Mach number applications · RMPH 2016 #B3

Physics · fluid-mechanics

Problem

For classifying the speed performances of bodies in a fluid (e.g. aircrafts), as well as the flow regimes of fluids, the Swiss aeronautical engineer Jakob Ackeret (1898 – 1981) – one of the leading authorities in the 20th century aeronautics – proposed in 1929 that the ratio of the body or of the fluid's speed vv and the local sound speed cc in that fluid to be called Mach's number M=vcM = \frac{v}{c} after the name of the great Czech (then in the Austrian empire) physicist and philosopher Ernst Mach (1838 – 1916). Primarily, the value of this non-dimensional quantity delimitates the incompressible from the compressible behavior of a flowing fluid, in aeronautics this limit being settled to M=0.3M = 0.3. B3.1 Find the relative variation of the gas density Δρρ\frac{\Delta \rho}{\rho} as a function of Mach's number, when its motion is slowed down to a stop, its initial velocity being vcv_c, and calculate its maximum value for a flow to be considered incompressible. B3.2 The pressure at the nose of an aircraft in flight was found to be p=105 kPap = 105\text{ kPa} and the speed of air relative to the aircraft was zero at this point. The pressure and temperature of the undisturbed air were p0=95 kPap_0 = 95\text{ kPa} and T0=280 KT_0 = 280\text{ K} respectively. The adiabatic exponent for this temperature is γ=1.4\gamma = 1.4. Find the speed and the Mach number of the aircraft. B3.3 When a gas is flowing through a pipe, it exerts a friction force on the fluid, which is not always negligible. If at the entrance of such a pipe the static pressure in the flowing fluid is p1=300 kPap_1 = 300\text{ kPa} and the Mach number is M1=0.2M_1 = 0.2, while at the exit M2=0.25M_2 = 0.25, find the expression and the numerical value of the force with which the fluid is acting on the pipe. The adiabatic exponent is γ=1.4\gamma = 1.4, the constant cross section of the pipe is A=0.01 m2A = 0.01\text{ m}^2, and the relative increase of the gas temperature through the pipe is T2−T1T1=0.06\frac{T_2 - T_1}{T_1} = 0.06.

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