Water flows steadily from an open tank into a pipe. The elevation of the top of the tank is 10.2 m, and the elevation at the pipe is 3.90 m. The initial cross-sectional area of the pipe is 6.40×10^−2 m; and at where the water is discharged from the pipe, it is 2.30×10^−2 m. The cross-sectional area of the tank is very large compared with the cross-sectional area of the pipe. Assuming that Bernoulli's equation applies, compute the volume of water that flows across the exit of the pipe in the time interval 2.40 seconds.
The average head is 10.2-3.9=6.3 m
g=9.8 m/s²
When the area of the orifice is small compared to the tank area, and if the tank is vented, pressure at the water surface equals that of the outlet, then the output velocity is approximately √(2gh).
Volume of discharge
=velocity * area * time
=√(2gh)*2.30×10^−2*2.4
=0.6 m³ approx.
For the derivation of the simplified formula, see:
http://www.engineeringtoolbox.com/bernouilli-equation-d_183.html
To compute the volume of water that flows across the exit of the pipe, we need to use Bernoulli's equation. Bernoulli's equation relates the pressure, velocity, and elevation of a fluid in a streamline flow. Here's how we can solve this problem step by step:
1. Determine the pressure difference between the top of the tank and the pipe exit. Since the cross-sectional area of the tank is large compared to the pipe, we can assume the pressure at the top of the tank is atmospheric pressure (Patm). The pressure at the pipe exit can be calculated using the equation: Pexit = Patm + ρgh, where ρ is the density of water (1000 kg/m³), g is the acceleration due to gravity (9.8 m/s²), and h is the elevation difference between the top of the tank and the pipe (10.2 m - 3.90 m = 6.30 m).
2. Calculate the velocity at the pipe exit using the continuity equation. The continuity equation states that the product of the velocity and the cross-sectional area is constant along a streamline. Therefore, we can write the equation as: A1v1 = A2v2, where A1 and A2 are the initial and final cross-sectional areas, and v1 and v2 are the corresponding velocities. Rearranging the equation, we can solve for v2: v2 = (A1v1) / A2. Plugging in the values, we have v2 = (6.40×10^−2 m)(v1) / (2.30×10^−2 m).
3. Use the equation for the volume flow rate to find the volume of water that flows across the exit of the pipe in 2.40 seconds. The volume flow rate (Q) is given by the equation Q = Av, where A is the cross-sectional area and v is the velocity. Multiplying the volume flow rate by the time interval will give us the volume of water. Therefore, the volume of water (V) is equal to Q multiplied by the time (t): V = Q × t = (A2v2) × t. Substituting the previously calculated values, we can solve for the volume of water.
By following these steps, you should be able to calculate the volume of water that flows across the exit of the pipe in the given time interval.
To compute the volume of water that flows across the exit of the pipe in the time interval of 2.40 seconds, we can use Bernoulli's equation:
P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂
Where:
- P₁ and P₂ are the pressures at points 1 and 2 respectively,
- ρ is the density of the fluid (water in this case),
- v₁ and v₂ are the velocities of the fluid at points 1 and 2 respectively,
- g is the acceleration due to gravity,
- h₁ and h₂ are the elevations at points 1 and 2 respectively.
Given values:
- h₁ = 10.2 m (elevation of the top of the tank)
- h₂ = 3.90 m (elevation at the pipe)
- A₁ = 6.40×10^−2 m² (initial cross-sectional area of the pipe)
- A₂ = 2.30×10^−2 m² (cross-sectional area at the exit of the pipe)
- Δt = 2.40 seconds
We can assume that the pressure at the top of the tank is atmospheric pressure, and the pressure at the pipe exit is also atmospheric pressure (since it is an open pipe). So P₁ = P₂ = 1 atm = 101325 Pa.
We also have the following relationship between the velocities and areas:
A₁v₁ = A₂v₂
Let's calculate the velocities first:
v₁ = (A₂/A₁) × v₂
Since the cross-sectional area of the tank is much larger than that of the pipe, we can assume that the velocity at the tank is negligible, so v₁ ≈ 0. Hence:
0 = (A₂/A₁) × v₂
v₂ = 0
Since v₂ = 0, the term ½ρv₂² in the Bernoulli's equation becomes zero.
Now we can compute the volume of water that flows across the exit of the pipe using the equation:
Volume = A₂ × v₂ × Δt
Substituting the known values:
Volume = (2.30×10^−2 m²) × 0 × 2.40 seconds
Volume = 0
Therefore, the volume of water that flows across the exit of the pipe in the time interval of 2.40 seconds is 0.