Answer to Question #239769 in Molecular Physics | Thermodynamics for Unknown346307

Question

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 diam￾eter 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

Answer 1

Applying Bernoulli’s equations at sections 1 and 2 for an adiabatic process, we have:

$\Big( \cfrac{\gamma}{\gamma-1} \Big) \dfrac{p_1}{\rho_1g} + \dfrac{V^2_1}{2g} +z_1=\Big( \cfrac{\gamma}{\gamma-1} \Big) \dfrac{p_2}{\rho_2g} + \dfrac{V^2_2}{2g} +z_2$

For the following situation we know that p₁=78 kN/m² and p₂=117 kN/m², $T_1=40\,{°C}=313\,K$, V₁= 300 m/s, R = 287 J/kg K and γ = 1.4. Since this is a horizontal pipe $z_1=z_2$ and we will remove $g$ from the equations to simplify the expression:

$\Big( \cfrac{\gamma}{\gamma-1} \Big) \dfrac{p_1}{\rho_1\cancel{g}} + \dfrac{V^2_1}{2\cancel{g}} =\Big( \cfrac{\gamma}{\gamma-1} \Big) \dfrac{p_2}{\rho_2\cancel{g}} + \dfrac{V^2_2}{2\cancel{g}}$

Following this, we separate the terms of velocity from the density and pressure relations:

$\Big( \cfrac{\gamma}{\gamma-1} \Big) \dfrac{p_1}{\rho_1} -\Big( \cfrac{\gamma}{\gamma-1} \Big) \dfrac{p_2}{\rho_2}= \dfrac{V^2_2}{2}-\dfrac{V^2_1}{2}$

We continue to simplify because we need to find V₂:

$V^2_2=V^2_1+\Big( \cfrac{2\gamma}{\gamma-1} \Big)\Big( \dfrac{p_1}{\rho_1} -\dfrac{p_2}{\rho_2} \Big) =V^2_1+\Big( \cfrac{2\gamma}{\gamma-1} \Big)\dfrac{p_1}{\rho_1}\Big(1 -\dfrac{\rho_1}{p_1}\dfrac{p_2}{\rho_2} \Big)$

From the adiabatic relations between the pressure and density, we can find the relation of densities in terms of the pressures:

$\dfrac{p_1}{\rho^{\gamma}_1} =\dfrac{p_2}{\rho^{\gamma}_2} \implies \dfrac{p_1}{p_2} =\Big(\cfrac{\rho_1}{\rho_2} \Big)^{\gamma}$

$\implies \cfrac{\rho_1}{\rho_2} =\Big(\dfrac{p_1}{p_2} \Big)^{1/\gamma}$

We substitute the last expression for $V^2_2$ to have:

$V^2_2=V^2_1+\big( \frac{2\gamma}{\gamma-1} \big)\dfrac{p_1}{\rho_1}\bigg(1 -\dfrac{p_2}{p_1} \Big(\dfrac{p_1}{p_2} \Big)^{1/\gamma} \bigg)$

Following simplification gives us a new expression for $V^2_2$ :

$V^2_2=V^2_1+\big( \frac{2\gamma}{\gamma-1} \big)\dfrac{p_1}{\rho_1}\bigg(1 - \Big(\dfrac{p_2}{p_1} \Big)^{1-1/\gamma} \bigg)$

Now we have an espression that will give us the value for V₂, but we need to know or calculate the following:

$\dfrac{p_1}{\rho_1}=RT_1=(287 \frac{J}{kg K})(313\,K)=89831 \frac{m^2}{s^2}$

$\dfrac{p_2}{p_1}=\dfrac{117\frac{kN}{m^2}}{78\frac{kN}{m^2}}=1.5$

After that we continue with the substitution:

$V^2_2=(300\frac{m}{s})^2+\big( \frac{2(1.4)}{1.4-1} \big)(89831\frac{m^2}{s^2})\bigg(1 - \Big(1.5 \Big)^{1-\frac{1}{1.4}} \bigg)$

At last, we have:

$V^2_2=12781.27\cfrac{m^2}{s^2}\implies V_2=113.05\cfrac{m}{s}$

In conclusion the velocity of the gas at th second section is $V_2=113.05\cfrac{m}{s}$ .

Reference:

• Rajput, R. K. (2005). A textbook of engineering thermodynamics. Laxmi Publications.