Homework 11 Solutions-1

Homework 11 Solutions-1 - ENGRD 2210 – Thermodynamics...

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Unformatted text preview: ENGRD 2210 – Thermodynamics PROBLEM SET 11 SOLUTIONS Problem 9.13: Problem Statement: Consider the modification of the air-standard Otto cycle in which the isentropic compression and expansion processes are each replaced with polytropic processes having n=1.3. The compression ratio is 9 for the modified cycle. At the beginning of compression, p 1 = 1 bar and T 1 = 300 K and V 1 = 2270 cm 3 . The maximum temperature during the cycle is 2000 K. Determine: (a) the heat transfer and work in KJ, for each process in the modified cycle (b) the thermal efficiency (c) the mean effective pressure, in bar Solution: The following p-v diagram summarizes the cycle described in the problem statement. (a) Before finding the work, we need to find the temperatures at each state in order to subsequently find the internal energy at each state. T 1 is given in the problem as 300 K and T 3 is given as the maximum temperature 2000 K. We use polytropic relations described by equations 3.56. 2 = 1 ( ? 1 ? 2 ) ?− 1 = 300 ?¡ (9) 1.3 − 1 = ??¢ £ (1) 4 = 3 ( ? 3 ? 4 ) ?− 1 = 2000 ?¡ ( 1 9 ) 1.3 − 1 = ¤¢?¢ £ (2) Where V 1 /V 2 = V 4 /V 3 is the compression ratio, 9, given in the problem. Now the temperatures are found, the internal energies at each state can be found using Table A-22. These values are summarized in the table below State Temperature (K) Internal Energy (kJ/kg) 1 300 214.07 2 580 419.55 3 2000 1678.7 4 1035 789.05 The mass is found through ideal gas law at state 1: ? = ?¡ ¢ = 1 ?£¤ (2270 ¥? 3 ) 8314 28.97 ¦ §¨© (300 © ) = 2.637 ª 10 − 3 §¨ For process 1 2: Work is found using equation 3.57 for a polytropic process « 12 ? = ??? 2 1 = ¢ ( 2 − 1 ) 1 −¬ = ­ 8.314 28.97 §¦ §¨© ®¯ 580 − 300 °© 1 − 1.3 = −±?? . ? ² W 12 = -0.706 kJ (3) Since the process is no longer isentropic, there is actually heat transfer to the system that can be calculated from an energy balance. ³ 12 ? = ¯´ 2 − ´ 1 ° + « 12 ? = ­ 419.55 §¦ §¨ − 214.07 §¦ §¨ ® − 267.9 §¦ §¨ = −?± . µ± ² (4) Q 12 = -0.1646 kJ For process 2 3: No work is done, so W 23 = 0. ³ 23 ? = ¯´ 3 − ´ 2 ° = ­ 1678.7 §¦ §¨ − 419.55 §¦ §¨ ® = ¶±·? . ¶· ² Q 23 = 3.320 kJ (5) For process 3 4: Following the same process as 1 2, the work and heat are « 34 ? = ??? 4 3 = ¢ ( 4 − 3 ) 1 −¬ = ­ 8.314 28.97 §¦ §¨© ®¯ 1035 − 2000 °© 1 − 1.3 = ?±¸ . ¶ ² W 34 = 2.434 kJ (6) ³ 34 ? = ¯´ 4 − ´ 3 ° + « 34 ? = ­ 789.05 §¦ §¨ − 1678.7 §¦ §¨ ® + 923.1 §¦ §¨ = ¸¸ . µ· ² (7) Q 34 = 0.0882 kJ For process 4 1: No work is done, so W 41 = 0 ?...
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Homework 11 Solutions-1 - ENGRD 2210 – Thermodynamics...

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