Otto Cycle | Its Important Relations and Formulas

Otto Cycle Definition

“An Otto cycle is an ideal thermodynamic cycle that explains the working of a typical spark ignition piston engine and this cycle specifically explains, what happens if mass of gas is subjected to changes due to pressure, temp, volume, heat input, and release of heat.”

TheOtto cycle engine | Valve timing diagram

1. Inlet valve opens at 5-100 before the Top Dead Center. This is to ensure that the inlet should open fully when piston reaches at TDC and fresh charge start entering to cylinder as early as possible after TDC.
2. Suction valve close at 20 – 300 after Bottom dead center BDC to take the advantage of momentum of moving gases.
3. The spark takes place 30 – 400 before TDC. This is to allow time delay between spark and completion of combustion.
4. Pressure at the end of power stroke is above atmospheric which increases the work to expel the exhaust gases. So exhaust valve opens at 20 – 300 before BDC so that at BDC pressure reduces to atmospheric pressure and useful work can be saved.
5. The exhaust valve closes at 15 – 200 after TDC so that inertia of exhaust gas has a tendency to to scavenge the cylinder which will increase volumetric efficiency.

Otto cycle efficiency| thermal efficiency of Otto CycleFormula

The efficiency of Otto cycle is specified by

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

Where r = compression ratio.

Otto, Diesel and Dual cycle | Comparison

Case 1: For similar compression ratio and similar heat i/p this relationship will be

[Qin]otto = [Qin]Diesel.

[QR]otto< [QR]Diesel.

$\\\eta=1-\frac{Q_R}{Q_{in}}\\\\ \eta_D<\eta_O$

In this case of same compression ratio and equal heat input it will be

$\eta_D<\eta_{dual}<\eta_O$

Case 2: In this case of, same compression ratio and same heat-rejection, this relationship will be

[Qin]otto> [Qin]Diesel.

[QR]otto= [QR]Diesel.

$\\\eta=1-\frac{Q_R}{Q_{in}}\\\\ \eta_D<\eta_O$

In this case of, same compression ratio and same heat-rejection.

$\eta_D<\eta_{dual}<\eta_O$

Case 3: In this case of, same Maximum Temperature and same heat-rejection.

[QR]otto= [QR]Diesel

[Qin]Diesel>[Qin]otto

$\\\eta=1-\frac{Q_R}{Q_{in}}\\\\ \eta_D>\eta_O$

For same Maximum Temperature and same heat rejection

$\eta_D>\eta_{dual}>\eta_O$

Compression ratio of Otto cycle

Compression ratio of Otto cycle is defined as the ratio of volume before expansion to volume after expansion

$r=\frac{V_s+V_c}{V_s}=\frac{V_1}{V_2}$

Where Vs = Swept volume of cylinder

Vc = Clearance volume of the cylinder

In this cycle Compression ratio is generally 6 – 10. It is limited to 10 because of knocking in the engine.

Mean effective pressure formula for Otto cycle

Usually Pressure inside the cylinder on an IC engine is continuously changing; mean effective pressure is an imaginary pressure which is assumed to be constant throughout the process.

$P_m=\frac{P_1 r(r_p-1)(r^{\gamma-1}-1)}{(\gamma-1)(r-1)}$

Where rp = Pressure ratio = P3/P2 = P4/P1

Otto cycle analysis | Otto cycle calculations | Otto cycle efficiency derivation

Consider an air standard Otto cycle with initial Pressure, Volume and temperature as P1, V1, T1 respectively.

$\frac{T_2}{T_1}=[\frac{V_1}{V_2}]^{\gamma-1}$

Where,

r is the compression ratio.

Process 2 -3: Heat addition at constant Volume is calculated as,

Qin = m Cv [T­3-T2].

Process 3-4: Reversible adiabatic expansion is calculated as

$\frac{T_3}{T_4}=[\frac{V_4}{V_3}]^{\gamma-1}=r^{\gamma-1}$

Process 4 -1: Heat-rejection at constant Volume will be

QR = m Cv [T­4-T1]

Work done = Qin – QR.

Efficiency of the Otto cycle is represented as.

$\eta=1-\frac{Q_R}{Q_{in}}$

$\\\eta=1-\frac{[T_4-T_1]}{[T_3-T_2]}\\\\ \frac{T_2}{T_1}=\frac{T_3}{T_4}\\\\ \frac{T_4}{T_1}=\frac{T_3}{T_2}\\\\ \eta=1-\frac{1}{r^{\gamma-1}}$

Where r = compression ratio.

Working of Two stroke Engine

Two strokes’ engines work on both Otto cycle as well as diesel cycle.

Brayton cycle vs Otto cycle

• This cycle has more thermal efficiency in comparison to diesel and dual cycle for identical compression ratio and equal heat input rate and same compression ratio and same heat rejection.
• This cycle engine required less maintenance and are simple and light-weight in design.
• For complete combustion pollutant emissions are low for Otto engines.

