A steam power plant and a gas-turbine plant have thermal efficiencies of 35 per cent and 48 per cent respectively. Which statement best explains why the steam plant is nevertheless far less sensitive to turbine and compressor inefficiency?

The steam plant pumps a liquid, so its back-work ratio is a fraction of a per cent instead of about a half

Steam turbines are inherently more efficient than gas turbines

The steam plant pumps a liquid, so its back-work ratio is a fraction of a per cent instead of about a half

The steam plant operates at a lower peak temperature

Water has a higher specific heat than air

Which of the following are correct about Euler’s turbomachine equation `w = u₂V_w2 − u₁V_w1`?

Only the tangential (whirl) component of absolute velocity contributes to the work

It applies equally to a steam turbine, a centrifugal pump and a fan

For a pump with radial inlet flow it reduces to `w = u₂V_w2`

Only the tangential (whirl) component of absolute velocity contributes to the work

It applies equally to a steam turbine, a centrifugal pump and a fan

For a pump with radial inlet flow it reduces to `w = u₂V_w2`

It gives the actual work after accounting for hydraulic losses

For a Diesel cycle at a fixed compression ratio, as the cut-off ratio is reduced towards 1, the air-standard efficiency

approaches the Otto efficiency at the same compression ratio

stays constant, since it depends only on the compression ratio

exceeds the Carnot efficiency between the same temperature limits

In a Brayton cycle with fixed inlet and peak temperatures, the pressure ratio that maximises net work per kg of air is

lower, and equal to `(T₃/T₁)^(γ/2(γ−1))`

the same as the one that maximises thermal efficiency

lower, and equal to `(T₃/T₁)^(γ/2(γ−1))`

unbounded — net work rises indefinitely with pressure ratio

independent of the temperature ratio

Thermodynamics and Its Applications: Cycles, Availability and Turbomachines

Thermodynamics in this paper is two laws and a great deal of careful bookkeeping. The first law is a conservation statement and it never fails; the second law is a direction statement and it is where every interesting result comes from. The most useful way to hold the second law for exam purposes is as entropy generation: any real process generates entropy, S_gen ≥ 0, and the work you could have had but did not is exactly T₀ S_gen. That one identity turns vague statements about irreversibility into a number. Then Applications, which the syllabus lists as its own heading and which is the same two laws drawn as cycles. Air-standard cycles reduce to one line each — Otto is η = 1 − r^(1−γ) and depends on nothing but the compression ratio — and the honest reading of that is that it is an upper bound no real engine reaches. The Rankine and Brayton cycles look similar on a diagram and behave completely differently in one respect that decides how they are built: the back-work ratio, which is 0.4 per cent for a steam plant pumping a liquid and 48 per cent for a gas turbine compressing a gas. That single number is why a gas turbine cannot start itself and why a steam plant’s pump is an afterthought. Refrigeration is the same cycle read backwards, with the two coefficients of performance differing by exactly 1. And turbomachinery adds one equation, Euler’s, which says the work per unit mass is the change in V_w u across the rotor and nothing else.

The Second Law as a Number: Entropy Generated, Work Lost

The Carnot limit η = 1 − T_L/T_H is usually taught as a ceiling, which is true but not very actionable. The more useful form is the entropy balance: for any process, S_gen = ΔS_system + ΔS_surroundings ≥ 0, with equality only for a reversible process, and the work destroyed is W_lost = T₀ S_gen where T₀ is the dead-state temperature. Work an example that sounds harmless. Move 1000 J of heat straight from a reservoir at 800 K to one at 300 K — a bare conduction, no engine. The hot reservoir loses 1000/800 = 1.25 J/K and the cold one gains 1000/300 = 3.3333 J/K, so S_gen = 2.0833 J/K. The destroyed work is 300 × 2.0833 = 625 J. Now compare that with what was available: a Carnot engine between the same reservoirs would have delivered 1000 × (1 − 300/800) = 625 J. The two numbers are identical, and not by coincidence — transferring heat directly to the dead state destroys the entire available work, every joule of it. That is the sharpest statement of why heat exchangers are designed for small temperature differences and why a boiler firing at 1800 K into steam at 600 K is throwing away most of the fuel’s availability before the cycle even begins.

