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Old 11-11-2008, 02:42 AM   #27 (permalink)
ConnClark
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Quote:
Originally Posted by instarx View Post
Sorry ConnClark, but you are mistaken. In a closed system as we are discussing,
But we are talking about an open loop system. An intercooler reduces the work required by the compressor to compress a given mass of air. It does not reduce the energy recovered by the turbine. This results in an increased mass of air compressed to balance the energy of what is recovered by the turbine. By definition this is an Open loop cycle.
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temperature does effect pressure, and it is easily measured because it is constant. A room is not a closed system so it is harder to measure,
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a room or a container is a closed loop system as there is no mass entering or leaving.
but in fact if you suddenly cooled one end of a hot room there would be airflow to the cold end because of the pressure differential. True, the pressure would equalize as the air mixed, but so would the temperature. If you took 10 pressure readings in the room you are in right now they would all be slightly different - and all directly related to the slightly different temperatures at each sample point.
However they would all have the same pressure. Pressure also equalizes far faster than temperature.
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As for magnitude, it seems reasonable to me. Three psi is only 1/10 of an atmosphere,
Since mean pressure at sea level is 14.7, 3 psi is about 20.4% or about 1/5th of an atmosphere.
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and reducing the temperature of air from 250F to 150F can easily induce a change of 0.10 bar (particularly when it is pressurized to begin with). Trust me on this - I spent a lifetime correcting air sample measurements to STP (standard temperature and pressure) for human exposure monitoring studies. Note that I made my biggest assumption in this paragraph - that the air was cooled about 100F by the IC. If this is way off let me know and I will recalculate the pressure drop that would be expected across the IC.
I will save you this trouble as it sounds like you already have a life time of work to recalculate since you assumed 1 atmosphere equaled 30 psi.
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Static pressure and dynamic pressure (also known as velocity pressure, VP) are independent of each other. VP is a measurement used to measure the inertia of moving air (usualy in a duct), and has no purpose when determining the pressure drop across a restriction such as, say, a filter or IC. In this situation VP values do not complicate the situation because they are simply not applicable.
If this is so then please explain the attached picture.
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I read the post that he was getting a pressure drop measured across the IC, hence he blamed restrictive airflow in the IC. If the pressure measurements were taken other places it might change things, but those were not the parameters of the problem, and the pressure drop across the IC is almost surely due to the temperature drop.
Once again I disagree. Any drop in pressure would be due to air flow restriction compounded by differences in velocity of the air flow at he points of measurements. Try playing around with a few simple computational fluid dynamics simulations that vary in temperature in a closed system. You will find that pressure is equal or does equalize very quickly. If you wish to invest the time, OpenFoam is a good package it just is very complex and has a steep learnung curve.
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