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Magazine Chapter 9: Problem 3
At the beginning of the compression process of an airstandard Otto cycle, \(p_{1}=1\) bar, \(T_{1}=290 \mathrm{~K}, V_{1}=400 \mathrm{~cm}^{3}\). The maximum temperature in the cycle is \(2200 \mathrm{~K}\) and the compression ratio is 8. Determine (a) the heat addition, in \(\mathrm{kJ}\). (b) the net work, in \(\mathrm{kJ}\). (c) the thermal efficiency. (d) the mean effective pressure, in bar. (e) Develop a full accounting of the exergy transferred to the air during the heat addition, in \(\mathrm{kJ}\). (f) Devise and evaluate an exergetic efficiency for the cycle. Let \(T_{0}=290 \mathrm{~K}, p_{0}=1\) bar.
Short Answer
Expert verified
q_in = 1120.25 kJ, W_net = 618.10 kJ, 畏_th = 55.2%, MEP 鈮� 17.72 bar, e_q = 973.13 kJ, 畏_exergetic 鈮� 63.54%
Step by step solution
01
Define given parameters
Given values are: \[ p_{1}=1 \text{ bar}, T_{1}=290 \text{ K}, V_{1}=400 \text{ cm}^3 \text{ or } 0.0004 \text{ m}^3 \] \[ T_{max}=2200 \text{ K}, r=8 \] The environmental conditions are \[ T_{0} = 290 \text{ K} \text{ and } p_{0} = 1 \text{ bar} \text{ or } 100 \text{ kPa} \]
02
Calculate intermediate state parameters
First, determine state parameters at the end of compression (state 2): Use the relation \[ V_{2} = \frac{V_{1}}{r} = \frac{0.0004}{8} = 0.00005 \text{ m}^3 \] For an air-standard Otto cycle, we also use the isentropic relations for temperature and pressure during the compression process from state 1 to state 2. For an ideal gas with specific heat ratio \( \gamma = 1.4 \): \[ T_{2}=T_{1}\left(\frac{V_{1}}{V_{2}}\right)^{\gamma-1}=290\left(\frac{0.0004}{0.00005}\right)^{0.4}=645.84 \text{ K} \] \[ \frac{p_{2}}{p_{1}}=\left(\frac{T_{2}}{T_{1}}\right)^{\frac{\gamma}{\gamma-1}}=\left(\frac{645.84}{290}\right)^{\frac{1.4}{0.4}}=\frac{p_{2}}{1}\] Therefore, \[ p_{2}\approx 15.69 \text{ bar} \]
03
Calculate parameters at maximum temperature
Using the relation for the heat addition process (isochoric process from state 2 to 3): \[ V_{3} = V_{2}=V_{1}/r \] \[ T_{3}=2200 \text{ K} \text{ (given)} \text{ and thus from the ideal gas law: } \frac{p_{3}}{p_{2}}= \frac{T_{3}}{T_{2}} \] \[ p_{3}=p_{2}\left(\frac{T_{3}}{T_{2}}\right)=15.69\left( \frac{2200}{645.84} \right) \text { bar } \approx 53.42 \text{ bar} \]
04
Calculate parameters at state 4
At state 4, we assume the isentropic expansion process: \( T_{4} \) and \( p_{4} \) can be found using similar relations as state 2: \[ T_{4}=T_{3}\left(\frac{V_{3}}{V_{4}}\right)^{\gamma-1} = T_{3}\left(\frac{V_{2}}{V_{1}}\right)^{\gamma-1}\] Substituting values: \[ T_{4}=2200 \left( \frac{1}{8} \right)^{0.4} \approx 989.37 \text{K} \] \[ \frac{p_{4}}{p_{3}}=\left(\frac{T_{4}}{T_{3}}\right)^{\frac{\gamma}{\gamma-1}} \] \[ p_{4} = p_{3}\left(\frac{T_{4}}{T_{3}}\right)^{\frac{\gamma}{\gamma-1}} = 53.42\left( \frac{989.37}{2200} \right)^{3.5} \approx 3.41 \text{ bar} \]
05
Compute the heat addition
Using specific heat values for air (constant values): \( c_{v} = 0.718 \text{ kJ/kg.K} \) and noting that \[ q_{in} = c_{v}(T_{3}-T_{2}) \] \[ q_{in}=(0.718)(2200-645.84) \text{ kJ/kg} \text{ } = 1120.25 \text{ kJ} \]
06
Compute the net work
The net work output \( W_{net} \) is the difference between heat added and rejected. Heat rejected \( q_{out} = c_{v}(T_{4}-T_{1}) \): \[ q_{out} = c_{v}(989.37-290) = 0.718 \times 699.37 = 502.15 \text{ kJ} \] Therefore, \[ W_{net}= q_{in} - q_{out} = 1120.25 - 502.15 = 618.10 \text{ kJ} \]
07
Calculate thermal efficiency
The thermal efficiency \( \eta_{th} \) for the Otto cycle is given by: \[ \eta_{th} = 1- \frac{q_{out}}{q_{in}} = 1 - \frac{502.15}{1120.25} = 0.552 \text{ or } 55.2% \]
08
