A tank which is raised above the ground level, has a hole in its flat base, through which water flows and falls to the ground below. If the depth of water in the tank is 0.5 m and the tank base is 0.7 m above the ground, what is the fluid velocity (u) (m/s) just before it reaches the ground? Assume a discharge coefficient of unity.
To find the fluid velocity (u) just before it reaches the ground, we can use the principle of conservation of energy.
The potential energy of the water in the tank when it is at a height of 0.7 m above the ground is converted into kinetic energy just before it reaches the ground.
The potential energy (PE) of the water in the tank can be calculated using the formula:
PE = m * g * h
Where:
m = mass of the water
g = acceleration due to gravity (approx. 9.8 m/s^2)
h = height of the water above the ground (0.7 m)
The mass (m) of the water can be calculated using the formula:
m = density * volume
The density of water is approximately 1000 kg/m^3, and the volume of the water can be calculated by multiplying the area of the hole and the depth of the water in the tank.
The area of the hole can be calculated using the formula:
A = π * (r^2)
Where:
r = radius of the hole
The radius of the hole can be determined by assuming it is a circular hole and using the equation:
A = π * (r^2) = 0.5 m^2
Solving for r, we get:
r^2 = 0.5 / π
r ≈ 0.4 m
Now, we can calculate the area of the hole:
A ≈ π * (0.4^2) ≈ 0.5 m^2
The volume (V) of the water can be calculated using the formula:
V = A * h
V = 0.5 m^2 * 0.5 m = 0.25 m^3
Now, we can calculate the mass of the water:
m ≈ density * volume ≈ 1000 kg/m^3 * 0.25 m^3 ≈ 250 kg
With the mass (m) of the water known, we can calculate the potential energy (PE) of the water:
PE = m * g * h ≈ 250 kg * 9.8 m/s^2 * 0.7 m ≈ 1715 J
The potential energy (PE) is equal to the kinetic energy (KE) just before the water reaches the ground:
KE = PE = 1715 J
The kinetic energy (KE) can be calculated using the formula:
KE = 0.5 * m * u^2
Where:
u = fluid velocity just before it reaches the ground
Rearranging the equation, we get:
u^2 = (2 * KE) / m
Substituting the values, we have:
u^2 = (2 * 1715 J) / 250 kg
Simplifying:
u^2 ≈ 13.72 m^2/s^2
Finally, taking the square root of both sides, we find:
u ≈ 3.7 m/s
Therefore, the fluid velocity (u) just before it reaches the ground is approximately 3.7 m/s.