22-Mec-B6 Advanced Fluid Mechanics · December 2018
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
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.
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.
The difference between the two machines is not one of degree but of kind, and it comes down to the direction of the pressure gradient the boundary layer on the blade has to survive. In a compressor rotor the passage between adjacent blades diverges, the relative velocity falls and the static pressure rises: the flow is diffusing. A boundary layer running into rising pressure is losing momentum at the wall at the same time as it is being pushed backwards by the pressure field, and beyond a modest amount of diffusion it separates. Separation in a compressor is not a gentle loss of efficiency; it is stall, in which a blade passage ceases to pass flow, the blockage moves round the annulus as a rotating stall cell, and in the limit the whole machine surges — the flow through the engine reverses in a violent, damaging oscillation. Every limit on compressor stage design is a defence against that outcome.
Three limits bite in practice. The first is the diffusion limit itself, expressed most simply by the de Haller criterion that the relative velocity at rotor exit should not fall below about 0.72 of its inlet value; the tip section of Question 2 sits at 0.886 and the root at 0.830, both comfortably inside it, and the designer has deliberately kept them there. The second is the limit on fluid deflection: because the flow is diffusing, a compressor blade row can turn the air by only some 20 to 45 degrees before the passage diffusion becomes excessive, whereas the same blade row in a turbine may turn the gas through 90 degrees or more. The third is a limit on blade speed, set jointly by centrifugal stress in the blade root and disc, and by the relative Mach number at the tip — the first-stage tip of Question 2 already sees a relative velocity of 423 m/s, and a transonic tip brings shock losses and shock-boundary-layer interaction with it.
Taken together these limits cap the temperature rise a single compressor stage can produce at something like 20 to 45 K, which corresponds to a stage pressure ratio of roughly 1.15 to 1.35. The LP spool of this engine must produce an overall pressure ratio of 3, and the HP spool a ratio of 5; at a stage ratio of about 1.17 and 1.26 respectively, seven stages each is what it takes. There is also an off-design reason for using many lightly loaded stages rather than a few heavily loaded ones: a compressor is only ever matched exactly at one operating point, and at all others the incidence on each row drifts. Lightly loaded stages have more incidence margin before they stall, so a multi-stage machine has a usefully wider surge-free operating range — which for a supersonic transport that must accelerate from a standstill to Mach 2 is not a refinement but a requirement.
The turbine faces the opposite situation in every respect. Its passages converge, the gas accelerates and the static pressure falls, so the boundary layer is running downhill into falling pressure and is being thinned and stabilised rather than thickened and destabilised. There is no diffusion limit to respect, so a turbine blade row can be given very high camber and can turn the gas through 90 to 120 degrees in a single row. Since the Euler work is $U\,\Delta C_Y$, and the achievable change of whirl is several times larger than in a compressor row at the same blade speed, one turbine stage can absorb the work of many compressor stages before any aerodynamic limit is approached.
A second effect roughly doubles the advantage. The turbine works on gas leaving the combustion chamber at perhaps 1400 K, while the compressor works on air entering at 353 K. The acoustic velocity varies as $\sqrt{T}$, so at the same blade speed the turbine's relative Mach numbers are far lower and it can be run at higher blade speed, or at higher gas velocity, before choking or shock losses appear. In addition the enthalpy drop available for a given pressure ratio is proportional to the inlet temperature, $\Delta T = T_{\text{in}}[1-(p_{\text{out}}/p_{\text{in}})^{(k-1)/k}]$, so the same pressure ratio yields several times more energy per kilogram in the hot turbine than it costs in the cold compressor. This is the quantitative reason the turbine can be smaller as well as fewer-staged: on this engine the compressor absorbs about 55 MW (Question 1), and one HP stage plus one LP stage return it comfortably.
The limits on the turbine are therefore not aerodynamic but material: creep and oxidation of the blade at metal temperatures that may exceed the melting point of the alloy without cooling, and the centrifugal stress at the blade root. That is exactly why the fifth-stage HP bleed of Question 1 exists — the 728 K air tapped there is fed through internal passages and film-cooling holes so that the single-stage turbine can survive the gas temperature that makes its high work output possible. The cost of that cooling air, which is compressed but never burned, is one of the real prices paid for a highly loaded turbine.
Each spool is free to run at whatever speed matches its own compressor to its own turbine, and the optimum is different for the two because the annulus geometry is different. The LP compressor handles cold, low-density air, so it needs a large annulus — its first-stage tip diameter is 1.12 m — and the blade speed at that radius is already 381 m/s at 6500 rev/min. Turning it faster would raise the tip relative Mach number and the root centrifugal stress beyond what the design allows. The HP compressor, working on air already compressed threefold and therefore three times denser, needs a much smaller annulus to pass the same mass flow; at the smaller radius, 8500 rev/min is required to reach a blade speed capable of producing the required work per stage, since the Euler work scales with $U^2$ for a given stage geometry.
There is a second, equally important benefit. Running the spools independently is what makes a twin-spool engine tolerant of off-design operation. At low power, or during acceleration, the LP spool tends to run relatively fast for the flow it is passing and the HP relatively slow, so the front stages are pushed towards stall. With separate shafts each spool is free to find its own speed and unload itself, which keeps the incidence on the front stages within their stall margin without variable geometry. A single shaft joining all fourteen stages would have to compromise between the two optima at every condition and would need extensive blow-off valves or variable stators to stay out of surge.
