Question

The water level in a tank z1, is 20 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:

the maximum height h to which the the water steam could rise is 40.65m

Explanation:

Given;

$Z_{1}$ = 20m,

$P_{1gage}$ = 2atm ≈ 20265 N/m

Density of water ρ = 1000 kg/$m^{2}$

Note: we take point 1 at the free surface of the water in the tank and point 2 at the top of the water trajectory. we also take reference level at the bottom of the tank.

fluid velocity at the free surface of tank is very low ($V_{1}$ ≅ 0) and at the top of the water trajectory $V_{2}$ = 0

Step 1: Applying Bernoulli equation between poin 1 and point 2

$P_{1}$/ρg + $V_{1} ^{2}$/2g + $Z_{1}$ = $P_{2}$/ρg + $V_{2} ^{2}$/2g + $Z_{2}$

$P_{1}$/ρg + $Z_{1}$ = $P_{atm}$/ρg + $Z_{2}$

$Z_{2}$ = ($P_{1}$ - $P_{atm}$)/ρg + $Z_{1}$

Substituting values into $Z_{2}$ we have,

$Z_{2}$ = $\frac{2atm}{(1000 kg/m^{3} )(9.81 m/s^{2} )} (\frac{101325N/m^{2} }{1atm} )(\frac{1 kg.m/s^{2} }{1N} ) + 20 = 40.65 m$