The question is about a diffuser with upstream and downstream diameters of 40 cm and 80 cm, a length of 4 meters, and involves calculating the local and convective accelerations of an incompressible fluid flow. ### Let's break down the information: - **Length of diffuser:** 4 m - **Upstream diameter:** 40 cm = 0.4 m - **Downstream diameter:** 80 cm = 0.8 m - **Discharge (Q):** 100 liters/s = 0.1 m³/s - **Discharge increasing at:** 20 liters/s = 0.02 m³/s² - **Location for calculation:** 2.5 m from upstream end ### The options are: 1. 0.0602 m/s² and -0.028 m/s² respectively 2. 0.0546 m/s² and -0.036 m/s² respectively 3. 0.0781 m/s² and -0.046 m/s² respectively 4. 0.0647 m/s² and -0.016 m/s² respectively ### **Calculation steps:** #### 1. **Area at x = 2.5 m from upstream** The diameter changes linearly from 0.4 m to 0.8 m over 4 m. Let \( d(x) \) be the diameter at distance \( x \) from upstream: \[ d(x) = d_1 + \frac{d_2 - d_1}{L} x \] Where: - \( d_1 = 0.4 \) m - \( d_2 = 0.8 \) m - \( L = 4 \) m At \( x = 2.5 \) m: \[ d(2.5) = 0.4 + \frac{0.8 - 0.4}{4} \times 2.5 = 0.4 + 0.1 \times 2.5 = 0.4 + 0.25 = 0.65 \text{ m} \] Area at \( x = 2.5 \): \[ A = \frac{\pi}{4} d^2 = \frac{\pi}{4} (0.65)^2 \approx 0.332 \text{ m}^2 \] #### 2. **Velocity at \( x = 2.5 \) m:** \[ V = \frac{Q}{A} = \frac{0.1}{0.332} \approx 0.301 \text{ m/s} \] #### 3. **Local acceleration (\( \frac{\partial V}{\partial t} \)):** \[ \frac{\partial V}{\partial t} = \frac{1}{A} \frac{dQ}{dt} = \frac{0.02}{0.332} \approx 0.0602 \text{ m/s}^2 \] #### 4. **Convective acceleration (\( V \frac{dV}{dx} \)):** First, find \( \frac{dV}{dx} \): Since \( Q \) and thus \( V \) is a function of \( x \): \[ V(x) = \frac{Q}{A(x)} \] \[ \frac{dV}{dx} = \frac{d}{dx} \left( \frac{Q}{A(x)} \right) = -\frac{Q}{A(x)^2} \frac{dA}{dx} \] Area: \( A = \frac{\pi}{4} d^2 \) \[ \frac{dA}{dx} = \frac{\pi}{2} d \frac{dd}{dx} \] \[ \frac{dd}{dx} = \frac{d_2 - d_1}{L} = \frac{0.4}{4} = 0.1 \] \[ \frac{dA}{dx} = \frac{\pi}{2} \times 0.65 \times 0.1 \approx 0.102 \text{ m}^2/\text{m} \] Now, \[ \frac{dV}{dx} = -\frac{0.1}{(0.332)^2} \times 0.102 \approx -0.0925 \text{ m/s} \cdot \text{m}^{-1} \] \[ \text{Convective acceleration} = V \frac{dV}{dx} = 0.301 \times (-0.0925) \approx -0.0278 \text{ m/s}^2 \] --- ### **Final Answers (rounded to match options):** - **Local acceleration:** \( 0.0602 \) m/s² - **Convective acceleration:** \( -0.028 \) m/s² So, the correct answer is: **0.0602 m/s² and −0.028 m/s² respectively** (first option).