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22-Mec-A1 Applied Thermodynamics and Heat Transfer · May 2015

Question 3 of 8: Brayton Gas-Turbine Power Plant

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

Notes on this paper

Paper: National Examinations — 07-Mec-A1 Applied Thermodynamics and Heat Transfer, May 2015. Open-book, 3-hour paper. 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, all of equal value. Full worked solutions to all eight questions are given below.

Reference texts: Çengel & Boles, Thermodynamics: An Engineering Approach (9th ed., McGraw-Hill) — closed- and open-system energy balances, steam tables, vapour and gas power cycles 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) — composite-wall conduction, internal-flow and cross-flow convection correlations, natural convection with radiation, and the ε–NTU / LMTD-correction heat-exchanger methods. Freon-12 property data are taken from the appendix supplied with the exam; steam, air and water data from standard tables.

Question 3: Brayton Gas-Turbine Power Plant (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 Brayton cycle (cold-air-standard: $c_p=1.005\ \text{kJ/kg}\cdot\text{K}$, $k=1.4$); pressure ratio $r_p=P_\text{max}/P_\text{min}=380/95=4$; $T_1=290\ \text{K}$, $T_3=1200\ \text{K}$; net electrical output $\dot W_\text{net}=20{,}000\ \text{kW}$.

Find. (a) turbine power, (b) compressor back-work fraction, (c) mass and volumetric flow of inlet air.

030060090012000.000.250.500.751.00entropy rise s − s₁ (kJ/kg·K)T (K)1 (290 K)2 (431 K)3 (1200 K)4 (808 K)2→3 heat addition at 380 kPathe two vertical legs are the isentropic compression 1→2 and expansion 3→44→1 heat rejection at 95 kPa
Figure 2 — T–s diagram of the ideal Brayton cycle: 1→2 isentropic compression ($r_p=4$), 2→3 constant-pressure heat addition to 1200 K, 3→4 isentropic expansion, 4→1 constant-pressure heat rejection.

Approach. Fix the compressor and turbine exit temperatures from the isentropic pressure ratio, form the specific turbine and compressor works, use the net work to get the mass flow, then scale up to the turbine power and inlet volumetric flow.

  1. Isentropic end temperatures. With $r_p^{(k-1)/k}=4^{0.2857}=1.486$,$$T_2=T_1r_p^{(k-1)/k}=290(1.486)=430.9\ \text{K},\qquad T_4=\frac{T_3}{1.486}=807.5\ \text{K}.$$
  2. Specific works. $$w_t=c_p(T_3-T_4)=1.005(392.5)=394.5\ \tfrac{\text{kJ}}{\text{kg}},\quad w_c=c_p(T_2-T_1)=1.005(140.9)=141.6\ \tfrac{\text{kJ}}{\text{kg}}.$$ Net specific work $w_\text{net}=394.5-141.6=252.9\ \text{kJ/kg}$.
  3. (c) Mass flow. The generator receives the net work, so$$\dot m=\frac{\dot W_\text{net}}{w_\text{net}}=\frac{20{,}000}{252.9}=\boxed{79.1\ \text{kg/s}}.$$
  4. (a) Turbine power. $$\dot W_t=\dot m\,w_t=79.1(394.5)=\boxed{3.12\times10^{4}\ \text{kW}\ (31.2\ \text{MW})}.$$
  5. (b) Back-work fraction. $$\frac{\dot W_c}{\dot W_t}=\frac{w_c}{w_t}=\frac{141.6}{394.5}=\boxed{0.359\ (35.9\%)}.$$
  6. (c) Inlet volumetric flow. $v_1=RT_1/P_1=0.287(290)/95=0.876\ \text{m}^3/\text{kg}$, so$$\dot V_1=\dot m\,v_1=79.1(0.876)=\boxed{69.3\ \text{m}^3/\text{s}}.$$
QuantityResult
$T_2$ / $T_4$431 K / 808 K
Turbine power≈ 31.2 MW
Compressor back-work fraction≈ 35.9 %
Air mass flow≈ 79.1 kg/s
Inlet volumetric flow≈ 69.3 m³/s
Check
The back-work ratio of about 36 % is characteristic of gas turbines — a large share of the turbine output merely runs the compressor, which is why component efficiencies matter so much in practice. Here the cycle is ideal (no efficiencies stated), so isentropic compression and expansion are assumed.