Lecture 5
Download as PPTX, PDF0 likes148 views
Thermodynamics II course taught at air university Islamabad by Sijal Ahmed Memon. Dual combustion cycle and solved examples
More Related Content
More from Sijal Ahmed (13)
Recently uploaded (20)
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