Question

The water level in a tank is 18 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 3 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. (Round the final answer to one decimal place.)

Answer 1

Answer:

48.98623 m

Explanation:

Given;

Height of the water level above the ground = 18 m

Pressure at the water surface level = 3 atm = 3 × 101325 N/m²

Density of water = 1000 kg/m³

let the maximum rise in the water stream be 'x'

Now, By  the conservation of energy at the water surface and the maximum height of the water stream

$\frac{P_1}{\rho g}+\frac{V_1^2}{2g}+h=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+x$

here,

V₁ and V₂ are the velocities at the surface level and the maximum level of water stream rise = 0 m/s

P₂ = 0 (atmospheric pressure)

on substituting the respective values, we get

$\frac{3\times101325}{1000\times9.81}+\frac{0^2}{2g}+18=\frac{0}{\rho g}+\frac{0^2}{2g}+x$

or

x = $\frac{3\times101325}{1000\times9.81}+18$

or

x = 30.98623 + 18

or

x = 48.98623 m