Lecture 5
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Thermodynamics II course taught at air university Islamabad by Sijal Ahmed Memon. Dual combustion cycle and solved examples
Lecture 5
- 1. THERMODYNAMICS II LECTURE 4 DUAL COMBUSTION CYCLE AND SOME SOLVED EXAMPLES 1
- 2. Contents General Discussion of dual combustion cycle Examples on Diesel cycle Home assignment 2
- 4. Example 4 1 = = 287288 10000 = 0.82656 3 1 = 288 , 1 = 205.478 情 1 =688.1 2 = 2 1 1 = 1 12 688.1 = 57.3416 2 = 751.37 2 = 768.79 Method 1
- 5. Example 5 3 = 1100 + 273 = 1373 2 2 = 3 3 3 = 2 2 3 = 1 2 1 3 2 = 0.82656 1 12 1373 751.37 = 0.126 3 3 = 9.492, 3 = 1483.1 4 = 4 3 3 = 0.82565 0.126 9.492 = 62.198 駒 ＄, = 3 2 = 0.126 0.82656 1 12 = 1.829 4 = 729.716 , 4 = 537.3753
- 6. Example 6 Heat input to the cycle = 3 2 = 1483.1 768.79 = 714.31 Heat input to the cycle = 4 1 = 537.3753 205.478 = 331.89 Net work output of cycle = = 714.31 331.89 = 382. 42 Cycle Efficiency = = . . = . . %
- 8. Example 8
- 9. Example 9
- 10. Example 10 Method 3 Cold Air Standard Cycle Efficiency = . . . . . = . . . = . % = . . . . . = . . . = . % Cut off ratio = 1.829 from method 1 Cut off ratio = 1.829 from method 2