Question

The pressure and temperature at the beginning of compression of an air- standard Diesel cycle are \(95 \mathrm{kPa}\) and \(300 \mathrm{~K}\), respectively. At the end of the heat addition, the pressure is \(7.2 \mathrm{MPa}\) and the temperature is \(2150 \mathrm{~K}\). Determine (a) the compression ratio. (b) the cutoff ratio. (c) the thermal efficiency of the cycle. (d) the mean effective pressure, in \(\mathrm{kPa}\).

Short answer

Compression ratio is found from isentropic relations; cutoff ratio by volume at end/start of combustion; thermal efficiency is determined using standard Diesel cycle efficiency formula, and mean effective pressure using net work and displaced volume.

Step-by-step solution

Identify Given Data

Record all the given values from the problem: Initial pressure: \(P_1 = 95 \text{kPa}\)Initial temperature: \(T_1 = 300 \text{K}\)Pressure after heat addition: \(P_3 = 7.2 \text{MPa} = 7200 \text{kPa}\)Temperature after heat addition: \(T_3 = 2150 \text{K}\)

Calculate Compression Ratio

The compression ratio \(r_c\) can be calculated using the relationship between pressure and temperature:\[P_2 = P_1 \left(\frac{V_1}{V_2}\right) = P_1 \cdot r_c^{\gamma}\]However, we use this to find the volume ratios in the cycles. First, calculate the intermediate pressures and temperatures at state point 2 using isentropic relations.Assume isentropic compression: \[\frac{T_2}{T_1} = r_c^{\gamma - 1}\]Rearrange to find \(r_c\):\[r_c = \left(\frac{T_2}{T_1}\right)^{\frac{1}{\gamma - 1}}\]Estimate \(\gamma = 1.4\) for air.

Calculate State Properties at Point 2

Use the isentropic relations again to get properties at state point 2:\[\frac{T_2}{T_1} = r_c^{\gamma - 1}\]Assuming ideal gas behavior:\[T_2 = T_1 \cdot r_c^{\gamma - 1}\]

Determine Cutoff Ratio

The cutoff ratio \(r_c\) is the ratio of the volumes at the end and start of combustion. It is calculated using temperatures and pressures:\[r_c = \frac{V_3}{V_2} = \frac{T_3/T_2 \cdot \left(P_2/P_1\right)}\]

Calculate Thermal Efficiency

The thermal efficiency \(\eta_{th}\) of a Diesel cycle is given by:\[\eta_{th} = 1 - \frac{1}{r_c^{\gamma - 1}} \cdot \left[\frac{r_c^{\gamma - 1} - 1}{r_v - 1}\right]\]Replace the values of \(r_c\) and \(\gamma = 1.4\) to get \(\eta_{th}\).

Calculate Mean Effective Pressure

The mean effective pressure \(MEP\) is calculated using:\[MEP = \frac{W_{net}}{V_{displaced}}\]Here, \(W_{net}\) is the net work done per cycle and \(V_{displaced}\) is the displaced volume, which can be calculated using the volume ratio.

Key concepts

compression ratio

The compression ratio in a Diesel cycle is a fundamental metric that tells us how much the air-fuel mixture is compressed during the compression stroke. It is denoted as \(r_c\) and is calculated as the ratio of the initial volume to the final volume in the compression phase.
To calculate it, we typically assume air behaves like an ideal gas and use isentropic relations. The formula used is:
\[ r_c = \left(\frac{T_2}{T_1}\right)^{\frac{1}{\gamma - 1}} \]
Here, \(T_1\) is the initial temperature, \(T_2\) is the temperature after compression, and \(\gamma\) (gamma) is the specific heat ratio. For air, we usually assume \(\gamma = 1.4\). This equation allows us to find the compression ratio when temperatures at the start and end of compression are known.
Higher compression ratios lead to better efficiency but also require stronger engine materials to handle the increased pressures.

cutoff ratio

The cutoff ratio in a Diesel cycle is denoted by \(r_v\). It is the ratio of the cylinder volume at the end of combustion to the cylinder volume at the start of combustion (not to be confused with compression volume).
This ratio determines how much fuel is injected and thus how long the combustion lasts. The formula used to calculate the cutoff ratio is:
\[ r_v = \frac{T_3}{T_2} \]
where \(T_3\) is the temperature at the end of heat addition and \(T_2\) is the temperature immediately before heat addition. Alternatively, it can be found using:
\[ r_v = \frac{V_3}{V_2} \]
Where \(V_3\) is the volume after combustion, and \(V_2\) is the volume before combustion.
Smaller cutoff ratios increase efficiency as they lead to higher thermal efficiency for the cycle. However, it also reduces the power output for each cycle.

thermal efficiency

Thermal efficiency \(\eta_{th}\) of a Diesel cycle quantifies how well the engine converts heat from fuel into useful work. It is given by the equation:
\[ \eta_{th} = 1 - \frac{1}{r_c^{\gamma - 1}} \cdot \left[\frac{r_v^{\gamma} - 1}{r_v - 1}\right] \]
In this formula, \(r_c\) is the compression ratio, \(\gamma\) is the specific heat ratio (1.4 for air), and \(r_v\) is the cutoff ratio.
This formula shows that higher compression ratios \(r_c\) and lower cutoff ratios \(r_v\) lead to higher thermal efficiencies. It means less fuel is wasted as heat and more is converted into mechanical work.
Therefore, enhancing the thermal efficiency involves optimizing the cycle parameters, but care must be taken to balance efficiency with practicality and engine durability.

mean effective pressure

Mean effective pressure (MEP) represents the average pressure in the cylinder during the power stroke and is a measure of the engine's ability to convert fuel energy into mechanical work. It is calculated using the formula:
\[ MEP = \frac{W_{net}}{V_{displaced}} \]
Here \(W_{net}\) is the net work done per cycle and \(V_{displaced}\) is the volume displaced by the piston. By dividing work done over a cycle by the displaced volume, we get a pressure measure that indicates efficiency and power output.
The MEP gives a simplified way to compare different engines' performances. Higher MEP values indicate more power output and higher efficiency, making it a key parameter in engine design and analysis.
Understanding this concept helps in assessing how well an engine performs relative to its size and fuel consumption, making it a critical metric for mechanical engineers and automotive designers.