Compressors and turbines

Welcome to part three, inside the compressor and the turbine. We derive the Euler equation, fix the conventions, analyse a compressor stage and the repeating-stage design with its chart, solve exercises 2 A and 2 B, and then do the same for turbine stages, with the design map, the Zweifel coefficient and the blade stress check of exercises 3 A and 3 B. Exercise 3 A is also your group 13 deliverable due on Wednesday the twenty-eighth at 17:00, so pay special attention to it.

Everything starts with Euler's equation. The torque on the fluid equals the change of its angular momentum, m dot times r v at exit minus r v at inlet, where v is the tangential velocity, and the power is omega times the torque. Equating with the first law, the change of total enthalpy equals omega times the change of r v. At the mean radius, with the blade speed U equal to omega times r m, the compressor work per unit mass is U times the increase of swirl, and the turbine work is U times the decrease of swirl. So the work depends only on the blade speed and the tangential velocities, not on the axial velocity or the blade shape, and high blade speed is the most direct route to high stage work, until blade stress stops you.

Most mistakes in turbomachinery are about conventions, so fix them now. A compressor stage is a rotor followed by a stator, while a turbine stage is a stator, or nozzle, followed by a rotor, so station 2 is the rotor exit in a compressor but the rotor inlet in a turbine, and these local stations have nothing to do with the cycle stations. Absolute and relative velocities are related by V equals V R plus U, the axial component u is the same in both frames, and all angles are measured from the axial direction, alpha absolute and beta relative. Static properties are the same in both frames, but total temperature, total pressure and Mach number are not, which is why the rotor must always be analysed in its own frame. In a compressor the tangential components satisfy v plus v R equals U; in a turbine, following Mattingly, v 2 is measured along U and v 3 against it, so v 2 R equals v 2 minus U and v 3 R equals v 3 plus U, and the two swirls add in the work.

In a compressor stage, only the rotor does work: c p delta T t equals U times v 2 minus v 1, which can be written with absolute angles, as adding swirl, or with relative angles, as removing relative swirl, which is paid for with diffusion. Across the rotor, in its own frame, nothing does work, so the relative total temperature is conserved, while the relative total pressure drops by the rotor loss ratio L R. Across the stator, total temperature is conserved and total pressure drops by L S. The losses are quantified by loss coefficients: the total pressure drop divided by the inlet dynamic pressure in the row's frame, which converts into L R equals one minus phi times gamma P 1 M 1 R squared over two P t 1 R, the form used in exercise 2 A. Both rows decelerate the flow in their own frame, against an adverse pressure gradient, and that is why a compressor needs many stages while one or two turbine stages can drive it.

A stage is characterised by a few dimensionless groups. The stage pressure and temperature ratios give its isentropic and polytropic efficiencies with the same formulas as for a whole compressor. The degree of reaction is the fraction of the static temperature rise that happens in the rotor; 0.5 shares the diffusion equally, and Farokhi argues for about 0.6 because rotating boundary layers tolerate adverse gradients slightly better. The stage loading coefficient Psi is c p delta T t over U squared, typically 0.30 to 0.35 in a high pressure compressor, and the flow coefficient Phi is u over U, typically 0.45 to 0.55. The diffusion factor, due to Lieblein, measures the deceleration on the suction side and must stay below about 0.6, while the de Haller ratio of 0.72 is only a rough sanity check. And the blade stress parameter, A N squared, links the annulus area and the shaft speed to the specific strength the material must provide.

The mean-line analysis gives flow angles, but the blades are drawn with metal angles, and the cascade vocabulary connects the two. The solidity sigma is the chord over the pitch, where the pitch is two pi r m over the number of blades; referred to the axial chord it is the axial solidity, sigma times the cosine of the stagger angle. The flow does not follow the blade exactly, so we distinguish the flow angles, alpha in the stator or beta in the rotor, from the metal angles gamma, drawn dotted in the sketch. At the leading edge the difference is the incidence, alpha i minus gamma i, which is nearly zero at design. At the trailing edge the flow under-turns by the deviation, alpha e minus gamma e: Carter's rule gives the camber over four root sigma for a compressor, while in a turbine the accelerating flow keeps it small, the sum of the metal angles over eight root sigma. Packing the blades more closely reduces deviation and the diffusion factor, at the price of more blades, weight and friction; the Zweifel criterion of the turbine fixes exactly this solidity.

