Question 12.3: An aircraft engine is fitted with a single-sided centrifugal......

The given data are

d_{\text{eh}}=14 \text{ cm}, \quad d_{\text{et}}=28 \text{ cm}, \quad N=270 \text{ rps}, \quad \dot{m}=3.6 \text{ kg/s}

At altitude 35,000 ft, the ambient conditions are

P_{\text{a}}=23.844 \text{ kPa} \quad T_{\text{a}}=218.67 \text{ K} \\ T_{01}=T_{\text{a}}\left[1+\frac{\gamma-1}{2}M^2_{\text{flight}} \right] \\ T_{01}=218.67\left[1+\frac{0.4}{2}(0.85)^2 \right] =250.67 \text{ K} \\ P_{01}=P_{\text{a}}\left(\frac{T_{01}}{T_{\text{a}}} \right) ^{\gamma/(\gamma-1)}=23,844.24\left(\frac{250.27}{218.67} \right)^{1.4/0.4} =38,243 \text{ Pa} \\ U_{\text{1t}}=\pi D_{\text{t}}\frac{N}{60} \\ U_{\text{1t }}=\pi \times 0.28 \times 270=237.5 \text{ m/s} \\ M_{\text{1t }}=\frac{W_{1_\text{T}}}{\sqrt{\gamma RT_1}} \\ W^2_{\text{1t}} =M^2_{\text{1t}} \gamma RT_1

From Figure 12.22, the velocity triangle at inlet with slip

C_{\text{a1}}^2+(U_{1_{\text{t}}}-C_{\text{a1}}\tan \zeta)^2=M^2_{\text{1t}}\gamma RT_1 \\ (U_{\text{1t}}-C_{\text{a1}} \tan \zeta)^2=M^2_{\text{1t}}\gamma RT_1-C^2_{\text{al}} \\ U_{\text{1t}}-C_{\text{a1}} \tan \zeta=\sqrt{M^2_{\text{1t}}\gamma RT_1-C_{\text{al}}^2} \\ C_{\text{a1}}\tan \zeta=U_{\text{1t}}-\sqrt{M^2_{\text{1t}}\gamma RT_1-C^2_{\text{al}}} \\ \tan \zeta=\frac{U_{1_{\text{t}}}}{C_{\text{a1}}} -\sqrt{\frac{M^2_{\text{1t}}\gamma RT_1}{C^2_{\text{a1}}}-1 } \\ \zeta=\tan^{-1}\left(\frac{U_{1_\text{t}}}{C_{\text{a1}}} -\sqrt{\frac{M^2_{\text{1t}}\gamma RT_1}{C^2_{\text{a1}}} -1}\right) \\ \zeta= \tan^{-1}\left(\frac{237.5}{C_{\text{a1}}} -\sqrt{\frac{(0.8)^2 \times 1.4 \times 287 \ \text{T}_1}{C_{\text{a1}}^2} -1}\right) \\ \zeta=\tan^{-1}\left(\frac{237.5}{C_{\text{a1}}}-\sqrt{\frac{257.15T_1}{C^2_{\text{a1}}} -1} \right) \\ A_1=\frac{\pi}{4} (D_{\text{t}}^2-D^2_{\text{h}})=\frac{\pi}{4} \left[(0.28)^2-(0.14)^2\right] =0.0461814 \text{ m}^2

As a first assumption, calculate the inlet density from the total temperature and pressure, thus

\rho_1=\frac{P_{01}}{RT_{01}} =\frac{38,243}{287 \times 250.27} =0.532428461 \text{ kg/m}^3 \quad \quad \quad \text{(a)}

Now, calculate the axial inlet velocity

C_{\text{a1}}=\frac{\dot{m}}{\rho_1A_1} =\frac{3.5}{0.53242861 \times 0.0461814} =142.344 \text{m/s}

The prewhirl angle is now calculated from the relation

\zeta=\tan^{-1}\left(\frac{237.5}{C_{\text{a1}}}-\sqrt{\frac{257.15T_{01}}{C_{\text{a1}}^2} -1} \right) \\ \zeta=\tan^{-1}\left(\frac{237.5}{142.344}-\sqrt{\frac{257.15 \times 250.27}{(142.344)^2} -1} \right) =10.94^\circ

The inlet absolute velocity is now calculated as

C_1=\frac{C_{\text{a1}}}{\cos \zeta} =\frac{142.244}{\cos 10.94} =144.98 \text{m/s}

The static inlet temperature is

T_1=T_{01}-\frac{C^2_1}{2C_{\text{p}}} =250.27-\frac{(144.98)^2}{2 \times 1005} =239.81 \text{ K}

The static inlet pressure is calculated as

P_1=P_{01}\left(\frac{T_1}{T_{01}} \right) ^{\gamma/(\gamma-1)}=32,840.86\text{ Pa}

The density is recalculated from the static temperature and pressure just obtained as

\rho_1=\frac{P_1}{RT_1} =0.47755 \text{ kg/m}^3 \quad \quad \quad (\text{b})

A large difference is noticed between this value and the value obtained in (a); thus iteration is performed until equal values for the density are obtained in two successive iterations as shown.

Eleven iterations were needed for convergence. Only some were displayed in the above table.
The final results are

C_{\text{a1}} =167.04 \text{m/s}, \quad \zeta=18.94^\circ, \quad C_1=176.6 \text{ m/s} \\ P_1=30,567.8 \text{ Pa}, \quad T_1=234.75 \text{ K}, \quad \rho_1=0.4537 \text{ kg/m}^3