(III) The operation of a diesel engine can be idealized by...

(III) The operation of a diesel engine can be idealized by the cycle shown in Fig. 18 . Air is drawn into the cylinder during the intake stroke (not part of the idealized cycle). The air is compressed adiabatically, path ab. At point b diesel fuel is injected into the cylinder which immediately burns since the temperature is very high. Combustion is slow, and during the first part of the power stroke, the gas expands at (nearly) constant pressure, path bc. After burning, the rest of the power stroke is adiabatic, path cd. Path da corresponds to the exhaust stroke. $(a)$ Show that, for a quasistatic reversible engine undergoing this cycle using an ideal gas, the ideal efficiency is $$e=1-\frac{\left(V_{\mathrm{a}} / V_{\mathrm{c}}\right)^{-\gamma}-\left(V_{\mathrm{a}} / V_{\mathrm{b}}\right)^{-\gamma}}{\gamma\left[\left(V_{\mathrm{a}} / V_{\mathrm{c}}\right)^{-1}-\left(V_{\mathrm{a}} / V_{\mathrm{b}}\right)^{-1}\right]}$$ where $V_{\mathrm{a}} / V_{\mathrm{b}}$ is the "compression ratio", $V_{\mathrm{a}} / V_{\mathrm{c}}$ is the "expansion ratio", and $\gamma$ is defined by $\gamma=\frac{C_{P}}{C_{V}}$ (b) If $V_{\mathrm{a}} / V_{\mathrm{b}}=16$ and $V_{\mathrm{a}} / V_{\mathrm{c}}=4.5$ calculate the efficiency assuming the gas is diatomic (like $\mathrm{N}_{2}$ and $\mathrm{O}_{2} )$ and ideal.

(III) The operation of a diesel engine can be idealized by the cycle shown in Fig. 18 . Air is drawn into the cylinder during the intake stroke (not part of the idealized cycle). The air is compressed adiabatically, path ab. At point b diesel fuel is injected into the cylinder which immediately burns since the temperature is very high. Combustion is slow, and during the first part of the power stroke, the gas expands at (nearly) constant pressure, path bc. After burning, the rest of the power stroke is adiabatic, path cd. Path da corresponds to the exhaust stroke. $(a)$ Show that, for a quasistatic reversible engine undergoing this cycle using an ideal gas, the ideal efficiency is
$$e=1-\frac{\left(V_{\mathrm{a}} / V_{\mathrm{c}}\right)^{-\gamma}-\left(V_{\mathrm{a}} / V_{\mathrm{b}}\right)^{-\gamma}}{\gamma\left[\left(V_{\mathrm{a}} / V_{\mathrm{c}}\right)^{-1}-\left(V_{\mathrm{a}} / V_{\mathrm{b}}\right)^{-1}\right]}$$ where $V_{\mathrm{a}} / V_{\mathrm{b}}$ is the "compression ratio", $V_{\mathrm{a}} / V_{\mathrm{c}}$ is the "expansion ratio", and $\gamma$ is defined by $\gamma=\frac{C_{P}}{C_{V}}$ (b) If $V_{\mathrm{a}} / V_{\mathrm{b}}=16$ and $V_{\mathrm{a}} / V_{\mathrm{c}}=4.5$
calculate the efficiency assuming the gas is diatomic (like $\mathrm{N}_{2}$ and $\mathrm{O}_{2} )$ and ideal.

Key Concepts

Thermodynamic Efficiency

Thermodynamic efficiency refers to the ratio of the work output of a cycle to the heat input, providing a measure of how effectively the cycle converts heat into work. In the Diesel cycle, the efficiency is determined by the distinct processes (adiabatic and constant pressure) and by key parameters like the compression and expansion ratios, making efficiency calculation a central aspect of engine performance analysis.