A powerful simplification is the repeating-row, repeating-stage design: the stage exit velocity and angle equal those at its inlet, using mirror-image rows, so stages can be stacked indefinitely. Then v 1 plus v 2 equals the blade speed, and the work becomes V 2 squared minus V 1 squared over c p: the same temperature rise for every stage, a consequence of the geometry, not an extra assumption. The stage temperature ratio depends only on the inlet Mach number and the two flow angles. The ratio of wheel speed to inlet velocity, and the loading and flow coefficients, become purely geometric functions of alpha 1 and alpha 2. The degree of reaction is exactly one half, and the polytropic efficiency of a stage equals that of the whole compressor, which is why it is the efficiency carried from stage to machine.

Because the repeating stage depends on only a few numbers, it is tabulated as a design chart, like this one from exercise 2 A. Against the inlet angle alpha 1, it shows the exit angle alpha 2 on the right axis, the wheel speed ratio omega r over V 1 in green, and the stage pressure ratio for several inlet Mach numbers in blue. There are two ways to use it: if the blade speed is fixed by the shaft and the radius, choose alpha 1, read omega r over V 1, get V 1 and the Mach number, and then the stage pressure ratio and alpha 2. If instead a pressure ratio is the target, every intersection with a Mach number line is a valid design, which fixes the angles and the Mach number, and the chart then tells you the blade speed needed. The trends: a higher inlet Mach number gives more pressure ratio per stage, and a large alpha 1 requires a high wheel speed; keep M 1 between 0.4 and 0.6.

A multistage compressor stacks identical repeating stages, each adding the same total temperature rise. So the exit total temperature is the inlet one plus n s times delta T t, and the overall pressure ratio follows from the polytropic efficiency: the temperature ratio raised to gamma e c over gamma minus one. Because the inlet temperature of each stage grows, its temperature ratio and pressure ratio fall along the machine, so the overall ratio is not the first stage ratio to the power n s; in the manual's example the stage ratio falls from 1.25 to 1.15. Inverting the formula gives the number of stages for a target pressure ratio, rounded up to an integer. The manual warns that Mattingly also writes pi c equals pi c s to the N, but that is only a device for relating efficiencies and is inconsistent with real repeating triangles.

So far everything happened at the mean radius, but a blade spans from hub to tip. If the same swirl change were imposed at every radius, Euler's equation would give a work proportional to the radius, much larger at the tip, and the resulting pressure gradients would drive recirculation at both walls. So the design keeps the work uniform along the span, r times the swirl change constant, combined with the radial equilibrium equation for constant total enthalpy. The simplest solution is the free vortex, r v constant, which keeps the axial velocity uniform and gives a reaction of one minus one minus the mean reaction, times r m over r squared. As the plot shows, the reaction is lowest at the hub and highest at the tip, and on a long compressor blade it can become negative at the hub, which is why other swirl distributions are used when the hub to tip ratio is low. In a turbine the blades are shorter, so the variation is milder, and the usual criterion is to keep the hub reaction above about minus 0.15. Finally, the hub and tip radii follow from the annulus area given by the mass flow parameter and the chosen hub to tip ratio, typically 0.6 to 0.75 at a compressor inlet.

