Problem 1.26

# Problem 1.26 - ⎝ ⎞ ⎠ 3 × = V prop 953 in 3 ⋅ = The...

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Problem 1.26 [Difficulty: 2] Given: Geometry of tank, and weight of propane. Find: Volume of propane, and tank volume; explain the discrepancy. Solution: Use Table G.2 and other sources (e.g., Google) as needed. The author's tank is approximately 12 in in diameter, and the cylindrical part is about 8 in. The weight of propane specified is 17 lb. The tank diameter is D1 2 i n = The tank cylindrical height is L8 i n = The mass of propane is m prop 17 lbm = The specific gravity of propane is SG prop 0.495 = The density of water is ρ 998 kg m 3 = The volume of propane is given by V prop m prop ρ prop = m prop SG prop ρ = V prop 17 lbm 1 0.495 × m 3 998 kg × 0.454 kg 1 lbm × 1in
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Unformatted text preview: ⎝ ⎞ ⎠ 3 × = V prop 953 in 3 ⋅ = The volume of the tank is given by a cylinder diameter D length L, π D 2 L/4 and a sphere (two halves) given by π D 3 /6 V tank π D 2 ⋅ 4 L ⋅ π D 3 ⋅ 6 + = V tank π 12 in ⋅ ( ) 2 ⋅ 4 8 ⋅ in ⋅ π 12 in ⋅ ( ) 3 6 ⋅ + = V tank 1810 in 3 ⋅ = The ratio of propane to tank volumes is V prop V tank 53 % ⋅ = This seems low, and can be explained by a) tanks are not filled completely, b) the geometry of the tank gave an overestimate of the volume (the ends are not really hemispheres, and we have not allowed for tank wall thickness)....
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