Question 13.5: Nitrogen flows isentropically through the pipe in Fig. 13–10......
Nitrogen flows isentropically through the pipe in Fig. 13–10 such that its gage pressure is p = 200 kPa, the temperature is 80°C, and the velocity is 150 m/s. Determine the stagnation temperature and stagnation pressure for this gas. The atmospheric pressure is 101.3 kPa.
c = \sqrt{kRT} = \sqrt{1.40 (296.8 J/kg · K)(353 K)} = 383.0 m/sFluid Description. The Mach number for the flow is first determined. The speed of sound for nitrogen at T = (273 + 80) K = 353 K is
Thus,
M = \frac{V}{c} = \frac{150 m/s}{383.0 m/s} = 0.3917 < 1Here we have steady subsonic compressible flow.
Stagnation Temperature. Applying Eq. 13–26, we have
T_0 = 363.8 K = 364 K
Stagnation Pressure. The static pressure p = 200 kPa is measured relative to the flow. Applying Eq. 13–27 and reporting the result as an absolute stagnation pressure, we have
p_0 = p(1 + \frac{k – 1}{2} M^2)^{k/(k – 1)}p_0 = [(101.3 + 200) kPa] (1 + \frac{1.4 – 1}{2} (0.3917)^2)^{1.4/(1.4 – 1)}
p_0 = 334.91 kPa = 335 kPa
Since k = 1.4 for nitrogen, these values for T_0 and p_0 can also be determined by using Table B–1 with linear interpolation. For example, the temperature ratio in Table B–1 is determined as follows: M = 0.39, T/T_0 = 0.9705, and M = 0.40, T/T_0 = 0.9690. Therefore,
for M = 0.3917
0.9690 – T/T_0 = -0.001251
T/T_0 = 0.97025
Thus,
NOTE: Because we have an isentropic process, realize there will be no change in the entropy. This can be shown by applying Eq. 13–15.
s – s_0 = c_p \ln \frac{T}{T_0} – R \ln \frac{p}{p_0}\Delta s = [\frac{1.4 (296.8 J/kg · K)}{1.4 – 1}] \ln (\frac{353 K}{363.82 K}) – (296.8 J/kg · K) \ln (\frac{301.3 kPa}{334.9 kPa})
\Delta s = 0