Question

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 $50^{\circ} \mathrm{F}$.

Short answer

Based on the calculations, the distances from the leading edge at which the transition from laminar to turbulent flow occurs for the given fluids at 50°F are: - Air: 64.25 ft - Liquid Water: 5.495 ft - Isobutane: 32 ft - Engine Oil: 220 ft - Mercury: 3.805 ft

Step-by-step solution

Understand the formula for the Reynolds number

Reynolds number (\(Re\)) is a dimensionless quantity which is used to predict flow patterns in different fluid flow situations. It is defined as the ratio of inertial forces to viscous forces within a fluid, and it can be expressed as: \(Re = \frac{U L}{\nu}\) where \(U\) is the flow velocity, \(L\) is the characteristic length, and \(\nu\) is the kinematic viscosity of the fluid.

Find the kinematic viscosity

For each fluid, we need to find the kinematic viscosity at \(50^{\circ} \mathrm{F}\). For this, we can use reference tables or online resources to find the values. After obtaining these values, we can proceed to calculate the distance for each fluid. The kinematic viscosity values used are as follows: - Air: \(\nu_{air} = 1.285 \times 10^{-4} \frac{ft^2}{s}\) - Liquid water: \(\nu_{water} = 1.099 \times 10^{-5} \frac{ft^2}{s}\) - Isobutane: \(\nu_{isobutane} = 6.40 \times 10^{-5} \frac{ft^2}{s}\) - Engine oil: \(\nu_{oil} = 4.40 \times 10^{-4} \frac{ft^2}{s}\) - Mercury: \(\nu_{mercury} = 7.61 \times 10^{-6} \frac{ft^2}{s}\)

Calculate the characteristic length

To calculate the characteristic length, we can rearrange the Reynolds number equation for \(L\) as follows: \(L = \frac{Re \times \nu}{U}\) And applying the given conditions, we have the critical Reynolds number (\(Re_{c}\)) as \(5 \times 10^{5}\), and the free stream velocity (\(U_s\)) as \(1 \frac{ft}{s}\). Therefore, we have the expression for the length as \(L = \frac{Re_{c} \times \nu}{U_s}\). We can calculate the transition distance for each fluid, the equations are: - \(L_{air} = \frac{Re_{c} \times \nu_{air}}{U_s}\) - \(L_{water} = \frac{Re_{c} \times \nu_{water}}{U_s}\) - \(L_{isobutane} = \frac{Re_{c} \times \nu_{isobutane}}{U_s}\) - \(L_{oil} = \frac{Re_{c} \times \nu_{oil}}{U_s}\) - \(L_{mercury} = \frac{Re_{c} \times \nu_{mercury}}{U_s}\)

Calculate the transition distance for each fluid

By plugging the values into the equations from Step 3, we can find the transition distances: - \(L_{air} = \frac{(5 \times 10^{5}) \times (1.285 \times 10^{-4} \frac{ft^2}{s})}{1 \frac{ft}{s}} = 64.25 ft\) - \(L_{water} = \frac{(5 \times 10^{5}) \times (1.099 \times 10^{-5} \frac{ft^2}{s})}{1 \frac{ft}{s}} = 5.495 ft\) - \(L_{isobutane} = \frac{(5 \times 10^{5}) \times (6.40 \times 10^{-5} \frac{ft^2}{s})}{1 \frac{ft}{s}} = 32 ft\) - \(L_{oil} = \frac{(5 \times 10^{5}) \times (4.40 \times 10^{-4} \frac{ft^2}{s})}{1 \frac{ft}{s}} = 220 ft\) - \(L_{mercury} = \frac{(5 \times 10^{5}) \times (7.61 \times 10^{-6} \frac{ft^2}{s})}{1 \frac{ft}{s}} = 3.805 ft\) As a result, the distances from the leading edge at which the transition from laminar to turbulent flow occurs for the given fluids at \(50^{\circ} \mathrm{F}\) are: - Air: \(64.25 ft\) - Liquid Water: \(5.495 ft\) - Isobutane: \(32 ft\) - Engine Oil: \(220 ft\) - Mercury: \(3.805 ft\)