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p0330 - atmosphere necessarily p ∆ z = p a Thus ρ 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 = 1.53 slug/ft 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 ρ Hg = 26.30 slug/ft 3 denotes the density of mercury. Since the tube is open to the
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Unformatted text preview: atmosphere, necessarily p ( ∆ z ) = p a . Thus, ρ Hg g ∆ z = p b − p a + ( N ρ E + 2 ρ H 2 O ) gh Solving for p b , we have p b = p a + ( ρ Hg ∆ z − N ρ E h − 2 ρ 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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