An Idealized Model of a Throttled Four-Stroke Reciprocating Spark Ignition Engine

Abstract

Four-stroke spark ignition (SI) engines are the workhorses of the automotive industry and several other niche markets. Traditional thermodynamic textbooks develop closed-form working equations to approximate the performance of these engines based on the Otto cycle using the cold air standard model, which treats the cycle as a closed thermodynamic system (i.e., no mass crosses the boundary) with the combustion and blowdown processes as equivalent heat additions and the gas as a calorically perfect gas (constant specific heats). Actual engines operate on a mechanical cycle over four linear strokes of a piston (over two revolutions), but the working fluid does not undergo a cyclic process. Rather, the analysis of an idealized throttled engine recognizes the engine as an open thermodynamic system. Each mechanical cycle requires fresh fuel and air to be drawn in (intake) and burned gases to be expelled (exhaust). A basic understanding of the gas exchange process (the intake, blowdown, and exhaust processes) is thereby attained. There is no external heat addition (as modeled traditionally), but rather a rapid conversion of one form of internal energy (chemical) into another (thermal) within the working fluid itself, that is, an internal combustion process, given an introductory treatment. The blowdown process (venting of high-pressure gases after the power stroke) is accomplished by driving mass out of the cylinder by the pressurized gases after the power stroke, not by a fictitious heat transfer process to the environment. Finally, the model includes the throttling process as the means to vary the mechanical load (torque and speed) of the cycle. Throttling reduces the mass throughput in the cycle at the expense of thermal efficiency. The calorically perfect gas model is retained, allowing for closed-form expressions for all processes. This model is still idealized and therefore will provide an upper limit on the performance of actual engines to which the performance of real engines could be compared, but it more closely resembles the nature of actual engines and is therefore a launching point for more thorough analyses of reciprocating internal combustion engines.

Appendices

Appendix 1: Compressible Flow Restriction on Piston Speed

One of the underlying assumptions made in the thermodynamic analysis of reciprocating engines is that the pressure is uniform within the cylinder at all times. In practice, there are two main effects that give rise to nonuniform pressures: the effect of the moving piston and undesired nonhomogeneous explosive combustion (called engine knock). This Appendix considers the first effect.

It might be expected that when the piston moves toward the gas (compression and exhaust strokes) the pressure on the piston face is somewhat higher than the average in cylinder pressure. When the piston moves away (power and intake), the piston face experiences a lower pressure than the cylinder average. These nonideal effects would increase the work required for gas exchange (pumping work) and reduce the work produced by the compression/expansion strokes.

Quantifying the effect requires application of compressible flow theory, in which dynamics of flows are influenced by temporal and/or spatial density variations. The physical mechanism by which the pressure tends toward uniformity is the speed at which pressure/density disturbances propagate, or the speed of sound, a thermodynamic property given by the relationship $c=\sqrt{kRT}$, not derived here. The Mach number of the piston is the ratio of the piston speed to the speed of sound in the gases. Therefore, a criterion for maintaining a uniform pressure is that the Mach number of the piston must be much less than some critical value (0.3 can be shown to be a reasonable criterion). If this criterion is met, any pressure disturbance is able to maintain a uniform pressure within the cylinder as the piston advances or withdraws. The pressure changes with time, but is the same at all locations.

To quantify the criterion, recall the expression for the piston velocity

$$ {v}_{\mathrm{piston}}=\frac{4{V}_{\mathrm{D}}N}{B^2}\left[\sin \theta +\tan \gamma \cos \theta \right] $$

The instantaneous Mach number of the piston is as follows:

$$ {\mathrm{Mach}}_{\mathrm{piston}}=\frac{\left|{v}_{\mathrm{piston}}\right|}{c}=\frac{4{V}_{\mathrm{D}}N\left|\left(\sin \theta +\tan \gamma \cos \theta \right)\right|}{B^2\sqrt{kRT}} $$

The Mach number is plotted as a function of crank angle for the base case in this chapter, with a very high engine speed of 10,000 RPM in Fig. 7.36. This result shows that the piston Mach number is highest during intake (because the temperature, hence speed of sound, is lowest), but that its maximum value midway during the stroke is approximately 0.13, which easily meets the criterion for uniform pressure.

Fig. 7.36

Piston Mach number versus crank angle for the default engine parameters, with an engine speed of 10,000 RPM

This base engine is a typical automobile engine, and this analysis demonstrates that pressure uniformity during the entire four-stroke cycle is reasonable for engines of this scale. However, for larger engines, it might be expected that the piston speeds are higher. A more general scale-dependent restriction can be met by rearranging the Mach number expression for the engine speed at its maximum (angle term set to unity) and with the temperature set at ambient (approximating its intake stroke value):

$$ {N}_{\mathrm{crit}}<{\mathrm{Mach}}_{\mathrm{crit}}\frac{B^2\sqrt{kRT_{\infty }}}{4{V}_{\mathrm{D}}} $$

