Question 12.8: In the early phase of design, two possibilities for the diff......

Figure 12.35 illustrates a vaned diffuser.
1. Isentropic vaned diffuser: Since

c_{\text{pd}}=\frac{P_3-P_2}{P_{02}-P_2} =\frac{(P_3/P_2)-1}{(P_{02}/P_2)-1} =0.65 \\ \frac{P_3}{P_2} =1+0.65\left[\left(\frac{P_{02}}{P_2} \right)-1 \right] =1+0.65\left[\left(1+\frac{\gamma-1}{2}M^2 \right)^{\gamma/\gamma-1}-1 \right] =1.449

Since

P_{03}=P_{02}, \quad \frac{P_3}{P_2} =\frac{P_3}{P_{03}} \frac{P_{02}}{P_2} =\frac{(1+((\gamma-1)/2)M_2^2)^{\gamma/(\gamma-1)}}{(1+((\gamma-1)/2)M_3^2)^{\gamma/(\gamma-1)}} =\left(\frac{1.162}{1+0.2M_3^2} \right) ^{3.5}

Or (1.449)^{1/3.5}=1.162/1+0.2M_3^2 \text{ or }1+0.2M^2_3=1.162/1.112=1.044964.
The outlet Mach number is M_3 = 0.474.
Since the flow is isentropic in the vaned diffuser, that is, T_{03}=T_{02} \text{ and }P_{03}=P_{02}

\frac{P_3}{P_2} =\frac{\rho_3}{\rho_2} \frac{T_3}{T_2} =\frac{\rho_3}{\rho_2} \frac{(T_3/T_{03})}{(T_2/T_{02})} =\frac{\rho_3}{\rho_2} \frac{1+((\gamma-1)/2)M_2^2}{1+((\gamma-1)/2)M_3^2} \\ \therefore 1.449=\frac{\rho_3}{\rho_2} \frac{1.162}{1.04496} \\ \frac{\rho_3}{\rho_2} = 1.303

From the continuity equation

\rho_2c_{\text{r2}}(\pi \ d_2 \ b_2)=\rho_3 c_{\text{r3}}(\pi \ d_3 \ b_3)

With b_2=b_3,

\frac{C_3}{C_2} =\frac{\rho_2 \ d_2}{\rho_3 \ d_3} \quad \quad \quad (1)

From the conservation of angular momentum in the vaned diffuser

r_2 \ C_{\text{u2}}=r_3 \ C_{\text{u3}}, \quad \text{thus}\frac{C_{\text{u3}}}{C_{\text{u2}}} =\frac{r_2}{r_3} =\frac{d_2}{d_3} \quad \quad \quad (2)

From Equations 1 and 2

\frac{C_{\text{u3}}/C_{\text{r3}}}{C_{\text{u2}}/C_{\text{r2}}} =\frac{\rho_3}{\rho_2}

or

\frac{\tan \alpha_3}{\tan \alpha_2} =\frac{\rho_3}{\rho_2} =1.303 \\ \tan \alpha_3=1.303 \ \tan \alpha_2=1.303 \times 2.4751 \\ \alpha_3=72.766^\circ

2. Vaneless diffuser: The following data are also assumed for the vaneless diffuser

M_2=0.9, \quad r_2=0.42 \text{ m}, \quad M_3=0.474, \quad \text{ and } \quad \alpha_2=68^\circ.

The flow is compressible and isentropic, and from Equation 12.30

\tan \alpha_3=\tan \alpha_2\left[\frac{1+((\gamma-1)/2)M_2^2}{1+((\gamma-1)/2)M_3^2} \right] ^{1/\gamma-1} \quad \quad \quad (12.30) \\ \alpha_3=72.7833

Also from Equation 12.31

r_3=r_2\frac{\sin \alpha_2}{\sin \alpha_3 } \frac{M_2}{M_3} \left[\frac{1+((\gamma-1)/2)M_3^2}{1+((\gamma-1)/2)M_2^2} \right] ^{1/2} \quad \quad \quad (12.31) \\ r_3=0.744 \text{ m}

It is clear from cases (1) and (2) that the outlet diameter of the vaneless diffuser (0.744 m) is much greater than the outlet diameter of the vaned diffuser (0.525 m); thus, vaneless diffuser enlarges the outer diameter of the compressor, which in turn increases the frontal area of aero engines and increases the drag force.
If P_{03}=0.9P_{02} \&T_{03}=T_{02}\text{ and }P_3/P_2=1.449

\therefore\frac{P_{02}/P_2}{P_{03}/P_3} =\left(\frac{1+((\gamma-1)/2)M_2^2}{1+((\gamma-1)/2)M_3^2} \right) ^{\gamma/(\gamma-1)}=\left(\frac{1.162}{1+0.2M_3^2} \right) ^{3.5}

But

\frac{P_{02}/P_2}{P_{03}/P_3} =\frac{P_{02}}{P_{03}} \frac{P_3}{P_2} =\frac{1.449}{0.9}=1.61

Then M_3 =0.2805

\frac{P_3}{P_2} =\frac{\rho_3}{\rho_2} \frac{T_3}{T_2} =\frac{\rho_3}{\rho_2} \frac{(T_3/T_{03})}{(T_2/T_{02})} =\frac{\rho_3}{\rho_2} \frac{1+((\gamma-1)/2)M_2^2}{1+((\gamma-1)/2)M_3^2} =\frac{\rho_3}{\rho_2} \frac{1.0157}{1.162} =\frac{\rho_3}{\rho_2} (0.87413) \\ \frac{\rho_3}{\rho_2} =\frac{1.449}{0.87413} =1.6576 \\ \text{Since}\frac{\tan \alpha_3}{\tan \alpha_2} =\frac{\rho_3}{\rho_2} ,\text{then }\tan \alpha_3=1.6576 \times 2.4751=4.1028 \text{ or } \alpha_3=76.3^\circ.

Figure 12.36 illustrates the vaned versus vaneless diffuser. Note that the outlet diameter of the vaneless diffuser is much greater than the vaned diffuser. In addition, the outlet angle from the radial direction is greater in the case of vaneless diffuser than vaned diffuser.