CHEMICAL ENGINEERING

88 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

The maximum rate of discharge is given by equation 6.29.

P 1 D 1. 5 ð 106 N/m^2

A 0 D 30 ð 10 ^6 m^2

kD 1 .4andCDD 0. 65

AtP 1 D 1 .5MN/m^2 andT 1 D293 K, the specific volumev 1 is:

v 1 D 22. 4 / 29  293 / 273  0. 1013 / 1. 5 D 0 .056 m^3 /kg

Substituting,GmaxD 0. 65 ð 30 ð 10 ^6

√

 1. 4 ð 1. 5 ð 106 / 0. 056  2 / 2. 4 ^2.^4 /^0.^4

D 0 .069 kg/s

PROBLEM 6.14

A gas cylinder containing air discharges to atmosphere through a valve whose charac-
teristics may be considered similar to those of a sharp-edged orifice. If the pressure in
the cylinder is initially 350 kN/m^2 , by how much will the pressure have fallen when the
flowrate has decreased to one-quarter of its initial value? The flow through the valve may
be taken as isentropic and the expansion in the cylinder as isothermal. The ratio of the
specific heats at constant pressure and constant volume is 1.4.

Solution

From equation 4.43:

the critical pressure ratio,wc D[2/kC 1 ]k/k^1 D 2 / 2. 4 ^1.^4 /^0.^4 D 0. 528

If the cylinder is discharging to atmospheric pressure, sonic velocity will occur until
the cylinder pressure has fallen to 101. 3 / 0. 528 D192 kN/m^2
The maximum discharge when the cylinder pressure exceeds 192 kN/m^2 is given by:

GmaxDCDA 0

√

kP 1

v 1

(

2

kC 1 

)kC 1 /k 1 

(equation 6.29)

IfPaandvaare the pressure and specific volume at atmospheric pressure, then:

1 /v 1 DP 1 /Pava

and: GmaxDCDA 0

√

kP^21

Pava

(

2

kC 1

)kC 1 /k 1 

DCDA 0 P 1

√

[k/Pava 2 /kC 1 ]kC^1 /k^1 

IfG 350 andG 192 are the rates of discharge at 350 and 192 kN/m^2 respectively, then:

G 350 /G 192 D 350 / 192 D 1. 82

or: G 192 D 0. 55 G 350