Question 5 of 5: Scaling a compressor stage by dimensional similarity

Question 5: Scaling a compressor stage by dimensional similarity (25 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.

Find. The rotor diameter and rotational speed of the geometrically similar stage that meets the client's duty, and the total pressure and total temperature it delivers.

Approach. Two dimensionless groups fix the whole compressible performance map of a geometrically similar family — a flow coefficient and a blade Mach number. Matching both transfers the tested pressure ratio and efficiency to the new machine, and the outlet state then follows from the isentropic temperature rise divided by the efficiency.

1. Identify the similarity groups for a compressible machine. Dimensional analysis of a compressor family gives $$\frac{\Delta P_o}{P_{o1}},\ \eta_{tt} =f\!\left(\frac{\dot m\sqrt{R T_{o1}}}{D^{2}P_{o1}},\ \frac{ND}{\sqrt{R T_{o1}}},\ \gamma,\ \mathrm{Re}\right)$$ For a fixed gas and at high Reynolds number the last two drop out. Because the duty here is quoted as an inlet volume flow, convert the first group using $ \dot m=\rho_{o1}Q $ and $ \rho_{o1}=P_{o1}/(RT_{o1}) $: $$\begin{aligned} \frac{\dot m\sqrt{RT_{o1}}}{D^{2}P_{o1}} &=\frac{P_{o1}}{RT_{o1}}\cdot\frac{Q\sqrt{RT_{o1}}}{D^{2}P_{o1}}\ &=\frac{Q}{D^{2}\sqrt{RT_{o1}}}\ \propto\ \frac{Q}{D^{2}\sqrt{T_{o1}}} \end{aligned}$$ so the two conditions to impose are simply $$\boxed{\;\frac{Q}{D^{2}\sqrt{T_{o1}}}=\text{const}\qquad\text{and}\qquad \frac{N D}{\sqrt{T_{o1}}}=\text{const}\;}$$ The inlet pressure cancels out of the flow group entirely, which is convenient here because both machines happen to breathe at 101.3 kPa.
2. Evaluate the flow group on the tested stage and solve for the new diameter. $$\begin{aligned} \frac{Q}{D^{2}\sqrt{T_{o1}}}\bigg|_{\text{model}} &=\frac{50}{(0.5)^{2}\sqrt{293}}=\frac{50}{0.25(17.117)}\ &=11.684\ \text{m}^{2}\,\text{s}^{-1}\text{K}^{-1/2} \end{aligned}$$ Imposing the same value on the client duty, $$D_2^{2}=\frac{200}{11.684(15.811)}=1.0826\ \text{m}^{2} \quad\Longrightarrow\quad\boxed{\;D_2=1.040\ \text{m}\;}$$ Four times the volume flow at a slightly lower inlet temperature calls for a rotor almost exactly twice the diameter — the flow group scales as D², so the area, not the diameter, tracks the duty.
3. Match the blade-speed group for the operating speed. $$\begin{aligned} \frac{ND}{\sqrt{T_{o1}}}\bigg|_{\text{model}}&=\frac{5000(0.5)}{\sqrt{293}}=146.05\ N_2&=146.05\times\frac{\sqrt{250}}{D_2}=\frac{146.05(15.811)}{1.0405} \end{aligned}$$ $$\boxed{\;N_2=2219\ \text{rpm}\;}$$ The tip speed falls from $ \pi D_1N_1/60=130.9 $ m/s to $ \pi D_2N_2/60=120.9 $ m/s, in the ratio $ \sqrt{250/293}=0.924 $ — exactly what constant blade Mach number demands when the inlet air is 43 K colder and the speed of sound is correspondingly lower.
4. Carry the pressure ratio across. Similarity preserves the non-dimensional performance, so the total-to-total pressure ratio is unchanged at 1.05 and $$\boxed{\;P_{o2}=1.05\,(101.3)=106.4\ \text{kPa}\;}$$ This is the whole point of the exercise: nothing about the aerodynamics has to be recomputed, because the two machines operate at the same point on the same non-dimensional map.
5. Compute the ideal (isentropic) temperature rise. For an isentropic compression through the same pressure ratio from the client's inlet temperature, $$T_{o2s}=T_{o1}\left(\frac{P_{o2}}{P_{o1}}\right)^{\frac{\gamma-1}{\gamma}} =250\,(1.05)^{0.2857}=250(1.01404)=253.51\ \text{K}$$ so $ \Delta T_{o,s}=3.51 $ K. Note that the isentropic rise is not transferable from the model — it scales with the inlet temperature, so the colder client machine does less ideal work per kilogram even at the same pressure ratio.
6. Divide by the total-to-total efficiency for the real rise. By definition $ \eta_{tt}=(T_{o2s}-T_{o1})/(T_{o2}-T_{o1}) $, so $$\Delta T_o=\frac{\Delta T_{o,s}}{\eta_{tt}}=\frac{3.509}{0.93}=3.774\ \text{K} \quad\Longrightarrow\quad\boxed{\;T_{o2}=250+3.77=253.8\ \text{K}\;}$$ The 7 % shortfall against the ideal is the irreversibility the 0.93 efficiency represents; it appears as extra temperature for no extra pressure.
7. Size the drive as a closing check. The client's inlet density is $ \rho_{o1}=P_{o1}/(RT_{o1})=101\,300/(287\times250)=1.412 $ kg/m³, so the mass flow is $ \dot m=\rho_{o1}Q=1.412(200)=282.4 $ kg/s and the shaft power is $$\dot W=\dot m\,c_p\,\Delta T_o=282.4(1004.5)(3.774)=1.07\ \text{MW}$$ The tested stage handles 60.2 kg/s and absorbs 268 kW, so the power ratio is 4.00 — exactly the volume-flow ratio, which is the consistency check worth quoting: at equal pressure ratio and equal inlet pressure the power must scale with the volume flow.