Question 11.SP.1: Air at 20°C is flowing through the pipe shown in Fig. S11.1,......

Table A.5 for air: R=287 \mathrm{~m}^{2} /\left(\mathrm{s}^{2} \cdot \mathrm{K}\right)

(a)   p_{\mathrm{abs}}=p_{\mathrm{at}}+p_{\mathrm{gage}}=101.32(684 / 760)+70.2=161.4 \mathrm{kPa}

Eq. (2.5):                  \quad \gamma=\frac{g p}{R T}=\frac{9.81(161.4)}{287(273+20)}=0.01883 \mathrm{kN} / \mathrm{m}^{3}

Eq. (11.6):                 \quad u=\sqrt{2(9.81)\left(\frac{71.1-70.2}{0.01883}\right)}=30.6 \mathrm{~m} / \mathrm{s} \quad

(b) Sec. 11.3, footnote 3: We need \mathbf{M}<0.1

Sec. 5.4: \mathbf{M}=V / c; at 20^{\circ} \mathrm{C}, c=345 \mathrm{~m} / \mathrm{s} [or use Eq. (13.15)].

So we need V<0.1 c=u_{\max }=34.5 \mathrm{~m} / \mathrm{s}

From Eq. (11.5):      \Delta p=p_{0}-p=\gamma \frac{u^{2}}{2 g}

So                             \Delta p_{\max }=\frac{\gamma\left(u_{\max }\right)^{2}}{2 g}=\frac{0.01883(34.5)^{2}}{2(9.81)}=1.142 \mathrm{kPa}

Note: This \Delta p_{\max } compares with the observed \Delta p of 71.1-70.2=0.9 \mathrm{kPa}. Therefore it was safe to neglect compressibility effects in part (a).

\gamma=\frac{g p}{R T}       (2.5)

u=C u_i=C \sqrt{2 g\left(\frac{p_0}{\gamma}-\frac{p}{\gamma}\right)}        (11.6)

c=\sqrt{k R T}=\sqrt{\frac{k p}{\rho}}        (13.15)

h_{L_{1-2}}=\left(\frac{1}{C_v^2}-1\right) \frac{V_2^2}{2 g}             (11.15)