9 - Prob 6.24 PROBLEM 6.24 PROBLEM STATEMENT: A...

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Prob 6.24 PROBLEM 6.24 PROBLEM STATEMENT: A frictionless piston-cylinder device undergoes a reversible adiabatic process from an initial condition of 200 kPa and 100 o C, to a final state of 200 kPa and 250 o C. The medium is helium. In what form did the energy transfer take place (heat transfer or work)? How much energy (kJ/kg) was transferred? How much boundary work was there? How much shaft work? DIAGRAM DEFINING SYSTEM AND PROCESS: w boundary w shaft State 1: P 1 =200 kPa T 1 =100 o C State 2: P 2 =200 kPa T 2 =250 o C GIVEN: Helium, with adiabatic and reversible process from state 1 to state 2 State 1: P 1 =200 kPa, T 1 =100 ° C State 2: P 2 =200 kPa, T 2 =250 ° C FIND: a). What form, heat transfer or work? b). Amount of energy transferred c). w boundary d). w shaft ASSUMPTIONS: Ideal gas, constant specific heats, NKEPE, reversible adiabatic process. GOVERNING RELATIONS: 1. First Law for a control mass system: q+ NET,1-2 IN,1-2 OUT,1-2 2 1 w - w = u u ) v 2. Boundary work: w= for constant pressure process ( 2 1 v OUT,1-2 1 2 1 v P d v = P v - 3. Second law: Adiabatic reversible process is isentropic. PROPERTY DATA: Since helium is monatomic, v V 3R c = constant = 2 R 8.3144 kJ/kmol K kJ R = = = 2.077 kJ/kg K c = 3.1156 M 4.003 kg/kmol kg K ⋅→ ± QUANTITATIVE SOLUTION: a) In an adiabatic process, q = 0, i.e. energy transfer is work.
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Prob 6.24 b) ( ) () ( ) IN,1-2 OUT,1-2 2 1 v 2 1 w- w = u - u = c T - T kJ kJ = 3.1156 523 373 K = 467.3 kg K kg  ×−   ± c) , 2 1 v OUT,boundary 1 2 1 v wP d v = P = v - v 12 12 1 RT RT v, v , P PP 2 P = == ( ) 21 OUT,boundary 1 2 1 11 RT RT wP R T T kJ kJ 2.077 523 373 K 311.6 kg K kg  =− =   = ± d) IN,shaft 2 1 OUT, boundary kJ kJ kJ w = u - u + w = 467.3 + 311.6 = 778.9 kg kg kg DISCUSSION OF RESULTS: Self-explanatory PROBLEM 6.26 PROBLEM STATEMENT: Nitrogen gas enters an adiabatic compressor at 20°C and 0.1 MPa. The outlet pressure is 10 times the inlet pressure. If the compressor is reversible, determine the outlet temperature and the work per unit mass required, assuming (a) constant specific heats (evaluated at the inlet temperature) and (b) variable specific heats. DIAGRAM DEFINING SYSTEM AND PROCESS: w IN,C 1 P 2 = 10 P 1 Comp. Comp. P 1 = 0.1 MPa T 1 = 20°C 2 GIVEN: N 2 , adiabatic and reversible compressor State 1: P 1 = 0.1 MPa, T 1 = 20°C = 293 K State 2: P 2 = 10 P 1 = 1.0 MPa FIND: T 2 (K) and w IN,C (kJ/kg) for constant and variable specific heats. ASSUMPTIONS: Ideal gas, SFSS, NKEPE. Adiabatic and reversible=isentropic. GOVERNING RELATIONS: 4. Isentropic P-T relation for constant specific heats: ( ) (k-1) k ba ss TT =PP 5. Compressor work for constant specific heats, w IN , C = c P (T 2 T 1 ) 6. Entropy change for isentropic process with variable specific heats:
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Prob 6.24 00 21 2 1 s(T)-s(T)-R ln (P P)=0 7. Compressor work for variable specific heats, w IN,C = h 2 h 1 QUANTITATIVE SOLUTION: a). From Table 6s (at T=20°C), the interpolated value of k is 1.3996. Thus, using the pressure-temperature relation, the temperature at State 2 is: () (k-1) k (1.3996-1) 1.3996 2 1 s 2 T = T P P = 293 10 T = 565.4 K The value of c P for N 2 at State 1 can be interpolated from Table 6s at T=20°C: P c =1.0402 kJ/kg K ± Now, the work per unit mass is: IN,C p 2 1 w = c (T - T ) =1.0402 (kJ/kg K) (565.4 - 293)(K) = 283.35 kJ/kg × ± b). The turbine is adiabatic and reversible, i.e. isentropic. From Table 6s:
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This note was uploaded on 03/19/2008 for the course ME 326 taught by Professor Schmidt during the Spring '07 term at University of Texas at Austin.

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9 - Prob 6.24 PROBLEM 6.24 PROBLEM STATEMENT: A...

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