Question 13.12: Air enters the 30-mm-diameter pipe with a velocity of 153 m/......
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, L_{max}, 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 L_{max}, and at the location L = 0.8 m?
Fluid Description. We assume that adiabatic steady compressible (Fanno) flow occurs along the pipe.
Maximum Pipe Length. The critical length, L_{max}, is determined using Eq. 13–45 \frac{fL_{max}}{D}=\frac{1-M^2}{kM^2}+\frac{k+1}{2k}ln[\frac{[\frac{1}{2}(k+1)]M^2}{1+[\frac{1}{2}(k-1)M^2}] or Table B–2.* First we need to determine the initial Mach number.
M_1 = 0.4408 < 1 subsonic flow
Using Eq. 13–45 or Table B–2, we get (f/D) (L_{max}) = 1.6817, so that
L_{max} = (\frac{0.03 m} {0.040}) (1.6817) = 1.2613 m = 1.26 m
At this exit M = 1. The velocity of the gas is determined from Eq. 13–48 \frac{V}{V^*}=\frac{M\sqrt{KRT}}{(1)\sqrt{KRT^*}}=M[\frac{\frac{1}{2}(k+1)}{1+\frac{1}{2}(k11)M^2}] or from the tabulated ratio for M_1 = 0.4408. The result is
V* = 322.98 m/s = 323 m/s
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,
\frac{f}{D} L` = \frac{0.04}{0.03 m}(1.2613 m – 0.8 m) = 0.6150This time using the table with interpolated values of the ratio for V/V*, we have
V = \frac{V}{V*} V* = (0.6133)(322.98 m/s) = 198 m/sAs 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.
