/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 97 A gasoline engine has a volumetr... [FREE SOLUTION] | 91Ó°ÊÓ

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A gasoline engine has a volumetric compression ratio of 10 and before compression has air at \(290 \mathrm{K}, 85 \mathrm{kPa},\) in the cylinder. The combustion peak pressure is 6000 kPa. Assume cold air properties. What is the highest temperature in the cycle? Find the temperature at the beginning of the exhaust (heat rejection) and the overall cycle efficiency.

Short Answer

Expert verified
The highest temperature in the cycle corresponds to the temperature after the combustion stroke, the next temperature corresponds to the start of the exhaust phase after the expansion stroke, and the overall cycle efficiency is calculated by applying the formula for Otto cycle efficiency.

Step by step solution

01

Find the Maximum Temperature in the Cycle

Using the equation for an isentropic process: \(T2 = T1 \cdot (CR)^{(k-1)}\), where k is the ratio of specific heats for air \(= 1.4\), T1=290K is the initial temperature and CR=10 is the compression ratio, we can calculate the temperature at the end of the compression stroke (T2). The highest temperature in the cycle is after the constant volume heat addition (combustion) in step 2, we use \(T3 = T2 \cdot (P3 / P2)\), where P2 and P3 are pressure at the end of compression stroke and peak pressure respectively.
02

Find the Temperature at the beginning of the exhaust

We know that after the gasoline burns, an isentropic expansion takes place. The temperature at the end of this expansion is calculated by using the formula for an isentropic process again: \(T4 = T3 / (CR)^{(k-1)}\), where CR=10, T3 was calculated in the previous step, and k=1.4. This is the temperature at the beginning of heat rejection.
03

Determine the Overall Cycle Efficiency

We'll use the formula for Otto cycle efficiency: \(\eta = 1 - (1 / CR^{(k-1)})\), where CR=10 is the compression ratio and k, the heat capacity ratio of air is 1.4. The result is the efficiency of the otto cycle expressed as a percentage.

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Key Concepts

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

Isentropic Process
When studying thermodynamics and engine cycles, an 'isentropic process' is a theoretical process where entropy remains constant. In other words, it's an idealized process that assumes no energy loss due to heat transfer or friction. In reality, every physical process experiences some form of energy loss, but assuming an isentropic process allows us to simplify complex calculations and understand the fundamental workings of a system like the Otto cycle.

For the Otto cycle, which is the ideal cycle for many spark-ignition internal combustion engines, the isentropic process comes into play during two stages: the compression of the air-fuel mixture and the expansion of the combustion gases. By using formulas such as \(T2 = T1 \cdot (CR)^{(k-1)}\) for compression and \(T4 = T3 / (CR)^{(k-1)}\) for expansion, where \(T1\) and \(T3\) are temperatures at specific points in the cycle, \(CR\) is the compression ratio, and \(k\) is the specific heat ratio, we can calculate temperatures throughout the engine cycle assuming these processes are isentropic.
Compression Ratio
The 'compression ratio' (\(CR\)) of an engine is a critical factor that influences its power, efficiency, and design. It is defined as the ratio of the maximum to minimum volume within the combustion chamber of an engine. In simpler terms, it compares the size of the space where the air-fuel mixture is drawn in, against the space available when that mixture is fully compressed before ignition.

Engines with higher compression ratios tend to be more efficient because this means the mixture is compressed more significantly, leading to better thermal efficiency according to the Otto cycle efficiency formula, \(\eta = 1 - (1 / CR^{(k-1)})\). In the given exercise, a compression ratio of 10 indicates that the volume is reduced to one-tenth of its original size, which significantly raises the temperature and pressure, making the subsequent combustion process more efficient.
Specific Heats for Air
In thermodynamics, the 'specific heats for air' refer to the amount of heat energy required to raise the temperature of a unit mass of air by one degree. For air, there are two specific heat values of interest: \(c_p\) and \(c_v\). The former is the specific heat at constant pressure, and the latter is the specific heat at constant volume.

