A large storage tank, open to the atmosphere
at the top and filled with water, develops a
small hole in its side at a point 18 m below
the water level. The rate of flow of water from
the leak is 1.6 × 10−3 m3
/min.
Determine the speed at which the water
leaves the hole. The acceleration of gravity is
9.81 m/s
2
.
Answer in units of m/s
To determine the speed at which the water leaves the hole, we can apply Bernoulli's equation, which relates the pressure, velocity, and height of a fluid in motion.
Given:
- The water level in the tank is 18 m above the hole.
- The rate of flow of water from the leak is 1.6 × 10^(-3) m^3/min.
- The acceleration due to gravity is 9.81 m/s^2.
Let's break down the steps to find the speed:
Step 1: Convert flow rate to m^3/s
To make the units consistent, we need to convert the given flow rate from m^3/min to m^3/s. Since there are 60 seconds in a minute, we can use the conversion factor of 1 min = 60 s.
Flow rate in m^3/s = (1.6 × 10^(-3) m^3/min) / (60 s/min)
Step 2: Calculate the velocity of water at the hole
We'll use Bernoulli's equation to find the velocity of the water at the hole. Bernoulli's equation is given as:
P + (0.5)ρv^2 + ρgh = constant
Where:
- P is the pressure of the fluid,
- ρ is the density of the fluid,
- v is the velocity of the fluid,
- g is the acceleration due to gravity,
- h is the height of the fluid.
In this case, the tank is open to the atmosphere, so the pressure at the hole is atmospheric pressure (which we can assume to be constant). The density of water is also constant.
Therefore, we can simplify Bernoulli's equation to:
(0.5)ρv^2 + ρgh = constant
To find the velocity (v) at the hole, we can set the constant to be zero for convenience:
(0.5)ρv^2 + ρgh = 0
Solve this equation for v:
v^2 = -2gh
v = √(-2gh)
Since velocity cannot be negative in this context, we take the positive square root.
Step 3: Calculate the speed at which the water leaves the hole
Using the formula v = √(-2gh), plug in the given values:
v = √(-2 * 9.81 m/s^2 * 18 m)
Simplifying:
v = √(-352.008) m/s
Since the negative square root is not physically meaningful, we discard it, and the speed at which the water leaves the hole is:
v = √(352.008) m/s
v ≈ 18.74 m/s
Therefore, the speed at which the water leaves the hole is approximately 18.74 m/s.