Dimensional Analysis of a Sudden Contraction: 22-Mec-A6 Fluid Machinery, May 2017, Q5

Question 5: Dimensional Analysis of a Sudden Contraction (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $\Delta p=f(D_1,\,D_2,\,U,\,\rho,\,\mu)$ — six variables in three primary dimensions (M, L, T); repeating variables $D_1$, $U$, $\mu$.

Find. (a) a complete set of dimensionless $\Pi$ groups; (b) the reason $U_2$ is not an independent variable.

Approach. Count variables and dimensions ($6-3=3$ groups), check that the repeating set spans M, L, T, then form each $\Pi$ by combining one non-repeating variable with powers of the repeating set and solving for zero net dimensions.

Variables and dimensions. $[\Delta p]=ML^{-1}T^{-2}$, $[D_1]=[D_2]=L$, $[U]=LT^{-1}$, $[\rho]=ML^{-3}$, $[\mu]=ML^{-1}T^{-1}$. The repeating trio $D_1,U,\mu$ carries L (from $D_1$), T (from $U$) and M (from $\mu$), so it is dimensionally independent. Buckingham gives $6-3=3$ groups.
Group with $\Delta p$. $\Pi_1=\Delta p\,D_1^{a}U^{b}\mu^{c}$; requiring $M,L,T$ exponents to vanish gives $c=-1,\ b=-1,\ a=1$: $$\Pi_1=\frac{\Delta p\,D_1}{\mu\,U}.$$
Group with $D_2$. $\Pi_2=D_2 D_1^{a}U^{b}\mu^{c}\Rightarrow c=0,\ b=0,\ a=-1$: $$\Pi_2=\frac{D_2}{D_1}\quad(\text{diameter ratio}).$$
Group with $\rho$. $\Pi_3=\rho\,D_1^{a}U^{b}\mu^{c}\Rightarrow c=-1,\ b=1,\ a=1$: $$\Pi_3=\frac{\rho\,U D_1}{\mu}=Re\quad(\text{Reynolds number}).$$ Hence $$\boxed{\frac{\Delta p\,D_1}{\mu U}=\phi\!\left(\frac{D_2}{D_1},\ \frac{\rho U D_1}{\mu}\right).}$$ Recombining $\Pi_1/\Pi_3$ recovers the more familiar pressure coefficient: $\dfrac{\Delta p}{\rho U^{2}}=\phi'\!\left(\dfrac{D_2}{D_1},\,Re\right)$.

(b) Why not include $U_2$. The velocity in the smaller pipe is not independent: incompressible continuity fixes it, $U_2=U\,(D_1/D_2)^2$. It is already determined by the two variables $U$ and $D_2/D_1$ that are in the list, so adding it would not introduce new physical information. Including a dependent variable violates the Buckingham requirement that the chosen quantities be independent; it would merely generate a redundant group ($U_2/U=(D_1/D_2)^2$) that is a pure function of $\Pi_2$, over-counting the physics rather than describing it.

Question 5 — dimensionless groups
Group	Expression	Meaning
$\Pi_1$	$\Delta p\,D_1/(\mu U)$	viscous-scaled pressure drop
$\Pi_2$	$D_2/D_1$	contraction (diameter) ratio
$\Pi_3$	$\rho U D_1/\mu$	Reynolds number

Topic: The Buckingham Pi theorem and selection of independent variables.

Key relations: number of groups $=n-k$; each $\Pi$ formed from one non-repeating variable times powers of the repeating set; recombination of groups is always permissible.

Why this works: physical laws must be dimensionally homogeneous, so any relation among $n$ dimensional variables collapses to a relation among $n-k$ dimensionless groups, where $k$ is the number of independent dimensions spanned by the variables. Choosing $\mu$ (rather than $\rho$) as a repeating variable simply shifts which group looks “natural”; recombining $\Pi_1$ and $\Pi_3$ recovers the Euler/pressure-coefficient form, showing the two descriptions are equivalent.

Common pitfall: including dependent variables (like $U_2$ here) or choosing a repeating set that does not span all dimensions or that itself forms a dimensionless group. Repeating variables must be dimensionally independent and should exclude the primary dependent variable $\Delta p$.

Source: Munson, Young & Okiishi, Fundamentals of Fluid Mechanics, Ch. 7; White, Fluid Mechanics, 8th ed., §5.3–5.4.

Question 5 of 6: Dimensional Analysis of a Sudden Contraction

Question 5: Dimensional Analysis of a Sudden Contraction (20 marks)