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\(\bullet\) (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

Expert verified
(a) The efficiency is 67.1%. (b) The engine discards 3,290 J of heat.

Step by step solution

01

Understanding Efficiency of Otto Cycle

In an Otto cycle, theoretical efficiency, \( \eta \), is calculated using the formula \( \eta = 1 - \frac{1}{r^{(\gamma-1)}} \), where \( \gamma \) is the heat capacity ratio and \( r \) is the compression ratio.
02

Plug in Values for Efficiency

For this problem, we are given \( \gamma = 1.40 \) and \( r = 9.50 \). Substitute these values into the formula to get:\[ \eta = 1 - \frac{1}{9.50^{(1.40-1)}} = 1 - \frac{1}{9.50^{0.40}} \]
03

Calculate Efficiency

Calculate \( 9.50^{0.40} \) and then compute the efficiency:\[ 9.50^{0.40} \approx 3.038 \]\[ \eta = 1 - \frac{1}{3.038} \approx 1 - 0.329 = 0.671 \]Thus, the theoretical efficiency is approximately \( 67.1\% \).
04

Use Efficiency to Find Heat Discarded

The efficiency \( \eta \) is also defined as the ratio of work done by the engine to the heat taken in, \( Q_{in} \). We can write the equation as:\[ \eta = 1 - \frac{Q_{out}}{Q_{in}} \]Rearranging gives:\[ Q_{out} = Q_{in} \cdot (1 - \eta) \]
05

Calculate Heat Discarded

The engine takes in \( Q_{in} = 10,000 \) J of heat. Using the efficiency, calculate the heat discarded, \( Q_{out} \):\[ Q_{out} = 10,000 \cdot (1 - 0.671) = 10,000 \cdot 0.329 = 3,290 \text{ J} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat Capacity Ratio
In thermodynamics, the heat capacity ratio, denoted as \( \gamma \), is a crucial concept that reflects how a substance responds to the addition or subtraction of energy. It is defined as the ratio of the heat capacity at constant pressure \( C_p \) to the heat capacity at constant volume \( C_v \).
This ratio is crucial in the analysis of thermodynamic cycles because it affects how efficiently the cycle can convert heat into work. For diatomic gases like nitrogen and oxygen, which are common in many applications, the heat capacity ratio \( \gamma \) typically takes a value around 1.4.
Understanding \( \gamma \) helps explain how much work can be derived from energy added to the system and is crucial when calculating the efficiency of cycles like the Otto cycle.
Compression Ratio
The compression ratio, denoted by \( r \), is another vital element in analyzing engines, particularly those following the Otto cycle. It is defined as the ratio of the total volume at bottom dead center (BDC) to the volume at top dead center (TDC).
Mathematically, it can be expressed as \( r = \frac{V_{BDC}}{V_{TDC}} \).
The compression ratio provides insight into how much the air-fuel mixture is compressed during the engine cycle. Higher compression ratios typically result in higher thermal efficiency because the engine can convert more of the heat from the fuel into useful work. For the Otto cycle, the efficiency is directly related to the compression ratio, making it one of the critical parameters for assessing engine performance.
Theoretical Efficiency
The theoretical efficiency of an Otto cycle, denoted by \( \eta \), can tell us how well the cycle converts the energy from fuel into mechanical work. The theoretical calculation leverages both the heat capacity ratio and the compression ratio to provide a benchmark efficiency.
It is given by the formula \( \eta = 1 - \frac{1}{r^{(\gamma - 1)}} \).
This means that as the compression ratio \( r \) increases or the heat capacity ratio \( \gamma \) increases, so does the efficiency, assuming all else remains constant.
These calculations are important because they allow engineers to design more efficient engines and predict the performance of different engine setups before they are physically tested.
Heat Discarded
In any real-world engine cycle, not all the heat absorbed from fuel combustion can be converted into work. The heat discarded, denoted here as \( Q_{out} \), represents the portion of energy that is "lost" to the environment, typically through cooling systems or exhaust gases.
To find the amount of heat discarded in an Otto cycle, we use the relationship between the efficiency \( \eta \), the heat input \( Q_{in} \), and the heat output \( Q_{out} \):
  • Start with \( \eta = 1 - \frac{Q_{out}}{Q_{in}} \)
  • Rearrange to find \( Q_{out} = Q_{in} \cdot (1 - \eta) \)
Heat discarded calculations are essential for assessing the environmental impact of an engine and for designing more efficient systems that minimize energy loss, maximizing mechanical work output instead.

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Most popular questions from this chapter

\(\bullet\) A cooling unit for chilling the water of an aquarium gives specifications of 1\(/ 10\) hp and 1270 \(\mathrm{Btu} / \mathrm{h}\) . Assuming the unit produces its 1\(/ 10\) hp at 70.0\(\%\) efficiency, calculate its perform- ance coefficient.

.. You are designing a Carnot engine that has 2 mol of \(\mathrm{CO}_{2}\) as its working substance; the gas may be treated as ideal. The gas is to have a maximum temperature of \(527^{\circ} \mathrm{C}\) and a maximum pressure of 5.00 atm. With a heat input of 400 \(\mathrm{J}\) per cycle, you want 300 \(\mathrm{J}\) of useful work. (a) Find the temperature of the cold reservoir. (b) For how many cycles must this engine run to melt completely a 10.0 -kg block of ice originally at \(0.0^{\circ} \mathrm{C},\) using only the heat rejected by the engine?

\(\cdot\) A window air-conditioner unit absorbs \(9.80 \times 10^{4} \mathrm{J}\) of heat per minute from the room being cooled and in the same period deposits \(1.44 \times 10^{5} \mathrm{J}\) of heat into the outside air. What is the power consumption of the unit in watts?

A diesel engine performs 2200 \(\mathrm{J}\) of mechanical work and discards 4300 \(\mathrm{J}\) of heat each cycle. (a) How much heat must be supplied to the engine in each cycle? (b) What is the thermal efficiency of the engine?

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