This expression contains the geometric size parameters bore and displacement volume. Expressing the geometry in terms of the bore to stroke ratio (B/S), the critical engine speed above which nonuniform pressures would be experienced is

$$ {N}_{\mathrm{crit}}<{\mathrm{Mach}}_{\mathrm{crit}}\sqrt{kRT_{\infty }}{\left(\frac{{\left(B/s\right)}^2}{4{\pi}^2{V}_{\mathrm{D}}}\right)}^{\frac{1}{3}} $$

This function is plotted in Fig. 7.37 for a wide range of engine displacement volumes and bore to stroke ratios of 0.1, 1, and 10. A “square engine” is one for which the bore to stroke ratio is unity. A long narrow cylinder is one for which B/s = 0.1. The maximum speed for uniform pressure automobile engines (with displacement volumes of order 1 liter and nearly square) is approximately 20,000 RPM. The largest reciprocating engines (used in container ships) have per-cylinder displacement volumes on the order of 2000 liters, and their maximum engine speeds are approximately 1500 RPM. In practice, other mechanically related criterion (i.e., maximum dynamic stresses) places tighter restrictions on the engine speed.

Fig. 7.37

Critical reciprocating engine speed required for uniform in-cylinder pressure as a function of displacement volume for three bore to stroke ratios

Appendix 2: Combustion Processes

Traditionally, the combustion process in reciprocating engines is treated as an “equivalent heat addition” to a working fluid. However, the actual process does not involve heat, the transfer of energy from a high to low temperature. Rather, it is an internal combustion process that involves a chemical reaction and the associated conversion of one form of internal energy to another, from chemical energy to thermal energy. Therefore, the working fluid is not strictly a pure substance. However, the properties of the working fluid are closely approximated by that of nitrogen (N₂) , the dominant chemical species in air-breathing combustion devices. Table 7.3 lists several chemical properties of interest for several representative fuels. Brief derivations and their explanations are given in this Appendix, with an eye toward their application to engineering systems that involve combustion processes, such as reciprocating engines considered in this chapter.

Table 7.3 Chemical properties of typical fuels used in engineering systems

Mass Considerations: Combustion Stoichiometry

The chemical reaction for complete combustion of a lean mixture of 1 mol of an oxygenated hydrocarbon (with chemical formula C_nH_mO_p) in air is as follows:

$$ {\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{\mathrm{m}}{\mathrm{O}}_{\mathrm{p}}+\left(\frac{i_{\mathrm{st}}}{\phi}\right)\left({\mathrm{O}}_2+3.76{\mathrm{N}}_2\right)\overset{\phi <1}{\to }{\mathrm{n}\mathrm{CO}}_2+\frac{m}{2}{\mathrm{H}}_2\mathrm{O}+{i}_{\mathrm{st}}\left(\frac{1}{\phi }-1\right){\mathrm{O}}_2+\left(\frac{3.76{i}_{\mathrm{st}}}{\phi}\right){\mathrm{N}}_2 $$

where the stoichiometric coefficient (i_st) represents the moles of O₂ in the air required to completely burn the carbon in the fuel to CO₂ and the hydrogen to H₂O and is given by

$$ {i}_{\mathrm{st}}=n+\frac{m}{4}-\frac{p}{2} $$

The equivalence ratio, ϕ, is the actual fuel to air ratio of the mixture to the stoichiometric fuel to air ratio. If ϕ <1, there is excess air and not all of the O₂ can possibly react chemically. If ϕ >1, the mixture is rich and there is not sufficient O₂ to burn all the fuel. The rich case is not considered in this introductory treatment of combustion.

The air to fuel mass ratio is calculated (using the molecular weights of air and fuel) as

$$ {\mathrm{AFR}}_{\mathrm{mass}}=\frac{\mathrm{Moles}\kern0.5em \mathrm{of}\kern0.5em \mathrm{air}\ \left(\mathrm{mass}\kern0.5em \mathrm{of}\kern0.5em \mathrm{air}/\mathrm{moles}\kern0.5em \mathrm{of}\kern0.5em \mathrm{air}\right)}{\mathrm{Moles}\kern0.5em \mathrm{of}\kern0.5em \mathrm{fuel}\ \left(\mathrm{mass}\kern0.5em \mathrm{of}\kern0.5em \mathrm{fuel}/\mathrm{moles}\kern0.5em \mathrm{of}\kern0.5em \mathrm{fuel}\right)}=\frac{4.76{i}_{\mathrm{st}}/\phi\ {\mathrm{MW}}_{\mathrm{air}}}{{\mathrm{MW}}_{\mathrm{fuel}}} $$

For a stoichiometric mixture:

$$ {\mathrm{AFR}}_{\mathrm{mass},\mathrm{stoich}}=\frac{4.76\ \left(n+m/4-p/2\right)\ (28.9)}{\left(12n+m+16p\right)} $$

The results of this calculation, shown in Table 7.3, make it clear that the mass of air is generally an order of magnitude greater than the mass of fuel, approximately 15 for almost all hydrocarbon fuels. For that reason, the mass of the working fluid is dominated by air, and the air is dominated by N₂. The added fuel is considered to contribute chemical energy with little mass. It is reasonable therefore to model the properties of the gas as that of air as a first (and maybe last) approximation in thermodynamic analysis of engineering systems.

