Question

2) An ideal gas flows through a horizontal tube at steady state. No heat is added and no shaft work is done. The cross- sectional area of the tube changes with length and this causes the velocity to change. Derive an equation relating the temperature to the velocity of the gas. If nitrogen at 150 °C flows past one section of the tube at a velocity of 2.5 m/s, what its temperature at another section where its velocity is 50 m/s? Cp = 7/2 R.

Answer 1

Answer:

The temperature of the nitrogen gas at another section is $148.8^{o}C$

Explanation:

Energy balance equation for steady state flow of gas under negiligible potential energy.

The negligible heat transfer and no shaft work is as follows.

$\Delta H+\frac{\Delta u^{2}}{2}=Q+W$

$\Delta H+\frac{\Delta u^{2}}{2}=0..........(1)$

$\Delta H$ is enthalphy of gas and it is changes with the temperature.

$\Delta H=C_{p}(T_{2}-T_{1}).................(2)$

$C_{p}$=  Molar heat capacity of the gas at constant pressure.$T_{1}$= Initial temperature at section 1

$T_{2}$ = Final temperature at section 2

Substitute the equation (2) in equation (1)

$C_{p}(T_{2}-T_{1})+\frac{u_{2}^2-u_{1}^2}{2}=0$

Solve the above equation is as follows.

$T_{2}=T_{1}-\frac{u_{2}^{2}-u_{1}^{2}}{2}=0...............(3)$

From the given,

$T_{1}=150+273=423K$

$C_{p}=\frac{7R}{2}$

$u_{1}=2.5\,m/s$

$u_{2}=50\,m/s$

Molar mass of nitrogen gas = 0.02802 kg/mol

Substitute the all values in the equation (3)

$T_{2}=423K-\frac{(50m/s)^{2}-(2.5m/s)^{2}}{2\times \frac{7}{2}\times8.314\,J\,mol^{-1}K^{-1}}\times \frac{J/kg}{m^{2}/s^{2}}\times \frac{0.02802\,kg}{mol}$

=421.8K=148.8^{o}C

Therefore,The temperature of the nitrogen gas at another section is $148.8^{o}C$.