Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper: National Examinations — 07-Mec-A6 Fluid Machinery, May 2013. Closed-book, three hours. Section A (Calculative) Q1–Q5, Section B (Descriptive) Q6–Q8; the candidate answers four from A and two from B, six questions of equal value (10 marks each). All eight questions are worked here as a complete study resource.
Reference texts: R. K. Turton, Principles of Turbomachinery; S. L. Dixon & C. A. Hall, Fluid Mechanics and Thermodynamics of Turbomachinery; R. W. Fox, A. T. McDonald, Introduction to Fluid Mechanics; Douglas, Gasiorek & Swaffield, Fluid Mechanics. Constants as printed on the exam attachment pages (g = 9.81 m/s², patm = 100 kPa, pvapour = 2.34 kPa at 20°C, ρwater = 1000 kg/m³).
Given. Per turbine ṁ = 278/2 = 139 kg/s; Tin = 955 K, Tex = 756 K; ηt = 0.85; cp = 1148, cv = 861 J/kg·K so R = 287 J/kg·K, k = 1.333; Dtip = 1.5 m, Droot = 1.05 m, N = 3000 rev/min; patm = 100 kPa.
Find. Inlet pressure, inlet axial velocity, output power, mean blade speed, total whirl change, and the first-stage 50%-reaction velocity diagram with blade angles.
Approach. Get the isentropic temperature drop from the actual drop and efficiency to fix the pressure ratio; density and the annulus area give axial velocity; cpΔT gives power; πDmN gives blade speed; the Euler equation gives the whirl change; symmetry of a 50% reaction stage fixes the angles.
(a) Inlet pressure. Actual drop ΔT = 199 K, so isentropic drop $\Delta T_s=\Delta T/\eta_t=234.1$ K and $T_{2s}=720.9$ K. Then $\dfrac{p_{in}}{p_{atm}}=\left(\dfrac{T_{in}}{T_{2s}}\right)^{k/(k-1)}=\left(\dfrac{955}{720.9}\right)^{4}=3.08$, so $\boxed{p_{in}=308\ \text{kPa}}$.
(b) Inlet axial velocity. Annulus area $A=\tfrac{\pi}{4}(D_{tip}^2-D_{root}^2)=0.901\ \text{m}^2$; density $\rho=\dfrac{p_{in}}{RT_{in}}=\dfrac{308000}{287(955)}=1.124\ \text{kg/m}^3$; $C_a=\dfrac{\dot m}{\rho A}=\dfrac{139}{1.124(0.901)}$, so $\boxed{C_a=137\ \text{m/s}}$.
(c) Output from actual temperature change. $P=\dot m\,c_p\,\Delta T=139(1.148)(199)$, giving $\boxed{P=31.8\ \text{MW}}$ (consistent with 30.4 MW net after mechanical/generator losses).
(d) Mean blade velocity. Mean diameter $D_m=\tfrac12(1.5+1.05)=1.275$ m; $U=\dfrac{\pi D_m N}{60}=\dfrac{\pi(1.275)(3000)}{60}$, so $\boxed{U=200\ \text{m/s}}$.
(e) Required total whirl change. From Euler's equation $P=\dot m\,U\,\Delta C_w\Rightarrow \Delta C_w=\dfrac{P}{\dot m\,U}=\dfrac{31.8\times10^{6}}{139(200.3)}$, so $\boxed{\Delta C_w\approx1140\ \text{m/s (summed over all stages)}}$.
(f)–(g) First-stage 50%-reaction diagram. A single stage carrying the whole whirl change would need C1 ≈ 684 m/s (Mach 1.13 at 955 K) — impossible — so the power turbine must be multi-stage. Assuming three equal stages, per stage ΔCw = 380 m/s. For 50% reaction the triangles are symmetric with $C_{w1}=\tfrac12(\Delta C_w+U)=290$ m/s, giving nozzle angle $\alpha_1=\tan^{-1}(C_{w1}/C_a)=64.7^\circ$ and rotor inlet angle $\beta_1=\tan^{-1}[(C_{w1}-U)/C_a]=33.2^\circ$; by symmetry $\beta_2=\alpha_1=64.7^\circ$ and $\alpha_2=\beta_1=33.2^\circ$. Absolute inlet velocity C1 = 321 m/s (Mach 0.53).
First-stage 50%-reaction velocity diagram (three-stage assumption). The absolute (C) and relative (W) triangles are mirror images, the signature of 50% reaction; U is the common base and the axial component C_a is constant.
Check: the number of stages is not given. A single stage is supersonic and therefore ruled out; three equal 50%-reaction stages keep every velocity comfortably subsonic and give realistic blade angles. With a different stage count the per-stage whirl and angles scale accordingly, but the method and parts (a)–(e) are unchanged.
