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QFlow

Civil/Environmental
Sep 27, 2004
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An oil pipeline was constructed to transport an oil with SG=0.8 for a distance of 10 miles. The country was hilly and the line makes many ups and downs. The rolling terrain can be considered 10 rises at 200 ft, followed by descents of 200 ft. When the pipe was completed, water was pumped through with an inlet pressure of 150 psi to test it. Upon satisfactory flow of water, oil was slowly fed into the pipe. As the oil flowed, the pressure at the inlet began to rise, and the flow rate began to fall. Finally, the flow stopped altogether, while the pressure at the inlet side remained at 150 psi. Explain what caused this. (hint: manometer problem.) PLEASE HELP!!
 
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Sound like the viscosity of the oil is the problem. Viscosity depends on the type oil (crude? refined? etc.) and the temperature.
Kinematic viscosity of water at say 80 deg F. is about 0.85 centistokes. At that same temperature crude oil could easily be 10 to 50 times more viscous.
Water is great for a static pressure (hydro) test for leaks but gives no real indication of the flow characteristics of oil.

See if you can find a copy of "Cameron Hydraulic Data" for a summary of information on the subject of viscosity, pipeline friction & pressure loss.
 
I must be missing something here.

The terrain has 200ft rises or 60.96m. Water rising to the top of a 200ft peak must require a pressure of at least 6 bar (1 bar = 10m of water pressure and 1 bar = 145 psi). Thus pumping water at 150 psi gives us a pressure of no more higher than 11m water. Thus the oil pipeline have never been tested beyond the height above 11m, let alone the full length.

Pumping oil doesn't change that.

Have I missed something?

Is this just somebody's homework again?
 
Bbird - I can't confirm your math. The static pressure of 1 foot of water is 0.433 psi (62.4 lb/ft^3 / 144 in^2/ft^2). Therefore the static pressure for 200 ft head is about 87 psi (0.433 psi/ft x 200 ft).
 
I like RWF7437's air locks.

If the line is filled completely in the 200ft uphill sections, on the downhill sections an airlock may form if the more viscous oil can not flow fast enough to fill the void before the bottom of the "valley" is filled, so there may not be enough oil in the downhill sections to balance the next uphill section.

So you need about 6 bar to push the oil uphill, but may only get only get 3 bar back downhill so for each up/down section you need 3 bar.

Ten up/down sections are going to need 30 bar, your pump only supplies 150 psi or about 10 bar.

With the lower vicosity of the water, the downhill sections must have filled quickly and displaced the air.

Jeff
 
Can I ask one really fundamental question? In the sizing of the line and pump, inconsideration of the oil product, who calculated the losses in the system? I say this because in this part of the world, pipelining bituminous oil is a way of life, but we allow the experts to do so. Typically the oil is very warm and the pressures are very high. Just a thought.

KRS Services
 
Been away for a month and come back to this thread again.

I think notnats has got it. The system acts like a series of U-tubes. Each summit will have an air lock which must be compressed in order to push the oil column along.

After flushing with water the first downhill section will be full of air which is gradually compressed and escape into the subsequent summit because it will float to the top for being lighter than the oil and water.

Starting from the far end each summit must be balanced by a head of oil/water column of up to 200ft high plus the compression of the air lock. Thus each successive air lock will consume a portion of the pump head until the system is held in equilibrium with zero flow.

The oil/water column should be moving again if the system is de-aired, as liquid is nearly incompressible.
 
I do not really believe the air pocket theory. I have the same doubts as KRS. Perhaps water just flowed due to syphoning action and pump really was not working and just dancing?
 
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