• Has lower compression ratio thus it is Poor at moving heavy loads at low speed.
• Will not be able to withstand higher stresses and strains as compared with diesel engine

Q.1] A spark Ignition engine designed to have a compression ratio of 10. This is operating at low temperature and pressure at value 2000C and 200 kilopascals respectively. If Work O/P is 1000 kilo-Joule/kg, compute maximum possible efficiency and mean effective pressure.

Efficiency of this cycle is given by

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

Where r = compression ratio = 10

$\eta =1-\frac{1}{10^{1.4-1}}=0.602=60.2\%$

For compression process

$\frac{T_2}{T_1}=r^{\gamma-1}$

$\frac{T_2}{473}=10^{1.4-1}$

$T_2=1188 \;K$

For expansion process, we can assume that

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

$\frac{T_3}{T_4}=10^{1.4-1}$

$T_3=2.512T_4$

Net work done can be computed by the formula

$W=C_v [T_3-T_2 ]-C_v [T_4-T_1]$

$\\1000=0.717*[473-1188+T_3-T_4]\\\\ 1000=0.717*[473-1188+2.512 T_4-T_4]\\\\ T_4=1395 K$

$T_3=2.512*1395=3505 K$

According to ideal gas theory, we know

P1v1 = RT1

v1=(RT1)/(P1)=(0.287*473)/200=0.6788 m3/kg

$mep=\frac{W}{v_1-v_2}=\frac{1000}{0.6788-\frac{0.6788}{10}}=1636.87\;kPa$

Q.2] what will be the effect on the efficiency of an Otto cycle having a compression ratio 6, if Cv increases by 20%. For the purpose of calculation Assume, that  Cv is 0.718 kJ/kg.K.

$\\\frac{\mathrm{d} C_v}{C_v}=0.02\\\\ \eta=1-\frac{1}{r^{\gamma -1}}=1-\frac{1}{6^{1.4 -1}}=0.511\\\\ \gamma -1=\frac{R}{C_v}\\\\ \eta=1-[\frac{1}{r}]^\frac{R}{C_v}$

Taking log on both sides

$ln(1-\eta)=\frac{R}{C_v} ln\frac{1}{r}$

Differentiating both sides

$\\\frac{d\eta}{1-\eta}=\frac{-R}{C_v^2}*dC_v*ln[1/r]\\\\ \frac{d\eta}{1-\eta}=\frac{-R}{C_v}*\frac{dC_v}{C_v}*ln[1/r]\\\\ \frac{d\eta}{\eta}=\frac{1-\eta}{\eta}*\frac{-R}{C_v}*\frac{dC_v}{C_v}*ln[1/r]\\\\ \frac{d\eta}{\eta}=\frac{1-0.511}{0.511}*\frac{-0.287}{0.718}*0.02*ln[1/6]\\\\ \frac{d\eta}{\eta}=-0.0136\\\\ \frac{d\eta}{\eta}*100=-0.0136*100=-1.36\%$

I.e. If Cv increases by 2% then η decrease by 1.36%.

What is the difference between Otto and Diesel cycle?

In Otto cycle heat addition takes place at constant volume while in diesel cycle, heat addition at constant pressure takes place and  Otto-cycle has lower compression ratio below 12 while, diesel cycle has higher compression ratio up to 22.  Otto-cycle uses spark plug for ignition while diesel cycle needs no assistance for ignition. Otto-cycle has lower efficiency as compared to diesel cycle.

Which fuel is used in Otto cycle?| What is 4-stroke fuel?

Generally Petrol or gasoline mixed with 3-5% ethanol is used in Otto engine. In air standard Otto-cycle, air is assumed to be as a fuel.

Which is more efficient Otto or Diesel cycle?

The normal range of compression ratio for diesel cycle is 16-20 while in Otto-cycle compression ratio is 6 – 10 and due to higher compression ratio used in diesel cycle, the efficiency of diesel cycle is greater than Otto-cycle.

How does Otto cycle works?

1. Inlet valve opens at 5-100 before the Top Dead Center. This is to ensure that the inlet should open fully when piston reaches at TDC and fresh charge start entering to cylinder as early as possible after TDC.
2. Suction valve close at 20 – 300 after Bottom dead center BDC to take the advantage of momentum of moving gases.
3. The spark takes place 30 – 400 before TDC. This is to allow time delay between spark and completion of combustion.
4. Pressure at the end of power stroke is above atmospheric which increases the work to expel the exhaust gases. So exhaust valve opens at 20 – 300 before BDC so that at BDC pressure reduces to atmospheric pressure and useful work can be saved.
5. The exhaust valve closes at 15 – 200 after TDC so that inertia of exhaust gases tends to scavenge the cylinder which will increase volumetric efficiency.

$\frac{T_2}{T_1}=[\frac{V_1}{V_2}]^{\gamma-1}=r^{\gamma-1}$

Where r = compression ratio

Process 2 -3: Heat additions at constant Volume

Qin = m Cv [T­3-T2]

$\frac{T_3}{T_4}=[\frac{V_4}{V_3}]^{\gamma-1}=r^{\gamma-1}$

Process 4 -1: Heat-rejection at constant Volume will be

QR = m Cv [T­4-T1]

Work done = Qin – QR.

Efficiency of the Otto-cycle is represented as.

