Discussion Forum

A water tank kept on the ground has an orifice of 2 mm diameter on the vertical side. What is the minimum height of the water above the orifice for which the output flow of water is found to be turbulent? [ Assume  g=10 m/s², $\rho_{water}$= 10³ kg/m³, viscosity =1 centi-poise]

Answer:

Explanation:

 Here, D=2mm, $\eta=1 cent-poise= 10^{-3} Pa-s$

 and density of the water , $\rho$ = $10^{-3} kg/m^{3}$

 for flow to be just turbulent , $R_{e}=3000$

 $\therefore$      $v= \frac{R_{e} \eta}{ \rho D}= \frac{ 3000 \times 10^{-3}}{10^{3} \times 2 \times 10^{-3}}=1.5$

 We know that the velocity head, $h= \frac{v^{2}}{2g}$

 $\Rightarrow$            $h = \frac{(1.5)^{2}}{2 \times 10}=0.1125=11 cm$

 SO, no option is matched