# #1 - Solutions to Homework # 1 Problem 1.24: Water vapor V...

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Unformatted text preview: Solutions to Homework # 1 Problem 1.24: Water vapor V = 2m . P = 0.3 MPa, T = 160 C, v = 0.651 m / kg V = m v ; Hence m = V/v = 2 / 0.651 = 3.07 kg. Number of moles n = m / M = 3.07 kg / 18.02 (kg/kmol) = 0.17 kmol. (b) Number of molecules in 1 gm mole = 6.022 x 10 23 23 3 o 3 In 0.17 kmol, number of molecules = 10 x 6.022 x 0.17 x 1000 = 1.02 x 10 molecules. 26 Problem 1.28 m = 2lbm; pV = constant; p1 = 20 lbf/in ; V1 = 10 ft ; p2 = 100 lbf/in ; V2 = 2.9 ft . a) 2 3 n 2 3 p1 V1 = p2 V2 ; p1/p2 = (V2/V1) ; 0.2 = (2.9/10) take logarithms on both sides. n n n n log (0.2) = n log (0.29); n = ­1.609 / ­1.2378 = 1.3 (b) v1 = V 1 / m = 10 / 2 = 5 ft /lbm; v2 = V2 / m = 2.9/2 = 1.45 ft /lbm 3 3 Problem 1.33 p 1, vac = 4.2 in. of manometer fluid; p atm = 14.5 lbf/in ; g = 32.2 ft/s ; ρ f = 49.94 lbm/ft 3 2 2 p 1, vac = ρ f g L = (49.94 lbm/ft ) (32.2 ft/s ) (4.2/12 ft.) x (1/32.2) x (1/144) = 0.1214 lbf/in 2 3 2 p 1, abs = patm – p 1, vac = 14.5 – 0.124 = 14.379 lbf/in Problem 1.38 2 The pressure acting on the vehicle at a depth of 1000 ft. is the sum of the atmospheric pressure and the pressure exerted by the water column. The total pressure = patm + ρ g L = 1 atm + (62.4 lbm/ft ) (32.2 ft/s ) (1000ft. ) x (1/32.2 lbf / lbm . ft/s ) x (1/144 ft /in ) x (1/14.696 atm/lbf/in ) = (1 + 29.49) atm = 30.49 atm 2 2 2 3 2 2 Problem 1.53 40 C = 1.8 x 40 + 32 = 104 F 37 C = 1.8 x 37 + 32 = 98.6 F o o o o ...
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