Engines and fuels

Otto cycle

• Gasoline engines
• 4-stroke cycle
  1. Intake
  2. Compression
     1. Combustion at the top
  3. Expansion/power
  4. Exhaust

Idealized cycle

• 1$\rightarrow$ 2: ideal compression
  • Adiabatic, PV work required by the engine.
• 2$\rightarrow$ 3: sensible energy added through burning of the fuel/air mixture
  • Combustion assumed to occur quickly, at constant volume. Pressure rises.
• 3$\rightarrow$ 4: ideal expansion
  • Adiabatic, PV work produced by the engine.
• 4$\rightarrow$ 1: heat rejection
  • drop in pressure as the valve is opened at the bottom of the stroke.
  • exhaust and intake are the horizontal blue and green lines.

Net work

• The net work produced is the area under curve 3$$4 minus the area under curve 1$$2.
• Since the compression and expansion are adiabatic, the work is equal to differences in internal energy $u$.
• Work required for compression: $W_c=u_2-u_1$.
• Work done by expansion: $W_e = u_3-u_4$.
• Net work performed by the engine = $W_e-W_c = (u_3-u_4)-(u_2-u_1)$.

Note that $u_2=u_3$ since process 2$$3 is adiabatic and constant volume. The “heat added” step is really sensible energy generated by combustion. But the internal energy of the mixture remains fixed.

Adiabatic compression and expansion.

• Assume ideal gas and adiabatic compression/expansion.

• $du = c_vdT$

  $$c_vdT = -PdV$$

• $P = RT/V$, rearrange

• $R = c_p-c_v$, $\gamma = c_p/c_v$.
• $\rightarrow$ $R/c_v = \gamma -1$.

* Assume $\gamma$ is constant. Then integrate to

* Rewrite as

• Using $P=RT/V$, and expressions for $\gamma$ above, we have

  $$\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{(\gamma-1)/\gamma}$$

• Similarly

  $$P_1V_1^\gamma = P_2V_2^\gamma$$

• If we relax the assumption that $\gamma$ is constant, we have

  $$\ln\frac{V_2}{V_1} = -\int_{T_1}^{T_2}\frac{c_v(T)}{RT}dT$$

Example

• Compute the work and efficiency on a lower heating value basis for an Otto cycle with a stoichiometric methane-air mixture.
• Use a compression ratio (CR) of 10. That is $V_1/V_2 = V_4/V_3 = 10$
• code below requires streams.py and cantera

from streams import streams

strm = streams({"O2":1,"N2":3.76}, {"CH4":1}, 300, 300, 101325, "gri30.yaml")
CR = 10

#----------- state 1: cold, atmospheric reactants
strm.set_gas_mixing_state(strm.ξst)
P1 = strm.gas.P
T1 = strm.gas.T
u1 = strm.gas.int_energy_mass

#----------- state 2: compress reactants
strm.set_gas_state_adiabatic_compression_expansion(1/CR)
P2 = strm.gas.P
T2 = strm.gas.T
u2 = strm.gas.int_energy_mass

#----------- state 3: burnt products
strm.gas.equilibrate("UV")
P3 = strm.gas.P
T3 = strm.gas.T
u3 = strm.gas.int_energy_mass

print(f'NO = {strm.gas.Y[strm.gas.species_index("NO")]*1000000:0f}')

#----------- state 4: expanded products
strm.set_gas_state_adiabatic_compression_expansion(CR)
P4 = strm.gas.P
T4 = strm.gas.T
u4 = strm.gas.int_energy_mass
print(T4, P4)

strm.gas.equilibrate("UV")
print(strm.gas.T, strm.gas.P)
print(f'NO = {strm.gas.Y[strm.gas.species_index("NO")]*1000000:.1f}')

