Question 2 of 6: Locomotive Gas Turbine Cycle

Question 2: Locomotive Gas Turbine Cycle (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.

Find. Part I: the four cycle temperatures, the T–s diagram, the net power and the thermal efficiency of the simple ideal Brayton cycle. Part II: the flow and T–s diagrams, the fuel flow, power and efficiency after an ideal regenerator is added, and the merits of the change.

Approach. With no machine losses the two pressure ratios are isentropic, so all four temperatures follow from \(T_1\), the pressure ratio and the fuel heat; the work terms are then simple \(c_p \Delta T\) products. Adding the regenerator changes only where the heat comes from, not the state points, so the work is untouched and the efficiency rises purely because less fuel is burnt.

1. Part I (a) — Compressor delivery temperature. The pressure ratio is \(r_p = 600/100 = 6\), and for an isentropic compression of an ideal gas $$\frac{T_2}{T_1} = r_p^{(k-1)/k} = 6^{0.2857} = 1.6684$$ $$T_2 = 293.15 \times 1.6684 = 489.1\ \text{K} = 216.0\,{}^{\circ}\text{C}$$
2. Part I (a) continued — Turbine inlet temperature. All of the fuel's calorific value is released into the air stream (the fuel mass is neglected), so $$\dot Q_{in} = \dot m_f\,CV = 0.25 \times 40\,000 = 10\,000\ \text{kW}$$ $$T_3 = T_2 + \frac{\dot Q_{in}}{\dot m_a c_p} = 489.1 + \frac{10\,000}{12 \times 1.005} = 489.1 + 829.2$$ $$T_3 = 1318.3\ \text{K} = 1045.2\,{}^{\circ}\text{C}$$
3. Part I (a) concluded — Turbine exhaust temperature. The expansion is through the same pressure ratio and is likewise isentropic: $$T_4 = \frac{T_3}{r_p^{(k-1)/k}} = \frac{1318.3}{1.6684} = 790.1\ \text{K} = 517.0\,{}^{\circ}\text{C}$$ $$\boxed{T_1 = 293\ \text{K},\quad T_2 = 489\ \text{K},\quad T_3 = 1318\ \text{K},\quad T_4 = 790\ \text{K}}$$ A turbine inlet temperature of 1045 °C is realistic for an uncooled industrial machine; the 517 °C exhaust is what makes regeneration worthwhile in Part II.
4. Part I (b) — T–s diagram. Figure 2.1 above shows the cycle. The two isentropes are vertical lines and the two heat-transfer legs run along the 600 kPa and 100 kPa isobars, which diverge as temperature rises — the geometric reason a Brayton cycle produces net work at all.
5. Part I (c) — Power output. Turbine and compressor work are evaluated separately and differenced: $$\dot W_T = \dot m_a c_p (T_3 - T_4) = 12 \times 1.005 \times (1318.3 - 790.1) = 6370\ \text{kW}$$ $$\dot W_C = \dot m_a c_p (T_2 - T_1) = 12 \times 1.005 \times (489.1 - 293.15) = 2363\ \text{kW}$$ $$\boxed{\dot W_{net} = 6370 - 2363 = 4007\ \text{kW} = 4.01\ \text{MW}}$$ The back-work ratio is 37 %, and 4 MW is a sensible rating for a mainline locomotive.
6. Part I (d) — Thermodynamic efficiency. $$\eta_{th} = \frac{\dot W_{net}}{\dot Q_{in}} = \frac{4007}{10\,000} = 0.401$$ $$\boxed{\eta_{th} = 40.1\ \%}$$ This must agree with the closed form for the ideal Brayton cycle, which depends on the pressure ratio alone: $$\eta_{th} = 1 - \frac{1}{r_p^{(k-1)/k}} = 1 - \frac{1}{1.6684} = 0.4007 \;\checkmark$$
7. Part II (a) — Flow diagram of the regenerative arrangement. A counter-flow heat exchanger is inserted so that turbine exhaust at state 4 preheats compressor delivery from state 2 to state x before the combustor; see Figure 2.2. Regeneration is possible only because \(T_4 = 790\ \text{K}\) exceeds \(T_2 = 489\ \text{K}\); had the pressure ratio been high enough to reverse that inequality the exchanger would cool the air and the modification would be worse than useless.
8. Part II (b) — T–s diagram of the modified cycle. Figure 2.3 shows the same four corner states with two new legs along the isobars: 2→x on the 600 kPa line (heat gained) and 4→y on the 100 kPa line (heat given up). With an ideal regenerator the air reaches \(T_x = T_4 = 790.1\ \text{K}\) and the exhaust falls to \(T_y = T_2 = 489.1\ \text{K}\).
9. Part II (c) — Fuel required at the same turbine inlet temperature. The combustor now has to lift the air only from \(T_x\) to \(T_3\): $$\dot Q_{in}' = \dot m_a c_p (T_3 - T_x) = 12 \times 1.005 \times (1318.3 - 790.1) = 6370\ \text{kW}$$ $$\dot m_f' = \frac{6370}{40\,000}$$ $$\boxed{\dot m_f' = 0.159\ \text{kg/s}}$$ a saving of 36.3 % on the original 0.25 kg/s.
10. Part II (d) — Power output. The compressor still runs between 293 K and 489 K, and the turbine still expands from 1318 K to 790 K, so both machines are working between identical states: $$\boxed{\dot W_{net}' = \dot W_{net} = 4007\ \text{kW}}$$ The regenerator moves heat about inside the cycle; it does no work and takes none.
11. Part II (e) — Thermal efficiency. $$\eta_{th}' = \frac{4007}{6370}$$ $$\boxed{\eta_{th}' = 0.629 = 62.9\ \%}$$ which again checks against the closed form for an ideal regenerator, $$\eta_{th}' = 1 - \frac{T_1}{T_3}\,r_p^{(k-1)/k} = 1 - \frac{293.15}{1318.3}\times 1.6684 = 0.629 \;\checkmark$$
12. Part II (f) — Advantages of the new arrangement. The single decisive advantage is fuel: the same 4 MW at the drawbar for 36 % less fuel, which for a locomotive translates directly into range between fuelling stops, operating cost, and greenhouse-gas and particulate emissions per tonne-kilometre. Secondary gains are an exhaust discharged at 489 K rather than 790 K — quieter, safer and less of a lineside fire risk — and a smaller combustor duty. Against these must be set the regenerator itself: it is bulky and heavy on a vehicle where both matter, it adds pressure drop on both sides (each 1 % of pressure loss costs roughly 0.4 % of cycle efficiency, so the ideal 62.9 % would fall to perhaps 55–58 % in practice), it slows the machine's response to load changes, and it introduces a fouling and thermal-fatigue maintenance burden. Note also that the benefit is confined to low pressure ratios; at any ratio for which \(T_4 \le T_2\), regeneration is impossible.