22-Mec-B6 Advanced Fluid Mechanics · December 2019

Question 2 of 8: Velocity Diagram for the Gas Turbine

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

Notes on this paper

Paper format: National Examinations, December 2019 — 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); a candidate answers 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 in the Attachments on pages 12–17, and the nomenclature, general constants and reference equations on pages 18–22. All eight questions are solved below, because this set is a study resource rather than a three-hour sitting.

Check — angle conventions are taken from the paper's own attachments, and on this sitting they are uniform. The gas-turbine velocity diagram on page 13 strikes $\theta$, $\phi$, $\gamma$ and $\delta$ off the tangential direction (the plane of rotation); the pump diagram on page 14 strikes $\alpha_1$, $\alpha_2$, $\beta_1$ and $\beta_2$ off the tangential direction as well; and the Francis diagram on page 15 defines $\alpha_1$ between $V_1$ and $u_1$, again tangential. Every angle below is therefore measured from the direction of blade motion. Every constant used is the paper's own page-19 value: $g = 9.81\ \text{m}\,\text{s}^{-2}$, $\rho_{\text{water}} = 1000\ \text{kg}\,\text{m}^{-3}$, $p_{\text{atm}} = 100\ \text{kPa}$ and $p_{\text{vapour}} = 1.71\ \text{kPa}$ at $15^{\circ}\text{C}$.

Reference texts

Dixon, S. L. and Hall, C. A., Fluid Mechanics and Thermodynamics of Turbomachinery, 7th ed. — Ch. 1 (specific speed), Ch. 4 (axial turbines), Ch. 7 (centrifugal pumps), Ch. 9 (hydraulic turbines, Thoma cavitation parameter). The efficiency chart on page 15 and the cavitation chart on page 17 of the examination paper are reproduced from this text.
Cohen, H., Rogers, G. F. C. and Saravanamuttoo, H. I. H., Gas Turbine Theory, 7th ed. — Ch. 5 (axial compressors, stage loading and stall), Ch. 7 (axial turbines, stage work and blade angles).
Turton, R. K., Principles of Turbomachinery, 2nd ed. — Ch. 2 (the Euler equation), Ch. 5 (radial machines, vane number and slip), Ch. 6 (hydraulic turbines).
White, F. M., Fluid Mechanics, 8th ed. — Ch. 11 (turbomachinery: pump characteristics, specific speed, Pelton, Francis and Kaplan turbines, cavitation). The pump characteristic curves reproduced on page 11 of the examination paper come from this text.
Fox, R. W., McDonald, A. T. and Pritchard, P. J., Introduction to Fluid Mechanics, 10th ed. — Ch. 10 (fluid machinery, Euler turbomachine equation, cavitation and NPSH).
Douglas, J. F., Gasiorek, J. M. and Swaffield, J. A., Fluid Mechanics, 6th ed. — Ch. 23 (rotodynamic machines, impeller velocity triangles).

Question 2 — Velocity Diagram for the Gas Turbine (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. The same machine as Question 1, but now with the geometry already settled: a mean rotor diameter of 0.617 m, four stages, 3 600 rev/min, an axial velocity of 100 m/s everywhere, and impulse blading with no friction.

Quantity	Symbol	Value
Rotor diameter at blade mid-height	$D_m$	$0.617\ \text{m}$
Number of stages	$n$	4
Rotational speed	$N$	$3600\ \text{rev}\,\text{min}^{-1}$
Inlet / exhaust temperature	$T_1$, $T_2$	$490^{\circ}\text{C}$, $341^{\circ}\text{C}$
Mass flow rate	$M$	$1937\ \text{kg}\,\text{s}^{-1}$
Specific heat at constant pressure	$c_p$	$844\ \text{J}\,\text{kg}^{-1}\text{K}^{-1}$
Axial velocity, and gas velocity entering each stage	$V_A = V_0$	$100\ \text{m}\,\text{s}^{-1}$

Find. The enthalpy drop per stage and the resulting nozzle-exit velocity, the blade velocity, the complete first-stage velocity diagram drawn to scale, the three blade angles read from it, the four gas velocities, and the power of the whole four-stage turbine.

Approach. Share the total enthalpy drop equally among the four stages and put all of it through the fixed blades (that is what "pure impulse" means), use the nozzle equation to get $V_{S1}$, close the inlet triangle against the blade speed, mirror it to get the outlet triangle, and finally take the work from the whirl change.

