MECH3260 Chap.6 Power Cycles: Otto, Diesel, Brayton and Rankine
Power Cycles: Otto, Diesel, Brayton and Rankine
Week 6 covers the four benchmark idealised cycles behind real thermal machines: Otto for spark-ignition engines, Diesel for compression-ignition engines, Brayton for gas turbines and Rankine for steam plant.
Each cycle's efficiency turns on one governing parameter, and the chapter pairs every formula with the practical limit that stops the parameter being pushed further, from spark knock to turbine metallurgy to blade erosion by wet steam.
It also covers the performance measures a question is more likely to hand you than a temperature, namely the cut-off ratio and the mean effective pressure, and the five standard ways of lifting a steam plant's efficiency: lower condenser pressure, superheat, higher boiler pressure, reheat and regenerative feedwater heating. All five are the same move in different hardware, and combined-cycle plant is the limiting case of it.
What this chapter covers
- 01
Otto cycle: constant-volume heat addition, eta = 1 - 1/r^(k-1), efficiency depends on compression ratio r alone
- 02
Diesel cycle: constant-pressure heat addition with cut-off ratio r_c; ideal eta always lower than Otto at the same r, but Diesel tolerates a much higher r in practice (no knock limit)
- 03
Brayton cycle: constant-pressure heat addition and rejection in a flow system, eta = 1 - 1/rp^((k-1)/k), independent of turbine-inlet temperature T3
- 04
Regeneration in the Brayton cycle: recovering exhaust heat to preheat compressor discharge, worthwhile at low pressure ratio and moderate T3 (stationary power) but not at high rp (aircraft)
- 05
Rankine cycle: raising average TH (superheat, limited by metallurgy to ~620°C) or lowering average TL (lower condenser pressure, at the cost of turbine-exit moisture) both raise eta
Otto-cycle efficiency and net specific work
- +2eta = 1 - 1/9.5^0.4 = 1 - 1/2.4609 = 0.5936 = 59.4%.
- +2w_net = eta * q_in = 0.5936 x 750 = 445.2 kJ/kg.
- +2Cross-check: q_out = 750 - 445.2 = 304.8 kJ/kg; eta = 1 - 304.8/750 = 0.5936 -- matches.
- +1Interpret: nearly 60% of heat added converts to net work in the ideal cycle; a real engine recovers substantially less.
Key terms
- Compression ratio
- The ratio of cylinder volume before compression to volume after it. It is the single parameter in the ideal Otto efficiency, and the knock limit is what stops a spark-ignition engine from raising it indefinitely.
- Cut-off ratio
- The volume where constant-pressure combustion finishes divided by the volume where it began, needed because the Diesel cycle burns at roughly constant pressure. Efficiency falls as it grows, and at a value of one the Diesel expression collapses onto the Otto one.
- Pressure ratio
- The governing parameter of the Brayton cycle, defined on pressures rather than volumes. Substituting it into a piston-cycle efficiency formula produces a plausible number that is simply wrong, so check which quantity a symbol denotes before it enters an exponent.
- Mean effective pressure
- Net work divided by the swept volume, giving the constant pressure that would produce the same work over one stroke. It is the fair way to compare engines of different sizes, so it appears in performance questions rather than efficiency questions.
- Reheat
- Taking partially expanded steam out of the turbine, returning it to the boiler for a second dose of heat and sending it back to the low-pressure stages. Its purpose is to allow a high boiler pressure without an unacceptably wet exhaust.
- Regeneration
- Preheating the water entering the boiler using steam bled part way through the turbine, so less fuel is needed to reach the same superheat temperature. Gas turbines do the equivalent with a heat exchanger on the exhaust.
- Combined cycle
- A gas turbine whose exhaust, still hotter than the hottest steam a Rankine plant can use, boils the steam for a second cycle underneath it. The pair reaches efficiencies above fifty percent, higher than either cycle alone.
Power Cycles: Otto, Diesel, Brayton and Rankine FAQ
If the ideal Diesel efficiency is lower than Otto at the same compression ratio, why are diesel engines efficient?
Because they are not restricted to the same compression ratio. A spark-ignition engine is limited by knock to roughly eight to twelve, while compression ignition has no such limit and runs at roughly fourteen to twenty-two. That much higher achievable ratio more than makes up for the lower efficiency at any fixed value, which is why real diesel engines generally come out ahead.
The comparison only favours Otto when it is made at a compression ratio a diesel would never use.
Why does lowering the condenser pressure not simply keep improving a steam plant?
It does improve the ideal efficiency, because it lowers the temperature at which heat is rejected. What stops it is what happens inside the turbine: expanding further into the wet region leaves the steam wetter through the last turbine stages, and liquid droplets erode the blades. Three of the five standard improvements push the exhaust wetter, which is exactly the problem reheat exists to solve.
When is a regenerator worth fitting to a gas turbine?
When the pressure ratio is low. The regenerator recovers heat from the exhaust into the compressor discharge air, and the amount available is the temperature gap between the two. A low pressure ratio leaves a large gap and a high one leaves almost none.
So a regenerative cycle is at its best in stationary power generation, while an aircraft engine chasing power density runs a high pressure ratio and gains nothing from the extra mass.
What does a cycle question actually want when it gives me component efficiencies?
It wants the isentropic-efficiency sequence from Chapter 1, applied once per compressor or pump and once per turbine, before any cycle efficiency is computed. Solve the reversible machine first because entropy fixes its exit state, then scale to the real one, then recover the actual exit enthalpy.
Expect the cycle efficiency to move by a few percentage points rather than collapse, and treat a result that barely moves as a sign that an efficiency was left out somewhere.
Exam move
Learn the four efficiency formulas as one family sharing an isentropic compression and expansion skeleton, differing only in how heat is added and rejected, rather than as four results to memorise separately.
Then attach each formula to its own practical limit, so that a design-comparison question is never bare formula recall: knock for Otto, cut-off for Diesel, turbine metallurgy for Brayton, and blade erosion by wet steam for Rankine. Next make sure you can say, for any symbol in front of you, whether it is a volume ratio or a pressure ratio, because that mix-up produces answers that look reasonable and are not.
Work at least one full cycle question with component efficiencies included, since that is where the Chapter 1 method earns its keep. Finally, be able to state each of the five steam-plant improvements together with the penalty that limits it, because the penalty is usually the part being examined.
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