Assignment__7

# Assignment__7 - Assignment#7 PROBLEM 6.20 PROBLEM STATEMENT...

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Assignment #7 PROBLEM 6.20 PROBLEM STATEMENT: An adiabatic steam turbine operates with inlet steam conditions of 500°C, 2 MPa, and a measured outlet condition of 10 kPa, x = 0.95. (a) How much work per unit mass (kJ/kg) will this turbine produce? (b) If the turbine were adiabatic and reversible, what would the outlet quality be, and how much work per unit mass could it produce for the same outlet pressure? DIAGRAM DEFINING SYSTEM AND PROCESS: P 1 = 2 MPa T 1 = 500°C Comp. P 2 = 10 kPa x 2 = 0.95 2 1 Turbine w OUT,1-2 GIVEN: H 2 O, adiabatic turbine State 1: P 1 = 2 MPa, T 1 = 500°C State 2: P 2 = 10 kPa, x 2 = 0.95 FIND: w OUT,1-2 (kJ/kg), x 2,s , w OUT,1-2s (kJ/kg) ASSUMPTIONS: No change in kinetic and potential energy. GOVERNING RELATIONS: 1. First Law for control volume: 2. Second law: Adiabatic and reversible process is isentropic, s 2 =s 1 QUANTITATIVE SOLUTION: a). From First Law equation:

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From the property tables for H 2 O, the following values can be obtained: Thus, work produced by the turbine is b). The turbine is adiabatic and reversible, i.e. isentropic. From Table 12s: The quality at State 2 for s 2,s =7.4318 kJ/kg-K and P 2 =10kPa can be computed from Table 11s as follows: Now, from Table 11s, the enthalpy at State 2s is Thus, work produced by adiabatic and reversible turbine is DISCUSSION OF RESULTS: Irreversibilities in a device due to fluid friction and turbulence reduce the potential of the device to produce work, so a reversible turbine produces the maximum possible work. Performance of real devices is usually compared with that of isentropic devices as the ultimate standard. 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: 1 P 1 = 0.1 MPa
T 1 = 20°C w IN,C Comp. Comp. 2 P 2 = 10 P 1 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: 3. Isentropic P-T relation for constant specific heats: 4. Compressor work for constant specific heats, w IN , C = c P (T 2 – T 1 ) 5. Entropy change for isentropic process with variable specific heats: 6. 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:

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## This note was uploaded on 09/01/2010 for the course ME 320 taught by Professor Dr.kinne during the Spring '07 term at University of Texas.

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Assignment__7 - Assignment#7 PROBLEM 6.20 PROBLEM STATEMENT...

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