Solution Guide and Solutions of the Problems
Problems of Chapter 2
The nominal density at 70 km altitude is рж = pnom = 8.283-10-5 kg/m3, Table 2.1. The actual density is pactuai = 7.4Б47-10-5 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
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: 10.1007/978-3-319-14373-6 _11
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.
Determine the density in the test section, take the simple approximation for the viscosity of air, Section 4.2, p = 0.702-10-7 T, find A facility = 0.00003 m, and Dfacility = 0.0001 m.