Question 12.4: Isentropic Flow of Air in a Nozzle Air enters a nozzle at 20...

Air enters a nozzle at specified temperature, pressure, and velocity.
The exit pressure, exit temperature, and exit-to-inlet area ratio are to be determined for a Mach number of Ma = 1 at the exit.
Assumptions   1  Air is an ideal gas with constant specific heats at room temperature. 2  Flow through the nozzle is approximated as steady, one-dimensional, and isentropic.
Properties   The properties of air are k = 1.4 and c_p = 1.005  kJ/kg·K (Table A–1).
Analysis   The properties of the fluid at the location where Ma = 1 are the critical properties, denoted by superscript *. We first determine the stagnation temperature and pressure, which remain constant throughout the nozzle since the flow is isentropic.

T_0=T+\frac{V_i^2}{2 c_p}=350  K +\frac{(135  m / s )^2}{2(1.005  kJ / kg \cdot K )}\left(\frac{1  kJ / kg }{1,000  m ^2 / s ^2}\right)=359.1  K

P_0=P_i\left(\frac{T_0}{T_i}\right)^{k /( k  –  1)}=(200  kPa )\left(\frac{359.1  K }{350  K }\right)^{1.4 /(1.4  –  1)}=218.8  kPa

From Table A–13 (or from Eqs. 12–18 and 12–19) at Ma = 1, we read

\frac{T_0}{T}=1+\left(\frac{k-1}{2}\right) Ma ^2         (12.18)

\frac{P_0}{P}=\left[1+\left(\frac{k-1}{2}\right) Ma ^2\right]^{k /(k-1)}             (12.19)

T/T_0 = 0.8333
P/P_0 = 0.5283

Thus,

T = 0.8333T_0 = 0.8333(359.1  K) = 299.2  K
P = 0.5283P_0 = 0.5283(218.8  kPa) = 115.6  kPa

Also,

c_i=\sqrt{k R T_i}=\sqrt{(1.4)(0.287  kJ / kg \cdot K )(350  K )\left(\frac{1000  m ^2 / s ^2}{1  kJ / kg }\right)}
= 375.0 m/s

Ma _i=\frac{V_i}{c_i}=\frac{135  m / s }{375  m / s }=0.3600

From Table A–13 at this Mach number we read A_i /A^* = 1.7681. Thus the ratio of the throat area to the nozzle inlet area is

\frac{A^*}{A_i}=\frac{1}{1.7681}=0.566

Discussion   If we solve this problem using the relations for compressible isentropic flow, the results would be identical to at least three significant digits.