#----------- work and efficiency

W   = (u3-u4) - (u2-u1)   # J/kg mixture
LHV = strm.get_LHV_pCC()  # J/kg fuel

η = W/strm.ξst/LHV        # efficiency; ξst is used to convert W from J/kg mix to J/kg fuel

print(f"η  = {η:.2f}")

print(f'P1 = {P1/101325:.2f} atm,  T1 = {T1:.0f} K')
print(f'P2 = {P2/101325:.2f} atm, T2 = {T2:.0f} K')
print(f'P3 = {P3/101325:.2f} atm, T3 = {T3:.0f} K')
print(f'P4 = {P4/101325:.2f} atm,  T4 = {T4:.0f} K')

NO = 7249.681704
1581.0497304280193 541668.9741244206
1834.455938236 619897.2037123361
NO = 193.1
η  = 0.43
P1 = 1.00 atm,  T1 = 300 K
P2 = 23.02 atm, T2 = 691 K
P3 = 96.24 atm, T3 = 2846 K
P4 = 5.35 atm,  T4 = 1581 K

• A compression ratio of 5 gives $\eta=0.33$
  • $P_2=9$ atm, $T_2=546$ K
• A compression ratio of 10 gives $\eta=0.43$
  • $P_2=23$ atm, $T_2=690$ K
• At high CR, the high $T_2$ can lead to auto-ignition and engine knock, which can cause damage.
• For methane-air, $\gamma=1.37$

Octane rating

• See Wikipedia for lots of information
• The octane number is the percent of iso-octane (by volume) in a mixture of iso-octane and n-heptane, that has the same knock characteristics as the given fuel.
• A higher octane number indicates a fuel that is more resistant to knock.
• There are several different measurement methods, including the “Research Octane Number” and the “Motor Octane Number”, an average of which is used in the U.S.
• It is possible to have an octane number greater than 100.
• Higher octane numbers can tolerate higher compression ratios.
• Elevation matters.

Review article on impact of fuel molecular structure on auto-ignition

Diesel cycle

• 4 strokes
• Compress only air $\rightarrow$ hot
• Then inject fuel $\rightarrow$ autoignition, but the injection timing is controlled.
• No knock issue, allows cheaper fuels.
• High compression radio gives higher efficiency.
  • CR = 12-24.
• In the diagram above, the combustion step 2$\rightarrow$ 3 occurs with some piston travel $\approx$ constant pressure.
• Heat steps $Q$ again refer to release of sensible energy, and exhaust/intake.
• In the code above, change the equilibrate("UV") to equilibrate("HP")

Other cycles

• Brayton cycle (gas turbines)
• Rankine cycle (steam cycle, e.g., coal plants)

Fuels

• Compare diesel fuel and gasoline

• Diesel: about 25% aromatics, $C_{12}H_{23}$, ($C_{10}H_{20}$ - $C_{15}H_{28}$)

• Gasoline: about 35% aromatics, $C_8H_{18}$, ($C_4$ - $C_{12}$)

• $\rho_D\approx 0.85$ kg/L, versus $\rho_G\approx 0.72$ kg/L

• Heating value: $HV_D\approx 38.6$ MJ/L versus $HV_G\approx 34.9$ MJ/L

• Heating value: $HV_D\approx 45.4$ MJ/kg versus $HV_G\approx 48.5$ MJ/kg

• On an equal mass basis, the large differences in efficiency are due to diesel having higher compression ratios.

• Ethanol additives reduces hydrocarbon emissions (volatile organic compounds, VOCs)

Exhaust gas recirculation (EGR)

• $NO_x$ is a pollutant formed at high temperatures.
• To reduce $NO_x$, reduce peak flame temperatures.
• One way to do this is by recirculating cool exhaust gases into the reaction mixture. This increases the heat capacity and reduces the flame temperature.
  • Also reduces efficiency.
• Very common in diesel engines.
  • Up to 50% EGR
• In SI engines, only 5-15% EGR

Recuperation/Regeneration

• Raise flame temperatures by preheating reactants (air).
• Exchange heat with hot combustion products.
• High temperature applications, glass furnaces.

1-(T1/T2)*(T4/T1-1)/(T3/T2-1)

$\displaystyle 0.405768344464695$