Exercise 2 A gives the inlet angle, the static inlet state and the shaft: 800 millimetres mean radius at 2 500 rpm. The blade speed is omega r: 2 500 times two pi over 60 times 0.8, which is 209.44 metres per second. Entering the chart at alpha 1 equal to 30 degrees, the green line gives omega r over V 1 of about 1.6, so V 1 is 130.9 metres per second. With the inlet speed of sound of 346 metres per second, the inlet Mach number is 0.38. Now the triangle: the axial velocity is V 1 cos 30, 113.36 metres per second, the inlet swirl is 65.45, and the repeating-row condition v 1 plus v 2 equals U gives v 2 equal to 143.99. So alpha 2 is 51.79 degrees, consistent with the 50 degrees read on the chart, and V 2 is 183.26 metres per second, all as in the official solution.

Here are the two triangles drawn to scale, with the tangential direction horizontal and the axial direction pointing down. At the rotor inlet, the absolute velocity V 1 and the relative velocity V 1 R start from the same point, and the blade speed U joins their tips, because V equals V R plus U. At the rotor exit the absolute velocity has gained swirl, 144 metres per second against 65, and the relative velocity has lost it: the rotor added work by turning the flow. Notice the mirror symmetry of the repeating stage: the relative inlet velocity equals the absolute exit velocity, 183.3 metres per second, and the relative exit velocity equals the absolute inlet velocity, so only one triangle really has to be solved.

Now the thermodynamics, row by row. The inlet total temperature is 306.53 kelvin and the total pressure 1.104 bar; in the rotor frame, with V 1 R equal to 183 metres per second, the relative totals are 314.72 kelvin and 1.211 bar. The rotor loss coefficient of 0.02 drops the relative total pressure only to 1.207 bar, and since the relative total temperature is conserved, the rotor exit static state is 306.19 kelvin and 1.096 bar. The stage work is V 2 squared minus V 1 squared over c p, 16.38 kelvin, so the absolute total temperature becomes 322.91 kelvin and the total pressure 1.320 bar. The stator loses a little pressure, to 1.316 bar, and decelerates the flow back to V 1, raising the static temperature to 314.38 kelvin and the static pressure to 1.198 bar. So the stage pressure ratio is 1.192 with an isentropic efficiency of 96.4 %, the reaction is exactly 0.5, the polytropic efficiency is 96.5 %, and nine such stages give an overall pressure ratio of 3.77, the official 3.76 to rounding.

The exercises ask for a T s sketch of the stage, so here is a template drawn to scale with the 2 A numbers. Start with the three static states on their isobars: the static temperature rises through the rotor and again through the stator, because both rows diffuse. Each total state sits above its static state by V squared over two c p, at the same entropy, because a total state is reached isentropically. In the rotor frame, the relative total temperature is the same at 1 R and 2 R, a horizontal line, while the relative total pressure drops a little by the loss ratio. In the absolute frame the rotor adds work: the total temperature rises from t 1 to t 2 by 16.4 kelvin. Through the stator the absolute total temperature is constant and the total pressure falls slightly, while the static temperature climbs back as the flow decelerates to V 1. And every step moves to the right: each loss is an entropy increase, which is how you show the losses on the sketch.

Exercise 2 B works the other way: the target is an overall pressure ratio of 5 in nine stages, with a polytropic efficiency of 90 %; careful, the statement gives an inlet total temperature of 316.18 kelvin, but the official solution uses 306.53, the value of 2 A, and so do we. On the 2 B chart, which plots the nine-stage pressure ratio against alpha 1, any intersection of the pi c equal to 5 level with a Mach line is valid; choosing M 1 equal to 0.45, in the recommended range, gives alpha 1 about 26 degrees, and the dashed line gives alpha 2 about 50 degrees. The static inlet temperature is 306.53 over 1.0405, 294.6 kelvin, so V 1 is 0.45 times 344, 154.82 metres per second. The axial velocity is V 1 cos 26, 139.15, and the repeating-stage condition gives the blade speed needed, u times tan 26 plus tan 50, 233.71 metres per second. V 2 is u over cos 50, 216.48 metres per second, with swirls of 67.87 and 165.84. And the stage work is V 2 squared minus V 1 squared over c p, 22.79 kelvin.

