Question 2 of 8: Compressor Design

Question 2: Compressor Design (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. $\dot m=50\ \text{kg/s}$; $T_1=293\ \text{K}$; $p_1=100\ \text{kPa}$; $p_2=500\ \text{kPa}$; $C_a=160\ \text{m/s}$ (constant, all stages); $N=8000\ \text{rev/min}$; $D_{hub}=0.6\,D_{tip}$; $\alpha_1=30^{\circ}$; degree of reaction $\Lambda=0.5$; $R=287\ \text{J/kg K}$, $k=1.4$, $c_p=1.005\ \text{kJ/kg K}$.

Find. Hub, tip and mean blade diameters at inlet; mean blade velocity; the inlet/outlet velocity triangle of the first stage; the work done per stage; and the number of stages needed for the overall pressure ratio.

Approach. Get the inlet annulus area from continuity (mass flow, inlet density, axial velocity), split it into hub/tip diameters using the given ratio, find the mean blade speed from the mean diameter and shaft speed, construct the 50%-reaction (symmetric) velocity triangle at that mean diameter to get the stage work, then divide the ideal overall enthalpy rise for the given pressure ratio by the per-stage work.

1. (a) Annulus area and diameters. Inlet density from the ideal gas law, $\rho_1=\dfrac{p_1}{RT_1}=\dfrac{100\,000}{287(293)}=1.189\ \text{kg/m}^3$. Volume flow $\dot Q=\dot m/\rho_1=50/1.189=42.05\ \text{m}^3/\text{s}$, so the annulus area is $A=\dot Q/C_a=42.05/160=0.2628\ \text{m}^2$. With $A=\tfrac{\pi}{4}(D_{tip}^2-D_{hub}^2)$ and $D_{hub}=0.6D_{tip}$, $A=\tfrac{\pi}{4}D_{tip}^2(1-0.36)$, so$$D_{tip}=\sqrt{\frac{4A}{\pi(1-0.6^2)}}=\sqrt{\frac{4(0.2628)}{\pi(0.64)}}=0.723\ \text{m}.$$Then $D_{hub}=0.6(0.723)=0.434\ \text{m}$ and the mean diameter $D_m=(D_{tip}+D_{hub})/2$,$$\boxed{D_{tip}=0.723\ \text{m},\ D_{hub}=0.434\ \text{m},\ D_m=0.578\ \text{m}}$$
2. (b) Mean blade velocity. $\omega=2\pi N/60=2\pi(8000)/60=837.8\ \text{rad/s}$, so$$U=\omega\,\frac{D_m}{2}=837.8\left(\frac{0.578}{2}\right)=242.3\ \text{m/s}.$$$$\boxed{U=242\ \text{m/s}}$$
3. (c) First-stage velocity triangle (50% reaction, symmetric). With $C_a$ constant and $\alpha_1=30^{\circ}$ measured from axial, the whirl component entering the rotor is $C_{w1}=C_a\tan\alpha_1=160\tan30^{\circ}=92.4\ \text{m/s}$, giving relative whirl $W_{w1}=U-C_{w1}=242.3-92.4=149.9\ \text{m/s}$ and $\beta_1=\tan^{-1}(W_{w1}/C_a)=43.1^{\circ}$. For 50% reaction the triangles are symmetric ($\beta_2=\alpha_1$, $\alpha_2=\beta_1$), so $W_{w2}=C_a\tan\beta_2=160\tan30^{\circ}=92.4\ \text{m/s}$ and $C_{w2}=U-W_{w2}=149.9\ \text{m/s}$. The four velocity magnitudes follow from $C_a$ and these whirl components:$$C_1=\sqrt{C_a^2+C_{w1}^2}=184.8,\ \ W_1=\sqrt{C_a^2+W_{w1}^2}=219.3,$$$$C_2=\sqrt{C_a^2+C_{w2}^2}=219.3,\ \ W_2=\sqrt{C_a^2+W_{w2}^2}=184.8\ \ (\text{all m/s}).$$

4. (d) Work done by the first stage. From the velocity diagram the change in whirl velocity across the rotor is $\Delta C_w=C_{w2}-C_{w1}=149.9-92.4=57.5\ \text{m/s}$, so by Euler's turbomachine equation,$$w_{stage}=U\,\Delta C_w=242.3(57.5)=13\,940\ \text{J/kg}.$$$$\boxed{w_{stage}=13.9\ \text{kJ/kg}}$$
5. (e) Number of stages. No stage/compressor efficiency is supplied for this preliminary design, so the ideal (isentropic) enthalpy rise for the required overall pressure ratio is used as the target: $r_p=p_2/p_1=5.0$, $r_p^{(k-1)/k}=5^{0.2857}=1.584$, $\Delta T_{0,ideal}=T_1(1.584-1)=293(0.584)=171.1\ \text{K}$, so$$w_{total}=c_p\,\Delta T_{0,ideal}=1.005(171.1)=171.9\ \text{kJ/kg}.$$Dividing by the per-stage work,$$n=\frac{w_{total}}{w_{stage}}=\frac{171.9}{13.9}=12.3\ \rightarrow\ \boxed{n=13\ \text{stages}}$$(rounded up, since a fractional stage cannot be built).