The cycle involved in the operation of an internal combustion engine is called the Otto cycle. Air can be considered to be the working substance and can be assumed to be a perfect gas. The cycle consists of the following steps: (1) reversible adiabatic compression from \(\mathrm{A}\) to \(\mathrm{B},(2)\) reversible constantvolume pressure increase from \(\mathrm{B}\) to \(\mathrm{C}\) due to the combustion of a small amount of fuel, (3) reversible adiabatic expansion from \(\mathrm{C}\) to \(\mathrm{D}\), and \((4)\) reversible and constant-volume pressure decrease back to state A. Determine the change in entropy (of the system and of the surroundings) tor each step of the cycle and determine an expression for the efficiency of the cycle, assuming that the heat is supplied in Step \(2 .\) Evaluate the efficiency for a compression ratio of \(10: 1\). Assume that, in state \(\mathrm{A}, \mathrm{V}=4.00 \mathrm{dm}^{3}, p=1.00 \mathrm{~atm}\), and \(\mathrm{T}=300 \mathrm{~K}\), that \(\mathrm{V}_{\mathrm{A}}==10 \mathrm{~V}_{\mathrm{B}}, \mathrm{p}_{\mathrm{C}} / \mathrm{p}_{\mathrm{B}}=5\), and that \(C_{p m}=\frac{2}{2} R\).

Short Answer

Expert verified
The entropy change for the system is zero for steps 1 and 3 (\text{adiabatic processes}), and the change for step 2 and 4 requires more data (temperatures at states \text{B} and \text{C}) to calculate. For the surroundings, entropy changes are zero for steps 1 and 2, and would require heat exchanged values for steps 3 and 4. The efficiency expression for the Otto cycle is \(\eta=1-\frac{1}{r^{\gamma-1}}\), with \(\gamma=C_{p m}/(C_{p m}-R)=\frac{2.5R}{1.5R}=\frac{5}{3}\) for the given data; the efficiency for a compression ratio of 10:1 is \(\eta=1-\frac{1}{10^{(5/3-1)}}\), which simplifies to \(\eta=1-\frac{1}{10^{2/3}}\).

Step by step solution

01

Analyze the Adiabatic Compression (A to B)

Since the process from A to B is adiabatic and reversible, there is no heat exchange with the surroundings, and the entropy change \(\Delta S_{1}\) is zero for this step. Use the adiabatic condition for a perfect gas, \(PV^{\gamma}=constant\), where \(\gamma=C_{p}/C_{v}=C_{p m}/(C_{p m}-R)\), to find the relation between pressure and volume or temperature for states A and B.
02

Compute the Constant-Volume Pressure Increase (B to C)

During the constant-volume process from B to C, the entropy change for the system \(\Delta S_{2}\) is given by \(\Delta S_{2}=n C_{v m} \ln\left(\frac{T_C}{T_B}\right)\), where \(T_C\) and \(T_B\) are the temperatures at points C and B, respectively. As \(C_{v m}\) is the molar heat capacity at constant volume, \(C_{v m}=C_{p m}-R\). The entropy change for the surroundings \(\Delta S_{sur,2}\) is zero since the system is insulated and no heat is exchanged with the surroundings.
03

Analyze the Adiabatic Expansion (C to D)

This step is similar to the compression from A to B. Since the process from C to D is adiabatic and reversible, there is no heat exchange with the surroundings, resulting in an entropy change \(\Delta S_{3}\) of zero for this step. Use the same adiabatic condition as Step 1 to relate the states C and D.
04

Compute the Constant-Volume Pressure Decrease (D to A)

For the constant-volume process from D to A, the system returns to its initial state, hence the total entropy change over one full cycle for the system is zero as the system is undergoing a cyclic process. Given that \(T_A=T_D\) from the adiabatic relations, \(\Delta S_{4}\) is zero for the system. For the surroundings, because heat is rejected by the system, \(\Delta S_{sur,4}=-\frac{Q_{out}}{T_A}\), where \(Q_{out}\) is the heat rejected and \(T_A\) is the temperature of the surroundings which is constant.
05

