Question 5.4: An orifice meter with orifice diameter 15 cm is inserted in ......

Area A_{o}=\frac{\pi }{4}\times (15)^{2}

= 176.714 cm²

Area A_{p}=\frac{\pi }{4}\times (30)^{2}

= 706.838 cm²

and                                  h=(\frac{13.6}{0.9}-1 )\times 50=705.555  cm  of  oil

Using the discharge equation,

Q=C_{d}\frac{A_{1}A_{2} }{\sqrt{A^{2}_{1}-A^{2}_{2} } }\sqrt{2gh}

or                   Q=0.64\frac{176.714 \times 706.838}{\sqrt{(706.838^{2}) – (176.714^{2}) } }\sqrt{2×9.81×100×705.555}

or                          Q = 137414 cm³/s

or                          Q = 137.414 1/s