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Ch 12: Fluid Mechanics

Chapter 12, Problem 12

A closed container is partially filled with water. Initially, the air above the water is at atmospheric pressure (1.01×10^5 Pa) and the gauge pressure at the bottom of the water is 2500 Pa. Then additional air is pumped in, increasing the pressure of the air above the water by 1500 Pa. (a) What is the gauge pressure at the bottom of the water?

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Hi everyone in this particular problem, we are asked to determine the reading on the pressure gauge connected to the lowest point of the water column. Along experimental container is placed with its length being vertical and is filled with seawater. In one particular instance, the air above the water column has the pressure equal to the atmospheric pressure. While a pressure gauge connected to the lowest point of the water column reads 11,760 pascal's. The container is connected to a pressurized air supply, adding air and raising the pressure above the water column by pascal's. And in this particular problem, we want to start with just identifying what 7 11,060 pascal's is and what 2500 pascal. So recall that to calculate pressure in a pressure gauge. We want to use this formula right here, which is B will equals to the atmospheric pressure plus H rho G. Which is the height difference. And then the gauge pressure is Ashley going to equal to the pressure difference. So the gauge pressure is going to equal to p minus P A. T. M, which B is from here. We can substitute api with this formula. So it's P A. T. M plus H rho G minus P A. T. M. So essentially delta P. Or the gauge pressure. It's just H multiplied by ro, multiplied by G. So initially it is mentioned that the lowest point of the water column reads 11,760 pascal's and that will be our delta P. That will be our delta p. So delta P equals 7 11 7 60 pascal's. And this is going to be only the pressure due to the height of the water column in the container. Next we know that in the final state it is going to be raising the pressure. The pressure is going to be raised by 2500 pascal's. So in final stage or state the pressure is gonna be the atmospheric pressure plus 2500 pascal's. Don't forget in a pressure gauge that will automatically should be automatically added with this delta P. R. D. H. O. G. Here. So I'm gonna write plus delta P. And like so so this will essentially be pressure equals P A. T. M. Plus 2500 P pascal's plus 7 60 pascal's just like so I'm gonna leave this as it is because what we are asked is determine the reading on the pressure gauge connected to the lowest point of the water column. And when reading a pressure gauge then what is being read is the delta P. Value. So the delta P. Is can can be calculated with this. It's the new P. Which is this P. In the final state minus the atmospheric p. So the new P. Is gonna be P. A. T. M. Plus 25:00 pascal's plus 11 60 pascal's minus P. A. T. M. Therefore we can cross off the P. A. T. M. And delta P. Or the pressure gauge reading is then going to be 14,260 pascal's just like so. And that will be the answer to this problem, which is 14,000 to 60 pascal's that will correspond to option E just like so. And that will be all. We actually do not need the value of the atmospheric pressure because we are handling changes in pressure in the pressure gauge itself. So let me know if you guys still have any questions and there will be a lot of other videos similar to this particular topic that you guys should check out if you guys still have any sort of confusion. So that will be all for this particular problem. And yeah.
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