17-Phys-B6 Applied Thermodynamics and Heat Transfer · May 2016
Question 3 of 8: Brayton Cycle with Compressor/Turbine Losses and Pressure Drop
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 98-Phys-B6 Applied Thermodynamics and Heat Transfer,
National Examination May 2016 — a three-hour open-book examination;
candidates are expected to bring both a thermodynamics text and a heat-transfer text to make use
of the property tables and charts. A complete examination is five questions — either three
from Part A (Thermodynamics, Q1–Q4) and two from Part B (Heat Transfer, Q5–Q8), or two
from Part A and three from Part B — every question carrying equal value; all eight are
solved below as a complete study set. Where the exam's own "state your assumptions" licence
applies (Part A cold-air-standard properties in Question 3; the rectangular-case geometry read
from the printed illustration in Question 7), the assumption is flagged explicitly in a check
callout rather than hedged inside the answer.
Reference texts. Y. A. Çengel and M. A. Boles, Thermodynamics: An
Engineering Approach, 8th ed. (ideal-gas processes, vapour power cycles, gas-turbine/Brayton
cycles, vapour-compression refrigeration); F. P. Incropera and D. P. DeWitt, Fundamentals of
Heat and Mass Transfer, 7th ed. (conduction with convective boundaries, internal/external
convection correlations, natural convection, radiation exchange, cross-flow heat-exchanger
effectiveness–NTU analysis).
Question 3: Brayton Cycle with Compressor/Turbine Losses and Pressure Drop
Given. $T_1=20\,{}^{\circ}\text{C}$, $P_1=100$ kPa (compressor inlet);
$P_2=475$ kPa (compressor exit); $T_3=870\,{}^{\circ}\text{C}$ (turbine inlet, the metallurgical
limit); $\eta_c=82\%$; $\eta_t=85\%$; combustor pressure drop $\Delta P=13.7$ kPa; turbine exhausts
back to $P_1=100$ kPa (open cycle).
Check: no air-property table is printed with this paper (only steam and
R-12 tables), so this is solved with the standard cold-air-standard assumption
($c_p=1.005\text{ kJ/kg}\cdot\text{K}$, $\gamma=1.4$, constant) — the reading a candidate
without a gas-turbine air table would use under the exam's "state your assumptions" licence.
Given data
Quantity
Symbol
Value
Compressor inlet temperature
$T_1$
$20\,{}^{\circ}\text{C}$ (293.15 K)
Compressor inlet / exit pressure
$P_1,\,P_2$
100 kPa / 475 kPa
Turbine inlet temperature
$T_3$
$870\,{}^{\circ}\text{C}$ (1143.15 K)
Compressor / turbine isentropic efficiency
$\eta_c,\,\eta_t$
82% / 85%
Combustor pressure drop
$\Delta P$
13.7 kPa
Fig. 3 — T-s diagram (air-standard Brayton with losses): dashed lines
are the isentropic paths $1\to2s$, $3\to4s$; solid lines are the actual compression/expansion.
Find. Pressure and temperature at states 1–4, the net specific work, the
thermal efficiency, and the percentage of turbine work consumed by the compressor.
Approach. Find each isentropic exit temperature from
$T_{2s}/T_1=(P_2/P_1)^{(\gamma-1)/\gamma}$ (and similarly for the turbine), then apply the
compressor/turbine efficiencies to get the actual exit temperatures; the combustor pressure drop
only shifts $P_3$, not $T_3$ (which is the stated metallurgical limit).
State 1 (compressor inlet). $T_1=293.15$ K, $P_1=100$ kPa.
State 2 (compressor exit), isentropic then actual.
$$T_{2s}=T_1\left(\frac{P_2}{P_1}\right)^{(\gamma-1)/\gamma}=293.15(4.75)^{0.2857}=457.6\text{ K}\;(184.4\,{}^{\circ}\text{C})$$
$$T_2=T_1+\frac{T_{2s}-T_1}{\eta_c}=293.15+\frac{457.6-293.15}{0.82}=493.7\text{ K}$$
$$\boxed{T_2 \approx 220.5\,{}^{\circ}\text{C}},\qquad P_2=475\text{ kPa}$$
State 3 (turbine inlet). $T_3=1143.15$ K ($870\,{}^{\circ}\text{C}$, the given
limit); with the 13.7 kPa combustor drop,
$$P_3=P_2-\Delta P=475-13.7=\boxed{461.3\text{ kPa}}$$
State 4 (turbine exit), isentropic then actual.
$$T_{4s}=T_3\left(\frac{P_4}{P_3}\right)^{(\gamma-1)/\gamma}=1143.15\left(\frac{100}{461.3}\right)^{0.2857}=738.7\text{ K}\;(465.4\,{}^{\circ}\text{C})$$
$$T_4=T_3-\eta_t(T_3-T_{4s})=1143.15-0.85(1143.15-738.7)=799.4\text{ K}$$
$$\boxed{T_4 \approx 526.1\,{}^{\circ}\text{C}},\qquad P_4=100\text{ kPa}$$
Specific work terms.
$$w_{\text{comp}}=c_p(T_2-T_1)=1.005(493.7-293.15)=201.5\text{ kJ/kg}$$
$$w_{\text{turb}}=c_p(T_3-T_4)=1.005(1143.15-799.4)=345.6\text{ kJ/kg}$$
$$w_{\text{net}}=w_{\text{turb}}-w_{\text{comp}}=345.6-201.5=\boxed{144.1\text{ kJ/kg}}$$
Thermal efficiency and compressor's share of turbine work.
$$q_{\text{in}}=c_p(T_3-T_2)=1.005(1143.15-493.7)=652.8\text{ kJ/kg}$$
$$\eta_{\text{th}}=\frac{w_{\text{net}}}{q_{\text{in}}}=\frac{144.1}{652.8}=\boxed{22.1\%}$$
$$\frac{w_{\text{comp}}}{w_{\text{turb}}}\times100\%=\frac{201.5}{345.6}\times100\%=\boxed{58.3\%}$$
