Diesel Cycle

Introduction

1

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  Diesel Cycle

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  Diesel Cycle

Ratio of specific heats (heat capacity ratio) is defined as $\gamma ={\frac {C_{p}}{C_{v}}}$

pV diagram

1. Isentropic (adiabatic) expansion
2. Isochoric cooling (Qout): Heat rejection. Power stroke ends, heat rejection starts.
3. Isobaric compression: Exhaust
4. Isobaric expansion: Intake
5. Isentropic (adiabatic) compression
6. Isobaric heating (Qin): Combustion of fuel (heat is added in a constant pressure;)

Engine displacement is the cylinder volume swept by all of the pistons of a piston engine, excluding the combustion chambers. A combustion chamber is part of an internal combustion engine in which the fuel/air mix is burned.

Only air is compressed, and then diesel fuel is injected directly into that hot, high-pressure air.

• Cylinder pressure: ~30–80 bar
• Injection pressure: ~1,000–2,500+ bar

Diesel Cycle and Ideal Gas

There are three (3) different processes:

1. Isentropic $dU=nC_{v}dT=-pdV$
2. Isobaric $Q=\Delta U+p\Delta V$
3. Isochoric $\Delta Q=mc_{v}\Delta T$

For a closed system, the total change in energy of a system is the sum of the work done and the heat added $dU=\delta W+\delta Q$, and the reversible work done on a system by changing the volume is $\delta W=-pdV$.

If the system is reversible and adiabatic (isentropic) $\delta Q=0$, which gives

$dU=\delta W+\delta Q=-pdV+0$

Furthermore, for any transformation of an ideal gas, it is always true that $dU=nC_{v}dT$, giving

$dU=nC_{v}dT=-pdV$

For the constant pressure (isobaric process) $\Delta p=0$ we have $W=\int pdV=p\Delta V$, and by applying the ideal gas law, we get $W=nR\Delta T$.

${\displaystyle }$

Real gas: Air

Basic ideal gas model (good first approximation). Air is usually approximated as an ideal gas with:

• Gas constant:
  • R≈287 J/(kg\cdotpK)
  • R≈287J/(kg\cdotpK)
• Equation of state:
• p=ρRT

This works well at:

• pressures near atmospheric
• temperatures roughly 200–500 K

“Generalized” ideal gas → temperature-dependent properties. To go beyond the simple model, you allow properties like heat capacity to vary with temperature:

Heat capacity. For air, a common approximation is a polynomial $c_{p}(T)=a+bT.$ Typical coefficients (for dry air, ~200–1000 K range):

• a≈1005
• b≈0.1

More accurate forms come from NASA polynomials: ${\frac {c_{p}}{R}}=a_{1}+a_{2}T+a_{3}T^{2}+a_{4}T^{3}+a_{5}T^{4}$ Which are widely used in CFD and thermodynamics.

Compressibility factor Z (real gas correction). If you want a generalized ideal gas, you often introduce $p=Z\rho RT$. Where

• Z=1 → ideal gas
• Z≠1 → real gas behavior

For air:

• At normal conditions: Z≈1
• At high pressure: use virial expansion

$Z=1+{\frac {B(T)}{V}}+{\frac {C(T)}{V^{2}}}+\cdots \approx 1+{\frac {B(T)}{RT}}p.$

Mixture-based formulation (more fundamental). Air is a mixture mainly of:

• N₂ (~78%)
• O₂ (~21%)
• Ar (~1%)

${\begin{aligned}R&=\sum _{i}y_{i}R_{i}\\c_{p}&=\sum _{i}y_{i}c_{p,i}(T)\end{aligned}}$

Realistic Diesel Cycle

• 

  The size of the combustion chamber of MB W211.

Ratio of specific heats γ

$\gamma ={\frac {\sum y_{1}c_{p,i}}{\sum y_{1}(c_{p,i}-R_{u})}}$

Caption text

Condition	γ (approx.)
Stoichiometric	~1.30–1.33
Moderate lean	~1.33–1.37
Very lean (diesel)	~1.37–1.40

Injection pressures

• Older systems: 200–500 bar
• Modern common-rail: 1,000–2,500 bar
• Latest systems: up to ~3,000 bar

• Air pressure in cylinder: ~30–80 bar, which is the pressure of the combustion chamber.
• Temperature: ~700–1000 K

Diesel-air Mixture

Only diesel is ejected into the cylinder. Clean air comes in during the intake stroke.

The heat capacity ratio (known as the adiabatic index) for a diesel-air mixture is typically around 1.4.

Air-fuel ratio

14.5:1	Near-stoichiometric; good combustion efficiency but higher emissions.
16:1	Balanced performance; good power output and efficiency.
18:1	Lean burn; improved fuel economy but potential for higher NOx emissions.

Diesel is a complicated compound, but it’s commonly approximated as a hydrocarbon like C₁₂H₂₃ (or C₁₂H₂₆). Air is about 21% O₂ and 79% N₂. Without nitrogen, the stoichiometric ratio is about

C₁₂​H₂₃​+17.75O₂​→12CO₂​+11.5H₂​O

and by including nitrogen, we get

C₁₂​H₂₃​+17.75( O₂ + 3.76 N₂ )​→12CO₂​+11.5H₂​O + 66.74 N₂.

The molar masses

• Fuel: 167 g/mol
• Air: 137.28 g/mol

And the total air needed is 17.72 x 137.28 = 2436 g, which gives air-to-fuel-ratio to

$AFR={\frac {2436}{167}}=14.6:1$

MB data

Motor: OM648.

Mercedes Benz W211 (2003)

• Engine displacement: 3222 cm³ = 0.003222 m³
• Bore x Stroke: 88.0 x 88.4 mm³
• Compression Ratio: 18.0

Bore x stroke gives V = 6xπ(8.8/2)² x 8.84 cm³ = 3225.95cm³, which is rather close.

$p_{1}V_{1}^{\gamma }={\text{constant}}_{1}$