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An analysis of Theoretical Internal Combustion Engines Cycles

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Internal Combustion Cycles

The Working Substance is treated as pure air.
Expansions and Compressions are Reversible and Adiabatic.
Heat can be added instantaneously if desired.

Air Standard Cycles :-

Constant Volume or Otto Cycle - Spark Ignition.
Constant Pressure or Joule Cycle - Gas Turbines.
Diesel Cycle - Slow Speed Compression Ignition.

THE CONSTANT VOLUME CYCLE

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The four Stages of this Cycle are:-

Adiabatic Compression 1 - 2
Heating at Constant Volume 2 - 3
Adiabatic Expansion 3 - 4
Cooling the WS to it's Original Condition 4 - 1

$Thermal\; Efficiency\;=\;\frac{Work\;out put}{Heat\;Supplied}$

(1)

$=\;\frac{Q_S\;-\;Q_R}{Q_S}$

(2)

$Heat\;Supplied\;=\;w\;C_V(T_3\;-\;T_2)$

(3)

$Heat\;Rejected\;=\;w\;C_V(T_4\;-\;T_1)$

(4)

$\therefore\;\;\;\;\;Thermal\;\eta \;=\;1\;-\;\frac{(T_4\;-\;T_1)}{(T_3\;-\;T_2)}$

(5)

$But\;\;\;\;\;T_1V_1^{\gamma \;-\;1}\;=\;T_2V_2^{\gamma \;-\;1}$

(6)

$\therefore\;\;\;\;\;T_2\;=\;T_1\left(\frac{V_1}{V_2} \right)^{\gamma \;-\;1}\;=\;T_1\;r^{\gamma \;-\;1}\;\;\;\;where\;\;\;\frac{V_1}{V_2}\;=\;r$

(7)

$Similarly\;\;\;\;\;T_3\;=\;T_4\;r^{\gamma \;-\;1}$

(8)

From equations (7) and (8)

$T_3\;-\;T_2\;=\;r^{\gamma \;-\;1}\;(T_4\;-\;T_1)$

(9)

$\therefore\;\;\;\;\;\frac{(T_4\;-\;T_1)}{(T_3\;-\;T_2)}\;=\;\frac{1}{r^{\gamma \;-\;1}}$

(10)

$\therefore\;\;\;\;\;Thermal\;\eta \;=\;1\;-\;\frac{1}{r^{\gamma \;-\;1}}$

(11)

NOTE "r" is called the Compression Ratio . For Carburettor Engines it's value is about 7. whilst for Fuel Injector Engines this figure is increased to around 10. Using these figures and a value for $\gamma$ of 1.25, the Thermal efficiency of a petrol/Gasoline engine is between 39% and 44%.

THE CONSTANT PRESSURE CYCLE

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The four stages of this Cycle are:-

Adiabatic reversible Compression 1 - 2
Heating at Constant Pressure 2 - 3
Adiabatic reversible Expansion 3 - 4
Cooling at Constant Pressure 4 - 1

$Thermal \;\eta \;=\;1\;-\;\frac{Q_R}{Q_S}$

(12)

$=\;1\;-\;\frac{wC_P(T_4\;-\;T_1)}{wC_P(T_3\;-\;T_2)}\;=\;1\;-\;\frac{(T_4\;-\;T_1)}{(T_3\;-\;T_2)}$

(13)

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For the Adiabatic Compression 1 - 2

$\frac{P_1}{T_1^{\frac{\gamma }{\gamma \;-\;1}}}\;=\;\frac{P_2}{T_2^{\frac{\gamma }{\gamma \;-\;1}}}$

(14)

$\therefore\;\;\;\;\;T_2\;=\;T_1\;J^{\frac{\gamma \;-\;1}{\gamma }}\;\;\;\;\;where\;J\;is\;the\;Pressure\;Ratio$

(15)

Likewise for the Adiabatic expansion 3 - 4

$\therefore\;\;\;\;\;T_3\;=\;T_4\;J^{\frac{\gamma \;-\;1}{\gamma }}$

(16)

$Subtracting\;\;\;\;\;\;\;(T_3\;-\;T_2)\;=\;J^{\frac{\gamma \;-\;1}{\gamma }}(T_4\;-\;T_1)$

