Post on 07-Jul-2018
Scilab Code forIntroduction to Fluid Mechanics
by Fox and McDonald1
Created byEswar Prasad
4th Year StudentB.E. (Mech. Engg.)
National Institute of Technology, Trichy
College Teacher and ReviewerShivraj Deshmukh
Ph.D studentIIT Bombay
29 June 2010
1Funded by a grant from the National Mission on Education through ICT,http://spoken-tutorial.org/NMEICT-Intro. This Textbook companion and scilabcodes written in it can be downloaded from the website www.scilab.in
Book Details
Authors: Robert W. Fox and Alan T. McDonald
Title: Introduction to Fluid Mechanics
Publisher: John Wiley & Sons
Edition: 5th
Year: 2001
Place: New Delhi
ISBN: 9971-51-355-2
1
Contents
List of Scilab Code 4
1 Introduction 101.1 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.2 Scilab Code . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
2 Fundamental Concepts 122.1 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.2 Scilab Code . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
3 Fluid Statics 143.1 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.2 Scilab Code . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
4 Basic Equations in Integral form for a Control Volume 234.1 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234.2 Scilab Code . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
5 Introducton to Differential Analysis of Fluid Motion 385.1 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385.2 Scilab Code . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
6 Incompressible Inviscid Flow 426.1 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 426.2 Scilab Code . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
7 Dimensional Analysis and Simlitude 497.1 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 497.2 Scilab Code . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
2
8 Internal Incompressible Viscous Flow 558.1 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 558.2 Scilab Code . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
9 External Incompressible Viscous Flow 669.1 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 669.2 Scilab Code . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
10 Fluid Machinery 7510.1 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7510.2 Scilab Code . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
11 Introduction to Compressible Flow 9511.1 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9511.2 Scilab Code . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
12 Steady One-Dimensional Compressible Flow 10012.1 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10012.2 Scilab Code . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
3
List of Scilab Code
1.01 1.01.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.01d 1.01-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 101.02 1.02.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.02d 1.02-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 112.02 2.02.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.02d 2.02-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 133.01 3.01.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.01d 3.01-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 153.03 3.03.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153.03d 3.03-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 173.04 3.04.sci . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173.04d 3.04-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 183.05 3.05.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193.05d 3.05-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 203.06 3.06.sci . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203.06d 3.06-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 203.07 3.07.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.07d 3.07-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 214.01 4.01.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234.01d 4.01-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 244.02 4.02.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244.02d 4.02-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 254.03 4.03.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254.03d 4.03-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 264.04 4.04.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264.04d 4.04-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 274.05 4.05.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274.05d 4.05-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 27
4
4.06 4.06.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284.06d 4.06-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 284.07 4.07.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294.07d 4.07-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 294.08 4.08.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304.08d 4.08-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 304.09 4.09.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304.09d 4.09-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 304.10 4.10.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314.10d 4.10-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 314.11 4.11.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324.11d 4.11-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 324.12 4.12.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344.12d 4.12-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 344.14 This is some example . . . . . . . . . . . . . . . . . . . . . 344.14d 4.14-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 354.16 4.16.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354.16d 4.16-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 364.17 4.17.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 364.17d 4.17-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 365.02 5.02.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385.02d 5.02-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 385.07 5.07.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 395.07d 5.07d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 395.08 5.08.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 395.08d 5.08-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 405.09 5.09.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415.09d 5.09-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 416.01 6.01.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 426.01d 6.01-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 426.02 6.02.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 436.02d 6.02-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 436.03 6.03.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 436.03d 6.03-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 446.04 6.04.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 446.04d 6.04-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 446.05 6.05.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 456.05d 6.05-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 45
5
6.06 6.06.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 456.06d 6.06-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 466.08 6.08.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 466.08d 6.08-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 476.09 6.09.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 476.09d 6.09-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 487.04 7.04.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 497.04d 7.04-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 507.05 7.05.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 507.05d 7.05-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 517.06 7.06.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 527.06d 7.06-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 538.01 8.01.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 558.01d 8.01-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 568.02 8.02.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 568.02d 8.02-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 578.04 8.04.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 578.04d 8.04-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 578.05 8.05.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 588.05d 8.05-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 588.06 8.06.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 598.06d 8.06-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 598.07 8.07.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 608.07d 8.07-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 608.08 8.08.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 608.08d 8.08-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 618.09 8.09.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 628.09d 8.09-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 628.10 8.10.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 638.10d 8.10-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 638.11 8.11.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 648.11d 8.11-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 649.01 9.01.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 669.01d 9.01-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 669.04 9.04.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 679.04d 9.04-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 689.05 9.05.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 689.05d 9.05-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 69
6
9.06 9.06.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 699.06d 9.06-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 709.07 9.07.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 709.07d 9.07-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 709.08 9.08.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 719.08d 9.08-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 729.09 9.09.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 729.09d 9.09-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . . 7410.01 10.01.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7510.01d 10.01-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . 7610.02 10.02.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7610.02d 10.02-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . 7710.03 10.03.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7710.03d 10.03-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . 7810.06 10.06.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7910.06d 10.06-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . 8010.07 10.07.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8110.07d 10.07-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . 8110.08 10.08.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8210.08d 10.08-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . 8510.11 10.11.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8610.11d 10.11-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . 8810.12 10.12.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9110.12d 10.12-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . 9210.14 10.14.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9210.14d 10.14-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . 9310.16 10.16.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9310.16d 10.16-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . 9411.01 11.01.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9511.01d 11.01-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . 9611.03 11.03.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9611.03d 11.03-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . 9711.04 11.04.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9711.04d 11.04-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . 9912.01 12.01.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10012.01d 12.01-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . 10112.02 12.02.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10212.02d 12.02-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . 103
7
12.03 12.03.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10312.03d 12.03-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . 10412.04 12.04.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10412.04d 12.04-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . 10512.05 12.05.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10612.05d 12.05-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . 10612.06 12.06.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10712.06d 12.06-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . 10812.07 12.07.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10812.07d 12.07-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . 10912.08 12.08.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11012.08d 12.08-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . 11112.09 12.09.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11112.09d 12.09-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . 11312.10 12.10.sce . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11312.10d 12.10-data.sci . . . . . . . . . . . . . . . . . . . . . . . . . 115
8
List of Figures
3.1 Output graph of S 3.01 . . . . . . . . . . . . . . . . . . . . . 16
4.1 Output graph of S 4.11 . . . . . . . . . . . . . . . . . . . . . 33
7.1 Output graph of S 7.05 . . . . . . . . . . . . . . . . . . . . . 52
9.1 Output graph of S 9.08 . . . . . . . . . . . . . . . . . . . . . 73
10.1 Output graph of S 10.03 . . . . . . . . . . . . . . . . . . . . 7910.2 Output graph of S 10.07 . . . . . . . . . . . . . . . . . . . . 8310.3 Output graph-1 of S 10.08 . . . . . . . . . . . . . . . . . . . 8610.4 Output graph-2 of S 10.08 . . . . . . . . . . . . . . . . . . . 8710.5 Output graph-1 of S 10.11 . . . . . . . . . . . . . . . . . . . 8910.6 Output graph-2 of S 10.11 . . . . . . . . . . . . . . . . . . . 8910.7 Output graph-3 of S 10.11 . . . . . . . . . . . . . . . . . . . 90
11.1 Output graph of S 11.03 . . . . . . . . . . . . . . . . . . . . 98
9
Chapter 1
Introduction
1.1 Discussion
When executing the code from the editor, use the ’Execute File into Scilab’taband not the ’Load in Scilab’tab
The .sci files of the respective problems contain the input parameters ofthe question
1.2 Scilab Code
Example 1.01 1.01.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 1 . 0 1 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 1.01−data . s c i ’3 exec ( f i l ename )4 / / H e a t a d d e d d u r i n g t h e p r o c e s s ( i n k J ) :
5 Q12=m∗cp ∗(T2−T1)6 printf ( ”\n\nRESULTS\n\n” )7 printf ( ”\n\nHeat added during the proce s s : %f kJ\n\n” ,
Q12/1000)
Example 1.01d 1.01-data.sci
1 / / M a s s o f o x y g e n p r e s e n t ( i n k g ) :
2 m=0.95;
10
3 / / I n i t i a l t e m p e r a t u r ( i n K ) :
4 T1=300;5 / / F i n a l t e m p e r a t u r e o f o x y g e n ( i n K ) :
6 T2=900;7 / / P r e s s u r e o f o x y g e n ( i n k P a ) :
8 p=150;9 / / S p e c i f i c h e a t a t c o n s t a n t p r e s s u r e ( i n J / k g −K ) :
10 cp =909.4;
Example 1.02 1.02.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 1 . 0 2 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 1.02−data . s c i ’3 exec ( f i l ename )4 / / S p e e d a t w h i c h t h e b a l l h i t s t h e g r o u n d ( i n m/ s e c ) :
5 V=sqrt (m∗g/k∗(1−%eˆ(2∗k/m∗(−y0 ) ) ) )6 / / T e r m i n a l s p e e d ( i n m/ s e c ) :
7 Vt=sqrt (m∗g/k )8 / / R a t i o o f a c t u a l s p e e d t o t h e t e r m i n a l s p e e d :
9 r=V/Vt ;10 printf ( ”\n\nRESULTS\n\n” )11 printf ( ”\n\nSpeed at which the b a l l h i t s he ground : %f
m/ sec \n\n” ,V)12 printf ( ”\n\nRatio o f a c tua l speed to the te rmina l speed
: %f\n\n” , r )
Example 1.02d 1.02-data.sci
1 / / M a s s o f b a l l ( i n k g ) :
2 m=0.2;3 / / H e i g h t f o m w h i c h b a l l i s d r o p p e d ( i n m ) :
4 y0=500;5 / / V a l u e o f k :
6 k=2∗10ˆ−4;7 / / A c c l e r a t i o n d u e t o g r a v i t y ( i n m/ s e c ˆ 2 ) :
8 g =9.81;
11
Chapter 2
Fundamental Concepts
2.1 Discussion
When executing the code from the editor, use the ’Execute File into Scilab’taband not the ’Load in Scilab’tab
The .sci files of the respective problems contain the input parameters ofthe question
2.2 Scilab Code
Example 2.02 2.02.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 2 . 0 2 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 2.02−data . s c i ’3 exec ( f i l ename )4
5 / / V i s c o s i t y i n u n i t s o f l b f − s / f t ˆ 2 :
6 u1=u /100/454/32 .2∗30 .57 / / K i n e m a t i c v i s c o s i t y ( i n m/ s e c ˆ 2 ) :
8 v=u1/SG/d ∗ ( 0 . 305 ) ˆ29 / / S h e a r s t r e s s o n t h e u p p e r p l a t e ( l b f / f t ˆ 2 ) :
10 tu=u1∗U/D∗100011 / / S h e a r s t r e s s o n t h e l o w e r p l a t e ( i n Pa )
12 t l=tu ∗4 .45/0 .305ˆ213 printf ( ”\n\nRESULTS\n\n” )
12
14 printf ( ”\n\ nVi s co s i ty in un i t s o f lb f−s / f t ˆ2 : %1. 8 f$ lb f−s / f t ˆ2\n\n” , u1 )
15 printf ( ”\n\nKinematic v i s c o s i t y : %1. 8 f m/ sec ˆ2\n\n” , v )16 printf ( ”\n\nShear s t r e s on the upeer p l a t e : %f l b f / f t
ˆ2\n\n” , tu )17 printf ( ”\n\nSear s t r e s s on the lower p l a t e : %f Pa\n\n” ,
t l )
Example 2.02d 2.02-data.sci
1 / / M a s s o f o x y g e n p r e s e n t ( i n k g ) :
2 m=0.95;3 / / I n i t i a l t e m p e r a t u r ( i n K ) :
4 T1=300;5 / / F i n a l t e m p e r a t u r e o f o x y g e n ( i n K ) :
6 T2=900;7 / / P r e s s u r e o f o x y g e n ( i n k P a ) :
8 p=150;9 / / S p e c i f i c h e a t a t c o n s t a n t p r e s s u r e ( i n J / k g −K ) :
10 cp =909.4;
13
Chapter 3
Fluid Statics
3.1 Discussion
When executing the code from the editor, use the ’Execute File into Scilab’taband not the ’Load in Scilab’tab
The .sci files of the respective problems contain the input parameters ofthe question
When we execute S 3.01, we get Fig. 3.1.
3.2 Scilab Code
Example 3.01 3.01.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 3 . 0 1 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 3.01−data . s c i ’3 exec ( f i l ename )4 / / T u b e d i a m e t e r ( i n mm ) :
5 D=1:25;6 D1=D/10007 [m n]= s ize (D1)8 for i =1:n9 / / C h a n g e i n l i q u i d l e v e l f o r w a t e r ( i n mm ) :
10 dhw( i )=4∗STw∗ cosd ( thetaw ) /dw/g/D1( i ) ;11 / / C h a n g e i n l i q u i d l e v e l f o r m e r c u r y ( i n mm ) :
12 dhm( i )=4∗STm∗ cosd ( thetam ) /dm/g/D1( i ) ;
14
13 end ;14
15 / / P l o t t i n g t u b e d a i m e t e r a n d w a t e r l e v e l :
16 plot (D1∗1000 ,dhw , ’−o ’ )17 / / P l o t t i n g t u b e d a i m e t e r a n d m e r c u r y l e v e l :
18 plot (D1∗1000 ,dhm, ’−∗ ’ )19 l egend ( [ ’ Water ’ ; ’ Mercury ’ ] ) ;20 xt i t le ( ’ Liquid l e v e l vs Tube diameter ’ , ’ L iquid l e v e l ( in
mm) ’ , ’Tube diameter ( in mm) ’ )
Example 3.01d 3.01-data.sci
1 / / S u r f a c e t e n s i o n o f w a t e r ( i n mN /m ) :
2 STw=72.8∗10ˆ−3;3 / / S u r f a c e T e n s i o n o f m e r c u r y ( i n mN /m ) :
4 STm=375∗10ˆ−3;5 / / C o n t a c t a n g l e f o r w a t e r :
6 thetaw =0;7 / / C O n t a c t a n g l e f o r m e r c u r y :
8 thetam =140;9 / / D e n s i t y o f w a t e r ( i n k g /m ˆ 3 ) :
10 dw=1;11 / / D e n s i t y o f m e r c u r y ( i n k g /m ˆ 3 ) :
12 dm=13.6;13 / / A c c e l e r a t i o n d e t o g r a v i t y ( i n m/ s e c ) :
14 g =9.81;
Example 3.03 3.03.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 3 . 0 3 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 3.03−data . s c i ’3 exec ( f i l ename )4 / / P r e s s u r e d i f f e r e n c e ( i n l b f / i n ˆ 2 ) :
5 dp=g∗d∗(−d1+SGm∗d2−SGo∗d3+SGm∗d4+d5 ) /12/1446 printf ( ”\n\nRESULTS\n\n” )7 printf ( ”\n\nPressure d i f f e r e n c e between A and B: %f l b f
/ in ˆ2\n\n” ,dp)
15
Figure 3.1: Output graph of S 3.01
16
Example 3.03d 3.03-data.sci
1 / / A c c e l e r a t i o n d u e t o g r a v i t y ( i n f t / s e c ˆ 2 ) :
2 g =32.2;3 / / S p e c i f i c g r a v i t y o f m e r c u r y :
4 SGm=13.6;5 / / S p e c i f i c g r a v i t y o f o i l :
6 SGo=0.88;7 / / S p e c i f i c g r a v i t y o f w a t e r :
8 SGw=1;9 / / D e n s i t y o f w a t e r ( i n s l u g / f t ˆ 3 ) :
10 d=1.94;11 / / H e i g h t s o f l i q u i d i n v a r i o u s t u b e s ( i n i n c h e s ) :
12 d1=10;13 d2=3;14 d3=4;15 d4=5;16 d5=8;
Example 3.04 3.04.sci
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 3 . 0 4 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 3.04−data . s c i ’3 exec ( f i l ename )4 / / A s s u m i n g t e m p e r a t u r e v a r i e s l i n e a r l y w i t h a l t i t u d e :
5 / / T e m p e r a t u r e g r a d i e n t ( i n F / f t ) :
6 m=(T1−T2) /( z2−z1 )7 / / V a l u e o f g / ( m ∗R ) :
8 v=g/m/R/32 .29 / / P r e s s u r e a t V a i l P a s s ( i n i n c h e s o f Hg ) :
10 p12=p1 ∗ ( (T2+460) /(T1+460) ) ˆv11 / / P e r c e n t a g e c h a n g e i n d e n s i t y :
12 pc1=(p12/p1∗(T1+460) /(T2+460)−1)∗10013 / / A s s u m i n g d e n s i t y i s c o n s t a n t :
14 / / P r e s s u r e a t V a i l P a s s ( i n i n c h e s o f Hg ) :
15 p22=p1∗(1−(g∗( z2−z1 ) /(R∗32 .2 ) /(T1+460) ) )
17
16 / / P e r c e n t a g e c h a n g e i n d e n s i t y :
17 pc2=0;18 / / A s s u m i n g t e m p e r a t u r e i s c o n s t a n t :
19 / / P r e s s u r e a t V a i l P a s s ( i n i n c h e s o f Hg ) :
20 p32=p1∗%eˆ(−g∗( z2−z1 ) /(R∗32 .2 ) /(T2+460) )21 / / P e r c e n t a g e c h a n g e i n d e n s i t y :
22 pc3=(p32/p1∗(T1+460) /(T1+460)−1)∗10023 / / F o r a n a d i a b a t i c a t m o s p h e r e :
24 p42=p1∗((62+460) /(80+460) ) ˆ( k /(k−1) )25 / / P e r c e n t a g e c h a n g e i n d e n s i t y :
26 pc4=(p42/p1∗(T1+460) /(T2+460)−1)∗10027 printf ( ”\n\nRESULTS\n\n” )28 printf ( ”\n\n1 ) I f temperature v a r i e s l i n e a r l y with
a l t i t u d e \n\n” )29 printf ( ”\n\n\ tAtmospheric p r e s su r e at Vai l Pass : %f
inche s o f Hg\n\n” , p12 )30 printf ( ”\n\n\ tPercentage change in dens i ty wrt Denver :
%f percent \n\n” , pc1 )31 printf ( ”\n\n2 ) I f dens i ty i s constant \n\n” )32 printf ( ”\n\n\ tAtmospheric p r e s su r e at Vai l Pass : %f
inche s o f Hg\n\n” , p22 )33 printf ( ”\n\n\ tPercentage change in dens i ty wrt Denver :
%f percent \n\n” , pc2 )34 printf ( ”\n\n3 ) I f temperature i s constant \n\n” )35 printf ( ”\n\n\ tAtmospheric p r e s su r e at Vai l Pass : %f
inche s o f Hg\n\n” , p32 )36 printf ( ”\n\n\ tPercentage change in dens i ty wrt Denver :
%f percent \n\n” , pc3 )37 printf ( ”\n\n4 ) For an a d i a b a t i c atmosphere\n\n” )38 printf ( ”\n\n\ tAtmospheric p r e s su r e at Vai l Pass : %f
inche s o f Hg\n\n” , p42 )39 printf ( ”\n\n\ tPercentage change in dens i ty wrt Denver :
%f percent \n\n” , pc4 )
Example 3.04d 3.04-data.sci
1 / / E l e v a t i o n o f D e n v e r ( i n f t ) :
2 z1 =5280;
18
3 / / P r e s s u r e a t D e n v e r ( i n mm o f Hg ) :
4 p1 =24.8 ;5 / / T e m p e r a t u r e a t D e n v e r ( i n F ) :
6 T1=80;7 / / E l e v a t i o n a t V a i l P a s s ( i n f t ) :
8 z2 =10600;9 / / T e m p e r a t u r e a t V s i l P a s s ( i n F ) :
10 T2=62;11 / / V a l u e o f R i n f t − l b f / l bm −R ) :
12 R=53.3;13 / / A c c e l e r a t i o n d u e t o g r a v i t y ( i n f t / s e c ˆ 2 ) :
14 g =32.2;15 / / V a l u e o f a d i a b a t i c c o n s t a n t :
16 k =1.4 ;
Example 3.05 3.05.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 3 . 0 5 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 3.05−data . s c i ’3 exec ( f i l ename )4 / / N e t f o r c e o n t h e g a t e ( i n kN ) :
5 Fr=d∗g∗w∗(D∗L+Lˆ2/2∗ s ind ( theta ) )6 / / C e n t r e o f p r e s s u r e :
7 / / C a l c u l a t i o n f o r y c o o r d i n a t e :
8 yc=D/ s ind ( theta )+L/29 / / A r e a ( i n m ˆ 2 ) :
10 A=L∗w11 / / M o m e n t o f i n e r t i a o f r e c t a n g u l a r g a t e ( i n m ˆ 4 ) :
12 Ixx=w∗Lˆ3/1213 / / y c o o r d i n a t e ( i n m ) :
14 y=yc+Ixx /A/yc15 / / C a l c u l a t i o n f o r x c o o r d i n a t e :
16 Ixy=017 xc=w/218 / / x c o o r d i n a t e ( i n m ) :
19 x=xc+Ixy /A/xc20 printf ( ”\n\nRESULTS\n\n” )21 printf ( ”\n\nNet f o r c e on the gate : %f kN\n\n” , Fr /1000)
19
22 printf ( ”\n\nCoordinate o f c en t r e o f p r e s su r e : (%0. 1 f ,%0. 1 f )\n\n” ,x , y )
Example 3.05d 3.05-data.sci
1 / / L e n g t h o f g a t e ( i n m ) :
2 L=4;3 / / W i d t h o f g a t e ( i n m ) :
4 w=5;5 / / D e p t h o f g a t e u n d e r w a t e r ( i n m ) :
6 D=2;7 / / D e n s i t y o f w a t e r ( i n k g /m ˆ 3 :
8 d=999;9 / / A c c e l e r a t i o n d u e t o g r a v i t y ( i n m/ s e c ˆ 2 ) :
10 g =9.81;11 / / A n g l e o f g a t e w i t h h o r i z o n t a l :
12 theta =30;
Example 3.06 3.06.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 3 . 0 6 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 3.06−data . s c i ’3 exec ( f i l ename )4 / / F o r c e r e q u i r e d t o k e e p t h e d o o r s h u t ( i n l b f ) :
5 function y=f ( z ) , y=b/L∗p0∗z+d∗b/L∗(L∗z−z ˆ2) , endfunction6 Ft=intg (0 ,L , f )7 printf ( ”\n\nRESULTS\n\n” )8 printf ( ”\n\nForce r equ i r ed to kep the door shut : %. 1 f
l b f \n\n” , Ft )
Example 3.06d 3.06-data.sci
1 / / P r e s s u r e a p l l i e d o n t h e d o o r ( i n p s f g ) :
2 p0=100;3 / / L e n g t h o f d o o r ( i n f e e t ) :
4 L=3;5 / / B r e a d t h o f t h e d o o r ( i n f e e t ) :
20
6 b=2;7 / / D e n s i t y o f l i q i u i d ( i n l b f / f t ˆ 3 ) :
8 d=100;
Example 3.07 3.07.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 3 . 0 7 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 3.07−data . s c i ’3 exec ( f i l ename )4 / / H o r i z o n t a l c o m p o n e n t o f r e s u l t a n t f o r c e ( i n kN ) :
5 Frh=0.5∗d∗g∗w∗Dˆ26 / / L i n e o f a c t i o n o f F r h ( i n m ) :
7 y1=0.5∗D+w∗Dˆ3/12/(0 .5∗D) /(w∗D)8 / / V e r t i c a l c o m p o n e n t o f r e s u l t a n t f o r c e ( i n kN ) :
9 function y=q ( x ) , y=d∗g∗w∗(D−sqrt ( a∗x ) ) , endfunction10 Frv=intg (0 ,Dˆ2/a , q )11 / / L i n e o f a c i o n o f F r v ( i n m ) :
12 function k=f ( x ) , k=d∗g∗w/Frv∗x∗(D−sqrt ( a∗x ) ) ,endfunction
13 xa=intg (0 ,Dˆ2/a , f )14 / / F o r c e r e q u i r e d t o k e e p t h e g a t e i n e q u i l i b r i u m ( i n kN )
:
15 Fa=1/ l ∗( xa∗Frv+(D−y1 )∗Frh )16 printf ( ”\n\nRESULTS\n\n” )17 printf ( ”\n\nForce r equ i r ed to keep the gate at
equ i l i b r i um : %f kN\n\n” ,Fa/1000)
Example 3.07d 3.07-data.sci
1 / / W i d t h o f g a t e ( i n m ) :
2 w=5;3 / / D e p t h o f w a t e r ( i n m ) :
4 D=4;5 / / D e n s i t y o f w a t e r ( i n k g /m ˆ 3 ) ;
6 d=999;7 / / A c c e l r a t i o n d e t o g r a v i t y ( i n m/ s e c ˆ 2 ) :
8 g =9.81;9 / / V a l u e o f a ( i n m ) :
21
10 a=4;11 / / P o i n t w h e r e f o r c e a c t s ( i n m ) :
12 l =5;
22
Chapter 4
Basic Equations in Integralform for a Control Volume
4.1 Discussion
When executing the code from the editor, use the ’Execute File into Scilab’taband not the ’Load in Scilab’tab
The .sci files of the respective problems contain the input parameters ofthe question
When we execute S 4.11, we get Fig. 4.1.
