Question 1 of 6: Regenerative Gas Turbine

Question 1: Regenerative Gas Turbine (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.

Given. A single-shaft open-cycle gas turbine on cold air-standard assumptions:

Find. The four cycle temperatures (ideal and actual), the net power and thermal efficiency of the simple cycle, the duty of a 75 % effective regenerator, the regenerative cycle efficiency, and the air/fuel ratio in each case.

Approach. Fix the two isentropic end states from the pressure ratio, apply the machine efficiencies to obtain the actual states, form the specific works from $c_p\,\Delta T$, and then re-use the same temperatures with the regenerator effectiveness — only the heat input changes, never the work.

1. Part (a) — establish the property set and the state numbering. Cold air-standard means air with constant properties throughout, so from the constants on page 16, $c_p = 1.005\ \text{kJ/kg}\cdot\text{K}$, $R = 0.287\ \text{kJ/kg}\cdot\text{K}$ and $k = 1.4$. State 1 is the compressor inlet, 2 its delivery, 3 the turbine inlet and 4 the turbine exhaust; primed and s-subscripted points are the ideal states. The T-s diagram above carries all of them.
2. Part (b) — ideal compressor delivery temperature. For an isentropic change between the two isobars, the page-18 gas-turbine relation gives $$\frac{T_{2s}}{T_1} = \left(\frac{p_2}{p_1}\right)^{(k-1)/k} = 6^{0.2857} = 1.6685$$ so that $T_{2s} = 288.15 \times 1.6685 = 480.8\ \text{K}$ (207.6 °C).
3. Part (b) — actual compressor delivery. The compressor efficiency is the ratio of the ideal to the actual temperature rise, so the real rise is larger: $$T_2 = T_1 + \frac{T_{2s} - T_1}{\eta_c} = 288.15 + \frac{480.8 - 288.15}{0.86} = \boxed{512.1\ \text{K}}$$ that is 239.0 °C. The extra 31.4 K above the ideal delivery is irreversible compression work that ends up as temperature rather than pressure.
4. Part (b) — ideal and actual turbine exhaust. Expanding from 0.6 MPa back to atmosphere over the same pressure ratio, $$T_{4s} = \frac{T_3}{6^{0.2857}} = \frac{1033.15}{1.6685} = 619.2\ \text{K}$$ and the turbine efficiency is the ratio of actual to ideal drop, so $$T_4 = T_3 - \eta_t (T_3 - T_{4s}) = 1033.15 - 0.89\times 413.9 = \boxed{664.7\ \text{K}}$$ or 391.6 °C. Question 2 prints this exhaust temperature as 392 °C, which confirms the whole compressor-and-turbine chain to better than half a degree.
5. Part (c) — specific works and net power. With equal mass flow through both machines and constant $c_p$, $$w_c = c_p (T_2 - T_1) = 1.005 \times 224.0 = 225.1\ \text{kJ/kg}, \qquad w_t = c_p (T_3 - T_4) = 1.005 \times 368.4 = 370.3\ \text{kJ/kg}$$ so the net specific work is $w_{net} = 370.3 - 225.1 = 145.1\ \text{kJ/kg}$ and the plant output is $$P = \dot m\, w_{net} = 1000 \times 145.1 = \boxed{145.1\ \text{MW}}$$ Note that the compressor absorbs 61 % of the turbine work — the high back-work ratio characteristic of a modest-pressure-ratio machine.
6. Part (c) — simple-cycle efficiency. Heat is added in the combustor between states 2 and 3: $$q_{in} = c_p (T_3 - T_2) = 1.005 \times 521.0 = 523.6\ \text{kJ/kg}$$ $$\eta_{simple} = \frac{w_{net}}{q_{in}} = \frac{145.1}{523.6} = \boxed{27.7\ \%}$$ The paper itself quotes 28 % as the simple-cycle efficiency in Question 2, so the two agree.
7. Parts (d) and (e) — the regenerator duty. The largest possible transfer would bring the compressed air all the way up to the exhaust temperature, a rise of $T_4 - T_2 = 152.6\ \text{K}$; the effectiveness scales that down: $$\Delta T_{reg} = \varepsilon\,(T_4 - T_2) = 0.75 \times 152.6 = 114.4\ \text{K}$$ so the air enters the combustor at $T_{2'} = 512.1 + 114.4 = 626.6$ K (353.4 °C) and, because both streams carry the same flow and the same $c_p$, the gas leaves the regenerator 114.4 K cooler, at 277.1 °C. The duty is $$\dot Q_{reg} = \dot m\, c_p\, \Delta T_{reg} = 1000 \times 1.005 \times 114.4 = \boxed{115.0\ \text{MW}}$$
8. Part (f) — regenerative cycle efficiency and comment. The turbine and compressor are untouched, so $w_{net}$ is unchanged at 145.1 kJ/kg; only the combustor duty falls, to $$q_{in}' = c_p (T_3 - T_{2'}) = 1.005 \times 406.6 = 408.6\ \text{kJ/kg}$$ $$\eta_{regen} = \frac{145.1}{408.6} = \boxed{35.5\ \%}$$ That is a gain of 7.8 percentage points, a 28 % relative improvement in fuel economy for no change whatever in output. The addition is clearly worthwhile here because the exhaust at 392 °C is far hotter than the compressor delivery at 239 °C, leaving a large temperature margin to exploit; the cost is a bulky gas-to-gas exchanger and a pressure drop on both sides that erodes part of the gain in practice.
9. Part (g) — air/fuel ratios. Per kilogram of air the fuel burned is $q_{in}/\mathrm{HV}$, so the air/fuel ratio is simply the heating value divided by the combustor duty: $$\left(\frac{A}{F}\right)_{simple} = \frac{40\,000}{523.6} = \boxed{76.4:1}, \qquad \left(\frac{A}{F}\right)_{regen} = \frac{40\,000}{408.6} = \boxed{97.9:1}$$ On 1000 kg/s of air the first corresponds to 13.1 kg/s of fuel, against the 13 kg/s printed in Question 2 — a further independent check. Regeneration lets the same machine run 28 % leaner for the same output.