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Problem 4.13 [Difficulty: 3] Given: Data on velocity field and control volume geometry Find: Volume flow rate and momentum flux Solution: First we define the area and velocity vectors kdydyxidydzAd ˆˆ +=r jbyiaxV ˆˆ +=r or jyixV ˆˆ +=r We will need the equation of the surface: xz 4 33−= or zx 3 44 −= Then a) Volume flow rate ( ) ( ) ( ) s m30 s m3060 3 245 3 445ˆˆˆˆ 33 3 0 2 3 0 3 0 5 0 =−= ⎟⎠ ⎞⎜⎝ ⎛ −=⎟⎠ ⎞⎜⎝ ⎛ −==+⋅+=⋅= ∫∫ ∫∫∫ Q zzdzzdzxdykdxdyidydzjyixAdVQ AA rr b) Momentum flux ( ) ( )( ) ( ) ( )( ) ( ) ( ) Nˆ75ˆ80flux Momentum ˆ75ˆ1648485 ˆ612 2 25ˆ 27 16 3 16165ˆ 3 24 2 25ˆ 9 16 3 32165 ˆ 3 44 2 ˆ 3 445ˆˆ ˆˆˆˆˆˆˆˆ 3 0 32 3 0 2 3 0 2 3 0 5 0 23 0 23 0 5 0 3 0 5 0 2 ji ji jizzzjzzidzzz jdzzyidzzjdzxydyidzdyx xdydzjyixkdxdyidydzjyixjyixAdVV AAA += ++−= −+⎟⎠ ⎞⎜⎝ ⎛ +−==⎟⎠ ⎞⎜⎝ ⎛ −+⎟⎠ ⎞⎜⎝ ⎛ +−= ⎟⎠ ⎞⎜⎝ ⎛ −+⎟⎠ ⎞⎜⎝ ⎛ −=+= +=+⋅++=⋅ ∫ ∫∫∫ ∫∫ ∫ ∫∫∫ ρρ ρρρ rrr x y 5 m 4 m 3 m z
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