Problem 21 An air-standard Diesel cycle has... [FREE SOLUTION]

Chapter 9: Problem 21

An air-standard Diesel cycle has a compression ratio of 16 and a cutoff ratio of 2 . At the beginning of compression, \(p_{1}=100 \mathrm{kPa}, V_{1}=0.01 \mathrm{~m}^{3}\), and \(T_{1}=300 \mathrm{~K}\). Calculate (a) the heat added, in \(\mathrm{kJ}\). (b) the maximum temperature in the cycle, in \(\mathrm{K}\). (c) the thermal efficiency. (d) the mean effective pressure, in \(\mathrm{kPa}\) (e) To investigate the effects of varying compression ratio, plot each of the quantities calculated in parts (a) through (d) for compression ratios ranging from 5 to 18 and for cutoff ratios of \(1.5,2\), and \(2.5\).

Short Answer

Expert verified

The heat added is approximately 7.6 kJ, the maximum temperature is 1855.04 K, the thermal efficiency is 46.9%, and the mean effective pressure is approximately 374.87 kPa.

Step by step solution

- Identify Known Values and Key Parameters

The given values are: Initial pressure: \(p_{1}=100 \,\mathrm{kPa}\), Initial volume: \(V_{1}=0.01 \,\mathrm{m}^{3}\), Initial temperature: \(T_{1}=300 \,\mathrm{K}\), Compression ratio: \(r=16\), Cutoff ratio: \(r_c=2\).

- Calculate Volume at the End of Compression

The volume at the end of compression, \(V_2\), can be found using the compression ratio: \[ V_2 = \frac{V_1}{r} \] \[ V_2 = \frac{0.01\,\mathrm{m}^{3}}{16} = 0.000625\,\mathrm{m}^{3} \]

- Calculate Temperature and Pressure at the End of Compression

Using the isentropic relation and the ideal gas law: \[ T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma-1} \] Assume air-standard values: \(\gamma = 1.4\), \[ T_2 = 300 \times 16^{0.4} \approx 927.52\,\mathrm{K} \] \[ p_2 = p_1 \left(\frac{V_1}{V_2}\right)^\gamma \] \[ p_2 = 100 \times 16^{1.4} \approx 3799.5\, \mathrm{kPa} \]

- Calculate Volume after Heat Addition

The volume after heat addition, \(V_3\), is found using the cutoff ratio: \[ V_3 = r_c \times V_2 \] \[ V_3 = 2 \times 0.000625 = 0.00125\, \mathrm{m}^{3} \]

- Calculate Temperature and Pressure after Heat Addition

Using the ideal gas law, compute temperature \(T_3\) and pressure \(p_3\): \[ T_3 = T_2 \left(\frac{V_3}{V_2}\right) \] \[ T_3 = 927.52 \times 2 = 1855.04\, \mathrm{K} \] The pressure after heat addition can be found using: \[ p_3 = p_2 \times \left(\frac{T_3}{T_2}\right) \] \[ p_3 = 3799.5 \times \left(\frac{1855.04}{927.52}\right) \approx 7599\,\mathrm{kPa} \]

- Calculate Heat Added

Using the properties of air and specific heats: Specific heat at constant volume, \(c_v\): \[ Q_{added} = m c_v (T_3 - T_2) \] Find mass, \(m\), using the ideal gas law at \(T_1\), \(p_1\), and \(V_1\): \[ m = \frac{p_1 V_1}{R T_1} \] Where \(R\) is the specific gas constant for air \(R = 0.287\,\mathrm{kJ/kg}\, \mathrm{K}\). \[ m = \frac{100 \times 0.01}{0.287 \times 300} \approx 0.0116\, \mathrm{kg} \] \[ Q_{added} = 0.0116 \times 0.718 \times (1855.04 - 927.52) \approx 7.6\, \mathrm{kJ} \]

- Calculate Maximum Temperature in the Cycle

\[ T_{max} = T_3 = 1855.04\, \mathrm{K} \]

