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Unformatted text preview: Chapter 5 The First Law of Thermodynamics Review Problems 5168 A cylinder is initially filled with saturated R134a vapor at a specified pressure. The refrigerant is heated both electrically and by heat transfer at constant pressure for 6 min. The electric current is to be determined, and the process is to be shown on a Tv diagram. Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are negligible. 2 The thermal energy stored in the cylinder itself and the wires is negligible. 3 The compression or expansion process is quasiequilibrium. Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as E E E Q W W U Q Q W m h h Q VI t m h h in out in e in b out in e in in = + = + = + = Net energy transfer by heat, work, and mass system Change in internal, kinetic, potential, etc. energies (since = KE = PE = 0) ∆ ∆ ∆ , , , ( ) ( ) ( ) 2 1 2 1 since ∆ U + W b = ∆ H during a constant pressure quasiequilibrium process. The properties of R134a are (Tables A11 through A13) kJ/kg 314.02 C 70 kPa 200 kJ/kg 241.30 vapor sat. kPa 200 2 1 2 kPa 200 @ 1 1 = = = = = = h T P h h P g Substituting, ( 29 ( 29 ( 29 ( 29 ( 29 A 15.72 =  = × + I I kJ/s 1 VA 1000 kJ/kg 241.30 314.02 kg 12 s 60 6 V 110 VA 250,000 5158 T v 2 1 R134a P= const. W Chapter 5 The First Law of Thermodynamics 5169 A cylinder is initially filled with saturated liquidvapor mixture of R134a at a specified pressure. Heat is transferred to the cylinder until the refrigerant vaporizes completely at constant pressure. The initial volume, the work done, and the total heat transfer are to be determined, and the process is to be shown on a Pv diagram. Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are negligible. 2 The thermal energy stored in the cylinder itself is negligible. 3 The compression or expansion process is quasiequilibrium. Analysis ( a ) Using property data from R134a tables (Tables A11 through A13), the initial volume of the refrigerant is determined to be ( 29 ( 29 ( 29 3 m 0.005078 = = = = × + = + = = × + = + = = = = = = = /kg m 0.02539 kg 0.2 kJ/kg 82.88 36.69 221.43 0.25 36.69 /kg m 0.02539 0.0007532) (0.0993 0.25 0.0007532 kJ/kg 221.43 , 69 . 36 /kg m 0.0993 , 0007532 . 25 . kPa 200 3 1 1 1 1 3 1 1 3 1 1 mv V u x u u v x v v u u v v x P fg f fg f g f g f ( b ) The work done during this constant pressure process is kJ/kg 221.43 /kg m 0.0993 . kPa 200 kPa 200 @ 2 3 kPa 200 @ 2 2 = = = = = g g u u v v vapor sat P ( 29 ( 29 ( 29 ( 29 kJ 2.96 = ⋅ = = = = ∫ 3 3 1 2 1 2 2 1 , m kPa 1 kJ 1 /kg m 0.02539 0.0993 kPa 200 kg 0.2 ) ( v v mP V V P dV P W out b ( c ) We take the contents of the cylinder as the system. This is a closed system since no mass enters or ) We take the contents of the cylinder as the system....
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 Spring '08
 Chung
 Thermodynamics, Energy, Heat Transfer, The First Law of Thermodynamics

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