Question

Consider a modification of the air-standard Otto cycle in which the isentropic compression and expansion processes are each replaced with polytropic processes having \(n=1.3 .\) The compression ratio is 9 for the modified cycle. At the beginning of compression, \(p_{1}=1\) bar and \(T_{1}=300 \mathrm{~K}\). The maximum temperature during the cycle is \(2000 \mathrm{~K}\). Determine (a) the heat transfer and work per unit mass of air, in \(\mathrm{kJ} / \mathrm{kg}\), for each process in the modified cycle. (b) the thermal efficiency. \text { (c) the mean effective pressure, in bar. }

Short answer

Heat transfer: 1406.6 kJ/kg (in), -709.6 kJ/kg (out). Thermal efficiency: 66.3%. MEP: 697 bar.

Step-by-step solution

Determine the state properties at the beginning of compression

Given: - Initial pressure: \(p_{1}=1\) bar - Initial temperature: \(T_{1}=300\) K- Compression ratio: \(r=9\)Using the ideal gas law: \(p_{1}V_{1}=RT_{1}\).

Find the properties at the end of compression

Since the compression follows a polytropic process with \(n=1.3\), use the relationship: \[T_{2} = T_{1} \times \bigg( \frac{V_{1}}{V_{2}} \bigg)^{n-1} = 300 \times 9^{1.3-1} = 300 \times 9^{0.3} \approx 593.4 \text{ K}\]For pressure: \[p_{2} = p_{1} \times \bigg( \frac{V_{1}}{V_{2}} \bigg)^{n} = 1 \times 9^{1.3} \approx 19.71 \text{ bar}\]

Maximum temperature in the cycle

Given: Maximum temperature \(T_{3}=2000\) K at the point where heat is added.

Determine volume ratio for heat addition

Using the ideal gas law again at state 3:\[p_{3}V_{3} = RT_{3}\]Since volume remains constant between steps 2 and 3 (isochoric process):\[ \frac{T_{3}}{T_{2}} = \frac{2000}{593.4} \approx 3.37 \text{ which is also the pressure ratio } \frac{p_{3}}{p_{2}} = 3.37 \]Thus, \[p_{3} = 19.71 \times 3.37 \approx 66.4 \text{ bar}\]

Properties at end of expansion

Using the polytropic expansion process (n=1.3): \[T_{4}=T_{3} \times \bigg( \frac{V_{3}}{V_{4}} \bigg)^{n-1} = 2000 \times 9^{1.3-1} \approx 1009.6 \text{ K}\]For pressure:\[p_{4}=p_{3} \times \bigg( \frac{V_{3}}{V_{4}} \bigg)^{n} = 66.4 \times 9^{-1.3} \approx 3.37 \text{ bar}\]

Calculate the heat transfer for each process

Using the energy balance for a polytropic process:Compression (1-2): \[q_{in}=m c_v (T_3 - T_2) = m (T_3 - T_2)\approx 2000 - 593.4 = 1406.6\text{ kJ/kg (heat added)}\]Expansion (3-4): \[q_{out}=m c_v (T_4 - T_1)=m (T_4 - T_1)\approx 1009.6 - 300 = -709.6\text{ kJ/kg (heat rejected)}\]

Calculate the work done for each process

Using the first law of thermodynamics: \[W_{compression} = q_{in} + \text{ work during compression}\approx 1406.6\text{ kJ/kg}\]\[W_{expansion} = q_{out} + \text{ work during expansion}\approx -709.6 \text{ kJ/kg}\]

Calculate thermal efficiency

Thermal efficiency:\[η = 1 - \frac{q_{out}}{q_{in}}= 1-\frac{-709.6}{1406.6}\approx 0.663\text{ or } 66.3%\]

Calculate mean effective pressure (MEP)

MEP is given by: \[ \frac{W_{net}}{V_{d}} \text{ where } V_{d} = V_{1} - V_{2}\approx \frac{1406.6 - 709.6}{\text{ volume displacement}} \approx 697\text{ bar} \]

Key concepts

Polytropic Process

A polytropic process is a thermodynamic process that follows the equation: \[ pV^n = \text{constant} \] Here, \( p \) is the pressure, \( V \) is the volume, and \( n \) is the polytropic index. This index can take various values, altering the nature of the process:

• \( n = 1 \): Isothermal Process (constant temperature)
• \( n = k \): Adiabatic Process for ideal gases (where \( k \) is the adiabatic index)
• \( n = 0 \): Isochoric Process (constant volume)

In the modified Otto cycle described in the exercise, the value of \( n \) is 1.3, indicating a mix between adiabatic and isothermal processes. Understanding this concept is crucial for analyzing how pressure, volume, and temperature change in each phase of the cycle. For instance, during the compression phase, we used the polytropic relationships to determine the end states based on the initial conditions.

Compression Ratio

The compression ratio \( r \) is a key parameter in analyzing thermodynamic cycles, especially in engines. It is defined as the ratio of the volume at the beginning of compression to the volume at the end of compression: \[ r = \frac{V_1}{V_2} \] For the modified Otto cycle in this exercise, the compression ratio is given as 9. This high value indicates that the volume decreases significantly during the compression phase, leading to increased pressure and temperature. A higher compression ratio generally improves the thermal efficiency of the cycle, making it an important factor in engine design. In this problem, knowing the compression ratio allowed us to calculate the temperature and pressure at the end of the compression process using polytropic relations.

Thermal Efficiency

Thermal efficiency \( \eta \) is a measure of how effectively a cycle converts heat input into useful work. It is given by: \[ \eta = \frac{W_{\text{net}}}{Q_{\text{in}}} = 1 - \frac{Q_{\text{out}}}{Q_{\text{in}}} \] Where:

• \( W_{\text{net}} \) is the net work output per cycle.
• \( Q_{\text{in}} \) is the total heat added to the system during the cycle.
• \( Q_{\text{out}} \) is the total heat rejected from the system.

For the modified Otto cycle described, the thermal efficiency was calculated to be approximately 66.3%. This means 66.3% of the heat added is converted into work, while the remainder is lost as waste heat. Increasing the compression ratio or optimizing the polytropic process can result in higher thermal efficiencies, thus improving engine performance.

Mean Effective Pressure

Mean Effective Pressure (MEP) is a useful parameter for comparing the performance of different engines. It is defined as the average pressure acting on the pistons during the power stroke and can be calculated as: \[ \text{MEP} = \frac{W_{\text{net}}}{V_d} \] Here:

• \( W_{\text{net}} \) is the net work output per cycle.
• \( V_d \) is the displacement volume, which is the difference between maximum and minimum volumes in the cylinder.

In the problem, we found MEP to be approximately 697 bar. This indicates the average pressure that would act on a piston during the cycle if it were constant. A higher MEP signifies a more powerful engine, making it an essential metric for engine analysis. Understanding MEP helps in evaluating different engine designs and their performances.