- Calculate Thermal Efficiency

Thermal efficiency for a Diesel cycle: \[ \eta = 1 - \frac{1}{r^{\gamma-1}} \times \left( \frac{r_c^{\gamma} - 1}{\gamma (r_c-1)} \right) \] Substitute the values: \[ \eta = 1 - \frac{1}{16^{0.4-1}} \times \left( \frac{2^{1.4} - 1}{1.4(2-1)} \right) \approx 0.469 \] \[ \eta = 46.9\% \]

- Calculate Mean Effective Pressure

The mean effective pressure \(MEP\) can be expressed as: \[ MEP = \frac{W_{cycle}}{V_1 - V_2} \] The work done per cycle, \(W_{cycle}\), can be approximated by: \[ W_{cycle} = Q_{added} \times \eta \] \[ W_{cycle} = 7.6 \, \mathrm{kJ} \times 0.469 \approx 3.56 \, \mathrm{kJ} \] \[ MEP = \frac{3.56 \, \mathrm{kJ}}{0.01 - 0.000625} \approx 374.87 \, \mathrm{kPa} \]

- Plot Variations of Quantities for Different Compression Ratios

To analyze the effects of varying compression ratio from 5 to 18 for cutoff ratios of 1.5, 2, and 2.5, repeat the calculations in steps 2-9 for each combination. A graphing tool can then plot the results for comparison.

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Key Concepts

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

compression ratio

The compression ratio is a crucial concept in Diesel cycle analysis. It is defined as the ratio of the volume of the combustion chamber when the piston is at the bottom of its stroke (volume at the beginning of compression, usually denoted as \( V_1 \)) to the volume when the piston is at the top of the stroke (volume at the end of compression, usually denoted as \( V_2 \)). The formula for compression ratio \( r \) is given by:
\[ r = \frac{V_1}{V_2} \]
A higher compression ratio results in higher thermal efficiency but also requires stronger engine components to withstand the increased pressure.

thermal efficiency

Thermal efficiency is a measure of how well an engine converts the heat from fuel into work. For a Diesel cycle, the thermal efficiency \( \eta \) can be expressed using the compression ratio \( r \), the specific heat ratio \( \gamma \), and the cutoff ratio \( r_c \). The formula is:
\[ \eta = 1 - \frac{1}{r^{\gamma - 1}} \times \left( \frac{r_c^{\gamma} - 1}{\gamma (r_c - 1)} \right) \]
It demonstrates that increasing the compression ratio and decreasing the cutoff ratio boost the engine's efficiency. However, there is a trade-off in terms of higher thermal stresses on the engine.

mean effective pressure

Mean effective pressure (MEP) is an important parameter that provides a measure of the engine's capability to do work. It is defined as the average pressure in the cylinders of the engine if it were to produce the same net work per cycle. The MEP can be found using the work done \( W_{cycle} \) and the displacement volume, given by:
\[ MEP = \frac{W_{cycle}}{V_1 - V_2} \]
Here, the displacement volume is the difference in volume before and after compression. The MEP helps in comparing the performance of different engines regardless of their size.

isentropic process

An isentropic process is a thermodynamic process that is both adiabatic and reversible, meaning there is no heat transfer and entropy remains constant. In the Diesel cycle, both the compression and expansion processes are ideally considered as isentropic processes. This means that for these processes:
\[ T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma - 1} \]
and
\[ p_2 = p_1 \left(\frac{V_1}{V_2}\right)^{\gamma} \]
These relations are used to calculate the pressure and temperature at the end of the isentropic compression and expansion stages in the Diesel cycle.

cutoff ratio

The cutoff ratio \( r_c \) in a Diesel cycle is the ratio of the cylinder volume after combustion to the cylinder volume at the start of combustion. It is a critical parameter as it affects the heat addition process. The cutoff ratio is given by:
\[ r_c = \frac{V_3}{V_2} \]
Here, \( V_3 \) is the volume after heat addition, and \( V_2 \) is the volume at the end of compression. A higher cutoff ratio means that more fuel is added for a longer portion of the piston movement, increasing the power output but also leading to lower thermal efficiency.

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