p0330 - atmosphere, necessarily p ( z ) = p a . Thus, Hg g...

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280 CHAPTER 3. EFFECTS OF GRAVITY ON PRESSURE 3.30 Chapter 3, Problem 30 Figure 3.15: Ethanol tank with water. The pressure at the ethanol-water interface is p (2 h )= p b + ρ E g ( Nh ) where ρ E =1 .53s lug / f t 3 denotes the density of ethanol. Also, p (0) = p (2 h )+ ρ H 2 O g (2 h ) where ρ H 2 O = 1.94 slug/ft 3 denotes the density of water. Combining these two equations yields p (0) = p b +( N ρ E +2 ρ H 2 O ) gh Finally, in the manometer, we have p ( z )+ ρ Hg g z = p (0) where
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Unformatted text preview: atmosphere, necessarily p ( z ) = p a . Thus, Hg g z = p b p a + ( N E + 2 H 2 O ) gh Solving for p b , we have p b = p a + ( Hg z N E h 2 H 2 O h ) g Finally, substituting the given values, p b = 2116 . 8 psf + (26 . 30 slug / ft 3 )(1 ft) 10(1 . 53 slug / ft 3 )(2 ft) 2(1 . 94 slug / ft 3 )(2 ft) = X 32 . 174 ft sec 2 ~ = 2116 . 8 psf 388 . 0 psf = 1728 . 8 psf...
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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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