Consider fluid flowing with a free stream velocity of \(1 \mathrm{ft} / \mathrm{s}\) over a flat plate, where the critical Reynolds number is \(5 \times 10^{5}\). Determine the distance from the leading edge at which the transition from laminar to turbulent flow occurs for air (at 1 atm), liquid water, isobutane, and engine oil, and mercury. Evaluate all properties at \(50^{\circ} \mathrm{F}\).

The Reynolds number (Re) is a dimensionless quantity that compares inertial forces to viscous forces in a fluid flow and can be calculated as: Re = \(\frac{\rho V L}{\mu}\) Where: - Re is the Reynolds number - \(\rho\) is the fluid density (in \(\frac{lb_m}{ft^3}\)) - V is the free stream velocity (in \(\frac{ft}{s}\), given as 1 \(\frac{ft}{s}\)) - L is the characteristic length (the distance from the leading edge in our case, in ft) - \(\mu\) is the fluid dynamic viscosity (in \(\frac{lb_m}{ft s}\)). We need to find L, so we rearrange the equation to have L as follows: L = \(\frac{Re \ \mu}{\rho V}\) However, it's more convenient to work with kinematic viscosity (\(\nu\)), which is defined as the ratio between dynamic viscosity and density: \(\nu = \frac{\mu}{\rho}\) So, we can rewrite the equation for L as: L = \(\frac{Re \ \nu}{V}\)

We need to look up the kinematic viscosity values for air, liquid water, isobutane, engine oil, and mercury at 50°F (10°C) from a fluid properties table or reference material. Here are the kinematic viscosity values (approximate) at 50°F for each fluid: 1. Air: \(\nu_{air}\) = 1.63 x \(10^{-4} \frac{ft^2}{s}\) 2. Liquid Water: \(\nu_{water}\) = 1.31 x \(10^{-5} \frac{ft^2}{s}\) 3. Isobutane: \(\nu_{isobutane}\) = 7.07 x \(10^{-6} \frac{ft^2}{s}\) 4. Engine oil: \(\nu_{oil}\) = 1.42 x \(10^{-4} \frac{ft^2}{s}\) 5. Mercury: \(\nu_{mercury}\) = 6.30 x \(10^{-7} \frac{ft^2}{s}\)

We can now substitute the values for the critical Reynolds number (Re = \(5 \times 10^5\)), the free stream velocity (V = 1 \(\frac{ft}{s}\)), and each kinematic viscosity into the L equation to find the distance for each fluid: 1. Air: \(L_{air}\) = \(\frac{(5 \times 10^5) \times (1.63 \times 10^{-4} \frac{ft^2}{s})}{1 \frac{ft}{s}}\) = 81.5 ft 2. Liquid Water: \(L_{water}\) = \(\frac{(5 \times 10^5) \times (1.31 \times 10^{-5} \frac{ft^2}{s})}{1 \frac{ft}{s}}\) = 6.55 ft 3. Isobutane: \(L_{isobutane}\) = \(\frac{(5 \times 10^5) \times (7.07 \times 10^{-6} \frac{ft^2}{s})}{1 \frac{ft}{s}}\) = 3.54 ft 4. Engine oil: \(L_{oil}\) = \(\frac{(5 \times 10^5) \times (1.42 \times 10^{-4} \frac{ft^2}{s})}{1 \frac{ft}{s}}\) = 71 ft 5. Mercury: \(L_{mercury}\) = \(\frac{(5 \times 10^5) \times (6.30 \times 10^{-7} \frac{ft^2}{s})}{1 \frac{ft}{s}}\) = 0.315 ft So, the distances from the leading edge where the transition from laminar to turbulent flow occurs for each fluid at 50°F are as follows: 1. Air: 81.5 ft 2. Liquid Water: 6.55 ft 3. Isobutane: 3.54 ft 4. Engine oil: 71 ft 5. Mercury: 0.315 ft