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Chapter 20: Problem 6

20.6. (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 dis- card to the outside air?

Short Answer

Expert verified
(a) 60.1% theoretical efficiency. (b) 3,990 J is discarded.

Step by step solution

01

Understand the Otto Cycle Efficiency Formula

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.
02

Calculate the Efficiency

Substitute the given values into the efficiency formula: \( r = 9.50 \) and \( \gamma = 1.40 \). So, the formula becomes \( \eta = 1 - \frac{1}{9.50^{1.40 - 1}} = 1 - \frac{1}{9.50^{0.40}} \). Calculate the result to find \( \eta \).
03

Compute the Theoretical Efficiency

Calculate \( 9.50^{0.40} \), which approximately equals 2.5097. Thus, \( \eta = 1 - \frac{1}{2.5097} \approx 0.601 \) or \( 60.1\% \). This is the theoretical efficiency of the engine.
04

Calculate the Heat Discarded

The heat discarded \( Q_c \) can be calculated using \( Q_c = Q_h (1 - \eta) \), where \( Q_h = 10,000 \) J is the heat taken in. Thus, \( Q_c = 10,000 \times (1 - 0.601) = 10,000 \times 0.399 \).
05

Compute the Heat Discarded

Perform the calculation: \( 10,000 \times 0.399 = 3,990 \) J. Thus, the engine discards 3,990 J of heat to the outside air.

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Key Concepts

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

Thermodynamics
Thermodynamics is a branch of physics that focuses on the relationship between heat and other forms of energy. It plays a vital role in understanding how engines, like the Otto cycle engine, work.
The laws of thermodynamics help us understand how energy is transferred and transformed in these systems. For instance, the first law of thermodynamics, also known as the law of energy conservation, states that energy cannot be created or destroyed, only transferred or converted from one form to another.
This is crucial in engine cycles where fuel combustion converts chemical energy into mechanical energy. The second law of thermodynamics, on the other hand, introduces the concept of entropy, stating that energy transformations are not 100% efficient and some energy is always lost as heat. This is why no engine can be perfectly efficient, making it essential to understand and calculate the efficiency of systems like the Otto cycle.
Compression Ratio
In the context of engines, the compression ratio is a key factor that influences performance and efficiency. It is defined as the ratio of the maximum volume to the minimum volume in the cylinder.
Mathematically, it's given by the formula \[ r = \frac{V_{max}}{V_{min}} \]where \( V_{max} \) is the maximum cylinder volume and \( V_{min} \) is the minimum cylinder volume or the volume at the top of the piston.
  • A higher compression ratio indicates that the engine compresses the air-fuel mixture to a smaller volume before ignition.
  • This often results in more efficient fuel usage and better performance since a higher compression ratio typically leads to a higher engine efficiency.
  • However, while high compression ratios can improve engine efficiency, they may also result in increased engine knocking, which needs to be managed.
Understanding this concept is important for assessing engine efficiency, as seen in the Otto cycle, where the compression ratio directly impacts the theoretical efficiency.
Heat Capacity Ratio
The heat capacity ratio, also known as the adiabatic index or \( \gamma \), is a significant parameter in thermodynamics, especially in processes involving gases. It is the ratio of heat capacity at constant pressure \( C_p \) to heat capacity at constant volume \( C_v \):\[ \gamma = \frac{C_p}{C_v} \]
  • Gases, when compressed or expanded adiabatically, change temperature more so than when they are heated or cooled at constant pressure or volume.
  • The value of \( \gamma \) usually ranges from 1.3 to 1.7 for most gases and plays a crucial role in determining how a gas behaves under compression and expansion processes.
  • In the Otto cycle, the heat capacity ratio helps determine the efficiency of the cycle through the formula \[ \eta = 1 - \frac{1}{r^{(\gamma - 1)}} \],where a higher \( \gamma \) contributes to greater efficiency.
Understanding \( \gamma \) is essential for calculating the efficiency of thermodynamic cycles and is integral in engine design and testing.

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Most popular questions from this chapter

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?

