Question 13.15: Outside air is drawn isentropically into the pipe having a d......
Outside air is drawn isentropically into the pipe having a diameter of 200 mm, Fig. 13–34. When it arrives at section 1, it has a velocity of 75 m/s, an absolute pressure of 135 kPa, and a temperature of 295 K. If the walls of the pipe supply heat at 100 kJ/kg · m, determine the properties of the air when it reaches section 2.
Fluid Description. We assume the air to be inviscid and to have steady compressible flow. Due to the heating, this is Rayleigh flow.
Air Properties at Critical Location. The air properties at location 2 can be determined using the ratios in Table B–3, provided we first know the properties at the critical location, where M = 1. We can find these using the properties at section 1, but first we need the Mach number at section 1.
M_1 = 0.2179 < 1 Subsonic
Using Table B–3,
p* = \frac{p_1}{p_1/p*} = \frac{135 KPa}{2.2504} = 59.99 KPA
V* = \frac{V_1}{V_1/V*} = \frac{75 m/s}{0.10683} = 702.03 m/s
Air Properties at Section 2. We can determine the Mach number at section 2 by using Eq. 13–64 \frac{T_0}{T^*_0}=\frac{2(k+1)M^2(1+\frac{k-1}{2}M^2)}{(1+kM^2)^2}. However, before we can do this or use the tables, we must find the stagnation temperatures (T_0)_2 and T_0^*.
First, (T_0)_1 can be determined using Eq. 13–26 T_0=T(1+\frac{k-1}{2}M^2), or Table B–1 for
isentropic flow. For M_1 = 0.2179, we get
Now, to determine (T_0)_2, we will use the energy equation, Eq. 13–57.
c_p = \frac{kR}{k – 1} = \frac{1.4 (286.9 J/kg · K)}{1.4 – 1} = 1004.15 J/kg · K\frac{ΔQ}{Δm} = c_p [(T_0)_2 – (T_0)_1]
\frac{100(10^3) J}{kg · m} (2 m) = [1.00415 (10^3) J/kg · K][(T_0)_2 – 297.80 K]
(T_0)_2 = 496.97 K
Also, from Table B–3, for M_1 = 0.2179, the stagnation temperature at the critical or reference location is therefore
T_0^* = \frac{(T_0)_1}{(T_0)_1/T_0^*} = \frac{297.80 K}{0.20225} = 1472.14 KFinally, we can find M_2 from the stagnation temperature ratio.
\frac{(T_0)_2}{T_0^*} = \frac{496.97 K}{1472.14 K} = 0.33758Using Table B–3, we get M_2 = 0.2949. The other ratios at M_2 give
T_2 = T*(\frac{T_2}{T*}) = 1227.01 K (0.39810) = 488 Kp_2 = p*(\frac{p_2}{p*}) = 59.99 KPa (2.13950) = 128 KPa
V_2 = V*(\frac{V_2}{V*}) = 702.03 m/s (0.18607) = 131 m/s
The results indicate that as the Mach number increases from M_1 = 0.2179 to M_2 = 0.2949, the pressure decreases from 135 kPa to 128 kPa, the temperature increases from 295 K to 488 K, and the velocity increases from 75 m/s to 131 m/s. These changes follow the trend shown by the curves for Rayleigh subsonic flow in Fig. 13–32.
