Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examination 07-Mec-A6-1 Fluid Machinery (May 2014) — closed book, three hours, 60 marks. Section A is calculative (Q1–Q5) and Section B is descriptive (Q6–Q8); the rubric asks for four of Section A plus two of Section B (six questions, each of equal value, 10 marks). All eight questions are solved in full as a study resource. General constants supplied with the paper: g = 9.81 m/s², patm = 100 kPa, pvapour = 2.34 kPa (20 °C), ρwater = 1000 kg/m³, ρair = 1.21 kg/m³ (15 °C), cp,air = 1.005, cv,air = 0.718 kJ/kg·K.
Reference texts. S. L. Dixon & C. A. Hall, Fluid Mechanics and Thermodynamics of Turbomachinery (7th ed.); R. K. Turton, Principles of Turbomachinery; H. Cohen, G. F. C. Rogers & H. I. H. Saravanamuttoo, Gas Turbine Theory (for the axial compressor stage); R. W. Fox, A. T. McDonald & P. J. Pritchard, Introduction to Fluid Mechanics (pump energy equation and affinity laws).
Given. Dhub = 480 mm, Dtip = 1120 mm (mean Dm = 800 mm); α₁ = 30°, β₂ = 40°; N = 6800 rev/min; M = 136 kg/s; T₁ = 288 K; ρair = 1.21 kg/m³, cp = 1.005 kJ/kg·K, k = cp/cv = 1.40.
Given data — Q4
Parameter
Value
Parameter
Value
Hub / tip diameter
480 / 1120 mm
Rotor speed N
6800 rev/min
IGV outlet angle α₁
30°
Air flow M
136 kg/s
Blade outlet angle β₂
40°
Inlet temp T₁
288 K
Find. (a) U and Cx1; (b) C₁, C₂, W₁, W₂; (c) work and power; (d) temperature rise; (e) pressure ratio.
First-stage rotor velocity triangles at the mean diameter (angles measured from the axial direction). Common blade speed U; axial velocity Cx constant through the rotor.
Approach. Get the blade speed from the mean diameter and rpm, the axial velocity from continuity through the annulus, build both triangles from α₁ and β₂, then apply the Euler work equation and the isentropic relation.
(a) Blade and axial velocities. At the mean diameter $D_m=0.80$ m,
$$U=\frac{\pi D_m N}{60}=\frac{\pi(0.80)(6800)}{60}=284.8\ \text{m/s}.$$
Continuity through the annulus $A=\tfrac{\pi}{4}(D_2^2-D_1^2)=0.804\ \text{m}^2$ gives the axial velocity
$$C_{x1}=\frac{M}{\rho A}=\frac{136}{1.21(0.804)}=139.8\ \text{m/s}.$$
(b) Inlet triangle. With α₁ measured from the axial direction, $C_1=C_{x1}/\cos\alpha_1=161.4$ m/s and whirl $C_{y1}=C_1\sin\alpha_1=80.7$ m/s. The relative inlet velocity uses $W_{y1}=U-C_{y1}=204.1$ m/s:
$$W_1=\sqrt{W_{y1}^2+C_{x1}^2}=\sqrt{204.1^2+139.8^2}=247.4\ \text{m/s}.$$
Outlet triangle. Axial velocity is unchanged; from β₂, $W_{y2}=C_{x1}\tan\beta_2=117.3$ m/s so
$$W_2=\frac{C_{x1}}{\cos\beta_2}=182.4\ \text{m/s},\qquad C_{y2}=U-W_{y2}=167.5\ \text{m/s},$$
$$C_2=\sqrt{C_{y2}^2+C_{x1}^2}=218.2\ \text{m/s}.$$
(c) Work and power. Euler work (change of whirl times blade speed):
$$w=U(C_{y2}-C_{y1})=284.8(167.5-80.7)=24.75\ \text{kJ/kg},$$
$$\boxed{P=wM=24.75\times136=3.37\ \text{MW}.}$$
(d) Enthalpy and temperature rise. All work raises the stagnation enthalpy, so $\Delta h=w=24.75$ kJ/kg and
$$\Delta T=\frac{w}{c_p}=\frac{24.75}{1.005}=24.6\ \text{K}.$$
(e) Stage pressure ratio. Isentropic, with $T_2/T_1=312.6/288=1.085$ and $k/(k-1)=3.5$:
$$\frac{p_2}{p_1}=\left(\frac{T_2}{T_1}\right)^{k/(k-1)}=1.085^{3.5}=1.33.$$
Question 4 — results
Quantity
Value
(a) Blade speed U / axial Cx1
284.8 / 139.8 m/s
(b) C₁, C₂
161.4, 218.2 m/s
(b) W₁, W₂
247.4, 182.4 m/s
(c) Work / power
24.75 kJ/kg / 3.37 MW
(d) Temperature rise
24.6 K
(e) Pressure ratio
1.33
Check: α₁ and β₂ are taken from the axial direction (consistent with the reference work equation $w=U\,\Delta C_y$ on the paper's formula sheet). The axial velocity is evaluated at the given inlet air density of 1.21 kg/m³ (15 °C). The whirl is applied at the mean blade diameter.
