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Engineering and Tech / Fluid Mechanics 2 / Chapter 4 / Problem 108P

Fluid Mechanics | 2nd Edition | ISBN: 9780071284219 | Authors: Yunus A. Cengel, John M. Cimbala

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Consider the vacuum cleaner of Prob. 11–54. For the case

Chapter 4, Problem 108P

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QUESTION:

Problem 108P

Consider the vacuum cleaner of Prob. 11–54. For the case where b = 2.0 cm, L = 35 cm, and = 0.1098 m3/s, create a velocity vector plot in the upper half of the xy-plane from x = − 3 cm to 3 cm and from y = 0 cm to 2.5 cm. Draw as many vectors as you need to get a good feel of the flow field.

Note:The velocity is infinite at the point (x, y) =(0,2.0 cm), so do not attempt to draw a velocity vector at that point.

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QUESTION:

Problem 108P

Consider the vacuum cleaner of Prob. 11–54. For the case where b = 2.0 cm, L = 35 cm, and = 0.1098 m3/s, create a velocity vector plot in the upper half of the xy-plane from x = − 3 cm to 3 cm and from y = 0 cm to 2.5 cm. Draw as many vectors as you need to get a good feel of the flow field.

Note:The velocity is infinite at the point (x, y) =(0,2.0 cm), so do not attempt to draw a velocity vector at that point.

ANSWER:

Solution 108P

Step 1 of 3:

In this question we need to draw velocity vector plot for a given velocity field

It is given

$b$ of the attachment above the floor

$L=35\ cm$ is the length of the attachment

$v'=0.1098\ {m}^{3}/s$ is the volume flow rate of air

The velocity components given are

$x=3.0\ cm\ to\ -3.0\ cm$

$y=0\ cm\ to\ -2.5\ cm$

$(x.y)=(0,2.0\ cm)$

Initial position

$u=\frac{-Vx}{π\ L}\frac{{x}^{2}+{y}^{2}+{b}^{2}}{{x}^{4}+2{x}^{2}{y}^{2}+2{x}^{2}{b}^{2}+{y}^{4}-2{y}^{2}{b}^{2}+{b}^{4}}$ -----------(1)

Final position

$v=\frac{-Vy}{π\ L}\frac{{x}^{2}+{y}^{2}-{b}^{2}}{{x}^{4}+2{x}^{2}{y}^{2}+2{x}^{2}{b}^{2}+{y}^{4}-2{y}^{2}{b}^{2}+{b}^{4}}$ -------------(2)

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