On November 16, 2017 the Keystone pipeline had a leak in Amherst, South Dakota. According to an NPR article, 210,000 gallons of oil spilled out of the pipeline before it was shut down. The spill covers approximately a 100 yard radius of flat grassland. According to a CNN article, the pipeline is 36 inches (.91m) in diameter and can carry 830,000 barrels of oil per day. There are 42 gallons in a barrel and one gallon is approximately0.004m^3. Thus 1.32x10^5 m^3 of oil flow each day, at a rate of 1.53m/s.
When the oil is flowing through the pipeline, it moves with laminar flow because the pipeline has the same height throughout its entirety.
A lot of oil spills are caused by corrosion at the bottom of the pipe to the point where a hole is created. I was curious what the pressure was at the bottom of the pipeline:
Pressure = (rho)(g)(h) + Patm
Density of oil = 900kg/m^3
Pressure = (900kg/m^3)(9.8m/s^2)(.91m) +1.013x10^5Pa = 1.09x10^5 Pa
If corrosion did occur all the way through and a hole developed on the bottom of the pipe, I wanted to know what the velocity of the oil would be when it exited:
P1 + rhogh1 + 1/2rhov1^2 = P2 + rhogh2 + 1/2rhov2^2
1.013x10^5Pa + (900kg/m3)(9.8m/s^2)(.91m) + 0 = 1.013x10^5Pa + 0 + .5(900kg.m^3)(v2)^2
1.09x10^5 = 1.013x10^5 + 450v2^2
8.03x10^3 = 450v2^2
v2^2 = 17.8m/s
v2 = 4.2m/s
The current crossectional area of the pipeline is (.91m/2)^2π = .65m^2. Periodically throughout the pipeline there are places where the line can be closed off to adjust the flow rate or in case of emergency. The pipeline company can control when this happens. If for example they close it half way, the new crossectional area would be .65m^2 / 2, or .325m^2. What would the resulting velocity of the oil be?
A1V1 = A2V2
(.65m^2)(1.53m/s) = .325m/s^2(V2)
V2 = 3.05m/s