CHEMICAL ENGINEERING

82 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

PROBLEM 6.6

A gas cylinder containing 30 m^3 of air at 6 MN/m^2 pressure discharges to the atmosphere
through a valve which may be taken as equivalent to a sharp edged orifice of 6 mm
diameter (coefficient of dischargeD 0 .6). Plot the rate of discharge against the pressure in
the cylinder. How long will it take for the pressure in the cylinder to fall to (a) 1 MN/m^2 ,
and (b) 150 kN/m^2? Assume an adiabatic expansion of the gas through the valve and that
the contents of the cylinder remain at 273 K.

Solution

Area of orificeD/ 4  0. 006 ^2 D 2. 828 ð 10 ^5 m^2.

The critical pressure ratiowcis:

wcD[2/kC 1 ]k/k^1  (equation 4.43)

TakingkDD 1 .4forair,wcD 0 .527.

Thus sonic velocity will occur until the cylinder pressure falls to a pressure of:

P 2 D 101. 3 / 0. 527 D 192 .2kN/m^2.

For pressures in excess of 192.2kN/m^2 , the rate of discharge is given by:

GDCDA 0

√

kP 1 /v 1  2 /kC 1 kC^1 /k^1  (equation 6.29)

IfkD 1 .4,GD 1. 162 ð 10 ^5

p
P 1 /v 1 
IfPaandvaare atmospheric pressure and the specific volume at atmospheric pressure
respectively,PavaDP 1 v 1 andv 1 DPava/P 1

PaD 101 ,300 N/m^2 andvaD 22. 4 / 29 D 0 .773 m^3 /kg

∴ v 1 D 101 , 300 ð 0. 773 /P 1 D 78 , 246 /P 1 

and: GD 1. 162 ð 10 ^5

√

P 12 / 78 , 246 D 4. 15 ð 10 ^8 P 1 kg/s

IfP 1 is expressed in MN/m^2 , then:GD 0 .0415 P 1 kg/s.

For pressures lower than 192.2kN/m^2 :

G^2 DA 0 CD/v 2 ^22 P 1 v 1 k/k 1 [1P 2 /P 1 k^1 /k] (equation 6.26)

v 2 DvaD 0 .773 m^3 /kg

P 2 DPaD 101 ,300 N/m^2

v 1 DPava/P 1

Substituting gives: G^2 D 2. 64 ð 10 ^4 [1Pa/P 1 ^0.^286 ]

Thus a table ofGas a function of pressure may be produced as follows: