Let's break down the question and analyze the given options: ### Given Data: - **Length of diffuser, L = 4 m** - **Upstream diameter, D₁ = 40 cm = 0.4 m** - **Downstream diameter, D₂ = 80 cm = 0.8 m** - **Discharge at a certain location, Q = 100 L/s = 0.1 m³/s** - **Discharge rate of increase, dQ/dt = 20 L/s = 0.02 m³/s²** - **Location from upstream, x = 2.5 m** ### To Find: - **Local acceleration** - **Convective acceleration** at 2.5 m from the upstream end. --- ### Step 1: **Calculate Area at x = 2.5 m** The diameter at 2.5 m can be found by linear interpolation: \[ \text{Diameter at } x = D_1 + \frac{(D_2 - D_1)}{L} \cdot x \] \[ D(x) = 0.4 + \frac{0.8 - 0.4}{4} \times 2.5 = 0.4 + 0.25 = 0.65 \text{ m} \] \[ A(x) = \frac{\pi}{4} \times (0.65)^2 = 0.3318 \text{ m}^2 \] --- ### Step 2: **Velocity at x = 2.5 m** \[ V(x) = \frac{Q}{A(x)} = \frac{0.1}{0.3318} = 0.301 \text{ m/s} \] --- ### Step 3: **Local Acceleration** \[ a_\text{local} = \frac{1}{A(x)} \frac{dQ}{dt} \] \[ a_\text{local} = \frac{1}{0.3318} \times 0.02 = 0.0602 \text{ m/s}^2 \] --- ### Step 4: **Convective Acceleration** Convective acceleration, \( a_c = V \frac{dV}{dx} \) First, express \( V(x) = \frac{Q}{A(x)} \). Since Q is changing with time, for convective acceleration, we only consider the spatial derivative (Q is instantaneously constant). \[ A(x) = \frac{\pi}{4}[D_1 + \frac{(D_2 - D_1)}{L}x]^2 \] Let \( D(x) = D_1 + \frac{(D_2 - D_1)}{L}x \) \[ A(x) = \frac{\pi}{4} D(x)^2 \] \[ V(x) = \frac{Q}{A(x)} = \frac{4Q}{\pi D(x)^2} \] Now, \[ \frac{dV}{dx} = \frac{d}{dx} \left( \frac{4Q}{\pi D(x)^2} \right) = -\frac{8Q}{\pi D(x)^3} \frac{dD(x)}{dx} \] But \[ \frac{dD(x)}{dx} = \frac{D_2 - D_1}{L} = \frac{0.8 - 0.4}{4} = 0.1 \] So, \[ \frac{dV}{dx} = -\frac{8Q}{\pi [0.65]^3} \cdot 0.1 \] \[ [0.65]^3 = 0.274625 \] \[ \frac{dV}{dx} = -\frac{8 \times 0.1}{\pi \times 0.274625} \cdot 0.1 \] \[ = -\frac{0.8}{0.8626} \cdot 0.1 \] \[ = -0.9276 \cdot 0.1 = -0.09276 \] So, \[ a_c = V(x) \cdot \frac{dV}{dx} = 0.301 \times (-0.09276) = -0.0279 \text{ m/s}^2 \] --- ### Step 5: **Match with Options** - **Option 1:** 0.0602 m/s² and -0.028 m/s² respectively This matches very closely with our calculations. --- ## **Final Answer:** **0.0602 m/s² and -0.028 m/s² respectively** **So, the correct option is the first one.**