Question

The water level in a tank z1, is 16 m above the ground. A hose is connected to the bottom of the tank, and the nozzle at the end of the hose is pointed straight up. The tank cover is airtight, and the air pressure above the water surface is 2 atm gage. The system is at sea level. Determine the maximum height to which the water stream could rise. Take the density of water to be 1000 kg/m3.

Answer 1

Answer:

40.7 m

Explanation:

Let point 1 represent the surface of the water, point 2 be the top of the water trajectory and the reference be the bottom of the tank. Hence:

$z_1=16\ m,P_1=2\ atm,V_1=0(velocity\ at\ the \ surface \ of\ water\ is\ low)\\V_2=0,P_2=P\\\\Using\ Bernoulli\ equation:\\\\\frac{P_1}{\rho g} +\frac{V_1^2}{2g}+z_1= \frac{P_2}{\rho g} +\frac{V_2^2}{2g}+z_2\\\\Sine\ V_1=0,V_2=0:\\\\\frac{P_1}{\rho g} +z_1=\frac{P}{\rho g} +z_2\\\\z_2=\frac{P_1}{\rho g} -\frac{P}{\rho g} +z_1\\\\z_2=\frac{P_1-P}{\rho g} +z_1\\\\z_2=\frac{P_{1.gage}}{\rho g} +z_1$

$z_2=\frac{2\ atm}{1000\ kg/m^3*9.81\ m/s^2}(\frac{101325\ N/m^2}{1\ atm} )(\frac{1\ kg.m/s^2}{1 \ N} ) +20\\\\z_2=20.7+20\\\\z_2=40.7\ m$