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25-Nav-A1 Fundamentals of Naval Architecture · May-98-Mar-A1 2016

Question 3 of 8: Brayton Cycle with Compressor/Turbine Losses

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

Notes on this paper

National Examinations, May 2016 — 98-Mar-A1 Applied Thermodynamics and Heat Transfer, 3 hours, open book (Part A: Thermodynamics, Part B: Heat Transfer; 5 of 8 questions required, all 8 answered below for full study coverage).

Reference texts: Cengel & Boles, Thermodynamics: An Engineering Approach; Sonntag, Borgnakke & Van Wylen, Fundamentals of Thermodynamics; Incropera & DeWitt, Fundamentals of Heat and Mass Transfer.

It is solved as the thermodynamics/heat-transfer exam it actually is.

Question 3: Brayton Cycle with Compressor/Turbine Losses

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. Open, simple Brayton (gas-turbine) cycle with non-ideal compressor and turbine and a combustor pressure drop.

Given data
QuantitySymbolValue
Compressor inlet$T_1,\,P_1$20°C, 100 kPa
Compressor exit pressure$P_2$475 kPa
Max. cycle temperature$T_3$870°C
Compressor / turbine efficiency$\eta_c,\,\eta_t$82% / 85%
Combustor pressure drop$\Delta P$13.7 kPa

Find. $P,T$ at each state, net work, thermal efficiency, and the fraction of turbine work consumed by the compressor. (Cold-air-standard assumption: $k=1.4$, $c_p=1.005\text{ kJ/kg}\cdot\text{K}$ — check: variable-property air tables would shift results by a few percent but not the method.)

Entropy s (kJ/kg·K) Temperature T (K) 1234 1→2 compressor · 2→3 combustor · 3→4 turbine · 4→1 exhaust
Cold-air-standard Brayton cycle sketch (T–s); actual states 2 and 4 lie to the right of the isentropic points 2s, 4s due to component losses.

Approach. Use isentropic $T$-$P$ relations to get ideal exit temperatures, then apply the stated compressor/turbine efficiencies to get actual states; take the turbine exhaust back to $P_1=100\text{ kPa}$ (open-cycle assumption).

  1. Compressor, isentropic then actual. $T_1=293.15\text{ K}$; $T_{2s}=T_1(P_2/P_1)^{(k-1)/k}=293.15(4.75)^{0.2857}=457.5\text{ K}$ ($184.4^\circ\text{C}$). With $\eta_c=0.82$: $$T_2=T_1+\frac{T_{2s}-T_1}{\eta_c}=293.15+\frac{457.5-293.15}{0.82}=493.6\text{ K}\quad\Rightarrow\quad \boxed{T_2\approx220.5^\circ\text{C}}\ \text{at } P_2=475\text{ kPa}$$
  2. Combustor exit / turbine inlet. $P_3=P_2-\Delta P=475-13.7=461.3\text{ kPa}$; $T_3=870^\circ\text{C}=1143.15\text{ K}$ (given maximum).
  3. Turbine, isentropic then actual. Expanding to $P_4=P_1=100\text{ kPa}$: $T_{4s}=T_3(P_1/P_3)^{0.2857}=1143.15(100/461.3)^{0.2857}=738.6\text{ K}$ ($465.4^\circ\text{C}$). With $\eta_t=0.85$: $$T_4=T_3-\eta_t(T_3-T_{4s})=1143.15-0.85(1143.15-738.6)=799.4\text{ K}\quad\Rightarrow\quad\boxed{T_4\approx526.1^\circ\text{C}}$$
  4. Work terms. $w_{comp}=c_p(T_2-T_1)=1.005(200.5)=201.5\text{ kJ/kg}$; $w_{turb}=c_p(T_3-T_4)=1.005(343.8)=345.5\text{ kJ/kg}$, so $$\boxed{w_{net}=w_{turb}-w_{comp}=345.5-201.5\approx144.1\text{ kJ/kg}}$$
  5. Efficiency and back-work ratio. $q_{in}=c_p(T_3-T_2)=652.8\text{ kJ/kg}$: $$\boxed{\eta_{th}=\frac{w_{net}}{q_{in}}=\frac{144.1}{652.8}\approx22.1\%},\qquad \boxed{\frac{w_{comp}}{w_{turb}}=\frac{201.5}{345.5}\approx58.3\%}$$
Question 3 — final results
State$P$ (kPa)$T$
1 (compressor inlet)10020°C
2 (compressor exit, actual)475220.5°C
3 (turbine inlet)461.3870°C
4 (turbine exit, actual)100526.1°C
Net work $w_{net}\approx144.1\text{ kJ/kg}$; $\eta_{th}\approx22.1\%$; back-work ratio $\approx58.3\%$