p0304 - 3 and ρ Oil = 870 kg/m 3 , we have ρ u = p − p...

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3.4. CHAPTER 3, PROBLEM 4 253 3.4 Chapter 3, Problem 4 Figure 3.1: Three-layer tank of fluid. The pressure at the bottom of the tank is the atmospheric pressure plus the weight of the fluid above, i.e., p = p a +( ρ Oil + ρ u + ρ Hg ) gh Now, we are given h = 60 cm = 0.6 m. Since the densities of mercury and SAE 10W Oil are ρ Hg = 13550 kg/m
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Unformatted text preview: 3 and ρ Oil = 870 kg/m 3 , we have ρ u = p − p a gh − ρ Oil − ρ Hg = (200000 − 101000) N / m 2 p 9 . 807 m / sec 2 Q (0 . 6 m) − 870 kg m 3 − 13550 kg m 3 = 16825 kg m 3 − 14420 kg m 3 = 2405 kg m 3 Therefore, the density of the unknown fluid is ρ u = 2405 kg m 3...
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This note was uploaded on 10/12/2009 for the course AME 309 at USC.

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