The ratio of these two values, denoted as \(k\) and sometimes called the heat capacity ratio or adiabatic index, becomes vital in analyzing thermodynamic cycles like the Otto cycle. For air, a common approximation is that \(k = 1.4\). This value of \(k\) is used in the calculation of temperatures and the efficiency of the isentropic processes within the cycle, as seen in the exercise.
Volumetric Compression
The term 'volumetric compression' refers to the process of reducing the volume of a gas within a confined space, such as a cylinder in an internal combustion engine. The level of volumetric compression is indicated by the compression ratio. Achieving effective volumetric compression is essential in engine performance because it directly affects the engine's ability to draw in, compress, and ignite the fuel-air mixture properly.

In the context of the Otto cycle and the exercise in question, a volumetric compression ratio of 10 suggests that the air is being reduced in volume by a factor of ten during the compression stroke. This level of compression leads to higher temperatures and pressures that result in better combustion and, subsequently, more power output from the engine.

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Most popular questions from this chapter

The power plant shown in Fig. 11.40 combines a gas-turbine cycle and a steam- turbine cycle. The following data are known for the gas-turbine cycle. Air enters the compressor at \(100 \mathrm{kPa}\) \(25^{\circ} \mathrm{C},\) the compressor pressure ratio is \(14,\) and the isentropic compressor efficiency is \(87 \%\); the heater input rate is \(60 \mathrm{MW}\); the turbine inlet temperature is \(1250^{\circ} \mathrm{C}\), the exhaust pressure is \(100 \mathrm{kPa},\) and the isentropic turbine efficiency is \(87 \%\); the cycle exhaust temperature from the heat exchanger is \(200^{\circ} \mathrm{C}\). The following data are known for the steam-turbine cycle. The pump inlet state is saturated liquid at \(10 \mathrm{kPa}\), the pump exit pressure is \(12.5 \mathrm{MPa}\), and the isentropic pump efficiency is \(85 \%\); turbine inlet temperature is \(500^{\circ} \mathrm{C}\), and the isentropic turbine efficiency is \(87 \% .\) Determine a. The mass flow rate of air in the gas-turbine cycle. b. The mass flow rate of water in the steam cycle c. The overall thermal efficiency of the combined cycle.

A closed feedwater heater in a regenerative steam power cycle heats \(40 \mathrm{lbm} / \mathrm{s}\) of water from \(200 \mathrm{F}, 2000 \mathrm{lbf} / \mathrm{in.}^{2}\) to \(450 \mathrm{F}, 2000\) Ibf/in. \(^{2}\). The extraction steam from the turbine enters the heater at \(500 \mathrm{lbf} / \mathrm{in.}^{2}, 550 \mathrm{F}\) and leaves as saturated liquid. What is the required mass flow rate of the extraction steam?

Consider an air-standard jet engine cycle operating in a \(280-\mathrm{K}, 100\) -kPa environment. The compressor requires a shaft power input of \(4000 \mathrm{kW}\) Air enters the turbine state 3 at \(1600 \mathrm{K}\) and 2 \(\mathrm{MPa}\), at the rate of \(9 \mathrm{kg} / \mathrm{s}\), and the isentropic efficiency of the turbine is \(85 \%\). Determine the pressure and temperature entering the nozzle.

A \(10-\mathrm{kg} / \mathrm{s}\) steady supply of saturated-vapor steam at \(500 \mathrm{kPa}\) is required for drying a wood pulp slurry in a paper mill (see Fig. \(\mathrm{P} 11.64\) ). It is decided to supply this steam by cogeneration; that is, the steam supply will be the exhaust from a steam turbine. Water at \(20^{\circ} \mathrm{C}\) and \(100 \mathrm{kPa}\) is pumped to a pressure of \(5 \mathrm{MPa}\) and then fed to a steam generator with an exit at \(400^{\circ} \mathrm{C}\). What is the additional heat-transfer rate to the steam generator beyond what would have been required to produce only the desired steam supply? What is the difference in net power?

A diesel engine has air before compression at 280 \(\mathrm{K}\) and \(85 \mathrm{kPa}\). The highest temperature is \(2200 \mathrm{K}\) and the highest pressure is 6 MPa. Find the volumetric compression ratio and the mean effective pressure using cold air properties at \(300 \mathrm{K}\).

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