Question

Dimensional Analysis

Is my reasoning correct?

*problem*
Oil with a given viscosity and density flows through a pipe with given diameter and flowrate. We want to make a scale model with water that flows through a pipe with a smaller given diameter. I have to find out what the speed should be in the scalemodel to achieve similar viscous forces.

*reasoning*
variables: v,mu,rho,D
units: kg,m,s

v = k*(mu^a * rho^b * D^c) (with k being some dimensionless constant)

[m/s] = [kg/m*s]^a * [kg/m^3]^b * [m]^c
=> a+b=0
=> a=1 <=> b=-1
=> 1=-a-3b+c <=> c=-2

so for the real model
v_r

Answer 1

If we consider the flowrate and
I call with \(\eta\) the viscosity, with \(\delta \) the water density, with \(D\) the diameter of the pipe, and with \(Q\) the flowrate of water. Then I can write this:
\[\Large k\; {\eta ^a}{\delta ^b}{D^c}{Q^d} = v\]
(where \(k\) is a dimensionless quantity)
so, replacing each quantity, with the corresponding unit of measure, I get:
\[\Large {\left( {\frac{{Kg}}{{m \cdot s}}} \right)^a}{\left( {\frac{{Kg}}{{{m^3}}}} \right)^b}{m^c}{\left( {\frac{{{m^3}}}{s}} \right)^d} = \frac{{m}}{s}\]
Finally, I can write the subsequent system:
\[\Large \left\{ \begin{gathered}
a + b = 0 \hfill \\
c + 3d - a - 3b = 1 \hfill \\
a + d = 1 \hfill \\
\end{gathered} \right.\]
nevertheless it is an undetermined system
Whereas if I don't consider the flowrate \(Q\), then I can write:
\[\Large \begin{gathered}
k\;{\eta ^a}{\delta ^b}{D^c} = v \hfill \\
\hfill \\
{\left( {\frac{{Kg}}{{m \cdot s}}} \right)^a}{\left( {\frac{{Kg}}{{{m^3}}}} \right)^b}{m^c} = \frac{m}{s} \hfill \\
\hfill \\
\left\{ \begin{gathered}
a + b = 0 \hfill \\
c - a - 3b = 1 \hfill \\
a = 1 \hfill \\
\end{gathered} \right. \hfill \\
\end{gathered} \]