Prelim_1_problem_3

# Prelim_1_problem_3 - ˙ Q cv = h 2-h 1 ˙ m(2 1 h 1 = h...

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Prelim-1 Solution December 2, 2004 1 Problem 3 Known p 1 = p 2 = 5 bar , ˙ m = 20 kg/s , | ˙ Q cv | = 21095 kJ/s , ˙ W cv = 0, x 2 = 0. Find x 1 Schematic and given data wet steam p 1 , ˙ m sat. liquid p 2 = p 1 Q W=0 Assumptions Analysis First, heat is transfering out of the system, so ˙ Q cv = - 21095 kJ/s . For state 2, from table A- 3, h 2 = 640 . 23 kJ/kg . Using the ±rst law for control volume problem 0 = ˙ Q cv - dotW cv + ( h 1 - h 2 + ( V 2 1 - V 2 2 ) / 2 + g ( z 1 - z 2 )) ˙ m, (1) Negnating kinetic and potential energy e²ects we ahve

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Unformatted text preview: ˙ Q cv = ( h 2-h 1 ) ˙ m (2) 1 h 1 = h 2-dotQ cv / ˙ m = 640 . 23 kJ/kg-(-21095 kJ/ ) / (20 kg/s ) = 1699 . 98 kJ/kg (3) Since h 1 = h f ( p 1 ) + x 1 h fg ( p 1 ) , (4) x 1 = h 1-h f h fg = 1699 . 98-640 . 23 2108 . 5 = 0 . 5 . (5) So x 1 = 50%. 2...
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Prelim_1_problem_3 - ˙ Q cv = h 2-h 1 ˙ m(2 1 h 1 = h...

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