Question

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 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 reached at the end of isentropic expansion (3-4a), and the state point 1 remains the same. Then}

• A. $T_{4a}<T_4$
• B. $T_{4a}>T_4$
• C. $T_{4a} = T_4$
• D. $T_{4a} = 2T_4$

Hint:

For the same final state, a constant-pressure heat rejection process always begins at a lower temperature than a constant-volume heat rejection process because constant-volume cooling removes more temperature per unit heat transfer.

Answer 1

The Correct Option is A

Step 1: Understand the two cycles.
In the original Diesel cycle, process 4–1 is a \emph{constant-volume heat rejection}. In the modified cycle, this process is replaced by a new process 4a–1, which is a \emph{constant-pressure heat rejection}. The compression process 1–2 and the isentropic expansion process starting at point 3 remain the same. State 1 is unchanged for both cycles. Step 2: Compare temperatures at 4 and 4a.
Point 4 and point 4a both lie on isentropic expansion paths: - In the original Diesel cycle: isentropic expansion 3–4. - In the modified cycle: isentropic expansion 3–4a. However, after reaching point 4 or 4a, the heat rejection path is different: - At point 4: heat is rejected at \emph{constant volume}. - At point 4a: heat is rejected at \emph{constant pressure}. Because state point 1 is the same in both cases, the final pressure and temperature after heat rejection are identical. That means both processes (4–1 and 4a–1) must bring the working fluid to the same pressure and temperature at state 1. Step 3: Understand the consequence of constant-pressure vs constant-volume heat rejection.
For the same initial state and same final state (point 1): - In a \emph{constant-volume} process (4–1), the temperature drop is large because internal energy change occurs without volume change. Thus, for the same amount of heat rejection, the initial temperature at point 4 must be comparatively \emph{higher}. - In a \emph{constant-pressure} process (4a–1), the temperature drop for the same heat rejection is smaller (since the process includes both internal energy decrease and work output). Therefore, to reach the same final state 1, the initial temperature at point 4a must be \emph{lower}. Step 4: Conclude the temperature relation.
To ensure that both cycles end at the same point 1 after heat rejection: \[ T_{4a}<T_4. \] Hence, the temperature at the end of isentropic expansion before constant-pressure heat rejection is always less than that before constant-volume heat rejection.