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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.
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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 r is the compression ratio, and \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 J is the pressure ratio, and \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 r is the compression ratio, \beta the expansion ratio during heating (also known as the cut-off ratio), and \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 \eta_{th}, is a measure of the ability to convert heat energy into work. Therefore, the thermal efficiency can be defined as:

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

where Q_R is the heat rejected in the WS and in losses (for a more detailed discussion see Thermodynamic Cycles ). The thermal efficiency from (2) becomes:

or furthermore,

Otto Cycle

The Otto cycle (also sometimes called the Constant Volume cycle) is diagramed on a pressure - volume (P-V) plot in Figure 1, and on a temperature - entropy (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 1-2), followed by an isochoric heating, i.e. heating at constant volume (step 2-3), then an adiabatic expansion of the WS (step 3-4), and ended with an isochoric cooling which reverts the system back to its original condition (step 4-1).

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

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

and the heat rejected Q_R is:

Therefore, the thermal efficiency of the Otto cycle becomes (see equation 5):

from which we obtain:

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

and for the adiabatic expansion 3-4 that:

In (11) and (12), \gamma is the heat capacity ratio defined as:

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

Equations (11) and (12) can also be written as:

and,

respectively. As V_2=V_3 and V_4=V_1 (steps 2-3 and 4-1 of the Otto cycle are isochoric - see Figure 1), we can rewrite equations (14) and (15) as:

and,

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

By subtracting equation (16) from equation (17), we get that:

or furthermore, that:

which leads to:

By using (20) in equation (10), we obtain the thermal efficiency of the Otto cycle as:

It can be noted that the compression ratio r for carburettor engines is about 7, while for fuel injector engines it increases to around 10. By using these figures, and also considering a value for the heat capacity ratio \gamma of 1.25, the thermal efficiency of a petrol/gasoline engine is found to be between 39\% and 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 PV plot in Figure 3, and on a 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 1-2), followed by an isobaric heating (i.e. heating at constant pressure) (step 2-3), then a reversible adiabatic (isentropic) expansion of the WS (step 3-4), and ended with an isobaric cooling which reverts the system back to its initial state (step 4-1).

Similar to equation (6), we can write the heat added or removed from the WS during a process at constant pressure as:

where n is the number of moles of the WS, C_P the heat capacity at constant pressure, and \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 5):

from which we obtain:

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

and for the isentropic expansion 3-4 that:

where \gamma is the heat capacity ratio (see equation 13).

Equations (27) and (28) can also be written as:

and,

respectively.

By raising equations (29) and (30) to the \frac{\gamma - 1}{\gamma} power, we respectively get that:

and,

As P_2=P_3 and P_4=P_1 (steps 2-3 and 4-1 of the Joule cycle are isobaric - see Figure 3), we can rewrite equations (31) and (32) as:

and,

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

By subtracting equation (33) from equation (34), we get that:

or furthermore, that:

which leads to:

By using (37) in equation (26), we obtain the thermal efficiency of the Joule cycle as:

Diesel Cycle

The Diesel cycle is diagramed on a PV plot in Figure 5, and on a 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 1-2), followed by an isobaric heating (step 2-3), then a reversible adiabatic (isentropic) expansion of the WS (step 3-4), and ended with an isochoric cooling which returns the system back to its original state (step 4-1).

Taking into account equations (6) and (22), we can write the heat supplied (at constant pressure, step 2-3) as:

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

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

Therefore, the thermal efficiency of the Diesel cycle becomes (see equation 5):

which can also be written as:

where \gamma is the heat capacity ratio (see equation 13).

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

for the isobaric heating 2-3 that:

and for the isentropic expansion 3-4 that:

Equation (43) leads to:

which can also be written as:

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

From equation (44) we obtain:

which can also be written as:

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

The temperature T_3 from (49) can further be expressed by considering the expression of T_2 from (47), as:

Equation (45) leads to:

which can also be written, by considering the expression of T_3 from (50), as:

We can write equation (52) in an analogous form as:

which leads to:

and furthermore, to:

Thus, we managed to express T_2, T_3, and T_4 all in function of T_1 (see equations 47, 50, and 55 respectively). By using these expression forms in equation (42), 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 642 \; psi (lb/in^2). The compression ratio is 18:1, and the air at the start of compression is at 15 \; psi and 100^\circ F. If 3 \; BTU are added during combustion and expansion in accordance to the law PV^{1.25} = const., the clearance volume is 3 \; in^3, the specific heat at constant pressure during combustion is 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 ft-lb);

C) the thermal efficiency of the cycle.

The corresponding Diesel cycle is diagramed on a 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 15 \; psi and 100^\circ F. Therefore, the initial pressure is:

while the initial temperature in kelvins (K) is:

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

we obtain the initial volume:

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

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

Stage 2

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

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

By taking into account that:

we can write T_2 as:

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

from which:

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

Stage 3

As step 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 2-3 of the cycle we can also write that:

where Q_S is the heat supplied (3 \; BTU), C_P the specific heat at constant pressure (0.027 \; BTU/ft^3 ^\circ F), and (T_3 - T_2) is the temperature difference expressed in ^\circ F, while V_S is the quantity of gas which is present at a pressure of P_S = 14.7 \; psi and at a temperature of T_S = 60^\circ F. In order to calculate 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 V_S from (19) in equation (14), and also considering that Q_S = 3 \; BTU, C_P = 0.027 \; BTU/ft^3 ^\circ F, and that 1 \; ft^3 = 1728 \; in^3, we get that:

which leads to:

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

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

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

from which V_3 becomes:

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

Equation (26) leads to:

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

Stage 4

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

we can write:

from which P_4 becomes:

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

From equation (33) we obtain:

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

Equation (35) can also be written as:

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

Hence, we can write that:

from which T_4 becomes:

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

From equation (40) we obtain:

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

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

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

which leads to:

By applying the 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 n_1=1.25, n_2=1.3 (see 48), and that 1 \; ft = 12 \; in, we get the work done W in ft-lb as:

from which we obtain:

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

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

As, in our case, W = 1030.85 \; ft-lb (see 50) and Q_S = 3 \; BTU, and also taking into account that 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\%