Question 12.7: The performance of a conical diffuser is illustrated in Figu......

From Figure 12.34, the pressure coefficient and area ratio are

C_{\text{P}_{\text{r}}}\approx 0.52 \quad \text{ and } \quad \text{Area ratio}=4.5

The static temperature and pressure are

T_2=T_{02}-\frac{C_2^2}{2C_{{\text{P}}}} =488-\frac{(356)^2}{2 \times 1005} =424.95 \text{ K} \\ P_2=P_{02}\left(\frac{T_2}{T_{02}} \right) ^{\gamma/(\gamma-1)}=3.376 \text{ bar}

The density is

\rho_2=\frac{P_2}{RT_2} =2.768 \text{ kg/m}^3

The mass flow rate \dot{m} is given by \dot{m} = \rho_2 C_2(A_{\text{d}} \times 12), \text{with }A_{\text{d}}=7.8 \times 10^{-4}\text{ m}^2

\dot{m}=2.768 \times 356 \times 7.8 \times 10^{-4} \times 12=9.223 \text{ kg/s}

An approximate value for C_3 can be evaluated based on the following arguments:
The continuity equation is \dot{m}=\rho_2 C_2A_2=\rho_3 C_3A_3 .
For incompressible flow A_2C_2=A_3C_3 \text{ or }C_3=C_2(A_2/A_3)=C_2/ \text{Area ratio}.
However, since the flow is compressible, the static pressure and consequently the density increase in the diffuser, that is, \rho_3 \gt \rho_2 , thus C_3 \lt C_2/\text{Area ratio}.
Now for our case, the area ratio is 4.5, then C_3 \lt 356/4.5, \text{ or }C_3 \lt 79.1 \text{ m/s}.

\text{Try }C_3=70 \text{ m/s} \\ C_{\text{P}_{\text{r}}}\approx \frac{2(P_{03}-P_{02})}{\rho_{02}C_2^2} +\left[1-\frac{P_{03}}{P_{02}}\left(\frac{C_3}{C_2} \right)^2 \right] \\ C_{\text{P}_{\text{r}}}=\frac{2P_{02}}{\rho_{02}C_2^2} \left[\frac{P_{03}}{P_{02}}-1 \right] +\left[1-\frac{P_{03}}{P_{02}} \left(\frac{C_3}{C_2} \right) ^2 \right] \\ = \frac{2RT_{02}}{C_2^2} \left[\frac{P_{03}}{P_{02}} -1\right] +\left[1-\frac{P_{03}}{P_{02}}\left(\frac{C_3}{C_2} \right)^2 \right]

Define \lambda=P_{03}/P_{02}, then

C_{\text{P}_{\text{r}}}=\frac{2 \times 287 \times 488}{(356)^2} [\lambda-1]+\left[1- \lambda\left(\frac{70}{356} \right)^2 \right]

or

0.52=2.21 (\lambda-1)+(1-0.039 \lambda)=2.171 \lambda-1.21

Thus, λ = 0.796 and P_{03}=4.366 \text{ bar}.

Then, the static properties at the diffuser outlet are

T_3=T_{03}-\frac{C^2_3}{2C_{\text{P}}} =488-\frac{(70)^2}{2 \times 1005} =485.6 \text{ K} \\ P_3=P_{03}\left(\frac{T_3}{T_{03}} \right) ^{\gamma/ \gamma-1}=4.366 \times \left(\frac{485.6}{488} \right) ^{3.5}=4.291 \text{ bar} \\ \rho_3=\frac{P_3}{RT_3} =\frac{4.291 \times 10^5}{287 \times 485.6 } =3.079 \text{ kg/m}^3

The outlet speed is

C_3=\frac{\dot{m}}{\rho_3 A_3} =\frac{9.223}{(4.079 \times 4.5 \times 7.8 \times 10^{-4} \times 12)} =71.1 \text{ m/s}.