Water from the dead sea flows into a pipe of diameter 18cm at a velocity of 5.0m s at pressure 4.0atm. The pipe reduces to a diameter of 10cm at a point 3.5m above.
a)Determine the rate of flow.
b)Determine the velocity at the top section.
c)Determine the pressure at the top section.
To solve this problem, we can use the principle of continuity, which states that the volume flow rate of an incompressible fluid is constant along a pipe of varying cross-sectional area.
a) To determine the rate of flow, we can use the equation:
Q = A1 * v1 = A2 * v2
where Q is the rate of flow, 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 the diameter at point 1 is 18 cm and the diameter at point 2 is 10 cm, we can calculate the areas:
A1 = π * (d1/2)²
A2 = π * (d2/2)²
where d1 and d2 are the diameters.
By substituting the given values into the equation, we get:
Q = π * (18/2)² * 5.0 = π * (10/2)² * v2
Simplifying, we find:
Q = 81π m²/s
Therefore, the rate of flow is 81π m²/s.
b) To calculate the velocity at the top section, we can rearrange the equation as follows:
v2 = (A1 * v1) / A2
Substituting the given values, we have:
v2 = (π * (18/2)² * 5.0) / (π * (10/2)²)
Simplifying, we find:
v2 = (81 * 5.0) / 25 = 16.2 m/s
Therefore, the velocity at the top section is 16.2 m/s.
c) To determine the pressure at the top section, we can use Bernoulli's equation, which states that the sum of the pressure energy, kinetic energy, and potential energy per unit volume of a fluid is constant along a streamline:
P1 + (1/2) * ρ * v1² + ρ * g * h1 = P2 + (1/2) * ρ * v2² + ρ * g * h2
where P1 and P2 are the pressures at points 1 and 2 respectively, ρ is the density of the fluid, v1 and v2 are the velocities at points 1 and 2 respectively, and g is the acceleration due to gravity. We can assume that the potential energy terms cancel out since the change in height is small compared to the other terms.
Given that the pressure at point 1 is 4.0 atm and the velocity at point 1 is 5.0 m/s, and assuming the density of water to be 1000 kg/m³, we can rearrange the equation to solve for the pressure at point 2:
P2 = P1 + (1/2) * ρ * (v1² - v2²)
Substituting the values into the equation, we get:
P2 = 4.0 atm + (1/2) * 1000 kg/m³ * (5.0² - 16.2²) m²/s²
Simplifying, we find:
P2 = 4.0 atm + 0.5 * 1000 kg/m³ * (-187.764) m²/s²
P2 ≈ 4.0 atm - 93882 atm
Therefore, the pressure at the top section is approximately -93878 atm. Note that the negative sign indicates the pressure is below atmospheric pressure.