Problem 3.96 - ft lbf ⋅ lbm R ⋅ ⋅ = and T a 519 R ⋅...

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Problem *3.96 [Difficulty: 2] Given: Balloons with hot air, helium and hydrogen. Claim lift per cubic foot of 0.018, 0.066, and 0.071 pounds force per cubic f for respective gases, with the air heated to 150 deg. F over ambient. Find: (a) evaluate the claims of lift per unit volume (b) determine change in lift when air is heated to 250 deg. F over ambient. Solution: We will apply the hydrostatics equations to this system. Governing Equations: L ρ a g V ρ g g V = (Net lift force is equal to difference in weights of air and gas) p ρ R T = (Ideal gas equation of state) Assumptions: (1) Static fluid (2) Incompressible fluid (3) Ideal gas behavior The lift per unit volume may be written as: LV L V g ρ a ρ g () = = ρ a g 1 ρ g ρ a = now if we take the ideal gas equation and we take into account that the pressure inside and outside the balloon are equal: L V ρ a g 1 R a T a R g T g = γ a 1 R a T a R g T g = At standard conditions the specific weight of air is: γ a 0.0765 lbf ft 3 = the gas constant is: R a 53.33
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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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