Jet Discharge Propulsion Force

Overview

Engineering reference data for Jet Discharge Propulsion Force in fluid mechanics.

Key Formulas

Reynolds Number

$Re = \frac{\rho v D}{\mu}$

Ratio of inertial to viscous forces — determines flow regime.

Bernoulli's Equation

$P + \frac{1}{2}\rho v^2 + \rho g h = \text{const}$

Conservation of energy for steady, inviscid, incompressible flow.

Continuity Equation

$A_1 v_1 = A_2 v_2$

Conservation of mass for incompressible flow.

Darcy-Weisbach

$\Delta P = f \frac{L}{D} \frac{\rho v^2}{2}$

Pressure drop due to friction in a pipe.

Variables

Worked Example: Garden Hose Propulsive Force

Consider water flowing through a garden hose with a diameter of 15 mm. The gauge pressure just before the outlet is 3×10⁵ Pa (absolute pressure is 4×10⁵ Pa) and the atmospheric pressure is 1×10⁵ Pa (absolute).

Calculation

1. Determine the Outlet Area (A): The cross-sectional area of the hose outlet is calculated using the diameter: $A = \pi \left( \frac{d}{2} \right)^2 = \pi \left( \frac{0.015 \, \text{m}}{2} \right)^2 = 1.77 \times 10^{-4} \, \text{m}^2$

2. Calculate the Propulsive Force (F): For a stationary jet ($v_1 = 0$), the force can be found directly from the pressure difference using Eq. (4): $F = 2 A (p_1 - p_2)$ $F = 2 \times (1.77 \times 10^{-4} \, \text{m}^2) \times ((4 \times 10^5 \, \text{Pa}) - (1 \times 10^5 \, \text{Pa}))$ $F = 2 \times 1.77 \times 10^{-4} \times 3 \times 10^5$ $F \approx 106 \, \text{N}$

This significant force, equivalent to about 24 lbf, demonstrates why a fire hose is difficult to hold and why the reaction thrust from a jet engine is so powerful.

Important Note on Fluid Applicability

The density $\rho$ is treated as constant in these incompressible flow equations. Therefore, these formulas are directly valid for liquids like water but not for gases like air, where density changes with pressure must be accounted for using compressible flow relations.

References

• Original Source