Chemical Energy

A bomb calorimeter is a device designed to measure the conversion of chemical energy available in fuels into thermal energy. It consists of a constant volume vessel (capable of withstanding high pressures) into which fuel and air (or pure O₂) are added, ignited, and then cooled (typically by a water bath). A schematic of the process is shown in Fig. 7.38. The calorimeter is initially charged with a lean fuel/air mixture. A spark or other ignition source initiates combustion that generally takes place quickly (so there is little time for heat transfer during the initial process), resulting in an intermediate state of high temperature (called the adiabatic flame temperature). Since the volume is constant, the pressure also rises dramatically according to the ideal gas law. The hot product gases are subsequently cooled (the heat transfer is negative in sign for an exothermic chemical reaction) until the product gases return to room temperature. The pressure will differ slightly from ambient due to the change in moles associated with the chemical reaction (Δn) from the original number of moles (n, which is generally much larger than Δn). For a 1 kg initial fuel charge, the heat transferred is equal to the negative of the lower heating value (LHV) of the fuel, which assumes the water vapor in the products is not condensed. The higher heating value (HHV) would be measured if all the water in the products were condensed (which does not generally occur during actual combustion processes). The LHV of fuels is approximately 40–50 MJ/kg_fuel for pure hydrocarbon fuels (Table 7.3). It is very high for pure H2 (due largely to its low molecular weight) and low for wood (due to the oxygen in the fuel).

Fig. 7.38

Schematic of bomb calorimeter, scaled to measure lower heating value (LHV)

The internal energy consists of thermal energy and chemical energy. An energy balance for a closed system (with negligible kinetic and potential energy) can therefore be written as

$$ \Delta {U}_{\mathrm{thermal}}+\Delta {U}_{\mathrm{chemical}}=Q-W $$

For a calorically perfect gas mixture of total mass m with a specific heat c_v,

$$ \Delta {U}_{\mathrm{thermal}}={mc}_{\mathrm{v}}\Delta T $$

and the temperature after a chemical reaction is

$$ {T}_2={T}_1+\frac{\left(-\Delta {U}_{\mathrm{chemical}}+Q\right)-W}{mc_{\mathrm{v}}} $$

The chemical energy change is grouped with the heat transfer to simply indicate how it is generally treated as being “equivalent” to a heat addition process, at least from a 1st Law perspective.

For an adiabatic, constant pressure process, the work of expansion can be grouped with the thermal internal energy as follows:

$$ \Delta {U}_{\mathrm{thermal}}-P\Delta V=\Delta \left({U}_{\mathrm{thermal}}- PV\right)=\Delta {H}_{\mathrm{thermal}}={mc}_{\mathrm{p}}\Delta T $$

If heat transfer is negligible (Q = 0), the final temperature is called the adiabatic flame temperature, an important metric in combustion analysis. The final temperature of an adiabatic, constant volume process is as follows:

$$ {T}_2={T}_1+\frac{\left(-\Delta {U}_{\mathrm{chemical}}\right)}{mc_{\mathrm{v}}}\kern0.75em \mathrm{constant}\kern0.5em \mathrm{volume},\kern0.5em \mathrm{adiabatic}\kern0.5em \mathrm{process} $$

The final temperature of an adiabatic constant pressure process is

$$ {T}_2={T}_1+\frac{\left(-\Delta {U}_{\mathrm{chemical}}\right)}{mc_{\mathrm{p}}}\kern0.5em \mathrm{constant}\kern0.5em \mathrm{pressure},\kern0.5em \mathrm{adiabatic}\kern0.5em \mathrm{process} $$

Since the enthalpy specific heat is larger than the internal energy specific heat (c_p = c_v + R), the adiabatic flame temperature of a constant pressure process is lower than that of a constant volume process due to the work of expansion done by the gases required to maintain pressure constant.

The change in chemical energy (related to the heating value) is given by the difference in the inherent energy of the products and reactants:

$$ \Delta {U}_{\mathrm{chemical}}=\sum \limits_{j,\mathrm{products}}{n}_j\Delta {u}_{\mathrm{f},j}^0-\sum \limits_{i,\mathrm{reactant}}{n}_i\Delta {u}_{\mathrm{f},i}^0 $$

Where each chemical species has an inherent chemical energy, called the “heat” of formation given by $\Delta {u}_{\mathrm{f}}^0$. A better name would be internal energy of formation and is closely related to the enthalpy of formation $\Delta {h}_{\mathrm{f}}^0$, which is typically listed in published data sets (such as the JANAF tables). If the chemical composition of a fuel is well defined, then the chemical energy change (and LHV or HHV) can be calculated from this relationship and from the stoichiometric chemical reaction (which defines the moles of fuel, components of air and products).

