PS_2%20soln - ChBE421 Problem Set #2 Solutions Fall 2009 1....

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
ChBE421 Problem Set #2 Solutions Fall 2009 1. 0 2 1 2 = + + + + m W lw ρ P z g v s ) ( where: 0 0 0 = = = lw P z , , ) ( 2 2 v m W s = = -0.5 x (1.2 kg/m 3 ) x (14 m/s) x ( π /4(25 m) 2 ) x [(5 m/s) 2 – (14 m/s) 2 ] = 0.705 MW 2. ndS u ρ ρ P E W Q edV ρ dt d S V = where: 0 0 0 0 = = = = z W Q dt d , , , , and no change in internal energy ndS u ρ ρ P u = 2 2 1 0 () + + = 2 1 2 2 2 1 2 2 1 1 2 1 2 1 2 1 2 1 0 m m ρ P U m ρ P u m ρ P u So, + + = 2 1 2 2 2 1 2 1 2 1 2 2 m m m u m u U ρ P P And, 0 = lw (viscous forces negligible) 3. 0 2 1 2 = + + + + m W η lw ρ P z g v s p ) ( where: 0 0 0 = = = lw P z , , UA V = = 75000 ft 3 /min Æ U 1 = 11.05 ft/s, U 2 = 63.66 ft/s ) ( 2 2 v η m W p s = where = m (0.076 lbm/ft 3 ) x (75000 ft 3 /min) x (1 min/60 s) = 95 lbm/s
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
= -0.5 x (95 lbm/s) / 0.82 x [(63.66 ft/s) 2 – (11.05 ft/s) 2 ] x (lbf s 2 /(32.2 lbm ft)) x x (hp/(550 lbf ft/s)) = -12.86 hp (work in) 4. 0 2 1 2 = + + + + T s η m W lw ρ P z g v ) ( 0.5 x ((24 ft/s) 2 – 0) x (lbf s 2 /(32.2 lbm ft)) + (32.2 ft/s
Background image of page 2
Background image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 03/18/2010 for the course CHBE 421 taught by Professor 1` during the Spring '10 term at University of Illinois, Urbana Champaign.

Page1 / 3

PS_2%20soln - ChBE421 Problem Set #2 Solutions Fall 2009 1....

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online