4.2 Scilab Code
Example 4.01 4.01.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 4 . 0 1 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 4.01−data . s c i ’3 exec ( f i l ename )4 / / I f I = i n t e g r a l o f ( pV . dA ) :
5 / / F o r s y s t e m : I c s = I A 1 + I A 2 + I A 3 + I A 4 .
6 / / F o r a r e a 1
7 IA1=−d∗V1∗A18 / / F o r a r e a 3 : I A 2 =d ∗ V3 ∗ A3=m3
9 IA3=m310 / / F o r a r e a 4 : I A 4 =−d ∗ V4 ∗ A4=−d ∗ Q4
23
11 IA4=−d∗Q412 / / F o r a r e a 2 :
13 IA2=−IA1−IA3−IA414 / / V e l o c i t y a t s e c t i o n 2 ( i n f t / s e c ) :
15 V2=IA2/d/A216 / / V2 i s i n t h e n e g a t i v e y d i r e c t i o n
17 printf ( ”\n\nRESULTS\n\n” )18 printf ( ”\n\ nVeloc i ty at s e c t i o n 2 : −%. 0 f j f t / s ec \n\n” ,
V2)
Example 4.01d 4.01-data.sci
1 / / A r e a o f 1 ( i n f t ˆ 2 ) :
2 A1=0.2;3 / / A r e a o f 2 ( i n f t ˆ 2 ) :
4 A2=0.5;5 / / A r e a o f 3 ( i n f t ˆ 2 ) :
6 A3=0.4;7 / / A r e a o f 4 ( i n f t ˆ 2 ) :
8 A4=0.4;9 / / D e n s i t y o f w a t e r ( i n s l u g / f t ˆ 3 ) :
10 d=1.94;11 / / M a s s f l o w r a t e o u t o f s e c t i o n 3 ( i n s l u g / s e c ) :
12 m3=3.88;13 / / V o l m e f l o w r a t e i n s e c t i o n 4 ( i n f t ˆ 3 / s e c ) :
14 Q4=1;15 / / V e l o c i t y a t 1 ( i n f t / s e c ) :
16 V1=10;
Example 4.02 4.02.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 4 . 0 2 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 4.02−data . s c i ’3 exec ( f i l ename )4 / / I f I = i n t e g r a l o f ( pV . dA ) :
5 / / F o r s y s t e m : I C S = I a b + I b c + I c d + I d a
6 / / B u t I C S =0
7
24
8 / / F o r A a b :
9 function p=f ( y ) ,p=−d∗U∗w∗y ˆ0 , endfunction10 IAab=intg (0 , t , f )11
12 / / F o r A c d :
13 function q=g ( y ) , q=d∗U∗w∗(2∗y/t−(y/ t ) ˆ2) , endfunction14 IAcd=intg (0 , t , g )15
16 / / M a s s f l o w r a t e a c r o s s s u r f a c e b c ( i n k g / s e c ) :
17 mbc=(−IAab−IAcd ) /100018 printf ( ”\n\nRESULTS\n\n” )19 printf ( ”\n\nMass f low ra t e a c r o s s s u r f a c e bc : %. 4 f kg/
sec \n\n” ,mbc)
Example 4.02d 4.02-data.sci
1 / / F l o w v e l o c i t y a h e a d o f t h e p l a t e ( i n m/ s e c ) :
2 U=30;3 / / B o u n d a r y l a y e r t c k n e s s a t l o c a t i o n d ( i n mm ) :
4 t =5;5 / / D e n s i t y o f f l u i d a i r ( i n k /m ˆ 3 ) :
6 d=1.24;7 / / P l a t e w d t h p e r p e n d i c u l a r t o t h e p l a t e ( i n m ) :
8 w=0.6;
Example 4.03 4.03.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 4 . 0 3 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 4.03−data . s c i ’3 exec ( f i l ename )4 / / R a t e o f c h a n g e o f a i r d e n s i t y i n t a n k ( i n ( k g /m ˆ 3 ) / s ) :
5 r=−d∗v∗A/V/10ˆ66 printf ( ”\n\nRESULTS\n\n” )7 printf ( ”\n\nRate o f change o f a i r dens i ty in tank : %. 3 f
kg/mˆ3\n\n” , r )8 printf ( ”\n\nThe dens i ty de c r ea s e s as i s i n d i c a t e d by
the negat ive s i gn \n\n” )
25
Example 4.03d 4.03-data.sci
1 / / V o l u m e o f t a n k ( i n m ˆ 3 ) :
2 V=0.05;3 / / P r e s s u r e o f a i r ( I n k P a ) :
4 p=800;5 / / T e m p e r a t u r e o f t a n k ( i n C ) :
6 T=15;7 / / V e l o c i t y o f l e a v i g a i r ( i n m/ s e c ) :
8 v=311;9 / / D e n s i t y o f a i r ( i n k g /m ˆ 3 ) :
10 d=6.13;11 / / A r e a o f v a l v e e x i t ( i n mm ˆ 2 ) :
12 A=65;
Example 4.04 4.04.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 4 . 0 4 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 4.04−data . s c i ’3 exec ( f i l ename )4 / / 1 ) C o n t r o l V o l u m e s e l e c t e d s o t h a t a r e a o f l e f t
s u r f a c e i s e q u a l t o t h e a r e a o f t h e r i g h t s u r f a c e
5 u1=15;6 / / F o r c e o f s u p p o r t o n c o n t r o l v o l u m ( i n kN ) :
7 function y=f (A) , y=−u1∗d∗V, endfunction8 Rx1=intg ( 0 , 0 . 0 1 , f )9 / / H o r i z o n t a l f o r c e o n s u p p o r t ( i n kN ) :
10 Kx=−Rx111 / / 2 ) C o n t r o l v o l u m e s a r e s e l e c t e d d o t h a t t h e a r e a o f
t h e l e f t a n d r i g h t s u r f a c e s a r e e q u i a l t o t h e a r e a
o f t h e p l a t e
12
13 function z=g (A) , z=−u1∗d∗V, endfunction14 Fsx=intg ( 0 , 0 . 0 1 , g )15 / / N e t f o r c e o n p l a t e : F x =0=−Bx − p a ∗ Ap+Rx
16 / / Rx= p a ∗ Ap+ Bx
17 / / F r o m t h e a b o v e , i t i s o b t a i n e d t h a t :
18 Rx2=−2.25
26
19 / / H o r i z o n t a l f o r c e o n s u p p o r t ( i n kN ) :
20 Kx2=−Rx221 printf ( ”\n\nRESULTS\n\n” )22 printf ( ”\n\nHor izonta l f o r c e on support : %. 3 f kN\n\n” ,
Kx/1000)
Example 4.04d 4.04-data.sci
1 / / V e l o c i t y o f w a t e r l e a v i n g t h e n o z l e ( i n m/ s e c ) :
2 V=15;3 / / A r e a o f n o z z l e ( i n m ˆ 2 ) :
4 A=0.01;5 / / D e n s i t y o f w a t e r ( i n k g /m ˆ 3 ) :
6 d=999;
Example 4.05 4.05.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 4 . 0 5 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 4.05−data . s c i ’3 exec ( f i l ename )4 / / W e i g h t o f w a t e r i n t h e t a n k ( i n l b f ) :
5 d1 =62.4 ;6 WH2O=d1∗A∗h7 v=−5;8 / / T o t a l b o d y f o r c e i n n e g a t i v e y d i r e c t i o n ( l b f ) :
9 function y=f (A) , y=−v∗d2∗V1 , endfunction10 F=intg (0 ,A1 , f )11 / / F o r c e o f s c a l e o n c o n t r o l v o l u m e ( i n kN ) :
12 Ry=W+WH2O−F13 printf ( ”\n\nRESULTS\n\n” )14 printf ( ”\n\nSca le Reading : %. 3 f l b f \n\n” ,Ry)
Example 4.05d 4.05-data.sci
1 / / H e i g h t o f t h e c o n t a i n e r ( i n f t ) :
2 l =2;3 / / A r e a o f c r o s s s e c t i o n ( i n f t ˆ 2 ) :
27
4 A=1;5 / / W e i g h t o f c o n t a i n e r ( i n l b f ) :
6 W=5;7 / / W a t e r d e p t h ( i n f t ) :
8 h=1.9;9 / / A r e a o f o p e n i n g 1 ( i n f t ˆ 2 ) :
10 A1=0.1;11 / / V e l o c i t y a t o p e n i n g 1 ( i n f t / s e c ) :
12 V1=−5;13 / / A r e a o f o p e n i n g 2 ( i n f t ˆ 2 ) :
14 A2=0.1;15 / / A r e a o f o p e n i n g 1 ( i n f t ˆ 2 ) :
16 A3=0.1;17 / / D e n s i t y o f w a t e r ( i n s l u g / f ˆ 3 ) :
18 d2 =1.94;
Example 4.06 4.06.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 4 . 0 6 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 4.06−data . s c i ’3 exec ( f i l ename )4 / / X− c o m p o n e n t o f r e a c t i o n f o r c e p e r u n i t w i d t h o f t h e
g a t e ( i n N /m ) :
5 Rxw=(d∗(V2ˆ2∗D2−V1ˆ2∗D1) )−(d∗g /2∗(D1ˆ2−D2ˆ2) )6 / / H o r i z o n t a l f o r c e e x e r t e d p e r u n t w i d t h o n t h e g a t e ( i n
N /m ) :
7 Kxw=−Rxw8 printf ( ”\n\nRESULTS\n\n” )9 printf ( ”\n\nHor izonta l f o r c e exer ted per unt width on
the gate : %. 3 f kN/m\n\n” ,Kxw/1000)
Example 4.06d 4.06-data.sci
1 / / D i a m e t e r o f c h a n n e l ( i n m ) :
2 D1=1.5;3 / / V e l c i t y o f f l o w i n c h a n n e l ( i n m/ s e c ) :
4 V1=0.2;5 / / D i a m e t e r a t s e c t i o n 2 ( i n m ) :
28
6 D2=0.0563;7 / / V e l o c i t y a s e c t i o n 2 ( i n m/ s e c ) :
8 V2=5.33;9 / / D e n s i t y o f w a t e r ( i n k g /m ˆ 3 ) :
10 d=999;11 / / A c c e l e r a t i o n d u e t o g r a v i t y ( i n m/ s e c 2 ) :
12 g =9.81;
Example 4.07 4.07.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 4 . 0 7 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 4.07−data . s c i ’3 exec ( f i l ename )4 / / V e l o c i t y a t s e c t i o n 1 ( i n m/ s e c ) :
5 V1=V2∗A2/A16 / / G a u g e p r e s s u r e ( i n k P a ) :
7 p1g=p1−patm8 u1=V1 ; u2=−V2 ;9 / / R e a c t i o n f o r c e c o m p o n e n t i n t h e x d i r e c t i o n ( i n N ) :
10 Rx=−p1g∗A1−u1∗d∗V1∗A111 / / R e a c t i o n f o r c e c o m p o n e n t i n t h e y d i r e c t i o n ( i n N ) :
12 Ry=u2∗d∗V2∗A213 printf ( ”\n\nRESULTS\n\n” )14 printf ( ”\n\nForce to hold elbow act ing to the l e f t : %. 3
f kN\n\n” ,Rx/1000)15 printf ( ”\n\nForce to hold elbow act ing downwards : %. 3 f
N\n\n” ,Ry)
Example 4.07d 4.07-data.sci
1 / / P r e s s u r e a t i n l e t t o t h e e l b o w ( i n N /m ˆ 2 ) :
2 p1 =2.21∗10ˆ5;3 / / A r e a o f c r o s s s e c t i o n ( i n m ˆ 2 ) :
4 A1=0.01;5 / / V e l o c i t y a t s e c t o n 2 ( i n m/ s e c ) :
6 V2=16;7 / / A r e a o f c r o s s s e c t i o n o f s e c t i o n 2 ( i n m ˆ 2 ) :
8 A2=0.0025;
29
9 / / A t m o s p h e r i c p r e s s u r e ( i n k P a ) :
10 patm=1.012∗10ˆ5;
Example 4.08 4.08.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 4 . 0 8 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 4.08−data . s c i ’3 exec ( f i l ename )4 / / T e n s i o n r e q u i r e d t o p u l l t h e b e l t ( i n l b f ) :
5 T=Vbelt∗m/32.26 printf ( ”\n\nRESULTS\n\n” )7 printf ( ”\n\nTension r equ i r ed to p u l l the b e l t : %. 3 f l b f
\n\n” ,T)
Example 4.08d 4.08-data.sci
1 / / V e l o c i t y o f c o n v e y o r b e l t ( i n f t / s e c ) :
2 Vbelt =3;3 / / V e l o c i t y o f s a n d a l l i n g o n t o b e l t ( i n f t / s e c ) :
4 Vsand=5;5 / / F l o w r a t e ( i n l b m / s e c ) :
6 m=500;
Example 4.09 4.09.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 4 . 0 9 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 4.09−data . s c i ’3 exec ( f i l ename )4 / / M i n i m u m g a u g e p r e s s u r e r e q u i r e d ( i n l b f / i n ˆ 2 ) :
5 pg=8/%piˆ2∗d/D1ˆ4∗Qˆ2∗ ( (D1/D2) ˆ4−1)∗1446 printf ( ”\n\nRESULTS\n\n” )7 printf ( ”Minimum gauge p r e s su r e r equ i r ed : %. 3 f l b f / in ˆ2”
, pg )
Example 4.09d 4.09-data.sci
1 / / N o z z l e i n l e t d i a m e t e r ( i n i n c h e s s ) :
30
2 D1=3;3 / / N o z z l e e x i t d i a m e t e r ( i n i n c h e s ) :
4 D2=1;5 / / D e s i r e d v o l u m e f l o w r a t e ( i n f t ˆ 3 / s e c ) :
6 Q=0.7;7 / / D e n s i t y o f w a t e r ( i n s l u g / f t ˆ 3 ) :
8 d=1.94;
Example 4.10 4.10.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 4 . 1 0 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 4.10−data . s c i ’3 exec ( f i l ename )4 u1=V−U5 u2=(V−U)∗ cosd ( theta )6 v2=(V−U)∗ s ind ( theta )7 V1=V−U8 V2=V19 / / X c o m p o n e n t o f m o m e n t e q u a t i o n ( i n N ) :
10 function y=f (A) , y=u1∗−(d∗V1) , endfunction11 function z=g (A) , z=u2∗d∗V2 , endfunction12 Rx=intg (0 ,A, f )+intg (0 ,A, g )13
14 / / Y c o m p o n e n t o f m o m e n t e q u a t i o n ( i n N ) :
15 function a=h(A) , a=v2∗d∗V1 , endfunction16 Ry=intg (0 ,A, h) / / T h i s i s a f t e r n e g l e c t i n g w e i g h t o f
v a n e a n d t h e w a t e r .
