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Internal Combustion Engine Cycles

A discussion on the thermodynamic cycles of internal combustion engines, including the Otto cycle, the Joule cycle, and the Diesel cycle.

Overview

Definitions

Working substance (WS). The WS is used as the carrier for heat energy. The heat engine carries out the conversion process by a series of changes of state of the WS. The state of the WS is defined by the values of its properties, e.g. pressure, volume, temperature, internal energy, enthalpy. These properties are also sometimes called functions of state.

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Key facts

Examples of internal combustion engine cycles are the Otto cycle (also known as the Constant Volume cycle), the Joule cycle (also known as the Brayton cycle, or the Constant Pressure cycle), and the Diesel cycle.

The thermal efficiency of the Otto cycle is given by:

\eta_O = 1 - \frac{1}{r^{\gamma -1}}

where \inline r is the compression ratio, and \inline \gamma the heat capacity ratio.

The thermal efficiency of the Joule cycle is given by:

\eta_J = 1 - \frac{1}{J^{\frac{\gamma - 1}{\gamma}}}

where \inline J is the pressure ratio, and \inline \gamma the heat capacity ratio.

The thermal efficiency of the Diesel cycle is given by:

\eta_D = 1 - \frac{1}{\gamma r^{\gamma - 1}} \frac{(\beta^\gamma - 1)}{(\beta - 1)}

where \inline r is the compression ratio, \inline \beta the expansion ratio during heating (also known as the cut-off ratio), and \inline \gamma the heat capacity ratio.

A thermodynamic cycle comprises a series of operations carried out on the working substance (WS) during which heat is supplied, and after which the WS is returned to its original state (for a more comprehensive introduction to thermodynamic cycles also see Thermodynamic Cycles ).

An example of a thermodynamic cycle is the internal combustion engine. In this case, the WS is treated as pure air, the expansions and compressions are reversible and adiabatic, and heat can be added instantaneously if desired. Applications of such cycles are the Otto cycle, the Joule cycle, and the Diesel cycle.

Before discussing these cycles, we first have to introduce the thermal efficiency of a cycle.

The thermal efficiency of a cycle, also denoted by \inline \eta_{th}, is a measure of the ability to convert heat energy into work. Therefore, the thermal efficiency can be defined as:

where \inline W is the work output and \inline Q_S the heat energy supplied. As the work done \inline W can also be expressed as:

where \inline Q_R is the heat rejected in the WS and in losses (for a more detailed discussion see Thermodynamic Cycles ). The thermal efficiency from (1) becomes:

or furthermore,

Otto Cycle

The Otto cycle (also sometimes called the Constant Volume cycle) is diagramed on a pressure - volume (\inline P-V) plot in Figure 1, and on a temperature - entropy (\inline T-S) plot in Figure 2.

746/img_therm30-1.png
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Figure 1
746/img_therm30-2.png
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Figure 2
This cycle consists of an adiabatic compression of the WS (step \inline 1-2), followed by an isochoric heating, i.e. heating at constant volume (step \inline 2-3), then an adiabatic expansion of the WS (step \inline 3-4), and ended with an isochoric cooling which reverts the system back to its original condition (step \inline 4-1).

We know that the heat added or removed from the WS during a process at constant volume can be written as:

where \inline n is the number of moles of the WS, \inline C_V the heat capacity at constant volume, and \inline \Delta T the change in temperature. Hence, for the Otto cycle depicted in Figure 1 and Figure 2, we have that the heat supplied \inline Q_S is:

and the heat rejected \inline Q_R is:

Therefore, the thermal efficiency of the Otto cycle becomes (see equation 4):

from which we obtain:

As in this case the WS is treated as pure air, we can write for the adiabatic compression \inline 1-2 that:

and for the adiabatic expansion \inline 3-4 that:

In (10) and (11), \inline \gamma is the heat capacity ratio defined as:

where \inline C_P is the heat capacity at constant pressure, and \inline C_V the heat capacity at constant volume.

