Question 4.P.1: A gas, having a molecular weight of 13 kg/kmol and a kinemat...
A gas, having a molecular weight of 13 kg/kmol and a kinematic viscosity of 0.25 cm²/s, flows through a pipe 0.25 m internal diameter and 5 km long at the rate of 0.4 m³/s and is delivered at atmospheric pressure. Calculate the pressure required to maintain this rate of flow under isothermal conditions. The volume occupied by 1 kmol at 273 K and 101.3 kN/m² is 22.4 m³. What would be the effect on the required pressure if the gas were to be delivered at a height of 150 m (i) above, and (ii) below its point of entry into the pipe?
From equation 4.57 and, as a first approximation, omitting the kinetic energy term:
\left(P_2-P_1\right) / v_m+4\left(R / \rho u^2\right)(l / d)(G / A)^2=0
At atmospheric pressure and 289 K, the density =(13 / 22.4)(273 / 289)=0.542 kg / m ^3
Mass flowrate of gas, G=(0.4 \times 0.542)=0.217 kg / s.
Cross-sectional area, A=(\pi / 4)(0.25)^2=0.0491 m ^2.
Gas velocity, u=(0.4 / 0.0491)=8.146 m / s
∴ G / A=(0.217 / 0.0491)=4.413 kg / m ^2 s
Reynolds number, \operatorname{Re}=\rho d u / \mu
=\left(0.25 \times 8.146 / 0.25 \times 10^{-4}\right)=8.146 \times 10^4
From Fig. 3.7, for e / d=0.002, R / \rho u^2=0.0031
v_2=(1 / 0.542)=1.845 m ^3 / kg
v_1=(22.4 / 13)(298 / 273)\left(101.3 / P_1\right)=190.5 / P_1 m ^3 / kg
and: v_m=\left(0.923 P_1+95.25\right) / P_1 m ^3 / kg
Substituting in equation 4.57:
P_1\left(P_1-101.3\right) 10^3 /\left(0.923 P_1+95.25\right)=4(0.0031)(5000 / 0.25)(4.726)^2
and: P_1=\underline{\underline{111.1 kN / m ^2}}
The kinetic energy term =(G / A)^2 \ln \left(P_1 / P_2\right)=(4.413)^2 \ln (111.1 / 101.3)
=1.81 kg ^2 / m ^4 s ^2
This is negligible in comparison with the other terms which equal 5539 kg ^2 / m ^4 s ^2 so that the initial approximation is justified. If the pipe is not horizontal, the term g dz in equation 4.49 must be included in the calculation. If equation 4.49 is divided by v², this term on integration becomes g \Delta z / v_m^2.
∴ v_m=(0.923 \times 111.1+95.25) / 111.1=1.781 m ^3 / kg
v_{\text {air }}=(24.0 / 29)=0.827 m ^3 / kg
As the gas is less dense than air, v_m is replaced by \left(v_{\text {air }}-v_m\right)=-0.954 m³/kg.
∴ g \Delta z / v_m^2=\left(9.81 \times 150 / 0.954^2\right)=1616 N/m² or 0.16 kN/m²
(i) If the delivery point is 150 m above the entry level, then since gas is less dense,
P_1=(111.1-0.16)=\underline{\underline{110.94 kN / m ^2}}
(ii) If the delivery point is 150 m below the entry level then,
P_1=(111.1+0.16)=\underline{\underline{111.26 kN / m ^2}}