Respuesta :
The continuity equation allows finding the results for the speed and flow of the outlet pipe are:
- Q = 3.14 kg / s
- v₂ = 0.4 m / s
Fluid mechanics studies the movement and fluid dynamics, it is described by two fundamental relationships:
- The Bernoulli equation that an expression of the energy conservation
[tex]P_1 + \frac{1}{2} \rho v_1^2 + \rho g y_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g y_2[/tex]
Where P is the pressure, v the velocity, y the height, the subscripts are two selected points of interest in the pipe.
- The continuity equation which is an expression that our principle of mass conservation for fluids
[tex]A_1v_1 = A_2 v_2[/tex]
Where A₁ and A₂ is the area of the inlet and outlet pipes and v₁ and v₂ are the velocities of the fluid at the inlet and outlet
They indicate the inlet and outlet diameters of the pipe are d₁ = 0.02 m and d₂ = 0.1 m, the inlet velocity v₁ = 10 m / s, with the continuity equation let's find the outlet velocity
[tex]v_2 = ( \frac{A_1}{A_2} )^2 \ v_1[/tex]
the cross section of the tavern is
A = [tex]\pi r^2 = \pi \ d^2/4[/tex]
Where r is the radius and d is the diameter
Let's substitutes
[tex]v_2 = (\frac{d_1}{d_2} )^2 \ v_1[/tex]
[tex]v_2 = (\frac{0.02}{0.1} )^2 \ 10[/tex]
v₂ = 0.4 m / s
The flow (Q) is defined as the amount of liquid that passes through the section of the pipe per unit of time, we can see that this is the same conservation of mass, therefore
Q = A v
Q = [tex]\pi \frac{d_2^2}{4} \ v_2[/tex]π d2 ^ 2 v2
Q = [tex]\pi \ \frac{0.1^2}{4} \ 0.4[/tex]
Q = 3.14 10⁻³ m ^ 3 / s
As the transported liquid is water
1 liter water = 1 kg
Consequently
1 m³ = 1000 liter = 1000 kg of water
Let's reduce the flow
Q = 3.14 10⁻³ m³/s ( [tex]\frac{10^3 kg}{1 m^3}[/tex] )
Q = 3.14 kg / s
In conclusion using the continuity equation we can find the results for the speed and flow of the outlet pipe are:
- v₂ = 0.4 m / s
- Q = 3.14 kg / s
Learn more here: brainly.com/question/14747327
