Problem 4.12 - ( ) ydy k y j dA V V 5 5 10 2 + = r r (e) (...

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Problem 4.12 [Difficulty: 3] Given: Data on velocity field and control volume geometry Find: Several surface integrals Solution: k wdy j wdz A d ˆ ˆ 1 = r k dy j dz A d ˆ ˆ 1 = r k wdy A d ˆ 2 = r k dy A d ˆ 2 = r ( ) k by j a V ˆ ˆ + = r ( ) k y j V ˆ 5 ˆ 10 + = r (a) ( ) ( ) ydy dz k dy j dz k y j dA V 5 10 ˆ ˆ ˆ 5 ˆ 10 1 = + = r (b) 5 . 7 2 5 10 5 10 1 0 2 1 0 1 0 1 0 1 1 = = = y z ydy dz dA V A r (c) ( ) ( ) ydy k dy k y j dA V 5 ˆ ˆ 5 ˆ 10 2 = + = r (d) ()
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Unformatted text preview: ( ) ydy k y j dA V V 5 5 10 2 + = r r (e) ( ) ( ) k j k y j y ydy k y j dA V V A 33 . 8 25 3 25 25 5 5 10 1 3 1 2 1 2 2 = = + = r r y z c d Control volume...
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