Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: National Examinations, May 2018 — 16-Mec-B6 Fluid Machinery. Closed book, three hours, 21 pages. Section A is calculative (Questions 1–5) and Section B descriptive (Questions 6–8); candidates answer four from Section A and two from Section B, six questions of ten marks each for a total of 60. Reference data for individual questions are supplied on pages 11–16 and the nomenclature, constants and reference equations on pages 17–21. All eight questions are solved below.
Check — angle conventions taken from the paper's own attachments. The compressor attachment on page 11 strikes $\alpha_1$ and $\beta_1$ off the axial component $C_{X1}$, so in Questions 1 and 2 all blade and vane angles are measured from the axial direction. The steam-turbine attachment on page 13 and the Francis attachment on page 14 strike $\theta$, $\phi$, $\gamma$, $\delta$, $\alpha$ and $\beta$ off the tangential (blade-motion) direction, so Questions 3 and 5 use the tangential reference. Mixing the two is the single most common way to lose all the marks on a velocity-diagram question. Every constant used below ($g = 9.81\ \text{m}\,\text{s}^{-2}$, $\rho_{\text{water}} = 1000\ \text{kg}\,\text{m}^{-3}$, $\rho_{\text{air}} = 1.21\ \text{kg}\,\text{m}^{-3}$ at 15 °C, $c_p = 1.005$ and $c_v = 0.718\ \text{kJ}\,\text{kg}^{-1}\text{K}^{-1}$) is taken from the paper's page 18 rather than from a textbook.
Reference texts
Dixon, S. L. and Hall, C. A., Fluid Mechanics and Thermodynamics of Turbomachinery, 7th ed. — Ch. 3 (two-dimensional cascades), Ch. 5 (axial compressors), Ch. 4 (axial turbines), Ch. 9 (hydraulic turbines).
Cohen, H., Rogers, G. F. C. and Saravanamuttoo, H. I. H., Gas Turbine Theory, 7th ed. — Ch. 3 (jet propulsion cycles), Ch. 5 (axial flow compressors), Ch. 6 (compressor and turbine matching, surge and stall).
Turton, R. K., Principles of Turbomachinery, 2nd ed. — Ch. 2 (Euler equation), Ch. 4 (pump characteristics), Ch. 8 (fans and their control).
White, F. M., Fluid Mechanics, 8th ed. — Ch. 11 (turbomachinery: specific speed, pump curves, Pelton and Francis turbines).
Fox, R. W. and McDonald, A. T., Introduction to Fluid Mechanics, 10th ed. — Ch. 10 (fluid machinery, similarity rules).
Çengel, Y. A. and Boles, M. A., Thermodynamics: An Engineering Approach, 9th ed. — Ch. 9 (gas power cycles, Brayton and turbojet).
Given. The same eight-stage machine as Question 1, now analysed thermodynamically: overall pressure ratio 6.8 split equally between the stages, 20 kg/s of air entering at 15 °C, 12.65 kN of static thrust with the engine at rest, and ideal (isentropic, loss-free) behaviour throughout.
Find. The stage pressure ratio, the first-stage and whole-compressor power, the exit jet velocity that produces the stated static thrust, the turbine power, the net power appearing in the gas flow, and a labelled temperature–entropy diagram of the cycle.
Figure 2.1 — Ideal (Brayton-type) turbojet cycle on temperature–entropy axes. Compression 1–2 and expansion 3–4–5 are isentropic (vertical); combustion 2–3 and exhaust 5–1 lie on constant-pressure lines. The turbine takes only enough work to drive the compressor; everything remaining is expanded in the nozzle to produce the jet.
Approach. Split the overall pressure ratio into eight equal stage ratios, apply the isentropic temperature relation to get the stage and overall temperature rises and hence the compressor power, then use the momentum equation for the jet velocity and the energy equation for the jet power.
Part (a)(i) — divide the pressure ratio between the stages. Equal ratios multiply to the overall ratio, so
$$r_{\text{stage}} = r_c^{1/n} = 6.8^{1/8}$$
$$\boxed{r_{\text{stage}} = 1.271}$$
Checking, $1.271^{8} = 6.80$. Note that equal pressure ratio is not the same as equal temperature rise; later stages, entering hotter air, gain more degrees for the same ratio.