🧠 Air-standard efficiencies are one line each, and all of them are ceilings

The Otto cycle gives η = 1 − r^(1−γ), and read what is absent: no temperature, no pressure, no fuel — the compression ratio is the only variable. At γ = 1.4, r = 8 gives 56.47 per cent, r = 10 gives 60.19 and r = 18 gives 68.53. The Diesel cycle adds a cut-off ratio: η = 1 − r^(1−γ)[(ρ^γ − 1)/(γ(ρ − 1))], and at r = 18 with cut-off 2 that is 63.16 per cent. Two comparisons are worth making carefully, because they are the ones examiners set. First, at the same compression ratio Otto always beats Diesel — 68.53 against 63.16 at r = 18 — and driving the cut-off ratio towards 1 makes the Diesel expression converge on Otto’s, which is the algebraic check that the two cycles are the same cycle in a limit. (Verified: at cut-off 1.0001 the Diesel formula returns 0.685298 against Otto’s 0.685304.) Second, that comparison is not the practical one, because a petrol engine cannot use r = 18 — it would knock — while a diesel engine routinely does. Real diesels are more efficient than real petrol engines despite being less efficient at equal compression ratio, and a candidate who has only the formula will get the practical question backwards. The honest frame for all of these: they are air-standard, meaning constant specific heats, no friction, no combustion irreversibility and air as the working fluid throughout. A real spark-ignition engine at r = 8 reaches perhaps 30 per cent, not 56.

Air-standard efficiencies at γ = 1.4, and the Carnot ceiling for comparison
Cycle and parameters	Efficiency	What it depends on
Otto, `r = 8`	56.47 %	Compression ratio only
Otto, `r = 18`	68.53 %	Unreachable in petrol — it would knock
Diesel, `r = 18`, cut-off 2	63.16 %	Compression and cut-off ratio
Brayton, `r_p = 10`	48.21 %	Pressure ratio only
Carnot, 800 K → 300 K	62.50 %	Reservoir temperatures only — the true ceiling

Rankine Against Brayton: The Back-Work Ratio Decides Everything

Both cycles are two isentropic devices and two constant-pressure devices — compress, heat, expand, cool — and on a p–v sketch they look almost the same. The difference is what gets compressed. Rankine compresses a liquid, and pumping a liquid costs almost nothing because its specific volume is tiny. Brayton compresses a gas, and that costs nearly half the turbine output. Put numbers on both. For Rankine with turbine inlet at h₁ = 3214 kJ/kg, turbine exit h₂ = 2144, condensate h₃ = 192 and pump work 4 kJ/kg: w_turb = 1070, w_net = 1066, q_in = 3018, so η = 35.32 per cent and the back-work ratio is 4/1070 = 0.37 per cent. For Brayton with r_p = 10, T₁ = 300 K and T₃ = 1200 K: T₂ = 579.2 K, T₄ = 621.5 K, w_comp = 280.6, w_turb = 581.4, so w_net = 300.8, q_in = 623.9 and η = 48.21 per cent — but the back-work ratio is 280.6/581.4 = 48.27 per cent. The gas turbine has the higher thermal efficiency here and spends 129 times more of its own output running its compressor.

🎯 What a 48 per cent back-work ratio actually costs you

A back-work ratio near a half is not an accounting curiosity; it decides how the machine is built and how it fails. It makes the plant hypersensitive to component efficiency. Suppose the Brayton turbine and compressor each fall from ideal to 85 per cent isentropic efficiency. The turbine now delivers 0.85 × 581.4 = 494.2 and the compressor now demands 280.6/0.85 = 330.1, so net work drops from 300.8 to 164.1 kJ/kg — a 45 per cent collapse from a 15 per cent loss in each device. Run the same 15 per cent loss through Rankine and net work falls from 1066 to about 905, a 15 per cent drop, because the pump term is too small to matter. That amplification is the reason gas-turbine development is an unending pursuit of a few points of compressor efficiency, while steam-plant engineering worries about boiler and condenser rather than about pumps. Two more consequences follow from the same number. A gas turbine cannot start itself — it needs an external starter to spin the compressor up to a speed where the turbine produces more than the compressor consumes, and below that speed the cycle simply does not close. And there is a pressure ratio beyond which net work falls even though efficiency keeps rising, because compressor work grows faster than turbine work; for T₃/T₁ = 4 the optimum for maximum net work is r_p = (T₃/T₁)^(γ/2(γ−1)) = 11.31, and past it you gain efficiency while losing output per kilogram of air. Choosing between those two optima is a real design decision, not a formula.

