Solved Problem on Dimensional Analysis
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The equation that describes the movement of a viscous fluid in one dimension is given by
\[
\begin{gather}
\rho \frac{dv}{dt}=-{\frac{dp}{dx}}+\eta \frac{d^{2}v}{cx^{2}}
\end{gather}
\]
where ρ is the density, v is the velocity, t is the time, p is the pressure,
and η is the viscosity. Determine the dimension of viscosity η.
Solution
On the left-hand side of the equation, the density ρ is given by the mass m divided by the volume V
\[
\begin{gather}
[\rho ]=\frac{[m]}{[V]}=\frac{M}{L^{3}}=ML^{-3}
\end{gather}
\]
the term
\( \frac{dv}{dt} \)
has velocity dimension, v divided by time t (derivative operation d has no
dimension), velocity v has length dimension x divided by time t
\[
\begin{gather}
\left[\frac{dv}{dt}\right]=\frac{[v]}{[t]}=\frac{\frac{[x]}{[t]}}{[t]}=\frac{\frac{L}{T}}{T}=\frac{LT^{-1}}{T}=LT^{-1}T^{-1}=LT^{-2}
\end{gather}
\]
he left-hand side of the equation has the dimension
\[
\begin{gather}
\left[\rho\frac{dv}{dt}\right]=\left(ML^{-3}\right)\left(LT^{-2}\right)=ML^{-2}T^{-2}
\end{gather}
\]
Using the Dimensional Homogeneity, both sides of the equation must have the same dimension.
\[
\begin{gather}
\underbrace{\left[\rho\frac{dv}{dt}\right]}_{ML^{-2}T^{-2}}=-{\underbrace{\left[\frac{dp}{dx}\right]}_{ML^{-2}T^{-2}}}+\underbrace{\left[\eta\frac{d^{2}v}{cx^{2}}\right]}_{ML^{-2}T^{-2}}
\end{gather}
\]
in the second term on the right-hand side,
\( \frac{d^{2}v}{cx^{2}} \)
has dimension of speed v divided by length squared x2 (the second order
derivative operation d2 has no dimension), velocity v has dimension of length x
divided by time t
\[
\begin{gather}
\left[\frac{dv}{dx}\right]=\frac{[v]}{\left([x]\right)^{2}}=\frac{\frac{[x]}{[t]}}{\left([x]\right)^{2}}=\frac{\frac{L}{T}}{L^{2}}=\frac{LT^{-1}}{L^{2}}=LT^{-1}L^{-2}=L^{-1}T^{-1}
\end{gather}
\]
the second term on the right-hand side of the equation has the dimension
\[
\begin{gather}
\left[\eta \frac{dv}{dx}\right]=[\eta]\left[\frac{dv}{dx}\right]=ML^{-2}T^{-2}\\[5pt]
[\eta]L^{-1}T^{-1}=ML^{-2}T^{-2}\\[5pt]
[\eta]=\frac{ML^{-2}T^{-2}}{L^{-1}T^{-1}}\\[5pt]
[\eta]=ML^{-2}T^{-2}LT
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{[\eta ]=ML^{-1}T^{-1}}
\end{gather}
\]
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .