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

Short Answer

Expert verified
The calculated efficiency of the Otto-cycle engine is \(η= \) Efficiency-value. The amount of heat discarded by this engine is \(q_{out} = \) Heat-value J.

Step by step solution

01

Calculate the Efficiency of the Otto-Cycle Engine

To calculate the efficiency (η) of the Otto-cycle engine, the provided formula \(1 - (1/r^{(γ-1)})\) can be used, where γ is the heat capacity ratio (1.40) and r is the compression ratio (9.50). Substituting the values provided should give the efficiency of the Otto-cycle engine.
02

Compute the Net Work Output of the Engine

The next step is to compute the net work output of the engine using the formula \(w_{net} = η × q_{in}\), where η is the efficiency computed from Step 1 and \(q_{in}\) is the total heat input provided (10,000J). Calculating this should give the net work output of the engine.
03

Determine the Heat Discarded by the Engine

The final step is to compute the heat \(q_{out}\) discarded by the engine using the formula \(q_{out} = q_{in} - w_{net}\), where \(q_{in}\) is the total heat input (10,000J) and \(w_{net}\) is the net work output from Step 2. This step gives the amount of heat discarded by the engine.

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Key Concepts

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

Heat Capacity Ratio
The heat capacity ratio, often symbolized as \( \gamma \), plays a crucial role in thermodynamic cycles like the Otto-cycle. 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 is represented as:
  • \( \gamma = \frac{C_p}{C_v} \)
For an ideal gas, \( \gamma \) typically ranges between 1.2 and 1.7. It indicates how a gas will react under compression or expansion.
This value is essential when determining the efficiency of an Otto-cycle engine because it affects how much heat energy from combustion can be converted into work. In our context, \( \gamma = 1.40 \) indicates a decent conversion efficiency, typically for engine-related applications.
Understanding \( \gamma \) helps in designing engines that are efficient and reliable, as it directly influences the temperature and pressure changes in the cycle.
Compression Ratio
The compression ratio \((r)\) in an Otto-cycle engine is a measure of how much the engine compresses the air-fuel mixture before ignition. It's defined as the ratio of the total volume of the cylinder when the piston is at the bottom of its stroke \((V_{max})\) to the volume when the piston is at the top \((V_{min})\):
  • \( r = \frac{V_{max}}{V_{min}} \)
In the problem at hand, the compression ratio is 9.50. This means that the volume is compressed from \(V_{max}\) to 1/9.5 of its original size at \(V_{min}\).
A higher compression ratio generally means better engine efficiency, as it can cause the fuel to burn at a higher pressure and temperature, which improves the conversion of heat into mechanical energy.
However, too high a compression ratio can lead to engine knocking — an inefficient and potentially damaging condition.
Net Work Output
Net work output \((w_{net})\) is the usable work produced by the engine, calculated by subtracting the heat discarded \((q_{out})\) from the heat input \((q_{in})\). For an Otto-cycle engine, the formula to determine \(w_{net}\) is:
  • \( w_{net} = \eta \times q_{in} \)
Here, \( \eta \) is the efficiency obtained from the earlier calculations. In our case, after determining efficiency, multiplying it by the heat input (10,000 J) gives the net work output.
This output represents the effective energy that the engine can use to perform work, such as turning a vehicle's wheels or running an electrical generator.
The design and material of the engine, along with the values of \(\gamma\) and compression ratio, significantly affect the net work output.
Heat Discarded
Heat discarded \((q_{out})\) by the engine is equally important to measure as it tells us how much energy is not used for doing work and is instead lost to the surroundings. The formula to compute this is:
  • \( q_{out} = q_{in} - w_{net} \)
By subtracting the net work output from the total heat input, we can determine how much heat energy was not converted to work but instead expelled. In this exercise, having a total input of 10,000 J and calculating \(w_{net}\) from the previous section helps in finding \(q_{out}\).
Understanding the heat discarded is crucial for improving an engine's design to reduce waste and improve efficiency. It involves fine tuning the combustion process and enhancing thermal insulation.
It's a constant challenge for engineers to minimize this discarded heat, ensuring that more of the input heat energy is converted into useful work.

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Most popular questions from this chapter

A lonely party balloon with a volume of \(2.40 \mathrm{~L}\) and containing \(0.100 \mathrm{~mol}\) of air is left behind to drift in the temporarily uninhabited and depressurized International Space Station. Sunlight coming through a porthole heats and explodes the balloon, causing the air in it to undergo a free expansion into the empty station, whose total volume is \(425 \mathrm{~m}^{3}\). Calculate the entropy change of the air during the expansion.

A Carnot engine is operated between two heat rescrvoirs at temperatures of \(520 \mathrm{~K}\) and \(300 \mathrm{~K}\). (a) If the engine receives \(6.45 \mathrm{~kJ}\) of heat energy Irom the reservoir at \(520 \mathrm{~K}\) in each eycle, how many joules per cycle does it discard to the reservoir at \(300 \mathrm{~K}^{\prime} ?\) (b) How much mechanical work is performed by the engine during each cycle? (c) What is the thermal efficiency of the engine?

The theorctical cfficicncy of an Otto cycle is \(63 \% .\) The ratio of the \(C_{p}\) and \(C_{V}\) heat capacities for the gas in the cngine is \(\frac{7}{5}\). (a) What is the compression ratio \(r ?\) (b) If the net work done per cycle is \(12.6 \mathrm{~kJ}\), how much heat is rejected and how much is absorbed by the engine during each cycle?

Entropy of Metabolism. An average sleeping person metabolizes at a rate of about \(80 \mathrm{~W}\) by digcsting food or burning fat. Typically, \(20 \%\) of this energy goes into bodily functions, such as cell repair, pumping blood, and other uses of mechanical energy, while the rest goes to heat. Most people get rid of all this excess heat by transferring it (by conduction and the flow of blood) to the surface of the body. where it is radiated away, The normal intemal temperature of the body (where the metabolism takes place) is \(37^{\circ} \mathrm{C},\) and the skin is typically \(7 \mathrm{C}^{3}\) cooler. By how much does the person's entropy change per sccond due to this heat transfer?

A cylinder contains oxygen at a pressure of \(2.00 \mathrm{~atm}\). The volume is \(4.00 \mathrm{I}_{2}\), and the temperature is \(300 \mathrm{~K}\). Assume that the oxygen may be treated as an ideal gas. The oxygen is carried through the following processes: (i) Heated at constant pressure from the initial state (state 1 ) to state 2 , which has \(T=450 \mathrm{~K}\). (ii) Cooled at constant volume to \(250 \mathrm{~K}\) (state 3 ). (iii) Compressed at constant temperature to a volume of \(4.00 \mathrm{~L}\) (state 4 ). (iv) Heated at constant volume to \(300 \mathrm{~K},\) which takes the system back to state 1 (a) Show these four processes in a \(p V\) -diagram, giving the numerical values of \(p\) and \(V\) in each of the four states. (b) Calculate \(Q\) and \(W\) for each of the four processes. (c) Calculate the net work done by the oxygen in the complete cycle. (d) What is the efficiency of this device as a heat engine? How does this compare to the efliciency of a Carnot-cycle engine operating between the same minimum and maximum temperatures of \(250 \mathrm{~K}\) and \(450 \mathrm{~K}\) ?

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