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.

Step 1: List the primary dimensions of each variable.

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

Step 2: Determine the number of variables and primary dimensions.

• 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 ($\pi$) = $n - k = 5 - 3 = 2$

Step 3: Choose repeating variables. 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.

Step 4: Form the dimensionless Pi groups. Each Pi group will be formed by combining the repeating variables with one of the remaining non-repeating variables.

For $\pi_1$ (using $F_D$): $\pi_1 = \rho^a V^b D^c F_D$ Substitute the dimensions: $[M^0 L^0 T^0] = [M L^{-3}]^a [L T^{-1}]^b [L]^c [M L T^{-2}]$ 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 $\mu$): $\pi_2 = \rho^x V^y D^z \mu$ Substitute the dimensions: $[M^0 L^0 T^0] = [M L^{-3}]^x [L T^{-1}]^y [L]^z [M L^{-1} T^{-1}]$ 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, $\pi_2 = \rho^{-1} V^{-1} D^{-1} \mu = \frac{\mu}{\rho V D}$.

Step 5: Express the functional relationship. According to the Buckingham Pi theorem, the relationship between the variables can be expressed as a function of the dimensionless Pi groups: $\pi_1 = f(\pi_2)$ Substituting the derived Pi groups: $\frac{F_D}{\rho V^2 D^2} = f\left(\frac{\mu}{\rho V D}\right)$ 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: \frac{F_D{\rho V^2 D^2} = f\left(\frac{\rho V D}{\mu}\right)}