Trans Alaska PipelineSome 800,000 barrels of oil per day are pumped 800 miles from Prudhome Bay to Valdez where it is loaded into tankers. At a location in the pipeline, a gage measures the average flow velocity of the oil passing through the four-foot diameter pipe. What is the average speed of the flowing oil?

♦  Let us imagine an event that starts right now, at the time, t = 0+, with us standing beside the flow gage. By this time tomorrow 800,000 barrels of oil will have passed. The pipe is four feet in diameter and there are 31.5gal/bbl.

At t = 0+, way upstream a distance L toward Prudhome, is a drop of oil moving in the pipe. That drop will be precisely here in 24 hours. The mass of oil contained by the pipe between here and the "drop" will have passed this location also. We know the volume of that mass, we use it and the pipe diameter mass to calculate the distance L.

This length is the magnitude of the position vector (at t = 0+) of the last drop of 800,000 barrels of oil pass this locaton by tomorrow. The actual pipe is not straight, but our answer will be correct. The initial and final position vectors of the drop are:

Uniform motion of a point in the oil is expressed as:

And upon substitution of our conditions,

At first it appears the answer is a negative speed while speed is zero or greater. The negative does not belong to the speed; it belongs to the direction ( - I ); meaning simply that the velocity is toward origin, the section of observation.

♦  A second way of determining the oil flow velocity is to visualize the filling of an empty oil barge. The image, the barge Douglas M., is empty and has the hold dimensions, 120 by 820 by 40 feet.

We take the mass of oil in the hold as our system. We write that mass symbolically at two times:
m( t* + Δt ) and at time, m( t = t*). Next substract the later from the former. That group is called a "difference." So divide the difference by the increment, Δt take the limit (as Δt → 0) and we have the derivative of mass which is the instantaneous rate of change of mass in the hold.

The rate of change of mass of oil within the tanker equals the constant mass rate (m-dot_in) of oil that flows into the tank. Our simple differential equation becomes:

We developed this first order differential equation because some readers might not be familiar with it. The mass flow across the system boundary, m-dot_in, equals the oil density time pipe area times flow speed. We now use constant hold geometry ( A_barge ) and constant density of oil to change the equation variable from mass to depth of oil, H(t).

The velocity of oil delivered in the pipe, V_in,avg is the answer we seek. As the tanker is filled, the depth of oil, H( t ), increases from zero to a second height. A simple calculation gives the oil depth after 800,000 barrels are loaded.

For this differential equation we have that the initial depth of oil in the tank, D( t = 0+) equals zero and the depth 24 hours later is D( t = 24hr) is 34.1 feet. So we separate variables and integrate:

Entering numbers...