(17)

$\therefore\;\;\;\;\;Thermal\;\eta \;=\;1\;-\;\frac{1}{J^{\frac{\gamma \;-\;1}{\gamma }}}$

(18)

THE DIESEL CYCLE

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The four stages of this cycle are :-

An Adiabatic reversible Compression. 1 - 2
Constant Pressure Heating. 2 - 3
Adiabatic reversible expansion. 3 - 4
Constant Volume cooling 4 - 1

$Thermal\;\eta \;=\;1\;-\;\frac{Heat\;rejected}{Heat\;Supplied}$

(19)

$Heat\;rejected\;=\;wC_V(T_4\;-\;T_1)$

(20)

$Heat\;Supplied\;=\;wC_P(T_3\;-\;T_2)$

(21)

$\therefore\;\;\;\;\;Thermal\;\eta \;=\;1\;-\;\frac{1}{\gamma }\times \frac{(T_4\;-\;T_1)}{(T_3\;-\;T_2)}$

(22)

For Stage 1 - 3

$T_1V_1^{\gamma \;-\;1}\;=\;T_2V_2^{\gamma \;-\;1}$

(23)

$\therefore\;\;\;\;\;T_2\;=\;T_1\left(\frac{V_1}{V_2} \right)^{\gamma \;-\;1}\;\;\;\;\;where\;r\;is\;the\;compression\;ratio\;\frac{V_1}{V_2}$

(24)

For Stage 2 - 3

$\frac{V_2}{T_2}\;=\;\frac{V_3}{T_3}$

(25)

$\therefore\;\;\;\;\;T_3\;=\;T_2\frac{V_3}{V_2}$

(26)

$=\;T_2\;\beta \;\;\;\;\;\;Where\;\beta \;is\;the\;expansion\;ratio\;during\;heating$

(27)

$=\;T_1\;r^{\alpha \;-\;1}\;\beta$

(28)

For Stage 3 - 4

$T_3V_3^{\gamma \;-\;1}\;=\;T_4\;V_4^{\gamma \:-\:1}$

(29)

$\therefore\;\;\;\;\;T_4\;=\;T_3\;\left(\frac{V_3}{V_4} \right)^{\gamma \;-\;1}$

(30)

$\therefore\;\;\;\;\;T_4\;=\;T_1\;r^{\gamma \;-\;1}\;\beta \;\left(\frac{V_3}{V_4} \right)^{\gamma \;-\;1}$

(31)

$=\;T_1\;r^{\gamma \;-\;1}\;\beta \;\left(\frac{V_3}{V_2} \frac{V_2}{V_4}\right)^{\gamma \;-\;1}$

(32)

$=\;T_1\;r^{\gamma \;-\;1}\;\beta \;\left(\beta \right)^{\gamma \;-\;1}\;\left(\frac{1}{r} \right)^{\gamma \;-\;1}$

(33)

$Thus\;\;\;\;\;T_4\;=\;T_1\;\beta ^\gamma$

(34)

$\therefore\;\;\;\;\;Thermal\;\eta \;=\;1\;-\;\frac{1}{\gamma }\;\left(\frac{\beta ^\gamma \;-\;1}{\beta \;-\;1} \right)\times \frac{T_1}{T_1\;r^{\gamma \;-\;1}}$

(35)

$=\;1\;-\;\frac{1}{\gamma }\left(\frac{\beta ^\gamma \;-\;1}{\beta \;-\;1} \right)\times \frac{1}{r^{\gamma \;-\;1}}$

(36)

THE AIR STANDARD EFFICIENCY

Unless otherwise stated it can be assumed that the Air Standard efficiency will be that of the Constant Volume Cycle.

Example 1

An engine operates under a diesel cycle with a top pressure of 642 psi, The compression ratio is 18 : 1 and the air at the start of compression is at 15 psi and 100 degrees F. 3 BTU are added during combustion and expansion in accordance to the law $P\;V^{1,25}\;-\;Constant$ . If the clearance volume is $3\;in.^3$ and the specific heat at Constant pressure during combustion is 0.027 BTU/cu.ft. Find ignoring the fuel mass.

1) Pressure; Volume;and Temperature at each change point of the cycle.

