Answer:
From the answer, the water level is falling (since rate of outflow is more than that of inflow), and the rate at which the water level in the storage tank is falling is
(dh/dt) = - 0.000753
Units of m/s
Explanation:
Let the volume of the system at any time be V.
V = Ah
where A = Cross sectional Area of the storage tank, h = height of water level in the tank
Let the rate of flow of water into the tank be Fᵢ.
Take note that Fᵢ is given in the question as 500 gal/min = 0.0315 m³/s
Let the rate of flow of water out of the storage tank be simply F.
F is given in the form of (cross sectional area of outflow × velocity)
Cross sectional Area of outflow = πr²
r = 2 inches/2 = 1 inch = 0.0254 m
Cross sectional Area of outflow = πr² = π(0.0254)² = 0.00203 m²
velocity of outflow = 60 ft/s = 18.288 m/s
Rate of flow of water from the storage tank = 0.0203 × 18.288 = 0.0371 m³/s
We take an overall volumetric balance for the system
The rate of change of the system's volume = (Rate of flow of water into the storage tank) - (Rate of flow of water out of the storage tank)
(dV/dt) = Fᵢ - F
V = Ah (since A is constant)
dV/dt = (d/dt) (Ah) = A (dh/dt)
dV/dt = A (dh/dt) = Fᵢ - F
Divide through by A
dh/dt = (Fᵢ - F)/A
Fi = 0.0315 m³/s
F = 0.0371 m³/s
A = Cross sectional Area of the storage tank = πD²/4
D = 10 ft = 3.048 m
A = π(3.048)²/4 = 7.30 m²
(dh/dt) = (0.0315 - 0.0370)/7.3 = - 0.000753
(dh/dt) = - 0.000753
