91Ó°ÊÓ

Understanding thermal efficiency helps you evaluate how well an engine converts heat into work. It is a crucial parameter in thermodynamics for assessing engine performance. Thermal efficiency, denoted by \( \eta \), is calculated using the formula:

• \( \eta = \frac{W}{Q_{in}} \)

where \( W \) represents the work output per cycle and \( Q_{in} \) stands for the heat input per cycle.
In the provided exercise, we are given \( W = 3700 \text{ J} \) and \( Q_{in} = 1.61 \times 10^4 \text{ J} \). Thus, substituting these values in the formula, we find that the thermal efficiency \( \eta \approx 0.229 \) or 22.9%.
This percentage means that around 22.9% of the input energy is converted into useful work, while the rest is lost as waste heat. Understanding this concept helps in the design and enhancement of engine performance, aiming for higher efficiency.

Heat of combustion plays a vital role in determining how much energy can be obtained from burning a specific quantity of fuel. In thermodynamics, it refers to the amount of heat released when a fuel fully combusts, which is crucial for calculating how much fuel an engine needs to produce a certain amount of energy.

• The heat of combustion is given in the problem as \( 4.60 \times 10^4 \text{ J/g} \).

To find out how much fuel is burned per cycle, the formula \( Q_{in} = m \times \text{heat of combustion} \) is used, where \( m \) represents the mass of fuel. By rearranging and solving for \( m \), we calculate:

• \( m = \frac{1.61 \times 10^4}{4.60 \times 10^4} \approx 0.35 \text{ grams} \).

Thus, approximately 0.35 grams of fuel is burned per engine cycle. This information is crucial for optimizing fuel consumption and reducing costs.

Power output defines the rate at which work is performed or energy is converted in a system. In the context of engines, it’s a measure of the overall performance and is typically expressed in units like kilowatts (kW) or horsepower (hp).
Power output can be calculated using the formula:

• \( P = \text{work per cycle} \times \text{cycles per second} \).

In the example, the engine performs \( 3700 \text{ J} \) of work per cycle and completes 60 cycles per second. To find the power output, multiply:

• \( P = 3700 \text{ J} \times 60 = 2.22 \times 10^5 \text{ J/s} = 222 \text{ kW} \).

To convert to horsepower, knowing that 1 hp = 746 watts:

• \( P = \frac{222000}{746} \approx 298 \text{ hp} \).

This reveals the engine delivers impressive power, critical for applications demanding high energy output.

Engine cycle analysis is a crucial aspect of thermodynamics, helping us to understand and evaluate engine performance comprehensively. It involves examining each component of energy input and output in a cycle to ascertain various characteristics, like efficiency and waste.

• In practice, an engine cycle includes phases where fuel is combusted, work is done, and heat is expelled.
• The analysis uses formulas to calculate values like thermal efficiency and power output.

For instance, in the provided problem, the engine absorbs \( 1.61 \times 10^4 \text{ J} \) of heat but only utilizes 22.9% of it for work. By determining the amount of heat discarded, \( Q_{out} = 1.24 \times 10^4 \text{ J} \), we understand that a significant portion of energy is lost.
Assessing how much fuel is consumed per cycle—here, 0.35 grams—also informs us about the efficiency and sustainability of the engine.
Through such analysis, engineers aim to design more efficient engines, minimizing losses and enhancing power.