ME 367 HW2 Solution

ME 367 HW2 Solution - Problem 1.35[1 Given Find...

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Problem 1.35 [1] Given: "Engineering" equation for a pump Find: SI version Solution: The dimensions of "1.5" are ft. The dimensions of "4.5 x 10 -5 " are ft/gpm 2 . Using data from tables (e.g. Table G.2), the SI versions of these coefficients can be obtained 1.5 ft 1.5 ft 0.0254 m 1 12 ft × = 0.457 m = 4.5 10 5 × ft gpm 2 4.5 10 5 ft gpm 2 0.0254 m 1 12 ft × 1 gal 4 quart 1quart 0.000946 m 3 60 s 1min 2 × = 4.5 10 5 ft gpm 2 3450 m m 3 s 2 = The equation is Hm ( ) 0.457 3450 Q m 3 s 2 =
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Problem 3.10 [2] Given: Properties of a cube suspended by a wire in a fluid Find: The fluid specific gravity; the gage pressures on the upper and lower surfaces Solution: From a free body analysis of the cube: Σ F0 = Tp L p U () d 2 + Mg = where M and d are the cube mass and size and p L and p U are the pressures on the lower and upper surfaces For each pressure we can use Eq. 3.7 pp 0 ρ g h + = Hence p L p U p 0 ρ g Hd + + p 0 ρ g H + = ρ g d = SG ρ H2O d = where H is the depth of the upper surface Hence the force balance gives SG T ρ H2O g d 3 = SG 2 slug 32.2 × ft s 2 lbf s 2 slug ft × 50.7 lbf
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ME 367 HW2 Solution - Problem 1.35[1 Given Find...

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