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The document discusses mass and momentum fluxes through surfaces in fluid dynamics. It provides the answers to two questions: 1) The mass flux through two surfaces must be the same, which is calculated to be 15.7 kg s-1. 2) Using the steady-state momentum principle and given values, the x-component of momentum flux on the fluid in the control volume is calculated to be 503 N.

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Karim Magdy
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39 views4 pages

Answers 1

The document discusses mass and momentum fluxes through surfaces in fluid dynamics. It provides the answers to two questions: 1) The mass flux through two surfaces must be the same, which is calculated to be 15.7 kg s-1. 2) Using the steady-state momentum principle and given values, the x-component of momentum flux on the fluid in the control volume is calculated to be 503 N.

Uploaded by

Karim Magdy
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Answers 1

Q1.
For S1:

This must be the same as the mass flux through S2. (Otherwise, mass would accumulate
between these surfaces.)

Answer: 15.7 kg s–1 (both surfaces).

To calculate the mass flux in general, use the component of velocity normal to the area or,
equivalently, the projected area normal to the velocity):

CFD Answers 1 – 1
Q2.
Steady-state momentum principle:
force (on fluid) = (momentum flux)out – (momentum flux)in
where, for a uniform velocity:
momentum flux = mass flux × velocity

For the x-component of momentum and the fluid in a control volume encompassing the
region shown:

Hence,

Answer: 503 N.

CFD Answers 1 – 2

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