A pressurized tank of water has a 10cm diameter orifice at the bottom, where water discharges to the atmosphere. the water level is 3 m above the outlet. the tank air pressure above the water level is 300 kpa (absolute)while the atmospheric pressure is 100 kpa. neglecting frictional effects. determine the velocity of the discharged flow and the volume flow rate of water from the tank.
0.168m^3/s
To determine the velocity of the discharged flow and the volume flow rate of water from the tank, we can use Bernoulli's equation and the principle of continuity.
1. Velocity of the discharged flow:
Bernoulli's equation states that the sum of the pressure, kinetic energy, and potential energy per unit volume of a flowing fluid remains constant along any streamline. In this case, we can assume steady and incompressible flow.
Bernoulli's equation can be expressed as:
P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂
Where:
P₁ = pressure at point 1 (initial pressure)
P₂ = pressure at point 2 (discharged pressure)
v₁ = velocity at point 1 (initial velocity)
v₂ = velocity at point 2 (discharged velocity)
ρ = density of the fluid (water)
g = acceleration due to gravity
h₁ = height at point 1 (initial height)
h₂ = height at point 2 (discharged height)
In this case, P₁ is the air pressure above the water level (300 kPa absolute), P₂ is the atmospheric pressure (100 kPa), v₁ and h₁ are zero since the water is at rest initially, and h₂ is the height of the water above the outlet (3 m).
The equation simplifies to:
300 kPa + (1/2)ρv₁² + ρgh₁ = 100 kPa + (1/2)ρv₂² + ρgh₂
As mentioned earlier, v₁ = 0 m/s and h₁ = 0 m, so the equation further simplifies to:
300 kPa = 100 kPa + (1/2)ρv₂² + ρgh₂
Rearranging the equation, we get:
(1/2)ρv₂² = 200 kPa - ρgh₂
Since the density of water is known (approximately 1000 kg/m³) and g = 9.81 m/s², we can substitute these values into the equation:
(1/2)(1000 kg/m³)v₂² = 200,000 Pa (1 kPa = 1000 Pa) - (1000 kg/m³)(9.81 m/s²)(3 m)
Solving for v₂², we get:
v₂² = (200,000 Pa - 29,430 Pa) / (500 kg/m³)
v₂² = 340.14 m²/s²
Finally, taking the square root of v₂², we find the velocity of the discharged flow (v₂):
v₂ ≈ √340.14 ≈ 18.43 m/s
Therefore, the velocity of the discharged flow is approximately 18.43 m/s.
2. Volume flow rate of water from the tank:
The volume flow rate (Q) can be calculated using the principle of continuity, which states that the mass flow rate of an incompressible fluid remains constant through a confined flow path.
The equation for the principle of continuity can be written as:
Q₁ = Q₂
Where:
Q₁ = volume flow rate at point 1 (initial flow rate)
Q₂ = volume flow rate at point 2 (discharged flow rate)
The volume flow rate (Q) is given by:
Q = A₁v₁ = A₂v₂
Where:
A₁ = cross-sectional area at point 1 (initial cross-sectional area)
A₂ = cross-sectional area at point 2 (discharged cross-sectional area)
In this case, the orifice at the bottom of the tank is circular with a diameter of 10 cm or radius (r) of 5 cm or 0.05 m. Therefore:
A₁ = πr₁² = π(0.05 m)² = 0.00785 m²
The cross-sectional area at point 2 is considered to be almost zero compared to the orifice area, as the orifice is very small compared to the tank. Therefore:
A₂ ≈ 0
Since A₂ ≈ 0, the equation Q = A₂v₂ becomes Q ≈ 0. Hence, there is no volume flow rate at point 2 as the water is being discharged to the atmosphere.
Hence, the volume flow rate of water from the tank is approximately zero.
To determine the velocity of the discharged flow and the volume flow rate of water from the tank, we can use Bernoulli's equation and the continuity equation.
Step 1: Calculate the pressure difference:
The pressure difference is given by:
ΔP = P1 - P2
Where:
P1 = Pressure above the water level (300 kPa)
P2 = Atmospheric pressure (100 kPa)
ΔP = 300 kPa - 100 kPa
ΔP = 200 kPa
Step 2: Convert the pressure difference to Pascals:
1 kPa = 1000 Pa
So, ΔP = 200 kPa = 200,000 Pa
Step 3: Calculate the velocity of the discharged flow:
Using Bernoulli's equation, we have:
P1 + 0.5ρv1^2 + ρgh1 = P2 + 0.5ρv2^2 + ρgh2
Assuming the outlet is at the same level as the water surface (h2 = 0), and neglecting the speed of the water at the surface (v2 ≈ 0), the equation simplifies to:
P1 + 0.5ρv1^2 + ρgh1 = P2
v1^2 = (2 / ρ) * (P2 - P1) - 2gh1
Where:
ρ = Density of water (1000 kg/m^3)
g = Acceleration due to gravity (9.8 m/s^2)
h1 = Height of water above the outlet (3 m)
Substituting the known values:
v1^2 = (2 / 1000) * (200,000 - 0) - 2 * 9.8 * 3
v1^2 = 400 - 58.8
v1^2 = 341.2
v1 ≈ √341.2
v1 ≈ 18.47 m/s
Step 4: Calculate the volume flow rate of water from the tank:
The volume flow rate (Q) is given by:
Q = A * v1
Where:
A = Cross-sectional area of the orifice
The diameter of the orifice is given as 10 cm, which means the radius (r) is half that value:
r = 10 cm / 2
r = 5 cm = 0.05 m
The cross-sectional area (A) is given by:
A = πr^2
Substituting the value of the radius:
A = π * (0.05 m)^2
A ≈ 0.00785 m^2
Finally, calculating the volume flow rate (Q):
Q = 0.00785 m^2 * 18.47 m/s
Q ≈ 0.1448 m^3/s
So, the velocity of the discharged flow is approximately 18.47 m/s, and the volume flow rate of water from the tank is approximately 0.1448 m^3/s.