Now the stage cycle, row by row, which the official sheet lists. The inlet static pressure is 0.983 bar; in the rotor frame the inlet velocity equals V 2, so the relative totals are 317.93 kelvin and 1.284 bar, and the rotor loss ratio of 0.9975 leaves 1.281 bar. With the relative total temperature conserved, the rotor exit static temperature is 305.99 kelvin and the pressure 1.120 bar, and in the absolute frame the totals are 329.32 kelvin and 1.449 bar. The polytropic efficiency gives the stage pressure ratio, 329.32 over 306.53 to the power 1.4 times 0.9 over 0.4, which is 1.2535, so the stage exit total pressure is 1.416 bar, the implied stator loss ratio 0.978, and the reaction 0.5. The stage isentropic efficiency is 89.67 %. The compressor exit temperature is 306.53 plus nine times 22.79, 511.66 kelvin, and the overall isentropic efficiency is 87.24 %, lower than the polytropic 90 %, exactly as part one predicts; with the statement's 316.18 kelvin the exit would be 527.8 kelvin and the efficiency the same.

The turbine is not a compressor run backwards. Its stage is a nozzle followed by a rotor, and the flow accelerates through both, under a favourable pressure gradient, so the boundary layers stay thin and there is nothing like compressor stall. That allows large turning and large pressure drops per stage, which is why one or two turbine stages drive a ten-stage compressor. Instead, the turbine is limited by temperature and stress: the gas is hotter than any blade material can bear, so the blades are cooled with compressor air, which is a direct penalty on the cycle. Euler's equation for the stage is c p delta T t equals U times v 2 plus v 3: with our sign convention the inlet and exit swirls add, and that is why a turbine stage extracts so much more work than a compressor stage absorbs. The nozzle conserves total temperature and loses total pressure, and the rotor conserves relative total temperature and loses relative total pressure.

Turbine stages are characterised by the loading coefficient, c p delta T t over U squared, and the flow coefficient, u over U, and with constant axial velocity both ratios Psi over Phi and R t over Phi depend only on the relative angles. In a zero-reaction, or impulse, stage the relative angles are equal and the rotor only turns the flow; with zero exit swirl it delivers Psi equal to 2. In a 50 % reaction stage the triangles are symmetric, like the repeating compressor stage, and with zero exit swirl Psi equals 1: half the work of the impulse stage at the same blade speed, but with much lower velocities and losses. In general, with zero exit swirl, R t equals one minus Psi over two: high loading means low reaction, which is the fundamental trade between weight, through fewer stages, and efficiency. Two hard limits: the nozzle exit angle alpha 2 should not exceed about 70 degrees, and Psi of about 2 is the ceiling, since beyond it the rotor would have to diffuse. Finally the manual’s choking check: the first stage nozzle must reach M 2 above one to meter the engine flow, while the relative Mach number at the rotor exit must stay subsonic, below about 0.9 at the tip; in a 50 % reaction stage V 3 R equals V 2 and T 3 is lower than T 2, so M 3 R is even larger than M 2 and the two conditions cannot hold together, which is why a 50 % stage cannot meter the flow; exercise 3 A does reach an M 2 of 1.11, but only with an M 3 R of 1.21, an artefact of its unrealistic blade speed.

With constant axial velocity, the whole design space collapses onto this map: the total turning epsilon, beta 2 plus beta 3, against beta 2, with contours of Psi over Phi and of R t over Phi. Choosing the stage type fixes a locus: a zero-reaction stage lives on the R t over Phi equal to zero line, where epsilon is twice beta 2. A 50 % reaction stage with zero exit swirl lives on the left edge, beta 2 equal to zero, where the turning equals alpha 2. Along that locus, each intersection with a Psi over Phi contour is a candidate, and the usual choice in the exercises is the largest Psi over Phi that respects alpha 2 below 70 degrees and the Zweifel limit; since the reaction already fixes Psi and the cycle fixes the work, this choice sets the flow coefficient and the shape of the triangles, with lower axial velocity and more turning. Then the flow coefficient follows from the angles, the blade speed from the temperature drop that the cycle demands, U equals the square root of c p delta T t over Psi, and the axial velocity is Phi times U.

