Water flows steadily from an open tank into a pipe as shown in the figure. The elevation of the top of the tank is 12.3 m, and the elevation at the pipe is 3.80 m. The initial cross-sectional area of the pipe (at point 2) is 5.60×10−2 m^2; and at where the water is discharged from the pipe (at point 3), it is 1.40×10−2 m^2. The cross-sectional area of the tank is very large compared with the cross-sectional area of the pipe.
Q: Assuming that Bernoulli's equation applies, compute the volume of water DeltaV that flows across the exit of the pipe in the time interval 4.40 s.
A similar question has been answered before:
http://www.jiskha.com/display.cgi?id=1259634522
If you need further help, please post what you have done, and explain where you need help.
the pipe has a cross sesctional are of 1.6cm^2.
The water comes out of the pipe at a speed of 14cm/s.
How long does it take to fill the tank? Give your answer in hours and minutes, correct to the nearest min
a rectangular when full of water takes 10 min to be emptied through an office in its bottom. how much time does it take to be emptied when half filled with water
To calculate the volume of water that flows across the exit of the pipe, we can use Bernoulli's equation. Bernoulli's equation relates the pressure, velocity, and elevation of a fluid flowing in a pipe or tube.
The equation is given by:
P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂
Where:
P₁ and P₂ are the pressures at points 1 and 2 respectively,
v₁ and v₂ are the velocities at points 1 and 2 respectively,
ρ is the density of the fluid,
g is the acceleration due to gravity, and
h₁ and h₂ are the elevations at points 1 and 2 respectively.
In this case, point 1 is at the top of the tank (elevation = 12.3 m) and point 2 is at the pipe (elevation = 3.80 m). Point 3 is where the water is discharged from the pipe.
We need to find the velocity at point 2 to solve the equation. We can use the principle of conservation of mass, which states that the mass flow rate at any point in a pipe is constant.
The mass flow rate is given by:
ρ₁A₁v₁ = ρ₂A₂v₂
Where:
ρ₁ and ρ₂ are the densities at points 1 and 2 respectively,
A₁ and A₂ are the cross-sectional areas at points 1 and 2 respectively, and
v₁ and v₂ are the velocities at points 1 and 2 respectively.
Given:
A₁ = 5.60×10⁻² m²,
A₂ = 1.40×10⁻² m².
Since the density of water is constant, ρ₁ = ρ₂ = ρ.
To find v₂, rearrange the equation:
v₂ = (A₁v₁) / A₂
Now we can substitute the values into Bernoulli's equation and solve for ΔV, the volume of water that flows across the exit of the pipe in the time interval 4.40 s.
P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρ[(A₁v₁)/A₂]² + ρgh₂
The pressure terms cancel out since the pipe is open to the atmosphere, so the equation becomes:
½ρv₁² + ρgh₁ = ½ρ[(A₁v₁)/A₂]² + ρgh₂
Simplifying further:
½ρv₁² + ρgh₁ = ½ρ(A₁/A₂)²v₁² + ρgh₂
Let's assume that the pressure terms are negligible, and solve for ΔV:
ΔV = A₂v₂Δt
Substitute the value of v₂ obtained earlier and solve for ΔV.