Problem 4.189 - R 2 V R 3 3 + A = For the surface integral...

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

View Full Document Right Arrow Icon
Problem 4.189 [Difficulty: 3] Given: Data on rotating spray system Find: Differential equation for motion; steady speed Solution: Basic equation: Rotating CV Assumptions: 1) No surface force; 2) Body torques cancel; 3) Steady flow; 5) Uniform flow; 6) L<<r The given data is Q1 5 L min = R 225 mm = d5 m m = ρ 999 kg m 3 = For each branch V 1 2 Q π 4 d 2 = V 6.37 m s = A π 4 d 2 = A 19.6 mm 2 = The basic equation reduces to a single scalar equation (FOR EACH BRANCH) V r 2 ω V × r × α r × + () × ρ d A r V xyz ⎯⎯ × ρ V xyz ⎯⎯ d = where α is the angular acceleration But r 2 ω V × r × α r × + () × 2 ω r V α r 2 + = (r and α perpendicular) The volume integral is then V r 2 ω V × r × α r × + () × ρ d
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: R 2 V R 3 3 + A = For the surface integral (FOR EACH BRANCH) A r V xyz V xyz d R V Q 2 = Combining R 2 V R 3 3 + A R V Q 2 = or 3 A R 2 V A R Q V 2 = (1) The steady state speed ( = 0 in Eq 1) is then when max V A R Q V 2 = or max Q 2 A R = max 28.3 1 s = max 270 rpm =...
View Full Document

Ask a homework question - tutors are online