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22-Mec-A1 Applied Thermodynamics and Heat Transfer · December 2014

Question 3 of 8: Gas-Turbine Cycle with Reheat and Ideal Regeneration

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

Notes on this paper

Open-book, 3-hour paper. Part A (Thermodynamics, Q1–Q4) and Part B (Heat Transfer, Q5–Q8); the rubric grades any five (three from one part, two from the other), all of equal value. All eight questions are solved in full. Freon-12 property values are read from the saturated and superheated tables printed in the exam appendix (pages 4–5). Reference texts: Çengel & Boles, Thermodynamics: An Engineering Approach (9th ed.); Çengel & Ghajar, Heat and Mass Transfer (6th ed.); Incropera et al., Fundamentals of Heat and Mass Transfer (8th ed.).

Question 3: Gas-Turbine Cycle with Reheat and Ideal Regeneration (equal value)

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. Air standard ($\gamma=1.4$, $c_p=1.005$); $P_1=100$, $T_1=300\ \text{K}$, $P_2=1000\ \text{kPa}$; turbine inlets at $1400\ \text{K}$; reheat at $300\ \text{kPa}$; regenerator 100 % effective.

Find. Thermal efficiency.

entropy sT12regen (to Tₖ)3456
Figure 4 — $T$–$s$ schematic: 1→2 compression, 2→3 ideal regeneration (heated to the exhaust temperature $T_6$), 3→4 and 5→6 the two turbine stages, 4→5 reheat. States 4 and 6 sit at the same 1400 K inlets after reheat.

Approach. Work out each isentropic temperature ratio, use the 100 %-effective regenerator to raise the compressor discharge to the turbine-exhaust temperature, then form $\eta=w_\text{net}/q_\text{in}$.

  1. Compressor exit. $T_2=T_1(P_2/P_1)^{(\gamma-1)/\gamma}=300(10)^{0.2857}=\boxed{579.2\ \text{K}}$; $w_C=c_p(T_2-T_1)=280.6\ \text{kJ/kg}$.
  2. Turbine stage 1. $T_4=1400(300/1000)^{0.2857}=\boxed{992.5\ \text{K}}$.
  3. Turbine stage 2 (after reheat to 1400 K). $T_6=1400(100/300)^{0.2857}=\boxed{1022.9\ \text{K}}$.
  4. Ideal regeneration. 100 % effective ⇒ compressor air is heated to $T_6=1022.9\ \text{K}$ before the combustor, so $q_\text{in}=c_p(T_3-T_6)+c_p(T_5-T_4)=1.005(377.1)+1.005(407.5)=788.6\ \text{kJ/kg}$.
  5. Net work and efficiency. $w_T=c_p(T_3-T_4)+c_p(T_5-T_6)=788.6\ \text{kJ/kg}$; $w_\text{net}=788.6-280.6=508.0\ \text{kJ/kg}$; $\eta=\dfrac{508.0}{788.6}=\boxed{0.644\ (64.4\%)}$.

Reheat between the two turbine stages raises the average temperature of heat addition, and the ideal regenerator recovers the still-hot exhaust to preheat the compressed air — together they lift the efficiency well above the simple Brayton value (about 48 % at this pressure ratio). Note the exhaust after stage 2 (1022.9 K) is hotter than the compressor discharge (579.2 K), which is exactly why regeneration pays off here.

StateTemperature
1 (compressor inlet)300 K
2 (compressor exit)579.2 K
3, 5 (turbine inlets)1400 K
4 (stage-1 exit)992.5 K
6 (stage-2 exit)1022.9 K
Net work / heat in508 / 789 kJ/kg
Thermal efficiency≈ 64.4 %