Unformatted text preview: 6 . 977 ( 3 3 tank = = ρ = V m Substituting, [ ] kg/s 0.0565 = ° ° × ° ° + hot hot C 80)C)(60 kJ/kg. kg)(4.18 (58.656 = s) 60 (8 C 70)C)(20 kJ/kg. (4.18 kJ/s 6 . 1 m m & & To determine the average temperature of the mixture, an energy balance on the mixing section can be expressed as C 44.2 ° = ° + = ° ° + ° ° + = + = mixture mixture mixture cold hot cold cold hot hot out in C) kJ/kg. kg/s)(4.18 0.06 (0.0565 C) C)(20 kJ/kg. kg/s)(4.18 (0.06 C) C)(70 kJ/kg. kg/s)(4.18 (0.0565 ) ( T T CT m m CT m CT m E E & & & & & & T mixture =? T in = 20 ° C m cold = m hot T cold = 20 ° C m cold =0.06 kg/s T 1 = 80 ° C T 2 = 60 ° C T out = 70 ° C m hot = ? W e =1.6 kW 173...
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This note was uploaded on 01/14/2012 for the course PHY 4803 taught by Professor Dr.danielarenas during the Fall '10 term at UNF.
 Fall '10
 Dr.DanielArenas
 Thermodynamics, Mass, Heat

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