22-Mec-A6 Fluid Machinery · December 2019

Question 5 of 6: Dimensional Analysis of the Pressure Drop at a Sudden Contraction

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams — December 2019, 16-Mec-A6 Advanced Fluid Mechanics. Open book, 3 hours. Six questions of equal value (20 marks); any five constitute a complete paper. All six are solved here as a study resource.

Reference texts. J.D. Anderson, Modern Compressible Flow, 3rd ed. (Q1); Kundu, Cohen & Dowling, Fluid Mechanics, 6th ed. and F.M. White, Fluid Mechanics, 8th ed. (Q2, Q5); F.M. White, Fluid Mechanics, 8th ed. (Q3); F.M. White, Viscous Fluid Flow, 3rd ed. (Q4, Q6); Schlichting & Gersten, Boundary-Layer Theory, 8th ed. (Q6); Fox & McDonald, Introduction to Fluid Mechanics, 10th ed. (general).

Question 5: Dimensional Analysis of the Pressure Drop at 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 governed by three primary dimensions ($M,L,T$); repeating variables $D_1$, $U$, $\mu$.

Variables and $MLT$ dimensions
Pressure drop $\Delta p$	$[M L^{-1} T^{-2}]$
Larger-pipe diameter $D_1$ (repeating)	$[L]$
Smaller-pipe diameter $D_2$	$[L]$
Velocity in larger pipe $U$ (repeating)	$[L T^{-1}]$
Density $\rho$	$[M L^{-3}]$
Viscosity $\mu$ (repeating)	$[M L^{-1} T^{-1}]$

Find. (a) a complete set of dimensionless $\pi$ groups; (b) why the smaller-pipe velocity $U_2$ must not be added.

Approach. With $n=6$ variables and $k=3$ independent dimensions, Buckingham’s theorem gives $n-k=3$ dimensionless groups. Form each non-repeating variable ($\Delta p$, $\rho$, $D_2$) with the repeating set $\{D_1,U,\mu\}$ and force the $M,L,T$ exponents to cancel.

Number of groups. $n-k=6-3=3$ — three $\pi$ groups.
Group from $\Delta p$. $\pi_1=\Delta p\,D_1^{a}U^{b}\mu^{c}$. Cancelling dimensions gives $c=-1$ (from $M$), $b=-1$ (from $T$), $a=1$ (from $L$): $$\pi_1=\frac{\Delta p\,D_1}{\mu\,U}.$$
Group from $\rho$. $\pi_2=\rho\,D_1^{a}U^{b}\mu^{c}$ gives $c=-1$, $b=1$, $a=1$, i.e. the Reynolds number $$\pi_2=\frac{\rho\,U D_1}{\mu}=Re .$$
Group from $D_2$. $\pi_3=D_2/D_1$ (already dimensionless — a geometric ratio).
Assemble the result. A complete dimensionless statement is $$\boxed{\frac{\Delta p\,D_1}{\mu U}=\phi\!\left(\frac{\rho U D_1}{\mu},\ \frac{D_2}{D_1}\right).}$$ Dividing $\pi_1$ by $\pi_2$ recovers the more familiar pressure-coefficient (Euler) form $\dfrac{\Delta p}{\rho U^{2}}=\phi_2\!\big(Re,\ D_2/D_1\big)$ — equally valid, since any product of $\pi$ groups is itself a $\pi$ group.

(b) Why $U_2$ must not be added. The velocity in the smaller pipe is not independent: continuity fixes it as $U_2=U\,(D_1/D_2)^2$, i.e. $U_2$ is already determined by $U$ and the ratio $D_2/D_1$ that are in the list. Adding it would introduce a variable that is a known function of the others, so the set would no longer be dimensionally independent; the Pi theorem would then return a spurious extra group $U_2/U=(D_1/D_2)^2$ that is merely $\pi_3^{-2}$ and carries no new physics. The analysis must use only independent quantities.

Question 5 — the three $\pi$ groups
Group	Expression	Meaning
$\pi_1$	$\dfrac{\Delta p\,D_1}{\mu U}$ (or $\dfrac{\Delta p}{\rho U^2}$)	dimensionless pressure drop
$\pi_2$	$\dfrac{\rho U D_1}{\mu}$	Reynolds number
$\pi_3$	$\dfrac{D_2}{D_1}$	contraction ratio

Topic. The Buckingham Pi theorem and the selection of independent variables in dimensional analysis.

Key relations. Number of groups $=n-k$; each $\pi=(\text{non-repeating})\times\prod(\text{repeating})^{\text{exp}}$ with $M,L,T$ exponents zero; continuity $U_1A_1=U_2A_2\Rightarrow U_2=U(D_1/D_2)^2$.

Why this works. Choosing $\{D_1,U,\mu\}$ as repeating (they span $M,L,T$ and none is the sought $\Delta p$) guarantees each remaining variable yields exactly one independent group. The result compresses a 6-variable problem to a 2-parameter chart, $\Delta p/\rho U^2$ versus $Re$ and $D_2/D_1$ — the essence of model testing.

Common pitfall. Padding the variable list with quantities derivable from the others (here $U_2$), or picking repeating variables that do not together span all three dimensions. Both corrupt the group count and produce redundant or ill-defined $\pi$’s.

Source. F.M. White, Fluid Mechanics, 8th ed., Ch. 5 (dimensional analysis and similarity); Fox & McDonald, Introduction to Fluid Mechanics, 10th ed., Ch. 7.

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