(a) In case of isentropic flow of a compressible fluid through a variable duct, show that, $$ \frac{A}{A_{c}}=\frac{1}{M}\left[\frac{1+\frac{1}{2}(\gamma-1) M^{2}}{\frac{1}{2}(\gamma+1)}\right]^{\frac{\gamma+1}{2(\gamma-1)}} \\ $$ where $\gamma$ is the ratio of specific heats, $M$ is the Mach number at a section whose area is $A$ and $A_{c}$ is the critical area of flow. (b) A supersonic nozzle is to be designed for air flow with Mach number 3 at the exit section which is $200 \mathrm{~mm}$ in diameter. The pressure and temperature of air at the nozzle exit are to be $7.85 \mathrm{kN} / \mathrm{m}^{2}$ and $200 \mathrm{~K}$ respectively. Determine the reservoir pressure, temperature and the throat area. Take : $\gamma=1.4$.

UPSC Exam

Solution:

Mach number, $ \quad M=3 \\ $

Area at the exit section, $ A=\pi / 4 \times 0.2^{2}=0.0314 \mathrm{~m}^{2} \\ $

Pressure of air at the nozzle, $ \quad(p)_{\text {nozzle }}=7.85 \mathrm{kN} / \mathrm{m}^{2} \\ $

Temperature of air at the nozzle, $ (T)_{\text {nozzle }}=200 \mathrm{~K} \\ $

Reservoir pressure, $ (\mathbf{p})_{\text {res. }} \\ $ :

$$ \quad(p)_{\text {res. }}=(p)_{\text {nozzle }}\left[1+\left(\frac{\gamma-1}{2}\right) M^{2}\right]^{\left(\frac{\gamma}{\gamma-1}\right)} \\ $$

$$or,$$

$$ (p)_{\text {res. }}=7.85\left[1+\left(\frac{1.4-1}{2}\right) \times 3^{2}\right]^{\left(\frac{1.4}{1.4-1}\right)}=288.35 \mathbf{k N} / \mathbf{m}^{2} \\ $$

Reservoir temperature, $ (\mathrm{T})_{\text {res }^{*}} \\ $ :

$$ \quad(T)_{\text {res. }}=(T)_{\text {nozzle }}\left[1+\left(\frac{\gamma-1}{2}\right) M^{2}\right] \\ $$

$$or,$$

$$ (T)_{\text {res. }}=200\left[1+\left(\frac{1.4-1}{2}\right) \times 3^{2}\right]=560 \mathbf{K} \quad \text { (Ans.) } \\ $$

Throat area (critical), $ \mathbf{A}_{\mathrm{c}} \\ $ :

$$ \quad \frac{A}{A_{c}}=\frac{1}{M}\left[\frac{2+(\gamma-1) M^{2}}{\gamma+1}\right]^{\frac{\gamma+1}{2(\gamma-1)}} \\ $$

$$or,$$

$$ \begin{aligned} \frac{0.0314}{A_{c}} &=\frac{1}{3}\left[\frac{2+(14-1) 3^{2}}{1.4+1}\right]^{\frac{1.4+1}{2(1.4-1)}} \text { or } \\\\ A_{c} &=\frac{0.0314}{4.23}=0.00742 \mathbf{~ m}^{2} \quad \text { (Ans.) } \end{aligned} \\ $$