Air enters the 30-mm-diameter pipe with a velocity of 153 m/ s, and a temperature of 300 K, Fig. 13–28. If the average friction factor is f = 0.040, determine how long, Lmax, the pipe should be so that sonic flow occurs at the exit. Also, what is the velocity of the air in the pipe at the exit Lmax

Question

Air enters the 30-mm-diameter pipe with a velocity of 153 m/ s, and a temperature of 300 K, Fig. 13–28. If the average friction factor is f = 0.040, determine how long, [latex] L_{\max } [/latex], the pipe should be so that sonic flow occurs at the exit. Also, what is the velocity of the air in the pipe at the exit [latex] L_{\max } [/latex], and at the location L = 0.8 m?

Accepted Answer

Fluid Description. We assume that adiabatic steady compressible (Fanno) flow occurs along the pipe.

Maximum Pipe Length. The critical length, [latex] L_{\max } [/latex], is determined using Eq. 13–45 or Table B–2.* First we need to determine the initial Mach number.

[latex] \frac{f L_{\max }}{D}=\frac{1- M ^2}{k M ^2}+\frac{k+1}{2 k} \ln \left[\frac{\left[\frac{1}{2}(k+1) ight] M ^2}{1+\frac{1}{2}(k-1) M ^2} ight] [/latex] (13-45)

[latex] \begin{array}{ll} V= M \sqrt{k R T} ; \quad 153 m / s = M _1 \sqrt{1.4(286.9 J / kg \cdot K )(300 K )} \ M _1=0.4408<1 ext { subsonic flow } \end{array} [/latex]

Using Eq. 13–45 or Table B–2, we get [latex] (f / D)\left(L_{\max } ight)=1.6817 [/latex], so that

[latex] L_{\max }=\left(\frac{0.03 m }{0.040} ight)(1.6817)=1.2613 m =1.26 m [/latex]

At this exit M = 1. The velocity of the gas is determined from Eq. 13–48 or from the tabulated ratio for [latex] M _1=0.4408 [/latex]. The result is

[latex] \frac{V}{V^*}=\frac{ M \sqrt{k R T}}{(1) \sqrt{k R T^*}}= M \left[\frac{\frac{1}{2}(k+1)}{1+\frac{1}{2}(k-1) M ^2} ight]^{1 / 2} [/latex] (13-48)

[latex] V^*=322.98 m / s =323 m / s [/latex]

Flow Properties at L = 0.8 m. Since the equations (and table) are referenced from the critical location, we must calculate (f/D)L from this location, Fig. 13–28. Thus,

[latex]\frac{f}{D}L^{\prime}=\frac{0.04}{0.03 m}(1.2613 m-0.8 m)=0.6150[/latex]

This time using the table with interpolated values of the ratio for [latex] V>V^*[/latex], we have

[latex]V=\frac{V}{V^*}V^*=(0.6133)(322.98 m/s)=198 m/s[/latex]

As the air travels 0.8 m down the pipe, notice how the velocity has increased from 153 m/s to 196 m/s. As an exercise, show that the temperature decreases from 300 K to 293 K at 0.8 m. The values of V and T follow the trend shown by the curves in Fig. 13–26.

*More accurate results can be obtained from the equation or from an Internet website, rather than using linear interpolation from the table.

Question 13.12: Air enters the 30-mm-diameter pipe with a velocity of 153 m/...

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