Let's go through each part step-by-step: --- # Given Data: - Pipe, \(L = .\, \text{m}\) - Radius, \(R = 100\, \text{mm} = .1\, \text{m}\) - Maximum velocity at outlet, \(u_{max} = 5\, \text{m/s}\) - Water density, \(\rho = 100\, \text{kg/m}^3\) - Velocity profile: \[ u(r) = u_{max} \left( 1 - \frac{r^2}{R^2} \right) \] --- ## Part 1: Determine the magnitude of the uniform velocity \(U\) at the inlet ### Step 1: Understand the velocity profile The velocity distribution across the pipe's cross-section is parabolic: \[ u(r) = u_{max} \left( 1 - \frac{r^2}{R^2} \right) \] ### Step 2: Find the average velocity \(U\) The **average velocity** \(U\) is obtained by integrating the velocity across the cross-section: \[ U = \frac{1}{A} \int_A u(r) \, dA \] where the cross-sectional area: \[ A = \pi R^2 \] ### Step 3: Express the differential area element In cylindrical coordinates: \[ dA = 2\pi r dr \] ### Step 4: Write the integral for the average velocity \[ U = \frac{1}{\pi R^2} \int_^{R} u(r) \, 2\pi r dr = \frac{2}{R^2} \int_^{R} u(r) r dr \] Substituting \(u(r)\): \[ U = \frac{2}{R^2} \int_^{R} u_{max} \left( 1 - \frac{r^2}{R^2} \right) r dr \] Factor out \(u_{max}\): \[ U = \frac{2 u_{max}}{R^2} \int_^{R} \left( 1 - \frac{r^2}{R^2} \right) r dr \] ### Step 5: Compute the integral Let’s evaluate: \[ I = \int_^{R} \left( 1 - \frac{r^2}{R^2} \right) r dr \] Split the integral: \[ I = \int_^{R} r dr - \frac{1}{R^2} \int_^{R} r^3 dr \] Calculate each: \[ \int_^{R} r dr = \frac{1}{2} R^2 \] \[ \int_^{R} r^3 dr = \frac{1}{4} R^4 \] So, \[ I = \frac{1}{2} R^2 - \frac{1}{R^2} \times \frac{1}{4} R^4 = \frac{1}{2} R^2 - \frac{1}{4} R^2 = \frac{1}{4} R^2 \] ### Step 6: Calculate the average velocity \(U\) \[ U = \frac{2 u_{max}}{R^2} \times \frac{1}{4} R^2 = \frac{2 u_{max}}{R^2} \times \frac{1}{4} R^2 = \frac{1}{2} u_{max} \] **Answer for Part 1:** \[ \boxed{ U = \frac{1}{2} u_{max} = \frac{1}{2} \times 5\, \text{m/s} = 2.5\, \text{m/s} } \] --- ## Part 2: Volume flow rate \(Q\) \[ Q = U \times A = U \times \pi R^2 \] Calculate: \[ A = \pi \times (.1)^2 = \pi \times .01 = .01\pi\, \text{m}^2 \] \[ Q = 2.5\, \text{m/s} \times .01 \pi\, \text{m}^2 = .025\pi\, \text{m}^3/\text{s} \] **Numerical value:** \[ Q \approx .025 \times 3.1416 \approx .0785\, \text{m}^3/\text{s} \] --- ## Part 3: Mass flow rate \(\dot{m}\) \[ \dot{m} = \rho Q \] \[ \dot{m} = 100\, \text{kg/m}^3 \times .0785\, \text{m}^3/\text{s} = 78.5\, \text{kg/s} \] --- # **Final Answers:** 1. **Uniform velocity at inlet**: \(\boxed{U = 2.5\, \text{m/s}}\) 2. **Volume flow rate**: \(\boxed{Q \approx .0785\, \text{m}^3/\text{s}}\) 3. **Mass flow rate**: \(\boxed{\dot{m} \approx 78.5\, \text{kg/s}}\) Let me know if you'd like further explanations!