Interactive Lecture
Momentum Equation Applied to Jets and Turbine Blades
by Dr Arnaldo Delli Carri
When a fluid jet strikes a solid surface, it exerts a force on that surface due to the change in momentum of the fluid. This momentum exchange is exactly what turbines, jet thrusters, and hydraulic actuators exploit to do useful work.
The force a fluid jet exerts on a surface equals the rate of change of the fluid's momentum. This is Newton's second law applied to a control volume: \(F = \dfrac{dp}{dt} = \dot{m}(v_1 - v_2)\), where \(\dot{m}\) is the mass flow rate.
The momentum equation for steady flow states that the sum of external forces acting on a control volume equals the net rate of momentum flux leaving the control volume.
Force on Deflector:
\[F = \rho Q (v_1 - v_2 \cos\theta)\]
Where:
Different deflector geometries produce different force characteristics:
For moving deflectors, the relative velocity between the jet and deflector must be considered:
Moving Deflector Force:
\[F = \rho A (v - u)^2 (1 - \cos\theta)\]
Where:
Power and Efficiency:
\[P = F \cdot u = \rho A (v - u)^2 u (1 - \cos\theta)\]
\[\eta = \frac{P}{\frac{1}{2}\rho A v^3} = \frac{2(v-u)^2 u (1-\cos\theta)}{v^3}\]
The kinetic-energy flux of the jet \(\left(\tfrac{1}{2}\rho A v^3\right)\) does not depend on \(u\), so maximum power and maximum efficiency coincide. Setting \(\mathrm{d}P/\mathrm{d}u = 0\) gives \((v-u)(v-3u) = 0\), so both peak at \(u = v/3\). For a flat plate (θ = 90°) this is a peak efficiency of \(\eta_{\max} = 8/27 \approx 29.6\%\).
Adjust the jet velocity, deflector type, and deflector velocity to see how the force and power change. Watch the jet interact with different deflector geometries.
| Concept | Equation | Description |
|---|---|---|
| Mass Flow Rate | \(\dot{m} = \rho A v\) | Mass per unit time (kg/s) |
| Momentum Flux | \(\dot{m} v = \rho A v^2\) | Rate of momentum transfer (N) |
| Force (Stationary Flat Plate) | \(F = \rho A v^2\) | Normal force on perpendicular plate (N) |
| Force (Stationary Curved Vane) | \(F = \rho A v^2 (1 - \cos\theta)\) | Force with deflection angle θ (N) |
| Force (Moving Deflector) | \(F = \rho A (v-u)^2 (1-\cos\theta)\) | Force on deflector moving at velocity u (N) |
| Power | \(P = F \cdot u\) | Power delivered to moving deflector (W) |
| Kinetic Energy of Jet | \(KE = \frac{1}{2}\rho Q v^2\) | Available kinetic-energy flux in the jet (W) |
| Efficiency | \(\eta = \frac{P}{KE}\) | Ratio of extracted to available power (dimensionless) |
A fire hose has a nozzle diameter of 50 mm and discharges water at 25 m/s. The water jet hits a wall perpendicularly. Calculate the force exerted on the wall. (Density of water = 1000 kg/m³)
Show Solution
Solution:
Given:
Step 1: Calculate jet area:
\[A = \frac{\pi D^2}{4} = \frac{\pi (0.05)^2}{4} = 0.0019635 \text{ m}^2\]
Step 2: Apply force equation for flat plate:
\[F = \rho A v^2 = 1000 \times 0.0019635 \times 25^2\]
\[F = 1000 \times 0.0019635 \times 625 = 1227 \text{ N}\]
Answer: The force on the wall is 1227 N (approximately 125 kg force)
A water jet with velocity 30 m/s and diameter 75 mm strikes a flat plate moving at 10 m/s in the same direction as the jet. Calculate: (a) the force on the plate, (b) the power delivered, and (c) the efficiency.
Show Solution
Solution:
Given:
Step 1: Calculate jet area:
\[A = \frac{\pi (0.075)^2}{4} = 0.00442 \text{ m}^2\]
Step 2: Calculate force (moving deflector):
\[F = \rho A (v-u)^2 (1-\cos 90°) = \rho A (v-u)^2\]
\[F = 1000 \times 0.00442 \times (30-10)^2 = 1000 \times 0.00442 \times 400\]
\[F = 1768 \text{ N}\]
Step 3: Calculate power:
\[P = F \cdot u = 1768 \times 10 = 17680 \text{ W} = 17.68 \text{ kW}\]
Step 4: Calculate efficiency:
\[Q = A v = 0.00442 \times 30 = 0.1326 \text{ m}^3\text{/s}\]
\[KE = \frac{1}{2}\rho Q v^2 = 0.5 \times 1000 \times 0.1326 \times 900 = 59670 \text{ W}\]
\[\eta = \frac{P}{KE} = \frac{17680}{59670} = 0.296 = 29.6\%\]
Answers: (a) F = 1768 N, (b) P = 17.68 kW, (c) η = 29.6%
A water jet with velocity 20 m/s and cross-sectional area 0.01 m² strikes a stationary curved vane that deflects the jet through 180°. Calculate the force on the vane in the direction of the jet.
Show Solution
Solution:
Given:
Apply force equation for curved vane:
\[F = \rho A v^2 (1 - \cos\theta)\]
\[F = 1000 \times 0.01 \times 20^2 \times (1 - \cos 180°)\]
\[F = 1000 \times 0.01 \times 400 \times (1 - (-1))\]
\[F = 4000 \times 2 = 8000 \text{ N}\]
Note: The force is twice that of a flat plate because the jet is reversed completely.
Answer: The force on the vane is 8000 N in the direction of the incoming jet