At some section in the convergent-divergent nozzle, in which air is flowing, pressure, velocity, temperature and cross-sectional area are $200 \mathrm{kN} / \mathrm{m}^{2}, 170 \mathrm{~m} / \mathrm{s}, 200^{\circ} \mathrm{C}$ and $1000 \mathrm{~mm}^{2}$ respectively. If the flow conditions are isentropic, determine :

(i) Stagnation temperature and stagnation pressure,

(ii) Sonic velocity and Mach number at this section,

(iii) Velocity, Mach number and flow area at outlet section where pressure is $110 \mathrm{kN} / \mathrm{m}^{2}$,

(iv) Pressure, temperature, velocity and flow area at throat of the nozzle.

Take for air $: R=287 \mathrm{~J} / \mathrm{kg} K, c_{p}=1.0 \mathrm{~kJ} / \mathrm{kg} K$ and $\gamma=1.4$.

Solution:

Let subscripts 1, 2 and t refers to the conditions at given section, outlet section and throat section of the nozzle respectively.

Pressure in the nozzle, $ \quad p_{1}=200 \mathrm{kN} / \mathrm{m}^{2} \\ $

Velocity of air, $ \quad V_{1}=170 \mathrm{~m} / \mathrm{s} \\ $

Temperature, $ \quad T_{1}=200+273=473 \mathrm{~K} \\ $ Cross-sectional area, $ \quad A_{1}=1000 \mathrm{~mm}^{2}=1000 \times 10^{-6}=0.001 \mathrm{~m}^{2} \\ $

For air : $\quad R=287 \mathrm{~J} / \mathrm{kg} \mathrm{K}, c_{p}=1.0 \mathrm{~kJ} / \mathrm{kg} \mathrm{K}, \gamma=1.4 \\ $

(i) Stagnation temperature, $ \left(\mathrm{T}_{\mathrm{s}}\right) $ and stagnation pressure, $ (\left.\mathbf{p}_{\mathrm{s}}\right) $ :

Stagnation temperature, $ \quad T_{s}=T_{1}+\frac{V_{1}^{2}}{2 \times c_{p}} \\ $

Also,

$$ =473+\frac{170^{2}}{2 \times(1.0 \times 1000)}=487.45 \mathrm{~K}\left(\text { or } 214.45^{\circ} \mathrm{C}\right) \\ $$

$\therefore$ Stagnation pressure, $ \quad p_{s}=200 \times 1.111=222.2 \mathrm{kN} / \mathbf{m}^{2} \quad \\ $ (Ans.)

(ii) Sonic velocity and Mach number at this section :

Sonic velocity, $ \quad C_{1}=\sqrt{\gamma R T_{1}}=\sqrt{1.4 \times 287 \times 473}=\mathbf{4 3 5 . 9} \mathbf{~ m} / \mathbf{s} \quad \\ $ (Ans.)

Mach number, $ \quad M_{1}=\frac{V_{1}}{C_{1}}=\frac{170}{435.9}=\mathbf{0 . 3 9} \quad \\ $ (Ans.)

(iii) Velocity, Mach number and flow area at outlet section where pressure is, $ 110 \mathrm{kN} / \mathrm{m}^{2} \\ $ :

Pressure at outlet section, $ p_{2}=110 \mathrm{kN} / \mathrm{m}^{2}...(Given) \\ $

From eqn, $ (16.17), \quad \frac{p_{s}}{p_{1}}=\left[1+\left(\frac{\gamma-1}{2}\right) M_{2}^{2}\right]^{\frac{\gamma}{\gamma-1}} \\ $

$$ \frac{222.2}{110}=\left[1+\left(\frac{1.4-1}{2}\right) M_{0}^{2}\right]^{\frac{1.4}{1.4-1}}=\left(1+0.2 M_{0}^{2}\right)^{3.5} \\ $$

$$or$$

$$ \begin{gathered} \left(1+0.2 M_{2}^{2}\right)=\left(\frac{222.2}{110}\right)^{\frac{1}{3.5}}=1.222 \\\\ \left.M_{2}=\left(\frac{1.222-1}{0.2}\right)^{1 / 2}=1.05 \quad \text { (Ans. }\right) \\ \end{gathered} $$

Also,

$$ \frac{T_{2}}{T_{s}}=\left(\frac{p_{2}}{p_{s}}\right)^{\frac{\gamma-1}{\gamma}}=\left(\frac{110}{222.2}\right)^{\frac{1.4-1}{1.4}}=0.818 \\ $$

$$or,$$

$$ T_{2}=0.818 \times 487.45=398.7 \mathrm{~K} \\ $$

Sonic velocity at outlet section, $$ C_{2}=\sqrt{\gamma R T_{2}}=\sqrt{1.4 \times 287 \times 398.7}=400.25 \mathrm{~m} / \mathrm{s} \\ $$

$\therefore$ Velocity at outlet section, $ V_{2}=M_{2} \times C_{2}=1.05 \times 400.25=\mathbf{4 2 0 . 2 6} \mathbf{m} / \mathbf{s} \\ $. (Ans.)

Now, mass flow at the given section = mass flow at outlet section (exit) continuity equation,

$$ \rho_{1} A_{1} V_{1}=\rho_{2} A_{2} V_{2} \quad \text { or } \quad \frac{p_{1}}{R T_{1}} A_{1} V_{1}=\frac{p_{2}}{R T_{2}} A_{2} V_{2} \\ $$

$\therefore$ Flow area at the outlet section, $$ A_{2}=\frac{p_{1} A_{1} V_{1} T_{2}}{T_{1} p_{2} V_{2}}=\frac{200 \times 0.001 \times 170 \times 398.7}{473 \times 110 \times 420.26}=6.199 \times 10^{-4} \mathrm{~m}^{2} \\ $$

Hence, $ A_{2}=6.199 \times 10^{-4} \mathrm{~m}^{2}$ or $\mathbf{6 1 9 . 9} \mathbf{m m}^{2} \\ $. (Ans.)

(iv) Pressure, $\left(\mathbf{p}_{\mathbf{t}}\right)$, temperature, $\left(\mathbf{T}_{\mathbf{t}}\right)$, velocity, $\left(\mathbf{V}_{\mathbf{t}}\right)$, and flow area, $\left(\mathbf{A}_{\mathbf{t}}\right)$ at throat of the nozzle :

At throat, critical conditions prevail, i.e. the flow velocity becomes equal to the sonic velocity and Mach number attains a unit value.

$$ \quad \frac{T_{s}}{T_{t}}=\left[1+\left(\frac{\gamma-1}{2}\right) M_{t}^{2}\right] \\ $$

$$or,$$

$ \quad \frac{487.45}{T_{t}}=\left[1+\left(\frac{1.4-1}{2}\right) \times 1^{2}\right]=1.2$ or $T_{t}=406.2 \mathrm{~K} \\ $

Also,

$ T_{t}=406.2 \mathrm{~K}\left(\right.$ or $\left.133.2^{\circ} \mathrm{C}\right) \\ $ (Ans.)

$ \frac{p_{t}}{T_{s}}=\left(\frac{T_{t}}{T_{s}}\right)^{\frac{\gamma}{\gamma-1}} \\ $ or

$ \frac{p_{t}}{222.2}=\left(\frac{406.2}{487.45}\right)^{\frac{1.4}{1.4-1}}=0.528 \\ $

or,

$ p_{t}=222.2 \times 0.528=117.32 \mathbf{k N} / \mathbf{m}^{2} \\ $. (Ans.)

Sonic velocity (corresponding to throat conditions),

$$ C_{t}=\sqrt{\gamma R T_{t}}=\sqrt{1.4 \times 287 \times 406.2}=404 \mathrm{~m} / \mathrm{s} \\ $$

$\therefore$ Flow velocity, $ \quad V_{t}=M_{t} \times C_{t}=1 \times 404=404 \mathrm{~m} / \mathrm{s} \\ $

By continuity equation, we have : $$ \rho_{1} A_{1} V_{1}=\rho_{t} A_{t} V_{t} \\ $$

$$or,$$

$$ \frac{p_{1}}{R T_{1}} A_{1} V_{1}=\frac{p_{t}}{R T_{t}} A_{t} V_{t} \\ $$

$\therefore$ Flow area at throat, $$ A_{t}=\frac{p_{1} A_{1} V_{1} T_{t}}{T_{1} p_{t} V_{t}}=\frac{200 \times 0.001 \times 170 \times 406.2}{473 \times 117.32 \times 404}=6.16 \times 10^{-4} \mathrm{~m}^{2} \\ $$

Hence,

$$ A_{t}=6.16 \times 10^{-4} \mathrm{~m}^{2} \text { or } 616 \mathrm{~mm}^{2} \quad \text { (Ans.) } \\ $$