{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

PS5_solution

PS5_solution - AAE 334 PS-5 SOLUTIONS Note In general the...

This preview shows pages 1–5. Sign up to view the full content.

AAE 334 PS-5 SOLUTIONS Note: In general, the final values of flow variables are presented with 5 significant digits .The numbers used in the calculations are of 7 significant digits. The symbol * represents arithmetic multiplication. These solutions make use of Appendix B and the chart for problems 1a-3a and the functions obshk and nshk, for the rest. PROBLEM 1a The turning angle , since the flat plate is aligned with the flow. The Mach number corresponding to different wave angles is given below in table-1. Table-1 Wave angle Mach number 60 1.15 30 2.00 15 3.85 We see that as the wave angle decreases, the upstream Mach number of the flow increases. We also notice that , the Mach angle. PROBLEM 1b From the chart, for and M=3.6, The normal component ( ) From appendix B, for , ( ) ( )

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
We have from standard isentropic relations for M1=3.6, ( ) ( ) Therefore, The entropy increase can be calculated using the formula ( ) ( ) PROBLEM 1c From the chart, for and M=3.6, The normal component ( ) ( ) From appendix B, for , We have from standard isentropic relations for M1=3.6, ( )
( ) Therefore, Also from appendix B, we have This value of P02 is much lower than that computed from the ratio , and this value is incorrect. PROBLEM 1d The strongest weak shock solution is found by locating the maximum deflection angle for the M=2.8 curve on the chart, . The corresponding maximum wave angle is determined to be The normal component of the Mach number M1 is then ( ) ( ) From appendix B for this Mach number we find Hence PROBLEM 1e From the chart, for and M=3.0 , we find The normal component ( ) ( ) From appendix B, for , ( ) ( )

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Also The downstream flow behind the oblique shock is supersonic, and hence a bow shock is formed in front of the pitot probe. We hence make use of the Rayleigh-Pitot formula in Appendix B.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}