Equations (10) and (11) can also be written as:

and,

respectively. As \inline V_2=V_3 and \inline V_4=V_1 (steps \inline 2-3 and \inline 4-1 of the Otto cycle are isochoric - see Figure 1), we can rewrite equations (13) and (14) as:

and,

respectively, where \inline \displaystyle r=\frac{V_1}{V_2}=\frac{V_4}{V_3} is called the compression ratio.

By subtracting equation (15) from equation (16), we get that:

or furthermore, that:

which leads to:

By using (19) in equation (9), we obtain the thermal efficiency of the Otto cycle as:

It can be noted that the compression ratio \inline r for carburettor engines is about \inline 7, while for fuel injector engines it increases to around \inline 10. By using these figures, and also considering a value for the heat capacity ratio \inline \gamma of \inline 1.25, the thermal efficiency of a petrol/gasoline engine is found to be between \inline 39\% and \inline 44\%. It can also be noted that for an Otto cycle engine, the ignition is usually performed by using a spark plug ("spark ignition").

Joule Cycle

The Joule cycle (also sometimes called the Brayton cycle, or the Constant Pressure cycle) is a thermodynamic cycle that describes the workings of the gas turbine engine. The Joule cycle is depicted on a \inline PV plot in Figure 3, and on a \inline TS plot in Figure 4.

746/img_therm31-1.png
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Figure 3
746/img_therm31-2.png
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Figure 4
This cycle consists of a reversible adiabatic (i.e. isentropic) compression of the WS (step \inline 1-2), followed by an isobaric heating (i.e. heating at constant pressure) (step \inline 2-3), then a reversible adiabatic (isentropic) expansion of the WS (step \inline 3-4), and ended with an isobaric cooling which reverts the system back to its initial state (step \inline 4-1).

Similar to equation (5), we can write the heat added or removed from the WS during a process at constant pressure as:

where \inline n is the number of moles of the WS, \inline C_P the heat capacity at constant pressure, and \inline \Delta T the change in temperature. Therefore, for the Joule cycle depicted in Figure 3 and Figure 4, the heat supplied becomes:

and the heat rejected:

Hence, the thermal efficiency of the Joule cycle becomes (see equation 4):

from which we obtain:

As for the Joule cycle the WS is also treated as pure air, we can write for the isentropic compression \inline 1-2 that:

and for the isentropic expansion \inline 3-4 that:

where \inline \gamma is the heat capacity ratio (see equation 12).

Equations (26) and (27) can also be written as:

and,

respectively.

By raising equations (28) and (29) to the \inline \frac{\gamma - 1}{\gamma} power, we respectively get that:

and,

As \inline P_2=P_3 and \inline P_4=P_1 (steps \inline 2-3 and \inline 4-1 of the Joule cycle are isobaric - see Figure 3), we can rewrite equations (30) and (31) as:

and,

respectively, where \inline \displaystyle J=\frac{P_2}{P_1}=\frac{P_3}{P_4} is called the pressure ratio.

By subtracting equation (32) from equation (33), we get that:

or furthermore, that:

which leads to:

By using (36) in equation (25), we obtain the thermal efficiency of the Joule cycle as:

Diesel Cycle

The Diesel cycle is diagramed on a \inline PV plot in Figure 5, and on a \inline TS plot in Figure 6.

746/img_therm32-1.png
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Figure 5
746/img_therm32-2.png
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Figure 6
This cycle consists of a reversible adiabatic (isentropic) compression of the WS (step \inline 1-2), followed by an isobaric heating (step \inline 2-3), then a reversible adiabatic (isentropic) expansion of the WS (step \inline 3-4), and ended with an isochoric cooling which returns the system back to its original state (step \inline 4-1).

Taking into account equations (5) and (21), we can write the heat supplied (at constant pressure, step \inline 2-3) as:

and the heat rejected (at constant volume, step \inline 4-1) as:

where \inline n is the number of moles of the WS, \inline C_P the heat capacity at constant pressure, and \inline C_V the heat capacity at constant volume.