The chemical energy change divided by the total mass is an unconventional measure of the chemical energy, but one that is most relevant to thermodynamic analysis of combustion systems. It can be defined as follows:

$$ \frac{{\left(-\Delta {U}_{\mathrm{chemical}}\right)}_{\mathrm{stoich}}}{m}=\Delta {u}_{\mathrm{comb}}=\frac{\mathrm{LHV}}{1+\mathrm{AFR}}\left[=\right]\frac{\mathrm{J}}{{\mathrm{kg}}_{\mathrm{mix}}} $$

It is listed in Table 7.3 and is remarkably similar for hydrocarbons and approximately equal to 2.7 MJ/kg_mix and a bit higher for H2. In the table, it is listed as being divided by the equivalence ratio, ϕ, as it represents a maximum possible per-unit total mass chemical energy release of a stoichiometric mixture. Actual combustion processes can be overall lean (for diesel engines, gas turbines, boilers) and the chemical energy release can be adjusted accordingly.

The heating value is a mass-based chemical energy density of a fuel. However, an important practical consideration of a fuel is the space (volume) it occupies. Therefore, a volume-based energy density is calculated (simply the heating value multiplied by the density of the fuel) in Table 7.3, which shows that gaseous fuels have a low volume energy density, at least when stored at atmospheric pressure. Gaseous fuels would be pressurized, which would increase their volumetric energy density, but the increased mass of the storage vessel should be factored into their mass-based energy density.

A final metric associated with fuels is related to the controversial social issue of global warming due to the emission of greenhouse gases. The combustion of hydrocarbon fuels produces carbon dioxide (CO₂) , which is not considered to cause direct health effects, but its gradual build-up in the atmosphere is cause for concern as a gas that traps radiant energy that would otherwise be emitted to outer space. A good metric for comparing the contribution of greenhouse gases of various fuels is the mass of CO₂ released normalized by the chemical energy release (Table 7.3). Of the fossil fuels, coal is approximately 50% higher than natural gas and oil-based fuels are in between. Interestingly, wood has the highest value (due to its low LHV), but wood is not a fossil fuel and trees absorb CO₂ from the atmosphere, making no net contribution to greenhouse gas emissions if the total biomass remains constant.

Knocking in Spark Ignition Engines

Gasoline is a complex blend of a mixture of hydrocarbons (carbon numbers ranging between 4 and 12, roughly) generally obtained from the distillation of crude oil. It is a liquid fuel, which has the advantage over a gaseous fuel in that it can be stored in an ambient pressure vessel. It also has a higher chemical energy volume density (heating value per unit volume). A liquid fuel has the advantage over a solid fuel in that it can be pumped, and it is volatile. The fuel is injected as a liquid, but evaporated prior to induction into the cylinder.

The octane rating refers to an industry standardized method for determining the tendency of a fuel/air mixture to knock or autoignite (react chemically without a spark) relative to two standard fuels: n-heptane (C₇H₁₆) and an isomer of octane (2-2-4- trimethyl pentane, which has the chemical formula C₈H₁₈). Heptane has somewhat weaker bonds than iso-octane and autoignites at a lower compression ratio. The critical compression ratio above which the engine knocks is measured for both fuels in a laboratory engine under one of two standard protocols (research method or motor method). A scale is established with 0 being the compression ratio for n-heptane and 100 for the iso-octane. Then, the fuel whose octane rating is sought is tested for its critical compression ratio. The octane rating is determined by its location on the scale. The (R + M)/2 method refers to the test done on two different engine types (research and motor) with different operating conditions and the results averaged.

The octane rating of natural gas (primarily methane in composition) is 120. That means that automobiles designed to run on natural gas can be designed with significantly higher compression ratios than those designed for gasoline, and therefore will be more thermodynamically efficient. Also, the greenhouse gas emissions per mass of fuel burned is lower. Natural gas is an attractive automotive fuel for those reasons as a transition fuel to a hydrogen economy.

The octane rating of hydrogen (H2) is greater than 130. Ultimately, it would seem that a complete shift to an energy economy based on H2 is the logical direction for the evolution of man. The mass of the fuel would come from water, and the energy to convert would come from renewable sources (solar, wind, hydroelectric, tidal, etc.).

One of the arguments against H2 is its safety because it is “too explosive.” This perception may be largely based on a single historical event, namely, the Hindenburg disaster. However, that disaster was due to the use of H2 as a buoyant medium, not as a fuel.

It would be easy to create an industry standard hydrogen-based fuel blend that is less explosive than pure H2 by simply adding an inert component. While N2 would be abundant, a better thermodynamic choice would be argon, which is plentiful (1 mol% of the Earth’s enormous atmosphere) and has a specific heat ratio of 1.667, and that is independent of temperature. A fuel/air mixture of a H2/argon fuel would have a higher k and, therefore, more efficient reciprocating engine. Sounds like a good deeper dive workshop in the making … .

Appendix 3: Blowdown Process

Traditionally, the blowdown process is modeled as being due to a heat rejection to the environment, which is nothing of the sort. In practice, it is the first stage of the gas exchange process needed to expel burned gases and induct a fresh fuel/air mixture. As the piston approaches BDC during the power stroke, the pressure is above ambient. When the exhaust valve opens, mass is driven out of the cylinder in an open system venting, with a single exit stream. In the ideal cycle, the process is modeled to take place instantaneously at BDC.

