## Diesel Cycle – Problem with Solution

**Diesel Cycle – Problem with Solution**

Let assume the Diesel 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**. Also the cut-off ratio V

_{3}/V

_{2}, which is the ratio of volumes at the end and start of the combustion phase.

In this example let assume the Diesel cycle with compression ratio of CR = 20 : 1 and cut-off ratio α = 2. The air is at 100 kPa = 1 bar, 20 °C (293 K), and the volume of the chamber is 500 cm³ prior to the compression stroke.

- 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**_{2}**the pressure p**_{2}**the temperature T**_{3}**the amount of heat added by burning of fuel-air mixture****the thermal efficiency of this cycle****the MEP**

**Solution:**

1)

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**

**2)**

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

- V
_{1}= V_{4}= V_{max}= 500×10^{-6}m^{3}(0.5l) - V
_{2}= V_{min}= V_{max}/ CR = 25 ×10^{-6}m^{3}

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

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

thus

T_{2} = T_{1} . CR^{κ – 1} = 293 . 20^{0.4} = 971 K

3)

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

**p**** _{2}** = mR

_{specific}T

_{2}/ V

_{2}= 5.95×10

^{-4}x 287.1 x 971 / 25 ×10

^{-6}= 6635000 Pa =

**66.35 bar**

4)

Since process 2 → 3 occurs at constant pressure, the ideal gas equation of state gives

T_{3} = (V_{3}/V_{2}) x T_{2} = 1942 K

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 isobaric process, which states:

Q_{add} = mc_{p} (T_{3} – T_{2}) = 5.95×10^{-4} x 1010 x 971 = **583.5 J**

5)

Thermal efficiency for this Diesel cycle:

As was derived in the previous section, the thermal efficiency of Diesel cycle is a function of the compression ratio, the cut-off ratio and κ:

- η
_{Diesel}is the maximum thermal efficiency of a Diesel cycle - α is the cut-off ration V
_{3}/V_{2}(i.e. the ratio of volumes at the end and start of the combustion phase) - CR is the compression ratio
**κ = c**_{p}**/c**_{v}**= 1.4**

For this example:

η_{Diesel} = 0.6467 = 64.7%

6)

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}* = **583.5 x 0.6467 = 377.3 J*

*MEP **= 377.3 / (*500×10^{-6} – 25 ×10^{-6}*) = 794.3 kPa = 7.943 bar*

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