Question 13.9: Figure 13.47 illustrates the air flow pattern through an axi......
Figure 13.47 illustrates the air flow pattern through an axial fan. The fan blading and velocity triangles at the tip of a transonic fan rotor are shown in Figure 13.47a, b, d and e while the oblique shock pattern is shown in Figure 13.47c. The inlet and outlet angles of flow relative to the blades are \beta_1 = 68.8^\circ \text{ and } \beta_2=65^\circ, respectively. The flow approaches the rotor in the axial direction with an absolute velocity C_1 = 200 m/s. The inlet stagnation temperature and pressure are 300 K and 1.15 bar, respectively (radially uniform). The hub-to-tip ratio is 0.4 and the rotor tip diameter is 2.0 m. Inside the rotor blade passage two oblique shocks are generated. Experimental measurements of the static pressure over the blade-to-blade surface indicated that the turning angles of both oblique shocks are, respectively, \delta_1=10^\circ \text{ and } \delta_2=10^\circ.
At the outlet to the rotor, the relative Mach number equals that in region 3. The absolute inlet speed (C_1) and the specific work (W_{\text{s}}) are assumed constant at all radii.
1. Calculate the static and total pressure ratios across the rotor.
2. Calculate the absolute and relative velocities at the outlet to the rotor, then draw the velocity triangles.
3. Define which portion of the blade is subjected to supersonic relative inlet speed, then draw the velocity triangles at its minimum radius.
The given data are T_{01}=300 \text{ K},P_{01}=1.15 \text{ bar, and }C_1=200 \text{ m/s}. The sonic speed is then a_1=\sqrt{\gamma RT_1}=335 \text{ m/s}.
The static conditions are then
T_1=T_{01}-\frac{C_1^2}{2 Cp} =300-\frac{200^2}{2 \times 1005} =280 \text{ K} \\ P_1=P_{01}\left(\frac{T_1}{T_{01}} \right) ^{3.5}=1.15(0.933)^{3.5}=0.90319 \text{ bar} \\ \rho_1 =\frac{P_1}{RT_1} =1.1239 \text{ kg/m}^3
1. The first shock wave:
The inlet velocity triangle is shown in Figure 13.47b
W_1=\frac{C_1}{\cos \beta_1} =\frac{200}{\cos 68.8} =553 \text{ m/s} \\ M_{\text{W}1} =\frac{W_1}{a} =1.65
since M_{\text{W}1}=1.65 \text{ and } \delta =10^\circ , \theta=50^\circ.
The normal component of the inlet relative Mach number is then
M_{\text{W}1n}=M_{\text{W}1} \sin \theta =1.65 \sin 50=1.2639
From normal shock wave tables, M_{\text{W}2n}=0.8052. As illustrated in Figure 13.47c
M_{\text{W}2}=\frac{M_{\text{W}2n}}{\sin(\theta-\delta)} =\frac{0.8052}{\sin 40^\circ} =1.2526
From normal shock wave tables also
\frac{P_2}{P_1} =1.6976, \quad \frac{\rho_2}{\rho_1} =1.4528, \quad \frac{T_2}{T_1} =1.1682, \quad \frac{P_{02}}{P_{01}} =0.9849
2. The second shock wave:
Since M_{\text{W}2}=1.2526 \text{ and } \delta=10^\circ, then from shock table [43], θ = 57°. The normal component of the Mach number M_{\text{W}2n}=1.2526 \sin57=1.0513. From normal shock wave tables, M_{\text{W}3n}=0.952.
The Mach number is then
M_{\text{W}3}=\frac{M_{\text{W}3n}}{\sin(\theta-\delta)} =\frac{0.952}{\sin(57-2)} =1.1621
From normal shock wave tables
\frac{P_3}{P_2} =1.1229, \quad \frac{\rho_3}{\rho_2} =1.08619 \\ \frac{T_3}{T_2} =1.03365, \quad \frac{P_{03}}{P_{02}} =0.9985 \\ P_{03}=\frac{P_{03}}{P_{02}} \frac{P_{02}}{P_{01}} P_{01}=(0.9985)(0.984918)(1.15)=1.1325 \text{ bar}
The static conditions at state 3 are
\frac{P_3}{P_1} =\frac{P_3}{P_2} \frac{P_2}{P_1} =1.906 \\ P_3=1.7217 \text{ bar} \\ T_3=\frac{T_3}{T_2} \frac{T_2}{T_1} T_1=338 \text{ K} \\ \rho_3 =\frac{\rho_3}{\rho_2} \frac{\rho_2}{\rho_1} \rho_1=1.7738 \text{ kg/m}^3
The sonic speed is a_3 = 368.5 \text{ m/s}.
If at the outlet to rotor the Mach number is M_{\text{W}3}=1.1621, then the rotor outlet relative speed is
W_2=1.1621 \times 368.5=428 \text{ m/s} \\ C_{\text{a}2}=W_2 \cos 65=428 \times 0.4226=180.8 \text{ m/s} \\ \frac{U}{C_{\text{a}2}} =\tan \beta_2+\tan \alpha_2 \\ \tan \alpha_2=\frac{515.6}{180.8} -\tan 65=0.7073 \\ \alpha_2=35.27 ^\circ \\ C_2=\frac{C_{\text{a}2}}{\cos \alpha_2} =221.4 \text{ m/s} \\ M_{\text{C}2}=\frac{221.4}{368.5} =0.6
The outlet velocity triangle at tip is plotted in Figure 13.47d, while the inlet and outlet velocity triangles having common apex are drawn in Figure 13.47e.
Then, to determine which part of the blade is subjected to supersonic flow, the location at which the inlet relative speed equals the sonic speed (W_1 = a_1=\sqrt{\gamma RT_1}=335 \text{ m/s}) is first determined.
Denoting the relative speed at this case with W_{1\text{s}}=335 \text{ m/s}, then
\cos \beta_{1\text{s}}=\frac{C_1}{W_{1\text{s}}}=\frac{200}{335} \quad \text{or} \quad \beta_{1\text{s}}=53.3^\circ
From the velocity triangle; similar to that drawn in Figure 13.47b, the rotational speed
U_{\text{s}}=C_1 \tan \beta_{1\text{s}}=200 \times 1.3416=268.32 \text{ m/s}
Since the ratio between rotational speeds at two positions is related to their radii ratio,
\frac{U_{\text{s}}}{U_{\text{t}}} =\frac{r_{\text{s}}}{r_{\text{tip}}} \\ r_{\text{s}}=r_{\text{tip}}\frac{U_{\text{s}}}{U_{\text{tip}}} =(1.0)\left(\frac{268.3}{515.6} \right) =0.52 \text{ m}
The hub radius is
r_{\text{h}}=\left(\frac{r_{\text{h}}}{r_{\text{t}}} \right) r_{\text{t}}=(0.4)(1.0)=0.4 \text{ m}
As shown in Figure 13.47f, the part of the blades extending from hub (radius equals 0.4 m) to a radius of (0.544 m) represents the subsonic part of the blade. The outer portion is then a supersonic part. The fan is thus a transonic fan.