Enthalpy drop per stage (part a). All four stages are identical, so each takes a quarter of the total drop: $$\Delta h_{\text{stage}} = \frac{c_p (T_1 - T_2)}{n} = \frac{844 \times 149}{4} = 31.44\ \text{kJ}\,\text{kg}^{-1}$$
Gas velocity entering the moving blades (part a, continued). In a pure impulse stage the whole stage drop is converted to kinetic energy in the fixed (nozzle) row, and the gas arrives at that row with the axial velocity $V_0 = 100\ \text{m}\,\text{s}^{-1}$. The page-21 nozzle equation $h_1 - h_2 = (V_2^2 - V_1^2)/2$ then gives $$\boxed{V_{S1} = \sqrt{2\,\Delta h_{\text{stage}} + V_0^{2}} = \sqrt{2 \times 31\,440 + 100^{2}} = 270.0\ \text{m}\,\text{s}^{-1}}$$ Note how closely this agrees with the 271.1 m/s that Question 1 obtained from an entirely different route — the $2.33\,V_B$ guideline. The two calculations are independent, and their agreement to within 0.4 per cent is the strongest single check available on this pair of questions.
Blade velocity (part b). At the stated mean diameter, $$V_B = \frac{\pi D_m N}{60} = \frac{\pi \times 0.617 \times 3600}{60} = 116.3\ \text{m}\,\text{s}^{-1}$$
Close the inlet triangle. The axial component of $V_{S1}$ is 100 m/s and page 13 measures $\theta$ from the plane of rotation, so $$\theta = \arcsin\!\left(\frac{100}{270.0}\right) = 21.7^{\circ},\qquad V_{w1} = V_{S1}\cos\theta = 250.8\ \text{m}\,\text{s}^{-1}$$ Subtracting the blade speed gives the tangential component of the relative velocity, and the relative velocity itself follows: $$V_{w1} - V_B = 250.8 - 116.3 = 134.5\ \text{m}\,\text{s}^{-1},\qquad V_{R1} = \sqrt{134.5^{2} + 100^{2}} = 167.6\ \text{m}\,\text{s}^{-1}$$ so the moving blade must be set at $\phi = \arctan(100/134.5) = 36.7^{\circ}$ to receive the gas without incidence.
Mirror the triangle to get the outlet. Impulse blading with no friction returns the relative velocity unchanged in magnitude with its whirl reversed, and symmetrical blades give $\gamma = \phi$. Hence $V_{R2} = V_{R1} = 167.6\ \text{m}\,\text{s}^{-1}$, and adding the blade velocity back gives the absolute exit velocity: $$V_{w2} = V_B - (V_{w1} - V_B) = 116.3 - 134.5 = -18.1\ \text{m}\,\text{s}^{-1}$$ $$V_{S2} = \sqrt{18.1^{2} + 100^{2}} = 101.6\ \text{m}\,\text{s}^{-1},\qquad \delta = 100.3^{\circ}$$ The negative sign matters physically: the gas leaves the stage with a small swirl against the direction of blade motion, which is why $\delta$ exceeds $90^{\circ}$. That is the normal signature of a stage running slightly faster than the impulse optimum, exactly as the $2.33$ guideline predicted in Question 1.
The scale drawing (part b). The diagram below is drawn to scale from the numbers above, with the inlet triangle above the common blade-velocity base line and the outlet triangle below it, as the page-13 nomenclature requires. At the suggested exam scale of 50 mm to 100 m/s, $V_{S1}$ would be a line 135 mm long and $V_B$ 58 mm.

First-stage velocity diagram. Blue is the absolute velocity entering the moving blades, green and orange the relative velocities entering and leaving, red the absolute velocity leaving. The two relative velocities are equal in length because the blading is pure impulse and frictionless; the outlet absolute velocity leans backwards because the exit whirl is small and negative.

