Otto vs. Diesel: Performance Comparison

A Quick Refresher: The Key Difference

Before we compare, let's lock in the one crucial difference that drives everything else. On a P-V diagram:

• Otto Cycle: Heat is added instantly at Top Dead Center. This is a constant volume process (a vertical line up).
• Diesel Cycle: Heat is added as the piston begins to move down. This is modeled as a constant pressure process (a horizontal line across).

This single change in the heat addition process is the root cause of all the performance differences we're about to explore.

Case 1: Same Compression Ratio and Same Heat Input

This is the comparison we hinted at last time. Let's imagine we build two engines, one gasoline and one diesel, with the exact same size and compression ratio (r). Then, we supply the exact same amount of heat (Q_in) to each one per cycle. Which one is more efficient?

To figure this out, we'll turn to the Temperature-Entropy (T-S) diagram, because it visually represents heat transfer as the area under the curve.

A fantastic and clear explanation of this is given by Praveen Kulkarni. Let's watch how he builds the two cycles on top of each other.

Let's walk through the key logic from the video together.

1. Shared Processes: Both cycles start at the same state 1 and are compressed to the same state 2 because their compression ratios are identical. The heat rejection process for both is also a constant volume process from state 4 back to state 1.

2. Heat Input (Q_in): On a T-S diagram, the constant-volume line (Otto) is steeper than the constant-pressure line (Diesel). Since we must supply the same amount of heat (Q_in), the area under the process 2 → 3 must be the same for both. To achieve this, the Diesel process has to stretch out, while the Otto process goes up to a higher peak temperature (T₃, Otto > T₃, Diesel).

3. Heat Rejection (Q_out): Now look at the heat rejection process (from state 4 to state 1). The video makes it visually clear from the T-S diagram that the area under the Diesel cycle's heat rejection curve (4_D → 1) is larger than the area for the Otto cycle (4_O → 1). This means the Diesel cycle has to dump more waste heat.
   $Q_{out,Diesel}>Q_{out,Otto}$

4. Efficiency (η): The efficiency formula is η = 1 - Q_out / Q_in. Since Q_in is the same for both, the cycle that has the lower Q_out will be more efficient.

Conclusion for Case 1: For the same compression ratio and heat input, the Otto cycle is more efficient and produces more net work.

• η_Otto > η_Diesel
• W_net, Otto > W_net, Diesel

Case 2: Same Compression Ratio and Same Heat Rejection

Let's try a different comparison. What if we again have the same compression ratio (r), but this time we adjust the fuel input so that both cycles reject the same amount of heat (Q_out)?

This scenario is also beautifully explained in the same video.

Here's the logical breakdown for this case:

1. Shared Processes: Like before, states 1 and 2 are the same due to the equal compression ratio. But now, since Q_out is the same, and it occurs over the same volume change (V₄ to V₁), this forces state 4 to be identical for both cycles as well. So, points 1, 2, and 4 are fixed and common to both.

2. Heat Input (Q_in): Looking at the T-S diagram, we can see the area under the heat addition curve for the Otto cycle (2 → 3) is clearly larger than the area under the curve for the Diesel cycle (2 → 3').
   $Q_{in,Otto}>Q_{in,Diesel}$

3. Efficiency (η): Again, η = 1 - Q_out / Q_in. This time, Q_out is the same. The cycle with the higher Q_in will have a smaller Q_out / Q_in ratio, and therefore higher efficiency.

Conclusion for Case 2: For the same compression ratio and heat rejection, the Otto cycle is still more efficient and produces more net work.

• η_Otto > η_Diesel
• W_net, Otto > W_net, Diesel

A Reality Check

At this point, you might be thinking, "Okay, the Otto cycle seems to win every time. Why do we even bother with diesel engines?"

The instructor in the video calls these first two comparisons "a bit ridiculous," and he's right. The core assumption—that the compression ratios are the same—is unrealistic. The very essence of a Diesel engine is its ability to compress only air, allowing it to use much, much higher compression ratios without fear of engine knock.

• Gasoline (Otto) Engines: r is typically 8 to 12.
• Diesel Engines: r is typically 15 to 24.

Comparing them at r=9 is an unfair fight that ignores the Diesel's main advantage. So, let's set up a final, more practical comparison.

Case 3: Same Peak Temperature, Peak Pressure, and Heat Rejection

This is the most realistic scenario. Real engines are limited by the materials they're made of. There's a maximum pressure (P_max) the cylinder can withstand and a maximum temperature (T_max) the components can tolerate before they start to melt or fail.

So, let's compare the two cycles assuming they have the same P_max and T_max (which occurs at state 3) and the same Q_out.

This final comparison is the most important one. Let's see how it plays out.

Let's summarize the game-changing results from this analysis.

1. Shared Processes: In this scenario, state 3 (the peak of the cycle) and the entire heat rejection process (4 → 1) are the same for both cycles.

2. Compression Ratio (r): To reach the same peak temperature and pressure (state 3) via their different heat addition paths, the cycles must have different compression processes. As shown on the T-S diagram, the Diesel cycle must compress the air to a much higher state before injecting fuel. This means, under these constraints, the Diesel cycle operates at a higher compression ratio.
   $r_{Diesel}>r_{Otto}$

3. Heat Input (Q_in): On the T-S diagram, the area under the Diesel cycle's constant-pressure heat addition curve (2'_D → 3) is larger than the area under the Otto cycle's constant-volume curve (2_O → 3).
   $Q_{in,Diesel}>Q_{in,Otto}$

4. Efficiency (η): We're back to η = 1 - Q_out / Q_in. Q_out is the same. The cycle with the greater Q_in will be more efficient.

Conclusion for Case 3: For the same peak pressure, peak temperature, and heat rejection, the Diesel cycle is more efficient and produces more net work.

• η_Diesel > η_Otto
• W_net, Diesel > W_net, Otto

This finally explains the real-world performance! Because Diesel engines can operate at higher compression ratios, in a practical scenario where both engines are built to withstand similar mechanical and thermal stresses, the Diesel cycle comes out on top.

Conclusion

Great work today. We've moved beyond simply analyzing the cycles in isolation and have now performed a sophisticated comparison that reveals the true engineering trade-offs between them.

Key Takeaways:

• The Comparison Matters: The answer to "Which cycle is more efficient?" depends entirely on the conditions of the comparison.
• Case 1 & 2 (Same r): If both cycles are forced to use the same (unrealistically low for a diesel) compression ratio, the Otto cycle is more efficient. This is a useful theoretical point.
• Case 3 (Same Peak P, T): In a more practical comparison limited by material stresses, the Diesel cycle is more efficient. This is because it must operate at a higher compression ratio to meet those limits, and a higher r is the king of efficiency.

You now understand not just that diesel engines are often more efficient, but the thermodynamic reasoning why.

Preview of the Next Lesson

We have now mastered the two fundamental cycles for piston-cylinder internal combustion engines. But what about engines that don't use pistons? Think about the giant engines on an airplane wing or the turbines in a power plant. These operate on a different principle altogether. In our next lesson, we will explore the Brayton cycle, the ideal model for gas turbines and jet engines. We'll see how continuous combustion and flow create power in a whole new way.