Linear Momentum Flux

\sum F = ρQ (v_{out} - v_{in})

Steady-flow x-momentum on a control volume: the net external force equals the net efflux of momentum. ρQ is the mass-flow rate. A jet striking a normal plate and stopping (vout,n = 0) delivers ρQv to the plate. Reversing a jet 180° doubles that force.

$\sum F$ (N): Net external force on the CV
$ρ$ (kg/m³): Density
$Q$ (m³/s): Volume flow rate
$v_{out}, v_{in}$ (m/s): Velocity leaving/entering the CV

Step-by-step solved example

A water jet, A = 0.020 m², v = 20 m/s, strikes a normal wall and stops. Force of the jet on the wall?

ΣF = ρQ(v_out − v_in) on a control volume (x or y).

1. Mass flow
Q = AV = 0.020×20 = 0.40 m³/s. ṁ = ρQ = 400 kg/s.
2. Momentum
ΣFx on fluid = ṁ(0 − 20) = −8000 N. Fluid on wall = +8.00 kN.
$F_{wall} = 8.00 kN$

Answer: 8.00 kN on the wall

10 practice questions

0/10 correct

1.The product ρQ is

2.x-momentum on a CV is

3.Jet hits a normal wall and stops. Force on the wall is

4.If v_out = v_in in the same direction, momentum-flux change is

5.Q = 0.050 m³/s, vin = 4 m/s, vout = 10 m/s, ρ = 1000. Δ momentum =

6.Turning a jet 180° (ideal, same speed) the force magnitude is

7.ρ Q v for a jet also equals

8.Body force on the CV includes

9.Water ṁ = 20 kg/s, Δvx = 15 m/s. Fx =

10.The momentum equation is