To analyze this open system process, the mass, energy, and entropy balance equations are written in a general differential form with an incremental amount of mass exiting (m_e) as the independent variable:

$$ {dM}_{\mathrm{CV}}=-{m}_{\mathrm{e}} $$

$$ {dE}_{\mathrm{CV}}=\delta q-\left( PdV+{m}_{\mathrm{e}}{P}_{\mathrm{e}}{v}_{\mathrm{e}}\right)-{m}_{\mathrm{e}}{e}_{\mathrm{e}} $$

$$ {dS}_{\mathrm{CV}}=\delta q/{T}_{\mathrm{B}}+\delta \Sigma -{m}_{\mathrm{e}}{s}_{\mathrm{e}} $$

The subscript CV on the left side refers to the average values within the control volume (not internal energy specific heat). Note that there are two forms of work: expansion work, which is zero during ideal blowdown, and flow work associated with pushing fluid out. With the model assumptions of no control volume change, negligible KE and PE and no heat transfer to the walls, and expressing the total energy and entropy in terms of their specific (per unit mass) properties, the energy equation reduces to

$$ d\left({M}_{\mathrm{CV}}{u}_{\mathrm{CV}}\right)=-{m}_{\mathrm{e}}\left({u}_{\mathrm{e}}+{P}_{\mathrm{e}}{v}_{\mathrm{e}}\right) $$

Using the chain rule, d(M_CVu_CV) = M_CVd(u_CV) + u_CVd(M_CV)_.

The internal energy within the (u_CV) represents the average specific internal energy within the cylinder, while the exiting internal energy (u_e) represents that of the mass as it exits, which in general could be quite different. However, in this case, the conditions within the cylinder are considered to be uniform, and they are equal, that is, u_CV = u_e. The assumption that the exit conditions are equal to the average conditions within a control volume is called the “well-mixed model” approximation. Along with the mass relation (dM_CV =  − m_e), the energy equation reduces to

$$ {M}_{\mathrm{CV}}d\left({u}_{\mathrm{CV}}\right)=-{m}_{\mathrm{e}}{P}_{\mathrm{CV}}{v}_{\mathrm{CV}} $$

This result shows that as mass exits the cylinder, flow work decreases the specific internal energy stored within the cylinder. Rearranging,

$$ \frac{d\left({u}_{\mathrm{CV}}\right)}{P_{\mathrm{CV}}{v}_{\mathrm{CV}}}=\frac{dM_{\mathrm{CV}}}{M_{\mathrm{CV}}} $$

Invoking ideal gas relations and using the mass relation:

$$ {M}_{\mathrm{CV}}=\frac{P_{\mathrm{CV}}{V}_{\mathrm{CV}}}{RT_{\mathrm{CV}}} $$

For a differential change:

$$ \frac{dM_{\mathrm{CV}}}{M_{\mathrm{CV}}}=\frac{dP_{\mathrm{CV}}}{P_{\mathrm{CV}}}+\frac{dV_{\mathrm{CV}}}{V_{\mathrm{CV}}}-\frac{dT_{\mathrm{CV}}}{T_{\mathrm{CV}}} $$

Substituting, with the volume constant during blowdown:

$$ \frac{c_{\mathrm{v}}}{R}\frac{dT_{\mathrm{CV}}}{T_{\mathrm{CV}}}=\frac{dP_{\mathrm{CV}}}{P_{\mathrm{CV}}}-\frac{dT_{\mathrm{CV}}}{T_{\mathrm{CV}}} $$

Rearranging:

$$ \frac{dP_{\mathrm{CV}}}{P_{\mathrm{CV}}}=\left(\frac{c_{\mathrm{v}}}{R}+1\right)\frac{dT_{\mathrm{CV}}}{T_{\mathrm{CV}}} $$

This relation is recognized as that of an adiabatic expansion of an ideal gas. Assuming constant specific heat and integrating from state 4 (the end of the power stroke) until the pressure decreases to the exhaust manifold pressure (state 5):

$$ \ln \left(\frac{P_{\mathrm{exhaust}}}{P_4}\right)=\left(\frac{c_{\mathrm{v}}}{R}+1\right)\ln \left(\frac{T_5}{T_4}\right) $$

The temperature at the end of the blowdown process is

$$ {T}_5={T}_4{\left(\frac{P_{\mathrm{exhaust}}}{P_4}\right)}^{\frac{R}{c_{\mathrm{v}}+R}} $$

The mass of gas that remains after blowdown can be calculated with the ideal gas law.

With the key assumptions that blowdown takes place instantaneously at BDC (no time for heat transfer or volume change), this analysis demonstrates formally that the gases that remain inside the cylinder after blowdown have undergone an adiabatic expansion.

To demonstrate formally that the adiabatic, reversible process on the cylinder gases is isentropic, consider the entropy balance:

$$ {dS}_{\mathrm{CV}}=\delta q/{T}_{\mathrm{B}}+\delta \Sigma -{m}_{\mathrm{e}}{s}_{\mathrm{e}} $$

If the process is considered to be adiabatic and reversible:

$$ d\left({M}_{\mathrm{CV}}{s}_{\mathrm{CV}}\right)={M}_{\mathrm{CV}}{ds}_{\mathrm{CV}}+{s}_{\mathrm{CV}}{dM}_{\mathrm{CV}}=-{m}_{\mathrm{e}}{s}_{\mathrm{e}} $$

Using mass (dM_CV =  − m_e) and the well-mixed model (s_CV = s_e) results in

$$ {ds}_{\mathrm{CV}}=0 $$

That is, the specific entropy within the cylinder remains constant. The gases that remain in the cylinder undergo an isentropic expansion process.

