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# Exam2 - UNIVERSITY OF CENTRAL FLORIDA Mechanical Materials...

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mdot 396.04 kg s = mdot ρ Vav A := b ( ) Vav 50.93 m s = Vav Vdot A := a ( ) The ± averagevelocity A 7.854 10 3 × m 2 = A π d 2 4 := d 0.1 m := For Vdot 0.4 m 3 s := ρ 990.099 kg m 3 = ρ 1 vf := Properties From Table A-1.1 vf 0.00101 m 3 kg := Solution UNIVERSITY OF CENTRAL FLORIDA Mechanical, Materials & Aerospace Engineering EGN 3358 – Thermo-Fluids-Heat Transfer, Fall 2007 Exam 2 (Open text book) November 08, 2006 from 10:30 to 11:45 1. ( 6 points ) Saturated liquid water at 45 o C flows through circular duct. The diameter of the duct changes along its length. If the flow rate through the duct is 0.4 m 3 /s calculate: (a) The average velocity of the water where the duct diameter is 0.1 m and (b) The mass flow rate of the water. Assume the flow is steady-state, steady-flow (SSSF).

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tw 0.092 Pa = tw 0.332 ρ U 2 Rex 1 2 := δ 3.24 10 3 × m = δ x ( ) 5.0 Rex 1 2 := Since Rex < 5.0x10^5, the boundary layer is laminar. From table 6-1 Rex 9.524 10 4 × = Rex U x ν := Analysis ν 18.9 10 6 m 2 s := ρ 1.0596 kg m 3 :=
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