17 printf ( ”\n\nRESULTS\n\n” )18 printf ( ”\n\nNet f o r c e on the vane : %. 3 f i+%. 2 f j kN\n\n
” ,Rx/1000 ,Ry/1000)
Example 4.10d 4.10-data.sci
1 / / V a n e t u r n i n g a n g l e :
2 theta =60;3 / / S p e e d o f v a n e ( i n m/ s e c ) :
4 U=10;5 / / A r e a o f n o z z l e ( i n m2 ) :
31
6 A=0.003;7 / / F l o w v e l o c i t y o f w a t e r ( i n m/ s e c ) :
8 V=30;9 / / D e n s i t y o f w a t e r ( i n k g /m ˆ 3 ) :
10 d=999;
Example 4.11 4.11.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 4 . 1 1 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 4.11−data . s c i ’3 exec ( f i l ename )4 / / E v a l u a t i n g t h e v a l u e o f Vb :
5 Vb=V∗(1− cosd ( theta ) )∗d∗A/M6 / / V a l u e o f U / V f o r v a r i o u s v a l u e s o f t
7 t =0:20;8 [m n]= s ize ( t )9 for i =1:n
10 U V( i )=Vb∗ t ( i ) /(1+Vb∗ t ( i ) ) ;11 end12
13 / / P l o t t i n g U / V v s t :
14 plot ( t , U V)15 xt i t le ( ’U/V vs t ’ , ’ t ( in sec ) ’ , ’U/V ’ )
Example 4.11d 4.11-data.sci
1 / / M a s s o f v a n e a n d c a r t ( i n k g ) :
2 M=75;3 / / T u r n i n g a n g l e o f v a n e :
4 theta =60;5 / / S p e e d o f w a t e r l e a v i n g n o z z l e h o r i z o n t a l l y ( i n m/ s e c ) :
6 V=35;7 / / E x i t a r e a o f n o z z l e ( i n m ˆ ) :
8 A=0.003;9 / / D e n s i t y o f w a t e r ( i n k g /m ˆ 3 ) :
10 d=999;
32
Figure 4.1: Output graph of S 4.11
33
Example 4.12 4.12.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 4 . 1 2 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 4.12−data . s c i ’3 exec ( f i l ename )4 / / A c c e l e r a t i o n o f r o c k e t a t t = 0 ( i n m/ s e c ˆ 2 ) :
5 Ve∗me/M0−g6 / / V e l o c i t y o f r o c k e t a t t = 1 0 ( i n m/ s e c ) :
7 function y=f ( t ) , y=Ve∗me/(M0−me∗ t )−g , endfunction8 Vcv=intg (0 , t , f )9 printf ( ”\n\nRESULTS\n\n” )
10 printf ( ”\n\ nVeloc i ty o f rocke t at t =10: %. 1 f m/ sec \n\n”,Vcv)
Example 4.12d 4.12-data.sci
1 / / I n i t i a l m a s s o f t h r o c k e t ( i n k g ) :
2 M0=400;3 / / R a t e o f f u e l c o n s u m p t i o n ( i n k g / s e c ) :
4 me=5;5 / / E x h a u s t v e l o c i t y ( i n m/ s e c ) :
6 Ve=3500;7 / / A c c e l e r a t i o n d u e t o g r a v i t y ( i n m/ s e c ˆ 2 ) :
8 g =9.81;9 / / T i m e a f t e r w h i c h v e l o c i t y i s t o b e c a l c u l a t e d ( i n s e c )
:
10 t =10;
Example 4.14 4.14.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 4 . 1 4 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 4.14−data . s c i ’3 exec ( f i l ename )4 / / A r e a o f j e t ( i n mm ˆ 2 ) :
5 Ajet=%pi/4∗Dˆ26 / / J e t s p e e d r e l a t i v e t o t h e n o z z l e ( i n m/ s e c ) :
7 Vrel=Q/2/ Ajet ∗10ˆ6/60/10008 / / V a l u e o f w ∗R i n m/ s e c :
34
9 wR=w∗R∗2∗%pi/60/100010 / / F r i c t i o n t o r q u e a t p i v o t ( i n N−m ) :
11 Tf=R∗( Vrel∗ cosd ( alpha )−wR)∗d∗Q/1000/60/100012 printf ( ”\n\nRESULTS\n\n” )13 printf ( ”\n\nJet speed r e l a t i v e to each nozz l e : %. 2 f m/
sec \n\n” , Vrel )14 printf ( ”\n\ nFr i c t i on torque at p ivot : %. 5 f N−m\n\n” , Tf )
Example 4.14d 4.14-data.sci
1 / / I n l e t g a u g e p r e s s u r e ( i n k P a ) :
2 p=20;3 / / V o l u m e f l o w r a t e o f w a t e r t h r o u g h t h e s p r i n k l e r ( i n l /
m i n ) :
4 Q=7.5;5 / / S p e e d o f r o t s t i o n o f s p r i n k l e r ( i n r p m ) :
6 w=30;7 / / D i a m e t e r o f j e t f s p r i n k l e ( i n mm ) :
8 D=4;9 / / R a d i u s o f s p r i n k l e r ( i n mm ) :
10 R=150;11 / / S u p p l y p r e s s u r e t o s p r i n k l e r ( i n k P a ) :
12 p=20;13 / / A n g l e a t w h i c h j e t i s s p r a y e d w r t h o r i z o n t a l :
14 alpha =30;15 / / D e n s i t y o f w a t e r ( i n k g /m ˆ ) :
16 d=999;
Example 4.16 4.16.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 4 . 1 6 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 4.16−data . s c i ’3 exec ( f i l ename )4 / / V e l o c i t y a t e x i t ( i n f t / s e c ) :
5 V2=m∗R∗(T2+460)/A2/p2/1446 / / A s p o w e r i n p u t i s t o CV , Ws = − 6 0 0
7 / / R a t e o f h e a t t r a n s f e r ( i n B t u / s e c ) :
8 Q=Ws∗550/778+m∗cp ∗(T2−T1)+m∗V2ˆ2/2/32.2/778
35
9 printf ( ”\n\nRESULTS\n\n” )10 printf ( ”\n\nRate o f heat t r a n s f e r : %. 3 f Btu/ sec \n\n” ,Q)
Example 4.16d 4.16-data.sci
1 / / P r e s s u r e a t e n t r y ( i n p s i a ) :
2 p1 =14.7 ;3 / / T e m p e r a t u r e a t e n t r y ( i n F ) :
4 T1=70;5 / / P r e s s u r e a t e x i t ( i n p s i a ) :
6 p2=50;7 / / T e m p r a t u r e a e x i t ( i n F ) :
8 T2=100;9 / / C r o s s s e c t i o n a l a r e a o f t h e p i p e a t e x i t ( i n f t ˆ 2 ) :
10 A2=1;11 / / M a s s f l o w r a t e ( i n l b f / s e c ) :
12 m=20;13 / / P o w e r i n p u t t o t h e c o m p r e s s o r ( i n h p ) :
14 Ws=−600;15 / / V a l u e o f c p ( i n B t u / l bm −R ) :
16 cp =0.24;17 / / V a l u e o f g a s c o n s t a n t ( i n f t − l b f / ( l bm −R ) )
18 R=53.3;
Example 4.17 4.17.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 4 . 1 7 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 4.17−data . s c i ’3 exec ( f i l ename )4 / / D e n s i t y o f t a n k ( i n k g /m ˆ 3 ) :
5 d=(p1+patm) /R/T6 / / M a s s f l o w r a t e o f a i r i n t o t h e t a n k ( i n k g / s e c ) :
7 m=d∗V∗cv∗ r /R/T∗10008 printf ( ”\n\nRESULTS\n\n” )9 printf ( ”\n\nMass f low ra t e o f a i r i n to the tank : %. 3 f g
/ sec \n\n” ,m)
Example 4.17d 4.17-data.sci
36
1 / / V o l u m e o f t a k ( i n m ˆ 3 ) :
2 V=0.1;3 / / T e m p e r a t u r e o f l i n e a n d t a n k ( i n K ) :
4 T=293;5 / / I n i t i a l t a n k g a u g e p r e s s u r e ( i n N /m ˆ 2 ) :
6 p1=1∗10ˆ5;7 / / A b s o l u t e l i n e p r e s s u r e ( i n N /m ˆ 2 ) :
8 p=2∗10ˆ6;9 / / R a t e o f r i s e o f t e m p e r a t u r e a f t e r o p e n i n g o f t h e
v a l v e ( i n C / s e c ) :
10 r =0.05;11 / / A t m o s p h e r i c p r e s s u r e ( i n N /m ˆ 2 ) :
12 patm=1.01∗10ˆ5;13 / / G a s C o n s t a n t ( i n N−m / ( k g −K ) ) :
14 R=287;15 / / V a l u e o f c v ( i n N−m/ kg −K ) :
16 cv =717;
37
Chapter 5
Introducton to DifferentialAnalysis of Fluid Motion
5.1 Discussion
When executing the code from the editor, use the ’Execute File into Scilab’taband not the ’Load in Scilab’tab
The .sci files of the respective problems contain the input parameters ofthe question
5.2 Scilab Code
Example 5.02 5.02.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 5 . 0 2 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 5.02−data . s c i ’3 exec ( f i l ename )4 / / R a t e o f c h a n g e o f d e n s i t y w i t h t i m e ( i n k g /mˆ3 − s ) :
5 r=−d∗V/L6 printf ( ”\n\nRESULTS\n\n” )7 printf ( ”\n\nRate o f change o f dens i ty with time : %. 1 f
kg/mˆ3−s\n\n” , r )
Example 5.02d 5.02-data.sci
38
1 / / D i s t a n c e f p i s t o n f r o m c l o s e d e n d o f t h e c y l i n d e r a t
t h e g i v e i n s t a n t ( i n m ) :
2 L=0.15;3 / / D e n s i t y o f g a s ( i n k g /m ˆ 3 ) :
4 d=18;5 / / V e l o c i t y o f p i s t o n ( i n m/ s e c ) :
6 V=12;
Example 5.07 5.07.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 5 . 0 7 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 5.07−data . s c i ’3 exec ( f i l ename )4 / / A t p o i n t b , u =3 mm/ s e c
5 u=3;6 / / D i s p l a c e m e t o f b ( i n mm ) :
7 xb=u∗ t8 / / R a t e o f a n g u l a r d e f o r m a t i o n ( i n s ˆ − 1 ) :
9 de f=U/h10 / / R a t e o f r o t a t i o n ( i n s ˆ − 1 ) :
11 ro t =−0.5∗U/h12 printf ( ”\n\nRSULTS\n\n” )13 printf ( ”\n\nRate o f angular deformation : %. 1 f / s ec \n\n”
, de f )14 printf ( ”\n\nRate o f r o t a t i o n : %. 1 f / s ec \n\n” , ro t )
Example 5.07d 5.07d
1 / / V a l u e o f ( i n mm/ s e c ) :
2 U=4;3 / / V a l u e o f h ( i n mm ) :
4 h=4;5 / / Tme a t w h i c h t o f i n d p o s i t i o n ( i n s e c ) :
6 t =1.5 ;
Example 5.08 5.08.sce
39
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 5 . 0 8 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 5.08−data . s c i ’3 exec ( f i l ename )4 / / V a l u e o f T :
5 T=log (3/2) /A6 x0 =1:2;7 y0 =1:2;8 for i =1:29 for j =1:2
10 / / F o r X c o o r d i n a t e :
11 X( i ) ( j )=x0 ( i )∗%eˆ(A∗T)12 / / F o r Y c o o r d i n a t e :
13 Y( i ) ( j )=y0 ( j )∗%eˆ(−A∗T)14 end15 end16 plot (X,Y)17 / / R a t e s o f l i n e a r d e f o r m a t i o n i n X d i r e c t i o n :
18 Ax=0.3;19 / / R a t e o f l i n e a r d e f o r m a t i o n i n t h e y d i r e c t i o n :
20 Ay=−0.3;21 / / R a t e o f v o l u m e d i l a t i o n ( s ˆ − 1 ) :
22 v=A−A23 / / A r e a o f a b c d :
24 A1=1;25 / / A r e a o f a ’ b ’ c ’ d ’ :
26 A2=(3−3/2)∗(4/3−2/3)27 printf ( ”\n\nRESULTS\n\n” )28 printf ( ”\n\nRates o f l i n e a r deformation in X and Y
d i r e c t i o n : %. 1 f / s , %. 1 f / s\n\n” ,Ax,Ay)29 printf ( ”\n\nRate o f volume d i l a t i o n : %. 0 f / s ec \n\n” , v )30 printf ( ”\n\nArea o f abcd and a , b , c , d :%. 1 f mˆ2 , %. 1 f mˆ\
n\n” ,A1 , A2)
Example 5.08d 5.08-data.sci
1 / / V a l u e o f A ( i n s e c ˆ − 1 ) :
2 A=0.3;
40
Example 5.09 5.09.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 5 . 0 9 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 5.09−data . s c i ’3 exec ( f i l ename )4 / / V o l u m e f l o w r a t e ( i n m ˆ 3 / s e c ) :
5 Q=d∗g∗ s ind ( theta )∗b∗(h/1000) ˆ3∗1000/u/36 printf ( ”RESULTS” )7 printf ( ”\n\nVolume f low ra t e : %. 4 f mˆ3/ sec \n\n” ,Q)
Example 5.09d 5.09-data.sci
1 / / T h i c k n e s s o f w a t e r f i l m ( i n mm ) :
2 h=1;3 / / W i d t h o f s u r f a c e ( i n m ) :
4 b=1;5 / / A n g l e o f i n c l i n a t i o n o f s u r f a c e :
6 theta =15;7 / / D e n s i t y o f w a t e r ( i n k g /m ˆ 3 ) :
8 d=999;9 / / A c c e l e r a t i o n d u t o g r a v i t y ( i n m/ s e c ˆ 2 ) :
10 g =9.81;11 / / V i s c o s i t y ( k g /m− s ) :
12 u=10ˆ−3;
41
Chapter 6
Incompressible Inviscid Flow
6.1 Discussion
When executing the code from the editor, use the ’Execute File into Scilab’taband not the ’Load in Scilab’tab
The .sci files of the respective problems contain the input parameters ofthe question
6.2 Scilab Code
Example 6.01 6.01.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 0 6 . 0 1 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 06.01−data . s c i ’3 exec ( f i l ename )4 / / V e l o c i t y o f f l o w ( i n m/ s e c ) :
5 V=sqrt (dw/ log ( ( r+w) / r )∗g/da∗p/1000)6 / / V o l u m e f l o w r a t e ( i n m ˆ 3 / s e c ) :
7 Q=V∗(d∗w)8 printf ( ”\n\nRESULTS\n\n” )9 printf ( ”\n\nVolume f low ra t e : %. 3 f mˆ3/ sec \n\n” ,Q)
Example 6.01d 6.01-data.sci
1 / / D e p t h o f t h e d u c t ( i n m ) :
42
2 d=0.3;3 / / W i d t h o f t h e d u c t ( i n m ) :
4 w=0.1;5 / / I n n e r r a d i u s o f t h e b e n d ( i n m ) :
6 r =0.25;7 / / P r e s s u r e d i f f e r e n c e b e t w e e n t h e t a p s ( i n mm o f Hg ) :
8 p=40;9 / / D e n s i t y o f w a t e r ( i n k g /m ˆ 3 ) :
10 dw=999;11 / / A c c e l e r a t i o n d u e t o g r a v i t y ( i n m/ s e c ˆ 2 ) :
12 g =9.8 ;13 / / D e n s i t y o f a i r ( i n k g /m ˆ 3 ) :
14 da =1.23;
Example 6.02 6.02.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 0 6 . 0 2 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 06.02−data . s c i ’3 exec ( f i l ename )4 / / V e l o c i t y o f f l o w ( i n m/ s e c ) :
5 V=sqrt (2∗dw∗g∗p/1000∗SG/da )6 printf ( ”\n\nRESULTS\n\n” )7 printf ( ”\n\ nVeloc i ty o f f low : %. 3 f m/ sec \n\n” ,V)
Example 6.02d 6.02-data.sci
1 / / P r e s s u r e d i f e r e n c e ( i n mm o f m e c u r y ) :
2 p=30;3 / / D e n s i t y o f w a t e r ( i n k g /m ˆ 3 ) :
4 dw=1000;5 / / A c e l e r a t i o n d u e t o g r a v i t y ( i n m/ s e c ˆ 2 ) :
6 g =9.81;7 / / D e n s i t y o f a i r ( i n k g /m ˆ 3 ) :
8 da =1.23;9 / / S p e c i f i c g r a v i t y o f m e r c u r y :
10 SG=13.6;
Example 6.03 6.03.sce
43
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 0 6 . 0 3 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 06.03−data . s c i ’3 exec ( f i l ename )4 / / V e l o c i t y o f f l w a t t h e i n l e t ( i n m/ s e c ) :
5 V1=Ae/Ai∗V26 / / G a u g e p r e s s u r e r e q u i r e d a t t h e i n l e t ( i n k P a ) :
7 p=0.5∗da∗(V2ˆ2−V1ˆ2)8 printf ( ”\n\nRESULTS\n\n” )9 printf ( ”\n\nGauge pr s su r e r equ i r ed at the i n l e t : %. 3 f
kPa\n\n” ,p/1000)
Example 6.03d 6.03-data.sci
1 / / A r e a o f n o z z l e a t i n p u t ( i n m ˆ 2 ) :
2 Ai =0.1 ;3 / / A r e a o f n o z z l e a t e x i t ( i n m ˆ 2 ) :
4 Ae=0.02;5 / / O u t l e t v e l o c i t y o f f l o w ( i n m/ s e c ) :
6 V2=50;7 / / D e n s i t y o f a i r ( i n k g /m ˆ 3 ) :
8 da =1.23;
Example 6.04 6.04.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 0 6 . 0 4 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 06.04−data . s c i ’3 exec ( f i l ename )4 / / S p e e d o f w a t e r a t e x i t ( i n m/ s e c ) :
5 V2=sqrt (2∗ g∗z )6 / / P r e s s u r e a t p o i n t A i n t h e f l o w ( k P a ) :
7 pA=p1+d∗g∗(0− l )−0.5∗d∗V2ˆ28 printf ( ”\n\nRESULTS\n\n” )9 printf ( ”\n\nSpeed o f water at e x i t : %. 3 f m/ sec \n\n” ,V2)
10 printf ( ”\n\nPressure at po int A in the f low : %3f kPa\n\n” ,pA/1000)
Example 6.04d 6.04-data.sci
44
1 / / L e n g t h o f t u b e a b o v e s u r f a c e ( i n m ) :
2 l =1;3 / / D e p t h o f e x i t b e l o w w a t e r s u r f a c e ( i n m ) :
4 z=7;5 / / A c c e l e r a t i o n d u e t o g r a v i t y ( i n m/ s e c ˆ 2 ) :
6 g =9.81;7 / / D e n s i t y o f w a t e r ( i n k g /m ˆ 3 ) :
8 d=999;9 / / A t m o s p h e r i c p r e s s u r e ( i n N /m ˆ 2 ) :
10 p1 =1.01∗10ˆ5;
Example 6.05 6.05.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 0 6 . 0 5 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 06.05−data . s c i ’3 exec ( f i l ename )4 / / V e l o c i t y o f f l o w a t t h e e x i t ( i n f t / s e c ) :
5 V2=sqrt (2∗ g∗(Du−Dd/12) )6 / / V o l u m e f l o w r a t e / w i d t h ( f t ˆ 2 / s e c ) :
7 Q=V2∗Dd/128 printf ( ”\n\nRESULTS\n\n” )9 printf ( ”\n\ nVeloc i ty o f f low at the e x i t : %. 3 f f t / s ec \n
\n” ,V2)10 printf ( ”\n\nVolume f low ra t e /width : %. 3 f f t ˆ2/ sec \n\n” ,
Q)
Example 6.05d 6.05-data.sci
1 / / D e p t h o f w a t e r a t t h e u p s t r e a m ( o n f e e t ) :
2 Du=1.5;3 / / D e p t h o f w a t e r a t t h e v e n a c o n t r a c t a d o w n s t r e a m f r o m
t h e g a t e ( i n i n c h e s ) :
4 Dd=2;5 / / A c c e l e r a t i o n d u e t o g r a v i t y ( i n f t / s e c ˆ 2 ) :
6 g =32.2;
Example 6.06 6.06.sce
45
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 0 6 . 0 6 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 06.06−data . s c i ’3 exec ( f i l ename )4 / / P r e s s u r e o f a i r a t 1 0 0 0 m ( i n N /m ˆ 2 ) :
5 p=P1∗pa6 / / D e n s i t y o f a i r a t 1 0 0 0 m ( i n k g /m ˆ 3 ) :
7 d=D1∗da8 / / S t a g n a t i o n p r e s s u r e a t A ( i n k P a ) :
9 p0A=p+0.5∗d∗(V∗1000/3600) ˆ210 / / S t a t i c p r e s s u r e a t B ( i n k P a ) :
11 pB=p+d /2∗ ( (V∗1000/3600)ˆ2−Vbˆ2)12 printf ( ”\n\nRESULTS\n\n” )13 printf ( ”\n\nStagnat ion pr e s su r e at A: %. 3 f kPa\n\n” ,p0A
/1000)14 printf ( ”\n\ nSta t i c p r e s su r e at B: %. 3 f kPa\n\n” ,pB
/1000)
Example 6.06d 6.06-data.sci
1 / / S p e e d o f p l a n e ( i n km / h r ) :
2 V=150;3 / / S p e e d a t p o i n t B r e l a t i v e t o t h e w i n g ( i n m/ s e c ) :
4 Vb=60;5 / / D e n s i t y o f a i r ( i n k g /m ˆ 3 ) :
6 da =1.23;7 / / A t m o s p h e r i s p r e s s u r e ( i n N /m ˆ 2 ) :
8 pa =1.01∗10ˆ5;9 / / A t 1 0 0 0 m ,
10 / / p / p S L :
11 P1=0.8870;12 / / d / d S L :
13 D1=0.9075;
Example 6.08 6.08.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 0 6 . 0 8 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 06.08−data . s c i ’3 exec ( f i l ename )
46
4 / / V e l o c i t y o f f l o w a t e x i t ( i n f t / s e c ) :
5 V4=sqrt (2∗ g∗( z3−0) )6 / / M a s s f l o w r a t e o f w a t e r ( i n s l u g / s e c ) :
7 m=d∗V4∗A4/1448 / / R i s e i n t e m p e r a t u r e b e t w e e n p o i n t s 1 a n d 2 ( i n R ) :
9 T=Q∗3413/3600/m/32 .210 printf ( ”\n\nRESULTS\n\n” )11 printf ( ”\n\nRise in temperature between po in t s 1 and 2 :
%. 3 f R\n\n” ,T)
Example 6.08d 6.08-data.sci
1 / / A r e a o f c r o s s s e c t i o n o f t h e n o z z l e ( i n i n ˆ 2 ) :
2 A4=0.864;3 / / C a p a c i t y o f h e a t e r ( i n kW ) :
4 Q=105 / / A c c e l e r a t i o n d u e t o g r a v i t y ( i n f t / s e c ˆ 2 ) :
6 g =32.2;7 / / W a t e r l e v e l i n r e s e r v o i r a b o v e d a t u m l i n e ( i n f t ) :
8 z3 =10;9 / / D e n s i t y o f w a t e r ( I n s l u g / f t ˆ 3 ) :
10 d=1.94;
Example 6.09 6.09.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 0 6 . 0 9 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 06.09−data . s c i ’3 exec ( f i l ename )4 t =0:55 / / V a l u e o f s q r t ( 2 g h ) :
6 x=sqrt (2∗ g∗h)7 / / V a l u e o f 1 / 2 L ∗ s q r t ( 2 g h ) :
8 y=1/2/L∗x9 [m n]= s ize ( t )
10 i =1:n ;11 / / V e l o c i t y ( i n m/ s e c ) :
12 V2=x∗tanh ( y∗ t ( i ) )13 plot ( t , V2) ;
47
14 xt i t le ( ’ St reaml ine f low from 1 to 2 ’ , ’Time( in s ) ’ , ’V2(in m/ sec ) ’ )
Example 6.09d 6.09-data.sci
1 / / D e p t h t o w h i c h w a t e r i s f i l l e d ( i n m ) :
2 h=3;3 / / L e n g t h o f p i p e ( i n m ) :
4 L=6;5 / / D i a m e t e r o f p i p e ( i n mm ) :
6 D=150;7 / / A c c e l e r a t i o n d u e t o g r a v i t y ( i n m/ s e c ˆ 2 ) :
8 g =9.81;
48
Chapter 7
Dimensional Analysis andSimlitude
7.1 Discussion
When executing the code from the editor, use the ’Execute File into Scilab’taband not the ’Load in Scilab’tab
When we execute S 7.05, we get Fig. 7.1.The .sci files of the respective problems contain the input parameters of
the question
7.2 Scilab Code
Example 7.04 7.04.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 7 . 0 4 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 7.04−data . s c i ’3 exec ( f i l ename )4 / / V e l o c i t y o f p r o t o t y p e i n f t / s e c
5 Vp1=Vp∗6080/36006 / / R e y n o l d s n u m b e r o f p r o t o t y p e :
7 Rep=Vp1∗Dp/vp8 / / R e p =Rem
