how fast does the level of water in a full 2m high, 10cm diameter tank reduces when a 4cm diameter hole is opened at the bottom?
To determine how fast the level of water in the tank decreases when a hole is opened at the bottom, we need to apply the principles of fluid dynamics and use Bernoulli's equation.
First, let's find the area of the hole at the bottom of the tank. The area of a circle can be calculated using the formula A = π * r^2, where A is the area and r is the radius. The radius of the 4cm diameter hole is half of the diameter, which is 2cm or 0.02 meters. Hence, the area of the hole is A = π * (0.02)^2 = 0.00126 square meters.
Now, let's consider Bernoulli's equation, which relates the pressure, velocity, and height of a fluid in a system:
P + 1/2 * ρ * v^2 + ρ * g * h = constant
Where:
P is the pressure of the fluid,
ρ (rho) is the density of the fluid,
v is the velocity of the fluid,
g is the acceleration due to gravity, and
h is the height of the fluid.
Since the tank is open at the top and atmospheric pressure is acting on the water surface, we can assume the pressure term cancels out in our calculations. Thus, we are left with:
1/2 * ρ * v^2 + ρ * g * h = constant
The density of water, ρ, is approximately 1000 kg/m^3, and the acceleration due to gravity, g, is approximately 9.8 m/s^2.
Initially, when the tank is full, the height of the water is 2 meters. At this point, the velocity of the water at the hole is zero since the water is not flowing out yet. Thus, we can simplify the equation to:
ρ * g * h1 = constant
When the water level decreases to h2, the velocity at the hole increases. Let's assume the velocity is v2. Now we have:
1/2 * ρ * v2^2 + ρ * g * h2 = constant
Since the pressure at the hole and the top surface of the water are the same (due to the hole being relatively small), we can equate the two constant expressions:
ρ * g * h1 = 1/2 * ρ * v2^2 + ρ * g * h2
Canceling out common terms, we get:
h1 = 1/2 * v2^2 + h2
Now, let's calculate the velocity, v2, using the equation of Bernoulli's equation. Rearranging it, we have:
1/2 * ρ * v2^2 = ρ * g * (h1 - h2)
Now substituting the known values, we have:
1/2 * 1000 kg/m^3 * v2^2 = 1000 kg/m^3 * 9.8 m/s^2 * (2 m - h2)
Simplifying further:
v2^2 = 2 * 9.8 m/s^2 * (2 m - h2)
We can solve this equation to find v2. Once we know v2, we can calculate the volumetric flow rate through the hole, Q, using the formula:
Q = A * v2
where A is the area of the hole we calculated earlier (0.00126 m^2).