Question

Consider an imaginary method to generate a supersonic flow in a two-step process of heating and cooling. Heat is first added to air at a low Mach number, increasing the Mach number until the flow is thermally choked at M = 1. Heat is then removed from the air, which further increases the Mach number (this is the imaginary part, that you were able to jump over M = 1 to the supersonic branch). Air at a pressure of 100 kPa and temperature of 293 K flows steadily in a frictionless pipe with constant area of 0.1 m2, with an inlet Mach number of 0.25. Calculate:

(a) The heat addition rate (J=s) required to reach M = 1

(b) The heat extraction rate (J=s) required to go from M = 1 to M = 1:5.

3. Consider an imaginary method to generate a supersonic flow in a two-step process of heating and cooling. Heat is first added to air at low Mach number, increasing the Mach number until the flow is thermally choked at M = 1. Heat is then removed from the air, which further increases the Mach number (this is the imaginary part, that you were able to jump over M = 1 to the supersonic branch). Air at a pressure of 100 kPa and temperature of 293 K flows steadily in a frictionless pipe with constant area of 0.1 m², with an inlet Mach number of 0.25. Calculate (a) The heat addition rate (J/s) required to reach M = 1 (b) The heat extraction rate (J/s) required to go from M = 1 to M = 1.5.

Answer 1

na = Heat added + air Album LLLLLLLL LUNCU Annnnnnnnnnnnnnnnny Exit Entry M2=1 My = 0.25 pressure (A)=100 kPa .fricffonless pipe · Steady flow Teusperature (T) = 293K (Properties #f(t)) Area of duck (A) = 0.1 m2 (a) frous Energy balance; (stagnaffon enthalpy), + Heat added ) =(stagnation enthalpy), milhitus) + Q= m Khatuny Pů = MRT, rů volume flow rate on I us= mass flow date f U = velocity of apy | c= velocity of sound / V TPRT Vi= 0.25 x 11.9x287* 293 = 85.78 usis V = AV, = 8.578 us 3s 0.25-

Now, ima iu RT, i m m = 10088.578 0.2874293 [m= 10.2 kg/ (GPT. + ) = wis (CPTz tuz) test -03 M21 ve 3 V2 20 = SORT (V2=343.11 usja * Tia Tz (Steady flow) :: wish tops is used 82 m (52_m2) Q = 10,2 (343-12_88.782) Q = 562867.945 3/8 Q = 562-8679 k Te H Heat addition rate o (6) AiX - ↑ IAE = 1 Heat removed M2=105

k, = - 1 = -c = RT M2 = 15= g Vs = 115C =154J VRT UUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUU V2 = 115*17.4x2877 293 TV2 = 514.67 usia 7 spustellarly frous Energy balance, Q = us ( Vz2_112) on 10-2 (cian 0= 10,2 (514-672-343:112) ( .consexved) o n Limass us 1 .. is = 10-2kg Q = 750519.75 3/5 la = 750.519 kils Heat extraction rate