Air (gama = 1.4. molecular weight=28.97) in a large tank is to be accelerated th

Question

Air (gama = 1.4. molecular weight=28.97) in a large tank is to be accelerated through a converging- diverging nozzle to an exit Mad, number of 3.0. The air in the tank is at a stagnation pressure of 400 kPa and Agnation temperature of 300K. The nozzle exit area is 60 cm2. Assume the process is isentropic and determine the exit temperature (K) and exit pressure (kpa) the throat area (cm2), and the mass flow rate (kg/sec) If the stagnation pressure is raised to 4000 kPa, what happens to the exit velocity (higher, lower, same)?

Explanation / Answer

a)inlet velocity Vi =0 [stagnent condition]

exit velocity V0 = mach no. * speed of sound in air =3*343 =1029 m/s

cross sectional area A=60cm2

density at the inlet = 4.7124 kg/m3 [using table T=300 K , P=400 kPa]

assuming air density will remain constant

using bernoullie equation

0.5Vi2 + Pi= 0.5V02 + Po [ height is same at inlet and outlet so the height term will get cancelled]

P0=0.5[Vi2 -V02] + Pi

Pi = pressure at inlet =400 kPa = 4x105 Pa

putting the values

outlet pressure P0 =4x105 -0.5*4.7124*(10292 -0) = 200 kPa

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