**Find the ideal efficiency of the heat engine. Utilize**$\gamma = 1.40$.**The Dodge Viper GT2 engine has a compression ratio of**$9.6$**. With this increase in the compression ratio, how much does the ideal efficiency increase?**

This problem aims to familiarize us with **ratios** and **efficiency.** The concept required to solve this problem is related to the **ratio, proportion,** and **efficiency** of an **otto cycle.** The **otto Cycle** defines how **heat engines shift fuel** into **motion.**

A **standard fuel engineĀ **has an **operational thermal** efficiency of around $25\%$ to $30\%$. The rest of $70-75\%$ is abandoned as **scrap heat** which means it is not used in **deriving** the **wheels.**

Similar to other **thermodynamic cycles,** this **cycle** transforms **chemical energy** into **thermal heat** and consequently into **motion.** As an outcome of this information, we can specify the **thermal efficiency,** $\eta_{th}$, as the **ratio** of the **work** being done by the heat engine $W$, to the **heat infusion** at the increased **temperature,** $Q_H$. The formula for **thermal efficiency** helps in deriving the formula for **efficiency** of the **otto cycle,**

\[\eta_{th} = \dfrac{W}{Q_H}\]

The standard **Otto cycle efficiency** is just a function of the **compression ratio** given as:

\[\eta_{th} = 1- \dfrac{1}{r^{\gamma – 1}}\]

Where $r$ is the **compression** ratio and,

$\gamma$ is the **thermodynamic compression** equal to $\dfrac{Const_{pressure}}{Const_{volume}}$.

## Expert Answer

**Part a:**

In this part, we are required to **calculate** the **ideal efficiency** of the **heat engine** when the **ratio** of **thermodynamics compression** is $\gamma = 1.40$. Then the **ideal efficiency** $(e)$ of the **otto cycle** can be expressed as:

\[\eta_{th}=1- \dfrac{1}{r^{\gamma – 1}}\]

Now **substituting** the values of $r$ and $\gamma$ into the above **equation** gives us:

\[\eta_{th}=1- \dfrac{1}{8.8^{1.40 – 1}}\]

\[\eta_{th}=1- \dfrac{1}{8.8^{0.40}}\]

\[\eta_{th}=1- \dfrac{1}{2.38}\]

\[\eta_{th}=\dfrac{2.38 – 1}{2.38}\]

\[\eta_{th}=0.578\]

**OR,**

\[\eta_{th} = 58\%\]

So the **ideal efficiency** of **Mercedes-Benz SLK230** comes out to be $\eta_{th} = 58\%$.

**Part b:**

The **Dodge Viper GT2** engine has a negligibly **higher compression ratio** of $r = 9.6$. We are required to **calculate** the increase in **ideal efficiency** after this increase in the **compression ratio.** So using the equation of **thermal efficiency** for the **otto cycle** with $r = 9.6$ gives us:

\[\eta_{th}=1- \dfrac{1}{9.6^{1.40 – 1}}\]

\[=1- \dfrac{1}{9.6^{0.40}} \]

\[=1- \dfrac{1}{2.47} \]

\[=\dfrac{2.47 – 1}{2.47} \]

\[\eta_{th}=0.594 \]

**OR,**

\[\eta_{th} = 59.4\%\]

So the **increase** in the **ideal efficiency** is $\eta_{th} = 59.4\% – 58\% = 1.4\%$.

The **ideal efficiency** gets **increased** as the compression ratio **increases.**

## Numerical Result

**Part a**: The **ideal efficiency** of Mercedes-Benz $SLK230$ is $\eta_{th} = 58\%$.

**Part b:** The **increase** in the ideal efficiency is $1.4\%$.

## Example

Suppose an **Otto cycle** has $r = 9 : 1$. The **pressure** of the **air** is $100 kPa = 1 bar$, and at $20^{\circ}$ C and $\gamma = 1.4$. Calculate the **thermal efficiency** of this cycle.

We are required to calculate the **thermal efficiency** with the **compression ratio** $\gamma=1.4$. So using the equation of **thermal efficiency** for the otto cycle gives us:

\[\eta_{th} = 1- \dfrac{1}{9^{1.40 – 1}} \]

\[= 1- \dfrac{1}{9^{0.40}} \]

\[= 0.5847 \]

**OR**

\[\eta_{th} = 58\%\]