$\eta=1-\frac{Q_R}{Q_{in}}$

$\\\eta=1-\frac{[T_4-T_1]}{[T_3-T_2]}\\\\ \frac{T_2}{T_1}=\frac{T_3}{T_4}\\\\ \frac{T_4}{T_1}=\frac{T_3}{T_2}\\\\ \eta=1-\frac{1}{r^{\gamma-1}}$

Where r = compression ratio.

Otto cycle formula

The efficiency of Otto-cycle is given by the equation

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

Where r = compression ratio = 10

Otto cycle with polytropic process example

An SI engine has a compression ratio 8 while operating with low temperature of 3000C and a low pressure of 250 kPa. If Work o/p is 1000 kilo-Joule/kg, then compute highest efficiency. The compression and expansion takes place polytropically with polytropic index (n = 1.33).

Solution: The efficiency of Otto- cycle is given by the equation

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

Here γ = n

$\eta =1-\frac{1}{r^{n-1}}=1-\frac{1}{8^{1.33-1}}=49.65\%$

Why the Otto cycle is known as a constant volume cycle?

For this cycle, heat-addition and rejection happens at the fixed volume and the amount of work done is proportionate on heat addition and heat rejection rate, because of this reason Otto-cycle is termed as constant volume cycle.

What are the limitations of the Otto cycle?

• It has lower compression ratio thus it is Poor at moving heavy loads at low speed.
• Cannot withstand higher stresses and strains as compared with diesel engine.
• Overall fuel efficiency is lower than diesel cycle.

Are two stroke engines considered to be Otto cycle engines?

Two strokes engines work on both Otto-cycle as well as diesel cycle. The working of 2-stroke engine is given below:

1. Piston moves down and useful power is obtained. The downward motion of piston compresses the fresh charge stored in crankcase.
2. Near the end of expansion stroke, the piston will reveal the exhaust-port at first. Then the cylinder pressure will drop to atmospheric pressure as during that time combustion materialwill leave from the cylinder.
3. Further motion of piston reveals the transfer port allowing the slightly compressed charge in crank case to be entered the engine’s cylinder.
4. Projection in piston prevents the fresh charge from passdirectly to the exhaust port and scavenging the combustion materials.
5. When piston move up from bottom dead centre to top dead center and the transfer port closes at first then exhausts port will close and compression will happen. At the same time vacuum is created in crankcase and fresh charge enter into crankcase for next cycle.

Why is the Atkinson cycle more efficient even though it produces lower compression and pressure than the Otto cycle?

In Atkinson cycle, for isentropic expansion process in Otto cycle is further allowed to proceed and extend to lower cycle pressure in order to increase the work output and we know that efficiency increases for increase in work produced. That is, why the Atkinson cycle is more efficient even though it produces lower compression and pressure than the Otto cycle.

What is the compression ratio of the Otto cycle

Compression ratio of this cycle is elaborated  as

$r=\frac{V_s+V_c}{V_s}=\frac{V_1}{V_2}$

Where,

Vs = Swept volume of cylinder.

Vc = Clearance volume of the cylinder.

Generally in Otto cycle compression ratio is 6 – 10. It is limited to 10 because of knocking in the engine.

Otto cycle vs diesel cycle efficiency

The normal range of compression ratio for diesel cycle is 16-20 while in Otto cycle compression ratio is 6 – 10 and for more compression ratio used in diesel cycle, the efficiency of diesel cycle is greater than Otto cycle.

Case 1: For same compression ratio and exactly identical heat input, the relationship will be

[Qin]otto = [Qin]Diesel.

[QR]otto< [QR]Diesel.

$\\\eta=1-\frac{Q_R}{Q_{in}}\\\\ \eta_D<\eta_O$

In this case of same compression ratio and equal heat input it will be

$\eta_D<\eta_{dual}<\eta_O$

Case 2: In this case of, same compression ratio and same heat-rejection, this relationship will be

[Qin]otto> [Qin]Diesel.

[QR]otto= [QR]Diesel.

$\\\eta=1-\frac{Q_R}{Q_{in}}\\\\ \eta_D<\eta_O$

In this case of, same compression ratio and same heat-rejection.

$\eta_D<\eta_{dual}<\eta_O$

Case 3: In this case of, same Maximum Temperature and same heat-rejection.

[QR]otto= [QR]Diesel

[Qin]Diesel>[Qin]otto

$\\\eta=1-\frac{Q_R}{Q_{in}}\\\\ \eta_D>\eta_O$

For same Maximum Temperature and same heat rejection

$\eta_D>\eta_{dual}>\eta_O$

Under which condition the efficiency of Brayton cycle and Otto cycle are going to be equal.

The efficiency of Otto cycle is given by the equation

Solution: The efficiency of Otto cycle is given by the equation

$\eta_o =1-\frac{1}{r^{\gamma-1}}$

r = compression ratio = V1/V2

The efficiency of Brayton cycle is given by the equation

$\eta_B =1-\frac{1}{r^{\gamma-1}}$

r = compression ratio = V1/V2

For same compression ratio of the Brayton and Otto cycle, their efficiency will be equal.