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A closed tank has an orifice $0.025m$ diameter in one of its vertical sides.The tank contains oil to a depth of 0.61m above the centre of the orifice and the pressure in the air space above oil is maintained at $13780N/m^2$ above atmospheric.What is the discharge from the orifile.(Coefficient of discharge of the orifice is 0.61 relative density of oil is 0.9)

$\begin{array}{1 1}(A)\;0.0018m^3/s\\(B)\;0.00195m^3/s\\(C)\;0.012m^3/s\\(D)\;0.0013m^3/s\end{array} $

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$\sigma=0.9=\large\frac{\rho_0}{\rho_{\omega}}$
Apply Bernoulli's principle
$\large\frac{\rho_1}{\rho_g}+\frac{u_1^2}{2g}$$+z_1=\large\frac{\rho_z}{\rho_g}+\frac{u_2^2}{2g}$$+z_2$
$\large\frac{13780}{\rho_0g}$$+0.61=\large\frac{u_2^2}{2g}$
$u_2=6.63m/s$
$Q=0/61\times 6.53\times \pi(\large\frac{0.025}{2})^2$
$\;\;\;=0.00195m^3/s$
Hence (B) is the correct answer.
answered May 12, 2014 by sreemathi.v
 

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