With reference to alkalinesolution1 Post #6047, 01/30/14
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AOT Technical Question:
“A” equals the VOLUME of oil in the pipeline upstream of the AOT.
“B” equals the VOLUME of oil in the pipeline downstream of the AOT.
Even though the flow rate is increased by the AOT, how is the VOLUME of oil “B” increased by the AOT if the volume of oil “A” coming into the equipment is constant?
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The answer to this question : The flow rate through A is equal to the flow rate through B
This is simply the principle of conservation of mass. Because no fluid can leave through the walls and there are no “sources” or “sinks” wherein the fluid can be created or destroyed, the mass crossing section A and then section B of the tube per unit time must be the same.
We know that the oil entering the AOT is more viscous. We also know that when you reduce the diameter of a pipe, the velocity of a liquid increases. So it is logical that the incoming oil, moving at a slower velocity, will enter the AOT through a slightly larger pipe and exit through a slightly narrower pipe, maintaining the flow rate while experiencing reduced viscosity and increased velocity. I think this also explains why the AOT midstream is set up in four pack (5,000 X 4 gpm) each of the four outlets are sized to carry 5,000 gpm back to the main pipeline, while maintaining the constant flow rate. The key word here is 'sized'. The oil on arrival at the next pump station has increased in viscosity and slowed down in velocity, so it is gradually fed into a larger pipe that will maintain the flow rate (volume) while adjusting to a slower velocity.
I examined the AOT midstream Four Pack photos, I am not imaging things here, I can detect a slightly larger pipe size on inlet compared to the outlet pipe size.
