22-Mec-B6 Advanced Fluid Mechanics · December 2018

Question 1 of 8: Compressor Stage Performance

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

Notes on this paper

Paper format: National Examinations, December 2018 — 16-Mec-B6 Fluid Machinery. Closed book, three hours, 22 pages. Section A is calculative (Questions 1–5) and Section B descriptive (Questions 6–8); candidates answer four questions from Section A and two from Section B — six questions of ten marks each, 60 marks in all. Reference data for individual questions are supplied on pages 11–17, and the nomenclature, general constants and reference equations on pages 18–22. All eight questions are solved below.

Check — angle conventions are taken from the paper's own attachments, and they are not the same across the paper. The compressor velocity diagram on page 11 strikes $\alpha_1$ and $\beta_1$ off the axial component $C_{X1}$, so every blade and vane angle in Questions 1 and 2 is measured from the axial direction. The steam-turbine diagram on page 16 strikes $\theta$, $\phi$, $\gamma$ and $\delta$ off the tangential (blade-motion) direction, and the Francis diagram on page 15 defines $\alpha_1$ between $V_1$ and $u_1$ — also tangential. Questions 4 and 5 therefore use the tangential reference. Every constant below is the paper's own page-19 value: $g = 9.81\ \text{m}\,\text{s}^{-2}$, $\rho_{\text{water}} = 1000\ \text{kg}\,\text{m}^{-3}$, $c_p = 1005\ \text{J}\,\text{kg}^{-1}\text{K}^{-1}$ and $c_v = 718\ \text{J}\,\text{kg}^{-1}\text{K}^{-1}$, from which $R = c_p - c_v = 287\ \text{J}\,\text{kg}^{-1}\text{K}^{-1}$ and $k = c_p/c_v = 1.3997$ — not a textbook 1.40.

Reference texts

Dixon, S. L. and Hall, C. A., Fluid Mechanics and Thermodynamics of Turbomachinery, 7th ed. — 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. 5 (axial flow compressors), Ch. 6 (compressor and turbine matching, surge and stall), Ch. 3 (jet propulsion cycles).
Turton, R. K., Principles of Turbomachinery, 2nd ed. — Ch. 2 (Euler equation), Ch. 8 (fans, blade form and control).
White, F. M., Fluid Mechanics, 8th ed. — Ch. 11 (turbomachinery: specific speed, Francis and Kaplan turbines, cavitation).
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).

Question 1 — Compressor Stage Performance (10 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.

Quantity	Symbol	Value
LP compressor stages / pressure ratio	$n_{LP}$, $r_{LP}$	7 / 3.0 (whole spool)
HP compressor stages / pressure ratio	$n_{HP}$, $r_{HP}$	7 / 5.0 (whole spool)
Isentropic efficiency, each spool	$\eta_{is}$	0.90
Mach number at compressor inlet	$M_A$	0.49
Effective flow area at compressor inlet	$A_1$	$0.80\ \text{m}^2$
Air pressure and temperature at inlet	$p_1$, $T_1$	$80\ \text{kPa}$, $80\ {}^{\circ}\text{C} = 353.15\ \text{K}$
Air properties (page 19)	$c_p$, $c_v$	$1005$, $718\ \text{J}\,\text{kg}^{-1}\text{K}^{-1}$

Find. The T–s picture of the two-spool compression, the temperatures at LP exit, HP exit and at the fifth-stage cooling bleed, the axial air velocity entering the compressor, the mass flow and the shaft power absorbed by the whole compressor.

Part (a) — the T–s diagram. The sketch below carries every point the later parts refer to: the compressor face 1, the ideal and real LP delivery states 2s and 2, the ideal and real HP delivery states 3s and 3, and the bleed tapping b after the fifth HP stage. The three dashed curves are the constant-pressure lines through 80 kPa, 240 kPa and 1200 kPa.

The compression drawn on temperature–entropy axes. Point 1 is the compressor face, 2 the LP delivery, 3 the HP delivery, and b the bleed tapping after the fifth HP stage. The dashed verticals 1–2s and 2–3s are the ideal (isentropic) compressions; the solid lines are the real ones, which drift to the right because the losses generate entropy.

