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22-Mec-A6 Fluid Machinery · December 2016

Question 2 of 8: Compressor Blade Angles

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, December 2016 — closed book, 3 hours. Section A (Calculative, Q1–Q5) and Section B (Descriptive, Q6–Q8); candidates do four of A and two of B for 60 marks. All eight questions are solved as a study resource. Values marked “from the figure” are read from the examination attachment drawings.

Reference texts: Dixon & Hall, Fluid Mechanics and Thermodynamics of Turbomachinery (7th ed.); Cohen, Rogers & Saravanamuttoo, Gas Turbine Theory (6th ed.); Turton, Principles of Turbomachinery; Çengel & Boles, Thermodynamics (9th ed.); Fox & McDonald, Introduction to Fluid Mechanics (9th ed.). Constants from the paper: $g=9.81\ \text{m/s}^2$, $c_p=1.005\ \text{kJ/kg\,K}$, $k=1.4$, $R=0.287\ \text{kJ/kg\,K}$, $\rho_{water}=1000\ \text{kg/m}^3$, $p_{atm}=100\ \text{kPa}$, $p_{vap}=2.34\ \text{kPa}$.

Question 2: Compressor Blade Angles (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. First LP stage with $D_{tip}=1.12\ \text{m}$, $D_{hub}=0.38\ \text{m}$, $N=6500\ \text{rev/min}$, axial velocity $C_a=184\ \text{m/s}$, no inlet whirl ($C_{y1}=0$). The LP spool (Q1) has overall actual rise $\Delta T_{LP}=144.7\ \text{K}$ over 7 stages.

Find. Blade speeds at tip and hub, first-stage exit temperature, stage work and whirl velocities, and the inlet/outlet blade angles at tip and hub.

U=381 m/sC1W1C2W2UTIP (U=381 m/s)U=129 m/sC1W1C2W2UHUB (U=129 m/s)Ca=184 m/s axial (const), no inlet whirl (Cy1=0). Beta from axial.
Fig. 2.1 — First-stage velocity triangles at tip and hub (axial vertical, blade motion horizontal). C = absolute, W = relative, U = blade speed. Angles are measured from the axial direction.

Approach. Blade speed is $U=\pi D N/60$; equal work per stage sets the temperature rise; Euler work with zero inlet whirl gives $C_{y2}=w/U$ (a free-vortex distribution); the blade angles follow from the velocity triangles with constant axial velocity.

  1. Blade speeds (a). $U=\pi D N/60$ with $N=6500$: $$U_{tip}=\pi(1.12)(6500)/60=\boxed{381.2\ \text{m/s}},\qquad U_{hub}=\pi(0.38)(6500)/60=\boxed{129.3\ \text{m/s}}$$
  2. First-stage exit temperature (b). Equal work per stage over 7 LP stages: $\Delta T_{stage}=144.7/7=20.67\ \text{K}$, so $$T_{exit,1}=T_1+\Delta T_{stage}=353.15+20.67=\boxed{373.8\ \text{K}\ (100.7\,{}^\circ\text{C})}$$
  3. Stage work and whirl (c). $w=c_p\Delta T_{stage}=1005\times20.67=\boxed{20.77\ \text{kJ/kg}}$. With $C_{y1}=0$, Euler work $w=U\,C_{y2}$ gives $C_{y2}=w/U$ at each radius (a free-vortex distribution, $C_{y2}r=\text{const}$): $$C_{y2,tip}=\frac{20773}{381.2}=\boxed{54.5\ \text{m/s}},\qquad C_{y2,hub}=\frac{20773}{129.3}=\boxed{160.6\ \text{m/s}}$$
  4. Blade angles (d). Measured from axial, $\tan\beta_1=U/C_a$ and $\tan\beta_2=(U-C_{y2})/C_a$ with $C_a=184\ \text{m/s}$: $$\text{Tip: }\beta_1=\tan^{-1}\tfrac{381.2}{184}=\boxed{64.2^\circ},\quad \beta_2=\tan^{-1}\tfrac{381.2-54.5}{184}=\boxed{60.6^\circ}$$ $$\text{Hub: }\beta_1=\tan^{-1}\tfrac{129.3}{184}=\boxed{35.1^\circ},\quad \beta_2=\tan^{-1}\tfrac{129.3-160.6}{184}=\boxed{-9.7^\circ}$$

Check (hub loading): At the hub the required exit whirl $C_{y2,hub}=160.6\ \text{m/s}$ exceeds the blade speed $U_{hub}=129.3\ \text{m/s}$, so the relative flow leaves turned slightly past axial (negative $\beta_2$). A single rotor cannot in practice impart more tangential velocity than its own speed, so a strict constant-work / free-vortex design over-loads this very small hub (hub/tip ratio 0.34). Real first stages relieve the root by using a different vortex law and heavy twist — the point of part (e).

(e) Assessment. The tip blade is very highly staggered ($\beta_1\approx64^\circ$) while the root is much flatter ($\beta_1\approx35^\circ$); the $29^\circ$ change in inlet angle from hub to tip is exactly the pronounced twist visible in the blade photograph, so the tip and root inlet angles are reasonable. The near-axial / slightly negative hub exit angle flags that a simple constant-work assumption over-loads the root, which is why the actual blade is twisted and the root section is unloaded relative to the mean.

Question 2 results
QuantityTipHub
Blade speed $U$ (m/s)381.2129.3
Exit whirl $C_{y2}$ (m/s)54.5160.6
Inlet angle $\beta_1$64.2°35.1°
Outlet angle $\beta_2$60.6°−9.7°
First-stage exit temperature 373.8 K; stage work 20.77 kJ/kg; $C_{y1}=0$.