Question

A tank contains 3L of water and is held at a constant pressure by regulated air pressure. The tank has a nozzle at the bottom with a diameter of 5mm that discharges to the atmosphere. If the tank is to be drained of water in 10 seconds, what is the minimum gage air pressure that must be provided? The tank is cylindrical and the diameter is the same as the length. a. 56 kPa b. 117 kPa c. 79 kPa d. 154 kPa

Answer 1

Answer:(b) 117 KPa

Explanation:

Given

Voulme$\left ( V\right )$=3 L

diameter of nozzle$\left ( d\right )=5 mm$

time$\left ( t\right )=10 s$

Flow rate$\left ( \dot{Q}\right )=3\times 10^{-4] m^3/s$

Exit velocity at nozzle$\left ( v\right )$=$\frac{\dot{Q}}{A}$

v=$\frac{3\times 4\times 10^2}{\pi 5^2}$=15.27m/s

Applying bernoulli's equation

1 is top point and 2 is bottom point

$P_1+\frac{\rho v^2}{2}+Z_1gh=P_2+\frac{\rho v_{exit}^2}{2}+Z_2gh$

neglecting height variation as it is very small

$P_2=P_{atm}$

$P_1-P_2=P_{guage}$

$P_{gauge}$=$\frac{10^3\times 15.27^2}{2}$

$P_{gauge}=116.67 KPa\approx 117KPa$