Mean effective pressure
Mean effective pressure (MEP) is given by: \[ MEP = \frac{W_{net}}{V_{1}-V_{2}} = \frac{618.10}{0.0004-0.00005}\, \text{ Pa}\approx 1,771,714 \text{ Pa} \text{ or } 17.72 \text{ bar} \]
09
Exergy transferred during heat addition
Exergy transferred: \[ e_{q} = q_{in} \left(1 - \frac{T_{0}}{T_{3}} \right) = 1120.25 \left(1-\frac{290}{2200} \right) = 973.13 \text{ kJ} \]
10
Exergetic efficiency
Exergy transferred to work output: \[ \eta_{exergetic} = \frac{W_{net}}{e_{q}} = \frac{618.10}{973.13} \approx 63.54% \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Heat Addition in Otto Cycle
In the Otto cycle, heat addition occurs during the combustion process. This takes place at constant volume. Here, the air-fuel mixture is ignited, raising the temperature within the cylinder. The amount of heat added, denoted as \( q_{in} \), can be calculated using the specific heat capacity at constant volume \( c_v \), and the temperature change during the process. In our example: \[ q_{in} = c_v (T_3 - T_2) \]. This ensures we have all required parameters to determine the precise heat addition in kilojoules (kJ). \[ q_{in} = 1120.25 \text{ kJ} \].
Net Work in Thermodynamic Cycles
Net work in thermodynamic cycles is crucial for understanding the efficiency of an engine. The net work output \( W_{net} \) is the difference between the work done during heat addition and the work lost during heat rejection. This is represented formally as: \[ W_{net} = q_{in} - q_{out} \]. In our example, the heat rejected \( q_{out} \) is calculated using the specific heat capacity and temperature difference during the heat rejection phase. Subsequently, the net work can be computed: \[ W_{net} = 618.10 \text{kJ} \].
Thermal Efficiency Calculation
Thermal efficiency measures how effectively an engine converts heat into work. For an Otto cycle, the thermal efficiency \( \eta_{th} \) can be determined by the formula: \[ \eta_{th} = 1 - \frac{q_{out}}{q_{in}} \]. The higher the thermal efficiency, the more efficient the engine is. In our case, substituting the values of heat added and rejected yields: \[ \eta_{th} = 55.2% \].
Mean Effective Pressure
Mean effective pressure (MEP) represents the average pressure in the cylinder over one cycle, and it's an important indicator of an engine's performance. Calculated as: \[ \text{MEP} = \frac{W_{net}}{V_{1} - V_{2}} \], where \( W_{net} \) is the net work, and \( V_{1} - V_{2} \) is the change in volume. Here, the MEP is found to be: \[ 17.72 \text{ bar} \]. This helps engineer gauge how effectively the engine can perform useful work.
Exergy in Thermodynamic Cycles
Exergy quantifies the maximum useful work possible during a process that brings the system into equilibrium with its environment. During the heat addition process in the Otto cycle, exergy transferred, denoted as \( e_q \) is calculated by: \[ e_q = q_{in} \(1 - \frac{T_0}{T_3} \) \]. Using given and determined values, the exergy transferred during heat addition is: \[ 973.13 \text{ kJ} \]. Understanding exergy helps optimize the thermodynamic process by evaluating losses and inefficiencies.
Exergetic Efficiency
Exergetic efficiency assesses how efficiently a system converts available energy (exergy) into useful work. This efficiency is crucial for definitive performance improvement and environmental considerations. It is determined by: \[ \eta_{exergetic} = \frac{W_{net}}{e_q} \]. For our specific Otto cycle, the exergetic efficiency is: \[ \eta_{exergetic} = 63.54% \]. This indicates how efficiently the system transforms exergy into work. Enhancements to this metric can lead to significant improvements in overall engine performance and sustainability.
