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Rotary Tech - General Rotary Engine related tech section.. Tech section for general Rotary Engine... This includes, building 12As, 13Bs, 20Bs, Renesis, etc... |
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#30 | |||
RCC Loves Me Not You
Join Date: Jul 2008
Location: Influx.
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Back to the proposed comment "Water does not flow like air" is absolutely false. Point in fact the original exploration of air flow was done with water and the results extrapolated out to air, which was then tested and confirmed. Now-a-days with the proper instruments and technical ability we can test air independent of water, however, what we learn from air only reinforces what we already have known. That is, if you solve for a particular flow using the Reynolds number you can match any medium you want to the same exact flow. This means I can not only match water and air, but I could also match oil, nitrogen, helium, oxygen, etc, etc, etc. Here's something to consider: Air (mostly composed of nitrogen, and so nitrogen is used), has a molecular weight of 14.0067 g/mol, and water has a molecular weight of (1.00794 g/mol)*2+ 15.9994 g/mol (or 18.01528 g/mol), flow exactly the same. However if we take the hypothesis of "water does not flow like air" we by extension must emphatically state that the components of water do not flow like air. This means hydrogen and oxygen do not follow the same laws as nitrogen. This is obviously false, as both hydrogen and oxygen flow exactly the same as nitrogen when one matches the Reynolds number. Since water is a Newtonian fluid as well one can do the same. Another thought experiment; what if water was turned gaseous? Would the flow be any different? No. The medium is water, but just in its gaseous state. One need only match the Reynolds number and the flows will be exact. Now for turbulent flows: Turbulent flows are a bit harder to model as they are equivalent to white noise in radio. They therefore remain random when looked at molecular level Eulerian perspective. This means statistical analysis is done on the medium using various fluid properties (and other boundary conditions) to determine the numerical basis of the turbulent region of concern. What I'm basically saying is that all Newtonian fluids that experience a turbulent region of flow will behave exactly the same. This means that when the Re is matched, the flow (regardless of density) remains the same. Turbulent Boundary regions, Turbulent flows, Laminar Flows, they flow the same. This is how fluids work. The exceptions to this rule however do exist. They include super fluids, or fluids that have almost no viscosity. These fluids are never used inside of commercial engines and play no part in the discussion at hand, but they do bear mentioning. They do not behave the same as regular Newtonian fluids. For more information concerning these fluids I suggest reading the wiki-article concerning them.
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