for a=b=0)
 
Pengwuino said:
Ignore length real quick. [tex][MT^{ - 1} ]^a [MT^{ - 2} ]^b[/tex] can NEVER be dimensionless (except for a=b=0)

Okay cool!

Also, though longer and more tiresome, my method above works right? For the same reason.

I just want a general method just in case inspection is not that obvious.
 
Yes, that method works.
 

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