Let's solve the problem step by step. --- # **Step 1: Write the Reynolds Number Formula** \[ Re = \frac{\rho V D}{\mu} \] where: - \(\rho\) = density of fluid - \(V\) = velocity of fluid - \(D\) = diameter of pipe - \(\mu\) = dynamic viscosity --- # **Step 2: Dimensional Analysis** ## **A. Using the MLT System (Mass, Length, Time)** Let's write the dimensions of each term: - \(\rho\): density = \(\frac{\text{M}}{\text{L}^3}\) - \(V\): velocity = \(\frac{\text{L}}{\text{T}}\) - \(D\): length = \(\text{L}\) - \(\mu\): dynamic viscosity = \(\frac{\text{M}}{\text{L}\,\text{T}}\) Plugging these into the Reynolds number expression: \[ Re = \frac{(\frac{M}{L^3}) (\frac{L}{T}) (L)}{(\frac{M}{L\,T})} \] Simplify numerator: \[ (\frac{M}{L^3}) \cdot (\frac{L}{T}) \cdot (L) = \frac{M \cdot L \cdot L}{L^3 \cdot T} = \frac{M L^2}{L^3 T} = \frac{M}{L T} \] Denominator: \[ \frac{M}{L T} \] So, \[ Re = \frac{\frac{M}{L T}}{\frac{M}{L T}} = 1 \] **Conclusion:** Reynolds number is dimensionless in the MLT system. --- ## **B. Using the FLT System (Force, Length, Time)** First, recall: - \(\rho = \frac{\text{F} \cdot \text{T}^2}{\text{L}^4}\) (since \(F = MLT^{-2}\)) - \(V = \frac{L}{T}\) - \(D = L\) - \(\mu = \frac{\text{F} \cdot \text{T}}{\text{L}^2}\) Plug into Reynolds formula: Numerator: \[ \rho V D = \left(\frac{F T^2}{L^4}\right) \left(\frac{L}{T}\right) (L) = \frac{F T^2 \cdot L \cdot L}{L^4 \cdot T} = \frac{F T^2 L^2}{L^4 T} = \frac{F T L^2}{L^4} = \frac{F T}{L^2} \] Denominator: \[ \mu = \frac{F T}{L^2} \] So, \[ Re = \frac{\frac{F T}{L^2}}{\frac{F T}{L^2}} = 1 \] **Conclusion:** Reynolds number is dimensionless in the FLT system. --- # **Step 3: Calculate Reynolds Number for Ethyl Alcohol** **Given:** - \(V = 3 \ \text{m/s}\) - \(D = 2 \ \text{in.} = .0508 \ \text{m}\) - Properties of ethyl alcohol at about room temperature: - \(\rho \approx 789 \ \text{kg/m}^3\) - \(\mu \approx 1.20 \times 10^{-3} \ \text{kg/(m·s)}\) ## **Plug into Formula** \[ Re = \frac{\rho V D}{\mu} \] \[ Re = \frac{789 \ \text{kg/m}^3 \times 3 \ \text{m/s} \times .0508 \ \text{m}}{1.20 \times 10^{-3} \ \text{kg/(m·s)}} \] Calculate numerator: \[ 789 \times 3 \times .0508 = 789 \times .1524 \approx 120.27 \] So numerator ≈ **120.27 kg/(m²·s)** Now divide by viscosity: \[ Re = \frac{120.27}{1.20 \times 10^{-3}} \approx 100,225 \] --- # **Final Answers** ## **1. The Reynolds number is dimensionless** in both the MLT and FLT systems. ## **2. Its value for ethyl alcohol (at 3 m/s, 2-in pipe):** \[ \boxed{Re \approx 1. \times 10^5} \] --- **Let me know if you need more details or a breakdown of unit conversions!**