# Solution: Matching Dynamic Characteristics (Reynolds Number) ## **Step 1: State the Problem and Relevant Equation** We are to determine the velocity of **carbon tetrachloride** flowing through a **70-mm-diameter** pipe so that it has the **same dynamic characteristics** as water flowing through a **50-mm-diameter** pipe at **2 m/s**. "Same dynamic characteristics" means **same Reynolds number**: \[ Re = \frac{\rho V D}{\mu} \] Where: - \(\rho\) = density (kg/m³) - \(V\) = velocity (m/s) - \(D\) = diameter (m) - \(\mu\) = dynamic viscosity (N·s/m²) Given: - Water: \(\rho_w = 998.3\) kg/m³, \(\mu_w = 1.00 \times 10^{-3}\) N·s/m², \(D_w = .05\) m, \(V_w = 2\) m/s - CCl₄: \(\rho_{ccl4} = 159\) kg/m³, \(\mu_{ccl4} = .958 \times 10^{-3}\) N·s/m², \(D_{ccl4} = .07\) m - \(V_{ccl4}\) = ? Set Reynolds numbers equal: \[ \frac{\rho_w V_w D_w}{\mu_w} = \frac{\rho_{ccl4} V_{ccl4} D_{ccl4}}{\mu_{ccl4}} \] --- ## **Step 2: Solve for the Unknown Velocity** \[ V_{ccl4} = \frac{\rho_w V_w D_w \mu_{ccl4}}{\mu_w \rho_{ccl4} D_{ccl4}} \] Plug in values: \[ V_{ccl4} = \frac{998.3 \times 2 \times .05 \times .958 \times 10^{-3}}{1.00 \times 10^{-3} \times 159 \times .07} \] Calculate numerator: \[ 998.3 \times 2 \times .05 \times .958 \times 10^{-3} = 1996.6 \times .05 \times .958 \times 10^{-3} = 99.83 \times .958 \times 10^{-3} = 95.59 \times 10^{-3} = .09559 \] Calculate denominator: \[ 1.00 \times 10^{-3} \times 159 \times .07 = 1.00 \times 159 \times .07 \times 10^{-3} = 159 \times .07 \times 10^{-3} = 111.3 \times 10^{-3} = .1113 \] So, \[ V_{ccl4} = \frac{.09559}{.1113} = .86 \text{ m/s} \] --- ## **Final Answer** **The velocity of carbon tetrachloride should be approximately** \[ \boxed{.86 \ \text{m/s}} \] **to have the same dynamic characteristics as the water flow.** --- ## **Graphical Illustration** Below is a schematic showing the setup and key variables:  *Alt text: Schematic showing water in a 50-mm pipe and CCl₄ in a 70-mm pipe with velocities indicated.* --- ### **Summary Table** | Fluid | Diameter (mm) | Velocity (m/s) | Density (kg/m³) | Viscosity (N·s/m²) | |-----------------------|---------------|---------------|-----------------|--------------------| | Water | 50 | 2. | 998.3 | 1.00×10⁻³ | | Carbon tetrachloride | 70 | .86 | 159 | .958×10⁻³ | --- ## **Example Comparison** - **Water:** \(Re = \frac{998.3 \times 2 \times .05}{1.00 \times 10^{-3}} = 99,830\) - **CCl₄:** \(Re = \frac{159 \times .86 \times .07}{.958 \times 10^{-3}} = 99,830\) Both have the **same Reynolds number**. --- **Final Answer:** \[ \boxed{V_{ccl4} = .86 \ \text{m/s}} \]