Rankine against Brayton on the same page
	Rankine (given enthalpies)	Brayton (`r_p = 10`, 300 → 1200 K)
What is compressed	Liquid water	Air
Turbine work	1070 kJ/kg	581.4 kJ/kg
Compressor / pump work	4 kJ/kg	280.6 kJ/kg
Net work	1066 kJ/kg	300.8 kJ/kg
Thermal efficiency	35.32 %	48.21 %
Back-work ratio	0.37 %	48.27 % — 129× larger
Net work if both devices are 85 % efficient	≈ 905 (−15 %)	164.1 (−45 %)

Refrigeration Read Backwards, and One Equation for Every Turbomachine

Reverse a power cycle and you get a refrigerator or a heat pump — the same hardware, different accounting. COP_R = Q_L/W and COP_HP = Q_H/W, and since Q_H = Q_L + W by the first law, COP_HP = COP_R + 1 always, exactly, with no conditions attached. For a reversed Carnot between T_L = 273 K and T_H = 303 K, COP_R = 273/30 = 9.1 and COP_HP = 303/30 = 10.1 — a difference of exactly 1, which is the fastest check on any COP answer. Two readings worth keeping. First, a COP of 9.1 means moving nine joules of heat for one joule of work, which is why heat pumps are not "over 100 per cent efficient" but simply not heat engines: they move energy rather than converting it, and the first law is untroubled. Second, COP_R = T_L/(T_H − T_L) collapses as the temperature lift grows: widen the same span to T_L = 263, T_H = 313 and COP_R falls from 9.1 to 5.26. Every degree of extra lift is expensive, which is the reason a domestic freezer costs so much more to run than a refrigerator compartment at the same ambient.

🧠 Euler’s equation is the whole of turbomachinery you are asked for

Euler’s turbomachine equation says the work per unit mass exchanged with a rotor is w = u₂V_w2 − u₁V_w1, where u is the blade speed and V_w the whirl (tangential) component of absolute velocity. Read what is absent: no pressure, no density, no efficiency, and nothing about whether the fluid is a gas or a liquid. It is pure angular-momentum bookkeeping, which is why the same equation covers a steam turbine, a centrifugal pump, a fan and a torque converter. Three consequences worth carrying. First, only the tangential component does work — the axial and radial components of velocity move mass through the machine and contribute nothing to w, which is why blade angles matter and blade lengths mostly do not. Second, for a pump with radial inlet flow, V_w1 = 0 and the equation reduces to w = u₂V_w2, which is the form most questions actually use. Third, degree of reaction R = (static enthalpy change in the rotor)/(total enthalpy change in the stage) sorts the machines out: R = 0 is an impulse stage where all the pressure drop happens in the nozzle and the rotor only turns the flow, R = 0.5 is the symmetric reaction stage with identical blade shapes on rotor and stator, and R = 1 puts the whole drop in the rotor. And specific speed — N_s = N√Q/H^(3/4) for pumps, N√P/H^(5/4) for turbines — is a dimensionless-in-spirit group whose only job is to tell you which family of machine suits a duty: low specific speed means high head and low flow, so a radial machine; high specific speed means the opposite, so an axial one. It selects the type before any blade is drawn.

Key takeaways

The second law is most useful as a number: W_lost = T₀ S_gen. Moving 1000 J from 800 K to 300 K generates 2.083 J/K and destroys 625 J — which is exactly all the availability there was.
Otto depends on nothing but the compression ratio: 56.47 % at r = 8, 68.53 % at r = 18. At equal r, Otto always beats Diesel — but a petrol engine cannot reach r = 18, which is why real diesels win.
The back-work ratio is the difference between Rankine and Brayton: 0.37 per cent against 48.27 per cent. Rankine pumps a liquid; Brayton compresses a gas.
A high back-work ratio amplifies every component loss. Dropping both Brayton devices to 85 per cent isentropic efficiency cuts net work by 45 per cent, from 300.8 to 164.1 kJ/kg; the same loss costs Rankine only 15 per cent.
Efficiency and net work optimise at different pressure ratios. For T₃/T₁ = 4 the maximum-work ratio is 11.31; beyond it efficiency keeps rising while output per kilogram of air falls.
COP_HP = COP_R + 1, exactly and always. At 273 K against 303 K those are 10.1 and 9.1; widen the lift to 263 against 313 and COP_R falls to 5.26.
Euler’s equation w = u₂V_w2 − u₁V_w1 contains no pressure, density or efficiency — only the whirl components do work. Degree of reaction sorts impulse (R = 0) from reaction (R = 0.5), and specific speed picks the machine family before any blade is drawn.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

