Question 2 of 9: Regenerative Brayton Cycle, Exergy Destruction by Process
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examinations — 04-BS-10, Thermodynamics. Three hours. Closed-book exam with one
8.5×11 in. double-sided sheet of notes allowed and an approved Casio or Sharp calculator;
property tables and charts are supplied and interpolation is not necessary — the closest tabular
value may be used. Part A: two of three 20-mark questions; Part B: four of six 15-mark questions.
Every question is answered in full.
Reference texts. Cengel & Boles, Thermodynamics: An Engineering
Approach, 8th ed. (Vapor Power Cycles Ch. 10; Gas Power Cycles Ch. 9; Refrigeration Cycles
Ch. 11; Gas Mixtures Ch. 13; Psychrometrics Ch. 14; Exergy Ch. 8).
Check — paper format. The paper is headed "National Examination – May 2019 / 04-BS-10, Thermodynamics, 3 Hours Duration": four question pages followed by a 22-page appendix of property tables and a psychrometric chart. All nine questions are answered in full (Part A questions 1–3 at 20 marks each, Part B questions 4–9 at 15 marks each; the paper asks for two of Part A and four of Part B). Two wording notes from the printed page: Question 2's final sub-part is printed "(d) the second law efficiency" after an earlier "(d)", i.e. the paper repeats the letter, and Question 3(d) is printed "in kJ/K" for a quantity that is a rate of exergy destruction in kJ/s.
Question 2: Regenerative Brayton Cycle, Exergy Destruction by Process (Part A – 20 marks)
[The printed paper labels its last sub-part "(d)" a second time; it is
referred to below as (f), the sixth item asked.]
Given. Compressor inlet (state 1): $T_1=310$ K, $P_1=100$ kPa. Pressure ratio
$r_p=7$ ($P_2=700$ kPa). Turbine inlet (state 3): $T_3=1150$ K. $\eta_c=0.75$, $\eta_t=0.82$,
regenerator effectiveness $\varepsilon=0.65$. Variable-specific-heat air properties (ideal-gas
$s^\circ(T)$, root-solved rather than table-interpolated). Combustor heat source $T_H=1800$ K, sink
$T_0=310$ K. No mass flow rate is given, so the work and exergy quantities in parts (b), (d) and (e)
are all reported per unit mass of air (kJ/kg).
State
Description
T
P
h (kJ/kg)
s° (kJ/kg·K)
1
Compressor inlet
36.85°C (310 K)
100 kPa
436.36
3.9235
2
Compressor exit (actual)
337.23°C
700 kPa
744.34
4.0579
x
Regenerator cold exit / combustor inlet
450.04°C
700 kPa
864.71
4.2388
3
Turbine inlet
876.85°C (1150 K)
700 kPa
1345.67
4.7599
4
Turbine exit (actual)
509.63°C
100 kPa
929.52
4.8834
5
Regenerator hot exit
398.33°C
100 kPa
809.16
4.7176
Find. (a) $T_4$; (b) $w_{net}$ [kJ/kg]; (c) $\eta_{th}$; (d) $x_{dest}$ for each
process [kJ/kg]; (e) total $x_{dest}$; (f) $\eta_{II}$.
Use variable-specific-heat air properties with root-finding on temperature to satisfy each
isentropic pressure ratio, then apply the given component efficiencies to get the actual
compressor- and turbine-exit states. Size the regenerator from its effectiveness definition to fix
the combustor-inlet state, then evaluate net work, heat input, and thermal efficiency. Exergy
destruction in each process uses the full ideal-gas entropy $s(T,P)=s^\circ(T)-R\ln(P/P_{ref})$.
Compressor, actual exit (state 2). Isentropic exit: $T_{2s}$ solved from
$s^\circ(T_{2s})-s^\circ(T_1)=R\ln(P_2/P_1)$, giving $h_{2s}=667.35$ kJ/kg.
$$w_{c,a}=\frac{h_{2s}-h_1}{\eta_c}=\frac{667.35-436.36}{0.75}=307.99\ \text{kJ/kg}\Rightarrow h_2=744.34\ \text{kJ/kg}\ (T_2=337.2\ ^\circ\text{C}).$$
Turbine, actual exit — part (a). Isentropic exit from $T_3$: $h_{4s}=838.17$
kJ/kg.
$$w_{t,a}=\eta_t(h_3-h_{4s})=0.82\times(1345.67-838.17)=416.15\ \text{kJ/kg}\Rightarrow
h_4=h_3-w_{t,a}=929.52\ \text{kJ/kg}.$$
$$T_4=\boxed{509.6\ ^\circ\text{C}\ (782.8\ \text{K})}.$$
Regenerator sizing. $h_x=h_2+\varepsilon(h_4-h_2)=744.34+0.65(929.52-744.34)=864.71$
kJ/kg $\Rightarrow T_x=450.0\ ^\circ\text{C}$; energy balance gives the hot-side exit
$h_5=h_4-(h_x-h_2)=809.16$ kJ/kg $\Rightarrow T_5=398.3\ ^\circ\text{C}$.
Net work output (part b).
$$w_{net}=w_{t,a}-w_{c,a}=416.15-307.99=\boxed{108.2\ \text{kJ/kg}}.$$
Combustor heat input and thermal efficiency (part c). Heat is added only from
$T_x$ to $T_3$, since the regenerator has already preheated the compressor-exit air:
$$q_{in}=h_3-h_x=1345.67-864.71=480.96\ \text{kJ/kg}.$$
$$\eta_{th}=\frac{w_{net}}{q_{in}}=\frac{108.17}{480.96}=\boxed{0.2249\ (22.5\%)}.$$
Exergy destruction by process (part d). $x_{dest}=T_0\Delta s$ for the
adiabatic-but-irreversible compressor and turbine; the combustor and regenerator add the
heat-transfer $q/T$ terms:
$$x_{c}=T_0(s_2-s_1)=310\times(4.0579-3.9235)=\boxed{41.7\ \text{kJ/kg}}$$
$$x_{comb}=T_0\!\left[(s_3-s_x)-\frac{q_{in}}{T_H}\right]=310\times\left[(4.7599-4.2388)-\frac{480.96}{1800}\right]=\boxed{78.7\ \text{kJ/kg}}$$
$$x_{t}=T_0(s_4-s_3)=310\times(4.8834-4.7599)=\boxed{38.3\ \text{kJ/kg}}$$
$$x_{regen}=T_0\big[(s_x-s_2)+(s_5-s_4)\big]=310\times\big[0.1809+(-0.1658)\big]=\boxed{4.7\ \text{kJ/kg}}$$
$$x_{reject}=T_0\!\left[(s_1-s_5)+\frac{q_{out}}{T_0}\right],\quad q_{out}=h_5-h_1=372.80\ \text{kJ/kg}
\Rightarrow x_{reject}=\boxed{126.6\ \text{kJ/kg}}$$
Total exergy destruction (part e).
$$x_{dest,total}=41.7+78.7+38.3+4.7+126.6=\boxed{290.0\ \text{kJ/kg}}.$$
Second-law efficiency (part f). Exergy supplied to the cycle:
$$x_{in}=q_{in}\left(1-\frac{T_0}{T_H}\right)=480.96\times\left(1-\frac{310}{1800}\right)=398.13\ \text{kJ/kg}.$$
$$\eta_{II}=\frac{w_{net}}{x_{in}}=\frac{108.17}{398.13}=\boxed{0.2717\ (27.2\%)}
\quad(\text{check: }w_{net}+x_{dest,total}=108.2+290.0=398.1\approx x_{in}\ \checkmark).$$
