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 12, Problem 148P
To determine

To plot: the mass flow parameter.

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Answer to Problem 148P

Fluid Mechanics: Fundamentals and Applications, Chapter 12, Problem 148P , additional homework tip  1

Explanation of Solution

Given:

Mach number k=1.2,1.4 and 1.6 range of 0Ma1.

Calculation:

Equation is given by

  m. R T 0 AP0=PAMa k RT R T 0 AP0m. R T 0 AP0=Mak(P P 0 )( T 0 T )m. R T 0 AP0=Mak( ( 1+ k-1 2 M a 2 ) -k/k-1)(1+ k-1 2M a 2)1/2m. R T 0 AP0=Mak( ( 1+ k-1 2 M a 2 ) -( k+1 )/2( k-1 ))

  k=1.4Ma=0.2m. R T 0 AP0=0.21.4( ( 1+ 1.4-1 2 0. 2 2 ) -( 1.4+1 )/2( 1.4-1 ))m. R T 0 AP0=0.231

  k=1.4Ma=0.4m. R T 0 AP0=0.41.4( ( 1+ 1.4-1 2 0. 4 2 ) -( 1.4+1 )/2( 1.4-1 ))m. R T 0 AP0=0.4306

  k=1.4Ma=0.6m. R T 0 AP0=0.61.4( ( 1+ 1.4-1 2 0. 6 2 ) -( 1.4+1 )/2( 1.4-1 ))m. R T 0 AP0=0.57

  k=1.4Ma=0.8m. R T 0 AP0=0.81.4( ( 1+ 1.4-1 2 0. 8 2 ) -( 1.4+1 )/2( 1.4-1 ))m. R T 0 AP0=0.65

  k=1.4Ma=1.0m. R T 0 AP0=1.01.4( ( 1+ 1.4-1 2 1. 0 2 ) -( 1.4+1 )/2( 1.4-1 ))m. R T 0 AP0=0.68

  k=1.6Ma=0.2m. R T 0 AP0=0.21.6( ( 1+ 1.6-1 2 0. 2 2 ) -( 1.6+1 )/2( 1.6-1 ))m. R T 0 AP0=0.246

  k=1.6Ma=0.4m. R T 0 AP0=0.41.6( ( 1+ 1.6-1 2 0. 4 2 ) -( 1.6+1 )/2( 1.6-1 ))m. R T 0 AP0=0.457

  k=1.6Ma=0.6m. R T 0 AP0=0.61.6( ( 1+ 1.6-1 2 0. 6 2 ) -( 1.6+1 )/2( 1.6-1 ))m. R T 0 AP0=0.607

  k=1.6Ma=0.8m. R T 0 AP0=0.81.6( ( 1+ 1.6-1 2 0. 8 2 ) -( 1.6+1 )/2( 1.6-1 ))m. R T 0 AP0=0.69

  k=1.6Ma=1.0m. R T 0 AP0=1.01.6( ( 1+ 1.6-1 2 1. 0 2 ) -( 1.6+1 )/2( 1.6-1 ))m. R T 0 AP0=0.71

  k=1.2Ma=0.2m. R T 0 AP0=0.21.2( ( 1+ 1.2-1 2 0. 2 2 ) -( 1.2+1 )/2( 1.2-1 ))m. R T 0 AP0=0.214

  k=1.2Ma=0.4m. R T 0 AP0=0.41.2( ( 1+ 1.2-1 2 0. 4 2 ) -( 1.2+1 )/2( 1.2-1 ))m. R T 0 AP0=0.401

  k=1.2Ma=0.6m. R T 0 AP0=0.61.2( ( 1+ 1.2-1 2 0. 6 2 ) -( 1.2+1 )/2( 1.2-1 ))m. R T 0 AP0=0.541

  k=1.2Ma=0.8m. R T 0 AP0=0.81.2( ( 1+ 1.2-1 2 0. 8 2 ) -( 1.2+1 )/2( 1.2-1 ))m. R T 0 AP0=0.62

  k=1.2Ma=1.0m. R T 0 AP0=1.01.2( ( 1+ 1.2-1 2 1. 0 2 ) -( 1.2+1 )/2( 1.2-1 ))m. R T 0 AP0=0.64

The graph for the above values is given below:

Fluid Mechanics: Fundamentals and Applications, Chapter 12, Problem 148P , additional homework tip  2

Conclusion:

Therefore, for the values Ma and k are plotted but only one value of k and Ma.

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