PROBLEM 6.3. Using the appropriate table, determine the indicated property. In each case, locate the state on sketches of the T-v and T-s diagrams.

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1 PROBLEM 63 Using the appropriate table, determine the indicated property In each case, locate the state on sketches of the -v and -s diagrams (a) water at p = 040 bar, h = kj/kg K Find s, in kj/kg K (b) water at p = 10 bar, u = kj/kg Find s, in kj/kg K (c) Refrigerant 134a at = -22 C, x = 060 Find s, in kj/kg K (d) ammonia at = 28 C, s = kj/kg K Find u, in kj/kg (a) water at p = 040 bar, h = kj/kg K Find s, in kj/kg K able A-3: h f = kj/kg, h g = kj/kg, s f = kj/kg K, s g = kj/kg K 04 bar f x g v or s he quality is: x = h h f h g h f = = 05 s = s f + x (s g s f ) = ( ) = kj/kg K (b) water at p = 10 bar, u = kj/kg Find s, in kj/kg K able A-3: u f = kj/kg, u g = kj/kg, u = > u g therefore use able A-4 able A-4: s = kj/kg K 10 bar g u g = 25836kJ/kg v or s

2 (c) Refrigerant 134a at = -22 C, x = 060 Find s, in kj/kg K able A-10: s f = kj/kg K, s g = kj/kg K s = s f + x (s g s f ) = ( ) = kj/kg K (d) ammonia at = 28 C, s = kj/ kg K Find u, in kj/kg able A-13: u = u g (28 C) = kj/kg

3 PROBLEM 611

4 PROBLEM 618

5 PROBLEM 614 One kg of water in a piston-cylinder assembly undergoes the two internally reversible processes in series shown in Fig P614 For each process, determine, in kj, the heat transfer and work Schematic and Given Data: p 3 = 20 MPa 3 = constant 2 p 2 = 07 MPa s = constant Water m = 1 kg 1 p 1 = 01 MPa 1 = 100 C s Given: One kilogram of water in a closed system piston-cylinder undergoes two internally reversible processes See Figure P614 for given property data Find: Heat transfer and work, in kj, for each process Engineering Model: 1 he water is the closed system 2 For the system, kinetic and potential energy changes can be ignored 3 he processes are internally reversible 4 Assume saturated liquid values from able A-2 are used for states in compressed liquid region Analysis: Process 1-2: Using equation 623, Q 12 = 0 he energy balance then reduces to, ΔU 12 = W 12 simplifies to, W 12 = m(u 2 u 1 )

6 Using able A-3 at p 1 = 01 MPa = 1 bar, 1 > sat = 9963 C, therefore state 1 is in the SHR From able A-4: u 1 = kj/kg, s 1 = s 2 = kj/kg K For state 2, interpolate from able A-4: ( ) 2 = 3 = ( ) = C ( ) u 2 = ( ) ( ) ( ) W 12 = m(u 2 u 1 ) = 1 kg ( ) = kj Process 2-3: = kj/kg Using equation 623, Q 23 = m(s 3 s 2 ) and 2 = 3 = C Using able A-3 at p 3 = 02 MPa = 20 bar, 3 > sat = 2124 C, therefore state 3 is also in the SHR Interpolating able A-4 s 3 = ( ) ( ) = 6836 kj/kg K ( ) ( ) u 3 = ( ) = kj/kg ( ) Q 23 = m 3 (s 3 s 2 ) = 1 kg ( )( ) = kj An energy balance then reduces to read, W 23 = Q 23 m(u 3 u 2 ) W 23 = (1 kg)( ) = kj

