Question

An ideal diesel engine has a compression ratio of 20 and uses air as the working fluid. The state of air at the beginning of the compression process is \(95 \mathrm{kPa}\) and \(20^{\circ} \mathrm{C}\). If the maximum temperature in the cycle is not to exceed \(2200 \mathrm{K}\) determine \((a)\) the thermal efficiency and \((b)\) the mean effective pressure. Assume constant specific heats for air at room temperature.

Short answer

(a) The thermal efficiency of the diesel engine is 39.8%. (b) The mean effective pressure is \(MEP = \frac{536.9}{19 V_{2}} \mathrm{kPa}\).

Step-by-step solution

Determine the initial state of the air

We are given that the air is at \(95 \mathrm{kPa}\) and \(20^{\circ} \mathrm{C}\) initially. So, the initial pressure \(P_1 = 95 \mathrm{kPa}\) and initial temperature \(T_1 = 20 + 273.15 = 293.15 \mathrm{K}\).

Calculate the state at the end of the compression process

We are given the compression ratio \(r = 20\). Using the isentropic relations, we can find the pressure and temperature at the end of the compression process (state 2). $$ P_2 = P_1 \times r^{\gamma} $$ $$ T_2 = T_1 \times r^{\gamma - 1} $$ Here, \(\gamma\) is the specific heat ratio for air which is approximately 1.4. We can now calculate \(P_2\) and \(T_2\): $$ P_2 = 95 \times 20^{1.4} = 6124.87 \mathrm{kPa} $$ $$ T_2 = 293.15 \times 20^{0.4} = 861.77 \mathrm{K} $$

Determine the heat addition process

We are given the maximum temperature in the cycle is \(2200 \mathrm{K}\). So, \(T_3 = 2200 \mathrm{K}\). To find the heat addition, we can use the constant pressure process relationship: $$ q_{in} = C_p \times (T_3 - T_2) $$ Here, \(C_p\) is the specific heat at constant pressure for air, which is approximately \(1.005 \mathrm{kJ/(kg\cdot K)}\). Now, we calculate \(q_{in}\): $$ q_{in} = 1.005 \times (2200 - 861.77) = 1346.61 \mathrm{kJ/kg} $$

Determine the heat rejection process

To find the heat rejection, we need to calculate the temperature at the end of the heat rejection process (state 4). Since it's an isentropic expansion, we can use the following equation: $$ \frac{T_4}{T_3} = \left(\frac{P_4}{P_3}\right)^{\gamma-1} $$ We know that \(P_4=P_1\) and \(P_3=P_2\). Thus, we have: $$ \frac{T_4}{2200} = \left(\frac{95}{6124.87}\right)^{0.4} $$ Now, we can calculate \(T_4\): $$ T_4 = 2200 \times \frac{95}{6124.87}^{0.4} = 1421.46 \mathrm{K} $$ Next, we can find the heat rejection using the constant volume process relationship: $$ q_{out} = C_v \times (T_4 - T_1) $$ Here, \(C_v\) is the specific heat at constant volume for air, which is approximately \(0.7165 \mathrm{kJ/(kg\cdot K)}\). Now, we calculate \(q_{out}\): $$ q_{out} = 0.7165 \times (1421.46 - 293.15) = 809.71 \mathrm{kJ/kg} $$

Calculate the thermal efficiency

Now that we have the heat added (\(q_{in}\)) and heat rejected (\(q_{out}\)) in the cycle, we can calculate the thermal efficiency using the following formula: $$ \eta_{th} = 1 - \frac{q_{out}}{q_{in}} $$ We plug in the values to find the thermal efficiency: $$ \eta_{th} = 1 - \frac{809.71}{1346.61} = 0.398 = 39.8\% $$ (a) The thermal efficiency of the diesel engine is \(39.8\%\).

Calculate the mean effective pressure

To find the mean effective pressure (MEP), we first need to calculate the work done per cycle. Since it's an ideal cycle, the difference between the heat added and heat rejected is equal to the work done: $$ W_{cycle} = q_{in} - q_{out} $$ We can calculate the work done: $$ W_{cycle} = 1346.61 - 809.71 = 536.9 \mathrm{kJ/kg} $$ Now, we can use the following formula to find the mean effective pressure: $$ MEP = \frac{W_{cycle}}{V_{1} - V_{2}} $$ However, we know the compression ratio \(r = 20\), so \(V_1 = 20 V_2\). This leads to: $$ MEP = \frac{536.9}{20 V_{2} - V_{2}} = \frac{536.9}{19 V_{2}} $$ Since we only need to find the MEP value, we can keep it in terms of volume \(V_2\). (b) The mean effective pressure is \(MEP = \frac{536.9}{19 V_{2}} \mathrm{kPa}\).

Key concepts

Compression Ratio

The compression ratio is a key concept in the thermodynamics of diesel engines. It is defined as the ratio of the volume of the cylinder and combustion chamber when the piston is at the bottom of its stroke (maximum volume), to the volume when the piston is at the top of its stroke (minimum volume). In the given exercise, the compression ratio is 20, meaning the volume is reduced by a factor of 20 during the compression stroke.

This ratio is crucial because it influences the engine's efficiency and power. A higher compression ratio generally leads to better efficiency because it allows the engine to extract more mechanical energy from a given mass of air-fuel mixture. However, there are practical limits to increasing the compression ratio, such as engine knock and the structural limits of engine materials.

Thermal Efficiency

Thermal efficiency is a measurement of how well an engine converts heat from fuel into work. It is defined as the ratio of work output to heat input. For diesel engines, thermal efficiency is particularly important because these engines rely on compression ignition for efficiency gains.

The thermal efficiency is influenced by several factors, including the compression ratio, properties of the fuel, and the engine's design. In our exercise, the thermal efficiency is calculated by comparing the heat added to the engine (during fuel combustion) and the heat rejected (during the exhaust stroke). The result, given as a percentage, tells us what portion of the fuel's heat energy is converted into useful work—the higher the percentage, the more efficient the engine.

Mean Effective Pressure

Mean effective pressure (MEP) is another important term in engine thermodynamics. It represents the average pressure exerted on the pistons of an engine that produces the same amount of work as the actual cycle of pressure variation. It is a useful measure for comparing the performance of engines independently of their size.

To calculate mean effective pressure, one must find the work done per cycle and divide it by the displacement volume. In simpler terms, it's the average pressure that would produce the same power if the pressure remained constant throughout the expansion stroke. The mean effective pressure is an indicator of an engine’s capacity to do work; a higher MEP typically signifies a more powerful engine for its size.

Isentropic Process

An isentropic process is an idealization used in thermodynamics where the process is both adiabatic and reversible. In such a process, entropy remains constant. This concept is particularly relevant when analyzing the compression and expansion strokes of piston engines, where an isentropic process can serve as a model for efficiency.

In diesel engines, the compression stage is often approximated as isentropic. The isentropic relations for pressure and temperature, presented in the exercise, allow engineers to predict the state of the gas following compression or expansion, assuming no heat is lost to the surroundings. While real-world processes aren't perfectly isentropic due to factors like friction and heat transfer, the isentropic model is a valuable tool for understanding and optimizing engine cycles.