CHEMICAL ENGINEERING

62 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

From Fig. 3.7, forReD 3. 18 ð 106 ande/dD 0. 005 / 25 D 0 .0002,

R/u^2 D 0. 0017

In equation 4.57 and neglecting the first term:

P 2 
P 1 /vmC 4 R/u^2 l/dG/A^2 D 0

or: P 1 
P 2 D 4 vmR/u^2 l/dG/A^2

D 4 ð 0. 00885  0. 0017  30 / 0. 025  2540 ^2

D 466 ,000 N/m^2 or 0.466 MN/m^2

This is small in comparison with P 1 D12 MN/m^2 , and the average pressure of

10 MN/m^2 is seen to be too low. A mean pressure of 11.75 kN/m^2 is therefore selected

and the calculation repeated to give a pressure drop of 0.39 MN/m^2. The mean pressure

is then 12 C 11. 61 / 2 D 11 .8MN/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 termDG/A^2 lnP 1 /P 2 D 2540 ^2 ln 12 / 11. 61 D 2. 13 ð 105 kg2/m4s2

The termP 1 
P 2 /vm,wherevm is the specific volume at the mean pressure of

11 .75 MN/m^2 D 0. 39 ð 106 / 0. 00753 D 5. 18 ð 107 kg^2 /m^4 s.

Hence the omission of the kinetic energy term is justified

and the pressure dropD 0 .39 MN/m^2

PROBLEM 4.3

Hydrogen is pumped from a reservoir at 2 MN/m^2 pressure through a clean horizontal mild

steel pipe 50 mm diameter and 500 m long. The downstream pressure is also 2 MN/m^2

and the pressure of this gas is raised to 2.6MN/m^2 by a pump at the upstream end of

the pipe. The conditions of flow are isothermal and the temperature of the gas is 293 K.

What is the flowrate and what is the effective rate of working of the pump? Viscosity of

hydrogenD 0 .009 mN s/m^2 at 293 K.

Solution

Neglecting the kinetic energy term in equation 4.55, then:

P^22 
P^21 / 2 P 1 v 1 C 4 R/u^2 l/dG/A^2 D 0

whereP 1 D 2 .6MN/m^2 andP 2 D 2 .0MN/m^2.

Thus: v 1 D 22. 4 / 2  293 / 273  0. 1013 / 2. 6 D 0 .468 m^3 /kg

IfReD 107 ande/dD 0 .001, from Fig. 3.7,R/u^2 D 0 .0023.