The Ideal Otto Cycle: Analyzing and Calculating Thermal Efficiency

From a Real Engine to an Ideal Cycle

Before we can analyze an engine with our thermodynamic tools, we first need to understand how it works mechanically. You've probably heard of a "four-stroke engine." Let's walk through what those four strokes are.

Imagine a single cylinder in your car's engine. Inside it, a piston moves up and down.

To understand the mechanics and terminology, spend a few minutes reading about the real-world operation of a four-stroke engine.

Recapping the Real Engine

So, to summarize the mechanical process:

1. Intake: The piston moves down, sucking in a mix of air and gasoline vapor.
2. Compression: The piston moves up, squeezing the air-fuel mixture.
3. Power: A spark plug ignites the mixture. The resulting explosion violently pushes the piston down, creating the power that turns the wheels.
4. Exhaust: The piston moves back up, pushing the burnt gases out.

This real-world process is complex. There's friction, chemical reactions (combustion), and gas flowing in and out. To analyze it thermodynamically, we need to simplify it into an ideal cycle. This is a common and powerful engineering technique: make simplifying assumptions to create a model that captures the essential physics.

Here are the key assumptions we make, often called the air-standard assumptions, to get from the real engine to the ideal Otto cycle:

• Closed System: We ignore the intake and exhaust strokes. Instead of swapping out gases, we pretend the same fixed amount of air is inside the cylinder at all times.
• Ideal Gas: The working fluid is just air, which we treat as an ideal gas. We ignore the fuel and the complexities of combustion products.
• Heat Transfer, Not Combustion: The explosion from the spark plug is modeled as an instantaneous heat addition process (Q_in) from an external source. The exhaust stroke is modeled as an instantaneous heat rejection process (Q_out) to the surroundings.
• Fully Reversible: All processes are assumed to be internally reversible. This means no friction.
• Adiabatic Compression/Expansion: We assume no heat is lost to the cylinder walls during the compression and power strokes.

With these idealizations, we can now describe the engine's operation using four clean, analyzable thermodynamic processes.

The Four Processes of the Ideal Otto Cycle

Let's walk through the ideal cycle, keeping the P-V (Pressure-Volume) and T-S (Temperature-Entropy) diagrams in mind. These diagrams are our maps for navigating the cycle.

1. Process 1 → 2: Isentropic Compression

   • This models the compression stroke. The piston moves from its lowest point (BDC, maximum volume v_1) to its highest point (TDC, minimum volume v_2).
   • Work is done on the gas, so pressure and temperature increase.
   • Since it's isentropic (s = constant), this is a vertical line on the T-S diagram.

2. Process 2 → 3: Constant-Volume Heat Addition

   • This models the combustion. The piston is momentarily at TDC (v_2 = v_3), so the volume is constant.
   • Heat (Q_in) is added to the system, causing a rapid spike in temperature and pressure.
   • This is a vertical line on the P-V diagram.

3. Process 3 → 4: Isentropic Expansion (The Power Stroke)

   • This is where the magic happens. The high-pressure gas expands and pushes the piston down from TDC (v_3) back to BDC (v_4).
   • This produces a large amount of work (W_out). Temperature and pressure decrease as the gas expands.
   • Since it's isentropic (s = constant), this is another vertical line on the T-S diagram.

4. Process 4 → 1: Constant-Volume Heat Rejection

   • This models the exhaust process. The piston is momentarily at BDC (v_4 = v_1).
   • The cycle "resets" by rejecting waste heat (Q_out) to the surroundings, causing the temperature and pressure to drop back to their initial state.
   • This is another vertical line on the P-V diagram.

The net result is that the work produced during the power stroke (3→4) is greater than the work consumed during the compression stroke (1→2). The difference between them is the net work output of the engine, which is the area enclosed by the loop on the P-V diagram.

Calculating the Efficiency of the Otto Cycle

Our goal is to figure out how good this engine is at converting the heat we add (Q_in) into useful net work (W_net). This is the thermal efficiency, η_th.

From our very first lessons, we know:
${\eta }_{th}=\frac{\text{What we get}}{\text{What we pay for}}=\frac{W_{net}}{Q_{in}}$

By applying the First Law to the whole cycle, the net work is simply the heat in minus the heat out: W_net = Q_in - Q_out. So, our efficiency formula becomes:
${\eta }_{th}=\frac{Q_{in}-Q_{out}}{Q_{in}}=1-\frac{Q_{out}}{Q_{in}}$

Now, let's find expressions for Q_in and Q_out.

• Heat Addition (2→3): This happens at constant volume. For a closed system, Q = ΔU + W. Since volume is constant, work is zero. So, Q_in = ΔU = U_3 - U_2. For an ideal gas, this is m c_v (T_3 - T_2).
• Heat Rejection (4→1): This also happens at constant volume. The heat rejected is Q_out = U_4 - U_1. For an ideal gas, this is m c_v (T_4 - T_1).

