eml 3100 lec 9 eng dev problem

eml 3100 lec 9 eng dev problem - Thermodynamics 5-39 A...

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1 Thermodynamics ) ( ) ( 0 = PE = KE = since ) ( 1 2 in pw, 1 2 out b, in pw, in e, h h m W t I Q h h m W U W W - = + - = + = + V ( 29 kg 4.731 /kg m 0.001057 m 0.005 kJ/kg 1593.6 1 . 2213 5 . 0 01 . 487 5 . 0 kPa 175 /kg m 0.001057 kJ/kg 487.01 liquid sat. kPa 175 3 3 1 1 2 2 2 2 3 kPa 175 @ 1 kPa 175 @ 1 1 = = = = × + = + = = = = = = = = v V v v m h x h h x P h h P fg f f f V 223.9 = × = = - = + kJ/s 1 VA 1000 s) 60 A)(45 (8 kJ 4835 kJ 4835 kg 487.01)kJ/ kg)(1593.6 (4.731 kJ) (400 V V V t I t I H 2 O P = const. W pw W e v P 1 2 5-39 A cylinder is initially filled with saturated liquid water at a specified pressure. The water is heated electrically as it is stirred by a paddle-wheel at constant pressure. The voltage of the current source is to be determined, and the process is to be shown on a P- v diagram. Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 The cylinder is well-insulated and thus heat transfer is negligible. 3 The thermal energy stored in the cylinder itself is negligible. 4 The compression or expansion process is quasi-equilibrium. THIS IS AN EXAMPLE OF A CLOSED STATIONARY SYSTEM ENERGY ANALYSIS: since U + W b = H during a constant pressure quasi-equilibrium process. The properties of water are (Tables A-4 through A-6) Substituting,
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2 Thermodynamics m m m 1 2 = = /kg m 0.4525 kPa 300 ) K 473 )( K /kg m kPa 0.287 ( 3 3 1 1 1 = = = P RT v kg/s 0.5304 V = = = ) m/s 30 )( m 0.008 ( /kg m 0.4525 1 1 2 3 1 1 1 A v m E E E E E in out in out - = = = Rate of net energy transfer by heat, work, and mass system (steady) Rate of change in internal, kinetic,
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This note was uploaded on 08/03/2011 for the course EML 3100 taught by Professor Sherif during the Summer '08 term at University of Florida.

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eml 3100 lec 9 eng dev problem - Thermodynamics 5-39 A...

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