Question

The displacement volume of an internal combustion engine is \(5.6\) liters. The processes within each cylinder of the engine are modeled as an air-standard Diesel cycle with a cutoff ratio of \(2.4 .\) The state of the air at the beginning of compression is fixed by \(p_{1}=95 \mathrm{kPa}, T_{1}=27^{\circ} \mathrm{C}\), and \(V_{1}=6.0\) liters. Determine the net work per cycle, in \(\mathrm{kJ}\), the power developed by the engine, in \(\mathrm{kW}\), and the thermal efficiency, if the cycle is executed 1500 times per min.

Short answer

Net work is calculated considering state changes. Power is derived from net work and cycles per minute. Efficiency uses Diesel-cycle specific formula.

Step-by-step solution

Convert Initial Conditions to SI Units

Convert the given initial conditions to SI units. Temperature: \( T_1 = 27^{\text{°C}} = 27 + 273.15 = 300.15 \text{ K} \), Volume: \( V_1 = 6.0 \text{ liters} = 6.0 \times 10^{-3} \text{ m}^3 \) and Displacement Volume: \( 5.6 \text{ liters} = 5.6 \times 10^{-3} \text{ m}^3 \).

Define Diesel Cycle Relations

Use the Diesel cycle relations where the compression ratio (\(r\)) is given by \( r = \frac{V_1}{V_2} \) and the cutoff ratio (\( r_c \)) is \( \frac{V_3}{V_2} = 2.4 \). So, \( V_3 = 2.4V_2 \)

Calculate Compression Ratio

Use displacement volume to find volumes at different states. The volume at the beginning of compression, \( V_1 = V_3 \). Assuming the displacement volume occurs during compression, \(V_1 - V_2 = \text{ Displacement Volume} \implies V_2 = V_1 - 5.6 \text{ liters} \)

Calculate Temperatures at Other States

Use the ideal gas law to find temperatures and pressures at different states. Use the relation \( \frac{T_2}{T_1} = (r)^{\frac{\text{n-1}}{\text{n}}} \), and calculate accordingly. Here, for diesel cycle, \( \text{n} = 1.4 \)

Determine Net Work Per Cycle

Calculate the net work per cycle using \( W_{\text{net}} = mR(T_3 - T_2 - (T_4-T_1)) \)

Calculate Power Output

The power developed is given by \( P = \frac{W_{\text{net}} \times \text{cycles per minute}}{60 \text{ seconds}} \)

Find Thermal Efficiency

The thermal efficiency for the Diesel cycle is given by \( \text{Efficiency} = 1 - \frac{1}{r^{\text{k-1}}} \times \frac{\text{k}(r_c-1)}{\text{n}(r_c-\frac{1}{r})} \)

Conclusion

Use the formula output values to calculate the final net work, power, and efficiency.

Key concepts

internal combustion engine

An internal combustion engine is a type of heat engine where the combustion of fuel occurs inside the engine, more specifically in the combustion chamber. These engines operate based on various thermodynamic cycles, and in this exercise, we focus on the Diesel cycle. The main components of an internal combustion engine include cylinders, pistons, and a crankshaft. Understanding these components and how they work together is crucial for comprehending the complete processes within the engine. The transformation of chemical energy (fuel) into mechanical energy (work) is achieved through the cyclic processes of intake, compression, power, and exhaust strokes.

thermal efficiency calculation

The thermal efficiency of an internal combustion engine indicates how well the engine converts the heat from fuel into useful work. For a Diesel cycle, the thermal efficiency can be calculated using a specific formula. This formula considers the compression ratio (\r\frac{V_1}{V_2}\r), the cutoff ratio (\r\frac{V_3}{V_2}\r), and the specific heat ratio (\r\frac{c_p}{c_v}\r). The equation for thermal efficiency (\r\text{Efficiency}\r) in a Diesel cycle is given by: \[ \text{Efficiency} = 1 - \frac{1}{r^{\text{k-1}}} \times \frac{k(r_c-1)}{n(r_c-\frac{1}{r})}. \] This formula captures the importance of compression and cutoff ratios in determining the efficiency of the cycle, showing how changes in these ratios affect the overall performance of the engine.

compression ratio

The compression ratio (\r r \r) of an internal combustion engine is a measure of how much the air-fuel mixture is compressed before ignition. It is defined as the ratio of the volume of the combustion chamber from its largest capacity to its smallest. For example, in this exercise, the compression ratio (\r r \r) is calculated by: \[ r = \frac{V_1}{V_2}. \] Higher compression ratios typically lead to more efficient engines because they allow for better utilization of the fuel's energy. However, they must be managed carefully to avoid engine knocking and other issues.

ideal gas law

The ideal gas law, represented as \[ PV = nRT, \] is a fundamental equation in thermodynamics that relates the pressure (\r P \r), volume (\r V \r), and temperature (\r T \r) of an ideal gas. In the context of the Diesel cycle, it's crucial for calculating the state variables at different points in the cycle. For example, to find the temperature after compression or expansion, the ideal gas law is used in combination with specific cycle properties. This relationship enables us to determine how the gas behaves under different conditions, which is essential for calculating the engine performance parameters such as net work and thermal efficiency.

net work per cycle

Net work per cycle (\r W_{\text{net}} \r) refers to the total work produced by the engine during one complete cycle of operation minus the work consumed. For the Diesel cycle, it is calculated using the temperatures and specific gas constants through the formula: \[ W_{\text{net}} = mR(T_3 - T_2 - (T_4-T_1)). \] Here, \r m \r is the mass of the air, \r R \r is the gas constant, and \r T \r values are the temperatures at various state points. The goal is to find the net amount of energy converted into mechanical work per cycle, which is directly linked to the power output and efficiency of the engine.

power output calculation

Power output refers to the amount of work performed by the engine per unit of time. It can be determined from the net work per cycle and the number of cycles performed per minute. The formula used is: \[ P = \frac{W_{\text{net}} \times \text{cycles per minute}}{60}. \] By multiplying the net work per cycle by the number of cycles the engine completes every minute and then dividing by 60, we convert the work from per cycle to per second, leading to the power output in kilowatts (kW). This calculation is vital for understanding the engine's effectiveness in delivering usable energy.