Therefore, the thermal efficiency of the Diesel cycle becomes (see equation 4):

which can also be written as:

where \inline \gamma is the heat capacity ratio (see equation 12).

As for the Diesel engine the WS is again treated as pure air, we can write for the isentropic compression \inline 1-2 that:

for the isobaric heating \inline 2-3 that:

and for the isentropic expansion \inline 3-4 that:

Equation (42) leads to:

which can also be written as:

where \inline \displaystyle r = \frac{V_1}{V_2} is the compression ratio.

From equation (43) we obtain:

which can also be written as:

where \inline \displaystyle \beta = \frac{V_3}{V_2} is the expansion ratio during heating (also called the cut-off ratio).

The temperature \inline T_3 from (48) can further be expressed by considering the expression of \inline T_2 from (46), as:

Equation (44) leads to:

which can also be written, by considering the expression of \inline T_3 from (49), as:

We can write equation (51) in an analogous form as:

which leads to:

and furthermore, to:

Thus, we managed to express \inline T_2, \inline T_3, and \inline T_4 all in function of \inline T_1 (see equations 46, 49, and 54 respectively). By using these expression forms in equation (41), we get:

which leads to:

from which we obtain the thermal efficiency of the Diesel cycle as:

It can be noted that the ignition for the Diesel internal combustion engines is done by using a higher compression of the fuel, rather than by using a spark plug as in the case of gasoline powered Otto cycle engines. Therefore, the ignition for Diesel engines is sometimes called a "slow speed compression ignition", in contrast to the "spark ignition" of the Otto engines.

Example:
[imperial]
Example - Properties of a Diesel cycle
Problem
An engine operates under a Diesel cycle with a top pressure of \inline 642 \; psi (\inline lb/in^2). The compression ratio is \inline 18:1, and the air at the start of compression is at \inline 15 \; psi and \inline 100^\circ F. If \inline 3 \; BTU are added during combustion and expansion in accordance to the law \inline PV^{1.25} = const., the clearance volume is \inline 3 \; in^3, the specific heat at constant pressure during combustion is \inline 0.027 \; BTU/ft^3 ^\circ F, and also if we are to ignore the fuel mass, find:

A) the pressure, volume, and temperature at each change point in the cycle;

B) the work output of the cycle (in \inline ft-lb);

C) the thermal efficiency of the cycle.

The corresponding Diesel cycle is diagramed on a \inline PV plot in Figure E1.

746/img_therm33.png
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Figure E1
Workings
A) We are going to consider below each stage of the cycle separately.

Stage 1

We know from the hypothesis that the air at the start of compression is at \inline 15 \; psi and \inline 100^\circ F. Therefore, the initial pressure is:

while the initial temperature in kelvins (\inline K) is:

We also know from the hypothesis that the compression ratio is \inline 18:1. As the compression ratio \inline r is defined as:

we obtain the initial volume:

which becomes, by taking into account that \inline r=18 and that \inline V_2 is the clearance volume (\inline V_2 = 3 \; in^3):

Therefore \inline P_1 = 15 \; psi, \inline V_1 = 54 \; in^3, and \inline T_1 = 310.77 \; K.

Stage 2

We have from the hypothesis that the top pressure is \inline 642 \; psi. Hence:

Also from the hypothesis we have that the clearance volume is \inline 3 \; in^3. Thus:

By taking into account that:

we can write \inline T_2 as:

By replacing the numerical values (see 1, 2, 5, 6, and 7), we obtain:

from which:

Therefore \inline P_2 = 642 \; psi, \inline V_2 = 3 \; in^3, and \inline T_2 = 736.52 \; K.

