Question 5 of 6: Dimensional Analysis of Missile Lift
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams — December 2018, 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.
Given. The functional statement $F=f(L,V,D,\alpha,\rho,\mu,c)$ with eight variables, one of which ($\alpha$) is already dimensionless.
Variables and MLT dimensions
Variable
$M$
$L$
$T$
$F$ (lift)
1
1
−2
$L$ (length)
0
1
0
$V$ (velocity)
0
1
−1
$D$ (diameter)
0
1
0
$\rho$ (density)
1
−3
0
$\mu$ (viscosity)
1
−1
−1
$c$ (sound speed)
0
1
−1
$\alpha$ (angle)
0
0
0
Find. The complete set of independent $\pi$ groups and the reduced functional relation.
Approach. Apply the Buckingham $\pi$ theorem: count variables and independent dimensions to fix the number of groups, choose dimensionally independent repeating variables, and form each remaining variable into a dimensionless product.
Number of groups. There are $n=8$ variables and $k=3$ independent dimensions ($M,L,T$), so the number of independent dimensionless groups is
$$n-k = 8-3 = \boxed{5}.$$
Repeating variables. Choose $\rho,\ V,\ D$ — they contain all three dimensions and cannot form a dimensionless group among themselves.
Form the groups. Non-dimensionalising each remaining variable gives
$$\pi_1=\frac{F}{\rho V^{2} D^{2}}\ (\text{lift coefficient}),\quad
\pi_2=\frac{L}{D}\ (\text{slenderness}),\quad
\pi_3=\frac{\rho V D}{\mu}=Re,$$
$$\pi_4=\frac{V}{c}=Ma\ (\text{Mach number}),\quad \pi_5=\alpha\ (\text{already dimensionless}).$$
Reduced relation. The eight-variable law collapses to a five-group relation:
$$\boxed{\dfrac{F}{\rho V^{2} D^{2}}=\phi\!\left(\dfrac{L}{D},\ \dfrac{\rho V D}{\mu},\ \dfrac{V}{c},\ \alpha\right).}$$
So the lift coefficient depends on slenderness, Reynolds number, Mach number and angle of attack — all recognizable, named groups.
