Question

A cylindrical jet of water (density = 1000 kg/m³) impinges at the center of a flat, circular plate and spreads radially outwards, as shown in the figure. The plate is resting on a linear spring with a spring constant \( k = 1 \ \text{kN/m} \). The incoming jet diameter is \( D = 1 \ \text{cm} \). 

If the spring shows a steady deflection of 1 cm upon impingement of the jet, then the velocity of the incoming jet is ________ m/s (round off to one decimal place).
 

Hint:

The velocity of the jet is related to the spring deflection, as the force exerted by the jet is balanced by the spring force.

Answer 1

Correct Answer: 11.2

The force exerted by the water jet is equal to the force exerted on the spring due to the deflection. The spring force is given by: \[ F_{\text{spring}} = k \times \delta \] Where: - \( k = 1 \ \text{kN/m} = 1000 \ \text{N/m} \) is the spring constant, - \( \delta = 1 \ \text{cm} = 0.01 \ \text{m} \) is the deflection. Thus, the spring force is: \[ F_{\text{spring}} = 1000 \times 0.01 = 10 \ \text{N} \] This force is due to the momentum of the incoming jet. The force due to the jet is given by: \[ F_{\text{jet}} = \dot{m} \cdot v \] Where:
- \( \dot{m} = \rho \cdot A \cdot v \) is the mass flow rate,
- \( \rho = 1000 \ \text{kg/m}^3 \) is the density of water,
- \( A = \pi \left(\frac{D}{2}\right)^2 \) is the cross-sectional area of the jet,
- \( v \) is the velocity of the jet.
Substitute the values for \( A \) and \( \dot{m} \): \[ A = \pi \left(\frac{0.01}{2}\right)^2 = 7.854 \times 10^{-5} \ \text{m}^2 \] \[ \dot{m} = 1000 \times 7.854 \times 10^{-5} \times v = 0.07854 \times v \] Now, equating the spring force and jet force: \[ 10 = 0.07854 \times v^2 \] Solving for \( v \): \[ v^2 = \frac{10}{0.07854} = 127.4 \] \[ v = \sqrt{127.4} \approx 11.3 \ \text{m/s} \] Thus, the velocity of the incoming jet is: \[ \boxed{11.3 \ \text{m/s}} \]