Answer:
a. v1 = 5.06 m/s, v2 = 3.96 m/s , R = 1.27
b. t = 1 hr, 11 min, 26 sec
Explanation:
Using the Bernoulli's laws to use the conserved energy
a. Solve the speed and the radio of this speed of the tank is open to the air
p₀ + ρgh₀ + ½ρv₀² = p₁ + ρgh₁ + ½ρv₁²
5000Pa + 1000kg/m³ * 9.8m/s² * 0.800m + 0 = 0 + 0 + ½ * 1000kg/m³ * v²
v² = 25.68 m²/s²
v1 = 5.06 m/s
Because it is open the tank so P=0 pa so:
0 Pa + 1000kg/m³ * 9.8m/s² * 0.800m + 0 = 0 + 0 + ½ * 1000kg/m³ * v²
v² = 15.68 m²/s²
v2 = 3.96 m/s
The ratio on the air is solve using both velocities so:
R = v1/v2 = 5.06 m/s / 3.96 m/s
R = 1.27
b. Now to find the time it takes for the tank to drain if the tank is open to the air
dh/dt = -u
dh/dt = -v * A/A'
dh/dt = v*(.02m)²/(2.0m)² = -v / 10000
and we can further substitute for v:
dh/dt = -(1/1e4)*√[(p+9800h)/500]
Solve replacing
-(1000/49)*√(49000h) = t + C
-(1000/49)*√(49000*0.8) = 0 + C
C = - 4040.6
Then when h = 0,
t = 4286 s
t = 1 hr, 11 min, 26 sec