A good exercise (left to the reader) is to calculate the theoretical power that could be extracted by placing a turbine between the exit and ambient. The gases that exit during blowdown experience eXergy destruction as they are throttled across the valve (an isenthalpic process, with entropy generation). Note that the gases left behind in the cylinder undergo an isentropic process, while the gases that exit undergo an isenthalpic process. If a reversible turbine were placed there, the gases would expand isentropically as well as they extract useful power. The simpler way to calculate the total work extracted is to analyze a control mass.

Appendix 4: Exhaust Stroke

This Appendix formally shows the conditions for which the temperature remains constant during the exhaust and intake strokes (if the pressure drop across the valves is neglected). That is, the piston work is equal and opposite to the flow work.

With the control volume taken to be the in-cylinder gases, the differential mass, energy, and entropy balances are

$$ {dM}_{\mathrm{CV}}=-{m}_{\mathrm{e}} $$

$$ {dE}_{\mathrm{CV}}=\delta q-\left( PdV+{m}_{\mathrm{e}}{P}_{\mathrm{e}}{v}_{\mathrm{e}}\right)-{m}_{\mathrm{e}}{e}_{\mathrm{e}} $$

$$ {dS}_{\mathrm{CV}}=\delta q/{T}_{\mathrm{B}}+\delta \Sigma -{m}_{\mathrm{e}}{s}_{\mathrm{e}} $$

Assuming adiabatic, and neglecting kinetic and potential energy changes, the energy equation becomes

$$ d\left({M}_{\mathrm{CV}}{u}_{\mathrm{CV}}\right)={M}_{\mathrm{CV}}d\left({u}_{\mathrm{CV}}\right)+{u}_{\mathrm{CV}}d\left({M}_{\mathrm{CV}}\right)=-(PdV)-{m}_{\mathrm{e}}\left({P}_{\mathrm{e}}{v}_{\mathrm{e}}\right)-{m}_{\mathrm{e}}{u}_{\mathrm{e}} $$

Inserting the mass balance and rearranging:

$$ {M}_{\mathrm{CV}}d\left({u}_{\mathrm{CV}}\right)=-(PdV)-{m}_{\mathrm{e}}\left({P}_{\mathrm{e}}{v}_{\mathrm{e}}\right)+{m}_{\mathrm{e}}\left({u}_{\mathrm{CV}}-{u}_{\mathrm{e}}\right) $$

The first term on the right side represents the work on the piston on the gases (positive since the volume decreases). The second term represents the flow work of the gases as they exit the cylinder. The third term represents a difference between the average properties in the cylinder and the gases as they exit. This term is zero for the well-mixed model approximation.

The entropy balance is as follows:

$$ d\left({M}_{\mathrm{CV}}{s}_{\mathrm{CV}}\right)={M}_{\mathrm{CV}}d\left({s}_{\mathrm{CV}}\right)+{s}_{\mathrm{CV}}d\left({M}_{\mathrm{CV}}\right)=\delta \Sigma -{m}_{\mathrm{e}}{s}_{\mathrm{e}} $$

$$ {M}_{\mathrm{CV}}d\left({s}_{\mathrm{CV}}\right)=\delta \Sigma +{m}_{\mathrm{e}}\left({s}_{\mathrm{CV}}-{s}_{\mathrm{e}}\right) $$

If the in-cylinder process is assumed to be reversible, and the well-mixed model is invoked (s_CV = s_e)_,then the entropy balance becomes as follows:

$$ d\left({s}_{\mathrm{CV}}\right)=0 $$

That is, the specific entropy of the in-cylinder gases remains constant. Since the pressure remains equal to the exhaust runner pressure, the temperature must remain constant. Then, from the energy equation, the specific internal energy remains constant and the flow work is equal and opposite to the piston work.

Appendix 5: Second Law-Related Analysis

Pinpointing all sources of eXergy destruction through a thorough 2nd Law analysis of the engine cycle requires the entropy change associated with the combustion process, which is not developed here. Treating the combustion process as an “equivalent heat addition” is reasonable from a 1st Law perspective, but less so from a 2nd Law one, where the entropy transfer requires a thermal reservoir from which heat is drawn. For internal combustion engines, there is no such heat transfer. Actually, in a real engine, there is considerable heat transfer from the gases to the cylinder wall. A more advanced study of chemical reactions than done here can add an inherent eXergy destruction associated with the combustion reaction, similar to an internal energy change.

