Linear momentum and impulse-momentum equation

Momentum principle

Newton's second law: the rate of change of linear momentum of a body is equal to the resultant force acting on it, and takes place in the direction of that force.

$$\vec F = \frac{d(m\vec V)}{dt}$$

If the force $F$ acts for a short time $dt$, then $F\,dt = d(mV)$. The left side is the impulse and the right side is the change of momentum. This is the impulse-momentum equation.

Control volume form (steady flow)

Consider a fixed control volume with one inlet (section 1) and one outlet (section 2), through which fluid of density $\rho$ flows steadily. The mass flow rate is $m = \rho Q = \rho A_1 V_1 = \rho A_2 V_2$.

In time $dt$ the mass $dm = \rho Q\,dt$ enters with velocity $V_1$ and the same mass leaves with velocity $V_2$. Momentum change of the fluid in the given direction $x$:

$$dm\,(V_{2x} - V_{1x}) = \rho Q\,dt\,(V_{2x}-V_{1x})$$

Dividing by $dt$, the net external force on the fluid in the control volume (pressure forces, body force and force by the solid surface) is

$$\sum F_x = \rho Q\,(V_{2x} - V_{1x}), \qquad \sum F_y = \rho Q\,(V_{2y} - V_{1y})$$

Force of jet on a stationary body

For a free jet, the pressure is atmospheric everywhere on the control surface, so pressure forces cancel and the body force is negligible. The only force on the fluid is the force $R$ exerted by the body on the fluid. By Newton's third law the force of the jet on the body is equal and opposite:

$$F_x = \rho Q\,(V_{1x} - V_{2x})$$

Hence the force on a body equals the rate at which the fluid loses momentum in that direction. Direction of the force is found by the sign of $(V_{1x}-V_{2x})$.

Special results for a jet of area $a$ and velocity $V$:

• Flat plate normal to jet (fluid leaves along plate, so $V_{2x}=0$): $F = \rho a V^2$.
• Symmetrical curved vane turning the jet through angle $\theta$ (measured from the jet direction): $F = \rho a V^2(1-\cos\theta)$.

These expressions are the basis for the analysis of Pelton wheel buckets and other impulse machines.