7 PROBLEM 626 A gas initially at 28 bar and 60 o C is compressed to a final pressure of 14 bar in an isothermal internally reversible process Determine the work and heat transfer, each in kj per kg of gas, if the gas is (a) Refrigerant 134a, (b) air as an ideal gas Sketch the process on p-v and -s coordinates KNOWN: A gas is compressed isothermally with no internal irreversibilities from a specified initial state to a specified final pressure FIND: Determine the work and the heat transfer, per unit mass of gas, for (a) R-134a), (b) air as an ideal gas Sketch the process on p-v and -s coordinates SCHEMAIC AND GIVEN DAA: Gas = 60 o C p 1 = 28 bar p 2 = 14 bar ENGINEERING MODEL: (1) he gas is a closed system (2) he compression takes place isothermally and with no internal irreversibilities (3) Kinetic and potential energy effects are negligible (4) For part (b), the air behaves as an ideal gas ANALYSIS: Using modeling assumption (2), the definition of entropy change: ds = can be used to obtain Q = = m( s 2 s 1 ) Q/m = ( s 2 s 1 ) (*) he energy balance reduces to give: ΔU = Q W W/m = Q/m + (u 1 u 2 ) (**) (a) R-134a From able A-12: u 1 = kj/kg, s 1 = kj/kg K; u 2 = kj/kg, s 1 =09297kJ/kg K hus Q/m = ( )K( ) kj/kg K = kj/kg (out) W/m = (-5934 kj/kg) + ( )kJ/kg = kj/kg (in) p 16813bar 14 bar 16813bar 2 14 bar 1 28 bar bar v s

8 PROBLEM 626 (CONINUED) (b) Air Since 1 = 2, the specific entropy change of the air reduces to and s 2 s 1 = - R ln (p 2 /p 1 ) = ln = kj/kg K Q/m = ( s 2 s 1 ) = (333 K)( kj/kg K) = kj/kg (out) Since 1 = 2, the change in specific internal energy is zero, and W/m = Q/m + (u 1 u 2 ) = kj/kg (in) p 2 14 bar 1 28 bar 14 bar bar v s

9 PROBLEM 627 Nitrogen (N 2 ) undergoes an internally reversible process from 6 bar, 247 o C during which pv 120 = constant he initial volume is 01 m 3 and the work for the process is kj Assuming ideal gas behavior, and neglecting kinetic and potential energy effects, determine heat transfer, in kj, and the entropy change, in kj/k Show the process on a -s diagram KNOWN: Nitrogen at specified initial volume, pressure, and temperature undergoes a polytropic process with known work FIND: Determine heat transfer and the entropy change Show the process on a -s diagram SCHEMAIC AND GIVEN DAA: Nitrogen (N 2 ) W = kj p (1) 6 bar 520 K Q p 1 = 6 bar, 1 = 247 o C = 520 K, V 1 = 01 m 3 pv 120 = constant (2) p 2 2 ENGINEERING MODEL: (1)he nitrogen is a closed system (2) Nitrogen can be modeled as an ideal gas (2)he nitrogen undergoes an internally reversible polytropic process in which pv 120 =constant (3) he system is at an equilibrium state initially and finally (4) here is no change in kinetic or potential energy between the initial and final states ANALYSIS: For the polytropic process, the work can be expressed using the ideal gas equation of state as V he mass is W = = = 2 = + 1 m = = = kg and 2 = + (520 K) = 310 K Now, since the final temperature is known, the final pressure can be determined from Eq 356, as follows p 2 = p 1 = (6 bar) = bar he energy balance reduces to: ΔKE + ΔPE + ΔU = Q W With ΔU = m(u 2 u 1 ) Q = m(u 2 u 1 ) + W Values from able A-23 are expressed on a molar basis: = 10,848 kj/kmol and = 6,437 kj/kmol hus Q = (03887 kg) + (12114 kj) = 5993 kj

10 PROBLEM 627 (CONINUED) he change in entropy is S( 2,p 2 ) S( 1,p 1 ) = m From able A-23: = kj/kmol K and = kj/kmol K hus S( 2,p 2 ) S( 1,p 1 ) = (03887 kg) = kj/k he -s diagram is (1) 520 K 6 bar (2) bar 310 K s he positive entropy change indicates the entropy of the system increased during the process

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