Air flows through the 100-mm-diameter pipe in Fig. 13–23 having an absolute pressure of p1 = 90 kPa. Determine the diameter d at the end of the nozzle so that isentropic flow occurs out of the nozzle at M2 = 0.7. The air within the pipe is taken from a large reservoir at standard atmospheric

Question

Air flows through the 100-mm-diameter pipe in Fig. 13–23 having an absolute pressure of [latex] p_1 [/latex] = 90 kPa. Determine the diameter d at the end of the nozzle so that isentropic flow occurs out of the nozzle at [latex] M_2 [/latex] = 0.7. The air within the pipe is taken from a large reservoir at standard atmospheric pressure and temperature.

Accepted Answer

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

[latex] \dot{m}= ho_1 V_1 A_1= ho_2 V_2 A_2 [/latex] (1)

We must first find [latex] M_1 [/latex] and then find the densities and velocities at 1 and 2.

The stagnation values for atmospheric air are determined from Appendix A as [latex] p_0=101.3 kPa , T_0=15^{\circ} C ext {, and } ho_0=1.23 kg / m ^3 [/latex].
Knowing [latex]p_1[/latex] and [latex]p_0[/latex], we can now determine [latex] M_1 [/latex] at the entrance 1 of the nozzle using Eq. 13–27.

[latex] p_0=p_1\left(1+\frac{k-1}{2} M _1^2 ight)^{k /(k-1)} [/latex] (13-27)

[latex] \begin{gathered} 101.3 kPa =(90 kPa )\left(1+\frac{1.4-1}{2} M _1^2 ight)^{1.4 /(1.4-1)} \ M _1=0.4146 \end{gathered} [/latex]

As expected, [latex] M _1< M _2=0.7 [/latex].
Applying Eq. 13–26 to find the temperatures at the entrance and exit, we have

[latex] T_0=T_1\left(1+\frac{k-1}{2} M _1^2 ight) [/latex] (13-26)

[latex] \begin{gathered} (273+15) K =T_1\left(1+\frac{1.4-1}{2}(0.4146)^2 ight) \ T_1=278.43 K \ T_0=T_2\left(1+\frac{k-1}{2} M _2^2 ight) \ (273+15) K =T_2\left(1+\frac{1.4-1}{2}(0.7)^2 ight) \ T_2=262.30 K \end{gathered} [/latex]

Thus, the velocities at the entrance and exit are

[latex] \begin{aligned} & V_1= M _1 \sqrt{k R T_1}=(0.4146) \sqrt{1.4(286.9 J / kg \cdot K )(278.4 K )}=138.63 m / s \ & V_2= M _2 \sqrt{k R T_2}=(0.7) \sqrt{1.4(286.9 J / kg \cdot K )(262.3 K )}=227.21 m / s \end{aligned} [/latex]

The density of the air at the entrance and exit of the nozzle is determined using Eq. 13–28.

[latex] ho_0= ho_1\left(1+\frac{k-1}{2} M _1^2 ight)^{1 /(k-1)} [/latex]

[latex] \begin{aligned} & 1.23 kg / m ^3= ho_1\left[1+\frac{1.4-1}{2}(0.4146)^2 ight]^{1 /(1.4-1)} \ & ho_1=1.1304 kg / m ^3 \end{aligned} [/latex]

[latex] \begin{gathered} ho_0= ho_2\left(1+\frac{k-1}{2} M _2^2 ight)^{1 /(k-1)} \quad 1.23 kg / m ^3= ho_2\left[1+\frac{1.4-1}{2}(0.7)^2 ight]^{1 /(1.4-1)} \ ho_2=0.9736 kg / m ^3 \end{gathered} [/latex]

Finally, applying Eq. 1,

[latex] \begin{gathered} ho_1 V_1 A_1= ho_2 V_2 A_2 \ \left(1.1304 kg / m ^3 ight)(138.63 m / s )\left[\pi(0.05 m )^2 ight]=\left(0.9736 kg / m ^3 ight)(227.21 m / s ) \pi\left(\frac{d}{2} ight)^2 \ d=84.2 mm \end{gathered} [/latex]

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 [latex] A_1 ext { for } M _1=0.4146 [/latex] and [latex] A_2 ext { for } M _2=0.7 [/latex] to this reference.

[latex] \frac{A_2}{A_1}=\frac{A_2 / A^*}{A_1 / A^*} [/latex]

Thus,

[latex] A_2=A_1\left(\frac{A_2 / A^*}{A_1 / A^*} ight) [/latex]

Using Table B–1, we therefore have

[latex] \begin{gathered} \frac{1}{4} \pi d^2=\pi(0.05 m )^2\left(\frac{1.0944}{1.5451} ight) \ d=84.2 mm \end{gathered} [/latex]

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

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