TITLE.PM5

858 ENGINEERING THERMODYNAMICS dharm \M-therm\Th16-1.pm5 Differentiating the above equation, we get d(ρAV) = 0 or ρd(AV) + AVdρ ...

COMPRESSIBLE FLOW 859 dharm \M-therm\Th16-1.pm5 Dividing both sides by g, we get p ρg loge p + V g 2 2 + z = constant ...(16.6) ...

860 ENGINEERING THERMODYNAMICS dharm \M-therm\Th16-1.pm5 Sol. Section 1 : Velocity of the gas, V = 300 m/s Pressure, p 1 = 78 kN ...

COMPRESSIBLE FLOW 861 dharm \M-therm\Th16-1.pm5 Substituting the values in eqn. (ii), we get 1.4 1.4− F HG I 1 KJ × 89831 1 1 − ...

862 ENGINEERING THERMODYNAMICS dharm \M-therm\Th16-1.pm5 γ γρ γ γ − F HG I KJ − F HG I KJ R S | T | U V | W | − 1 (^11) 1 2 1 1 ...

COMPRESSIBLE FLOW 863 dharm \M-therm\Th16-1.pm5 Piston Wave front Rigid pipe dx = Vdt (dL – dx) (dL = Cdt) V C Fig. 16.1. One di ...

864 ENGINEERING THERMODYNAMICS dharm \M-therm\Th16-1.pm5 ∴ C = dp dρ ...(16.10) 16.3.2. Sonic velocity in terms of bulk modulus ...

COMPRESSIBLE FLOW 865 dharm \M-therm\Th16-1.pm5 16.3.4. Sonic velocity for adiabatic process For isentropic (reversible adiabati ...

866 ENGINEERING THERMODYNAMICS dharm \M-therm\Th16-1.pm5 (ii) If for any flow system the Mach number is less than about 0.4, the ...

COMPRESSIBLE FLOW 867 dharm \M-therm\Th16-1.pm5 Consider a tiny projectile moving in a straight line with velocity V through a c ...

868 ENGINEERING THERMODYNAMICS dharm \M-therm\Th16-1.pm5 the projectile at point ‘B’ is outside the spheres formed due to format ...

COMPRESSIBLE FLOW 869 dharm \M-therm\Th16-1.pm5 Speed of the aircraft, V : Refer Fig. 16.3. Let O represent the observer and A t ...

870 ENGINEERING THERMODYNAMICS dharm \M-therm\Th16-1.pm5 and ps, Vs, ρs and Ts are corresponding values of pressure, velocity de ...

COMPRESSIBLE FLOW 871 dharm \M-therm\Th16-1.pm5 For adiabatic process, the sonic velocity is given by, C = γγRT ρ p = Q p RT ρ = ...

872 ENGINEERING THERMODYNAMICS dharm \M-therm\Th16-1.pm5 Substituting the value of M 02 in eqn. (16.18), we get ps = p 0 + pV p ...

COMPRESSIBLE FLOW 873 dharm \M-therm\Th16-1.pm5 Substituting the values of ps and ρs from eqns. (16.17) and (16.18), we get Ts = ...

874 ENGINEERING THERMODYNAMICS dharm \M-therm\Th16-1.pm5 (ii) Stagnation temperature, Ts : The stagnation temperature is given b ...

COMPRESSIBLE FLOW 875 dharm \M-therm\Th16-1.pm5 Stagnation density, ρs : ρs = p RT s s = × × 131.27 10 287 358.4 3 = 1.276 kg/m^ ...

876 ENGINEERING THERMODYNAMICS dharm \M-therm\Th16-1.pm5 Speed of the aircraft, Va : p 0 V = mRT 0 = m × R M F 0 HG I KJ^ T^0 or ...

COMPRESSIBLE FLOW 877 dharm \M-therm\Th16-1.pm5 Substituting the value of dρ ρ in eqn. (16.24), we get dV V + dA A – VdV C^2 = 0 ...