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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

Ask a homework question - tutors are online