Specific Heats Ratio (gamma)

The specific heats ratio, commonly denoted as gamma (?), is defined as the ratio of the specific heat at constant pressure (Cp) to the specific heat at constant volume (Cv). This ratio is fundamental in the analysis of adiabatic processes, as it appears in the equations that describe the relationship between pressure, volume, and temperature during such processes in the Diesel cycle.

Compression Ratio

The compression ratio in an engine is defined as the ratio of the maximum to the minimum volume in the cylinder. It is a key parameter in engine design as it influences the thermal efficiency of the cycle by affecting the work done during the compression and expansion processes, thus playing a critical role in the performance of diesel engines.

Expansion Ratio

The expansion ratio is the ratio that relates the volume after combustion to the volume at the start of the combustion process. In the context of the Diesel cycle, this ratio determines how much the gas expands during the power stroke and is directly linked to the work output of the engine, making it a vital parameter for understanding engine efficiency.

Constant Pressure Process

In a constant pressure process, the pressure of the working fluid remains unchanged while other state properties such as volume and temperature vary. In the Diesel cycle, the heat addition phase is modeled as a constant pressure process, reflecting the characteristic of the diesel engine where fuel is injected and combusts as the piston begins its power stroke.

Adiabatic Process

An adiabatic process is one in which no heat is exchanged with the surroundings. During such a process, all changes in the internal energy of the gas are due solely to work done on or by the system. In both the compression and expansion strokes of the Diesel cycle, the processes can be idealized as adiabatic, meaning the temperature changes are a result of work interactions, which is central to analyzing engine performance.

Diesel Cycle

The Diesel cycle is an idealized thermodynamic cycle used to model the operation of diesel engines. It consists of four processes: an adiabatic (isentropic) compression, a constant pressure heat addition, an adiabatic expansion, and a constant volume heat rejection. This cycle serves as a theoretical model to analyze the performance and efficiency characteristics of diesel engines.

Ideal Gas Model

The ideal gas model assumes that the gas molecules do not interact except through elastic collisions and that the size of the molecules is negligibly small compared to the volume they occupy. This assumption simplifies the analysis of thermodynamic processes in the Diesel cycle, allowing the use of standard relationships between pressure, volume, and temperature during the various stages of the cycle.

Transcript

00:01 Using the definition of efficiency, so this is part a of the problem, efficiency is equal to 1 minus q low, heat low, divided by the heat high.

00:15 And this is equal to, in terms of a specific heat at a constant volume and pressure, is given by 1 minus n cv divided by cp times change in temperature here, in denominator is tc minus t b whereas in numerator the change in temperature is t t d minus t a minus t a so this is a expression and then simplifying it further we get one minus t t t d t d minus t a t d minus t a t a divided by tc tc minus t b t b whereas the c v or cp converts into the gamma and cancels out so this is the gamma downstairs here then writing temperature you know in terms of volume and pressure using ideal gas law then we can write efficiency what we obtain is 1 minus p d v d over nr minus so this is a bracket here p a v a divided by nr here here also and then in denominator the same thing repeating so gamma is here sitting and we have here pc vc n divided by r and here we have a p b b b divided by nr so nr cancels out so we are left with a simplified expression and then simplifying it in terms of pressure and volume then we have here pd divided by pc minus pa divided by pd divided by pd then in denominator we have here gamma vc over vc over va minus vb over va.

02:59 Then using adeptic relation, which is pv, pd, vd, gamma is equal to pc, pc, then we will convert our p and d's p d's and vc vas into in terms of the gamas so then our pd from this expression pd or pc becomes vc or vc or vd or vd whole gamma and this is also equal to vc or va over whole gamma.

04:04 Similarly, we can write pa over a pb using a same relation for a liberty case.

04:15 We can write this as a pb over va whole gamma.

04:21 So continues on the next page.

04:27 Our efficiency, result in efficiency by putting in the value of vc over vc over va becomes gamma minus vb over va whole power gamma divided by the gamma here, this factor va minus va minus, sorry, divided by vc, whole power minus gamma that comes up.

05:03 So minus va over vb over vb this is a minus gamma here...

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