Calculate the Efficiency of the Otto Cycle

The efficiency \(\eta\) of the Otto cycle can be expressed as \(\eta=1-\frac{Q_{out}}{Q_{in}}\), where \(Q_{in}\) is the heat added during step 2 and \(Q_{out}\) is the heat rejected during step 4. The efficiency can also be related to the temperatures and volumes by using the formula \(\eta=1-\frac{1}{r^{\gamma-1}}\), where \(r=V_A/V_B\) is the compression ratio, and \(\gamma=C_{p m}/(C_{p m}-R)\). Given \(V_A/V_B=10\) (compression ratio of 10:1), calculate the efficiency using the provided \(\gamma\) value.
06

Evaluate the Efficiency for the Given Compression Ratio

Substitute the given compression ratio \(r=10\) and \(C_{p m}\) value into the efficiency formula to calculate the actual efficiency value for the Otto cycle with these parameters.

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Key Concepts

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

Adiabatic Process
An adiabatic process is a thermodynamic process in which no heat is exchanged with the surroundings. In simpler terms, imagine it as an insulated system where thermal energy is trapped inside and can't escape. During the adiabatic compression and expansion steps in the Otto cycle, which powers internal combustion engines, energy within the system is conserved; it is transformed but not lost or gained through heat transfer.

Mathematically, this process is described by the equation \(PV^\gamma = \text{constant}\), where \(P\) represents pressure, \(V\) volume, and \(\gamma\) is the specific heat ratio (also known as adiabatic index), which is \(\gamma = C_p / C_v\). In the realm of an ideal gas, where air in the Otto cycle can be assumed as one, the temperature and volume relationship during adiabatic processes follow \(T V^{\gamma-1} = \text{constant}\).

Application in Otto Cycle

In the context of the Otto cycle, the adiabatic steps - the compression from state A to B and the expansion from C to D - require no heat transaction with the environment. Subsequently, these processes embody the adiabatic concept as the gas is compressed and expanded without any heat transfer, adhering to the principal equation mentioned above. This characteristic is crucial since it allows the cycle to conserve energy and contributes directly to the cycle's overall efficiency. Moreover, within this adiabatic scenario, the change in entropy, a measure of system disorder, remains constant, indicating that there's no increase or decrease in randomness during these steps.
Entropy Change
Entropy is a fundamental concept in thermodynamics, often referred to as the measure of disorder or randomness within a thermodynamic system. It is symbolized by the letter \(S\), and changes in entropy, denoted \(\Delta S\), indicate the degree to which energy within a system can be transformed into useful work.

Within the context of the Otto cycle, entropy change is a key factor when we analyze energy conversion efficiency. During reversible processes, the total entropy change over a complete cycle for a system in a cyclic process, like the Otto cycle, is zero. This is because, ultimately, the system returns to its initial state, and there is no net change in randomness or disorder.

Entropy and Otto Cycle Stages

Specifically, in the Otto cycle steps, we see the following regarding entropy:
  • The adiabatic steps (compression and expansion) exhibit no heat transfer with the environment; hence, \(\Delta S\) is zero.
  • During the constant-volume heat addition and removal steps, entropy does change in the system due to the increase or decrease of internal energy - heat exchange occurs here, albeit within the system itself or with its immediate surroundings.
Moreover, while the system's total entropy might not change after completing a full cycle, the surroundings might experience a net entropy increase due to irreversible real-world processes like friction and heat loss, which are not accounted for in the idealized Otto cycle.
Thermal Efficiency
Thermal efficiency is the proportion of heat energy converted into useful work in a thermodynamic cycle; it's the performance metric of an engine's effectiveness in transforming energy from fuel into motion. Represented by the Greek letter \(\eta\), efficiency is quantitatively expressed as a ratio of the work output divided by the heat input.

