Water Flows upward throw the pipe shown in the diagram at 96 L/Min. If the pressure at the lower end is 80kPa, find the velocity of the water is at both ends and the pressure at the upper end. Assume that the density of water remains constant throughout the tube and that h1= 10 m and h2 = 13m
To solve this problem, we can apply Bernoulli's principle, which states that the total energy of a fluid flowing in a pipe remains constant along a streamline. This principle can be used to relate the velocity and pressure of the fluid at different points in the pipe.
First, let's calculate the velocity of water at both ends of the pipe:
Given:
- Flow rate = 96 L/min
- Density of water remains constant
- h₁ = 10 m (height at the lower end of the pipe)
- h₂ = 13 m (height at the upper end of the pipe)
Step 1: Convert flow rate to m³/s
To apply Bernoulli's principle, we need the flow rate in m³/s instead of L/min.
1 L = 0.001 m³
So the flow rate in m³/s is:
96 L/min * 0.001 m³/L * 1 min/60 s = 0.0016 m³/s
Step 2: Calculate the velocity at both ends of the pipe
The flow rate (Q) is equal to the cross-sectional area (A) multiplied by the velocity (v).
Q = A * v
Let's assume the cross-sectional area at the lower end of the pipe is A₁ and at the upper end is A₂.
Q = A₁ * v₁ (equation 1)
Q = A₂ * v₂ (equation 2)
Since the flow rate remains constant, we can equate equations 1 and 2:
A₁ * v₁ = A₂ * v₂
We need to find v₁ and v₂, but we don't know the cross-sectional areas. However, we can relate the velocities to the heights using the equation of fluid pressure:
P + 1/2 * ρ * g * h + 1/2 * ρ * v² = constant
Where:
P = pressure
ρ = density of water
g = acceleration due to gravity
h = height
Step 3: Calculate the velocities
For the lower end of the pipe (Point 1):
P₁ + 1/2 * ρ * g * h₁ + 1/2 * ρ * v₁² = constant (equation 3)
For the upper end of the pipe (Point 2):
P₂ + 1/2 * ρ * g * h₂ + 1/2 * ρ * v₂² = constant (equation 4)
Since the density of water remains constant and we are comparing the points in the same pipe, the constants in equations 3 and 4 will be the same.
Step 4: Calculate the pressure at the upper end (P₂)
Given:
Pressure at the lower end (P₁) = 80 kPa
Substituting the known values into equation 3:
80,000 Pa + 1/2 * ρ * g * 10 m + 1/2 * ρ * v₁² = constant
Simplifying and rearranging:
1/2 * ρ * v₁² = constant - 80,000 Pa - 1/2 * ρ * g * 10 m
1/2 * ρ * v₁² = constant' (equation 5)
Substituting the known values into equation 4:
P₂ + 1/2 * ρ * g * 13 m + 1/2 * ρ * v₂² = constant' (equation 6)
Since the constant in equation 6 is the same constant' as in equation 5, we can subtract equation 5 from equation 6 to eliminate the constants:
P₂ - 80,000 Pa + 1/2 * ρ * g * 13 m - 1/2 * ρ * g * 10 m + 1/2 * ρ * v₂² - 1/2 * ρ * v₁² = 0
Simplifying:
P₂ - 80,000 Pa + 1/2 * ρ * g * (13 m - 10 m) + 1/2 * ρ * (v₂² - v₁²) = 0
Since the density of water remains constant, we can cancel it out:
P₂ - 80,000 Pa + 1/2 * g * (13 m - 10 m) + 1/2 * (v₂² - v₁²) = 0
Simplifying further:
P₂ - 80,000 Pa + 1/2 * 9.8 m/s² * 3 m + 1/2 * (v₂² - v₁²) = 0
We can now solve this equation to find P₂.