An Otto cycle operates with a compression ratio of 8 and γ = 1.4. Its air-standard efficiency, in per cent, is ______ (round to two decimals)
Numerical answer — type the value.
Show answer
Answer: 56.47
56.47 per cent. η = 1 − r^(1−γ) = 1 − 8^(−0.4) = 1 − 0.43528 = 0.56472. Read the formula for what it does not contain: no temperature, no pressure, no fuel, no heat input — the air-standard Otto efficiency depends on the compression ratio and γ alone. That is a strong statement and it is worth being clear that it is a statement about an idealisation, not about an engine. The air-standard cycle assumes constant specific heats, no friction, no heat loss to the cylinder walls, instantaneous combustion at constant volume and air as the working fluid throughout. A real spark-ignition engine at r = 8 reaches perhaps 30 per cent, so 56.47 is a ceiling and not a prediction. Two comparisons worth having ready. Raising r to 18 would give 68.53 per cent, but a petrol engine cannot use that ratio — the charge would auto-ignite, which is knock, and that limit is what confines spark ignition to about r = 10. And a Diesel cycle at r = 18 with a cut-off ratio of 2 gives 63.16 per cent, lower than Otto at the same r; real diesels are nevertheless more efficient than real petrol engines because they can actually run at that compression ratio. Confusing the equal-r comparison with the practical one is the most common conceptual error in this topic.
1000 J of heat is transferred directly from a reservoir at 800 K to one at 300 K. The entropy generated, in J/K, is ______ (round to four decimals)
Numerical answer — type the value.
Show answer
Answer: 2.0833
2.0833 J/K. Reservoirs are large enough that their temperatures do not change, so each entropy change is simply Q/T. The hot reservoir loses 1000/800 = 1.2500 J/K and the cold one gains 1000/300 = 3.3333 J/K, so S_gen = 3.3333 − 1.2500 = 2.0833 J/K. Positive, as the second law demands for a spontaneous process. Now the part that makes the number worth computing: the work destroyed is T₀ S_gen = 300 × 2.0833 = 625 J. Compare that with the work that was available — a Carnot engine between the same two reservoirs would have produced 1000 × (1 − 300/800) = 625 J. The two are identical, and that identity is exact rather than coincidental: letting heat fall to the dead state with no engine in the path destroys the entire available work. This is the sharpest available statement of why irreversibility is expensive, and it has a direct design consequence — heat exchangers are built for small temperature differences not out of caution but because every degree of unnecessary drop is availability that no downstream device can ever recover. The same reasoning is why a boiler firing at 1800 K into steam at 600 K has already lost most of the fuel’s work potential before the Rankine cycle starts.
An air-standard Brayton cycle has a pressure ratio of 10, compressor inlet 300 K, turbine inlet 1200 K, γ = 1.4. The back-work ratio (compressor work ÷ turbine work), in per cent, is ______ (round to two decimals)
Numerical answer — type the value.
Show answer
Answer: 48.27
48.27 per cent. The isentropic temperature factor is r_p^((γ−1)/γ) = 10^0.2857 = 1.9307. So T₂ = 300 × 1.9307 = 579.21 K and T₄ = 1200/1.9307 = 621.54 K. With c_p = 1.005 kJ/kg·K, the compressor takes 1.005(579.21 − 300) = 280.61 kJ/kg and the turbine delivers 1.005(1200 − 621.54) = 581.36 kJ/kg, so the back-work ratio is 280.61/581.36 = 48.27 per cent. Check the thermal efficiency the long way and the short way: w_net/q_in = 300.75/623.89 = 0.4821, and 1 − r_p^(−0.2857) = 0.4821 ✓ — the closed form agrees to six figures, which is worth doing once so you trust it thereafter. The back-work ratio is where the engineering lives. Nearly half the turbine’s output is consumed by its own compressor, and the consequence is hypersensitivity to component efficiency: drop both devices to 85 per cent isentropic and the turbine gives 0.85 × 581.36 = 494.2 while the compressor demands 280.61/0.85 = 330.1, so net work falls from 300.8 to 164.1 kJ/kg — a 45 per cent collapse from a 15 per cent loss each. A Rankine plant, whose back-work ratio is 0.37 per cent, loses only 15 per cent under the same treatment. This is also why a gas turbine needs an external starter, and why the pressure ratio that maximises net work, (T₃/T₁)^(γ/2(γ−1)) = 11.31, is not the one that maximises efficiency.
