Question 5 of 6: Buckingham Pi analysis of a sudden contraction

Question 5: Buckingham Pi 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. The functional statement Δp = f(D₁, D₂, U, ρ, μ), where U is the velocity in the larger pipe, with D₁, U and μ nominated as the repeating variables.

Find. A complete, independent set of dimensionless groups, and a reasoned explanation of why the velocity in the smaller pipe must not be added to the variable list.

Approach. Count variables and independent dimensions to fix the number of groups, confirm the nominated repeating set is dimensionally independent, then solve one small linear system per non-repeating variable.

1. Count. The variable list is Δp, D₁, D₂, U, ρ, μ, so n = 6. In the MLT system their dimensions are $$[\Delta p]=ML^{-1}T^{-2},\quad [D_{1}]=[D_{2}]=L,\quad [U]=LT^{-1}, \quad [\rho]=ML^{-3},\quad [\mu]=ML^{-1}T^{-1},$$ and the rank of the dimensional matrix is k = 3. Buckingham’s theorem therefore predicts $$n-k=6-3=3\ \text{dimensionless groups}.$$
2. Confirm the nominated repeating variables are admissible. The set {D₁, U, μ} must be dimensionally independent and must not by itself form a dimensionless group. Mass appears only in μ, time only in U and μ, and length in all three, and no product D₁^aU^bμ^c is dimensionless unless a = b = c = 0. The set is valid, and note in passing that it deliberately excludes the dependent variable Δp, as the method requires.
3. First group: the dependent variable. Seek Π₁ = Δp D₁^aU^bμ^c. Balancing exponents, $$M:\ 1+c=0,\qquad T:\ -2-b-c=0,\qquad L:\ -1+a+b-c=0,$$ which gives c = −1, b = −1, a = 1, so $$\boxed{\;\Pi_{1}=\frac{\Delta p\,D_{1}}{\mu U}\;}$$ a viscous-scaled pressure drop — the natural form when viscosity is a repeating variable.
4. Second group: the geometry. With Π₂ = D₂D₁^aU^bμ^c, the mass balance gives c = 0, the time balance b = 0, and the length balance a = −1: $$\boxed{\;\Pi_{2}=\frac{D_{2}}{D_{1}}\;}$$ the contraction ratio, which is the only geometric parameter the problem contains.
5. Third group: the remaining property. With Π₃ = ρD₁^aU^bμ^c, $$M:\ 1+c=0,\qquad T:\ -b-c=0,\qquad L:\ -3+a+b-c=0,$$ giving c = −1, b = 1, a = 1, so $$\boxed{\;\Pi_{3}=\frac{\rho U D_{1}}{\mu}=\mathrm{Re}_{D_{1}}\;}$$ the Reynolds number based on the upstream pipe.
6. State the result. The dimensionless statement of the problem is $$\frac{\Delta p\,D_{1}}{\mu U}=\phi\!\left(\frac{D_{2}}{D_{1}},\ \mathrm{Re}_{D_{1}}\right),$$ a reduction from a function of five variables to a function of two. Any independent recombination is equally valid, and the one used in practice divides out the viscosity: $$\frac{\Pi_{1}}{\Pi_{3}}=\frac{\Delta p}{\rho U^{2}}=\mathrm{Eu} =\phi'\!\left(\frac{D_{2}}{D_{1}},\ \mathrm{Re}_{D_{1}}\right),$$ the Euler number, which is preferred because at high Reynolds number the loss coefficient becomes almost independent of Re and collapses onto a single curve in D₂/D₁. A worked instance: with Δp = 4.2 kPa, D₁ = 0.20 m, D₂ = 0.10 m, U = 3 m/s and water, Π₁ = 2.79 × 10⁵, Π₂ = 0.5, Π₃ = 5.98 × 10⁵ and Eu = 0.468.
7. Part (b) — why the small-pipe velocity must be excluded. The velocity in the smaller pipe is not an independent quantity. Conservation of mass for an incompressible fluid fixes it completely from variables already on the list: $$U_{2}=U\left(\frac{D_{1}}{D_{2}}\right)^{2},\qquad\text{i.e.}\qquad \frac{U_{2}}{U}=\Pi_{2}^{-2}.$$ Buckingham’s theorem requires the variable list to be independent; a quantity that is an algebraic function of others is redundant.
8. What actually goes wrong if you add it. Adding U₂ raises n to 7 while the rank stays at k = 3, so the method dutifully produces a fourth group, U₂D₂²/(UD₁²). But that group is identically equal to unity for every physically realisable flow: it is not a free parameter, it is a constraint the fluid has already satisfied. Carrying it forward is misleading in two ways. It suggests the problem has one more degree of freedom than it does, so an experimenter might waste a whole test matrix trying to vary it independently; and it breaks the uniqueness of the correlation, because the same physical state can now be labelled by infinitely many nominally different parameter sets. The rule generalises: include the independent variables that define the problem, and let the conservation laws supply everything they determine.