Question

(a) Calculate the theoretical efficiency for an Otto-cycle engine with $\gamma$ = 1.40 and $r$ = 9.50. (b) If this engine takes in 10,000 J of heat from burning its fuel, how much heat does it discard to the outside air?

Short answer

(a) Efficiency is 61.5%. (b) Heat discarded is 3,850 J.

Step-by-step solution

Formula for Otto Cycle Efficiency

The efficiency of an Otto-cycle engine is given by the formula: $$\eta =1-\frac{1}{r^{\gamma -1}}$$where $\eta$ is the efficiency, $r$ is the compression ratio, and $\gamma$ is the heat capacity ratio.

Substitute the Given Values

Substitute $\gamma =1.40$ and $r=9.50$ into the efficiency formula:$$\eta =1-\frac{1}{{9.50}^{1.40-1}}$$

Calculate the Efficiency

Carry out the calculation:$${9.50}^{0.40}\approx 2.60$$Therefore, the efficiency $\eta$ is:$$\eta =1-\frac{1}{2.60}\approx 1-0.385=0.615$$So, the efficiency is approximately 61.5%.

Calculate Heat Discarded

The efficiency of an engine is also defined as the ratio of work output to heat input. If $Q_{input}=10,000$ J and the efficiency $\eta =0.615$, the work done $W$ is:$$W=\eta \times Q_{input}=0.615\times 10,000=6,150\text{ J}$$The heat discarded to the surrounding $Q_{output}$ is the remaining energy:$$Q_{output}=Q_{input}-W=10,000-6,150=3,850\text{ J}$$

Key concepts

Understanding Heat Capacity Ratio

The heat capacity ratio, often symbolized by $\gamma$, is a crucial factor in thermodynamics, particularly for the Otto cycle which is common in internal combustion engines. It is defined as the ratio of the heat capacity at constant pressure $C_p$ to the heat capacity at constant volume $C_v$. Mathematically, it's expressed as:$$\gamma =\frac{C_p}{C_v}$$This ratio provides insight into how a gas behaves when subjected to changes in pressure and temperature. A higher $\gamma$ typically indicates a larger change in pressure or temperature for a given energy change. In the context of an Otto cycle, a higher heat capacity ratio means that the gas can do more work on the piston, leading to higher efficiency. For most diatomic gases, $\gamma$ is around 1.4. This is why, in our problem, $\gamma =1.40$ is utilized to find the efficiency.

The Role of Compression Ratio

The compression ratio $r$ in an Otto cycle is defined as the ratio of the total volume when the piston is at the bottom of its stroke to the volume when the piston is at the top of its stroke. It is a critical parameter for determining engine efficiency:$$r=\frac{V_{bottom}}{V_{top}}$$A higher compression ratio means the gas is compressed into a smaller volume before combustion. This increase in compression can significantly improve the engine's thermal efficiency since it allows for more complete combustion and better use of the air-fuel mixture. It is important to note that while higher compression ratios can yield better efficiency, they also require engines that can handle higher pressures. For our problem, the given compression ratio is $r=9.50$, which indicates a relatively high level of efficiency suitable for gasoline engines, as evidenced in the calculated efficiency of approximately 61.5%.

Heat Engine Calculations Simplified

Heat engine calculations aim to quantify how effectively an engine converts heat from fuel into work. In the Otto cycle, the efficiency $\eta$ is expressed as:$$\eta =1-\frac{1}{r^{\gamma -1}}$$This formula highlights how the heat capacity ratio and compression ratio influence efficiency. When an engine receives a specific amount of heat energy, the work output can be calculated by multiplying the efficiency by the heat energy intake:$$W=\eta \times Q_{input}$$For example, with an input of 10,000 J and an efficiency of 61.5%, the work done is 6,150 J. The heat discarded, or the heat not converted to work, is simply the difference between input energy and work output:$$Q_{output}=Q_{input}-W$$In our scenario, this is 3,850 J, indicating the amount of energy lost to the environment. Understanding these calculations is crucial for designing more efficient engines and reducing energy waste in real-world applications.