How High Should the Compression Ratio of a Gasoline Engine Be?

A gasoline engine becomes more efficient when its compression ratio is increased. So why not simply make the compression ratio as large as possible?

There is a physical obstacle: increasing the compression ratio also increases the pressure inside the cylinder. An engine can withstand only a limited pressure.

This gives us a natural optimization problem:

For a fixed amount of heat released during combustion and a fixed maximum allowable cylinder pressure, what compression ratio gives the greatest possible efficiency?

The answer comes from combining a simple model of a gasoline engine with calculus.

The ideal Otto cycle

We use the ideal Otto cycle, the standard simplified model for a spark-ignition gasoline engine.

Let

$$r=\frac{V_1}{V_2}$$

be the compression ratio, where $V_1$ is the cylinder volume before compression and $V_2$ is the volume after compression.

Let $P_1$ and $T_1$ be the initial pressure and temperature.

For an ideal gas undergoing adiabatic compression,

$$T_2=T_1{r}^{\gamma -1}$$

and

$$P_2=P_1{r}^{\gamma }.$$

Here

$$\gamma =\frac{c_p}{c_v},$$

and for air we use the familiar approximation

$$\gamma \approx 1.4.$$

Efficiency increases with compression

The thermal efficiency of the ideal Otto cycle is

$$\eta (r)=1-\frac{1}{r^{\gamma -1}}.$$

Differentiate:

$${\eta }^{\prime }(r)=(\gamma -1)r^{-\gamma }.$$

Since $r>1$ and $\gamma >1$, we have

$${\eta }^{\prime }(r)>0.$$

Thus, according to the ideal model, efficiency always increases as the compression ratio increases.

So there is no unconstrained maximum. Mathematics would simply tell us to keep increasing $r$.

A real engine, however, cannot withstand unlimited pressure. This is where the optimization problem becomes interesting.

Adding a pressure constraint

Suppose combustion adds a fixed amount of heat $q$ per unit mass of air.

In the ideal Otto model, heat is added at constant volume. Therefore,

$$q=c_v(T_3-T_2).$$

Hence

$$T_3=T_2+\frac{q}{c_v}.$$

Because the volume does not change during combustion, the ideal-gas law gives

$$\frac{P_3}{P_2}=\frac{T_3}{T_2}.$$

Therefore,

$$P_3=P_2(1+\frac{q}{c_v{T}_2}).$$

Now substitute

$$P_2=P_1{r}^{\gamma }$$

and

$$T_2=T_1{r}^{\gamma -1}.$$

We obtain

$$P_3=P_1(r^{\gamma }+\frac{q}{c_v{T}_1}r).$$

The key equation

Define

$$B=\frac{q}{c_v{T}_1}.$$

Then the maximum pressure reached during the idealized cycle is

$$P_3=P_1(r^{\gamma }+Br).$$

Suppose the engine can safely withstand a maximum cylinder pressure $P_{\text{max}}$. Then

$$P_3\le P_{\text{max}}.$$

Define

$$A=\frac{P_{\text{max}}}{P_1}.$$

The pressure constraint becomes

$$r^{\gamma }+Br\le A.$$

Where does the maximum occur?

We already proved that the efficiency $\eta (r)$ is increasing.

Therefore, the most efficient engine uses the largest compression ratio permitted by the pressure constraint.

The optimum must occur when the pressure reaches its allowable maximum:

$$r^{\gamma }+Br=A.$$

This is an interesting kind of optimization problem. We do not find the optimum by solving ${\eta }^{\prime }(r)=0$. There is no critical point.

Instead, calculus tells us that efficiency is increasing, and the physical constraint tells us where we must stop.

A numerical example

Take

$$P_1=100\text{kPa},T_1=300\text{K}.$$

Use

$$c_v=0.718\text{kJ/(kg K)}$$

and suppose combustion supplies

$$q=1800\text{kJ/kg}.$$

Then

$$B=\frac{1800}{(0.718)(300)}\approx 8.36.$$

Suppose the maximum allowable cylinder pressure is

$$P_{\text{max}}=10\text{MPa}=10000\text{kPa}.$$

Therefore,

$$A=\frac{10000}{100}=100.$$

Using $\gamma =1.4$, the optimal compression ratio satisfies

$$r^{1.4}+8.36r=100.$$

Solving this equation numerically gives

$$r\approx 9.27.$$

Thus, in this simplified model, the greatest possible efficiency under the pressure restriction occurs at a compression ratio of approximately 9.27:1.

What efficiency does this give?

For $\gamma =1.4$, the ideal Otto-cycle efficiency is

$$\eta =1-\frac{1}{r^{0.4}}.$$

Using $r\approx 9.27$,

$$\eta \approx 1-\frac{1}{{9.27}^{0.4}}\approx 0.590.$$

So the theoretical efficiency is approximately

$$\eta \approx 59.0\%.$$

This is the efficiency of the idealized mathematical model, not the efficiency we should expect from a real gasoline engine. Real engines have friction, heat loss, pumping losses, finite combustion time, changing specific heats, and other effects that the ideal Otto cycle does not include.

An unexpected seventh-degree polynomial

There is one more mathematical surprise.

We used

$$\gamma =1.4=\frac{7}{5}.$$

Therefore the equation determining the optimal compression ratio has the form

$$r^{\frac{7}{5}}+Br=A.$$

Let

$$x=r^{\frac{1}{5}}.$$

Then

$$r=x^5$$

and

$$r^{\frac{7}{5}}=x^7.$$

So our engine-design equation becomes

$$x^7+B{x}^5-A=0.$$

For our numerical example,

$$x^7+8.36x^5-100=0.$$

A practical question about the design of a gasoline engine has led us to a seventh-degree polynomial.

We do not need to solve this polynomial symbolically. A numerical method gives the physically relevant positive root and therefore the optimal compression ratio.

The mathematical lesson

Without a pressure restriction, the ideal Otto model says

larger compression ratio → greater efficiency.

There is no finite optimum.

But an engine has to withstand the pressure produced inside its cylinder. Once we impose the constraint

$$P_3\le P_{\text{max}},$$

the optimization problem has a finite solution.

The optimum occurs precisely when increasing the compression ratio any further would violate the pressure constraint:

$$P_3=P_{\text{max}}.$$

This illustrates an important idea in applied calculus: sometimes the optimum is not created by a critical point of the function—it is created by the constraint.