Read the blade angles off the diagram (part c). The fixed-blade exit angle is the nozzle angle $\theta$, and the moving-blade entrance and exit angles are $\phi$ and $\gamma$: $$\boxed{\theta = 21.7^{\circ},\qquad \phi = 36.7^{\circ},\qquad \gamma = 36.7^{\circ}}$$ The absolute gas leaves at $\delta = 100.3^{\circ}$ from the direction of blade motion, but that is a flow direction, not a blade angle — it is set by the triangle, not by the metal.
Work and power of the whole turbine (part d). The work per unit mass per stage is the blade speed times the whirl change, taking due account of the sign of $V_{w2}$: $$w = V_B\left(V_{w1} - V_{w2}\right) = 116.3 \times \left(250.8 - (-18.1)\right) = 31.27\ \text{kJ}\,\text{kg}^{-1}$$ With four identical stages, $$\boxed{P = M\,n\,w = 1937 \times 4 \times 31\,270 = 2.423 \times 10^{8}\ \text{W} = 242.3\ \text{MW}}$$
Two free audits on the answer. First, the page-21 energy form of the blade work must return the same figure from the four velocities alone: $$w = \frac{\left(V_{S1}^{2} - V_{S2}^{2}\right) + \left(V_{R2}^{2} - V_{R1}^{2}\right)}{2} = \frac{\left(270.0^{2} - 101.6^{2}\right) + 0}{2} = 31.27\ \text{kJ}\,\text{kg}^{-1}$$ which checks every velocity in one line. Second, the velocity-diagram power of 242.3 MW must agree with the thermodynamic power $M c_p \Delta T = 243.6\ \text{MW}$ found in Question 1 — and it does, to 0.5 per cent. The small shortfall is real and instructive: the residual kinetic energy $V_{S2}^2/2$ leaving the last stage is not recovered as shaft work.

Question 2 — final results
Quantity	Symbol	Result
(a) Enthalpy drop per stage	$\Delta h_{\text{stage}}$	$31.44\ \text{kJ}\,\text{kg}^{-1}$
(a) Gas velocity entering moving blades	$V_{S1}$	$270.0\ \text{m}\,\text{s}^{-1}$
(b) Blade velocity	$V_B$	$116.3\ \text{m}\,\text{s}^{-1}$
(c) Fixed blade exit angle	$\theta$	$21.7^{\circ}$
(c) Moving blade entrance / exit angle	$\phi$, $\gamma$	$36.7^{\circ}$ / $36.7^{\circ}$
(d) Relative velocities in / out	$V_{R1}$, $V_{R2}$	$167.6$ / $167.6\ \text{m}\,\text{s}^{-1}$
(d) Absolute velocity leaving moving blades	$V_{S2}$ at $\delta = 100.3^{\circ}$	$101.6\ \text{m}\,\text{s}^{-1}$
(d) Work per stage	$w$	$31.27\ \text{kJ}\,\text{kg}^{-1}$
(d) Power of the whole four-stage turbine	$P$	$242.3\ \text{MW}$

Topic. Construction and interpretation of the combined velocity diagram for an axial impulse stage.

Key relations. Nozzle equation, $V_{S1} = \sqrt{2\Delta h_{\text{stage}} + V_0^{2}}$; blade speed, $V_B = \pi D N/60$; triangle closure, $\vec{V}_R = \vec{V}_S - \vec{V}_B$; impulse condition, $|V_{R2}| = |V_{R1}|$; Euler work, $w = V_B(V_{w1} - V_{w2})$; energy form, $w = \left[(V_{S1}^2 - V_{S2}^2) + (V_{R2}^2 - V_{R1}^2)\right]/2$.

Why this works. The velocity diagram is nothing more than the vector statement $\vec{V}_{\text{absolute}} = \vec{V}_{\text{relative}} + \vec{V}_{\text{blade}}$ drawn twice, once at inlet and once at outlet, and every blade angle in the machine is a direction read off that construction. What makes an impulse stage particularly easy is that the relative speed is unchanged across the rotor: the whole triangle at outlet is the mirror image of the one at inlet, so the diagram closes with no further information. The two expressions for the work are the same statement in different clothing — one counts momentum transfer at the blade, the other counts the kinetic energy the fluid gives up in the absolute frame plus what it gains in the rotating frame. Their agreement is a genuinely independent check, because they use different subsets of the four velocities.

Common pitfall. Forgetting that $V_{w2}$ can be negative and writing $w = V_B(V_{w1} - |V_{w2}|)$ instead of $V_B(V_{w1} + |V_{w2}|)$ understates the work by about 14 per cent here. The second trap is to assume the last stage's exit kinetic energy is recovered, and then to be alarmed that the diagram power is lower than $Mc_p\Delta T$; the discrepancy is physical, not arithmetic.

Source. Dixon & Hall, Fluid Mechanics and Thermodynamics of Turbomachinery, 7th ed., §4.2–4.5 (axial turbine velocity diagrams, degree of reaction, diagram efficiency); Cohen, Rogers & Saravanamuttoo, Gas Turbine Theory, 7th ed., §7.2 (vortex theory and stage velocity triangles).

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