Discussion
An air standard Diesel cycle consists of four processes: 1-2 (isentropic compression), 2-3 (constant pressure heat addition), 3-4 (isentropic expansion) and 4-1 (constant volume heat rejection). \(T_{4}\) is the temperature (in K) attained at the end of isentropic expansion (3-4) before constant volume heat rejection. The constant volume heat rejection process (4-1) is replaced by a constant pressure heat rejection process (4a-1) such that \(T_{4a}\) is the temperature (in K) reached at the end of isentropic expansion (3-4a), and the state point 1 remains the same. Then
Answer: A
The result is \(T_{4a} \lt T_{4}\).
With constant-volume heat rejection, expansion 3-4 stops at the volume \(v_{1}\). For constant-pressure rejection to return to state 1, expansion 3-4a must continue down to the pressure \(p_{1}\), which is below \(p_{4}\) (\(p_{4} \gt p_{1}\) because state 4 is hotter than state 1 at the same volume). Further isentropic expansion lowers the temperature, since along 3-4a
\[ \begin{aligned} \frac{T_{4a}}{T_{4}} &= (p_{1}/p_{4})^{(\gamma - 1)/\gamma} \\ &\lt 1 \end{aligned} \]
So \(T_{4a}\) is lower than \(T_{4}\), not equal to it, higher, or double.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development