Two more checks complete a turbine design. The Zweifel coefficient is the ratio of the actual tangential force on a blade to the maximum force a rectangular pressure distribution could give; for constant axial velocity it is two s over c x times cos squared beta 3 times tan beta 2 plus tan beta 3. Zweifel recommends values around 0.8 to 1, so for given angles it fixes the axial solidity; the chart of exercise 3 B plots Z r times c x over s against the two relative angles, with the allowed limit drawn on it. The second check is the centrifugal stress at the blade root, proportional to the density, the square of the speed and the annulus area, which rearranges into the A N squared parameter. The annulus area comes from the mass flow parameter at the rotor inlet, and the resulting specific strength must be compared with the alloys' curves at the blade temperature. If no material qualifies, lower the speed, change the blade taper, add a stage, or go back to the cycle.

Exercise 3 A, your group's deliverable, designs the single-stage low pressure turbine of the take-off engine of exercise 1 B. Constant axial speed, zero exit swirl and 50 % reaction mean beta 2 equals alpha 3, which is zero, beta 3 equals alpha 2, and the loading coefficient is exactly one. So we move along the left edge of the map, where the turning equals alpha 2, and tan alpha 2 equals one over Phi. We want the largest Psi over Phi with alpha 2 at most 70 degrees: Psi over Phi equal to 3 gives Phi of one third and alpha 2 of 71.6 degrees, which is too much, so it is rejected. The next line, 2.5, gives Phi equal to 0.4 and alpha 2 equal to 68.2 degrees, which is acceptable, and the official solution notes that any value below 3 could also be defended.

To size the triangles we need the temperature drop from the cycle: the low pressure turbine of exercise 1 B drops 347.8 kelvin. Careful: the official text says it comes from exercise 1 A, but 347.8 kelvin is the low pressure turbine drop of 1 B, the take-off engine; 1 A's is 294 kelvin. With Psi equal to one, the blade speed is the square root of 1 239 times 347.8, which is 656.4 metres per second, and the axial velocity is 0.4 times that, 262.6. At the rotor inlet, the nozzle has turned the flow to 68.2 degrees, so V 2 is 707 metres per second and its swirl equals U exactly, which makes the relative inlet velocity purely axial: beta 2 is zero. At the rotor exit the absolute velocity is axial, zero swirl, while the relative velocity carries the full blade speed backwards: by symmetry V 3 R equals V 2, 707 metres per second.

The thermodynamics run backwards from the stage exit, because the exit totals are fixed by the cycle of exercise 1 B. The static temperatures follow from the velocities: 1 090.46 kelvin after the nozzle, where the gas moves at 707 metres per second, and 916.57 kelvin at the exit. The relative total temperature at the rotor exit, with V 3 R of 707, is 1 118.28 kelvin, equal to the rotor inlet value; the exit static pressure is 1.265 bar and the relative total pressure 2.995 bar. Dividing by the rotor loss ratio 0.81 gives 3.698 bar at the rotor inlet in the relative frame, hence a static pressure of 3.315 bar and an absolute total pressure of 6.917 bar after the nozzle. Comparing with the 8.32 bar entering, the stator loss ratio implied is 0.831. The stage isentropic efficiency is 80.85 %, and 81.0 % with unrounded numbers: it must equal the 81 % assumed in exercise 1 B, because both end states were taken from that cycle, so it is a check, not a result; the genuinely new output of the stage design is the stator loss ratio.

