p0306 - ρ Hg = 13550 kg/m 3 ρ u = 1780 kg/m 3 and ρ...

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3.6. CHAPTER 3, PROBLEM 6 255 3.6 Chapter 3, Problem 6 Figure 3.3: Three-layer tank of fluid. Since 50 cm = 1 2 h in this problem, we seek the pressure at a point 1 2 h from the bottom ofthetank .Thepressurea tth ispo in tistheatmospheric pressure plus the weight of the fluid above, i.e., p = p a + w ρ Seawater + ρ u + 1 2
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Unformatted text preview: ρ Hg = 13550 kg/m 3 , ρ u = 1780 kg/m 3 and ρ Seawater = 1030 kg/m 3 , we have p = 1 . 01 · 10 5 N m 2 + X 1030 kg m 3 + 1780 kg m 3 + 1 2 · 13550 kg m 3 ~ w 9 . 807 m sec 2 W (1 m) = 101000 N m 2 + 94000 N m 2 = 101 kPa + 94 kPa = 195 kPa Therefore, the pressure 50 cm from the bottom is p = 195 kPa...
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This note was uploaded on 02/09/2012 for the course AME 309 taught by Professor Phares during the Spring '06 term at USC.

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