Let's plug these into the efficiency formula (the mass m and specific heat c_v will cancel out):
${\eta }_{th}=1-\frac{m{c}_v(T_4-T_1)}{m{c}_v(T_3-T_2)}=1-\frac{T_4-T_1}{T_3-T_2}$

This formula is correct, but not very useful. It depends on all four temperatures. We can simplify it dramatically. The key is to use the relationships for the two isentropic processes.

For an ideal gas undergoing an isentropic process, we know:
$\frac{T_2}{T_1}={\left(\frac{v_1}{v_2}\right)}^{k-1}\text{and}\frac{T_3}{T_4}={\left(\frac{v_4}{v_3}\right)}^{k-1}$
Where k is the specific heat ratio (c_p / c_v).

Now, let's bring in the compression ratio, r, which we defined earlier:
$r=\frac{\text{Maximum Volume}}{\text{Minimum Volume}}=\frac{v_1}{v_2}$
Looking at the P-V diagram, we see that the volume at the start of compression (v_1) is the same as the volume at the end of expansion (v_4). Also, the volume at the end of compression (v_2) is the same as the volume at the start of expansion (v_3).
So, v_1 = v_4 and v_2 = v_3. This means v_4 / v_3 is also equal to the compression ratio r.

Our isentropic relations become:
$\frac{T_2}{T_1}=r^{k-1}\text{and}\frac{T_3}{T_4}=r^{k-1}$
This means that T_2 / T_1 = T_3 / T_4. A little bit of algebra shows that T_4 / T_1 = T_3 / T_2.

Now look at our efficiency equation again:
${\eta }_{th}=1-\frac{T_1(T_4/T_1-1)}{T_2(T_3/T_2-1)}$
Since T_4/T_1 = T_3/T_2, the terms in the parentheses are identical and cancel out! We are left with:
${\eta }_{th}=1-\frac{T_1}{T_2}$
And since T_2 / T_1 = r^(k-1), we can substitute T_1 / T_2 = 1 / r^(k-1). This gives us our final, elegant result:

${\eta }_{th,Otto}=1-\frac{1}{r^{k-1}}$

To see a walkthrough of this derivation, the following video is quite helpful.

What Does This Mean?

This is a powerful result! It tells us that the maximum theoretical efficiency of a gasoline engine depends on only two things:

1. The compression ratio (r): The higher you squeeze the gas, the more efficient the engine becomes.
2. The working fluid (k): A gas with a higher specific heat ratio will yield a more efficient cycle.

In the real world, there's a limit to the compression ratio. If you squeeze the air-fuel mix too much, it gets so hot that it ignites on its own before the spark plug fires. This is called engine knock, and it's damaging to the engine. This is why high-performance cars require high-octane gasoline—it resists knocking, allowing for higher compression ratios and thus higher efficiency and power.

Worked Problem: Analyzing an Otto Cycle

Let's put this all together and analyze a full cycle from start to finish.

Problem Statement:
An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process (state 1), the air is at 100 kPa and 25°C. The heat added during the constant-volume heat addition process is 800 kJ/kg. Using the cold-air-standard assumptions (air is an ideal gas with constant specific heats at room temperature), determine:
a) The temperature and pressure at the end of compression (state 2).
b) The temperature and pressure at the end of heat addition (state 3).
c) The net work output per kilogram of air.
d) The thermal efficiency of the cycle.

For air, use $c_v=0.718$ kJ/kg·K and $k=1.4$.

Conclusion

Congratulations on working through your first full power cycle! This is a major milestone. We started with the greasy mechanics of a real car engine and, by making some smart idealizations, turned it into a thermodynamic model we could analyze precisely.

Key Takeaways:

• The Otto cycle is the ideal model for spark-ignition (gasoline) engines.
• It consists of four processes: isentropic compression, constant-volume heat addition, isentropic expansion, and constant-volume heat rejection.
• The efficiency of an ideal Otto cycle depends only on the compression ratio (r) and the specific heat ratio (k) of the gas.
• The simple and powerful efficiency formula is: ${\eta }_{th}=1-1/r^{k-1}$.
• Higher compression ratios lead to higher efficiency, but real engines are limited by engine knock.

Preview of the Next Lesson

Now that you understand the Otto cycle, a natural question arises: what about diesel engines? They don't have spark plugs, so how do they work?

In our next lesson, we will explore the ideal Diesel cycle, the thermodynamic model for compression-ignition engines. We will see how it differs from the Otto cycle (hint: it's all in how the heat is added) and compare the performance of these two fundamental engine types. See you then