Answer:
the maximum height h to which the the water steam could rise is 40.65m
Explanation:
Given;
[tex]Z_{1}[/tex] = 20m,
[tex]P_{1gage}[/tex] = 2atm ≈ 20265 N/m
Density of water ρ = 1000 kg/[tex]m^{2}[/tex]
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 ([tex]V_{1}[/tex] ≅ 0) and at the top of the water trajectory [tex]V_{2}[/tex] = 0
Step 1: Applying Bernoulli equation between poin 1 and point 2
[tex]P_{1}[/tex]/ρg + [tex]V_{1} ^{2}[/tex]/2g + [tex]Z_{1}[/tex] = [tex]P_{2}[/tex]/ρg + [tex]V_{2} ^{2}[/tex]/2g + [tex]Z_{2}[/tex]
[tex]P_{1}[/tex]/ρg + [tex]Z_{1}[/tex] = [tex]P_{atm}[/tex]/ρg + [tex]Z_{2}[/tex]
[tex]Z_{2}[/tex] = ([tex]P_{1}[/tex] - [tex]P_{atm}[/tex])/ρg + [tex]Z_{1}[/tex]
Substituting values into [tex]Z_{2}[/tex] we have,
[tex]Z_{2}[/tex] = [tex]\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[/tex]