9 / / T h e r e f o r e :
10 Rem=Rep ;
49
11 / / V e l o c i t y o f a i r f o r w i n d t u n n e l ( i n f t / s e c ) :
12 Vm=Rem∗vm/(Dm/12)13 / / D r a g f o r c e o n p r o t o t y p e ( i n l b f ) :
14 Fp=Fm∗(dp/dm) ∗(Vp1/Vm) ˆ2∗(Dp/(Dm/12) ) ˆ215 printf ( ”\n\nRESULTS\n\n” )16 printf ( ”\n\nTest speed in a i r : %. 3 f f t / sec \n\n” ,Vm)17 printf ( ”\n\nDrag f o r c e on prototype : %. 3 f l b f \n\n” ,Fp)
Example 7.04d 7.04-data.sci
1 / / D i a m e t e r o f t h e p r o t o t y p e ( i n f t ) :
2 Dp=1;3 / / S p e e d o f t o w i n g o f p r o t o t y p e ( i n k n o t s ) :
4 Vp=5;5 / / D i a m e t e r o f m o d e l ( i n i n c h e s ) :
6 Dm=6;7 / / D r a g f o r m o d e l a t t e s t c o n d i t i o n ( i n l b f ) :
8 Fm=5.58;9 / / D e n s i t y o f s e a w a t e r a t 5 C f o r p r o t o t y p e ( i n s l u g / f t
ˆ 3 ) :
10 dp=1.99;11 / / K i n e m a t i c v i s c o s i t y a t 5 C f o r p r o t o t y p e ( i n f t ˆ 2 / s e c )
:
12 vp=1.69∗10ˆ−5;13 / / D e n s i t y o f a i r a t STP f o r m o d e l ( i n s l u g / f t ˆ 3 ) :
14 dm=0.00238;15 / / K i n e m a t i c v i s c o s i t y o f a i r a t STP f o r m o d e l ( i n f t ˆ 2 /
s e c ) :
16 vm=1.57∗10ˆ−4;
Example 7.05 7.05.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 7 . 0 5 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 7.05−data . s c i ’3 exec ( f i l ename )4 / / W i d t h o f t h e m o d e l ( i n m ) :
5 wm=S∗wp∗0.30486 / / A r e a o f m o d e l ( i n m ˆ 2 ) :
50
7 Am=Sˆ2∗Ap∗0.305ˆ28 [m n]= s ize (V)9 i =1:n
10 / / A e r o d y n a m i c d r a g c o e f f i c i e n t ( ) :
11 Cd=2.∗Fd( i ) /d . / (V( i ) ) ˆ2/0.030512 / / R e y n o l d s n u m b e r :
13 Re=V( i )∗wm/v14 plot (Re ,Cd) ;15 a=gca ( )16 a . data bounds = [100000 , 0 . 4 ; 500000 , 0 . 6 ]17 xt i t le ( ’ Aerodynamic drag c o e f f i c i e n t vs drag f o r c e ’ , ’
Reynolds number ’ , ’ Model Drag Coef f . ’ )18 / / I t i s s e e n t h a t d r a g c o e f f i c i e n t b e c o m e s c o n s t a n t a t
CD = 0 . 4 6 a b o v e R e = 4 ∗ 1 0 ˆ 5 a t w h i c h s p e e d o f a i r i s 4 0 m/
s
19 CDc=0.46;20 Va=40;21 / / D r a g f o r c e ( i n N ) :
22 FDp=CDc/2∗d∗(Vp∗5/18) ˆ2∗Ap∗0.305ˆ223 / / P o w e r r e q u i r e d t o p u l l p r o t o t y p e a t 1 0 0 k m p h ( i n W)
24 Pp=FDp∗Vp∗5/1825 printf ( ”\n\nRESULTS\n\n” )26 printf ( ”\n\nSpeed above which Cd i s constant : %. 3 f m/
sec \n\n” ,Va)27 printf ( ”\n\nDrag Force : %. 3 f kN\n\n” ,FDp/1000)28 printf ( ”\n\nPower r equ i r ed to p u l l prototype at 100
kmph : %. 3 f kW\n\n” ,Pp/1000)
Example 7.05d 7.05-data.sci
1 / / W i d t h o f t h e p r o t o t y p e ( i n f t ) :
2 wp=8;3 / / F r o n t a l a r e a o f t h e p r o t o t y p e ( i n f t ˆ 2 ) :
4 Ap=84;5 / / M o d e l S c a l e :
6 S=1/16;7 / / D e n s i t y o f a i r ( i n k g /m ˆ 3 ) :
8 d=1.23;
51
Figure 7.1: Output graph of S 7.05
9 / / A i r s p e e d i n w i n d t u n n e l ( i n m/ s e c ) :
10 V=[18 21 .8 26 30 .1 35 38 .5 40 .9 44 .1 4 6 . 7 ] ;11 / / D r a g f o r c e ( i n N ) :
12 Fd=[3.1 4 .41 6 .09 7 .97 10 .7 12 .9 14 .7 16 .9 1 8 . 9 ] ;13 / / K i n e m a t i c v i s c o s i t y ( i n m ˆ 2 / s e c ) :
14 v=1.46∗10ˆ−5;15 / / D e n s i t y o f a i r ( i n k g /m ˆ 3 ) :
16 d=1.23;17 / / S p e e d o f p r o t o t y p e ( i n km / h r ) : \
18 Vp=100;
52
Example 7.06 7.06.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 7 . 0 6 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 7.06−data . s c i ’3 exec ( f i l ename )4 / / T h e s a m e pump i s u s e d f o r b o t h t h e c o n d i t i o n s . H e n c e :
5 D2=D1 ;6 / / T h e s a m e w a t e r i s u s e d f o r b o t h t h e c o n d i t i o n s . H e n c e
:
7 d2=d1 ;8 / / F l o w r a t e a t c o n d i t i o n 2 ( i n gpm ) :
9 Q2=Q1∗N2/N1∗(D2/D1) ˆ310 / / H e a d a t c o n d i t i o n 1 ( i n f t ) :
11 H1=(N1∗sqrt (Q1) /Nscu1 ) ˆ(4/3)12 / / H e a d a t c o n d i t i o n 1 ( i n f t ) :
13 H2=H1∗(N2/N1) ˆ2∗(D2/D1) ˆ214 / / Pump o u t p u t p o w e r a t c o n d i t i o n 1 ( i n h p ) :
15 P1=d1∗g∗Q1∗H1/7.48/60/55016 / / Pump o u t p u t p o w e r a t c o n d i t i o n 2 ( i n h p ) :
17 P2=P1∗( d2/d1 ) ∗(N2/N1) ˆ3∗(D2/D1) ˆ518 / / R e q u i r e d i n p u t p o w e r ( i n h p ) :
19 Pin=P2/ Effp20 / / S p e c i f i c s p e e d a t c o n d i t i o n 2 :
21 Nscu2=N2∗sqrt (Q2) /H2ˆ(3/4)22 printf ( ”\n\nRESULTS\n\n\n” )23 printf ( ”\n\nVolume f low ra t e at cond i t i on 2 : %. 3 f gpm\n
\n\n” ,Q2)24 printf ( ”\n\nHead at cond i t i on : %. 3 f f t \n\n\n” ,H2)25 printf ( ”\n\nPump output power at cond i t i on : %. 3 f hp\n\n
\n” ,P2)26 printf ( ”\n\nRequired input power : %. 3 f hp\n\n\n” , Pin )27 printf ( ”\n\ n S p e c i f i c speed at cond i t i on 2 : %. 3 f \n\n\n” ,
Nscu2 )
Example 7.06d 7.06-data.sci
1 / / E f f i c i n c o f p ump :
2 Effp =0.8 ;
53
3 / / D e s i g n s p e c i f i c s p e e d ( i n r p m ) :
4 Nscu1=2000;5 / / I m p e l l e r d i a m e t e r ( i n i n c h e s ) :
6 D1=8;7 / / O p e r t i o n s p e d a t e s i g n p o i n t f l o w c o n d i t i o n ( i n r p m ) :
8 N1=1170;9 / / F l o w r a t e a t d e s i g n p o i n t f l o w c o n d i t i o n ( i n gpm ) :
10 Q1=300;11 / / D e n s i t y o f w a t e r ( i n s l u g / f t ˆ 3 ) :
12 d1 =1.94;13 / / A c c e l e r a t i o n d u e t o g r a v i t y ( i n f t ˆ 2 / s e c ) :
14 g =32.2;15 / / W o r k i n g s p e e d 2 ( i n r p m ) :
16 N2=1750;
54
Chapter 8
Internal Incompressible ViscousFlow
8.1 Discussion
When executing the code from the editor, use the ’Execute File into Scilab’taband not the ’Load in Scilab’tab
The .sci files of the respective problems contain the input parameters ofthe question
8.2 Scilab Code
Example 8.01 8.01.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 8 . 0 1 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 8.01−data . s c i ’3 exec ( f i l ename )4 / / L e a k a g e f l o w r a t e ( i n mm ˆ 3 / s e c ) :
5 Q=%pi/12∗D∗a ˆ3∗(p1−p2 ) ∗10ˆ3/u/L6 / / V e l o c i t y o f f l o w ( i n m/ s e c ) :
7 V=Q/%pi/D/a/10008 / / S p e c i f i c g r a v i t y o f SAE 1 0W o i l :
9 SG=0.92;10 / / R e y n o l d s N u m b e r :
11 Re=SG∗dw∗V∗a/u/1000
55
12 / / A s Re < 1 4 0 0 , f l o w i s l a m i n a r .
13 printf ( ”\n\nRESULTS\n\n” )14 printf ( ”\n\nLeakage f low ra t e : %. 3 f mmˆ3/ sec \n\n” ,Q)
Example 8.01d 8.01-data.sci
1 / / O p e r a t i o n p r e s s u r e o f h y d r a u l i c s y s t e m ( i n k P a ) :
2 p1=20000;3 / / O p e r a t i o n t e m p e r a t u r e o f h y d r a u l i c s y s t e m ( i n C ) :
4 T=55;5 / / P i s t o n d i a m e t e r ( i n mm ) :
6 D=25;7 / / V i s c o s i t y o f SAE 1 0W a t 5 5 C ( i n k g / ( m− s ) :
8 u=0.018;9 / / M e a n r a d i a l c l e a r a n c e o f a c y l i n d e r ( i n mm ) :
10 a =0.005;11 / / G a u g e p r e s s u r e o n l o w e r p r e s s u r e s i d e o f p i s t o n ( i n
k P a ) :
12 p2=1000;13 / / L e n t h o f p i s t o n ( i n mm ) :
14 L=15;15 / / D e n i t y o f w a t e r ( i n k g /m ˆ 3 ) :
16 dw=1000;
Example 8.02 8.02.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 8 . 0 2 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 8.02−data . s c i ’3 exec ( f i l ename )4 / / S h e a r s t r e s ( i n l b f / f t ˆ 2 ) :
5 Tyx=u∗N∗2∗%pi/60∗D/2/( a /2)6 / / T o r q e ( i n i n c h e s − l b f ) :
7 T=%pi/2∗Tyx∗Dˆ2∗L/1448 / / P o w e r d i s s i p a t e d i n t h e b e a r i n g ( i n h p ) :
9 P=T∗N/60∗2∗%pi/12/55010 / / R e y n o l d s n u m b e r :
11 Re=SG∗p∗N∗2∗%pi /60∗1.5∗ a/2/u/14412 printf ( ”\n\nRESULTS\n\n” )
56
13 printf ( ”\n\nTorque : %. 3 f inches−l b f \n\n” ,T)14 printf ( ”\n\nPower d i s s i p a t e d in the bear ing : %. 3 f hp\n\
n” ,P)
Example 8.02d 8.02-data.sci
1 / / t e m p e r a t u r e f o o p e r a t i o n ( i n F ) :
2 T=210;3 / / D i a m e t e r o f t e b e a r i n g ( i n i n c h e s ) :
4 D=3;5 / / D i a m e t r a l c l e a r a n c e ( i n i n c h e s ) :
6 a =0.0025;7 / / L e n g t h o f s h a f t ( i n i n h e s ) :
8 L=1.25;9 / / S p e e d o f r o t a t i o n o f t h e s h a f t ( i n r p m ) :
10 N=3600;11 / / V i s c o s i t y o f t h e o i l ( i n l b f − s / f t ˆ 2 ) :
12 u=2.01∗10ˆ−4;13 / / S p e c i f i c g r a v i t y o f SAE 1 0W :
14 SG=0.92;15 / / D e n s i t y o f w a t e r ( i n s l u g / f t ˆ 3 )
16 p=1.94;
Example 8.04 8.04.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 8 . 0 4 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 8.04−data . s c i ’3 exec ( f i l ename )4 / / V i s c o s i t y o f t h e l i q u i d ( i n N− s /m ˆ 2 ) :
5 u=%pi/128∗p∗1000∗Dˆ4/Q/L/10006 / / V e l o c i t y ( i n m/ s e c )
7 V=Q/(%pi/4∗Dˆ2) /10008 / / R e y n o l d s n u m b e r :
9 Re=d∗V∗D/u/100010 printf ( ”\n\nRESULTS\n\n” )11 printf ( ”\n\ nVi s co s i ty o f f l u i d %. 3 f N−s /mˆ2\n\n” ,u)
Example 8.04d 8.04-data.sci
57
1 / / F l o w r a t e t h r o u g h c a p i l a r r y v i s c o m e t e r ( i n mm ˆ 3 / s e c ) :
2 Q=880;3 / / T u b e l e n g t h ( i n m ) :
4 L=1;5 / / T u b e d i a m e t e r ( i n mm ) :
6 D=0.5;7 / / P r e s s u r e d r o p ( i n k P a ) :
8 p=1000;9 / / D e n s i t y o f o i l ( i n k g /m ˆ 3 ) :
10 d=999;
Example 8.05 8.05.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 8 . 0 5 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 8.05−data . s c i ’3 exec ( f i l ename )4 / / R e s e r v o i r d e p t h r e q u i r e d t o m a i n t a i n f l o w ( i n m ) :
5 D1=8∗Qˆ2/(%pi ) ˆ2/Dˆ4/g∗( f ∗L/D+K+1)6 / / R e y n o l d s n u m b e r :
7 Re=4∗d∗Q/(( %pi )∗u∗D)8 printf ( ”\n\nRESULTS\n\n” )9 printf ( ”\n\ nReservo i r depth r equ i r ed to maintain f low :
%. 3 f m\n\n” ,D1)
Example 8.05d 8.05-data.sci
1 / / V o l m e f l o w r a t e o f w a t e r ( i n m ˆ 3 / s e c ) :
2 Q=0.0084;3 / / L e n g t h o f h o r i z o n t a l p i p e ( i n m ) :
4 L=100;5 / / D i a m e t e r o f p i p e ( i n m ) :
6 D=0.075;7 / / D e n s i t y o f w a t e r ( i n k g /m ˆ 3 ) :
8 d=999;9 / / F r i c t i o n f a c t o r :
10 f =0.017;11 / / M i n o r l o s s s e s c o e f f i c i e n t :
12 K=0.5;
58
13 / / V i s c o s i t y ( i n k g /m− s ) :
14 u=10ˆ−3;15 / / A c c e l e r a t i o n d u e t o g r a v i t y ( i n / s e c ˆ 2 ) :
16 g =9.8 ;
Example 8.06 8.06.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 8 . 0 6 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 8.06−data . s c i ’3 exec ( f i l ename )4 / / V e l o c i t y o f f l o w ( i n f t / s e c ) :
5 V=Q/24/3600/(%pi/4∗(D/12) ˆ2) ∗42/7.486 / / M a x i m u m s p a c i n g ( i n f t ) :
7 L=2/ f ∗D/12∗(p2−p1 ) /(SG∗d) /Vˆ2∗1448 / / P o w e r n e e d e d a t e a c h p ump ( i n h p ) :
9 Win=1/Effp ∗V∗%pi/4∗(D/12) ˆ2∗(p2−p1 ) /550∗14410 printf ( ”\n\nRESULTS\n\n” )11 printf ( ”\n\nMaximum spac ing : %. 3 f f e e t \n\n” ,L)12 printf ( ”\n\nPower needed at each pump : %. 3 f hp\n\n” ,Win
)
Example 8.06d 8.06-data.sci
1 / / F l o w r a t e o f c r u d e o i l ( i n b b l ) :
2 Q=1.6∗10ˆ6;3 / / I n s i d e d i a m e t e o f p i p e ( i i n c h e s ) :
4 D=48;5 / / M a x i m u m a l l o w a b l e p r e s s u r e ( i n p s i ) :
6 p2=1200;7 / / M i n i m u m p r e s s u r e r e q u i r e d t o k e e p g a s e s d i s s o l v e s ( i n
p s i ) :
8 p1=50;9 / / S p e c i f i c g r a v i t y o f c r d e o i l :
10 SG=0.93;11 / / V i s c o s i t y a t 1 4 0 F ( i n l b f − s / f t ˆ 2 ) :
12 u=3.5∗10ˆ−4;13 / / E f f i c i n c y o f p ump :
14 Effp =0.85;
59
15 / / D e n s i t y ( i n s l u g / f t ˆ 3 ) :
16 d=1.94;17 / / V i s c o s i t y ( i n l b f − s e c ) :
18 u=3.5∗10ˆ−4;19 / / F r i c t i o n f a c t o r :
20 f =0.017;
Example 8.07 8.07.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 8 . 0 7 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 8.07−data . s c i ’3 exec ( f i l ename )4 / / V e l o c i t y ( i n f t / s e c ) :
5 V2=sqrt (2∗ g∗ l /( f ∗ ( (L+l ) /D∗12+8)+1) )6 / / V o l u m e f l o w r a t e ( i n gpm ) :
7 Q=V2∗%pi∗(D/12) ˆ2/4∗7.48∗608 printf ( ”\n\nRESULTS\n\n” )9 printf ( ”\n\nVolume low ra t e : %. 3 f \n\n” ,Q)
Example 8.07d 8.07-data.sci
1 / / H e i g h t o f s t a n d p i p e ( i n f t ) :
2 l =80;3 / / L e n g t h o f l o n g e s t p i p e ( i n f t ) :
4 L=600;5 / / D i a m e t e r o f p i p e ( i n i n c h e s ) :
6 D=4;7 / / F r i c t i o n f a c t o r :
8 f =0.031;9 / / A c c e l e r a t i o n d u e t o g r a v i t y i n f t / s e c ˆ 2 ) :
10 g =32.2;
Example 8.08 8.08.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 8 . 0 8 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 8.08−data . s c i ’3 exec ( f i l ename )
60
4 / / V a l u e o f d P m a x ( i n p s i ) :
5 dPmax=p1−p26 / / Q i n c u b i c f e e t / s e c :
7 Q1=1500/60/7.48;8 / / I n i t i a l l y a s s u m e d i a m e t e r t o b e 4 i n c h e s :
9 D=4;10 / / R e y n o l d s n u m b e r :
11 Re=4∗Q1/%pi/v/D∗1212 / / F o r t h i s v a l u e ,
13 f =0.012;14 dP=8∗ f ∗L∗p∗Q1ˆ2/(%pi ) ˆ2/Dˆ5∗1728;15 while (dP>dPmax)16 dP=8∗ f ∗L∗p∗Q1ˆ2/(%pi ) ˆ2/Dˆ5∗1728;17 i f (dP<dPmax)18 break19 else20 D=D+1;21 end22 end23 printf ( ”\n\nRESULTS\n\n” )24 printf ( ”Minimum diameter that can be used :%. 1 f inche s \n
\n” ,D)
Example 8.08d 8.08-data.sci
1 / / L e n g t h o f A l t u b i n g ( i n f t ) :
2 L=500;3 / / V o l u m e f l o w r a t e o f p ump o u t p u t ( i n gpm ) :
4 Q=1500;5 / / D i s c h a r g e p r e s s u r e ( i n p s i g ) :
6 p1=65;7 / / S p r i n k l e r p r e s s u r e ( i n p s i g ) :
8 p2=30;9 / / K i n e m a t i c v i s c o s i t y ( i n f t ˆ 2 / s e c ) :
10 v=1.21∗10ˆ−5;11 / / D e n s i t y ( i n s l u g / f t ˆ 3 ) :
12 p=1.94;
61
Example 8.09 8.09.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 8 . 0 9 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 8.09−data . s c i ’3 exec ( f i l ename )4 / / A v e r a g e v e l o c i t y ( i n f t / s ) :
5 V2=4/%pi∗Q/Dˆ2∗1446 / / R e y n o l d s n u m b e r :
7 Re=V2∗D/v/128 / / F o r t h i s v a l u e ,
9 f =0.013;10 / / P o w e r l a w e x p o n e n t :
11 n=−1.7+1.8∗ log10 (Re)12 / / V a l u e o f V / U :
13 v u=2∗nˆ2/(n+1)/(2∗n+1)14 / / V a l u e o f a l p h a :
15 alpha =(1/v u ) ˆ3∗2∗nˆ2/(3+n) /(3+2∗n)16 / / L o s s C o e f f i c i e n t f o r a s q u a r e e d g e d e n t r a n c e :
17 K=2∗g∗h/V2ˆ2− f ∗L/D∗12−alpha ;18 printf ( ”\n\nRESULTS\n\n” )19 printf ( ”\n\nLoss C o e f f i c i e n t f o r a square edged
entrance : %. 3 f \n\n” ,K)
Example 8.09d 8.09-data.sci
1 / / L e n g t h o f c o p p e r w i r e ( i n f t ) :
2 L=10;3 / / I n n e r d i a m e e r o f p i p e ( i n i n c h e s ) :
4 D=1.5;5 / / D i s c h a r e ( i n f t ˆ 3 / s e c ) :
6 Q=0.566;7 / / L e v e l o f r e s e r v o i r a b o v e p i p e c e n t r e l i n e ( i n n f e e t ) :
8 h=85.1;9 / / K i n e m a t i c v i s c o s i t y a t 7 0 F ( i n f t ˆ 2 / s ) :
10 v=1.05∗10ˆ−5;11 / / A c c e l e r a t i o n d u e t o g r a v i t y ( i n f t / s e c ˆ 2 ) :
12 g =32.2;
62
Example 8.10 8.10.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 8 . 1 0 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 8.10−data . s c i ’3 exec ( f i l ename )4 / / V e l o c i t y V1 ( i n m/ s ) :
5 V1=sqrt (2∗ g∗z0 /1 . 04 )6 / / V o l u m e f l o w r a t e ( i n m ˆ 3 / s e c ) :
7 Q=V1∗%pi∗Dˆ2/48 Kdi f f=1−1/A Rˆ2−Cp9 / / F o r 2 n d c a s e :
10 / / V e l o c i t y ( i n m/ s ) :
11 V1=sqrt (2∗ g∗z0 /0 . 59 )12 / / V o l u m e f l o w r a t e ( i n m ˆ 3 / s ) :
13 Qd=V1∗%pi∗Dˆ2/414 / / I n c r e a s e i n d i s c h a r g e a f t e r a d d i t i o n o f d i f f u s e r i s :
15 dQ=(Qd−Q) /Q∗10016 printf ( ”\n\nRESULTS\n\n” )17 printf ( ”\n\nVolume f low ra t e in case1 : %. 3 f mˆ3/ sec \n\n
” ,Q)18 printf ( ”\n\nVolume f low ra t e in case 2 : %. 3 f mˆ3/ sec \n\
n” ,Qd)19 printf ( ”\n\ nInc rea se in d i s cha rge a f t e r add i t i on o f
d i f f u s e r i s : %. 3 f percent \n\n” ,dQ)
Example 8.10d 8.10-data.sci
1 / / N o z z l e e x i t d i a m e t e r ( i n mm ) :
2 D=25;3 / / N / R1 v a l u e v a l u e :
4 N R=3;5 / / AR v a l u e :
6 A R=2;7 / / S t a t i c h e a d a v a i l a b l e f r o m t h e m a i n ( i n m ) :
8 z0 =1.5 ;9 / / A c c e l e r a t i o n d u e t o g r a v i t y ( i n m/ s e c ˆ 2 ) :
10 g =9.8 ;11 / / V a l u e o f Cp :
63
12 Cp=0.45;
Example 8.11 8.11.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 8 . 1 1 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 8.11−data . s c i ’3 exec ( f i l ename )4
5 / / V a l u e o f K ∗ B ˆ 2 :
6 K B=Q/(%pi/4∗Dˆ2)∗sqrt ( 0 . 5∗ d1/g/d2/h)7 / / R e y n o d s n u m b e r :
8 ReD1=4/%pi∗Q/D/v9 / / By t r i a l a n d e r r o r m e t h o d , t h e v a l u e o f b e t a i s f i x e d
a t :
10 betta =0.66;11 / / K i s t h e n :
12 K=K B/ betta ˆ213 / / D i a m e t e r o f o r i f i c e p l a t e ( i n m ) :
14 Dt=betta ∗D15 / / V a l u e o f p 3 − p 2 ( i n N /m ˆ 2 ) :
16 P1=d1∗Qˆ2/(%pi/4∗Dˆ2) ˆ2∗ (1/0 .65/ betta ˆ2−1)17 / / V a l u e o f p 1 − p 2 ( i n N /m ˆ 2 ) :
18 P2=d2∗g∗h19 / / H e a d l o s s b e t w e e n s e c t i o n s 1 a n d 3 ( i n N−m/ k g ) :
20 hLT=(P2−P1) /d121 / / E x p r e s s i n g t h e p e r m a n e n t p r e s s u r e a s a f r a c t i o o f t h e
m e t e r d i f f e r e n t i a l :
22 C=(P2−P1) /P223 printf ( ”\n\n\nRESULTS\n\n” )24 printf ( ”\n\nDiameter o f the o r i f i c e : %. 3 f m\n\n” ,Dt)25 printf ( ”\n\nHead l o s s between s e c i o n s 1 and 3 : %. 3 f N−m
/kg\n\n” ,hLT)
Example 8.11d 8.11-data.sci
1 / / V o l u m e f l w r a t e o f a i ( i n m ˆ 3 / s e c ) :
2 Q=1;3 / / D i a m e t e r o f p i p e ( i n m ) :
64
4 D=0.25;5 / / D e n s i t y o f a i r ( i n k g /m ˆ 3 ) :
6 d1 =1.23;7 / / A c c e l e r a t i o n d u e t o g r a v i t y ( i n m/ s ˆ 2 ) :
8 g =9.8 ;9 / / D e n s i t y o f w a t e r ( i n k g /m ˆ 3 ) :
10 d2=999;11 / / Maxmum r a n g e o f m a n o m e t e r ( i n m ) :
12 h=0.3;13 / / K i n e m a t i c v i s c o s i t y ( i n m ˆ 2 / s ) :
14 v=1.46∗10ˆ−5;
65
Chapter 9
External IncompressibleViscous Flow
9.1 Discussion
When executing the code from the editor, use the ’Execute File into Scilab’taband not the ’Load in Scilab’tab
The .sci files of the respective problems contain the input parameters ofthe question
When we execute S 9.08, we get Fig. 9.1.
