Bernoulli’s Theorem
The pore-fluids (water, oil & water) associated with an oil reservoir have internal fluid friction (or viscosity). Hence, the work done on water, oil & gas particles would remain “greater” than the respective increases in their mechanical energy – because - a fraction of work done would have got converted into heat energy through the action of internal friction.
Whether this heat energy would remain sensitive (or significant) enough at the sub-pore-scale - in influencing the reduction in Inter Facial Tension (IFT) or Wettability Reversal - associated with Chemical/Microbial EOR?
Will the generated heat energy would vary significantly as a function of API Gravity of Crude?
Hi Suresh and all.
Strictly, the term “generated heat energy” is rather inadequate since energy is conserved. The meaning of the word “generated” is too close to “created” and, I believe, should be avoided. Energy is always transformed/converted from one form to another. A better term is “dissipated heat energy” or even better, “dissipated thermal energy”. Sorry for the academic preciosism. Mind you, a small fraction of the fluid mechanical energy lost, say, between two cross-sections in a pore chanell, remains in the fluid increasing its internal energy and, consequently, its temperature along the passage. The dissipated fraction will heat-up the solid matrix nearby.
The requirement for viscous dissipation is that the “no slip condition” hold, that is, fluid velocity is zero at the pore wall. How sensitive/relevant it will be in a given flow problem, depends on the proportion of the sub-pore-scale rock present in the reservoir. If the reservoir is located deep in the Earth crust, it will probably be too hot for dissipation effects to show. Be carefull with two-phase flow since interfacial tensions are present, affecting the flow and the mathematical modelling.
Known violations of the “no slip condition” relate to flow of polymer melts and molecular/Knudsen flow (gases under very low pressure).
Energy is conserved so the heat gained by the fluid as it passes through the pores is calculable. This heat will cause a temperature rise which will I believe be small unless the viscosity or the flow rate is very high. You can get the answer from Bernoulli's equation if you ignore any heat taken up by the formation (i.e. the rocks). Let's assume flow down a capillary pore with a pressure gradient of 0.1 bar/m i.e. 10 000 Pa/m. If the fluid is oil of density 900 kg/m3, the decrease in energy per unit mass is 10 000/900 = 11 J/kg. The increase in temperature would therefore be 11/specific heat capacity. The specific heat capacity of crude oil is about 2000 kJ/kg. Therefore the increase in T is about 0.005 degrees Celsius. Not enough to change the viscosity.
I know oil flows are radial and so a fixed pressure gradient is inappropriate but, these numbers show that it's quite difficult to get much change in T unless you have pressure gradients a couple of orders of magnitude larger than the one I chose.
The simple and nice numerical example given by R. B. Thorpe shows how insignificant is energy dissipation during fluid flow, in general. Since Bernoulli’s equation is valid for ideal/non viscous fluids, the example refers to the mechanical energy equation (or macroscopic energy balance) which hold for real/viscous fluids. The calculation assumes a horizontal pore channel (avoiding changes in gravitational potential energy) with constant cross-section (avoiding changes in kinetic energy) which would, otherwise, complicate the analysis. Actually, the result is 11 J/(m kg) since one meter of channel was considered. This is the well-known fluid head loss per meter during flow. Interestingly, if we express the fluid energy per unit weight (rather than mass) per unit length, the fluid head loss per meter is a dimensionless quantity characterizing the problem.
Excellent summary by Prof R B Thorpe and Prof R P Pecanha. Thanks for your time.
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