At section 2, further downstream, the pressure is 78.8 kPa(abs). Assuming isotropic flow. Calculate the mach no. at section 2. Also identify the type of nozzle.

Data:-

$P_1=60kPa=60\times 10^3 Pa,P_2=78.8\times 10^3 Pa$

$T_1=27^\circ=300 K,V_1=486 m/s,A_1=0.02 m^2$

To find:-

$(i) M_2=?$

$(ii) \text{Type of nozzle=?}$

Solution:-

At section (1) since velocity

$C_1=\sqrt{\gamma.R.T_1}=\sqrt{1.4\times 287\times 300}=347.188 m/sec$

Mach no. at section (1)

$M_1=\frac{v_1}{c_1}=\frac{486}{347.188}=1.3998$

Now, stagnation pressure at (1)

$P_{01}=P_1[1+\frac{\gamma-1}{2}M_1^2]^{\frac{\gamma}{\gamma-1}}=60\times 10^3[1 + \frac{1.4+1}{2}\times (1.3998)^2]^{\frac{1.4}{1.4-1}}$

$=190.887\times 10^3 Pa$

For sentropic flow through duct

$P_{01}=P_{02}=P_0=constant$

For section 02

$P_{02}=P_2[1+\frac{\gamma-1}{2}M_2^2]^{\frac{\gamma}{\gamma-1}}$

$190.887\times 10^3=78.8\times 10^3[1+\frac{1.4-1}{2}M_2^2]\frac{1.4}{1.4-1}$

$\therefore M_2=1.999$

As $P_1\gt P_2, M_1\gt M_2$ passage is supersonic diffuser.