9.2 Scilab Code
Example 9.01 9.01.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 9 . 0 1 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 9.01−data . s c i ’3 exec ( f i l ename )4 / / C h a n g e i n s t a t i c p r e s s u r e b e t w e e n s e c t i o n s 1 a n d 2 :
5 C=(((L−2∗d1 ) /(L−2∗d2 ) ) ˆ4−1) ∗100 ;6 printf ( ”\n\nRESULTS\n\n” )7 printf ( ”\n\nChange in s t a t i c p r e s su r e between the
s e c t i o n s 1 and 2 : %. 3 f percent \n\n” ,C)
Example 9.01d 9.01-data.sci
66
1 / / L e n g h o f s i d e o f t h e t e s t s e c t i o n ( i n mm ) :
2 L=305;3 / / F r e e s t e a m s p e e d a t s e c t i o n 1 ( i n m/ s e c ) :
4 U1=26;5 / / D i s p l a c e m e n t t h i c k n e s s a t s e c t i o n 1 ( i n mm ) :
6 d1 =1.5 ;7 / / D i s p l a c m e n t t h i c k n e s s a t s e c t i o n 2 ( i n mm ) :
8 d2 =2.1 ;
Example 9.04 9.04.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 9 . 0 4 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 9.04−data . s c i ’3 exec ( f i l ename )4 / / R e y n o l d s n u m b e r :
5 ReL=U∗L/v6 / / FOR TURBULENT FLOW
7 / / D i s t u r b a n c e t h i c k n e s s ( i n m ) :
8 dL1=0.382/ReLˆ0.2∗L9 / / D i s p l a c e m e n t t h i c k n e s s ( i n m ) :
10 function y=f (n) , y=dL1∗(1−n ˆ(1/7) )11 endfunction12 dl1=intg (0 , 1 , f )13 / / S k i n f r i c t i o n c o e f f i c i e n t :
14 Cf1 =0.0594/ReLˆ0 .215 / / W a l l s h e a r s t r e s s ( i n N /m ˆ 2 ) :
16 tw1=Cf1 ∗0.5∗d∗Uˆ217 / / F o r LAMINAR FLOW :
18 / / D i s t u r b a n c e t h i c k n e s s ( i n m )
19 dL2=5/sqrt (ReL)∗L20 / / D i s p l a c e m e n t t h i c k n e s s ( i n m ) :
21 dl2 =0.344∗dL222 / / S k i n f r i c t i o n c o e f f i c i e n t :
23 Cf2 =0.664/ sqrt (ReL)24 / / W a l l s h e a r s t r e s s ( i n N /m ˆ 2 ) :
25 tw2=Cf2 ∗0.5∗d∗Uˆ226 / / COMPARISON OF VALUES WITH LAMINAR FLOW
27 / / D i s t u r b a n c e t h i c k n e s s
67
28 D=dL1/dL229 / / D i s p l a c e m e n t t h i c k n e s s
30 DS=dl1 / dl231 / / W a l l s h e a r s t r e s s
32 WSS=tw1/tw233 printf ( ”\n\nRESULTS\n\n” )34 printf ( ”\n\nDisturbace t h i c k n e s s : %. 3 f m\n\n” ,dL1)35 printf ( ”\n\nDisplacement t h i c k n e s s : %. 3 f m\n\n” , d l1 )36 printf ( ”\n\nWall shear s t r e s s : %f N/mˆ2\n\n” , tw1 )37 printf ( ”\n\nCOMPARISON WIH LAMINAR FLOW\n\n\n” )38 printf ( ”\n\n Disturbance th i ckne s : %. 3 f \n\n” ,D)39 printf ( ”\n\nDisplacement t h i c k n e s s : %. 3 f \n\n” ,DS)40 printf ( ”\n\nWall shear s t r e s s : %. 3 f \n\n” ,WSS)
Example 9.04d 9.04-data.sci
1 / / V e o c i t y o f f l o w ( i n m/ s e c ) :
2 U=1;3 / / L e n g t h o f f l a t p l a t e ( i n m ) :
4 L=1;5 / / D e n s i t y o f w a t e r ( i n k g /m ˆ 3 ) :
6 d=999;7 / / K i n e m a t i c v i s c o s i t y o f w a t e r ( i n m ˆ 2 / s e c ) :
8 v=10ˆ−6;
Example 9.05 9.05.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 9 . 0 5 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 9.05−data . s c i ’3 exec ( f i l ename )4 / / S p e e d i n m/ s :
5 U=s ∗6076∗0.305/36006 / / R e y n o l d s n u m b e r :
7 Re=U∗L/v8 / / D r a g c o e f f i c i e n t :
9 Cd=0.455/ log10 (Re) ˆ2.58−1610/Re10 / / A r e a ( i n m ˆ 2 ) :
11 A=L∗(W+D)
68
12 / / D r a g f o r c e ( i n N )
13 Fd=Cd∗A∗0.5∗d∗Uˆ214 / / P o w e r r e q u i r e d t o o v e r c o m e s k i n f r i c t i o n d r a g ( i n W) :
15 P=Fd∗U16 printf ( ”\n\nRESULTS\n\n” )17 printf ( ”\n\nDrag f o r c e : %f N\n\n” ,Fd)18 printf ( ”\n\nPower r equ i r ed to overcome sk in f r i c t i o n
drag : %. 3 f W\n\n” ,P)
Example 9.05d 9.05-data.sci
1 / / L e n g t h o f t h e s u p e r t a n k e r ( i n m ) :
2 L=360;3 / / W i d t h o f s u p e r t a n k e r ( i n m ) :
4 W=70;5 / / D r a f t o f t h e s u p e r t a n k e r ( i n m ) :
6 D=50;7 / / C r u i s i n g s p e e d i n w a t e r ( i n k n o t s ) :
8 s =13;9 / / K i n e m a t i c v i s c o s i t y a t 1 0 C
10 v=1.37∗10ˆ−6;11 / / D e n s i t y o f s e a w a t e r ( i n k g /m ˆ 3 ) :
12 d=1020;
Example 9.06 9.06.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 9 . 0 6 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 9.06−data . s c i ’3 exec ( f i l ename )4 / / V e l o c i t y i n m/ s e c :
5 V=s ∗5/186 / / R e y n o l d s n u m b e r :
7 Re=d∗V∗D/u8 / / V a l u e o f Cd i s o b t a i n e d a s :
9 Cd=0.35;10 / / A r e a ( i n m ˆ 2 ) :
11 A=Lˆ2 ;12 / / M o m e n t a b o u t t h e c h i m n e y b a s e ( i n N−m ) :
69
13 M0=Cd∗A∗D/4∗d∗Vˆ214 printf ( ”\n\nRESULTS\n\n” )15 printf ( ”\n\nBending moment at the bottom of the chimney
: %. 3 f N−m\n\n” ,M0)
Example 9.06d 9.06-data.sci
1 / / D i a m e t e r o f c h i m n e y ( i n m ) :
2 D=1;3 / / H e i g h t o f c h i m n e y ( i n m ) :
4 L=25;5 / / S p e e d o f w i n d ( i n k m p h ) :
6 s =50;7 / / D e n s i t y o f a i r ( i n k g /m ˆ 3 ) :
8 d=1.23;9 / / V i s c o s i t y o f a i r ( i n k g / ( m− s ) ) :
10 u=1.79∗10ˆ−5;11 / / P r e s s u r e ( i n k P a ) :
12 p=101;
Example 9.07 9.07.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 9 . 0 7 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 9.07−data . s c i ’3 exec ( f i l ename )4 / / T i m e r e q u i r e d t o d e c e l e r a t e t o 1 0 0 mph ( i n s e c o n d s ) :
5 t=(s1−s2 ) ∗2∗w/( s1∗ s2 ) /Cd/d/A/g∗3600/52806 printf ( ”\n\nRESULTS\n\n” )7 printf ( ”\n\nTime requ i r ed to d e c e l e r a t e to 100 mph: %. 3
f seconds \n\n” , t )
Example 9.07d 9.07-data.sci
1 / / W e i g h t o f t h e d r a g s t e r ( i n l b f ) :
2 w=1600;3 / / S p e e d o f d r a g s t e r ( i n mph ) :
4 s1 =270;
70
5 / / A r e a o f d r a g c h u t e ( i n f t ˆ 2 ) :
6 A=25;7 / / S p e e d o f d r a g s t e r a f t e r d e c e l e r a t o n ( i n mph ) :
8 s2 =100;9 / / A c c e l e r a t i o n d u e t o g r a v i t y ( i n f t / s e c ˆ 2 ) :
10 g =32.2;11 / / D e n s i t y o f a i r ( i n s l u g / f t ˆ 3 ) :
12 d=0.00238;13 / / V a l u e o f c o e f f i c i e n t o f d r a g :
14 Cd=1.42;
Example 9.08 9.08.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 9 . 0 8 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 9.08−data . s c i ’3 exec ( f i l ename )4 / / P l o t t i n g v e l o c i t y w i t h d r a g f o r c e
5 V=175 :25 :455 ;6
7 [m n]= s ize (V) ;8 for i =1:n9 CL( i )=2∗W/p∗(3600/V( i ) /5280) ˆ2/A;
10 Cd( i )=Cd0+CL( i ) ˆ2/%pi/ ar ;11 Fd( i )=Cd( i ) /CL( i )∗W;12 FD( i )=Fd( i ) /1000 ;13 end14 plot (V,FD)15 xt i t le ( ’ F l i gh t speed vs th rus t ’ , ’ F l i gh t Speed ( in mph) ’ ,
’ Drag Force ( in 1000 l b f ) ’ )16 / / O p t i m u m c u i s e s p e e d a t s p e e d l e v e l i s o b t a i n e d t o b e
3 2 0 mph f r o m t h e g r a p h .
17 Vosl =320;18 / / R a t i o o f s p e e d s a t 3 0 0 0 0 f t a n d a t s e a l e v e l i s g i v e n
b y :
19 r=sqrt ( 1/0 . 375 ) ;20 / / S t a l l s p e e d a t 3 0 0 0 0 f t i s ( i n mph ) :
21 Vs3=Vss l ∗ r ;22 / / O p t i m u m C r u i s e s p e e d a t 3 0 0 0 0 f t ( i n mph ) :
71
23 Vo3=Vosl∗ r ;24 printf ( ”\n\nRESULTS\n\n” )25 printf ( ”\n\nOptimum c r u i s e speed at sea l e v e l : %. 3 f mph
\n\n” , Vosl )26 printf ( ”\n\ n S t a l l speed at 30000 f t : %. 3 f mph\n\n” , Vs3 )27 printf ( ”\n\nOptimum c r u i s e speed at 30000 f t : %. 3 f \n\n”
,Vo3)
Example 9.08d 9.08-data.sci
1 / / W i n g a r e a ( i n f t ˆ 2 ) :
2 A=1600;3 / / A s p e c t r a t i o :
4 ar =6.5 ;5 / / G r o o s w e i g h t o f a i r c r a f t ( i n l b f ) :
6 W=150000;7 / / C o e f f i c i e n t o f d r a g a t z e r o l i f t :
8 Cd0=0.0182;9 / / S o n i c s p e e d a t s e a l e v e l ( i n mph ) :
10 c =759;11 / / D e n s i t y o f a i r ( i n s l u g / f t ˆ 3 ) :
12 p=0.00238;13 / / S r a l l s p e e d a t s e a l e v e l ( i n mph ) :
14 Vss l =175;
Example 9.09 9.09.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 9 . 0 9 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 9.09−data . s c i ’3 exec ( f i l ename )4 / / R e y n o l d s n u m b e r :
5 / / V a l u e o f wD / 2 V :
6 W=0.5∗N∗D/1000/V∗2∗%pi/607 Red=V∗D/v ;8 / / F o r t h i s v a l u e , CL i s o b t a i n e d a s :
9 CL=0.3;10 / / A e r o d y n a m i c l i f t ( i n N ) :
72
Figure 9.1: Output graph of S 9.08
73
11 FL=%pi/8∗CL∗(D/1000) ˆ2∗d∗Vˆ2 ;12 / / R a d i u s o f c u r v a t u r e o f t h e p a t h i n t h e v e r t i c a l p l a n e
( i n m ) w i t h t o p s p i n :
13 Rts=Vˆ2/( g+FL/(m/1000) ) ;14 / / R a d i u s o f c u r v a t u r e w i t h o u t t o p s p i n ( i n m ) :
15 Rwts=Vˆ2/g ;16 printf ( ”\n\nRESULTS\n\n” )17 printf ( ”\n\nAerodynamic l i f t a c t ing on the b a l l :%. 3 f N\
n\n” ,FL)18 printf ( ”\n\nRadius o f curvature o f the path when b a l l
has topsp in :%. 3 f m\n\n” , Rts )19 printf ( ”\n\nRadius o f curvature o f the path when b a l l
has topsp in : %. 3 f m\n\n” ,Rwts )
Example 9.09d 9.09-data.sci
1 / / M a s s o f t h e t e n n i s b a l l ( i n g r a m s ) :
2 m=57;3 / / D i a m e t e r o f t h e b a l l ( i n mm ) :
4 D=64;5 / / V e l o c i t y w i t h w h i c h t e b a l l i s h i t ( i n m/ s ) : \
6 V=25;7 / / T o p s p i n g i v e n o n t h e b a l l ( i n r p m ) :
8 N=7500;9 / / A c c e l e r a t i o n d u e t o g r a v i t y ( i n m/ s ˆ 2 ) :
10 g =9.81;11 / / K i n e m a t i c v i s c o s i t y ( i n m ˆ 2 / s ) :
12 v=1.46∗10ˆ−513 / / D e s i t y o f a i r ( i n k g /m ˆ 3 ) :
14 d=1.23;
74
Chapter 10
Fluid Machinery
10.1 Discussion
When executing the code from the editor, use the ’Execute File into Scilab’taband not the ’Load in Scilab’tab
The .sci files of the respective problems contain the input parameters ofthe question
When S 10.03 is executed, we get Fig. 10.1. When S 10.07 is executed,we get Fig. 10.2. When S 10.08 is executed, we get Fig. 10.3 and Fig. 10.4.When S 10.11 is executed, we get Fig. 10.5 to Fig. 10.7.
