Question 11.SP.4: Find the discharge rate of 20°C water through the venturi tu......

Table A.1 for water at 20^{\circ} \mathrm{C}: \quad \nu=1.003 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}

Venturi size is 800 \mathrm{~mm} \times 400 \mathrm{~mm}=31.5 in \times 15.75 in

This is about midway between the 8 in \times 4 in and 200 in \times 100 in curves on Fig. 11.19. So, from Fig. 11.19: Maximum C \approx 0.988; assume this value. Eqs. (11.16) and (11.17):

Q=\frac{0.988 \pi(0.40 / 2)^{2}}{\sqrt{1-(400 / 800)^{4}}} \sqrt{2(9.81) 0.15\left(\frac{13.55}{1}-1\right)}=0.779 \mathrm{~m}^{3} / \mathrm{s}

V_{2}=\frac{Q}{A_{2}}=\frac{0.779}{\pi(0.40 / 2)^{2}}=6.20 \mathrm{~m} / \mathrm{s}

Eq. (7.6):                \mathbf{R}=\frac{D_{2} V_{2}}{\nu}=\frac{0.40(6.20)}{1.003 \times 10^{-6}}=2.47 \times 10^{6}

Check Fig. 11.19 for this \mathbf{R}: C=0.988. Thus Q=0.779 \mathrm{~m^3} / \mathrm{s}

\mathbf{R}=\frac{F_I}{F_V}=\frac{L^2 V^2 \rho}{L V \mu}=\frac{L V \rho}{\mu}=\frac{L V}{\nu}        (7.6)

Q=A_2 V_2=C A_2 V_{2 i}=\frac{C A_2}{\sqrt{1-\left(D_2 / D_1\right)^4}} \sqrt{2 g \Delta\left(\frac{p}{\gamma}+z\right)}               (11.16)

\Delta\left(\frac{p}{\gamma}+z\right)=R_m\left(\frac{s_M}{s_F}-1\right)             (11.17)