GATE AE 2017 • Q36 NAT • 2 Marks Propulsion

Propulsion

Inlets, Combustors & Nozzles

Subtopic: Convergent & Convergent-Divergent (C-D) Nozzles

Question Statement:

Water exits from a nozzle of diameter 50 mm into atmospheric pressure at a velocity of 20 m/s. The jet strikes a flat vertical plate held perpendicular to the jet. If the density of water is 1000 kg/m$^3$, the horizontal force (in N) required to hold the plate stationary is _________ (round off to the nearest integer).

Official Answer Key:

785.0

Step-by-Step Solution & Explanation

### **Detailed Step-by-Step Solution (GATE AE 2017 Q.36)** **1. Intuitive Physical Setup & Given Parameters**: - Water jet diameter: $d = 50\text{ mm} = 0.05\text{ m}$. - Jet exit velocity: $V = 20\text{ m/s}$. - Density of water: $\rho = 1000\text{ kg/m}^3$. - Target: Force exerted on the flat stationary normal plate. **2. Step-by-Step Fluid Mechanics Derivation**: 1. **Jet Cross-Sectional Area ($A$)**: A = \frac{\pi d^2}{4} = \frac{\pi (0.05)^2}{4} = \frac{0.0025\pi}{4} \approx 1.9635 \times 10^{-3}\text{ m}^2 2. **Mass Flow Rate ($\dot{m}$)**: \dot{m} = \rho A V = 1000 \times (1.9635 \times 10^{-3}) \times 20 = 39.27\text{ kg/s} 3. **Control Volume Linear Momentum Balance**: The fluid strikes the plate normally and deflects radially along the plate surface at $90^\circ$ (exit velocity normal to plate is zero): F_x = \dot{m}(V_{in, x} - V_{out, x}) = \dot{m}(V - 0) = \rho A V^2 4. **Evaluating Impact Force**: F_x = 1000 \times (1.9635 \times 10^{-3}) \times (20)^2 = 1.9635 \times 400 = 785.4\text{ N} Rounding to the nearest integer: F_x \approx 785\text{ N} Acceptable official range: **[780, 790] N**. **3. Comprehensive Options & Distractor Analysis**: - **Question Type**: Numerical Answer Type (NAT). - **Official Range**: **780 to 790 N**. **4. Official Master Key & Exam Takeaway**: - **Official Answer**: **785.0** (Acceptable Range: **780 to 790**) - **Key Formula**: $F = \rho A V^2 = \dot{m} V$ for normal stationary plate.
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