10.2 Scilab Code
Example 10.01 10.01.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 1 0 . 0 1 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 10.01−data . s c i ’3 exec ( f i l ename )4 / / I m p e l l e r e x i t w i d t h b 2 ( i n f e e t ) :
5 b2=Q∗12/(2∗%pi∗R2∗Vrb2 ∗7 .48∗60)6 / / T o r q u e o f t h e S h a f t , T s h a f t ( i n f t − l b f ) :
7 Tshaft=w∗R2ˆ2∗p∗Q∗2∗%pi /3600/7.48/1448 / / P o w e r , Wm( i n h p ) :
9 Wm=w∗Tshaft ∗2∗%pi/60/55010 printf ( ”\n\nRESULTS\n\n” )
75
11 printf ( ”\n\ nImpe l l e r e x i t width : %. 3 f f e e t \n\n” , b2 )12 printf ( ”\n\Torque input : %. 3 f f t−l b f \n\n” , Tshaft )13 printf ( ”\n\nPower : %. 3 f hp\n\n” ,Wm)
Example 10.01d 10.01-data.sci
1 / / V o l u m e f l o w r a t e i n gpm :
2 Q= 150 ;3 / / V a l u e o f V r b 2 i n f t / s e c :
4 Vrb2=10;5 / / R a d i u s o f o u t t e r i m p e l l e r i n i n c h e s :
6 R2=2;7 / / I m p e l l e r S p e e d i n r p m :
8 w=3450;9 / / D e n s i t y o f a i r i n s l u g / f t ˆ 3
10 p=1.94;
Example 10.02 10.02.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 1 0 . 0 2 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 10.02−data . s c i ’3 exec ( f i l ename )4 U=0.5∗(Dh+Dt) /2∗1200∗2∗%pi/605 k=tand ( alpha1 )+cotd ( betta1 )6 Vn1=U/k7 V1=Vn1/ cosd ( alpha1 )8 Vt1=V1∗ s ind ( alpha1 )9 Vrb1=Vn1/ s ind ( betta1 )
10 / / V o l u m e f l o w r a t e ( i n m ˆ 3 / s e c ) :
11 Q=%pi/4∗Vn1∗(Dtˆ2−Dhˆ2)12 k=(U−Vn1∗ cotd ( betta2 ) ) /Vn113 alpha2= atand ( k )14 V2=Vn1/ cosd ( alpha2 )15 Vt2=V2∗ s ind ( alpha2 )16 / / R o t o r T o r q u e ( i n N−m ) :
17 Tz=p∗Q∗(Dh+Dt) /4∗(Vt2−Vt1 )18 / / P o w e r r e q u i r e d ( i n W) :
19 Wm=w∗2∗%pi/60∗Tz
76
20 printf ( ”\n\nRESULTS\n\n” )21 printf ( ”\n\nVolume f low ra t e : %. 3 f mˆ3/ sec \n\n” ,Q)22 printf ( ”\n\nRotor Torque : %. 3 f N−m\n\n” ,Tz)23 printf ( ”\n\nPower r equ i r ed : %. 3 f W\n\n” ,Wm)
Example 10.02d 10.02-data.sci
1 / / T i p D i a m e t e r i n m e t r e s :
2 Dt=1.1;3 / / Hub D i a m e t e r i n m e t r e s :
4 Dh=0.8;5 / / O p e r a t i n g S p e e d i n r p m :
6 w=1200;7 / / A b s o l u t e i n l e t a n g l e i n d e g r e e s :
8 alpha1 =30;9 / / B l a d e i n l e t a n g l e i n d e g r e e s :
10 betta1 =30;11 / / B l a d e o u t l e t a n g l e i n d e g r e e s :
12 betta2 =60;13 / / D e n s i t y o f a i r i n k g /m ˆ 3
14 p=1.23;
Example 10.03 10.03.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 1 0 . 0 3 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 10.03−data . s c i ’3 exec ( f i l ename )4 [nQ mQ]= s ize (Q) ;5 [ nps mps]= s ize ( ps ) ;6 [ npd mpd]= s ize (pd ) ;7 [ nI mI]= s ize ( I ) ;8 / / C o r r e c t m e a s u r e d s t a t i c p r e s s u r e s t o h e p ump
c e n t r e l i n e p 1 , p 2 ( i n p s i g ) :
9 j =1:mps ;10 p1=ps ( j )+px∗g∗ zs /14411 j =1:mpd;12 p2=pd( j )+px∗g∗zd /14413 / / T h e v a l u e o f Pump h e a d ( i n f e e t ) :
77
14 j =1:mps ;15 Hp=(p2 ( j )−p1 ( j ) ) /( px∗g ) ∗14416 / / V a l u e s o f H y d r a u l i c P o w e r d e l i v e r e d ( i n h p ) :
17 j =1:mps ;18 Wh=Q( j ) . ∗ ( p2 ( j )−p1 ( j ) ) /7.48/60∗144/55019 / / V a l u e s o f m o t o r p o w e r o u t p u t ( i n h p ) :
20 j =1:mI ;21 Pin=Effm∗sqrt (3 ) ∗PF∗E∗ I ( j ) /74622 / / V a l u e s o f Pump E f f i c i e c y :
23 j =1:mI ;24 Effp= Wh( j ) . / Pin ( j ) ∗10025 / / P l o t t i n g p ump c h a r a c t e r i s t i c s :
26 plot (Q,Hp, ”−o” )27 plot (Q, Pin , ”−+” )28 plot (Q, Effp , ”−∗” )29 xt i t le ( ’Pump C h a r a c t e r i s t i c s ’ , ’ Volume f low ra t e ( in
gpm) ’ , [ ’Pump E f f i c i n c y (%) ’ , ’ Pump Head( inf e e t ) ’ , ’ Pump Power input ( in hp) ’ ] )
30 l egend ( ’Hp ’ , ’ Pin ’ , ’ Ef fp ’ )
Example 10.03d 10.03-data.sci
1 / / R a t e o f f l o w i n gm :
2 Q=[0 500 800 1000 1100 1200 1400 1 5 0 0 ] ;3 / / S u c t i o n p r e s s u r e i n p s i g :
4 ps=[ 0 .65 0 .25 −0.35 −0.92 −1.24 −1.62 −2.42 −2 .89 ] ;5 / / D i s c h a r g e p r e s s u r e i n p s i g :
6 pd=[53.3 48 .3 42 .3 36 .9 33 27 .8 15 .3 7 . 3 ] ;7 / / M o t o r C u r r e n t i n a m p s :
8 I =[18 26 .2 31 33 .9 35 .2 36 .3 38 3 9 ] ;9 / / A c c e l e r a t i o n d u e t o g r a v i t y i n f t / s ˆ 2 :
10 g =32.2;11 / / V a l u e o f Z s i n f e e t
12 zs =1;13 / / D e n s i t y o f a i r i n s l u g / f t ˆ 3 :
14 px =1.94;15 / / V a l u e o f ZD i n f e e t :
16 zd=3;
78
Figure 10.1: Output graph of S 10.03
17 / / D e n s i t y o f f l u i d i n s l u g / f t ˆ 3 :
18 py=1000;19 / / M o t o r E f f i c i e n c y :
20 Effm =0.9;21 / / M o t o r S u p p l y i n v o l t s :
22 E=460;23 / / P o w e r F a c t o r :
24 PF=0.875;
Example 10.06 10.06.sce
79
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 1 0 . 0 6 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 10.06−data . s c i ’3 exec ( f i l ename )4 / / S p e c i f i c s p e e d i n U s c u s t o m a r y u n i t s :
5 Nscu=N∗Qusˆ0 .5/ Hus ˆ0 .756 / / C o n v e r s i o n t o S I u n i t s :
7 w=1170∗2∗%pi /60 ;8 Qsi=Qus /7 .48/60∗0 .305ˆ3 ;9 Hsi=Hus ∗0 . 3 0 5 ;
10 / / E n e r g y p e r u n i t m a s s i s :
11 h=g∗Hsi ;12 / / S p e c i f i c s p e e d i n S I u n i t s :
13 Nss i=w∗Qsi ˆ0 .5/ h ˆ0 .7514 / / C o n v e r s i o n t o h e r t z :
15 whz=N/60 ;16 / / S p e c i f i c s p e e d i n E u r o p e a n u n i t s :
17 Nseu=whz∗Qsi ˆ0 . 5/65 . 5ˆ0 . 7518 / / R e l a t i o n b e t w e e n s p e c i f i c s p e e d s i n U s c u s t o m a r y
u n i t s a n d E u r o p e a n u n i t s :
19 Conver s i on fac to r1=Nscu/Nseu20 / / R e l a t i o n b e t w e e n s p e c i f i c s p e e d s i n U s c u s t o m a r y
u n i t s a n d S I u n i t s :
21 Conver s i on fac to r2=Nscu/ Nss i22 printf ( ”\n\nRESULTS\n\n” )23 printf ( ”\n\ n S p e c i f i c speed in US customary un i t s : %. 3 f
\n\n” , Nscu )24 printf ( ”\n\ n S p e c i f i c speed in SI un i t s : %. 3 f \n\n” , Nss i
)25 printf ( ”\n\ n S p e c i f i c speed in European un i t s : %. 3 f \n\n”
, Nseu )26 printf ( ”\n\nRelat ion between s p e c i f i c speeds in Us
customary un i t s and European un i t s : %. 3 f \n\n” ,Conver s i on fac to r1 )
27 printf ( ”\n\nRelat ion between s p e c i f i c speeds in Uscustomary un i t s and SI un i t s : %. 3 f \n\n” ,Conver s i on fac to r2 )
Example 10.06d 10.06-data.sci
80
1 / / H e a d i n U s c u s t o m a r y u n i t s :
2 Hus=21.9;3 / / V o l u m e f l o w r a t e i n US c u s t o m a r y u n i t s :
4 Qus=300;5 / / W o r k i n g s e e d i n r p m :
6 N=1170;7 / / A c e l e r a t i o n d u e t o g r a v i y i n m/ s ˆ 2
8 g =9.81;
Example 10.07 10.07.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 1 0 . 0 7 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 10.07−data . s c i ’3 exec ( f i l ename )4 / / V o l u m e f l o w r a t e ( i n gpm ) a t s h u t o f f c o n d i t i o n f o r N2
:
5 Q2so=N2/N1∗Q1so6 / / V o l u m e f l o w ( i n gpm ) r a t e a t b e s t e f f i c i e n c y f o r N2 :
7 Q2be=N2/N1∗Q1be8 / / R e l a t i o n b e t w e e n pump h e a d s :
9 h e a d r e l a t i o n =(N2/N1) ˆ210 / / H e a d ( i n f e e t ) a t s h u t o f f c o n d i t i o n f o r N2 :
11 H2so=(N2/N1) ˆ2∗H1so12 / / H e a d ( i n f e e t ) a t b e s t e f f i c i e n c y c o n d i t i o n f o r N2 :
13 H2be=(N2/N1) ˆ2∗H1be14 Q1=[Q1so Q1be ] ;15 Q2=[Q2so Q2be ] ;16 H1=[H1so H1be ] ;17 H2=[H2so H2be ] ;18 plot (Q1, H1 , ”−o” )19 plot (Q2, H2 , ”−∗” )20 xt i t le ( ’ Comparison o f head f o r both c o n d i t i o n s ’ , ’ Volume
Flow Rate ’ , ’Head ’ )21 l egend ( ’ 1170 ’ , ’ 1750 ’ )
Example 10.07d 10.07-data.sci
81
1 / / V o l u m e f l o w r a t e ( i n gpm ) a t s h u t o f f c o n d i t i o n f o r N1
:
2 Q1so=0;3 / / V o l u m e f l o w ( i n gpm ) r a t e a t b e s t e f f i c i e n c y f o r N1 :
4 Q1be=300;5 / / H e a d ( i n f e e t ) a t s h u t o f f c o n d i t i o n f o r N1 :
6 H1so=25;7 / / H e a d ( i n f e e t ) a t b e s t e f f i c i e n c y c o n d i t i o n f o r N1 :
8 H1be=21.99 / / O p e r a t i o n S p e e d 1 :
10 N1=1170;11 / / O p e r a t i o n s p e e d 2 :
12 N2=1750;
Example 10.08 10.08.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 1 0 . 0 8 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 10.08−data . s c i ’3 exec ( f i l ename )4 / / D i a m e t e r o f p i p e ( i n f e e t ) :
5 Df= Di/126 / / A r e a o f c r o s s e c t i o n o f p i p e ( i n f t ˆ 2 ) :
7 A=%pi/4∗Dfˆ28 / / V e l o c i t y o f f l o w ( i n f t / s e c ) :
9 V=Q/7.48/A/6010 / / F o r w a t e r a t T = 8 0 F , v i s c o s i t y = 0 . 9 2 7 e −5 f t ˆ 2 / s e c ,
R e y n o l d s n u m b e r :
11 Re=V∗Df/v12 / / F r i c t i o n l o s s C o e f f i c i e n t f o r t h i s v a l u e o f R e :
13 f =0.0237;14 / / F o r c a s t i r o n , r o u g h n e s s ( i n f e e t ) :
15 e =0.0008516 / / e / D i s :
17 e/Df18 / / T o t a l h e a d l o s s ( i n f e e t ) :
19 HL=K+f ∗(SE+OGV)+f ∗(L/Df )+120 / / T h e h e a d s a r e ( i n f e e t ) :
82
Figure 10.2: Output graph of S 10.07
83
21 H1=patm∗144/(p∗g )22 Vh=Vˆ2/2/g23 / / S u c t i o n h e a d ( i n f e e t ) :
24 Hs=H1+h−HL∗Vh25 / / NPSHA ( i n f e e t ) :
26 NPSHA=Hs+Vh−Hv127 / / F o r a f l o w r a t e o f 1 0 0 0 gpm , NPSHR ( i n f e e t ) f o r w a t e r
a t 8 0 F
28 NPSHR=1029 / / PLOTTING NPSHA AND NPSHR VERSUS VOLUME FLOW RATE :
30 / / F o r 8 0 F
31 Qp=0:100 :1500 ;32 [ nQp mQp]= s ize (Qp) ;33 for j =1:mQp;34 Vp( j )=Qp( j ) / (7 .48∗A∗60) ;35 Vhp( j )=(Vp( j ) ) ˆ2/2/g ;36 Hs( j )=H1+h−HL∗Vhp( j ) ;37 end38
39 for j =1:mQp;40 NPSHAp1( j )=Hs( j )+(Vhp( j ) )−Hv1 ;41 end42
43 plot (Qp,NPSHAp1, ”−+” )44 plot (Qh,NPSHRp, ”−o” )45 xt i t le ( ’ Suct ion head vs Flow ra t e ’ , ’ Volume f low ra t e (
gpm) ’ , ’ Suct ion Head( f e e t ) ’ ) ;46 printf ( ”\n\nType (Resume) to cont inue or ( abort ) to end
\n\n” )47 l egend ( ’NPSHA’ , ’NPSHR’ )48 pause49 c l f50
51 / / F o r 1 8 0 F
52 for j =1:mQp;53 NPSHAp2( j )=Hs( j )+(Vhp( j ) )−Hv2 ;54 end55 plot (Qp,NPSHAp2, ”−+” )
84
56 plot (Qh,NPSHRp, ”−o” )57 xt i t le ( ’ Suct ion head vs Flow ra t e ’ , ’ Volume f low ra t e (
gpm) ’ , ’ Suct ion Head( f e e t ) ’ ) ;58 l egend ( ’NPSHA’ , ’NPSHR’ )59 printf ( ”\n\nRESULTS\n\n” )60 printf ( ”\n\nNPSHA at Q=1000 gpm of water at 80 F : %. 2 f
f t \n\n” ,NPSHA)61 printf ( ”\n\nNPSHR at Q=1000 gpm of water at 80 F : %. 1 f
f t \n\n” ,NPSHR)
Example 10.08d 10.08-data.sci
1 / / F o r 5 i n c h n o m i n a l p i p e l i n e , d i a m e t e r D :
2 Di =5.047;3 / / L e n g t h o f p i p e l i n e ( i n f e e t ) :
4 L=6;5 / / O p e r a t n g s p p e d ( i n r p m ) :
6 N=1750;7 / / W a t e r l e v e l a b o v e p u m p c e n t r e l i n e ( i n f e e t ) :
8 h=3.5;9 / / T e m p e r a t u r e 1 o f w a t e r ( i n F a r e n h e i t ) :
10 T1=80;11 / / T e m p e r a t u r e 2 o f w a t e r ( i n F a r e n h e i t ) :
12 T2=180;13 / / V o l u m e f l o w r a t e o f w a t e r ( i n gpm ) :
14 Q=1000;15 / / M i n o r l o s s C o e f f i c i e n t s :
16 K=0.5; SE=30; OGV=8;17 / / A t m o s p h e r i c p r e s s u r e ( i n l b f / i n ˆ 2 ) :
18 patm=14.7;19 / / D e n s i t y o f a i r ( s l u g / f t ˆ 3 ) :
20 p=1.93;21 / / A c c e l e r a t i o n d u e t o g r a v i t y ( i n f t / s e c ˆ 2 ) :
22 g =32.2;23 / / H e a d ( i n f e e t ) d u e t o v a p o r p r e s s u r e o f w a t e r f o r T
= 8 0 F :
24 Hv1=1.17;
85
Figure 10.3: Output graph-1 of S 10.08
25 / / H e a d ( i n f e e t ) d u e t o v a p o r p r e s s u r e o f w a t e r f o r T
= 1 8 0 F :
26 Hv2=17.3;27 / / K i n e m a t i c v i s c o s i t y o f w a t e r a t 8 0 F :
28 v=0.927e−5;29 / / V a l u e o f d i s c h a r g e s f o r p l o t t i n g NPSHR ( i n gpm ) :
30 Qh=[500 700 900 1100 1300 ]31 / / V a l u e s o f NPSHR o b t a i n e d f r o m F i g . D3 o f a p p e n d i x D :
32 NPSHRp=[7 8 9 .5 12 16 ]
Example 10.11 10.11.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 1 0 . 1 1 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 10.11−data . s c i ’3 exec ( f i l ename )4 [nQ mQ]= s ize (Q1) ;5 [ np mp]= s ize ( p1 ) ;6 [ nP mP]= s ize (P1) ;7 / / V o l u m e f l o w r a t e f o r f a n 2 ( i n c f m ) :
86
Figure 10.4: Output graph-2 of S 10.08
8 j =1:mQ;9 Q2=Q1( j ) ∗(N2/N1) ∗(D2/D1) ˆ3
10 / / P r e s s u r e v a l u e s f o r f a n 2 ( i n i n c h e s o f H2O ) :
11 j =1:mp;12 p2=p1 ( j ) ∗( d2/d1 ) ∗ ( (N2/N1) ˆ2) ∗ ( (D2/D1) ˆ2)13 / / P o w e r v a l u e s f o r f a n 2 ( i n h p ) :
14 j =1:mP;15 P2=P1( j ) ∗( d2/d1 ) ∗ ( (N2/N1) ˆ3) ∗ ( (D2/D1) ˆ5)16 plot (Q2, p2 )17 xt i t le ( ’ Performance curves ’ , ’ Volume f low ra t e ( in cfm ) ’ ,
’ Pres sure head ( in inche s o f water ) ’ )18 printf ( ”\n\nType ( resume ) to cont inue or ( abort ) to
e x i t \n\n” )19 pause20 c l f21 plot (Q2, P2)22 xt i t le ( ’ Performance curves ’ , ’ Volume f low ra t e ( in cfm ) ’ ,
’ Power ( in hp) ’ )23 printf ( ”\n\nType ( resume ) to cont inue or ( abort ) to
87
e x i t \n\n” )24 pause25 c l f26 plot (Q2, E f f )27 xt i t le ( ’ Performance curves ’ , ’ Volume f low ra t e ( in cfm ) ’ ,
’ E f i c i e n c y ( in percentage ) ’ )28 / / S p e c i f i c s p e e d o f f a n ( i n US c u s t o m a r y u n i t s ) a t
o p e r a t i n g p o i n t :
29 Nscu= 1150∗110000ˆ0 .50∗0 .045ˆ0 .75/7 .4ˆ0 .7530 / / S p e c i f i c s p e e d o f f a n ( i n S I u n i t s ) a t o p e r a t i n g
p o i n t :
31 Nss i =120∗3110ˆ0.5∗0.721ˆ0 .75/1 .84 e3 ˆ0 .75
Example 10.11d 10.11-data.sci
1 / / D i a m e t e r o f f a n 1 ( i n i n c h e s ) :
2 D1=36;3 / / O p e r a t i n g s p e e d o f f a n 1 ( i n r p m ) :
4 N1=6005 / / D e n s i t y o f a i r u s e d i n f a n 1 ( i n l b m / f t ˆ 3 ) :
6 d1 =0.075;7 / / D i a m e t e r o f f a n 2 ( i n i n c h e s ) :
8 D2=42;9 / / O p e r a t i n g s p e e d o f f a n 2 ( i n r p m ) :
10 N2=1150;11 / / D e n s i t y o f a i f u s d i n f a n 2 ( i n l b m / f t ˆ 3 ) :
12 d2= 0 . 0 4 5 ;13 / / T h e f o l l o w i n g v a l u e s a r e o b t a i n e d f r o m t h e g i v e n
g r a p h
14 / / V a l u e s o f v o l u m e f l o w r a t e ( i n c f m ) t h r o u g h f a n 1 :
15 Q1= [ 0 10000 20000 30000 40000 50000 6 0 0 0 0 ] ;16 / / V a l u e s o f p r e s s u r e ( i n i n c h e s o f H2O ) :
17 p1=[ 3 .68 3 .75 3 .50 2 .96 2 .12 1 .02 0 ] ;18 / / V a l u e s o f p o w e r ( i n h p ) :
19 P1=[ 11 .1 15 .1 18 .6 21 .4 23 .1 23 .1 2 1 . 0 ] ;20 / / E f f i c i e n c y ( i n % ) :
21 Ef f =[0 37 59 65 57 34 0 ] ;
88
Figure 10.5: Output graph-1 of S 10.11
Figure 10.6: Output graph-2 of S 10.11
89
Figure 10.7: Output graph-3 of S 10.11
90
Example 10.12 10.12.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 1 0 . 1 2 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 10.12−data . s c i ’3 exec ( f i l ename )4 / / F r o m g i v e n g r a p h , f o r m a x i m u m d e l i v e r y c o n d i t i o n , Q
= 4 8 . 5 gpm .
