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 Δ T of 250 deg. F gives 45% more lift than air at Δ T of 150 deg.F!...
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 Fall '07
 Lear
 Fluid Dynamics, Fluid Mechanics, Ideal gas equation

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