Fluid Mechanics: Fundamentals and Applications
Fluid Mechanics: Fundamentals and Applications
4th Edition
ISBN: 9781259696534
Author: Yunus A. Cengel Dr., John M. Cimbala
Publisher: McGraw-Hill Education
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Chapter 9, Problem 65EP
To determine

(a)

The axial speeds (ft/s) at the nozzle entrance and at the nozzle exit.

Expert Solution
Check Mark

Answer to Problem 65EP

The axial speed ft/s at the nozzle entrance is 3.2688ft/s.

The axial speed ft/s at the nozzle exit is 41.7404ft/s.

Explanation of Solution

Given information:

The diameter at the entrance of the nozzle is 0.50in, the diameter at exit of the nozzle is 0.14in. The length of the nozzle is 0.20in. The volume flow rate through nozzle is 2gal/min.

Write the expression for axial speeds at the nozzle entrance.

  uz,entrance=4V˙πDentrance2   ....... (I)

Here, the volume flow rate is V˙, the axial speed at the entrance of the nozzle is uz,entrance, and the diameter at the entrance of the nozzle is Dentrance.

Write the expression for axial speeds at the nozzle exit.

  uz,exit=4V˙πDexit2   ....... (II)

Here, the axial speed at the exit of the nozzle is uz,exit, and the diameter at the exit of the nozzle is Dexit.

Calculation:

Substitute 0.50in for Dentrance and 2gal/min for V˙ in Equation (I).

  uz,entrance=4(2gal/min)π( 0.50in)2=8gal/min( 0.1337 ft 3 gal)0.7853in2=1.0696ft3/min( 144 in 2 ft 2 )0.7853in2=154.022ftin2/min( 1min 60s)0.7853in2

  uz,entrance=2.56704ftin2/s0.7853in2=2.56704ft/s0.7853=3.2688ft/s

Substitute 0.14in for Dexit and 2gal/min for V˙ in Equation (II).

  uz,exit=4(2gal/min)π( 0.14in)2=8gal/min( 0.1337 ft 3 gal)0.0.0615in2=1.0696ft3/min( 144 in 2 ft 2 )0.0.0615in2=154.022ftin2/min( 1min 60s)0.0.0615in2

  uz,exit=2.56704ftin2/s0.0615in2=2.56704ft/s0.0615=41.7404ft/s

Conclusion:

The axial speeds ft/s at the nozzle entrance is 3.2688ft/s.

The axial speeds ft/s at the nozzle exist is 41.7404ft/s.

To determine

(b)

The several streamlines in the rz -plane inside the plane and design the appropriate nozzle shape.

Expert Solution
Check Mark

Answer to Problem 65EP

The following figure represents the stream lines.

  Fluid Mechanics: Fundamentals and Applications, Chapter 9, Problem 65EP , additional homework tip  1

Explanation of Solution

Given information:

The diameter at the entrance of the nozzle is 0.50in, the diameter at exist of the nozzle is 0.14in. The length of the nozzle is 0.20in. The volume flow rate through nozzle is 2gal/min.

Write the expression for axial speeds at the nozzle entrance.

  uz,entrance=4V˙πDentrance2   ....... (I)

Here, the volume flow rate is V˙, the axial speed at the entrance of the nozzle is uz,entrance, and the diameter at the entrance of the nozzle is Dentrance.

Write the expression for axial speeds at the nozzle exit.

  uz,exit=4V˙πDexit2   ....... (II)

Here, the axial speed at the exit of the nozzle is uz,exit, and the diameter at the exit of the nozzle is Dexit.

Write the expression for stream function of the flow filed.

  ψ=r22(uz,entrance+uz,exituz,entranceLz)+C   ....... (III)

Here, the stream function of the flow field is ψ, the radius is r, the distance in z direction is z and the length of nozzle is L.

Calculation:

Substitute 0.50in for Dentrance and 2gal/min for V˙ in Equation (I).

  uz,entrance=4(2gal/min)π( 0.50in)2=8gal/min( 0.1337 ft 3 gal)0.7853in2=1.0696ft3/min( 144 in 2 ft 2 )0.7853in2=154.022ftin2/min( 1min 60s)0.7853in2

  uz,entrance=2.56704ftin2/s0.7853in2=2.56704ft/s0.7853=3.2688ft/s

Substitute 0.14in for Dexit and 2gal/min for V˙ in Equation (II).

  uz,exit=4(2gal/min)π( 0.14in)2=8gal/min( 0.1337 ft 3 gal)0.0.0615in2=1.0696ft3/min( 144 in 2 ft 2 )0.0.0615in2=154.022ftin2/min( 1min 60s)0.0.0615in2

  uz,exit=2.56704ftin2/s0.0615in2=2.56704ft/s0.0615=41.7404ft/s

Substitute 41.7404ft/s for uz,exit, 3.2688ft/s for uz,entrance, 0.20in for L and 0 for C in Equation (III).

  ψ=r22(3.2688ft/s+( 41.7404 ft/s 3.2688 ft/s 0.20in)z)ψ=r22(3.2688ft/s41.7404ft/s3.2688ft/s0.20in( 1ft 12in )z)ψ=r22(3.2688ft/s230.514zs1)r=2ψ( 3.2688 ft/s 230.514z s 1 )   ...... (IV)

Substitute 0 for z and 0.25in for r in Equation (IV).

  0.25in=2ψ( 3.2688 ft/s 230.514( 0 ) s 1 )0.25in(1ft12in)=2ψ3.2688ft/sψ=( 0.0208ft)22(3.2688ft/s)ψ=0.000707ft3/s

Substitute 0 for z and 0.07in for r in Equation (IV).

  0.07in=2ψ( 3.2688 ft/s 230.514( 0 ) s 1 )0.07in(1ft12in)=2ψ3.2688ft/sψ=( 0.0058334ft)22(3.2688ft/s)ψ=0.0000463ft3/s

The different values of stream function and radius are shown below the table.

      ψ(ft3/s)  z  r(in)  r
      0.00071  0  0.25  0.25
      0.00071  0.02  0.0014  0.0014
      0.0000463ft3/s  0  0.07  0.01796
      0.0000463ft3/s  0.02  0.011377  0.011377

The following figure represents the stream line with varying of length.

  Fluid Mechanics: Fundamentals and Applications, Chapter 9, Problem 65EP , additional homework tip  2

  Figure-(1)

For design, the shape of the nozzle is like the graph of the boundary layers.

Conclusion:

The following figure represents the stream lines.

  Fluid Mechanics: Fundamentals and Applications, Chapter 9, Problem 65EP , additional homework tip  3

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