NivaarExam PrepOfficial exam papers ↗

22-Mec-B3 Energy Conversion and Power Generation · December 2016

Question 5 of 6: Brayton Cycle Modifications

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. 07-Mec-B3 Energy Conversion and Power Generation, National Examinations, December 2016. Three hours, closed book. Two sections: Section A is calculative (Questions 1–4) and Section B is descriptive (Questions 5–6). Candidates answer three questions from Section A and one from Section B; four questions of 15 marks each constitute a complete paper (60 marks). Reference data for individual questions are bound in as attachments on pages 9–14, reference formulae and constants on pages 15–18, and Granet & Bluestein steam tables are supplied. All six questions are solved below, because the set is a study resource rather than a timed attempt.

Reference texts.

Where the paper's own attachments carry data that duplicate a computed result — the 392 °C exhaust in Question 2, the 177 °C boiler gas outlet, the 13 kg/s fuel flow, the torque curve on page 9 — those printed values are used as independent checks and the agreement is quoted in each answer.

Question 5: Brayton Cycle Modifications (15 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

(i) increased pressure ratiosThigher rp: narrower, more efficient loop(ii) regenerative heatingsTexhaust heat returned to the air(iii) compressor intercoolingsTcompression split, cooled back at constant p(iv) turbine reheatingsTexpansion split, reheated to the same T max(v) exhaust afterburningsTheat added after the turbine: thrust, not shaft work
Part (a): each panel carries the basic cycle in grey dashes and the modified cycle in blue (red for afterburning, which adds heat outside the work-producing loop). In every case the compressor inlet temperature and the maximum cycle temperature are held at the same values, as the question directs.

Part (a) — what each sketch shows. With $T_1$ and $T_{max}$ pinned, the five modifications move the cycle in five distinct ways. Increased pressure ratio raises the compressor delivery temperature and lowers the turbine exhaust temperature, so the loop becomes taller and narrower: heat is added over a shorter stretch of the upper isobar and rejected over a shorter stretch of the lower one. Regenerative heating leaves the loop itself untouched and simply transfers the shaded band of exhaust heat from the turbine exit back to the compressor delivery, shortening the combustion line. Compressor intercooling breaks the compression into two isentropic legs joined by a constant-pressure cooling line that pushes the state back toward the ambient isotherm, so the second leg starts colder and less total work is absorbed. Turbine reheating is the mirror image: expansion is broken into two legs joined by a constant-pressure heating line back up to $T_{max}$, so the second leg starts hotter and delivers more work. Exhaust afterburning adds a constant-pressure heating line downstream of the turbine, from state 4 outward along the atmospheric isobar — it lies wholly outside the closed loop, which is precisely why it produces no shaft work.

Given. To put numbers on the argument, take the Question 1 machine: $T_1 = 288$ K, $T_{max} = 1033$ K, $r_p = 6$, ideal components. Find. The ideal efficiency at the base and at a doubled pressure ratio, the pressure ratio for maximum specific work, and the ideal regenerative efficiency.

  1. Quantify the pressure-ratio trade. For the ideal cycle $\eta = 1 - r_p^{-(k-1)/k}$, so raising the ratio from 6 to 12 lifts efficiency from $40.1\ \%$ to $50.8\ \%$. Specific work, however, peaks and then falls, at $$r_{p,opt} = \left(\frac{T_{max}}{T_1}\right)^{k/2(k-1)} = \left(\frac{1033}{288}\right)^{1.75} = \boxed{9.3}$$ so beyond about 9 the machine gets more efficient but smaller for its size.
  2. Quantify regeneration, and its limit. With a perfect regenerator $\eta = 1 - (T_1/T_{max}) r_p^{(k-1)/k}$, which at $r_p = 6$ gives $\boxed{53.5\ \%}$ — better than the simple cycle at any pressure ratio in this range. But that expression falls as $r_p$ rises and crosses the simple-cycle line at the same $r_p = 9.3$, because there the compressor delivery has caught the turbine exhaust and there is nothing left to recover. Regeneration and high pressure ratio are therefore alternatives, not companions.

Part (b) — advantages, disadvantages and applications.

(i) Increased pressure ratio. The single most effective lever on efficiency, because it raises the mean temperature of heat addition without touching the metallurgical limit. Against it: specific work passes through a maximum and then declines, so a very high ratio buys efficiency at the cost of a physically larger machine for the same output; the compressor needs many more stages with variable inlet guide vanes and bleed valves to stay off surge at part speed; and the delivery air becomes too hot to be useful as turbine cooling air. It also destroys the case for regeneration. Applications: modern heavy-duty industrial machines run around 16–24:1 and aero-derivative units above 30:1, all of them without regenerators and usually in combined cycle so that the exhaust heat is recovered in a boiler instead.

(ii) Regenerative heating. Raises efficiency substantially at low to moderate pressure ratio — from 27.7 % to 35.5 % on the Question 1 machine — with no change in output at all, because it reduces fuel rather than increasing work. Against it: a large, expensive gas-to-gas exchanger operating across a big pressure difference, whose pressure drop on both sides claws back part of the gain; slower thermal response; a serious fire risk if it leaks; and no benefit whatever above the crossover pressure ratio. Applications: small industrial and mechanical-drive sets, microturbines (where recuperation is essential to reach useful efficiency), some marine and vehicular gas turbines, and closed-cycle helium plant.

(iii) Compressor intercooling. Cuts the compression work and so raises the net output markedly — the back-work ratio is the weakness of the Brayton cycle and intercooling attacks it directly. But on its own it lowers thermal efficiency, because the air now enters combustion colder and more fuel is needed to reach the same $T_{max}$; the extra heat added at the lower part of the temperature range is added inefficiently. It only pays for efficiency when combined with regeneration, which recovers precisely that extra low-grade heat. It also needs an intercooler and a heat sink. Applications: the intercooled and intercooled-recuperated marine gas turbines (WR-21 class), industrial machines specified for peak output, and the low- and high-pressure compressor split of large aero engines.

(iv) Turbine reheating. The mirror of intercooling: a large increase in specific work, because the second expansion begins at $T_{max}$ again and the isobars diverge, but a fall in efficiency on its own since the extra heat is added at a lower pressure. Combined with regeneration it is strongly positive, because the exhaust leaves hot and there is more heat to recover. Costs are a second combustor between turbines and a hotter exhaust. Applications: sequential-combustion heavy duty machines (the GT24/GT26 family) and, in aviation, the reheat or afterburning arrangement discussed next.

(v) Exhaust afterburning. Adds heat after the turbine, so it produces no shaft work at all — on a T-s diagram the added line lies outside the closed loop. It raises jet velocity and hence thrust dramatically, by 50 % or more, from a small and light addition to the engine. The penalties are severe: specific fuel consumption roughly doubles, exhaust temperature and infrared signature rise sharply, and a variable-area nozzle is required. In a shaft-power plant the same arrangement is only sensible when the exhaust heat is subsequently used — supplementary firing ahead of a heat recovery boiler, which raises steam production and is common on cogeneration plant. Applications: military aircraft afterburners for take-off and combat, and duct burners on combined-cycle and cogeneration heat recovery boilers.

Check: the numerical comparisons above are computed with ideal components on the Question 1 boundary conditions ($T_1 = 288$ K, $T_{max} = 1033$ K) purely to size the arguments. Real machine efficiencies of 86–89 % shift every figure downward and move the crossover pressure ratio, but they do not change the direction of any conclusion.