20.18. A Carnot device extracts 5.00 \(\mathrm{kJ}\) of heat from a body at \(-10.0^{\circ} \mathrm{C} .\) How much work is done if the device exhausts heat into the environment at \((a) 25.0^{\circ} \mathrm{C} ;(\mathrm{b}) 0.0^{\circ} \mathrm{C} ;(\mathrm{c})-25.0^{\circ} \mathrm{C} ;\) In each case, is the device acting as an engine or as a refrigerator?

20.25. A sophomore with nothing better to do adds heat to 0.350 \(\mathrm{kg}\) of ice at \(0.0^{\circ} \mathrm{C}\) until it is all melted. (a) What is the change in entropy of the water? (b) The source of heat is a very massive body at a temperature of \(25.0^{\circ} \mathrm{C}\) . What is the change in entropy of this body? (c) What is the total change in entropy of the water and the heat source?

20.50. A stirling-cycle Engine. the Otto cycle, except that the compression and expansion of the gas are done at constant temperature, not adiabatically as in the Otto cycle. The Stirling cycle is used in external combustion engines (in fact, burning fuel is not necessary; any way of producing a temperature difference will do -solar, geothermal, ocean temperature gradient, etc. \(.\) which means that the gas inside the cylinder is not used in the combustion process. Heat is supplied by burning fuel steadily outside the cylinder, instead of explosively inside the cylinder as in the Otto cycle. For this reason Stirling-cycle engines are quieter than Otto-cycle engines, since there are no intake and exhaust valves (a major source of engine noise). While small Stirling engines are used for a variety of purposes, Stiring engines for automobiles have not been successful because they are larger, heavier, and more expensive than conventional automobile engines. In the cycle, the working fluid goes through the following sequence of steps (Fig. 20.30\()\) : (i) Compressed isothermally at temperature \(T_{1}\) from the initial state \(a\) to state \(b\) , with a compression ratio \(r .\) (ii) Heated at constant volume to state \(c\) at temperature \(T_{2}\) . (iii) Expanded isothermally at \(T_{2}\) to state \(d\) . (iv) Cooled at constant volume back to the initial state \(a\) . Assume that the working fluid is \(n\) moles of an ideal gas (for which \(C_{V}\) is independent of temperature). (a) Calculate \(Q, W,\) and \(\Delta U\) for each of the processes \(a \rightarrow b, b \rightarrow c, c \rightarrow d,\) and \(d \rightarrow a\) . (b) In the Stirling cycle, the heat transfers in the processes \(b \rightarrow c\) and \(d \rightarrow a\) do not involve external heat sources but rather use regeneration: The same substance that transfers heat to the gas inside the cylinder in the process \(b \rightarrow c\) also absorbs heat back from the gas in the process \(d \rightarrow a\) . Hence the heat transfers \(Q_{b \rightarrow c}\) and \(Q_{d \rightarrow a}\) do not play a role in determining the efficiency of the engine. Explain this last statement by comparing the expressions for \(Q_{b \rightarrow c}\) and \(Q_{d \rightarrow a}\) calculated in part (a). (c) Calculate the efficiency of a Stirling-cycle engine in terms of the temperatures \(T_{1}\) and \(T_{2}\) . How does this compare to the efficiency of a Carnot-cycle engine operating between these same two temperatures? (Historically, the Stirling cycle was devised before the Carnot cycle.) Does this result violate the second law of thermodynamics? Explain. Unfortunately, actual Stirling-cycle engines cannot achieve this efficiency due to problems with the heat-transfer processes and pressure losses in the engine.

20.17. A Carnot refrigerator is operated between two heat reservoirs at temperatures of 320 \(\mathrm{K}\) and 270 \(\mathrm{K}\) (a) If in each cycle the refrigerator receives 415 \(\mathrm{J}\) of heat energy from the reservoir at 270 \(\mathrm{K}\) , how many joules of heat energy does it deliver to the reservoir at 320 \(\mathrm{K} ?\) (b) If the refrigerator completes 165 cycles each minute, what power input is required to operate it? (c) What is the coefficient of performance of the refrigerator?
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