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Pressure drop |
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| Feb23-05, 10:23 PM | #1 |
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Pressure drop
Let's say we have water flowing in a pipe under so and so pressure at a given point. Now the water flows through bends, turns and straight lines of pipe. Will the pressure after a distance be lower than it was at the starting point, due to friction in the pipes and bends and such? Or will the pressure be constant at all points? The pipe dimension is constant.
I'm asking 'cause we're learning about radioators in houses, and the fact that one must account for what I can translate to mean "pressure drop" as the water flows through the system. I'm just having a bit of trouble believing that the pressure is different at different locations in the system. If we measure 3 bars just after the pump, won't the pressure be 3 bars everywhere in the system...? Secondly I wonder about this; if we place water in a pipe at let's say 2 bars, does that mean that we have pushed more water into the pipe than there's actually room for? In other words; have we compressed the water? I didn't think that was possible...? I'm confused, please someone help me!
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| Feb24-05, 12:18 AM | #2 |
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Recognitions:
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AM |
| Feb24-05, 06:58 AM | #3 |
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Recognitions:
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For anything to flow in a pipe from point A to point B there MUST be a difference in pressure. In your example, water is flowing in pipes. Even if you had a straight pipe that had no restrictions or enlargements, bends or turns, there would still be a pressure drop along the length of the pipe. That is due to friction developed as a result of the fluid's viscocity. When you start to add things like elbows, tees and the like to go to all of the different items in a house, each one of those small items adds a bit more resistance to the flow. As a matter of facet, we equate the losses induced by these items in terms of an equivilent length of straight pipe.
Just remember, in the real world, for anything to flow, there must be a difference in pressure. |
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