Diesel Cycle Efficiency Analysis

1. The Ideal Diesel Cycle: Processes and Parameters

The Diesel cycle models the operation of a compression-ignition engine. Like the Otto cycle, it consists of four processes, but with one crucial change: heat is added at constant pressure, not constant volume.

Let's visualize the cycle on P-V and T-S diagrams.

The four processes are:

1. Process 1 → 2: Isentropic Compression. Pure air in the cylinder is compressed adiabatically. Work is done on the gas.
2. Process 2 → 3: Isobaric Heat Addition. At the point of maximum compression (minimum volume), fuel is injected. It ignites, and the combustion pushes the piston outward. This process is modeled as occurring at constant pressure as the gas expands. Heat $Q_H$ is added to the system.
3. Process 3 → 4: Isentropic Expansion (Power Stroke). The hot, high-pressure gas continues to expand adiabatically, doing work.
4. Process 4 → 1: Isochoric Heat Rejection. At maximum volume, a valve opens, and heat $|Q_C|$ is rejected at constant volume as the gas cools down to its initial state.

Two important ratios define this cycle:

• Compression Ratio (r): The ratio of the maximum volume to the minimum volume.
  $r=\frac{V_1}{V_2}$
• Cutoff Ratio ($\rho$): The ratio of the volume after combustion (fuel "cutoff") to the volume before combustion. This is unique to the Diesel cycle.
  $\rho =\frac{V_3}{V_2}$
  Since the volume increases during heat addition, $\rho >1$.

2. Deriving the Thermal Efficiency

Our goal is to find the efficiency, $\eta$, starting from the fundamental definition:

First, let's express the heat exchanged in terms of temperature changes. We assume we are working with $n$ moles of an ideal gas.

• Heat Addition (2 → 3, Constant Pressure):
  $Q_H=n{C}_p(T_3-T_2)$
• Heat Rejection (4 → 1, Constant Volume):
  $|Q_C|=n{C}_v(T_4-T_1)$

Substituting these into the efficiency formula, the $n$ terms cancel:

where $\gamma =C_p/C_v$. As with the Otto cycle, this expression is correct but not practical. We need to express it using the engine's physical ratios, $r$ and $\rho$.

The following derivation involves a sequence of substitutions using the ideal gas law and adiabatic relations. It's a classic problem that requires careful, sequential logic. To guide you through this, let's use a detailed video walkthrough.

3. Assembling the Final Formula

Let's recap the strategic steps from the video to relate the temperatures. This is an excellent exercise in applying thermodynamic principles, exactly the kind of multi-step thinking required for JEE Advanced.

1. For $T_2$: (Process 1→2, Isentropic Compression)
   $T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}⟹T_2=T_1{\left(\frac{V_1}{V_2}\right)}^{\gamma -1}=T_1{r}^{\gamma -1}$

2. For $T_3$: (Process 2→3, Isobaric Expansion)
   Using the Ideal Gas Law, since $P_2=P_3$, we have $\frac{V_2}{T_2}=\frac{V_3}{T_3}$.
   $T_3=T_2\left(\frac{V_3}{V_2}\right)=T_2\rho =(T_1{r}^{\gamma -1})\rho$

3. For $T_4$: (Process 3→4, Isentropic Expansion)
   $T_3{V}_3^{\gamma -1}=T_4{V}_4^{\gamma -1}⟹T_4=T_3{\left(\frac{V_3}{V_4}\right)}^{\gamma -1}$
   We know $V_4=V_1$. So, the volume ratio is $\frac{V_3}{V_1}=\frac{V_3}{V_2}\frac{V_2}{V_1}=\frac{\rho }{r}$.
   $T_4=T_3{\left(\frac{\rho }{r}\right)}^{\gamma -1}=(T_1{r}^{\gamma -1}\rho )\left(\frac{{\rho }^{\gamma -1}}{r^{\gamma -1}}\right)=T_1{\rho }^{\gamma }$

Now, we substitute these four temperature expressions ($T_1$, $T_2$, $T_3$, and $T_4$) into our efficiency formula:

Factor out the common terms:

This simplifies to the final expression for the efficiency of an ideal Diesel cycle:

4. Analysis and Comparison with the Otto Cycle

This formula is rich with physical meaning.

Let's compare this to the Otto cycle efficiency:
${\eta }_{\text{Otto}}=1-\frac{1}{r^{\gamma -1}}$
The Diesel cycle efficiency formula is identical except for the new bracketed term: $\left[\frac{{\rho }^{\gamma }-1}{\gamma (\rho -1)}\right]$.

Since the cutoff ratio $\rho >1$ and $\gamma >1$, it can be shown mathematically that this bracketed term is always greater than 1. (For those interested, this can be proven by analyzing the function $f(x)=\frac{x^k-1}{k(x-1)}$ for $x>1,k>1$).

This leads to a crucial theoretical point:
For the same compression ratio ($r$), the Otto cycle is more efficient than the Diesel cycle.
${\eta }_{\text{Diesel}}<{\eta }_{\text{Otto}}(\text{for the same }r)$

So why are diesel cars often more fuel-efficient in practice? The answer lies in the compression ratio. Because diesel engines compress only air, they can use much higher compression ratios ($r\approx 15-22$) without causing knocking. Gasoline engines are limited to lower ratios ($r\approx 8-12$). The significant gain in efficiency from the much higher $r$ in a diesel engine more than compensates for the inherent efficiency disadvantage from its cycle structure.

Conclusion

Congratulations on completing the analysis of the Diesel cycle. You have now mastered the two fundamental models for internal combustion engines.

Key Takeaways:

• The Diesel cycle is defined by four processes: isentropic compression, isobaric heat addition, isentropic expansion, and isochoric heat rejection.
• Its performance is characterized by the compression ratio $r$ and the cutoff ratio $\rho$.
• The thermal efficiency is given by the formula ${\eta }_{\text{Diesel}}=1-\frac{1}{r^{\gamma -1}}\left[\frac{{\rho }^{\gamma }-1}{\gamma (\rho -1)}\right]$.
• For a given compression ratio, the Diesel cycle is less efficient than the Otto cycle. However, practical diesel engines are often more efficient overall because they operate at much higher compression ratios.

Preview of the Next Lesson:
We have focused on devices that take heat and produce work (heat engines). What about devices that do the opposite? In the next lesson, we will explore refrigerators and heat pumps. We will define their measure of performance, the coefficient of performance (COP), and analyze how they work within the framework of thermodynamics.