5 / / V o l u m e o f o i l p e r r e v o l u t i o n d e l i v e r e d b y t h e p ump ( i n
i n ˆ 3 / r e v ) :
6 vc=Qe/N∗2317 / / V o l u m e t r i c E f f c i e n c y o f p ump a t max f l o w :
8 Effv=vc/va9 / / O p e r a t i n g p o i n t o f t h e p ump i s f o u n d t o b e a t 1 5 0 0
p s i g , Q = 4 6 . 5 gpm
10 / / P o w e r d e l i v e r e d b y t h e f l u i d ( i n h p ) :
11 Pf=Qo∗po1 /7.48/60∗144/55012 / / I n p u t p o w e r ( i n h p ) :
13 Pi=Pf/ Effp14 / / T h e p o w e r d e l i v e r e d t o t h e l o a d ( i n h p ) :
15 Pl=Q∗( po1 ) /7.48/60∗144/55016 / / P o w e r d i s s i p a t e d b y t h r o t t l i n g ( i n h p ) :
17 Pd=Pf−Pl18 / / T h e d i s s i p a t i o n w i t h t h e v a r i a b l e d i s p l a c e m e n t p ump (
i n h p ) :
19 Pvd=Q∗( po2−po1 ) /7.48/60∗144/55020 / / P o w e r r e q u i r e d f o r t e l o a d s e n s i n g p ump i f p ump
p r e s s u r e i s 1 0 0 p s i a b o v e t h a t r e q u i r e d b y t h e l o a d (
i n h p ) :
21 Pls=Q∗100/7.48/60∗144/55022 printf ( ”\n\nRESULTS\n\n” )23 printf ( ”\n\nVolume o f o i l per r e v o l u t i o n d e l i v e r e d by
the pump : %. 3 f in ˆ3/ rev \n\n” , vc )24 printf ( ”\n\nRequired pump power input : %. 3 f hp\n\n” , Pi )25 printf ( ”\n\nPower d e l i v e r d to the load : %. 3 f hp\n\n” , Pl
)26 printf ( ”\n\nPower d i s s i p a t e d by t h r o t t l i n g : %. 3 f hp\n\n
” ,Pd)27 printf ( ”\n\nThe d i s s i p a t i o n with the v a r i a b l e
91
disp lacement pump : %. 3 f hp\n\n” ,Pvd)28 printf ( ”\n\nPower r equ i r ed f o r te load s en s ing pump i f
pump pre s su r e i s 100 p s i above that r equ i r ed by theload : %. 3 f hp\n\n” , Pls )
Example 10.12d 10.12-data.sci
1 / / O p e r a t i o n s p e e d ( i n r p m ) :
2 N=2000;3 / / V o l u m e f l o w r a t e ( i n gpm ) :
4 Q=20;5 / / P r e s s u r e ( i n p s i g ) :
6 p=1500;7 / / A c t u a l Pump D i s p l a c e m e n t ( i n ˆ 3 / r e v ) :
8 va =5.9;9 / / V o l u m e f l o w r a t e a t o p e r a t i n g c o n d i t i o n ( i n gpm ) :
10 Qo=46.5 ;11 / / V o l u m e f l o w r a t e a t m a x i m u m d e l i v e r y ( i n gpm ) :
12 Qe=48.5;13 / / P r e s s u r e a t o p e r a t i o n c o n d i t i o n ( i n p s i ) :
14 po1=1500;15 / / E f f i c i e n c y o f p ump a t o p e r a t i n g c o n d i t i o n :
16 Effp =0.84;17 / / P r e s s u r e a t o p e r a t i n g c o n d i t i o n c a s e 2 ( i n p s i g ) :
18 po2=3000;
Example 10.14 10.14.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 1 0 . 1 4 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 10.14−data . s c i ’3 exec ( f i l ename )4 / / P r o p e l l e r T h r u s t ( i n MN ) :
5 Ft=P/V6 / / R e q u i r e d p o w e r i n p u t t o t h e p r o p e l l e r ( i n MW) :
7 Pin=P/ Ef f8 / / C a l c u l a t i n g v a l u e o f D ( i n m ) :
9 nD=V/J10 D=(Ft∗10ˆ6/p/(nD) ˆ2/Cf ) ˆ0 .5
92
11 / / O p e r a t i n g s p e e d ( i n r p m ) i s g i v e n b y :
12 n=nD/D∗6013 printf ( ”\n\nRESULTS\n\n” )14 printf ( ”\n\nDiameter o f the s i n g l e p r o p e l l e r r equ i r ed
to pwer the sh ip :%. 3 f m\n\n” ,D)15 printf ( ”\n\nOperating speed o f the p r o p e l l e r : %. 3 f rpm\
n\n” ,n)
Example 10.14d 10.14-data.sci
1 / / T o t a l p r o p u l s i o n p o w e r r e q u i r e m e n t ( i n MW) :
2 P=11.4;3 / / F r o m t h e g i v e n c u r v e s , V a l u e o f c o e f f i c i e n t s a t o p t i m u m
e f f i c i e n c y a r e a s f o l l o w s :
4 / / S p e e d o f a d v a n c e c o e f f i c i e n t :
5 J =0.85;6 / / T h r u s t C o e f f i c i e n t :
7 Cf =0.1;8 / / T o r q u e C o e f f i c i n t :
9 Ct=0.02;10 / / E f f i c i e n c y :
11 Ef f =0.66;12 / / V e l o c i t y o f s h i p ( i n m/ s e c ) :
13 V=6.69;14 / / D e n s i t y o f w a t e r ( i n k g /m ˆ 3 ) :
15 p=1025;
Example 10.16 10.16.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 1 0 . 1 6 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 10.16−data . s c i ’3 exec ( f i l ename )4 / / T i p s p e e d r a t i o o f w i n d m i l l :
5 X=N∗2∗%pi/60∗D/2/(V∗5/18)6 / / A c c o u n t i n g f o r w h i r l , max a t t a i n a b l e e f f i c i e n c y i s :
7 Efw=0.53;8 / / K i n e t i c e n e r g y f l u x ( i n W) i s g i v e n b y :
9 KEF=0.5∗p∗(V∗5/18) ˆ3∗%pi∗(D/2) ˆ2
93
10 / / A c t u a l E f f i c i e n c y :
11 Ef fa=Po/KEF12 / / T h e m a x i m u m p o s s i b l e t h r u s t o c c u r s f o r a n
i n t e r f e r e n c e f a c t o r o f :
13 amax=0.5;14 / / T h r u s t ( i n W) :
15 Kx=p∗(V∗5/18) ˆ2∗%pi∗(D/2) ˆ2∗2∗amax∗(1−amax)16 printf ( ”\n\nRESULTS\n\n” )17 printf ( ”\n\nTip speed r a t i o o f windmi l l :%. 3 f \n\n” ,X)18 printf ( ”\n\nActual E f f i c i e n c y : %. 3 f \n\n” , Ef fa )19 printf ( ”\n\nActual Thrust : %. 3 f W\n\n” ,Kx)
Example 10.16d 10.16-data.sci
1 / / D i a m e t e r o f w i n d m i l l ( i n m ) :
2 D=26;3 / / O p e r a t i n g s p e e d ( i n r p m ) :
4 N=20;5 / / W i n d s p e e d ( i n km / h r ) :
6 V=36;7 / / P o w e r O u t p u t ( i n W) :
8 Po=41000;9 / / M a x i m u m e f f i c i e n c y o c c u r s i n f o l l o w i n g c o n d i t i o n s :
10 / / E f f i c i e n c y :
11 Ef f =0.593;12 / / I n t e f e r e n c e F a c t o r :
13 a=1/3;14 / / D e n s i t y o f a i r ( i n k g /m ˆ 3 ) :
15 p=1.23;
94
Chapter 11
Introduction to CompressibleFlow
11.1 Discussion
When executing the code from the editor, use the ’Execute File into Scilab’taband not the ’Load in Scilab’tab
The .sci files of the respective problems contain the input parameters ofthe question
When S 11.03 is executed, we get Fig. 11.1.
11.2 Scilab Code
Example 11.01 11.01.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 1 1 . 0 1 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 11.01−data . s c i ’3 exec ( f i l ename )4 / / D e n s i t y o f a i r a t e n t r y :
5 d1=p1∗10ˆ3/R/T16 / / A r e a ( i n m ˆ 2 ) :
7 A=m/d1/V18 / / C h a n g e i n e n t h a l p y o f a i r ( i n k J / k g ) :
9 dh=cp ∗(T2−T1)10 / / C h a n g e i n i n t e r n a l e n e r g y o f a i r ( i n k J / k g ) :
95
11 du=cv ∗(T2−T1)12 / / C h a n g e i n e n t r o p y ( i n k J / ( k g −K ) ) :
13 ds=cp∗ log (T2/T1)−R/1000∗ log ( p2/p1 )14 printf ( ”\n\nRESULTS\n\n” )15 printf ( ”\n\nDuct Area : %. 3 f mˆ2\n\n” ,A)16 printf ( ”\n\nChange in enthalpy o f a i r : %. 3 f kJ/kg\n\n” ,
dh)17 printf ( ”\n\nChange in i n t e r n a l energy o f a i r :%. 3 f kJ/kg
\n\n” ,du)18 printf ( ”\n\nChange in entropy : %. 3 f kg−K\n\n” , ds )
Example 11.01d 11.01-data.sci
1 / / T e m p e r a t u r e o f a i r e n t e r i n g t h e c o l d s e c t i o n ( i n K ) :
2 T1=440;3 / / A b s o l u t e p r e s s u r e o f a i r e n t e r i n g t h e c o l d s e c t i o n ( i n
k P a ) :
4 p1=188;5 / / V e l o c i t y o f a i r e n t e r i n g t h e c o l d s e c t i o n ( i n m/ s e c ) :
6 V1=210;7 / / T e m p e r a t u r e o f a i r a t o u t l e t : ( i n K )
8 T2=351;9 / / A b s o l u t e p r e s s u r e o f a i r a t o u t l e t ( i n k P a ) :
10 p2=213;11 / / R a t e o f h e a t l o s s i n t h e s e c t i o n ( i n k J / s e c ) :
12 / / G a s C o n s t a n t ( i n N−m ) :
13 R= 287 ;14 / / M a s s f l o w r a t e o f a i r ( i n k g / s e c ) :
15 m=0.15;16 / / S p e c i f i c h e a t a t c o n s t a n t p r e s s u e ( i n k J / ( k g −K ) ) :
17 cp=1;18 / / S p e c i f i c e n e r g y a t c o n s t a n t v o l u m e ( i n k J / ( k g −K ) ) :
19 cv =0.717;
Example 11.03 11.03.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 1 1 . 0 3 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 11.03−data . s c i ’
96
3 exec ( f i l ename )4 / / V a l u e s o f a l t i t u d e ( i n m ) :
5 Al =0:1000:150006 [ nAl mAl]= s ize ( Al ) ;7 / / V a l u e s o f t e m p e r a t u r e a t g i v e n a l t i t u d e s ( i n K ) :
8 T=[288.2 281 .7 275 .2 268 .7 262 .2 255 .7 249 .2 242 .7236 .2 229 .7 223 .3 216 .8 216 .7 216 .7 216 .7 2 1 6 . 7 ] ;
9 [ nT mT]= s ize (T) ;10 / / V a l u e s o f s p e e d o f s o u n d a t t h e s e a l t i t u d e s ( i n m/ s e c )
:
11 j =1:mT;12 c=sqrt ( k∗R∗T( j ) )13 / / S p e e d o f s o u n d a t s e a l e v e l ( i n m/ s e c ) :
14 c1=sqrt ( k∗R∗T(1) )15 plot ( c , Al )16 xt i t le ( ’ Var ia t ion o f sound speed with a l t i t u d e ’ , ’ Speed
o f sound (m/ sec ) ’ , ’ A l t i tude (m) ’ )17 printf ( ”\n\nRESULTS\n\n” )18 printf ( ”\n\nSpeed o f sound at sea l e v e l : %. 3 f m/ sec \n\n
” , c1 )
Example 11.03d 11.03-data.sci
1 / / V a l u e o f k :
2 k =1.4 ;3 / / G a s C o n s t a n t ( i n K j / ( k g −K ) ) :
4 R=287;
Example 11.04 11.04.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 1 1 . 0 4 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 11.04−data . s c i ’3 exec ( f i l ename )4 / / M a c h n u m b e r a t e n t r y :
5 M1=V1/sqrt ( k∗R∗T1)6 / / S t a g n a t i o n p r e s s u r e a t e n t r y ( i n k P a ) :
7 p01=p1∗(1+(k−1)/2∗M1ˆ2) ˆ( k /(k−1) )
97
Figure 11.1: Output graph of S 11.03
98
8 / / S t a g n a t i o n t e m p e r a t u r e a t e n t r y ( i n K ) :
9 T01=T1∗(1+(k−1)/2∗M1ˆ2)10 / / S t a t i c p r e s s u r e a t e x i t ( i n k P a ) :
11 p2=p02/(1+(k−1)/2∗M2ˆ2) ˆ( k /(k−1) )12 / / T e m p e r a t u r e a t e x i t ( i n K ) :
13 T2=T02/(1+(k−1)/2∗M2ˆ2)14 / / C h a n g e i n e n t r o p y ( i n k J / k g −K ) :
15 ds=cp∗ log (T2/T1)−R/1000∗ log ( p2/p1 )16 printf ( ”\n\nRESULTS\n\n” )17 printf ( ”\n\nStagnat ion pr e s su r e at entry : %. 3 f kPa\n\n”
, p01 )18 printf ( ”\n\nStagnat ion temperature at entry : %. 3 f K\n\n
” ,T01)19 printf ( ”\n\ nSta t i c p r e s su r e at e x i t : %. 3 f kPa\n\n” , p2 )20 printf ( ”\n\nTemperature at e x i t : %. 3 f K\n\n” ,T2)21 printf ( ”\n\nChange in entropy : %. 3 f kJ/kg−K\n\n” , ds )
Example 11.04d 11.04-data.sci
1 / / P r e s s u r e a t e n t r y ( i n k P a ) :
2 p1=350;3 / / T e m p e r a t u r e a t e n t r y ( i n K )
4 T1=333;5 / / V e l o c i t y a t e n t r y ( i n m/ s ) :
6 V1=183;7 / / M a c h n o . a t e x i t :
8 M2=1.3;9 / / S t a g n a t i o n p r e s s u r e a t e x i t ( i n k P a ) :
10 p02=385;11 / / S t a g n a t i o n t e m p e r a t u r e a t e x i t ( i n K ) :
12 T02=350;13 / / V a l u e o f k :
14 k =1.4 ;15 / / G a s c o n s t a n t ( i n N−m/ kg −K )
16 R=287;17 / / S p e c i f i c h e a t a t c o n s t a n t p r e s s u r e ( k J / ( k g −K ) :
18 cp=1;
99
Chapter 12
Steady One-DimensionalCompressible Flow
12.1 Discussion
When executing the code from the editor, use the ’Execute File into Scilab’taband not the ’Load in Scilab’tab
The .sci files of the respective problems contain the input parameters ofthe question
12.2 Scilab Code
Example 12.01 12.01.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 1 2 . 0 1 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 12.01−data . s c i ’3 exec ( f i l ename )4 / / H e r e t h e s t a g n a t i o n q u a n t i t i e s a r e c o n s t a n t .
5 / / S t a g n a t i o n t e m p e r a t u r e ( i n K ) :
6 T0=T1∗(1+(k−1)/2∗M1ˆ2)7 / / S t a g n a t i o n p r e s s u r e ( i n k P a ) :
8 p0=p1∗((1+(k−1)/2∗M1ˆ2) ˆ( k /(k−1) ) )9 / / F i n d i n g T2 / T1 :
10 T=t2 / t111 / / T e m p e r a t u r e a t e x i t ( i n K ) :
100
12 T2=T∗T113 / / F i n d i n g p 2 / p 1 :
14 P=P2/P115 / / P r e s s u r e a t e x i t ( i n k P a ) :
16 p2=P2∗p117 / / D e n s i t y o f a i r a t e x i t ( i n k g /m ˆ 3 ) :
18 d2=p2∗10ˆ3/R/T219 / / V e l o c i t y o f a i r a t e x i t ( i n m/ s e c ) :
20 V2=M2∗sqrt ( k∗R∗T2)21 / / F i n d i n g A2 / A1 :
22 a=a2/a123 / / A r e a a t e x i t ( i n m ˆ 2 ) :
24 A2=a∗A125 printf ( ”\n\nRESULTS\n\n” )26 printf ( ”\n\nStagnat ion temperature : %. 3 f K\n\n” ,T0)27 printf ( ”\n\nStagant ion pr e s su r e : %. 3 f kPa\n\n” , p0 )28 printf ( ”\n\nTemperature a e x i t %. 3 f K\n\n” ,T2)29 printf ( ”\n\nPressure at e x i t : %. 3 f kPa\n\n” , p2 )30 printf ( ”\n\nDensity o f a i r at e x i t : %. 3 f kg/mˆ3\n\n” , d2
)31 printf ( ”\n\ nVeloc i ty o f a i r at e x i t : %. 3 f m/ sec \n\n” ,V2
)32 printf ( ”\n\nArea at e x i t : %. 3 f \n\n” ,A2)
Example 12.01d 12.01-data.sci
1 / / M a c h n u m b e r a t e n t r y :
2 M1=0.3;3 / / T e m p e r a t u r e a t e n t r y ( i n K ) :
4 T1=335;5 / / P r e s s u r e a t e n t r y ( i n k P a ) :
6 p1=650;7 / / A r e a a t e n t r y ( i n m ˆ 2 ) :
8 A1=0.001;9 / / M a c h n u m b e r a t e x i t :
10 M2=0.8;11 / / / V a l u e o f k :
12 k =1.4 ;
101
13 / / F o r t h e M a c h n o : 0 . 3 :
14 / / T / T0 :
15 t1 =0.9823 ,16 / / p / p 0 :
17 P1=0.9395;18 / / d / d 0 :
19 den1 =0.9564;20 / / A / A ∗ :
21 a1 =2.035;22 / / F o r t h e M a c h n o : 0 . 8 :
23 / / T / T0 :
24 t2 =0.8865;25 / / p / p 0 :
26 P2=0.6560;27 / / d / d 0 :
28 den2 =0.7400;29 / / A / A ∗ :
30 a2 =1.038;31 / / G a s C o n s t a n t ( i n N−m/ kg −K ) :
32 R=287;
Example 12.02 12.02.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 1 2 . 0 2 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 12.02−data . s c i ’3 exec ( f i l ename )4 / / C h e c k i n g f o r c h o c k i n g :
5 c=pb/p0 ;6 i f ( c <=0.528)7 / / c h o k e d
8 else9 / / N o t c h o k e d
10 / / T h e r e f o r e p r e s s u r e a t e x i t = b a c k p r e s s u r e
11 pe=pb ;12 / / M a c h n u m b e r a t e x i t :
13 Me=(((p0/pe ) ˆ ( ( k−1)/k )−1)∗ (2/( k−1) ) ) ˆ0 .514 / / T e m p e r a t u r e a t e x i t ( i n K ) :
15 Te=T0/(1+(k−1)/2∗Meˆ2)
102
16 / / V e l o c i t y a t e x i t ( i n m/ s e c ) :
17 Ve=Me∗sqrt ( k∗R∗Te)18 / / D e n s i t y a t e x i t ( i n k g /m ˆ 3 ) :
19 de=pe∗10ˆ3/R/Te20 / / M a s s f l o w r a t e o f a i r ( k g / s e c ) :
21 m=de∗Ve∗Ae22 end ;23 printf ( ”\n\nRESULTS\n\n” )24 printf ( ”\n\nMach number at e x i t : %. 3 f \n\n” ,Me)25 printf ( ”\n\nMass f low ra t e o f a i r : %. 3 f kg/ sec \n\n” ,m)
Example 12.02d 12.02-data.sci
1 / / T h r o a t a r e a o f n o z z l e ( i n m ˆ 2 ) :
2 Ae=0.001;3 / / B a c k p r e s s u r e o f a i r ( i n k P a ) :
4 pb=591;5 / / S t a g n a t i o n p r e s s u r e ( i n k P a ) :
6 p0=1000;7 / / S t a g n a t i o n t e m p e r a t u r e ( i n K ) :
8 T0=333;9 / / G a s C o n s t a n t ( i n N−m/ kg −K ) :
10 R=287;11 / / V a l u e o f k :
12 k =1.4 ;
Example 12.03 12.03.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 1 2 . 0 3 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 12.03−data . s c i ’3 exec ( f i l ename )4 / / S a t u r a t i o n p r e s s u r e ( i n p s i a ) :
5 p0=p1∗(1+(k−1)/2∗M1ˆ2) ˆ( k /(k−1) )6 / / C h e c k i n g f o r c h o k i n g :
7 x=pb/p0 ;8 i f (x>0.528)9 / / N o t c h o k e d
10 else
103
11 / / c h o k e d
12 end13 / / A s t h e r e i s c h o k i n g :
14 Mt=1;15 / / V e l o c i t y a t e n t r y :
16 V1=M1∗sqrt ( k∗R∗(T1+460) ∗32 .2 )17 / / D e n s i t y a t t h e e n t r y ( i n l b m / f t ˆ 3 ) :
18 d1=p1 /(R∗(T1+460) ) ∗14419 / / M a s s f l o w r a t e ( i n l b m / s e c ) :
20 m=d1∗V1∗A121 / / F i n d i n g t h e v a l u e o f A1 / A ∗ ;
22 A=1/M1∗((1+(k−1)/2∗M1ˆ2) /(1+(k−1)/2) ) ˆ ( ( k+1) /(2∗ ( k−1) ) )23 / / F o r c h o k e d f l o w , A t =A ∗