Approach. Take the two spools in series: for each, find the ideal temperature rise from the isentropic pressure relation, divide it by the isentropic efficiency to get the real rise, then use the compressor-face density and Mach number to fix the velocity and mass flow, and close with a steady-flow energy balance for the power.

Fix the gas properties from the paper's own constants. The specific gas constant and the ratio of specific heats follow from the page-19 values of $c_p$ and $c_v$: $$R = c_p - c_v = 1005 - 718 = 287\ \text{J}\,\text{kg}^{-1}\text{K}^{-1},\qquad k = \frac{c_p}{c_v} = \frac{1005}{718} = 1.3997$$ so the isentropic exponent is $\dfrac{k-1}{k} = 0.28557$. Using a memorised 1.400 instead shifts the delivery temperature by about a degree; it is quicker and safer to use the paper's numbers.
Ideal and real temperature rise across the LP spool (part b). For an isentropic compression through the pressure ratio $r_{LP}$, $$T_{2s} = T_1\,r_{LP}^{(k-1)/k} = 353.15 \times 3^{0.28557} = 483.29\ \text{K}$$ The isentropic efficiency is defined as the ideal rise divided by the real rise, so the real rise is larger: $$\Delta T_{LP} = \frac{T_{2s}-T_1}{\eta_{is}} = \frac{483.29 - 353.15}{0.90} = 144.60\ \text{K}$$ and the LP delivery temperature is $T_2 = 353.15 + 144.60 = 497.75\ \text{K}$.
Repeat for the HP spool, which breathes the LP delivery air. The HP inlet temperature is $T_2$, so $$\begin{aligned}T_{3s} &= T_2\,r_{HP}^{(k-1)/k} = 497.75 \times 5^{0.28557} = 788.17\ \text{K} \\ \Delta T_{HP} &= \frac{788.17-497.75}{0.90} = 322.69\ \text{K}\end{aligned}$$ Adding this to the HP inlet temperature gives the compressor delivery condition: $$\boxed{T_2 = 497.8\ \text{K} = 224.6\ {}^{\circ}\text{C}\qquad T_3 = 820.4\ \text{K} = 547.3\ {}^{\circ}\text{C}}$$ Note that the overall pressure ratio is $3 \times 5 = 15$ but the overall rise is not the sum of two independent rises calculated from $T_1$: each spool multiplies a hotter inlet, which is exactly why the HP spool contributes more than twice the LP rise for a comparable pressure ratio.
Temperature of the turbine-cooling bleed (part c). The paper gives no stage-by-stage pressure schedule, so take the standard preliminary assumption for a multi-stage axial machine — an equal temperature rise in each stage of the spool: $$\Delta T_{\text{stage},HP} = \frac{\Delta T_{HP}}{n_{HP}} = \frac{322.69}{7} = 46.10\ \text{K}$$ Air tapped downstream of the fifth of the seven HP stages has therefore received five of those rises: $$T_b = T_2 + 5\,\Delta T_{\text{stage},HP} = 497.75 + 5(46.10) = 728.25\ \text{K}$$ $$\boxed{T_b = 728.2\ \text{K} = 455.1\ {}^{\circ}\text{C}}$$ That is the temperature of the air handed to the turbine blade-cooling passages — hot by any ordinary standard, but several hundred degrees below the gas it must protect.
Axial velocity at the compressor face (part d). The inlet temperature is the static value at that station, so the local acoustic velocity and the flow velocity are $$a_1 = \sqrt{k\,R\,T_1} = \sqrt{1.3997 \times 287 \times 353.15} = 376.65\ \text{m/s}$$ $$\boxed{V_1 = M_A\,a_1 = 0.49 \times 376.65 = 184.6\ \text{m/s}}$$ This is a free check on the whole reading of the data: Question 2 states the inlet air velocity to the first stage as 184 m/s, and the two agree to 0.3 per cent.
Density, mass flow and shaft power (part e). The ideal-gas law fixes the density at the compressor face, and continuity through the stated effective area gives the mass flow: $$\begin{aligned}\rho_1 &= \frac{p_1}{R\,T_1} = \frac{80\,000}{287 \times 353.15} = 0.7893\ \text{kg}\,\text{m}^{-3} \\ \dot m &= \rho_1 A_1 V_1 = 0.7893 \times 0.80 \times 184.6 = 116.5\ \text{kg/s}\end{aligned}$$ Question 2 quotes an air mass flow of 116 kg/s, which confirms the area and Mach number have been read correctly. Neglecting the cooling extraction, the steady-flow energy equation for an adiabatic compressor makes the shaft power the enthalpy rise of the whole stream: $$P = \dot m\,c_p\,(T_3 - T_1) = 116.54 \times 1005 \times (820.44 - 353.15)$$ $$\boxed{\dot m = 116.5\ \text{kg/s}\qquad P = 54.7\ \text{MW}}$$ A single Olympus 593 absorbing 55 MW to drive its compressors is entirely credible: the turbines must deliver that much before a newton of thrust is produced.

