Question 4.P.2: Nitrogen at 12 MN/m² pressure is fed through a 25 mm diamete...

Molecular weight of nitrogen = 28 kg/kmol.
Assuming a mean pressure in the pipe of 10 MN/m², the specific volume, $v_m$ at 10 MN/m² and 298 K is:

$v_m=(22.4 / 28)\left(101.3 / 10 \times 10^3\right)(298 / 273)=0.00885  m ^3 / kmol$

Reynolds number, $\rho u d / \mu=d(G / A) / \mu)$.

$A=(\pi / 4)(0.025)^2=4.91 \times 10^{-3}  m ^2$.

∴ $(G / A)=\left(1.25 / 4.91 \times 10^{-3}\right)=2540  kg / m ^2 s$

and:                       $R e=\left(0.025 \times 2540 / 0.02 \times 10^{-3}\right)=3.18 \times 10^6$

From Fig. 3.7, for $R e=3.18 \times 10^6$ and $e / d=(0.005 / 25)=0.0002$,

$R / \rho u^2=0.0017$

In equation 4.57 and neglecting the first term:

$\left(P_2-P_1\right) / v_m+4\left(R / \rho u^2\right)(l / d)(G / A)^2=0$

or:      $P_1-P_2=4 v_m\left(R / \rho u^2\right)(l / d)(G / A)^2$

$=4 \times 0.00885(0.0017)(30 / 0.025)(2540)^2$

= 466,000 N/m² or 0.466 MN/m²

This is small in comparison with $P_1=12  MN / m ^2$, and the average pressure of 10 MN/m² is seen to be too low. A mean pressure of 11.75 kN/m² is therefore selected and the calculation repeated to give a pressure drop of 0.39 MN/m². The mean pressure is then $(12+11.61) / 2=11.8 MN / m ^2$ which is close enough to the assumed value.
It remains to check if the assumption that the kinetic energy term is negligible is justified.

Kinetic energy term $=(G / A)^2 \ln \left(P_1 / P_2\right)=(2540)^2 \ln (12 / 11.61)=2.13 \times 10^5 kg 2 / m 4 s 2$

The term $\left(P_1-P_2\right) / v_m$, where $v_m$ is the specific volume at the mean pressure of $11.75  MN / m ^2=\left(0.39 \times 10^6\right) / 0.00753=5.18 \times 10^7  kg ^2 / m ^4 s$.
Hence the omission of the kinetic energy term is justified

and the pressure drop $=\underline{\underline{0.39  MN / m ^2}}$