A steam power plant and a gas-turbine plant have thermal efficiencies of 35 per cent and 48 per cent respectively. Which statement best explains why the steam plant is nevertheless far less sensitive to turbine and compressor inefficiency?
1. Steam turbines are inherently more efficient than gas turbines
2. The steam plant pumps a liquid, so its back-work ratio is a fraction of a per cent instead of about a half
3. The steam plant operates at a lower peak temperature
4. Water has a higher specific heat than air
Show answer
Answer: B — The steam plant pumps a liquid, so its back-work ratio is a fraction of a per cent instead of about a half
The back-work ratio. Rankine compresses a liquid, whose specific volume is about a thousandth of a gas’s, so the pump costs about 4 kJ/kg against a turbine output of 1070 — a back-work ratio of 0.37 per cent. Brayton compresses a gas, and the compressor takes 280.6 kJ/kg against a turbine output of 581.4 — 48.27 per cent. That is 129 times larger, and it is what amplifies component losses. Put both through a 15 per cent isentropic loss in each device: Brayton’s net work falls from 300.8 to 164.1 kJ/kg, a 45 per cent collapse, because the turbine loses 15 per cent while the compressor’s demand rises 18 per cent and the two errors act on almost equal quantities. Rankine loses about 15 per cent, full stop, because the pump term is too small to amplify anything. Option A is a comparison of devices, not of cycles, and steam and gas turbines actually reach similar isentropic efficiencies. Option C is true as a fact and irrelevant as an explanation — a lower peak temperature would make the steam plant less efficient, which is consistent with the 35 versus 48 in the question, but it says nothing about sensitivity. Option D confuses a fluid property with a cycle property; what matters is specific volume at the compression stage, not specific heat. The same 48 per cent also explains two practical facts: a gas turbine cannot start itself, and its maximum-work pressure ratio differs from its maximum-efficiency one.
A reversed Carnot refrigerator works between 273 K and 303 K. If the same machine is used as a heat pump, its COP is ______ (round to one decimal)
Numerical answer — type the value.
Show answer
Answer: 10.1
10.1. COP_HP = T_H/(T_H − T_L) = 303/30 = 10.1, and as a refrigerator the same machine gives COP_R = T_L/(T_H − T_L) = 273/30 = 9.1. The relation is exact and unconditional: since Q_H = Q_L + W by the first law, COP_HP = COP_R + 1 always, so if you have one you have the other by inspection — the fastest check available on a COP answer. Two readings worth carrying. First, a COP above 1 is not a violation of anything: a heat pump moves energy rather than converting it, so comparing 10.1 with a heat engine’s 0.4 is a category error, and the honest statement is that one joule of work relocates ten joules of heat. Second, COP_R = T_L/(T_H − T_L) collapses as the lift widens — go from 273/303 to 263/313 and COP_R drops from 9.1 to 5.26 for a 20 K wider span. Every extra degree of temperature lift is expensive, which is why a freezer compartment costs several times more to run than a chiller at the same ambient, and why heat pumps are marketed on mild-climate performance. The common slip is to divide by the absolute temperature rather than the difference; both COPs have T_H − T_L in the denominator, and 30 K rather than 303 K is what makes the numbers large.
Which of the following are correct about Euler’s turbomachine equation w = u₂V_w2 − u₁V_w1?
1. Only the tangential (whirl) component of absolute velocity contributes to the work
2. It applies equally to a steam turbine, a centrifugal pump and a fan
3. For a pump with radial inlet flow it reduces to w = u₂V_w2
4. It gives the actual work after accounting for hydraulic losses
Show answer
Answer: A — Only the tangential (whirl) component of absolute velocity contributes to the work; B — It applies equally to a steam turbine, a centrifugal pump and a fan; C — For a pump with radial inlet flow it reduces to `w = u₂V_w2`
