Question 11.SP.5: A 2 -in ISA flow nozzle is installed in a 3 -in pipe carryin......

This requires a trial-and-error type of solution. First assume a reasonable value ofK. From Fig. 11.22, forD_{2} / D_{1}=0.67, and for the level part of the curve,K=1.06

A_{1}=\frac{\pi}{4}\left(\frac{3}{12}\right)^{2}=0.0491 \mathrm{ft}^{2} ; \quad A_{2}=\frac{\pi}{4}\left(\frac{2}{12}\right)^{2}=0.0218 \mathrm{ft}^{2}

This air-water manometer is like Fig.3.14 b withz_{B}-z_{A}=0. For air and water,s_{M} / s_{F} \approx 0.001, so it can be neglected in Eq. (3.13), and

\Delta\left(\frac{p}{\gamma}+z\right)=R_{m}=\frac{2}{12}=0.1667 \mathrm{ft}

Eq. (11.18):Q=1.06 \times 0.0218 \sqrt{2(32.2) \times 0.1667}=0.0746 \mathrm{cfs}

With this first determination of Q,

V_{1}=\frac{Q}{A}=\frac{0.0746}{0.0491}=1.519 \mathrm{fps}

Then                  D_{1}^{\prime \prime} V_{1}=3 \times 1.519=4.56

From Fig. 11.22, K=1.04 and

Q=\frac{1.04}{1.06} \times 0.0746=0.0732 \mathrm{cfs}

No further correction is necessary.

\frac{p_A}{\gamma}-\frac{p_B}{\gamma}=\left(z_B-z_A\right)+\left(1-\frac{s_M}{s_F}\right) R_m             (3.13a)

\Delta\left(\frac{p}{\gamma}+z\right)=\left(1-\frac{s_M}{s_F}\right) R_m       (3.13b)

Q=K A_2 \sqrt{2 g \Delta\left(\frac{p}{\gamma}+z\right)}         (11.18)