Solution Guide and Solutions of the Problems
Problems of Chapter 2
Problem 2.1
The nominal density at 70 km altitude is рж = pnom = 8.283105 kg/m3, Table 2.1. The actual density is pactuai = 7.4Б47105 kg/m3.
Assume that the RV can be approximated by a flat plate and determine the drag with the help of Newton’s theory, Section 6.7:
CD = 2 sin3 a.
This relation yields
CDnom = 0.707.
Because the actual drag is to be the nominal one, we have pnomCDnom = pactUaiCDcorr. We obtain from this CDcorr = 0.78Б.
Make a Taylor expansion around the actual drag coefficient and keep the linear term only:


From that equation the correction angle Aa is found to be
, dCD m
da
With






the correction angle is Aa = 0.037 rad, respectively Aa = 2.12°.
With the corrected angle acorr = 47.12° the value CDcorr = 0.787 is obtained.
(C Springer International Publishing Switzerland 2015 E. H. Hirschel, Basics of Aerothermodynamics,
DOI: 10.1007/9783319143736 _11
Problem 2.2
Remember the mass fractions of the undisturbed air and find eventually with eq. (2.6): a) pn2 = 912.98 Pa, po2 = 284.05 Pa, b) pn2 = 0.803 Pa, po2 = 0.250 Pa.
Problem 2.3
Determine the density in the test section, take the simple approximation for the viscosity of air, Section 4.2, p = 0.702107 T, find A facility = 0.00003 m, and Dfacility = 0.0001 m.
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