25-Nav-A1 Fundamentals of Naval Architecture · May-98-Mar-A1 2017

Question 3 of 8: Gas Turbine Jet Engine — Static Thrust

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

Notes on this paper

National Examinations, May 2017 — 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: Gas Turbine Jet Engine — Static Thrust (20 marks)

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.

Given data
Quantity	Symbol	Value
Ambient pressure, temperature	$P_1,T_1$	101.325 kPa, -20°C (253.15 K)
Compressor pressure ratio	$r_p$	8
Compressor, turbine isentropic eff.	$\eta_c,\eta_t$	0.88, 0.88
Maximum cycle temperature	$T_3$	1000°C (1273.15 K)
Nozzle efficiency	$\eta_n$	0.90
Mass flow through nozzle	$\dot m$	25 kg/s

Turbojet station numbering for the static test-bed cycle.

Find. Nozzle exit velocity and static thrust.

Approach. Walk the Brayton-cycle stations with actual (efficiency-corrected) compressor and turbine work, then expand the turbine-exit gas through the nozzle to ambient pressure using the nozzle efficiency; thrust follows from momentum (inlet velocity negligible on a static test bed).

Compressor (1→2, actual). $T_{2s}=T_1r_p^{(k-1)/k}=253.15\times8^{0.2857}=458.6\text{ K}$, so $\Delta T_s=205.4\text{ K}$ and, with $\eta_c=\Delta T_s/\Delta T_{actual}$, $$\Delta T_{2,actual}=205.4/0.88=233.4\text{ K}\ \Rightarrow\ T_2=253.15+233.4=486.6\text{ K}\ (213.4\,{}^{\circ}\text{C})$$ $P_2=P_1r_p=8(101.325)=810.6\text{ kPa}$; compressor work $w_c=c_p\Delta T_{2,actual}=1.005(233.4)=234.6\text{ kJ/kg}$.
Turbine (3→4). In a turbojet the turbine drives only the compressor, so $w_t=w_c=234.6\text{ kJ/kg}$, giving $\Delta T_{34,actual}=w_t/c_p=233.4\text{ K}$ and $T_4=1273.15-233.4=1039.7\text{ K}\ (766.6\,{}^{\circ}\text{C})$. With $\eta_t=\Delta T_{actual}/\Delta T_s$, the isentropic drop is $\Delta T_s=233.4/0.88=265.3\text{ K}$, so $T_{4s}=1007.9\text{ K}$ and $$P_4=P_2\left(\dfrac{T_{4s}}{T_3}\right)^{k/(k-1)}=810.6\times\left(\dfrac{1007.9}{1273.15}\right)^{3.5}=357.8\text{ kPa}$$
Nozzle (4→5, expansion to ambient). $T_{5s}=T_4(P_1/P_4)^{(k-1)/k}=1039.7\times(101.325/357.8)^{0.2857}=725.0\text{ K}$, so the isentropic drop is $314.6\text{ K}$ and, with $\eta_n=\Delta T_{actual}/\Delta T_s$, $$\Delta T_{noz,actual}=0.90(314.6)=283.2\text{ K}$$
Exit velocity and static thrust. With the nozzle inlet velocity neglected, $$V_5=\sqrt{2c_p\Delta T_{noz,actual}}=\sqrt{2(1005)(283.2)}=\boxed{754.5\text{ m/s}}$$ $$F=\dot m(V_5-V_1)=25(754.5-0)=\boxed{18{,}860\text{ N}=18.9\text{ kN}}$$

Question 3 — final results
Quantity	Value
Compressor exit temperature	$T_2 = 486.6\text{ K}\ (213.4\,{}^{\circ}\text{C})$
Turbine exit pressure	$P_4 = 357.8\text{ kPa}$
Nozzle exit velocity	$V_5 = 754.5\text{ m/s}$
Static thrust	$F = 18.9\text{ kN}$

Check: the nozzle is treated as expanding fully to ambient pressure (101.3 kPa) with the given 90% nozzle efficiency — the standard exam-level treatment implied by supplying $\eta_n$. A rigorous check shows the critical pressure ratio for air ($P_4/P_{crit}\approx1.89$) would actually choke this converging nozzle at a back-pressure of $\approx189\text{ kPa}$, above ambient; a fully choked analysis would add a pressure-thrust term $A_5(P_5-P_1)$ evaluated at sonic exit conditions. That refinement is beyond the scope implied by the given data and is flagged here rather than silently assumed away.

Topic: Brayton-cycle turbojet performance with component inefficiencies.

Key relations: isentropic $T_2/T_1=r_p^{(k-1)/k}$; compressor/turbine efficiency $\eta=\Delta T_{isentropic}/\Delta T_{actual}$ (compressor) or $\Delta T_{actual}/\Delta T_{isentropic}$ (turbine); turbojet work balance $w_t=w_c$; thrust $F=\dot m(V_e-V_0)+A_e(P_e-P_0)$.

Why this works: In a pure turbojet (no separate power turbine), the gas generator's turbine extracts just enough work to drive the compressor; all remaining enthalpy is available to accelerate the exhaust through the nozzle, which is exactly what produces static thrust when the vehicle itself is stationary.

Common pitfall: Forgetting that the turbine efficiency direction flips relative to the compressor ($\eta_t=\Delta T_{actual}/\Delta T_{isentropic}$, not the reciprocal), and expanding the nozzle from the isentropic turbine-exit state $T_{4s}$ instead of the actual state $T_4$.

Source: Cengel & Boles, Thermodynamics: An Engineering Approach, Ch. 9 (Brayton cycle, jet-propulsion cycles).

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