A horizontal pipe of diameter 0.991 m has a smooth constriction to a section of diameter 0.5946 m. The density of oil flowing in the pipe is 821 kg/m^3
If the pressure in the pipe is 7240 N/m^2? and in the constricted section is 5430 N/m^2, what is the rate at which oil is flowing?
Answer in units of m^3/s.
To solve this problem, we can apply the principle of continuity, which states that the mass flow rate of an incompressible fluid through a pipe is constant, given by the equation:
A1v1 = A2v2
where A1 and A2 are the cross-sectional areas of the pipe at the two sections, and v1 and v2 are the velocities of the oil flowing through the pipe at the two sections.
Since the oil is incompressible, the mass flow rate is given by:
m_dot = ρ * A1 * v1 = ρ * A2 * v2
where ρ is the density of the oil.
We can rearrange the equation for the mass flow rate to solve for v1:
v1 = (A2/A1) * v2
Using the equation of continuity:
A1v1 = A2v2
we can solve for v2:
v2 = (A1/A2) * v1
Given that the pressure in the pipe is 7240 N/m^2 and in the constricted section is 5430 N/m^2, we can use the Bernoulli's equation to relate the pressure difference to the velocity difference:
ΔP = 0.5 * ρ * (v2^2 - v1^2)
Substitute the values and solve for v1:
1810 = 0.5 * 821 * (v2^2 - v1^2)
And since v2 = (A1/A2) * v1, we have v2 = (0.991/0.5946) * v1
Substitute this into the previous equation and solve for v1:
1810 = 0.5 * 821 * (((0.991/0.5946) * v1)^2 - v1^2)
Solving this equation, we get:
v1 ≈ 2.24 m/s
Now, we can calculate the cross-sectional area of the pipe at section 1 and section 2:
A1 = π * (0.991/2)^2 ≈ 0.770 m^2
A2 = π * (0.5946/2)^2 ≈ 0.278 m^2
Finally, we can calculate the mass flow rate of the oil:
m_dot = ρ * A1 * v1 ≈ 821 * 0.770 * 2.24 ≈ 1480.53 kg/s
Convert this mass flow rate to volume flow rate by dividing by the density of the oil:
Volume flow rate = m_dot / ρ ≈ 1480.53 / 821 ≈ 1.80 m^3/s
Therefore, the rate at which oil is flowing through the pipe is approximately 1.80 m^3/s.