Lecture 5
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The document discusses the dual combustion cycle in thermodynamics, providing general insights and detailed examples related to the diesel cycle. It includes calculations for various parameters such as heat input, net work output, and cycle efficiency with specific examples and solutions. Additionally, it assigns homework related to the concepts covered in the lecture.
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 𝑣1 = 𝑅𝑇 𝑃 = 287×288 10000 = 0.82656 𝑚3 𝑘𝑔 𝑎𝑡 𝑇1 = 288 𝐾, 𝑢1 = 205.478 𝐾𝐽 𝑘 𝑎𝑛𝑑 𝑣 𝑟1 =688.1 𝑣 𝑟2 = 𝑣2 𝑣1 × 𝑣 𝑟1 = 1 12 × 688.1 = 57.3416 ⇒ 𝑇2 = 751.37 𝑎𝑛𝑑 ℎ2 = 768.79 Method 1
- 5. Example 5 𝑇3 = 1100 + 273 = 1373 𝐾 𝑣2 𝑇2 = 𝑣3 𝑇3 ⇒ 𝑣3 = 𝑣2 𝑇2 × 𝑇3 = 𝑣1 × 𝑣2 𝑣1 × 𝑇3 𝑇2 = 0.82656 × 1 12 × 1373 751.37 = 0.126 𝑚3 𝑘𝑔 𝑣 𝑟3 = 9.492, ℎ3 = 1483.1 𝑣 𝑟4 = 𝑣4 𝑣3 × 𝑣 𝑟3 = 0.82565 0.126 × 9.492 = 62.198 𝐶𝑢𝑡 𝑜𝑓𝑓 𝑟𝑎𝑡𝑖𝑜, 𝑟𝑐 = 𝑣3 𝑣2 = 0.126 0.82656 × 1 12 = 1.829 𝑇4 = 729.716 𝐾, 𝑢4 = 537.3753 𝑘𝐽 𝑘𝑔
- 6. Example 6 Heat input to the cycle 𝑞𝑖𝑛 = ℎ3 − ℎ2 = 1483.1 − 768.79 = 714.31 𝑘𝐽 𝑘𝑔 Heat input to the cycle 𝑞 𝑜𝑢𝑡 = 𝑢4 − 𝑢1 = 537.3753 − 205.478 = 331.89 𝑘𝐽 𝑘𝑔 Net work output of cycle 𝑤 𝑛𝑒𝑡 = 𝑞𝑖𝑛 − 𝑞 𝑜𝑢𝑡 = 714.31 − 331.89 = 382. 42 𝑘𝐽 𝑘𝑔 Cycle Efficiency 𝜼 𝒄𝒚𝒄𝒍𝒆 = 𝒘 𝒏𝒆𝒕 𝒒𝒊𝒏 = 𝟑𝟖𝟐. 𝟒𝟐 𝟕𝟏𝟒. 𝟑𝟏 = 𝟎. 𝟓𝟑𝟓𝟑 𝒐𝒓 𝟓𝟑. 𝟓𝟑%
- 8. Example 8
- 9. Example 9
- 10. Example 10 Method 3 Cold Air Standard Cycle Efficiency 𝜼 𝒄𝒚𝒄𝒍𝒆 = 𝟏 − 𝟏 𝟏𝟐 𝟏.𝟒−𝟏 𝟏. 𝟖𝟐𝟗 𝟏.𝟒 − 𝟏 𝟏. 𝟒 𝟏. 𝟖𝟐𝟗 − 𝟏 = 𝟏 − 𝟏 𝟐. 𝟕 𝟏. 𝟑𝟐𝟖𝟔 𝟏. 𝟏𝟔𝟎𝟔 = 𝟓𝟕. 𝟔𝟑% 𝜼 𝒄𝒚𝒄𝒍𝒆 = 𝟏 − 𝟏 𝟏𝟐 𝟏.𝟒−𝟏 𝟏. 𝟕𝟔𝟓 𝟏.𝟒 − 𝟏 𝟏. 𝟒 𝟏. 𝟕𝟔𝟓 − 𝟏 = 𝟏 − 𝟏 𝟐. 𝟕 𝟏. 𝟐𝟏𝟓𝟑𝟓 𝟏. 𝟎𝟕𝟏 = 𝟓𝟕. 𝟗𝟕% Cut off ratio = 1.829 from method 1 Cut off ratio = 1.829 from method 2