And here is the turbine template, with the 3 A numbers. Through the nozzle the total temperature stays at 1 292 kelvin while the total pressure drops a little; the static temperature falls by 174 kelvin, from 1 264 kelvin at the inlet, where the flow is axial at 263 metres per second, to 1 091 kelvin, as it accelerates to 707 metres per second. In the rotor frame, the relative total temperature is the same at 2 R and 3 R, while the relative total pressure falls by the rotor loss ratio of 0.81. In the absolute frame, the total temperature falls from t 2 to t 3 by the stage work, 347.7 kelvin. With 50 % reaction the two static drops are equal, T 1 minus T 2 equals T 2 minus T 3, 174 kelvin each, and every loss again moves the states to the right, where the isobars spread apart.

Last part of 3 A: the blade stress at 2 500 rpm with a taper ratio of 0.8. The nozzle exit Mach number is 707 over the local speed of sound, 1.11, and the mass flow parameter for gamma 1.3 at that Mach number is 0.661. With the core flow of 64.29 kilograms per second, the total conditions in pascal and kelvin, and the cosine of 68.2 degrees, the annulus area is 0.230 square metres; mind the units, pressure in pascal. The specific strength required is A N squared times pi times one plus the taper ratio over 3 600 times ten to the 3, which gives only 2.26 megapascals per gram per cubic centimetre. Now compare with Table 1 of the statement, whose ranges are room-temperature values: nickel alloys give at least 59 over 13.4, 4.4, steel at least 35 and titanium about 157, but the weakest aluminium alloys only 1.24 over 3.63, 0.34, so it is not true that any material in the table works. And temperature decides: the blades see a relative total temperature of about 1 118 kelvin, where aluminium and titanium are useless, their curves on the 3 B chart stop near 570 and 810 kelvin, and a low-alloy steel has lost its strength, so the answer is a nickel alloy, which on the chart still gives about 25 at that temperature; the official solution simply says it is possible to manufacture, and adds that otherwise you would lower the speed, change the taper, add stages or revisit the cycle.

Exercise 3 B, done by the other groups, designs the single-stage high pressure turbine of exercise 1 A as a zero-reaction stage, with a Zweifel limit. Zero reaction and zero exit swirl give beta 3 equal to beta 2, a loading coefficient of 2, and a nozzle swirl of twice the blade speed. Then Psi over Phi is twice tan beta 2, which is also tan alpha 2, and the Zweifel coefficient simplifies to four cos squared beta 2 tan beta 2, which is two sin two beta 2. Psi over Phi equal to one gives beta 2 of 26.6 degrees and a Zweifel value of 1.60, within the 1.8 limit, with Phi equal to 2 and alpha 2 of 45 degrees; the official sheet reads 27.5 and 45.6 degrees from the chart. Psi over Phi of 1.5 gives 1.92, above the limit, so it is rejected; in fact 1.8 allows at most about 1.25. With the relaxed limit of 2.0 the Zweifel coefficient never binds, since two sin two beta cannot exceed 2, so the real constraint becomes alpha 2 below 70 degrees: Psi over Phi of 2.5, Phi of 0.8 and alpha 2 of 68.2 degrees, with a Zweifel value of 1.95.

The 1 A cycle fixes the work: its high pressure turbine drops 260.3 kelvin, so with Psi equal to 2 the blade speed is the square root of 1 239 times 260.3 over 2, 401.6 metres per second. With Phi equal to 2 the axial velocity is twice that, 803.1 metres per second, and the nozzle swirl is also 2 U, so V 2 is 1 136 metres per second at 45 degrees. The relative inlet velocity, V 2 minus U, has a swirl equal to U and a magnitude of 898 metres per second at 26.6 degrees. At the rotor exit the absolute velocity is purely axial, 803.1 metres per second, and the relative velocity, with the same 898 metres per second, is the mirror image of the inlet one: the rotor only turns the flow, which is what zero reaction means.