Nevertheless, the overall 2nd Law efficiency of the cycle is defined as the 1st Law efficiency divided by the maximum possible thermal efficiency, namely, the Carnot efficiency for a heat engine operating between two thermal reservoirs (one at the dead state temperature, T_∞, one at the maximum experienced in the cycle, T_max):

$$ {\eta}_{\mathrm{I},\operatorname{MAX}}=1-\frac{T_{\infty }}{T_{\mathrm{max}}} $$

The 2nd Law efficiency is:

$$ {\eta}_{\mathrm{I}\mathrm{I}}=\frac{\eta_{\mathrm{I}}}{\eta_{\mathrm{I},\mathrm{mas}}} $$

eXergy Destruction Associated with Throttling

Most of the processes in the ideal throttled Otto cycle as defined here are reversible, and a 2nd Law analysis of them is not particularly insightful. On the other hand, the pressure drop associated with throttling in the intake system results in eXergy destruction. The steady-state, open system entropy balance, multiplied by the dead state temperature (T_inf), is:

$$ {T}_{\infty}\left[\frac{dS}{dt}=\dot{m}\left({s}_{\mathrm{in}}-{s}_{\mathrm{out}}\right)+\frac{\dot{Q}}{T_{\mathrm{b}}}+\overset{\overset{\dddot{}}{}}{\Sigma}\right] $$

where T_b is the temperature at the boundary of the control volume where heat is transferred and $\overset{\overset{\dddot{}}{}}{\Sigma}$ is the entropy generation rate. At steady state and rearranging for the eXergy destruction rate:

$$ e{\overset{\overset{\dddot{}}{}}{X}}_{\mathrm{D}}=\overset{\overset{\dddot{}}{}}{\Sigma}{T}_{\infty }=\dot{m}\left({s}_{\mathrm{out}}-{s}_{\mathrm{in}}\right){T}_{\infty }-\dot{Q}\left(\frac{T_{\infty }}{T_{\mathrm{b}}}\right) $$

For ideal gases, the entropy change is related to temperature and pressure changes by:

$$ {s}_{\mathrm{out}}-{s}_{\mathrm{in}}={c}_{\mathrm{p}}\ln \left(\frac{T_{\mathrm{out}}}{T_{\mathrm{in}}}\right)-R\ln \left(\frac{P_{\mathrm{out}}}{P_{\mathrm{in}}}\right) $$

The intake and exhaust processes are considered to be adiabatic, and an energy balance confirms that they are isenthalpic, and therefore isothermal (for ideal gases):

$$ \frac{dE}{dt}=0={\dot{Q}}_{\mathrm{in}}+\dot{m}\left({h}_{\mathrm{in}}-{h}_{\mathrm{out}}\right) $$

Therefore, the eXergy destruction rate is given by

$$ e{\overset{\overset{\dddot{}}{}}{X}}_{\mathrm{D}}=\overset{\overset{\dddot{}}{}}{\Sigma}{T}_{\infty }=-\dot{m}{RT}_{\infty}\ln\ \left(\frac{P_{\mathrm{intake}}}{P_{\infty }}\right)\kern0.5em \mathrm{for}\kern0.5em \mathrm{the}\kern0.5em \mathrm{intake}\kern0.5em \mathrm{process} $$

If pressure losses occur in the exhaust system, the eXergy destruction rate is:

$$ e{\overset{\overset{\dddot{}}{}}{X}}_{\mathrm{D}}=\overset{\overset{\dddot{}}{}}{\Sigma}{T}_{\infty }=-\dot{m}{RT}_{\infty}\ln\ \left(\frac{P_{i\infty }}{P_{\mathrm{exhaust}}}\right)\kern0.5em \mathrm{for}\kern0.5em \mathrm{the}\kern0.5em \mathrm{exhaust}\kern0.5em \mathrm{process} $$

Workshop 7.1: Otto and Diesel Cycle Comparison

The spark ignition (SI) engine is considered to be most closely modeled by the Otto cycle for first-order thermodynamic analysis. A smaller, but very large, market exists for the compression ignition (CI) engine, which is modeled by the Diesel cycle. This workshop compares the performance of these two idealized cycles using a value of the compression ratio that is typical of their practical use.

In an SI engine, a fuel/air mixture is inducted during intake, and the combustion process takes place by the propagation of a turbulent premixed flame. The compression ratio of the engine is limited by the chemical nature of the fuel/air mixture. Combustion will initiate prematurely (during the compression stroke) and give rise to violent nonuniform pressure waves if the compression ratio is too high. This complex phenomenon (sometimes called knocking or other descriptive names) is problematic from both a thermal efficiency and mechanical view. Also, the fuel/air ratio must be maintained near a constant value to produce a reliable propagating flame. Therefore, power output is varied by throttling in practice.

In a CI engine, only air is inducted during intake, and fuel is injected directly into the cylinder under high pressure during the power stroke. The temperature after compression is sufficiently high that combustion initiates without a spark and the combustion process occurs in a turbulent non-premixed flame, which has very different nature than the premixed combustion in an SI engine. In the Diesel cycle, a model for this practical engine, the combustion process is treated as being isobaric (which is difficult and not necessary to really accomplish in practice, requiring a very precise and variable rate of fuel injection related to the rate of volume change). The compression ratio is not limited by preignition, but rather by more mechanical restrictions such as high instantaneous torque needed to compress the gases into a small clearance volume (not investigated further here). The gas exchange process is the same as for the SI engine, in that there is blowdown followed by exhaust and intake strokes. However, there is no need for throttling a Diesel engine to vary the power output. Rather, the power output of a CI engine is varied by changing the total amount of fuel injected.

