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

Question

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.

Accepted Answer

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.

[latex] \begin{gathered} V_1= M _1 \sqrt{k R T_1} \quad 75 m / s = M _1 \sqrt{1.4(286.9 J / kg \cdot K )(295 K )} \ M _1=0.2179<1 \quad ext { Subsonic } \end{gathered} [/latex]

Using Table B–3,

[latex] \begin{aligned} T^* & =\frac{T_1}{T_1 / T^*}=\frac{295 K }{0.24042}=1227.01 K \ 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 \end{aligned} [/latex]

Air Properties at Section 2. We can determine the Mach number at section 2 by using Eq. 13–64. However, before we can do this or use the tables, we must find the stagnation temperatures [latex] \left(T_0 ight)_2 ext { and } T_0^* [/latex].
First, [latex] \left(T_0 ight)_1 [/latex] can be determined using Eq. 13–26, or Table B–1 for isentropic flow. For [latex] M _1 [/latex] = 0.2179, we get

[latex] \frac{T_0}{T_0^*}=\frac{2(k+1) M ^2\left(1+\frac{k-1}{2} M ^2 ight)}{\left(1+k M ^2 ight)^2} [/latex] (13-64)

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

[latex] \left(T_0 ight)_1=\frac{T_1}{T_1 /\left(T_0 ight)_1}=\frac{295 K }{0.9906}=297.80 K [/latex]

Now, to determine [latex] \left(T_0 ight)_2 [/latex], we will use the energy equation, Eq. 13–57.

[latex] \begin{gathered} c_p=\frac{k R}{k-1}=\frac{1.4(286.9 J / kg \cdot K )}{1.4-1}=1004.15 J / kg \cdot K \ \frac{\Delta Q}{\Delta m}=c_p\left[\left(T_0 ight)_2-\left(T_0 ight)_1 ight] \qquad (13-57) \ \frac{100\left(10^3 ight) J }{ kg \cdot m }(2 m )=\left[1.00415\left(10^3 ight) J / kg \cdot K ight]\left[\left(T_0 ight)_2-297.80 K ight] \ \left(T_0 ight)_2=496.97 K \end{gathered} [/latex]

Also, from Table B–3, for [latex] M _1=0.2179 [/latex], the stagnation temperature at the critical or reference location is therefore

[latex] T_0^*=\frac{\left(T_0 ight)_1}{\left(T_0 ight)_1 / T_0^*}=\frac{297.80 K }{0.20225}=1472.14 K [/latex]

Finally, we can find [latex] M _2 [/latex] from the stagnation temperature ratio

[latex] \frac{\left(T_0 ight)_2}{T_0^*}=\frac{496.97 K }{1472.14 K }=0.33758 [/latex]

Using Table B–3, we get [latex] M _2 [/latex] = 0.2949. The other ratios at [latex] M _2 [/latex] give

[latex] \begin{gathered} T_2=T^*\left(\frac{T_2}{T^*} ight)=1227.01 K (0.39810)=488 K \ p_2=p^*\left(\frac{p_2}{p^*} ight)=59.99 kPa (2.13950)=128 kPa \ V_2=V^*\left(\frac{V_2}{V^*} ight)=702.03 m / s (0.18607)=131 m / s \end{gathered} [/latex]

The results indicate that as the Mach number increases from [latex] M _1=0.2179 ext { to } M _2=0.2949 [/latex], 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.

Question 13.15: Outside air is drawn isentropically into the pipe having a d...

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