Respuesta :
Answer:
The temperature of the nitrogen gas at another section is [tex]148.8^{o}C[/tex]
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.
[tex]\Delta H+\frac{\Delta u^{2}}{2}=Q+W[/tex]
[tex]\Delta H+\frac{\Delta u^{2}}{2}=0..........(1)[/tex]
[tex]\Delta H[/tex] is enthalphy of gas and it is changes with the temperature.
[tex]\Delta H=C_{p}(T_{2}-T_{1}).................(2)[/tex]
[tex]C_{p}[/tex]= Molar heat capacity of the gas at constant pressure.[tex]T_{1}[/tex]= Initial temperature at section 1
[tex]T_{2}[/tex] = Final temperature at section 2
Substitute the equation (2) in equation (1)
[tex]C_{p}(T_{2}-T_{1})+\frac{u_{2}^2-u_{1}^2}{2}=0[/tex]
Solve the above equation is as follows.
[tex]T_{2}=T_{1}-\frac{u_{2}^{2}-u_{1}^{2}}{2}=0...............(3)[/tex]
From the given,
[tex]T_{1}=150+273=423K[/tex]
[tex]C_{p}=\frac{7R}{2}[/tex]
[tex]u_{1}=2.5\,m/s[/tex]
[tex]u_{2}=50\,m/s[/tex]
Molar mass of nitrogen gas = 0.02802 kg/mol
Substitute the all values in the equation (3)
[tex]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}[/tex]
=421.8K=148.8^{o}C
Therefore,The temperature of the nitrogen gas at another section is [tex]148.8^{o}C[/tex].
