Question 13.11: Air flows through the 100-mm-diameter pipe in Fig. 13–23 hav......

Fluid Description.   We assume steady isentropic flow through the nozzle.
Analysis.   The diameter d can be determined from continuity of the mass flow, which requires
\dot{m}  =  \rho_1 V_1 A_1  =  \rho_2 V_2 A_2                                     (1)
We must first find M_1 and then find the densities and velocities at 1 and 2.
The stagnation values for atmospheric air are determined from Appendix A [ table A.1 , table A.2 , table A.3 , table A.4 , table A.5 ] as p_0 = 101.3 kPa,T_0 = 15°C, and \rho_0 = 1.23 kg /m³.
Knowing p_1 and p_0, we can now determine M_1 at the entrance 1 of the nozzle using Eq. 13–27.

M_1 = 0.4146
As expected, M_1 < M_2 = 0.7.
Applying Eq. 13–26 to find the temperatures at the entrance and exit, we have

T_1 = 278.43 K

T_2 = 262.30  K

Thus, the velocities at the entrance and exit are
V_1  =  M_1\sqrt{kRT_1}  =  (0.4146)\sqrt {1.4 (286.9  J/kg · K) (278.4  K)}= 138.63 m/s

V_2  =  M_2\sqrt{kRT_2}  =  (0.7)\sqrt{1.4 (286.9  J/kg · K) (262.3  K)}  =  227.21  m/s
The density of the air at the entrance and exit of the nozzle is determined using Eq. 13–28. \rho _0  =  \rho(1  +  \frac{k  –  1}{2} M_1^2)^{1/(k-1)}

Finally, applying Eq. 1,
\rho_1 V_1 A_1  = \rho_2 V_2 A_2
(1.1304  kg/m^3) (138.63  m/s) [\pi (0.05  m)^2]  =  (0.9736  kg/m^3) (227.21 m/s) \pi (\frac{d}{2})^2
d = 84.2 mm

SOLUTION II
We can also solve this problem in a direct manner using Table B–1, even though subsonic flow occurs at the end of the nozzle. To do so, we will make reference to a phantom extension of the nozzle, where M = 1 and A = A*, and then relate the area ratios A_1 for M_1 = 0.4146 and A_2 for M_2 = 0.7 to this reference.

Thus,

Using Table B–1, we therefore have

d = 84.2 mm