This preview shows page 1. Sign up to view the full content.
Unformatted text preview: ft lbf lbm R = and T a 519 R = For helium: R g 386.1 ft lbf lbm R = T g T a = and therefore: LV He 0.0765 lbf ft 3 1 53.33 386.1 = LV He 0.0659 lbf ft 3 = For hydrogen: R g 766.5 ft lbf lbm R = T g T a = and therefore: LV H2 0.0765 lbf ft 3 1 53.33 766.5 = LV H2 0.0712 lbf ft 3 = For hot air at 150 degrees above ambient: R g R a = T g T a 150 R + = and therefore: LV air150 0.0765 lbf ft 3 1 519 519 150 + = LV air150 0.0172 lbf ft 3 = The agreement with the claims stated above is good. For hot air at 250 degrees above ambient: R g R a = T g T a 250 R + = and therefore: LV air250 0.0765 lbf ft 3 1 519 519 250 + = LV air250 0.0249 lbf ft 3 = LV air250 LV air150 1.450 = Air at T of 250 deg. F gives 45% more lift than air at T of 150 deg.F!...
View
Full
Document
 Fall '07
 Lear
 Fluid Mechanics

Click to edit the document details