Respuesta :
Answer:
Volume flow rate, Q, into the tank is 12.376 m³
Step-by-step explanation:
Here we have Bernoulli's equation presented as follows
[tex]\frac{P_{1}}{\rho _{1}g}+\frac{v_{1}^{2}}{2g} + z_{1} = \frac{P_{2}}{\rho _{2}g}+\frac{v_{2}^{2}}{2g} + z_{2}[/tex]
Where:
P₁ = Pressure, of the open tank = Atmospheric pressure = 0
P₂ - Pressure at the discharge of the pipe = Atmospheric pressure = 0
ρ₁ = Density of the fluid = ρ₂
v₁ = Velocity of the fluid in the tank
v₂ = Velocity of the fluid in the pipe
d₁ = Diameter of the tank = 1.5 m
d₂ = Diameter of the pipe = 15 cm = 0.15 m
A₁ = Cross sectional area of tank = π×d₁²/4 = 1.767 m²
A₂ = Cross sectional area of pipe= π×d₂²/4 = 0.01767 m²
z₁ = Level of water in the tank = 0 m
z₂ = Level of water in the pipe = 2.5 m
Whereby the level remains constant at 2.5 m, pressure at the base, P₂ = ρ₁×g×2.5
Also
v₁×A₁ = v₂×A₂ which gives;
v₁×1.767 = v₂×0.01767
v₂ = v₁×1.767/0.01767 = 100·v₁
Hence we have;
[tex]\frac{0}{\rho _{1}g}+\frac{v_{1}^{2}}{2g} + z_{1} = \frac{0}{\rho _{1}g}+\frac{(100v_{1})^{2}}{2g} + z_{2}[/tex]
[tex]z_{1} - z_{2} = \frac{(100v_{1})^{2}}{2g}- \frac{v_{1}^{2}}{2g}[/tex]
[tex](2.5)2g = 99v_{1}^{2}}[/tex]
v₁² = 2.5×2×9.81 = 49.05
v₁ = √49.05 = 7.0035 m/s
Therefore, since the rate of discharge = Rate of entry into the tank, we have;
Volume flow rate, Q, into the tank = v₁×A₁ = 7.0035 m/s × 1.767 m² = 12.376 m³.
