Given that the drag force experienced by a projectile which can move in water or in air is given by F = f(p, V, D, μ) where p= density of the fluid, V= Velocity of the projectile, D= Diameter of the projectile, μ = viscosity of the fluid. Show by dimensional analysis that F / (pv²D²) = f(pVD/μ). (b) A projectile having a diameter of 1000 mm is to move in air at 50 m/s and its drag force is to be estimated from tests in water on a half-scale model. Determine: (i) the speed of the model in water and (ii) the drag force acting on the full-scale projectile given that the drag force on the half-scale model is 1153N.

• Drag Force ($F_D$): $[M L T^{-2}]$
• Density ($\rho$): $[M L^{-3}]$
• Velocity ($V$): $[L T^{-1}]$
• Diameter ($D$): $[L]$
• Dynamic Viscosity ($\mu$): $[M L^{-1} T^{-1}]$

• Number of variables ($n$) = 5 ($F_D, \rho, V, D, \mu$)
• Number of primary dimensions ($k$) = 3 (Mass [M], Length [L], Time [T])
• Number of dimensionless Pi groups ($\pi$) = $n - k = 5 - 3 = 2$

We choose $\rho$, $V$, and $D$ as repeating variables because they collectively contain all three primary dimensions (M, L, T) and are fundamental to the flow.

Each Pi group will be formed by combining the repeating variables with one of the remaining non-repeating variables.

For $\pi_1$ (using $F_D$):

Substitute the dimensions:

Equate the exponents for each primary dimension:

• For M: $0 = a + 1 \implies a = -1$
• For T: $0 = -b - 2 \implies b = -2$
• For L: $0 = -3a + b + c + 1$ Substitute $a = -1$ and $b = -2$: $0 = -3(-1) + (-2) + c + 1$ $0 = 3 - 2 + c + 1$ $0 = 2 + c \implies c = -2$ So, $\pi_1 = \rho^{-1} V^{-2} D^{-2} F_D = \frac{F_D}{\rho V^2 D^2}$.

For $\pi_2$ (using $\mu$):

Substitute the dimensions:

Equate the exponents for each primary dimension:

• For M: $0 = x + 1 \implies x = -1$
• For T: $0 = -y - 1 \implies y = -1$
• For L: $0 = -3x + y + z - 1$ Substitute $x = -1$ and $y = -1$: $0 = -3(-1) + (-1) + z - 1$ $0 = 3 - 1 + z - 1$ $0 = 1 + z \implies z = -1$ So, $\pi_2 = \rho^{-1} V^{-1} D^{-1} \mu = \frac{\mu}{\rho V D}$.

According to the Buckingham Pi theorem, the relationship between the variables can be expressed as a function of the dimensionless Pi groups:

Substituting the derived Pi groups:

Since $f$ represents an arbitrary function, $f\left(\frac{\mu}{\rho V D}\right)$ is equivalent to $f\left(\frac{\rho V D}{\mu}\right)$ (as one is simply the inverse of the other, and the functional relationship can absorb this change).
Therefore, we have shown that: