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Find the velocity of the gas at this section if the flow of the gas is adiabatic.

A gas with a velocity of 300 m/s is flowing through a horizontal pipe at a section where pressure is 78 KN/m2 absolute and temperature 40°C. The pipe changes in diameter and at this section, the pressure is 117 kN/m2 absolute. Find the velocity of the gas at this section if the flow of the gas is adiabatic. Take R = 287 J/kg K and γ = 1.4.

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Solution:

Section 1:

Velocity of the gas, $V=300 \mathrm{~m} / \mathrm{s} \\$

Pressure, $p_{2}= 78 \mathrm{kN} / \mathrm{m}^{2} \\$

Temperature, $\quad T_{1}=40+273=313 \mathrm{~K} \\$

Section 2 :

Pressure, $p_{2}=117 \mathrm{kN} / \mathrm{m}^{2} \\$

$R=287 \mathrm{~J} / \mathrm{kg} \mathrm{K}, \gamma=1.4 \\$

Velocity of gas at section $2, V_{2}$ :

Applying Bernoulli's equations at sections 1 and 2 for adiabatic process, we have,

$$\left(\frac{\gamma}{\gamma-1}\right) \frac{p_{1}}{\rho_{1} g}+\frac{V_{1}^{2}}{2 g}=z_{1}=\left(\frac{\gamma}{\gamma-1}\right) \frac{p_{2}}{\rho_{2} g}+\frac{V_{2}^{2}}{2 g}+z_{2} \\$$

But $z_{1}=z_{2}$, since the pipe is horizontal.

$$\therefore \quad\left(\frac{\gamma}{\gamma-1}\right) \frac{p_{1}}{\rho_{1} g}+\frac{V_{1}^{2}}{2 g}=\left(\frac{\gamma}{\gamma-1}\right) \frac{p_{2}}{\rho_{2} g}+\frac{V_{2}^{2}}{2 g} \\$$

Cancelling ' $g$ ' on both sides, we get, $$\begin{array}{r} \left(\frac{\gamma}{\gamma-1}\right)\left(\frac{p_{1}}{\rho_{1}}-\frac{p_{2}}{\rho_{2}}\right)=\frac{V_{2}^{2}}{2}-\frac{V_{1}^{2}}{2} \\ \text { or, } \quad\left(\frac{\gamma}{\gamma-1}\right) \frac{p_{1}}{\rho_{1}}\left(1-\frac{p_{2}}{\rho_{2}} \times \frac{\rho_{1}}{p_{1}}\right)=\frac{V_{2}^{2}}{2}-\frac{V_{1}^{2}}{2} \\ \therefore \quad\left(\frac{\gamma}{\gamma-1}\right) \frac{p_{1}}{\rho_{1}}\left(1-\frac{p_{2}}{p_{1}} \times \frac{\rho_{1}}{\rho_{2}}\right)=\frac{V_{2}^{2}}{2}-\frac{V_{1}^{2}}{2}....(1) \end{array}\$$

$\frac{p_{1}}{\rho_{1}^{\gamma}}=\frac{p_{2}}{\rho_{2}^{\gamma}}$ or $\frac{p_{1}}{p_{2}}=\left(\frac{\rho_{1}}{\rho_{2}}\right)^{\gamma}$ or $\frac{\rho_{1}}{\rho_{2}}=\left(\frac{p_{1}}{p_{2}}\right)^{\frac{1}{\gamma}} \\$

Substituting the value of, $\frac{\rho_{1}}{\rho_{2}}$ in eqn (i), we get,

$$\left(\frac{\gamma}{\gamma-1}\right) \frac{p_{1}}{\rho_{1}}\left\{1-\frac{p_{2}}{p_{1}} \times\left(\frac{p_{1}}{p_{2}}\right)^{\frac{1}{\gamma}}\right\}=\frac{V_{2}^{2}}{2}-\frac{V_{1}^{2}}{2} \\$$

$$\left(\frac{\gamma}{\gamma-1}\right) \frac{p_{1}}{\rho_{1}}\left\{1-\left(\frac{p_{2}}{p_{1}}\right)^{1-\frac{1}{\gamma}}\right\}=\frac{V_{2}^{2}}{2}-\frac{V_{1}^{2}}{2} \\$$

$$\left(\frac{\gamma}{\gamma-1}\right) \frac{p_{1}}{\rho_{1}}\left\{1-\left(\frac{p_{2}}{p_{1}}\right)^{\frac{\gamma-1}{\gamma}}\right\}=\frac{V_{2}^{2}-V_{1}^{2}}{2}...(2) \\$$

At section 1 :

\begin{aligned} &\frac{p_{1}}{\rho_{1}}=R T_{1}=287 \times 313=89831 \\\\ &\frac{p_{2}}{p_{1}}=\frac{117}{78}=1.5, \text { and } V_{1}=300 \mathrm{~m} / \mathrm{s} \\ \end{aligned}

Substituting the values in eqn. (ii), we get,

$\left(\frac{1.4}{1.4-1}\right) \times 89831\left\{1-(1.5)^{\frac{1.4-1}{1.4}}\right\}=\frac{V_{2}^{2}}{2}-\frac{300^{2}}{2} \\$

$314408.5(1-1.1228)=\frac{V_{2}^{2}}{2}-45000$ or $-38609.4=\frac{V_{2}^{2}}{2}-45000 \\$

$V_{2}{ }^{2}=12781.2$ or, $\mathbf{V}_{2}=\mathbf{1 1 3 . 0 5} \mathbf{m} / \mathbf{s} . \quad$ (Ans.)