In an Otto cycle, which is the foundation of gasoline engines, thermal efficiency is critical as it determines how well the engine converts the heat from fuel combustion into usable mechanical energy. To determine the efficiency of the Otto cycle, one can use the relationship \(\eta = 1 - \frac{Q_{out}}{Q_{in}}\), where \(Q_{out}\) is the heat rejected and \(Q_{in}\) is the heat supplied to the engine.

Efficiency Relations and Compression Ratio

The efficiency of the Otto cycle can also be related to the compression ratio of the engine,\(r = V_A/V_B\), and the specific heat ratio, \(\gamma\). The formula \(\eta = 1 - \frac{1}{r^{\gamma-1}}\) highlights the importance of compression ratio—engines with a higher compression ratio tend to have better efficiency since more mechanical work can be extracted from the same amount of heat energy. The calculation provided for a 10:1 compression ratio in the exercise exemplifies the practical use of these relations to optimize engine design for improved fuel economy and performance.

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Most popular questions from this chapter

Consider a perfect gas contained in a cylinder and separated by a frictionless adiabatic piston into two sections \(\mathrm{A}\) and \(\mathrm{B}\). All changes in \(\mathrm{B}\) is isothermal; that is, a thermostat surrounds \(\mathrm{B}\) to keep its temperature constant. There is \(2.00 \mathrm{~mol}\) of the gas in each section. Initially, \(T_{\mathrm{A}}==T_{\mathrm{B}}=300 \mathrm{~K}, V_{\mathrm{A}}=\) \(V_{\mathrm{B}}\) \(=2.00 \mathrm{dm}^{3}\). Energy is supplied as heat to Section A and the piston moves to the right reversibly until the final volume of Section B is \(1.00 \mathrm{dm}^{3}\). Calculate (a) \(\Delta S_{\mathrm{A}}\) and \(\Delta S_{\mathrm{B}}\), (b) \(\Delta A_{\mathrm{A}}\) and \(\Delta \mathrm{A}_{\mathrm{B}}\), (c) \(\Delta G_{\mathrm{A}}\) and \(\Delta G_{\mathrm{B}}\), (d) AS of the total system and its surroundings. If numerical values cannot be obtained, indicate whether the values should be positive, negative, or zero or are indeterminate from the information given. (Assume \(C_{\mathrm{v}, \mathrm{m}}=20 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\).)

Given that \(s_{m}^{\circ}=29.79 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) for bismuth at \(100 \mathrm{~K}\) and the following tabulated heat capacities data (D.G. Archer, \(J\). Chem. Eng. Data 40 , 1015 (1995)), compute the standard molar entropy of bismuth at \(200 \mathrm{~K}\). $$ \begin{array}{lccccccl} \text { TIK } & 100 & 120 & 140 & 150 & 160 & 180 & 200 \\ C_{\mathrm{Am}} /\left(\mathrm{K} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right) & 23.00 & 23.74 & 24.25 & 24.44 & 24.61 & 24.89 & 25.11 \end{array} $$ Compare the value to the value that would be obtained by taking the heat capacity to be constant at \(24.44 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) over this range.

Show that, for a perfect gas, \((\partial U / \partial S)_{v}=T\) and \((\partial U / \partial V)_{\mathrm{S}}=-p\).

The heat capacity of chloroform (trichloromethane, \(\mathrm{CHCl}_{3}\) ) in the range \(240 \mathrm{~K}\) to \(330 \mathrm{~K}\) is given by \(\mathrm{C}_{\mathrm{p}, \mathrm{m}} /\left(\mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right)=91.47+7.5 \times 10^{-2}(T / \mathrm{K})\). In a particular experiment, \(1.00 \mathrm{~mol} \mathrm{CHCl}_{3}\) is heated from \(273 \mathrm{~K}\) to \(300 \mathrm{~K}\). Calculate the change in molar entropy of the sample.

The adiabatic compressibility, \(\kappa_{S}\), is defined like \(\kappa_{T}\) (eqn 2.44) but at constant entropy. Show that for a perfect gas \(p \gamma \kappa_{\mathrm{S}}=1\) (where \(\gamma\) is the ratio of heat capacities).

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