NivaarExam PrepOfficial exam papers ↗

22-Mec-A1 Applied Thermodynamics and Heat Transfer · May 2017

Question 3 of 8: Turbojet on a Static Test Bed — Nozzle Velocity and Thrust

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

Notes on this paper

Reference texts: Çengel & Boles, Thermodynamics: An Engineering Approach (9th ed., McGraw-Hill) — closed- and open-system energy balances, filling of evacuated vessels, air-standard dual and gas-turbine (turbojet) 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) — radial composite-cylinder conduction with convection, external cross-flow over a cylinder, internal-flow temperature decay, radiation between concentric spheres with a shield, and the ε–NTU cross-flow heat-exchanger method. Ammonia and ideal-gas air properties are read from the tables appended to the examination paper; the turbojet uses cold-air-standard constant specific heats.

Paper format: National Examination 16-Mec-A1, May 2017, 3 hours, open book. Part A — Thermodynamics (Q1–4); Part B — Heat Transfer (Q5–8). Each answer carries equal value; a complete paper is any five (three questions from one part and two from the other). All eight questions are solved in full below.

Question 3: Turbojet on a Static Test Bed — Nozzle Velocity and Thrust (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.

QuantityValue
Ambient (test bed, $V\approx0$)$T_1=-20$ °C $=253.15$ K, $P_1=101.325$ kPa
Compressor pressure ratio$r_p=8$; $\eta_c=\eta_t=0.88$
Turbine inlet (max) temperature$T_3=1000$ °C $=1273.15$ K
Nozzle efficiency / air flow$\eta_n=0.90$; $\dot m=25$ kg/s
Working fluid (cold-air standard)$c_p=1.005$ kJ/kg·K, $\gamma=1.4$

Given. The cycle and ambient data above. Find. the nozzle exit velocity and the static thrust.

inletcompressorcombustorturbinenozzleshaft: $w_t=w_c$$V_5$12345
Figure 3 — Turbojet stations: 1 inlet, 2 compressor exit, 3 turbine inlet (max $T$), 4 turbine exit, 5 nozzle exit. The turbine produces exactly the compressor work; the remaining enthalpy drop accelerates the jet in the nozzle.

Approach. Work the compressor and turbine with their isentropic efficiencies (the turbine work equals the compressor work on a single spool), find the turbine-exit pressure, then expand isentropically in the nozzle, apply the nozzle efficiency to the kinetic energy, and get thrust from momentum ($\dot m V_5$ on a static bed).

  1. Compressor (1→2). Ideal exit $T_{2s}=T_1r_p^{(\gamma-1)/\gamma}=253.15(8)^{0.2857}=458.6$ K; with $\eta_c$, $T_2=T_1+\dfrac{T_{2s}-T_1}{\eta_c}=486.6$ K, so $w_c=c_p(T_2-T_1)=234.6$ kJ/kg and $P_2=8P_1=810.6$ kPa.
  2. Turbine (3→4), work = compressor work. $c_p(T_3-T_4)=w_c\Rightarrow T_4=T_3-\dfrac{w_c}{c_p}=1273.15-233.4=1039.7$ K. The isentropic drop is $T_3-T_{4s}=\dfrac{T_3-T_4}{\eta_t}$, giving $T_{4s}=1007.9$ K.
  3. Turbine exit pressure. $P_4=P_2\left(\dfrac{T_{4s}}{T_3}\right)^{\gamma/(\gamma-1)}=810.6(0.7917)^{3.5}=357.8$ kPa.
  4. Nozzle isentropic expansion to ambient (4→5). $T_{5s}=T_4\left(\dfrac{P_1}{P_4}\right)^{(\gamma-1)/\gamma}=1039.7(0.2831)^{0.2857}=725.0$ K, so the ideal kinetic energy is $c_p(T_4-T_{5s})=316.3$ kJ/kg.
  5. Actual jet velocity. Applying the nozzle efficiency, $\tfrac12V_5^{2}=\eta_n\,c_p(T_4-T_{5s})=0.90(316.3)=284.6$ kJ/kg, hence $$V_5=\sqrt{2(284.6\times10^{3})}$$ $V_5\approx755$ m/s
  6. Static thrust. On a stationary bed the inlet momentum is zero and the nozzle exhausts at ambient pressure, so $F=\dot m\,V_5=25(754.5)$. $F\approx18.9$ kN
QuantityResult
Compressor exit temperature486.6 K
Turbine exit temperature / pressure1039.7 K, 357.8 kPa
Nozzle exit velocity≈ 755 m/s
Static thrust≈ 18.9 kN