Question 13.7: Air is drawn through the 50-mm-diameter pipe in Fig. 13–17 a......
Air is drawn through the 50-mm-diameter pipe in Fig. 13–17 and passes section 1 with a speed of V_1 = 150 m/s, while it has an absolute pressure of p_1 = 400 kPa and a temperature of T_1 = 350 K. Determine the required area of the throat of the nozzle to produce sonic flow at the throat. Also, if supersonic flow is to occur at section 2, find the velocity, temperature, and required pressure at this location.
Fluid Description. We assume isentropic steady flow through the nozzle.
Throat Area. The Mach number at section 1 is first calculated. For air k = 1.4, R = 286.9 J/kg · K. Therefore,
M_1 = \frac{V_1}{\sqrt{kRT_1}} = \frac{150 m/s}{\sqrt{1.4 (86.9 J/kg · K) (350 K)}} = 0.40
Although we can use Eq. 13–36 \frac{A}{A^*}=\frac{1}{M}[\frac{1+\frac{1}{2}(k-1)M^2}{\frac{1}{2}(k+1)}]^{\frac{k+1}{2(k-1)}} with M_1 and A_1 to determine the throat area A*, it is simpler to use Table B–1 since k = 1.4. Thus, for M_1 = 0.40, we find
\frac{A_1}{A*} = 1.5901A* = \frac{\pi (0.025 m)^2}{1.5901}
= 0.0012348 m² = 0.00123 m²
Properties at Section 2. Now that A* is known, the Mach number at A_2 can be determined from Eq. 13–36; however, again it is simpler to use Table B–1, with the ratio
We get M_2 = 2.8232 (approximately), because supersonic flow must occur at the end of the divergent section. (The other root, M_1 = 0.1645
refers to subsonic flow at the exit.)
The temperature and pressure at the exit can be determined using M = 2.8230 and Eqs. 13–26 and 13–27; however, first we must know
the stagnation values T_0 and p_0. To find them we can again use Eqs. 13–26 T_0=T(1+\frac{k-1}{2}M^2) and 13–27 p_0=p(1+\frac{k-1}{2}M^2)^{k/(k-1)} with M_1 = 0.40 and T_1 and p_1. A simpler method is to use Table B–1 by referencing the ratios T_2/T_1 and p_2/p_1 in terms of the stagnation ratios as follows:
\frac{p_2}{p_1} = \frac{p_2/p_0}{p_1/p_0} = \frac{0.03557}{0.8956} = 0.03972
Therefore, without having to find T_0 and p_0, we have
T_2 = 0.3978T_1 = 0.3978 (350 K) = 139.23 K = 139 K
p_2 = 0.03972p_1 = 0.03972 (400 kPa) = 15.9 kPa
This lower pressure at the exit is what draws the air through the 50-mm-diameter pipe at 150 m/s.
The average velocity of the air at section 2 is
V_2 = M_2 \sqrt{kRT_2} = 2.8232 \sqrt {1.41286.9 J/kg · K) (139.23 K)}
= 668 m/s