2)The work output of the cycle

3 The Thermal efficiency of the cycle.

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For convenience the following chart has been produced showing number given in the question in Roman numbers and calculate values in italics.

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Between 1 - 2

$\frac{P_1V_1}{T_1}\;=\;\frac{P_2V_2}{T_2}$

(37)

$\therefore\;\;\;\;\;T_2\;=\;\frac{P_2V_2}{P_1V_1}\;T_1$

(38)

$=\;\frac{642}{15}\times \frac{3}{54}\times 560\;=\;1335^0R$

(39)

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Between 2 - 3

Normally the heat supplied is shown as $w\;C_P\;dt$ but in this case;_

$Heat\;Supplied\;=\;V\;C_P\;\Delta T$

(40)

Where $V_S$ is the quantity of Gas present at 14.7 psi and $560^0R$

C_P is the Specific heat at Constant Pressure.

$V_S\;=\;54\times \frac{15}{14.7}\times \frac{(60\;+\;460)}{(100\;+\;460)}$

(41)

$=\;51.1\; in^3\;at\; 14,7 \;psi\;\;\;\; and \;\;\;\;60^0F.$

(42)

$\therefore\;\;\;\;\;3\;=\;\frac{51.1}{144\times 12}\times 0.025\left(T_3\;-\;T_2 \right)$

(43)

$\therefore\;\;\;\;\;(T_3\;-\;T_2)\;=\;\frac{3\times 1728}{51.1\times 0.027}\;=\;3750^0F.$

(44)

$\therefore\;\;\;\;\;T_3\;=\;3750\;+\;1335\;=\;5085^0R.$

(45)

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Also between 2 - 3

$\frac{V_2}{T_2}\;=\;\frac{V_3}{T_3}$

(46)

$\therefore\;\;\;\;\;V_3\;=\;V_2\times \frac{T_3}{T_2}\;=\;11.4\;ft^3.$

(47)

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For the expansion 3 - 4

$PV^{1.25}\;=\;Constant$

(48)

$TV^{0.25}\;=\;Constant$

(49)

$P_4\;=\;P_3\times \left(\frac{V_3}{V_2} \right)^{1.25}\;=\;642\times \left(\frac{11.4}{5.4} \right)^{1.25}\;=\;91.7 psi.$

(50)

$T_4\;=\;T_3\;\left(\frac{11.4}{54} \right)^{0.25}\;=\;\frac{5085}{(4.73)^{0.25}}\;\;3450^0R.$

(51)

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To Find the Work output of the Cycle.

$Work\;out put=(P_3V_3-P_2V_2)+\left(\frac{P_3V_3-P_4V_4}{n_1-1} \right)-\left(\frac{P_2V_2-P_1V_1}{n_2\:-\;1} \right)$

(52)

To find $n_2$

$P_1V_1^{n_2}=P_2V_2^{n_2}$

(53)

$\therefore\;\;\;\;\;n_2\;=\;\frac{Ln\frac{P_2}{P_1}}{Ln\frac{V_1}{V_2}}$

(54)

$\therefore\;\;\;\;\;n_2\;=\;\frac{Ln\frac{642}{15}}{Ln18}\;=\;1.3$

(55)

$Work\;Done\;Cycle\;=\; \left[(642\times 11.4)-(642\times 3) \right]+\left[\frac{(642\times 11.4)-(91.7\times 54)}{0.25} \right]-\left[\frac{(642\times 3)-(15\times 54)}{0.3} \right]$

(56)

$=\;11070\times \frac{144}{1728}\;ft.lbs.\;=\;922.5\.ft.lbs.$

(57)

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The Thermal efficiency of the Cycle.

$Thermal\;\eta \;=\;\frac{Work\;out put}{Heat\;Supplied}$

(58)

$=\;\frac{9225}{3\times 778}\;=\;0.395\;\;\;\;\;i.e. 39.5\%$

(59)

Internal Combustion Engine Cycles

Internal Combustion Cycles

Air Standard Cycles :-

THE CONSTANT VOLUME CYCLE

THE CONSTANT PRESSURE CYCLE

THE DIESEL CYCLE

THE AIR STANDARD EFFICIENCY

Example 1