## Otto Cycle – Problem with Solution

Let assume the **Otto cycle**, which is the one of most common **thermodynamic cycles** that can be found in **automobile engines**. One of key parameters of such engines is the change in volumes between top dead center (TDC) to bottom dead center (BDC). The ratio of these volumes (*V*_{1}* / V** _{2}*) is known as the

**compression ratio**.

The **compression ratio** in a gasoline-powered engine will usually not be much higher than 10:1 due to potential engine knocking (autoignition) and not lower than 6:1. For example, some sportscar engines can have compression ratio up to 12.5 : 1 (e.g. Ferrari 458 Italia).

In this example let assume an Otto cycle with **compression ratio** of **CR = 9 : 1**. The intake air is at 100 kPa = 1 bar, 20 °C, and the volume of the chamber is 500 cm³ prior to the compression stroke. The temperature at the end of adiabatic expansion is T_{4} = 800 K.

- Specific heat capacity at constant pressure of air at atmospheric pressure and room temperature:
**c**_{p}**= 1.01 kJ/kgK.** - Specific heat capacity at constant volume of air at atmospheric pressure and room temperature:
**c**_{v}**= 0.718 kJ/kgK.** **κ = c**_{p}**/c**_{v}**= 1.4**

**Calculate:**

**the mass of intake air****the temperature T**_{3}**the pressure p**_{3}**the amount of heat added by burning of fuel-air mixture****the thermal efficiency of this cycle****the MEP**

**Solution:**

1) **the mass of intake air**

At the beginning of calculations we have to determine the amount of gas in the cylinder before the compression stroke. Using the ideal gas law, we can find the mass:

*pV = mR*_{specific}*T*

where:

*p*is the absolute pressure of the gas*m*is the mass of substance*T*is the absolute temperature*V*is the volume*R*is the specific gas constant, equal to the universal gas constant divided by the molar mass (M) of the gas or mixture. For dry air R_{specific}_{specific}= 287.1 J.kg^{-1}.K^{-1}.

therefore

**m** = p_{1}V_{1}/R_{specific}T_{1} = (100000 × 500×10^{-6} )/(287.1 × 293) = **5.95×10**^{-4}** kg**

**In this problem all volumes are known: **

- V
_{1}= V_{4}= V_{max}**=**500×10^{-6}m^{3}(0.5l) - V
_{2}= V_{3}= V_{min}= V_{max}/ CR = 55.56 ×10^{-6}m^{3}

Note that (V_{max} – V_{min}) x number of cylinders = total engine displacement.

2) **the temperature T**_{3}

Since the process is adiabatic, we can use the following p, V, T relation for adiabatic processes:

thus

**T _{3}** = T

_{4}. CR

^{κ – 1}= 800 . 9

^{0.4}=

**1926 K**

3) **the pressure p**_{3}

Again, we can use the ideal gas law to find the pressure at the beginning of the power stroke as:

**p**** _{3}** = mR

_{specific}T

_{3}/ V

_{3}= 5.95×10

^{-4}x 287.1 x 1926 / 55.56 ×10

^{-6}= 5920000 Pa =

**59.2 bar**

4) **the amount of heat added**

To calculate the amount of heat added by burning of fuel-air mixture, Q_{add}, we have to use the first law of thermodynamics for isochoric process, which states the Q_{add} = ∆U, therefore:

**Q _{add} = mc_{v} (T_{3} – T_{2})**

the temperature at the end of the compression stroke can be determined using the p, V, T relation for adiabatic processes between points **1 → 2.**

**T _{2}** = T

_{1}. CR

^{κ – 1}= 293 . 9

^{0.4}=

**706 K**

then

**Q**** _{add}** = mc

_{v}(T

_{3}– T

_{2}) = 5.95×10

^{-4}x 718 x 1220 =

**521.2 J**

5) **the thermal efficiency**

**Thermal efficiency** for an Otto cycle:

As was derived in the previous section, the thermal efficiency of an Otto cycle is a function of compression ratio and κ:

6) **the mean effective pressure**

The **MEP** was defined as:

It this equation the displacement volume is equal to V_{max} – V_{min}. The net work for one cycle can be calculated using the heat added and the thermal efficiency:

*W*_{net}* = **Q*_{add}* . η*_{Otto}* = **521.2 x 0.5847 = 304.7 J*

*MEP**= 304.7 / (*500×10

^{-6}– 55.56 ×10

^{-6}

*) = 685.6 kPa =*

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