Check — the bleed temperature depends on how the seven HP stages are assumed to share the duty. Equal temperature rise per stage (used above) gives 728.2 K. If instead each stage is assumed to take an equal share of the pressure ratio, so that the pressure after five stages is $5^{5/7} = 3.157$ times the HP inlet pressure, the same isentropic-efficiency treatment gives $T_b = 712.7\ \text{K} = 439.5\ {}^{\circ}\text{C}$, about 15 K lower. Both are defensible for a two-mark part; state which assumption is used, as done here. Equal work per stage is the conventional preliminary-design assumption and is the one that follows directly from the equal-blade-speed, equal-loading layout of a constant-mean-diameter spool.

Quantity	Symbol	Result
LP compressor exit temperature	$T_2$	$497.8\ \text{K}$ ($224.6\ {}^{\circ}\text{C}$)
HP compressor exit temperature	$T_3$	$820.4\ \text{K}$ ($547.3\ {}^{\circ}\text{C}$)
Blade-cooling bleed after HP stage 5	$T_b$	$728.2\ \text{K}$ ($455.1\ {}^{\circ}\text{C}$)
Air velocity at compressor inlet	$V_1$	$184.6\ \text{m/s}$
Mass flow through the compressor	$\dot m$	$116.5\ \text{kg/s}$
Power required to drive the compressor	$P$	$54.7\ \text{MW}$

Topic. Multi-stage axial compression with a stated isentropic efficiency, and the compressor-face flow condition of a supersonic-intake turbojet.

Key relations. Isentropic temperature ratio $T_{2s}/T_1 = (p_2/p_1)^{(k-1)/k}$; isentropic efficiency $\eta_{is} = (T_{2s}-T_1)/(T_2-T_1)$; acoustic velocity $a = \sqrt{kRT}$ and $M_A = V/a$; ideal gas $\rho = p/RT$; continuity $\dot m = \rho A V$; steady-flow work $P = \dot m\,c_p\,\Delta T$.

Why this works. The isentropic pressure relation is a statement about a reversible adiabatic path, so it never gives the real delivery temperature directly; what it gives is a reference rise, which the efficiency then scales up to the real one. Because the real path ends hotter, its entropy is higher, and this is exactly what the rightward drift on the T–s diagram represents. Applying the same construction spool by spool, rather than once across the overall ratio of 15, matters because the second spool starts from the first spool's hot delivery and so needs more work per unit of pressure ratio — the divergence of the constant-pressure lines on the T–s diagram is the geometric statement of the same fact. On the flow side, the inlet Mach number and the static temperature between them fix the axial velocity without any reference to the flight condition, and the ideal-gas density closes the mass flow. The fact that both the velocity and the mass flow reproduce the values quoted independently in Question 2 shows the paper was designed around one consistent operating point.

Common pitfall. Dividing by the efficiency in the wrong sense — multiplying the ideal rise by 0.90 instead of dividing — gives a delivery temperature below the isentropic one, which is thermodynamically impossible for a compressor, and the error is easy to spot by asking whether the real machine should be hotter or cooler than the ideal one. The second trap is applying the overall pressure ratio of 15 in one step to a single isentropic efficiency: that gives a different (and lower) delivery temperature than the two-spool calculation, because it hides the reheat effect between the spools.

Source. Cohen, Rogers & Saravanamuttoo, Gas Turbine Theory, 7th ed., Ch. 5 (axial flow compressors) and §2.2 (efficiency definitions); Dixon & Hall, Fluid Mechanics and Thermodynamics of Turbomachinery, 7th ed., §5.2–5.4.

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