Part 3 is the thermodynamics, which the official sheet gets partly from exercise 1 B by mistake, so here it is recomputed consistently with 1 A. For the strict design, the statics follow from the velocities: with zero reaction the static enthalpy does not change in the rotor, so T 2 equals T 3, 1 079.4 kelvin, a horizontal static line on the T s sketch, even though the static pressure still falls, from 1.521 to 1.369 bar, because of the rotor loss. The relative total temperature is 1 404.8 kelvin on both sides of the rotor; from the exit totals, the static exit pressure is 1.369 bar and the relative total 4.288 bar, so the rotor inlet relative total is 4.288 over 0.9, 4.764 bar. Back to the absolute frame, P 2 is 1.521 bar and the nozzle exit total pressure 8.374 bar, an implied stator loss ratio of 0.895, and the stage efficiency is 80 %, equal to the 1 A turbine efficiency by construction. The relaxed design has a lower axial velocity, 321.3 metres per second, a nozzle exit velocity of 865 metres per second and static temperatures of 1 298 kelvin, with pressures of 3.38 and 3.04 bar. Its total pressures, stator loss ratio and efficiency are identical, because they depend only on the cycle end states and the rotor loss; the official sheet's 8.64 bar and 83 % come from mixing in 1 B data, so trust the method, not those numbers.

Part 4, the stress check, uses the same formula as 3 A, now at 25 000 rpm. In the strict design the nozzle exit Mach number is 1.79 and the mass flow parameter for gamma 1.3 is 0.452. With the 1 A core flow of 24.29 kilograms per second, the annulus area is 0.061 square metres. So the required specific strength is 60.2 megapascals per gram per cubic centimetre, close to the official 57.7, which corresponds to their 0.059 square metres; at the 20 000 rpm printed on the official sheet it would be only 38.5, so that number is a typo and 25 000 is right. The relaxed design has a lower Mach number, 1.25, hence a larger annulus, 0.083 square metres, and needs 81.4. The blades see a relative total temperature of about 1 405 kelvin, beyond the end of every curve on the chart, so only a cooled single-crystal nickel alloy works: the chart gives it about 55 to 57 near 1 200 kelvin, so the strict design is marginal and needs the metal below about 1 180 kelvin, while the relaxed one would need it near 1 050 kelvin; the official answer is single crystal.

Before leaving the turbines, a few sanity checks that the exercises do not ask for but that show real understanding. In 3 A, a blade speed of 656 metres per second at 2 500 rpm means a mean radius of U over omega, 2.51 metres, and with an annulus of 0.23 square metres the blades would be only about 1.5 centimetres tall, which is not a real turbine. Its nozzle exit Mach number of 1.11 also comes with a rotor exit relative Mach number of 707 over 584, 1.21, above the 0.9 the manual allows. The strict 3 B design is worse: M 2 of 1.79 would need convergent-divergent vanes, and the rotor exit relative Mach number is 1.42. The relaxed 3 B design, with M 2 of 1.25 and a rotor exit relative Mach number of 0.74, is aerodynamically better, but it needs a specific strength of 81 instead of 60, so a designer would keep it only with a lower speed or more cooling. The manual's section on the optimal nozzle exit swirl Mach number ties these together: with zero exit swirl, M 2 sin alpha 2 equals 2 M T times one minus R t, where M T is the blade speed over the speed of sound at the nozzle exit, so the reaction and the blade Mach number set the swirl, and the nozzle exit Mach number is chosen for maximum work per unit mass flow. In the exam, solve exactly what the statement asks, and then add one line of sanity check like these: it costs a minute and shows the examiner you understand the result.

To finish, here is what I would put on the second sheet, pages three and four, for the turbomachinery. Euler for both machines with the sign conventions, and the frame relation V equals V R plus U with angles from the axial direction. The compressor row relations with the loss coefficient conversion and the degree of reaction. The repeating-stage formulas and the multistage pressure ratio. The turbine loading and flow coefficients, the map relations, the reaction cases and the 70 degree limit, plus the Zweifel and A N squared formulas and the annulus area with the mass flow parameter. Add one line for the radial variation and the cascade vocabulary. And since the exam is about method, copy the station table of 1 A and the triangles and T s templates of 2 A and 3 A as worked templates; then practise each exercise once more without looking, against the clock. Good luck.