Question

Problem
A 20 x 10 cm venturimeter is provided in a vertical pipe line carrying oil of sp.gr
0.8. the flow being upwards. The difference in elevation of the throat and entrance section of the venturimeter is 50 cm. The differential U tube miercury
manometer shows a deflection of 40 cm of mercury. Calculate the discharge of oil
Assume cd= 0.98 and Sp.gr of mercury = 13​

Answer 1

Answer:

78725.75 cm³/s

Explanation:

Answer is in the document.

pls mark as braineliest

Answer 2

• Given data,

               $d_{1} =20cm\\d_{2}=10cm \\a_{1} =\frac{\pi }{4} d_{1} ^{2} \\ =\frac{\pi }{4} (20)^{2} \\ =314.16cm^{2}$

• $S.P.G$ of oil $S_{0} =$$0.8$

• $S.P.G$ of mercury $S_{m} =13.6$

          differential manometer reading, $x=30cm$

                $h=(\frac{P_{1} }{dg} +Z _{1} )-(\frac{P_{2} }{dg} - Z_{2} )-x(\frac{s_{g} }{s_{0} } -1)\\ =30[\frac{13.6}{0.8} -1)\\h=480cm of oil$

• we know that dicharge coefficient

             $C_{d} =\sqrt{{\frac{h-h_{l} }{h} } }$

• as losses are neglectd $h_{l}$ $=0$

           $C_{d} =\sqrt{\frac{480}{480} }$

             $C_{d} =1$

• Discharge of oil is

           $q=C_{d}$ × $\frac{a_{1}a_{2} }{y\sqrt{a_{1} ^{2} -a_{2}^{2} } }$  × $\frac{\sqrt{2gh} }{1}$

lets substitute all the values and the answer would be,

          $q=78725.7575cm^{3}/s$