Homework_Assignment_8__MAE103_Solutions

# Homework_Assignment_ - gd Q L h f = and 4 2 2 2 128 gd Q L h f = We can substitute head losses in the energy conservation equation above Noting

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HW#8 Solutions Problem 1. Problem 2.

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Problem 3.
Problem 4.

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Problem 5.
Problem 6. Problem 7.

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Problem 8 2 1 2 2 2 2 1 2 1 1 2 2 f f h h z g V p z g V p + + + + = + + γ where h f1 and h f2 are friction head losses in cylinder 1 (diameter 5 mm) and cylinder 2 (diameter 0.3 mm). From continuity 2 2 1 1 A V A V = , therefore, ( ) () s m A A V V / 5 4 0003 . 0 4 005 . 0 018 . 0 2 2 2 1 1 2 = = = π . 73 10 980 . 0 005 . 0 018 . 0 800 Re 3 1 1 1 = = = µ ρ d V - i.e. laminar flow in cylinder 1; 1224 10 980 . 0 0003 . 0 5 800 Re 3 2 2 2 = = = d V - laminar flow in cylinder 2 as well. Because there is a laminar flow in both cylinders we can use analytical expressions for h f : 4 1 1 1 128
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Unformatted text preview: gd Q L h f = and 4 2 2 2 128 gd Q L h f = . We can substitute head losses in the energy conservation equation above. Noting that z 1 =z 2 we can simplify to get: Pa d L d L d V V V p p p 525 , 104 128 2 2 4 2 2 4 1 1 1 1 2 1 2 2 2 1 + + = = . Required force F is: N d p A p F 25 . 2 4 2 1 = = =...
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## This note was uploaded on 10/12/2009 for the course MAE 210318200 taught by Professor Kulinsky during the Spring '09 term at UCLA.

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Homework_Assignment_ - gd Q L h f = and 4 2 2 2 128 gd Q L h f = We can substitute head losses in the energy conservation equation above Noting

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