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Chapter 5 | 251(b) Noting that the microscopic energies of flowing and nonflowing fluids arerepresented by enthalpy h and internal energy u, respectively, the mass andenergy balances for this uniform-flow system can be expressed asMass balance:m in m out ¢m system S m e 1m 2 m 1 2 CV orm e 1m 1 m 2 2 CVEnergy balance:E in E out ¢E system⎫⎪⎬⎪⎭⎫⎪⎬⎪⎭Net energy transferChange in internal, kinetic,by heat, work, and masspotential, etc., energiesQ in m e h e (m 2 u 2 m 1 u 1 ) CV (since W 0, ke pe 0)Combining the mass and energy balances givesThe amount of heat transfer during this process is found fromSteam leaves the pressure cooker as saturated vapor at 175 kPa at all times(Fig. 5–51). Thus,The initial internal energy is found after the quality is determined:Thus,andQ in (m 1 m 2 )h e (m 2 u 2 m 1 u 1 ) CVQ in Q # in ¢t 10.5 kJ/s2 130 60 s2 900 kJv 1 V m 1h e h g @ 175 kPa 2700.2 kJ/kg0.006 m31 kg 0.006 m 3 /kgx 1 v 1 v f 0.006 0.001v fg 1.004 0.001 0.00499u 1 u f x 1 u fg 486.82 10.00499212037.72 kJ/kg 497 kJ/kgU 1 m 1 u 1 (1 kg)(497 kJ/kg) 497 kJThe mass of the system at the final state is m 2 V/v 2 . Substituting thisinto the energy equation yieldsPSat. vaporSat. liquidSat. vaporh e = h g@175 kPaFIGURE 5–51In a pressure cooker, the enthalpy ofthe exiting steam is h g @ 175 kPa(enthalpy of the saturated vapor at thegiven pressure).Q in a m 1 V v 2b h e a V v 2u 2 m 1 u 1 bThere are two unknowns in this equation, u 2 and v 2 . Thus we need to relatethem to a single unknown before we can determine these unknowns. Assumingthere is still some liquid water left in the cooker at the final state (i.e.,saturation conditions exist), v 2 and u 2 can be expressed asv 2 v f x 2 v fg 0.001 x 2 11.004 0.0012 m 3 /kgu 2 u f x 2 u fg 486.82 x 2 12037.72 kJ/kgRecall that during a boiling process at constant pressure, the properties ofeach phase remain constant (only the amounts change). When these expressionsare substituted into the above energy equation, x 2 becomes the onlyunknown, and it is determined to bex 2 0.009

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