A large tank is filled with water to a depth of d = 18 m. A spout located h = 13 m above the bottom of the tank is then opened as shown in the drawing. With what speed will water emerge from the spout?
The pressure of water at the hole location is
P = (rho)gH + Po
where Po is atmospheric pressure
The Bernoulli equation tells you that
(1/2)(rho) V^2 = P - Po = (rho) g H
The water density (rho) cancels out, so
V = sqrt (2 g H)
H is this case the the height of water ABOVE the hole, which is 5 m, not h=13 m
To find the speed at which water will emerge from the spout, we can use the principle of conservation of energy. We can equate the potential energy at the bottom of the tank to the kinetic energy at the spout.
Step 1: Find the potential energy at the bottom of the tank.
The potential energy can be calculated using the formula: PE = mgh, where m is the mass of the water, g is the acceleration due to gravity (9.8 m/s^2), and h is the height. Since the density of water is 1000 kg/m^3, the mass of water can be calculated using the formula: m = ρV, where ρ is the density and V is the volume.
The volume of water can be calculated using the formula: V = A * d, where A is the cross-sectional area of the tank and d is the depth. Since the tank is not given a shape, let's assume it is a cylinder.
Step 2: Find the cross-sectional area of the tank.
The cross-sectional area of a cylinder can be calculated using the formula: A = πr^2, where r is the radius of the tank. Since the radius is not given, let's assume it is 1 meter for simplicity.
Step 3: Calculate the potential energy.
Using the given values, we can calculate the potential energy:
ρ = 1000 kg/m^3 (density of water)
g = 9.8 m/s^2 (acceleration due to gravity)
h = 18 m (height of the water)
d = 13 m (height of the spout above the bottom of the tank)
r = 1 m (radius of the tank)
V = A * d = πr^2 * d = π * (1)^2 * 18 = 18π m^3 (volume of water)
m = ρV = 1000 * 18π ≈ 56,548.67 kg (mass of water)
PE = mgh = 56,548.67 * 9.8 * 18 ≈ 9,766,333.96 J (potential energy at the bottom of the tank)
Step 4: Find the kinetic energy at the spout.
According to the conservation of energy principle, the potential energy at the bottom is equal to the kinetic energy at the spout. The kinetic energy can be calculated using the formula: KE = 1/2 mv^2, where m is the mass of the water and v is the velocity.
PE = KE
9,766,333.96 J = 1/2 mv^2
v^2 = (2 * PE) / m
v = √((2 * 9,766,333.96) / 56,548.67) ≈ 13.60 m/s
Therefore, the speed at which water will emerge from the spout is approximately 13.60 m/s.
To determine the speed at which the water will emerge from the spout, we can use the principle of conservation of energy. Here is how you can calculate it:
1. Find the potential energy of the water at the surface of the tank: The potential energy is given by the equation PE = mgh, where m is the mass of the water, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height of the water above the reference point (in this case, the bottom of the tank). In this case, the distance from the surface of the water to the bottom of the tank is (18 - 13) = 5 m.
PE = mg * h = ρVg * h = ρAhg * h = ρAh * g * h, where ρ is the density of water and A is the area of the water surface. Since the area of the water surface on the tank is constant and equals A, we can simplify it as:
PE = ρAhgh
2. Find the kinetic energy of the water as it exits the spout: The kinetic energy is given by the equation KE = 1/2 * mv^2, where m is the mass of the water and v is the velocity of the water as it exits the spout. The mass of the water can be expressed as the product of its density and volume: m = ρV.
KE = 1/2 * (ρV) * v^2 = 1/2 * ρAh * v^2
3. According to the principle of conservation of energy, the potential energy of the water at the surface of the tank is equal to its kinetic energy as it exits the spout.
PE = KE
ρAhgh = 1/2 * ρAh * v^2
4. Simplify and solve for v:
gh = 1/2 * v^2
2gh = v^2
v = √(2gh)
Substituting the given values, where g is approximately 9.8 m/s^2 and h is 13 m:
v = √(2 * 9.8 m/s^2 * 13 m)
v ≈ √254.8
v ≈ 15.97 m/s.
Therefore, the water will emerge from the spout with a speed of approximately 15.97 m/s.