The gas exchange process involves two strokes (exhaust followed by intake), but in the ideal cycle, the cylinder pressure equals ambient and there is no associated net work. Therefore, both ideal cycles consist of four internally reversible processes (in two strokes) in a piston/cylinder arrangement (Figs. 7.39 and 7.40). The only difference is process 2–3, the internal combustion process, which is constant volume (isochoric) in the Otto and constant pressure (isobaric) in the Diesel. They will be compared with the same displacement volume of 1.0 liters, but different compression ratios, the ratio of maximum to minimum volume (10 for Otto, 20 for Diesel, considered to be representative values). Also, the conversion of chemical energy to thermal energy per unit mass during combustion will differ (2730 kJ/kg for Otto and 583 kJ/kg for Diesel, which runs fuel lean). For both cycles, invoke the cold air standard approximation, that is, assume the properties of the working fluid are that of air as a calorically perfect gas (with c_v = 0.720 kJ/kg/ΔK). The initial state is at 1.0 atm, 298 K. Thermal efficiency is defined as the net work divided by chemical energy added during combustion (process 2–3). For power output, assume an eight-cylinder engine operating at 2000 revolutions per minute (RPM), recognizing that two full revolutions are required to complete the cycle (one for the intake and exhaust, one for the two strokes analyzed) but there is no net work associated with the gas exchange (wide open throttle for Otto engine). Complete the state and results tables, and sketch both cycles on the PV scale provided (Figs. 7.41, 7.42, and 7.43).

Fig. 7.39

Schematic of ideal Otto cycle

Fig. 7.40

Schematic of ideal diesel cycle

Fig. 7.41

State tables

Fig. 7.42

Results table. The sign of the work is that done by the working fluid

Fig. 7.43

Indicator diagram. Place the state properties according to the calculations for both cycles and sketch the processes in between. (You could calculate a few intermediate points for curved processes)

Workshop 7.2: Adding a Compressor and/or Turbine

This exercise builds on Workshop 7.1 (Otto and Diesel cycles) and is a good segue to the Brayton cycle, (Chap. 9). The thermodynamic performance of a reciprocating engine can be improved by adding a compressor (and possibly an intercooler) to the intake and/or a turbine to the exhaust. There are therefore several possible combinations. The schematic of Fig. 7.44 shows all components. The goal will be to determine how each combination affects the mass flow rate, power output, and First Law efficiency of the engine by completing the following table:

Fig. 7.44

Schematic of reciprocating engine turbo-charging and intercooling

Consider the ideal four-stroke Otto cycle from Workshop 7.1 to be the base case (compression ratio of 10, 8 cylinders, each with a displacement volume of 1 liter operating unthrottled at 2000 RPM, and a calorically perfect gas, with c_p = 1.0 kJ/kg/ΔK). The key concept is that while the cycle is executed as a series of processes, the engine as a whole operates as a quasi-steady-state open system, with a steady mass flow rate equal to the total mass exchanged during the intake and exhaust strokes divided by the cycle time (with four strokes per cycle). Note that the mass exchanged is the inlet density times the displacement volume, which is slightly lower than the mass contained in the cylinder. The difference is the “residual gases.”

With supercharging, a portion of the output shaft power is used to drive a compressor (assumed to be isentropic) that maintains a pressurized intake plenum, as shown. The net power output to the shaft is the power produced by the four strokes of cycle minus that used to drive the compressor. The temperature after the compressor is elevated, and an option is to cool these gases prior to entry into the engine cycle (intercooling). For this exercise, choose an intake plenum pressure of 2 atm for all cases that involve a compressor.

At the end of the power stroke, the exhaust valve opens and the gases are hot and above ambient pressure. This available energy is dissipated in the engine cycle without a turbine. However, placement of a turbine in the exhaust stream is a means to extract a portion of that available energy. The maximum possible power that can be extracted from the turbine is obtained by setting the exhaust plenum pressure to the pressure at the end of the power stroke. The net power output to the shaft is the power produced by the four strokes of cycle and the turbine, less that used to drive the compressor.

The turbo-compounded case is when only a turbine is used (no compressor), and its mechanical output is tied directly into the output shaft. Turbo-charging involves both a compressor and a turbine. In practice, all the mechanical output of the turbine is used to drive the compressor, and the intake plenum pressure would be whatever this coupling yields. In this exercise, the compressor and turbine are considered to be uncoupled, so that the compressor work does not equal the turbine work.

This workshop is a little tricky to get precisely right because of the subtle ways that the added components alter the performance of the main engine (i.e., pumping work and mass undergoing the cycle). Assume that the temperature at the start of compression equals the intake manifold pressure, which neglects the effect of mixing the fresh intake with the hot residual gases (and requires a more complex iteration calculation). The main points of this exercise are as follows:

• To observe how a reciprocating engine can be viewed as a steady-state engine

• To observe how much additional output power and/or increase in efficiency are theoretically possible with these modifications to the basic cycles