A, B and C. A is the physical content of the equation: it is an angular-momentum balance, so only the component of velocity in the direction of blade motion can exchange work — the axial and radial components move fluid through the machine and contribute nothing to w. That is why blade angles dominate turbomachinery design while blade lengths mostly set capacity rather than work. B is right and is the reason the equation is worth learning once: no pressure, no density and no property of the working fluid appears in it, so a steam turbine, a water pump, an air fan and a torque converter are all the same calculation. C is the standard simplification, and it holds because a radial inlet means the absolute velocity has no tangential component there, so V_w1 = 0. D is false, and importantly so. Euler’s equation is the ideal energy exchange between fluid and rotor; real machines lose additional energy to friction, secondary flows, leakage, shock at off-design incidence and disc friction, none of which the equation sees. The actual work is Euler’s work multiplied by a hydraulic efficiency, and treating the Euler head as delivered head will overstate a pump by 10 to 25 per cent. The related quantity that does classify machines is the degree of reaction: R = 0 is an impulse stage with the whole pressure drop in the nozzle, R = 0.5 the symmetric reaction stage, R = 1 puts the drop entirely in the rotor.
For a Diesel cycle at a fixed compression ratio, as the cut-off ratio is reduced towards 1, the air-standard efficiency
1. falls towards zero
2. approaches the Otto efficiency at the same compression ratio
3. stays constant, since it depends only on the compression ratio
4. exceeds the Carnot efficiency between the same temperature limits
Show answer
Answer: B — approaches the Otto efficiency at the same compression ratio
It approaches the Otto efficiency. The Diesel expression is η = 1 − r^(1−γ)[(ρ^γ − 1)/(γ(ρ − 1))], and as the cut-off ratio ρ → 1 the bracket tends to 1 — take the limit and (ρ^γ − 1)/(γ(ρ − 1)) is a 0/0 form whose limit is exactly 1 — leaving η = 1 − r^(1−γ), which is Otto’s formula. Verified numerically: at r = 18 with ρ = 1.0001 the Diesel formula returns 0.685298 against Otto’s 0.685304. The physical reading is cleaner than the algebra: ρ = 1 means heat is added over zero volume change, which is constant-volume heat addition, which is what makes a cycle Otto rather than Diesel. The two cycles are the same cycle in that limit, and this is exactly the kind of check worth running on any formula you are unsure you have remembered correctly. Option A confuses cut-off ratio with something like a load fraction; efficiency rises as ρ falls, it does not vanish. Option C states the Otto dependence, which is true of Otto and false of Diesel — the second parameter is precisely what distinguishes them. Option D is impossible: no air-standard cycle can beat Carnot between the same temperature extremes, and any derivation that suggests otherwise contains an error rather than a discovery.
In a Brayton cycle with fixed inlet and peak temperatures, the pressure ratio that maximises net work per kg of air is
1. the same as the one that maximises thermal efficiency
2. lower, and equal to (T₃/T₁)^(γ/2(γ−1))
3. unbounded — net work rises indefinitely with pressure ratio
4. independent of the temperature ratio
Show answer
Answer: B — lower, and equal to `(T₃/T₁)^(γ/2(γ−1))`
Lower, at (T₃/T₁)^(γ/2(γ−1)). Thermal efficiency 1 − r_p^(−(γ−1)/γ) rises monotonically with pressure ratio and has no interior maximum — so if efficiency were the only criterion you would raise r_p without limit. Net work does not behave that way. As r_p grows, T₂ climbs towards T₃ and the turbine has less temperature range to expand through, while the compressor work keeps growing; at the extreme where T₂ = T₃ no heat can be added at all and net work is zero. So net work rises, peaks and falls, and differentiating gives the optimum r_p = (T₃/T₁)^(γ/2(γ−1)) — for T₃/T₁ = 4 that is 4^1.75 = 11.31. Option A is the trap and it is worth being explicit about why: the two optima genuinely differ, so a gas-turbine designer must choose between fuel economy and output per kilogram of air, and the choice depends on the application — an aircraft engine values work per kilogram of air because that sets engine size and weight, while a stationary plant values efficiency because fuel is the lifetime cost. Option C ignores the zero at T₂ = T₃, which alone proves net work must have an interior maximum. Option D is contradicted by the formula: the optimum depends entirely on T₃/T₁, which is why raising the turbine inlet temperature — the whole history of gas-turbine development — moves the optimum pressure ratio up with it.