**Given:**

$N = 50\hspace{0.05cm}rpm, \hspace{0.25cm}Q_{act} =0.00736\hspace{0.05cm}m^3/s\\
D = 0.2\hspace{0.05cm}m, \hspace{0.25cm}L = 0.3\hspace{0.05cm}m, \hspace{0.25cm}H_s = 3.5\hspace{0.05cm}m,\hspace{0.25cm}H_d = 11.5\hspace{0.05cm}m$

**To Find:** $Q_{th} = \hspace{0.05cm}?,\hspace{0.25cm}C_d = \hspace{0.05cm}?,\hspace{0.25cm}\% slip\hspace{0.05cm} = \hspace{0.05cm}?$

**Solution**

Theoretical discharge:

$Q_{th} = \frac{A\hspace{0.05cm}\times\hspace{0.05cm}L\hspace{0.05cm}\times\hspace{0.05cm}N}{60} = \frac{\pi\hspace{0.05cm}\times\hspace{0.05cm}D^2\hspace{0.05cm}\times\hspace{0.05cm}L\hspace{0.05cm}\times\hspace{0.05cm}N}{4\hspace{0.05cm}\times\hspace{0.05cm}60}\\
\hspace{0.5cm}=\frac{\pi\hspace{0.05cm}\times\hspace{0.05cm}0.2^2\hspace{0.05cm}\times\hspace{0.05cm}0.3\hspace{0.05cm}\times\hspace{0.05cm}50}{4\hspace{0.05cm}\times\hspace{0.05cm}60}\\
\hspace{0.5cm}= 0.00785 m^3/s$

Coefficient of discharge:
$C_d = \frac{Q_{act}}{Q_{th}} = \frac{0.00736}{0.00785} =0.9375\\$
Percentage slip:$\%slip = (\frac{Q_{th}-Q_{act}}{Q_{th}})\hspace{0.05cm}\times\hspace{0.05cm}100\
\hspace{0.6cm}=\frac{0.00785 - 0.00736}{0.00785}\hspace{0.05cm}\times\hspace{0.05cm}100\
\hspace{0.6cm}=6.242\hspace{0.05cm}\%$