9.2 Diesel cycle

The Diesel cycle powers compression-ignition engines, offering higher efficiency than spark-ignition engines. It achieves this through higher compression ratios and constant-pressure heat addition, resulting in better fuel economy and strong torque characteristics.

Understanding the Diesel cycle is crucial for gas power cycle analysis. It demonstrates how combining different thermodynamic processes produces efficient power-generating systems, and it gives you a direct comparison point against the Otto cycle you've already studied.

Diesel Cycle Principles

Key Components and Stages

A Diesel engine uses the same basic hardware as other reciprocating engines: a cylinder, piston, connecting rod, crankshaft, fuel injector, and intake/exhaust valves. What makes it distinct is how combustion happens.

The Diesel cycle consists of four idealized stages:

1. Process 1→2: Isentropic compression of air alone (no fuel yet)
2. Process 2→3: Constant-pressure heat addition as fuel is injected and burns
3. Process 3→4: Isentropic expansion as hot gases push the piston down
4. Process 4→1: Constant-volume heat rejection as exhaust gases are expelled

Notice that the heat addition step is at constant pressure, not constant volume. This is the defining feature that separates the Diesel cycle from the Otto cycle.

Compression Ignition and Ratios

Diesel engines don't use a spark plug. Instead, air is compressed to such high pressures and temperatures that the fuel autoignites when injected. This compression-ignition approach requires much higher compression ratios than spark-ignition engines.

• Typical Diesel compression ratios range from 14:1 to 25:1, compared to roughly 8:1 to 12:1 for gasoline engines.
• These higher ratios directly increase thermal efficiency, but they also mean the engine components must be built stronger and heavier to handle the elevated pressures and temperatures.

Diesel vs. Otto Cycles

Similarities and Differences

Both cycles are idealized four-process models for internal combustion engines, and both include two isentropic (adiabatic, reversible) processes. The key difference is in how heat is added:

• Otto cycle: constant-volume heat addition (rapid combustion with the piston nearly stationary)
• Diesel cycle: constant-pressure heat addition (fuel burns more gradually as the piston moves)

Because the Diesel cycle operates at higher compression ratios, it achieves higher thermal efficiency for a given amount of heat input. On a P-v or T-s diagram, you can see this difference clearly: the Diesel cycle's heat addition step is a horizontal line on the P-v diagram, while the Otto cycle's is a vertical line.

Fuel Delivery and Engine Characteristics

In an Otto engine, fuel and air are premixed before compression. In a Diesel engine, only air is compressed, and fuel is injected near the end of the compression stroke directly into the hot, high-pressure air.

This injection approach gives Diesel engines precise control over fuel quantity and timing, which helps optimize combustion and reduce emissions. The tradeoff is that Diesel engines generally have lower power-to-weight ratios due to their heavier construction, but they compensate with better fuel economy and higher torque output.

Thermodynamic Processes in the Diesel Cycle

Isentropic Compression and Expansion

Process 1→2 (Isentropic Compression): Air is compressed adiabatically from low pressure/temperature to high pressure/temperature. Because the process is isentropic, there's no heat transfer and entropy stays constant. The relationship governing this process is:

$Pv^{\gamma} = \text{constant}$

You can also relate states 1 and 2 using:

$T_2 = T_1 \left(\frac{v_1}{v_2}\right)^{\gamma - 1} = T_1 \, r^{\,\gamma - 1}$

where $r = v_1 / v_2$ is the compression ratio.

Process 3→4 (Isentropic Expansion): The high-pressure, high-temperature combustion gases expand adiabatically, pushing the piston down and producing work. The same isentropic relation $Pv^{\gamma} = \text{constant}$ applies.

Constant-Pressure Heat Addition and Constant-Volume Heat Rejection

Process 2→3 (Constant-Pressure Heat Addition): Fuel is injected into the hot compressed air and burns at approximately constant pressure. The heat added per unit mass is:

$q_{\text{in}} = c_p (T_3 - T_2)$

The volume increases during this process, and the ratio of volumes defines the cutoff ratio:

$r_c = \frac{v_3}{v_2}$

This cutoff ratio tells you how long combustion continues as the piston moves. A larger cutoff ratio means more fuel is burned, but it also reduces efficiency.

Process 4→1 (Constant-Volume Heat Rejection): After expansion, the exhaust valve opens and pressure drops rapidly. This is modeled as heat rejection at constant volume. Since no boundary work occurs ($W = 0$), the heat rejected per unit mass is:

$q_{\text{out}} = c_v (T_4 - T_1)$

The constant-pressure heat addition in the Diesel cycle allows for more gradual combustion compared to the Otto cycle, which is one reason Diesel engines can tolerate lower-quality fuels.

Efficiency and Work Output of Diesel Engines

Thermal Efficiency Calculation

The thermal efficiency of an ideal Diesel cycle depends on the compression ratio $r$, the cutoff ratio $r_c$, and the specific heat ratio $\gamma$:

$\eta_{\text{Diesel}} = 1 - \frac{1}{r^{\,\gamma-1}} \left[\frac{r_c^{\,\gamma} - 1}{\gamma(r_c - 1)}\right]$

A few things to notice about this expression:

• As $r_c \to 1$ (meaning the constant-pressure process shrinks to zero duration), the bracketed term approaches 1, and the efficiency approaches the Otto cycle efficiency $\eta_{\text{Otto}} = 1 - 1/r^{\,\gamma-1}$.
• For any $r_c > 1$, the bracketed term is greater than 1, which means at the same compression ratio, the Diesel cycle is less efficient than the Otto cycle.
• However, Diesel engines can operate at much higher compression ratios than Otto engines (since there's no premixed fuel to knock), so in practice Diesel engines achieve higher overall efficiencies.

Factors affecting real Diesel engine efficiency include the compression ratio, cutoff ratio, fuel properties, and engine geometry (bore, stroke, valve timing).

Work Output and Power

The net work output per unit mass is the difference between heat added and heat rejected:

$w_{\text{net}} = q_{\text{in}} - q_{\text{out}} = c_p(T_3 - T_2) - c_v(T_4 - T_1)$

Equivalently, you can compute it as the difference between expansion work and compression work. Using enthalpy values at each state point:

$W_{\text{net}} = m(h_3 - h_4) - m(h_2 - h_1)$

where $m$ is the mass of the working fluid and $h$ is specific enthalpy at each state.

The power output of a Diesel engine equals the net work per cycle multiplied by the number of cycles per second (which depends on engine speed and whether it's a two-stroke or four-stroke design) and the number of cylinders.

Diesel engines are known for high torque at lower engine speeds, which makes them well-suited for heavy-duty applications like trucks, buses, and industrial machinery.