Stage 3

As step \inline 2-3 of the Diesel cycle is isobaric (see Figure E1), we have that:

from which we obtain, by taking into account (6), that:

For step \inline 2-3 of the cycle we can also write that:

where \inline Q_S is the heat supplied (\inline 3 \; BTU), \inline C_P the specific heat at constant pressure (\inline 0.027 \; BTU/ft^3 ^\circ F), and \inline (T_3 - T_2) is the temperature difference expressed in \inline ^\circ F, while \inline V_S is the quantity of gas which is present at a pressure of \inline P_S = 14.7 \; psi and at a temperature of \inline T_S = 60^\circ F. In order to calculate \inline V_S, we can write that:

from which:

Taking into account that:

and also replacing the other numerical values (see 1, 2, and 5), we get that:

from which we obtain:

By using the value of \inline V_S from (19) in equation (14), and also considering that \inline Q_S = 3 \; BTU, \inline C_P = 0.027 \; BTU/ft^3 ^\circ F, and that \inline 1 \; ft^3 = 1728 \; in^3, we get that:

which leads to:

or, by converting the temperature difference into kelvins, to:

As \inline T_2 = 736.52 \; K (see 11), we obtain \inline T_3 as:

Still for step \inline 2-3 of the cycle we can write that:

from which \inline V_3 becomes:

or, by replacing numerical values (see 7, 11, and 23):

Equation (26) leads to:

Therefore \inline P_3 = 642 \; psi, \inline V_3 = 12.54 \; in^3, and \inline T_3 = 3078.69 \; K.

Stage 4

As step \inline 4-1 of the Diesel cycle is isochoric (see Figure E1), we have that:

from which we obtain, by taking into account (5), that:

As for step \inline 3-4 of the cycle we have that:

we can write:

from which \inline P_4 becomes:

or, by replacing numerical values (see 13, 27, and 29):

From equation (33) we obtain:

We know that \inline \frac{PV}{T} = const. Therefore, we can divide equation (30) by \inline \frac{PV}{T} and still obtain a constant:

Equation (35) can also be written as:

from which we obtain, for step \inline 3-4 of the cycle, that:

Hence, we can write that:

from which \inline T_4 becomes:

or, by replacing numerical values (see 23, 27, and 29):

From equation (40) we obtain:

Therefore \inline P_4 = 102.71 \; psi, \inline V_4 = 54 \; in^3, and \inline T_4 = 2136.61 \; K.

B) The work output \inline W of the cycle is given by:

where \inline n_1 = 1.25 (as \inline P_3 V_3^{1.25} = P_4 V_4^{1.25}). In order to calculate \inline n_2, we can write that:

which leads to:

By applying the \inline ln function to equation (44), we get:

from which we obtain:

or, by replacing numerical values (see 1, 5, 6, and 7):

Equation (47) leads to:

By replacing all the numerical values in equation (42) (see 1, 5, 6, 7, 13, 27, 29, and 34), and also taking into account that \inline n_1=1.25, \inline n_2=1.3 (see 48), and that \inline 1 \; ft = 12 \; in, we get the work done \inline W in \inline ft-lb as:

from which we obtain:

C) We know that the thermal efficiency is defined as:

where \inline W is the work output, and \inline Q_S the heat supplied.

As, in our case, \inline W = 1030.85 \; ft-lb (see 50) and \inline Q_S = 3 \; BTU, and also taking into account that \inline 1 \; BTU = 778 \; ft-lb, we get the thermal efficiency of the Diesel cycle as:

from which we obtain:

or, expressed as percentage:

Solution
P_1 = 15 \; psi$, $V_1 = 54 \; in^3$, $T_1 = 310.77 \; K
P_2 = 642 \; psi$, $V_2 = 3 \; in^3$, $T_2 = 736.52 \; K
P_3 = 642 \; psi$, $V_3 = 12.54 \; in^3$, $T_3 = 3078.69 \; K
P_4 = 102.71 \; psi$, $V_4 = 54 \; in^3$, $T_4 = 2136.61 \; K
W = 1030.85 \; ft-lb
\eta_{th} = 44.1\%