Part (a)(ii) — first-stage temperature rise and power. For isentropic compression the paper's page 19 relation $T_2/T_1 = (p_2/p_1)^{(k-1)/k}$ gives, with $(k-1)/k = 0.2857$,
$$\Delta T_1 = T_1\left(r_{\text{stage}}^{(k-1)/k} - 1\right) = 288.15\left(1.271^{0.2857} - 1\right) = 20.41\ \text{K}$$
Multiplying by the flow and the specific heat,
$$P_{\text{stage}} = \dot{M}c_p\Delta T_1 = 20 \times 1.005 \times 20.41$$
$$\boxed{P_{\text{stage}} = 410.2\ \text{kW}}$$
This is the thermodynamic counterpart of the 412.1 kW obtained from the velocity diagram in Question 1(c); the two routes agree to 0.5 %, which is the accuracy of a scale drawing.
Part (a)(iii) — power for the whole compressor. Applying the same isentropic relation across all eight stages at once,
$$\Delta T_{\text{tot}} = T_1\left(r_c^{(k-1)/k} - 1\right) = 288.15\left(6.8^{0.2857} - 1\right) = 210.0\ \text{K}$$
so the delivery temperature is $T_1 + \Delta T_{\text{tot}} = 498.1\ \text{K}$ (225.0 °C) and
$$P_c = \dot{M}c_p\Delta T_{\text{tot}} = 20 \times 1.005 \times 210.0$$
$$\boxed{P_c = 4221\ \text{kW} = 4.221\ \text{MW}}$$
Eight times the first stage would be only 3282 kW. The 29 % shortfall is the whole point of the "equal pressure ratio" instruction: the rise per stage grows in proportion to the temperature at which each stage begins, so a stage-by-stage march from 288.15 K through eight ratios of 1.271 lands exactly on 498.1 K.
Part (b)(i) — jet velocity from the momentum equation. The page-21 thrust relation is $T = \dot{M}(V_{\text{jet}} - V_{\text{aircraft}})$, and the engine is stationary, so $V_{\text{aircraft}} = 0$:
$$V_{\text{jet}} = \frac{T}{\dot{M}} = \frac{12\,650}{20}$$
$$\boxed{V_{\text{jet}} = 632.5\ \text{m}\,\text{s}^{-1}}$$
Substituting back, $20 \times 632.5 = 12\,650\ \text{N}$, the stated static thrust. The 200 m/s quoted as "air inlet velocity" is the velocity at the compressor face, drawn in by the engine itself; it is not a free-stream velocity and must not be subtracted here.
Part (b)(ii) — turbine power. On a single-spool engine the turbine's only mechanical duty is to drive the compressor, and the problem specifies ideal conditions, so there are no bearing, windage or leakage losses to cover:
$$P_{\text{turbine}} = P_c$$
$$\boxed{P_{\text{turbine}} = 4.221\ \text{MW}}$$
Everything the turbine does not extract is left in the gas as pressure and temperature for the propelling nozzle — that division is what distinguishes a turbojet from a shaft-power gas turbine.
Part (b)(iii) — net power in the gas flow. The page-21 jet-power relation with the aircraft at rest reduces to the kinetic energy flux leaving the nozzle:
$$P_{\text{jet}} = \frac{\dot{M}\left(V_{\text{jet}}^{2} - V_{\text{aircraft}}^{2}\right)}{2} = \frac{20 \times 632.5^{2}}{2}$$
$$\boxed{P_{\text{jet}} = 4.001\ \text{MW}}$$
An independent form of the same result is $P_{\text{jet}} = \tfrac{1}{2}TV_{\text{jet}} = 0.5 \times 12\,650 \times 632.5 = 4.001\ \text{MW}$. Note the propulsive efficiency $\eta_P = 2V_{\text{aircraft}}/(V_{\text{jet}} + V_{\text{aircraft}})$ is identically zero on the test bed: 4.0 MW is being poured into the atmosphere and none of it is doing useful work, because the thrust acts through no distance.
Part (c) — place the results on the T–s diagram. Figure 2.1 carries the cycle with each number on the process that produces it: 4.221 MW consumed on 1–2 (of which 410.2 kW in the first stage, ending at 498.1 K), heat added at constant pressure on 2–3, 4.221 MW produced on 3–4 by the turbine, and the remaining expansion 4–5 in the nozzle delivering 4.001 MW of jet kinetic power at 632.5 m/s.
