ENSC2004 - Lecture 17

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ENSC2004 - Lecture 16
Impulse and Momentum of Fluid Flow - Slides

Steady Fluid Flow

• Principle of impulse and momentum can be applied to the steady flow of fluid into and out of a control volume
  • Flow in = flow out → steady flow

The change in direction of the fluid flow is caused by the impulse produced by the resultant external force exerted on the fluid volume by the boundaries

Steady Fluid Flow Linear Momentum

• Same flow of water at both A and B

$dm{v}_A+\sum Fdt=dm{v}_B$

• Change in linear momentum:
  $\sum \text{F}=\rho Q(v_B-v_A)$
• When calculating forces:
  $\dot{m}v=vQp$

Steady Fluid Flow Angular Momentum

• Angular momentum at A + Angular Impulse = Angular Momentum at B
  $r_A\times dm{v}_A+r^{\prime }\times \sum Fdt=r_B\times dm{v}_B$
• Change in angular momentum about $O$:
  $\sum {\text{M}}_O=\rho Q(r_A\times v_A-r_B\times v_B)$

Mass Flow Rate

• The term dm/dt is the mass flow rate and can be expressed as:
  $\frac{dm}{dt}=\rho vA=\rho Q$
  $\rho$ $-$ fluid density
  $A$ $-$ cross-sectional area
  $Q$ $-$ volumetric flow rate ($Q=vA$)

You will need to combine this equation with either equation above (usually linear momentum) to solve questions

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Propulsion of Control Volume that Loses Mass - Slides

Propulsion with a Control Volume that Loses Mass

• Analysis of a rocket is complicated as it loses mass over time
  i.e. fuel is burnt and expelled
• Problem can be analysed using impulse and momentum and considering control volume

$\sum {\text{F}}_{cv}=m\frac{dv}{dt}-v_{D/e}\frac{d{m}_e}{dt}$
$\frac{d{m}_e}{dt}$ $-$ rate at which mass is being ejected
$v_{D/e}=v+v_e$ $-$ velocity of ejected mass relative to ground
Remember $\frac{dv}{dt}$ is acceleration!

• The last term in this equation is also known as the thrust ($T$)

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ENSC2004 - Lecture 18