Internal Combustion Engines

Internal combustion (IC) engines are the second big block of the Industrial and Power Plant Engineering paper. The theory rests on three air-standard cycles (Otto, Diesel, Dual), and the practice rests on a handful of power and efficiency formulas. If you can compute an air-standard efficiency from a compression ratio and turn a torque and speed into brake power, you have most of the marks. Work in SI: pressure in kPa or Pa, power in kW, torque in N-m, speed in rev/s or rad/s.

The three air-standard cycles

Each cycle is an idealization of a real engine, using air as the working fluid with a constant specific-heat ratio k = 1.4.

• Otto cycle (spark-ignition, gasoline): heat is added at constant volume. Its efficiency depends only on compression ratio:

  eta_Otto = 1 minus 1/r^(k minus 1)
  

• Diesel cycle (compression-ignition): heat is added at constant pressure. Its efficiency adds a cut-off correction:

  eta_Diesel = 1 minus (1/r^(k minus 1)) x [(rc^k minus 1) / (k(rc minus 1))]
  

  where rc is the cut-off ratio (the volume ratio during constant-pressure heat addition).

• Dual (mixed) cycle: heat is added partly at constant volume and partly at constant pressure, the most realistic model of a modern high-speed diesel. Its efficiency lies between the Otto and Diesel values.

At the same compression ratio, the Otto cycle is the most efficient of the three, but real spark-ignition engines are limited to low compression ratios (about 8 to 11) by knock, while diesels run at 14 to 22, which is why diesels are more efficient in practice.

Compression ratio and air-standard efficiency

The compression ratio is the ratio of cylinder volume at bottom dead center to that at top dead center:

r = (Vc + Vs) / Vc

where Vc is the clearance (top dead center) volume and Vs is the swept (displacement) volume. A cylinder with Vc = 0.0001 m3 and Vs = 0.0009 m3 has r = (0.0001 + 0.0009)/0.0001 = 10.

Worked example (Otto): with r = 9 and k = 1.4, eta = 1 minus 1/9^0.4. Since 9^0.4 = 2.408, eta = 1 minus 0.415 = 0.585, or 58.5%. At r = 6 the same formula gives 51.2%: efficiency climbs with compression ratio.

Worked example (Diesel): with r = 16, cut-off ratio rc = 2, and k = 1.4, eta = 1 minus (1/16^0.4)[(2^1.4 minus 1)/(1.4(2 minus 1))]. Here 16^0.4 = 3.031, 2^1.4 = 2.639, so eta = 1 minus (0.330)(1.639/1.4) = 1 minus 0.330 x 1.171 = 1 minus 0.386 = 0.614, or 61.4%.

Indicated and brake power

Two power levels matter:

• Indicated power (IP) is the power developed inside the cylinders by the gas on the piston:

  IP = pmi x L x A x n x N_c
  

  where pmi is the indicated mean effective pressure, L the stroke, A the piston area, n the number of cylinders, and N_c the number of power strokes per second per cylinder (for a four-stroke engine, N_c = N/2 with N in rev/s).

• Brake power (BP) is the useful power at the output shaft:

  BP = 2 x pi x N x T = T x omega
  

  with N in rev/s, T the torque in N-m, and omega in rad/s.

Worked example: an engine delivers 200 N-m of torque at 2,400 rpm. Then omega = 2 x pi x 2,400/60 = 251.3 rad/s, so BP = 200 x 251.3 = 50,265 W, about 50.3 kW.

The gap between IP and BP is the friction power (FP): IP = BP + FP.

Mean effective pressure

The mean effective pressure (MEP) is the constant pressure that, acting over the swept volume, would do the same net work as the real cycle:

MEP = net work per cycle / swept volume

Worked example: a cylinder does 900 J of net work per cycle in a swept volume of 0.0006 m3. MEP = 900/0.0006 = 1,500,000 Pa = 1,500 kPa (1.5 MPa). MEP is a fair way to compare engines of different size because it removes the effect of displacement.

Thermal, mechanical, and overall efficiency

• Brake thermal efficiency is useful shaft power over the fuel's energy rate:

  eta_bth = BP / (mf x HHV)
  

  Worked example: BP = 50 kW, fuel rate mf = 12 kg/h = 0.003333 kg/s, HHV = 44,000 kJ/kg. Fuel power = 0.003333 x 44,000 = 146.7 kW, so eta_bth = 50/146.7 = 0.341, or 34.1%.

• Mechanical efficiency is brake power over indicated power:

  eta_mech = BP / IP
  

  If BP = 50 kW and IP = 60 kW, then eta_mech = 50/60 = 0.833, or 83.3%. The missing 10 kW is friction power.

Specific fuel consumption

Specific fuel consumption (SFC) is the fuel burned per unit of output, the standard measure of an engine's thirst:

SFC = mf / BP

Worked example: mf = 12 kg/h and BP = 50 kW give SFC = 12/50 = 0.24 kg per kWh. A lower SFC means a more economical engine, and SFC is inversely related to brake thermal efficiency.

Exam-day strategy

• Otto efficiency needs only r: eta = 1 minus 1/r^0.4 for air. Diesel adds the cut-off correction, which always makes it less efficient than an Otto of the same r.
• Compression ratio is (Vc + Vs)/Vc: total over clearance, not swept over clearance.
• Brake power from torque is BP = 2 x pi x N x T. Convert rpm to rev/s (divide by 60) first.
• Keep IP, BP, and FP straight: IP = BP + FP, and mechanical efficiency is BP/IP.
• SFC = mf/BP; brake thermal efficiency = BP/(mf x HHV). Watch units: put mf in kg/s and HHV in kJ/kg to get power in kW.