24 At=A1/A25 printf ( ”\n\nRESULTS\n\n” )26 printf ( ”\n\nMach number at throat : %. 3 f \n\n” ,Mt)27 printf ( ”\n\nMass f low ra t e : %. 3 f lbm/ sec \n\n” ,m)28 printf ( ”\n\nArea at throat : %. 3 f f t ˆ2\n\n” ,At)
Example 12.03d 12.03-data.sci
1 / / M a c h n u m b e r a t e n t r y :
2 M1=0.52;3 / / T e m p e r a t u r e a t e n t r y ( i n F ) :
4 T1=40;5 / / P r e s s u r e a t e n t r y ( i n p s i a ) :
6 p1=60;7 / / A r e a a t e n t r y ( i n f t ˆ 2 ) :
8 A1=0.013;9 / / B a c k p r e s s u r e ( i n p s i a ) :
10 pb=30;11 / / G a s C o n s a n t ( i n f t − l b f / l bm −R )
12 R=53.3;13 / / V a l u e o f k :
14 k =1.4 ;
Example 12.04 12.04.sce
104
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 1 2 . 0 4 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 12.04−data . s c i ’3 exec ( f i l ename )4 / / T e m p e r a t u r e a t t h e t h r o a t ( i n K ) :
5 Tt=T0/(1+(k−1)/2∗Mtˆ2)6 / / P r e s s u r e a t t h r o a t ( i n k P a ) :
7 pt=p0∗(Tt/T0) ˆ( k /(k−1) )8 / / D e n s i t y a t t h r o a t ( i n k g /m ˆ 3 ) :
9 dt=pt ∗1000/R/Tt10 / / V e l o c i t y a t t h e t h r o a t ( i n m/ s ) :
11 Vt=Mt∗sqrt ( k∗R∗Tt)12 / / V a l u e o f A t / A ∗ :
13 Ax=1/Mt∗((1+(k−1)/2∗Mtˆ2) /(1+(k−1)/2) ) ˆ ( ( k+1) /(2∗ ( k−1) ))
14 / / S t a g n a t i o n p r o p e r t i e s a r e c o n s t a n t
15 / / A s a r e s u l t p r e s s u r e a t e x i t ,
16 pe=pb ;17 / / T h e M a c h n u m b e r a t t h e e x i t i s t h e r e f o r e g i v e n b y
18 Me=sqrt ( ( ( p0/pe ) ˆ ( ( k−1)/k )−1)∗2/(k−1) )19 / / C a l c u l a t i n g t h e v a l u e o f A e / A ∗ :
20 Ay=1/Me∗((1+(k−1)/2∗Meˆ2) /(1+(k−1)/2) ) ˆ ( ( k+1) /(2∗ ( k−1) ))
21 / / V a l u e o f A ∗ ( i n m ˆ 2 ) :
22 A star=Ae/Ay23 / / A r e a a t t h r o a t ( i n m ˆ 2 ) :
24 At=Ax∗A star25 printf ( ”\n\nRESULTS\n\n” )26 printf ( ”\n\nTemperature at the throat : %. 3 f K\n\n” ,Tt )27 printf ( ”\n\nPressure at throat : %. 3 f kPa\n\n” , pt )28 printf ( ”\n\nDensity at throat : %. 3 f kg/mˆ3\n\n” , dt )29 printf ( ”\n\ nVeloc i ty at the throat : %. 3 f m/ sec \n\n” ,Vt)30 printf ( ”\n\nMach number at the e x i t : %. 3 f \n\n” ,Me)31 printf ( ”\n\nArea at throat : %. 3 f mˆ2\n\n” ,At)
Example 12.04d 12.04-data.sci
1 / / S t a g n a t i o n t e m p e r a t u r e ( i n K ) :
2 T0=350;
105
3 / / S t a g n a t i o n p r e s s u r e ( i n k P a ) :
4 p0=1000;5 / / B a c k P r e s s u r e ( i n k P a ) :
6 pb=954;7 / / M a c h n u m b e r a t t h r o a t :
8 Mt=0.68;9 / / A r e a a t e x i t ( i n m ˆ 2 ) :
10 Ae=0.001;11 / / V a l u e o f k :
12 k =1.4 ;13 / / G a s C o n s t a n t ( i n N−m/ kg −K ) :
14 R=287;
Example 12.05 12.05.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 1 2 . 0 5 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 12.05−data . s c i ’3 exec ( f i l ename )4 / / M a c h n u m b e r a t t h e e x i t :
5 Me=sqrt ( ( ( p0/pe ) ˆ ( ( k−1)/k )−1)∗2/(k−1) )6 / / T e m p e r a t u r e a t e x i t ( i n K ) :
7 Te=T0/(1+(k−1)/2∗Meˆ2)8 / / M a s s f l o w r a t e ( i n k g / s ) :
9 m=pe∗1000∗Me∗sqrt ( k/R/Te)∗Ae10 printf ( ”\n\nRESULTS\n\n” )11 printf ( ”\n\nMach number at the e x i t : %. 3 f \n\n” ,Me)12 printf ( ”\n\nMass f low ra t e : %. 3 f kg/ sec \n\n” ,m)
Example 12.05d 12.05-data.sci
1 / / S t a g n a t i o n t e m p e r a t u r e ( i n K ) :
2 T0=350;3 / / S t a g n a t i o n p r e s s u r e ( i n k P a ) \
4 p0=1000;5 / / P r e s s u r e a t e x i t ( i n k P a )
6 pe =87.5;7 / / B a c k P r e s s u r e ( i n k P a ) :
8 pb=50;
106
9 / / A r e a a t e x i t ( i n m ˆ 2 ) :
10 Ae=0.001;11 / / G a s C o n s t a n t ( i n N−m/ kg −K )
12 R=287;13 / / V a l u e o f k :
14 k =1.4 ;
Example 12.06 12.06.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 1 2 . 0 6 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 12.06−data . s c i ’3 exec ( f i l ename )4 / / M a c h u m b e r a t s e c t i o n 1 :
5 M1=sqrt ( ( 2/ ( k−1) ∗ ( ( p0/p1 ) ˆ ( ( k−1)/k )−1) ) )6 / / T e m p e r a t u r e a t s e c t i o n 1 ( i n K ) :
7 T1=T0/(1+(k−1)/2∗M1ˆ2)8 / / D e n s i t y a t s e c t i o n 1 ( i n k g /m ˆ 3 ) :
9 d1=p1∗1000/R/T110 / / V e l o c i t y a t s e c t i o n 1 ( i n m/ s e c ) :
11 V1=M1∗sqrt ( k∗R∗T1)12 / / A r e a a t s e c t i o n 1 ( i n m ˆ 2 ) :
13 A1=%pi/4∗Dˆ214 / / M a s s f l o w r a t e ( i n k g / s e c ) :
15 m=d1∗A1∗V116 / / M a c h n u m b e r a t s e c t i o n 2 :
17 M2=sqrt ( ( 2/ ( k−1) ) ∗ ( (T0/T2)−1) )18 / / V e l o c i t y a t s e c t i o n 2 ( i n m/ s e c ) :
19 V2=M2∗sqrt ( k∗R∗T2)20 / / D e n s i t y a t s e c t i o n 2 ( i n k g /m ˆ 3 ) :
21 d2=d1∗V1/V222 / / P r e s s u r e a t s e c t i o n 2 ( i n k P a ) :
23 p2=d2/1000∗R∗T224 / / S t a g n a t i o n p r e s s u r e a t s e c t i o n 2 ( i n k P a ) :
25 p02=p2∗(1+(k−1)/2∗M2ˆ2) ˆ( k /(k−1) )26 / / F o r c e e x e r t e d o n c o n t r o l v o l u m e b y d u c t w a l l ( i n N ) :
27 F=(p2−p1 ) ∗1000∗A1+m∗(V2−V1)28 printf ( ”\n\nRESULTS\n\n” )29 printf ( ”\n\nMass f low ra t e : %. 3 f kg/ sec \n\n” ,m)
107
30 printf ( ”\n\nLocal i s e n t r o p i c s tagnat i on pr e s su r e ats e c t i o n 2 :%. 3 f kPa\n\n” , p02 )
31 printf ( ”\n\nForce exer ted on c o n t r o l volume by ductwa l l :%. 3 f N\n\n” ,F)
Example 12.06d 12.06-data.sci
1 / / D i a m e t e r o f p i p e ( i n m ) :
2 D=7.16∗10ˆ−3;3 / / S t a g n a t i o n p r e s s u r e ( i n k P a ) :
4 p0=101;5 / / S t a g n a t i o n t e m p e r a t u r e ( i n K ) :
6 T0=296;7 / / P r e s s u r e a t s e c t i o n 1 ( i n k P a ) :
8 p1 =98.5 ;9 / / T e m p e r a t u r e a t s e c t i o n 2 ( i n K ) :
10 T2=287;11 / / G a s c o n s t a n t ( i n N−m/ kg −K ) :
12 R=287;13 / / V a l u e o f k :
14 k =1.4 ;
Example 12.07 12.07.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 1 2 . 0 7 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 12.07−data . s c i ’3 exec ( f i l ename )4 / / M a c h n u m b e r a t s e c t i o n 1 :
5 M1= sqrt (2/( k−1) ∗ ( ( p0 /( p0+p1 ) ) ˆ ( ( k−1)/k )−1) )6 / / T e m p e r a t u r e a t s e c t i o n 1 ( i n K ) :
7 T1=T0/(1+(k−1)/2∗(M1) ˆ2)8 V1=M1∗sqrt ( k∗R∗T1)9 / / P r e s s u r e a t s e c t i o n 1 ( i n k P a ) :
10 p1=g∗dHg∗(760−18.9)∗10ˆ−311 / / D e n s i t y a t s e c t i o n 1 ( i n k g /m ˆ 3 ) :
12 d1=p1/R/T113 / / A t M1 = 0 . 1 9 0 ,
14 / / ( p / p ∗ ) 1 :
108
15 P1=5.74516 / / ( f L m a x / Dh ) 1 :
17 F1=16.3818 / / V a l u e o f L 1 3 ( i n m ) :
19 L13=F1∗D/ f20 / / V a l u e o f ( p / p ∗ ) 2 :
21 P2=p2/p1∗P122 / / F o r t h i s v a l u e , V a l u e o f M2 i s o b t a i n e d a s 0 . 4
23 M2=0.4;24 / / F o r M = 0 . 4 , f LmX / D = 2 . 3 0 9
25 F2=2.30926 / / V a l u e o f L 2 3 ( i n m ) :
27 L23=F2∗D/ f28 / / L e n g t h o f d u c t b e t w e e n s e c t i o n 1 a n d 2 ( i n m ) :
29 L12=L13−L2330 printf ( ”\n\nRESULTS\n\n” )31 printf ( ”\n\nLength o f duct r equ i r ed f o r choking from
s e c t i o n 1 : %3f m\n\n” , L13 )32 printf ( ”\n\nMach number s e c t i o n 2 : %. 3 f \n\n” ,M2)33 printf ( ”\n\Length o f duct between s e c t i o n 1 and 2 : %. 3 f
m\n\n” , L12 )
Example 12.07d 12.07-data.sci
1 / / S t a g n a t i o n t e m p e r a t u r e ( i n K ) :
2 T0=296;3 / / S t a g n a t i o n p r e s s u r e ( i n mm o f Hg ) :
4 p0=760;5 / / G a u g e p r e s s u r e a t s e c t i o n 1 ( i n mm o f Hg ) :
6 p1=−18.9;7 / / G a u g e p r e s s u r e a t s e c t i o n 2 ( i n mm o f Hg ) :
8 p2=−412;9 / / M a c h n u m b e r a t 3 :
10 M3=1;11 / / G a s c o n s t a n t :
12 R=287;13 / / D e n s i t y o f m e r c u r y ( k g /m ˆ 3 ) :
14 dHg=13500;
109
15 / / A c c e l e r a t i o n d u e t o g r a v i t y ( i n m/ s e c ˆ 2 ) :
16 g =9.8 ;17 / / F r i c t i o n f a c t o r :
18 f =0.0235;19 / / D i a m e t e r o f t u b e ( i n m ) :
20 D=7.16∗10ˆ−3;21 / / V a l u e o f k :
22 k =1.4 ;
Example 12.08 12.08.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 1 2 . 0 8 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 12.08−data . s c i ’3 exec ( f i l ename )4 / / D e n s i t y a t s e c t i o n 1 ( i n l b m / f t ˆ 3 ) :
5 d1=p1∗144/R/T16 / / V e l o c i t y a t s e c t i o n 2 ( i n f t / s e c ) :
7 V2=(p1−p2 ) ∗144/d1/V1∗32.2+V18 / / D e n s i t y a t s e c t i o n 2 ( i n l b m / f t 3 ) :
9 d2=d1∗V1/V210 / / T e m p e r a t u r e a t s e c t i o n 2 ( i n R ) :
11 T2=p2/d2/R∗14412 / / M a c h n u m b e r a t s e c t i o n 2 :
13 M2=V2/sqrt ( k∗R∗32.16∗T2)14 / / S t a g n a t i o n T e m p e r a t u r e a t s e c t i o n 2 ( i n R ) :
15 T02=T2∗(1+(k−1)/2∗M2ˆ2)16 / / S t a g n a t i o n p r e s s u r e a t s e c t i o n 2 ( i n p s i a ) :
17 p02=p2∗(T02/T2) ˆ( k /(k−1) )18 / / M a c h N u m b e r a t s e c t i o n 1 :
19 M1=V1/sqrt ( k∗R∗32.16∗T1)20 / / S t a g n a t i o n t e m p e r a t u r e a t s e c t i o n 1 ( i n R ) :
21 T01=T1∗(1+(k−1)/2∗M1ˆ2)22 / / E n e r g y a d d e d ( i n B t u / l b m ) :
23 E=Cp∗(T02−T01)24 / / C h a n g e i n e n t r o p y ( i n B t u / ( l bm −R ) ) :
25 dS=Cp∗ log (T2/T1)−(Cp−Cv)∗ log ( p2/p1 )26 printf ( ”\n\nRESULTS\n\n” )27 printf ( ”\n\ nVeloc i ty at s e c t i o n 2 : %. 3 f f t / sec \n\n” ,V2)
110
28 printf ( ”\n\nDensity at s e c t i o n 2 : %. 3 f lbm/ f t ˆ3\n\n” , d2)
29 printf ( ”\n\nTemperature at s e c t i o n 2 : %. 3 f R\n\n” ,T2)30 printf ( ”\n\nStagnat ion Temperature at s e c t i o n 2 : %. 3 f R
\n\n” ,T02)31 printf ( ”\n\nStagnat ion pr e s su r e at s e c t i o n 2 : %. 3 f p s i a
\n\n” , p02 )32 printf ( ”\n\nEnergy added : %. 3 f Btu/lbm\n\n” ,E)33 printf ( ”\n\nChange in entropy : %. 3 f Btu /( lbm−R)\n\n” ,dS
)
Example 12.08d 12.08-data.sci
1 / / T e m p e r a t u r e a t s e c t i o n 1 ( i n R ) :
2 T1=600;3 / / P r e s s u r e a t s e c t i o n 1 ( i n p s i a ) :
4 p1=20;5 / / P r e s s u r e a t s e c t i o n 2 ( i n p s i a ) :
6 p2=10;7 / / V e l o c i t y a t s e c t i o n 1 ( i n f t / s ) :
8 V1=360;9 / / C r o s s − s e c t i o n a l a r e a o f t h e d u c t ( i n f t ˆ 2 ) :
10 A=0.25;11 / / G a s C o n s t a n t ( i n f t − l b f / l bm −R ) :
12 R=53.3;13 / / V a l u e o f k :
14 k =1.4 ;15 / / S p e c i f i c h e a t a t c o n s t a n t p r e s u r e ( i n B t u / l bm −R ) :
16 Cp=0.24;17 / / S p e c i f i c h e a t a t c o n s t a n t v o l u m e ( i n B t u / l bm −R ) :
18 Cv=0.171;
Example 12.09 12.09.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 1 2 . 0 9 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 12.09−data . s c i ’3 exec ( f i l ename )4 / / M a c h n u b e r a t s e c t i o n 1 :
111
5 M1=V1/sqrt ( k∗R∗T1)6 / / F o r t h e s e v a l u e o f M1 a n d M2 , t h e f o l l o w i n g v a l u e s a r e
o b t a i n e d :
7 / / ( To / T0 ∗ ) 1 :
8 t01 =0.7934;9 / / ( T0 / T0 ∗ ) 2 :
10 t02 =0.9787;11 / / ( p 0 / p 0 ∗ ) 1 :
12 P01=1.503;13 / / ( p 0 / p 0 ∗ ) 2 :
14 P02=1.019;15 / / ( T / T ∗ ) 1 :
16 t1 =0.5289;17 / / ( T / T ∗ ) 2 :
18 t2 =0.9119;19 / / ( p / p ∗ ) 1 :
20 P1=0.3636;21 / / ( p / p ∗ ) 2 :
22 P2=0.7958;23 / / ( V / V ∗ ) 1 :
24 v1 =1.455;25 / / ( V / V ∗ ) 2 :
26 v2 =1.146;27 / / V a l u e o f T2 / T1 :
28 t=t2 / t129 / / T e m p e r a t u r e a t s e c t i o n 2 ( i n K ) :
30 T2=t∗T131 / / V a l u e o f p 2 / p 1 :
32 p=P2/P133 / / P r e s s u r e a t s e c t i o n 2 ( i n k P a ) :
34 p2=p∗p135 / / V a l u e o f V2 / V1 :
36 v=v2/v137 / / V e l o c i t y a t s e c t i o n 2 ( i n m/ s e c ) :
38 V2=v∗V139 / / D e n s i t y a t s e c t i o n 2 ( i n k g /m ˆ 3 ) :
40 d2=p2∗1000/R/T241 / / A t M1 , T / T0 = 0 . 5 5 5 6
112
42 T01=T1/0.555643 / / A t M2 , T / T0 = 0 . 7 7 6 4
44 T02=T2/0.776445 / / H e a t a d d e d ( i n k J / k g ) :
46 E=Cp∗(T02−T01)47 / / C h a n g e i n e n t r o p y ( k J / k g −K ) :
48 dS=Cp∗ log (T2/T1)−R∗ log ( p2/p1 ) /100049 printf ( ”\n\nRESULTS\n\n” )50 printf ( ”\n\nTemperature at s e c t i o n 2 : %. 3 f K\n\n” ,T2)51 printf ( ”\n\nPressure at s e c t i o n 2 : %. 3 f kPa\n\n” , p2 )52 printf ( ”\n\ nVeloc i ty at s e c t i o n 2 : %. 3 f m/ sec \n\n” ,V2)53 printf ( ”\n\nDensity at s e c t i o n 2 : %. 3 f kg/mˆ3\n\n” , d2 )54 printf ( ”\n\nStagnat ion temperature at s e c t i o n 2 : %. 3 f K
\n\n” ,T02)55 printf ( ”\n\nHeat added : %. 3 f kJ/kg\n\n” ,E)56 printf ( ”\n\nChange in entropy : %. 3 f kJ/kg\n\n” ,dS)
Example 12.09d 12.09-data.sci
1 / / T e m p e r a t u r e a t s e c t i o n 1 ( i n K ) :
2 T1=333;3 / / P r e s s u r e a t s e c t i o n 1 ( i n k P a ) :
4 p1=135;5 / / V e l o c i t y a t s e c t i o n 1 ( i n m/ s e c ) :
6 V1=732;7 / / M a c h n u m b e r a t s e c t i o n 2 :
8 M2=1.2;9 / / G a s c o n s t a n t ( i n N−m/ kg −K ) :
10 R=287;11 / / V a l u e o f k :
12 k =1.4 ;13 / / S p e c i f i c h e a t a t c o n s t a n t p r e s u r e ( i n N−m/ kg −K ) :
14 Cp=1;
Example 12.10 12.10.sce
1 pathname=g e t a b s o l u t e f i l e p a t h ( ’ 1 2 . 1 0 . s ce ’ )2 f i l ename=pathname+f i l e s e p ( )+’ 12.10−data . s c i ’
113
3 exec ( f i l ename )4 / / D e n s i t y a t s e c t i o n 1 ( i n k g /m ˆ 3 ) :
5 d1=p1∗1000/R/T16 / / M a c h n u m b e r a t s e c t i o n 1 :
7 M1=V1/sqrt ( k∗R∗T1)8 / / S t a g n a t i o n t e m p e r a t u r e a t s e c t i o n 1 ( i n K ) :
9 T01=T1∗(1+(k−1)/2∗M1ˆ2)10 / / S t a g n a t i o n p r e s s u r e a t s e c t i o n 1 ( i n k P a ) :
11 p01=p1∗(1+(k−1)/2∗M1ˆ2) ˆ( k /(k−1) )12 / / T h e f o l l o w i n g v a l u e s a r e o b t a i n e d f r o m t h e a p p e n d i x :
13 / / p o 2 / p 0 1 :
14 p0 =0.7209;15 / / T2 / T1 :
16 T=1.687;17 / / p 2 / p 1 :
18 p=4.5;19 / / V2 / V1 :
20 V=0.3750;21 / / T e m p e r a t u r e a t s e c t i o n 2 ( i n K ) :
22 T2=T∗T123 / / P r e s s u r e a t s e c t i o n 2 ( i n k P a ) :
24 p2=p∗p125 / / V e l o c i t y a t s e c t i o n 2 ( i n m/ s e c ) :
26 V2=V∗V127 / / D e n s i t y a t s e c t i o n 2 ( i n k g /m ˆ 3 ) :
28 d2=p2∗1000/R/T229 / / S t a g n a t i o n p r e s s u r e a t s e c t i o n 2 ( i n k P a ) :
30 p02=p0∗p0131 / / S t a g n a t i o n t e m p e r a t u r e a t s e c t i o n 2 ( i n K ) :
32 T02=T01 ;33 / / C h a n g e i n e n t r o p y ( i n k J / ( k g −K ) ) :
34 dS=−R/1000∗ log ( p0 )35 printf ( ”\n\nRESULTS\n\n” )36 printf ( ”\n\nTemperature at s e c t i o n 2 : %. 3 f K\n\n” ,T2)37 printf ( ”\n\nPressure at s e c t i o n 2 : %. 3 f kPa\n\n” , p2 )38 printf ( ”\n\ nVeloc i ty at s e c t i o n 2 : %. 3 f m/ sec \n\n” ,V2)39 printf ( ”\n\nDensity at s e c t i o n 2 : %. 3 f kg/mˆ3\n\n” , d2 )40 printf ( ”\n\nStagnat ion pr e s su r e at s e c t i o n 2 : %. 3 f kPa\
114
n\n” , p02 )41 printf ( ”\n\nChange in entropy : %. 3 f kg−K\n\n” ,dS)42 printf ( ”\n\nStagnat ion temperature at s e c t i o n 2 : %. 3 f K
\n\n” ,T02)
Example 12.10d 12.10-data.sci
1 / / T e m p e r a t u r e a t s e c t i o n 1 ( i n K ) :
2 T1=278;3 / / P r e s s u r e a t s e c t i o n 1 ( i n k P a ) :
4 p1=65;5 / / V e l o c i t y a t s e c t i o n 1 ( i n m/ s e c ) :
6 V1=668;7 / / V a l u e o f g a s c o n s t a n t ( i n N−m/ kg −K ) :
8 R=287;9 / / V a l u e o f k :
10 k =1.4 ;
115