22-Mec-A1 Applied Thermodynamics and Heat Transfer · December 2018
Question 3 of 8: Regenerative automotive gas-turbine power plant
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: National Examination 16-Mec-A1, 3 hours, open book. Eight questions of equal value: Part A — Thermodynamics (Q1–Q4) and Part B — Heat Transfer (Q5–Q8). A complete paper is any five questions (three from one part and two from the other).
Reference texts: Çengel & Boles, Thermodynamics: An Engineering Approach (9th ed., McGraw-Hill) — ideal-gas mixtures, the air-standard Otto cycle, wet-region steam properties, the throttling calorimeter, the steady-flow energy equation, the regenerative gas-turbine (Brayton) cycle and vapour-compression refrigeration; Çengel & Ghajar, Heat and Mass Transfer (6th ed.) and Incropera, DeWitt, Bergman & Lavine, Fundamentals of Heat and Mass Transfer (8th ed., Wiley) — radial conduction through composite cylinders, conduction with internal heat generation, internal-flow convection with a constant surrounding-fluid temperature, and the effectiveness–NTU method for shell-and-tube exchangers. Steam properties are IAPWS-consistent (equivalent to the steam tables); ammonia properties are read from the saturated- and superheated-ammonia tables appended to the examination; air and combustion gases are treated as ideal gases with constant specific heats ($\gamma=1.4$, $R=0.287\ \text{kJ/kg}\cdot\text{K}$, $c_p=1.005\ \text{kJ/kg}\cdot\text{K}$).
Question 3 — Regenerative automotive gas-turbine power plant (Part A, equal value)
Given. A two-shaft regenerative gas turbine: a compressor driven by its own "compressor turbine," and a free "power turbine" that drives the wheels; a rotating regenerator preheats the compressor discharge with the turbine exhaust. Labelled temperatures (°C): intake $T_1=30$, compressor discharge $T_2=425$, combustor inlet (after regenerator) $T_3=1025$, combustor exit $T_4=1700$, exhaust to atmosphere $T_\text{exh}=500$. Air/gas treated as ideal with $c_p=1.005\ \text{kJ/kg}\cdot\text{K}$.
Find. (a) a plant schematic, (b) the T–s diagram, (c) the thermal efficiency, (d) the air flow to deliver 50 kW to the axle.
Parts (a) & (b) — Schematic and T–s diagram
Figure 3a — Plant schematic. Ambient air (30°) is compressed to 425°, preheated in the rotary regenerator to 1025° by the turbine exhaust, burned to 1700°, expanded through the compressor turbine (which drives the compressor) and then through the free power turbine that delivers 50 kW to the wheels; the exhaust gives up heat in the regenerator and leaves at 500°.
Figure 3b — T–s diagram. 1→2 isentropic compression; 2→3 constant-pressure preheat in the regenerator; 3→4 constant-pressure combustion (fuel heat, $q_\text{in}$); 4→5 expansion in the compressor turbine; 5→6 expansion in the power turbine (useful work); 6→exhaust constant-pressure cooling in the regenerator, then rejection to atmosphere at 500°.
Part (c) — Thermal efficiency
Approach. The regenerator only moves heat within the plant, so an overall energy balance around the whole plant needs just the fuel heat added in the combustor and the heat finally rejected with the exhaust. This side-steps the individual turbine/compressor works (and the diagram's internally inconsistent turbine labels).
Fuel heat supplied (combustor, 1025°→1700°). Per unit mass of gas,
$$q_\text{in}=c_p(T_4-T_3)=1.005(1700-1025)=678\ \text{kJ/kg}$$
Heat rejected with the exhaust (500°→ambient 30°).
$$q_\text{out}=c_p(T_\text{exh}-T_1)=1.005(500-30)=472\ \text{kJ/kg}$$
Net work and efficiency. By the first law around the plant, the net (useful) work is the difference:
$$w_\text{net}=q_\text{in}-q_\text{out}=678-472=206\ \text{kJ/kg}$$
$$\eta_\text{th}=\frac{w_\text{net}}{q_\text{in}}=1-\frac{T_\text{exh}-T_1}{T_4-T_3}=1-\frac{470}{675}=0.304$$
$\eta_\text{th}=30.4\%$
Part (d) — Air flow for 50 kW
Size the mass flow. The net plant work per unit mass drives the wheels, so
$$\dot m_\text{air}=\frac{\dot W_\text{axle}}{w_\text{net}}=\frac{50}{206}=0.243\ \text{kg/s}$$
$\dot m_\text{air}\approx 0.24\ \text{kg/s}$
Check — the diagram's turbine labels are internally inconsistent.
The labelled temperatures do not close a rigorous stage-by-stage balance (a compressor-turbine drop of $1700-1200=500$ °C exceeds the compressor rise of $425-30=395$ °C, and a regenerator cannot preheat air to 1025 °C from a 500 °C exhaust). This is a cutaway "illustrative" figure — almost certainly originally in °F, as the question hints ("assuming... expressed in degrees Celsius"). The overall energy balance used here is robust to those inconsistencies: it needs only the fuel-side rise (1025→1700) and the exhaust dumped to atmosphere (500→30), both unambiguous. A constant $c_p=1.005$ is used; real combustion-gas $c_p$ at these temperatures is nearer $1.1$–$1.15\ \text{kJ/kg}\cdot\text{K}$, which would lower $\dot m_\text{air}$ by a few percent while leaving $\eta_\text{th}$ (a temperature ratio) unchanged.
