Problems of Chapter 6

Problem 6.1

The relation for the maximum speed is eq. (6.15), which yields with cp = 1004.71 m2/s2K for air as perfect gas the maximum speed Vm = 1,736,12 m/s.

Problem 6.2

The critical speed is found from eq. (6.19) and is u* = 708.77 m/s. Problem 6.3

The pressure coefficient is found with eq. (6.37): cp = a) indeterminate solu­tion, b) 1.000025, c) 1.0025, d) 1.064, e) 1.276.

The result a) points to the fact that eq. (6.37) is not valid for incompress­ible flow, remember Problem 4.5. For incompressible flow eq. (6.38) with u = 0 at the stagnation point yields the correct value cp = 1.

Problem 6.4

The pressure coefficient is found with eq. (6.65): cp = a) 1.276, b) 1.809, c) 1.832, d) 1.839.

Problem 6.5

Eq. (6.120) in the simplified form must be formulated in terms of the body radius Rb:

The shock stand-off distances are A0 = a) 0.25 m, b) 0.21 m, c) 0.2 m. With increasing Mach number the stand-off distance becomes smaller.

Problem 6.6

A0 = a) 0.146 m, b) 0.11 m, c) 0.1 m. With decreasing Yeff the stand-off distance becomes smaller.

Problem 6.7

A0 = a) 0.042 m, b) 0.01 m, c) 0 m. With Yeff = 1, the stand-off distance becomes even smaller. For Ыж ^ ж the stand-off distance becomes zero.

Problem 6.8

At H = 30 km altitude the relevant free-stream parameters are Тж = 226.509 K and рж = 1.841-10-2 kg/m3. The dynamic pressure is qж = 30,165.62 Pa. The resultant forces and the moment are L = 1.22-106 N, D = 0.128-106 N, M = 48.8-106 Nm. The lift-to-drag ratio is L/D = 9.51.

Problem 6.9

At H = 40 km altitude the relevant free-stream parameters are Tж = 250.35 K and рж = 3.996-10~3 kg/m3. The dynamic pressure is qж = 7,236.54 Pa. The resultant forces and the moment are L = 0.292-106 N, D = 0.031-106 N, M = 11.7-106 Nm. The lift-to-drag ratio is L/D = 9.51.

Problem 6.10

At the higher altitude the density рж is smaller. Hence the forces are smaller. Increasing the angle of attack would enlarge the lift, but also the drag. In­creasing the wing area would increase the wing’s mass. This would make necessary a larger lift.

During a cruise flight the fuel mass decreases with flight time. If the angle of attack is kept constant this would mean an increase of the flight altitude with time. The flight altitude hence is not constant.

Problem 6.11

At higher altitudes we have в = 1.40845-10~4 1/m. The density at sea level is PH=0km = 1.225 kg/m3.

For Mж = 6 at T* = 242.55 K we find vж = 1873.28.8 m/s. With q= 60 kPa this gives рж = 0.0342 kg/m3. The altitude is found from the inverted eq. (2.3) to be

H=-~ InjW – « 25 km. в PH=0km