Question 11.SP.3: A 2-in circular orifice (not standard) at the end of the 3-i......

Define the inlet as section 1 and the throat as section 2 .

\frac{p_{1}}{\gamma}=10\left(\frac{144}{62.4}\right)=23.1 \mathrm{ft}

V_{1}=\frac{Q}{A_{1}}=\frac{0.60}{\pi(1.5 / 12)^{2}}=12.22 \mathrm{fps}, \frac{V_{1}^{2}}{2 g}=2.32 \mathrm{ft}

Express the ideal energy equation from 1 to 2 to determine the ideal velocity V_{2} at 2

\frac{p_{1}}{\gamma}+\frac{V_{1}^{2}}{2 g}=\frac{V_{2 i}^{2}}{2 g}

\begin{aligned}& \frac{V_{2 i}^{2}}{2 g}=23.1+2.32=25.4 ; \quad V_{2 i}=40.4 \mathrm{fps} \\& C_{v}=\frac{V_{2}}{V_{2 i}}=\frac{39.2}{40.4}=0.969 \end{aligned}

Area of jet            A_{2}=\frac{Q}{V_{2}}=\frac{0.60}{39.2}=0.01531 \mathrm{ft}^{2}

C_{c}=\frac{A_{2}}{A_{o}}=\frac{0.01531}{\pi(1 / 12)^{2}}=0.702

Hence                  C_{d}=C_{c} C_{v}=0.680

From Eq. (11.14):

\begin{aligned}h_{L_{1-2}} & =\left(\frac{1}{(0.969)^{2}}-1\right)\left[1-\left(\frac{2}{3}\right)^{4}\right] \frac{V_{2}^{2}}{2 g}=0.0517 \frac{V_{2}^{2}}{2 g} \\& =0.0517 \frac{(39.2)^{2}}{2(32.2)}=1.233 \mathrm{ft} \end{aligned}

Check: Calculate the actual velocity V_{2} at 2 by expressing the real energy equation from 1 to 2

\frac{p_{1}}{\gamma}+\frac{V_{1}^{2}}{2 g}-h_{L_{1-2}}=\frac{V_{2}^{2}}{2 g}

23.1+2.32-1.233=\frac{V_{2}^{2}}{2 g} ; actual V_{2}=39.4 \mathrm{fps}

This checks well with the measured velocity of 39.2 \mathrm{fps}.

h_{L_{1-2}}=\left(\frac{1}{C_v^2}-1\right)\left[1-\left(\frac{A_2}{A_1}\right)^2\right] \frac{V_2^2}{2 g}            (11.14)