CHEMICAL ENGINEERING

66 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

Outside area of pipeD 30 ðð 0. 025 D 2 .36 m^2.
Heat requiredD 7. 34 / 2. 36 D 3 .12 W/m^2
This low value of the heat required stems from the fact that the change in kinetic energy
is small and conditions are almost adiabatic. If the pipe were perfectly lagged, the flow
would be adiabatic and the pressure drop would then be calculated from equations 4.77
and 4.72. The specific volume at the low pressure endv 2 to be calculated from:

8 R/u^2 l/dD

[


1

2 

C

P 1

v 1

(

A

G

) 2 ][

1

(

v 1

v 2

) 2 ]

C 1



ln

(

v 2

v 1

)

(equation 4.77)

For nitrogen,D 1 .4 and hence:

8  0. 0028  30 / 0. 025 D

[

1. 4
1

2 ð 1. 4

C

12 ð 106

0. 00742

(

1

816

) 2 ][

1

(

0. 00742

v 2

) 2 ]

1. 4 C 1

1. 4

ln

( v 2

0. 00742

)

Solving by trial and error,v 2 D 0 .00746 m^3 /kg.

Thus:

1

2

(

G

A

) 2

v^21 C

(




1

)

P 1 v 1 D

1

2

(

G

A

) 2

v^22 C

(




1

)

P 2 v 2 (equation 4.72)

Substitution gives:

 816 ^2  0. 00742 ^2 / 2 C[1. 4 / 1. 4 
 1 ]12ð 106 ð 0. 00742

D 816 ^2  0. 00746 ^2 / 2 C[1. 4 / 1. 4 
 1 ]P 2 ð 106 ð 0. 00746

and:P 1 D 11 .94 MN/m^2
The pressure drop for adiabatic flowD 12. 0
11. 94 D 0 .06 MN/m^2 or 60 kN/m^2

PROBLEM 4.7

Air, at a pressure of 10 MN/m^2 and a temperature of 290 K, flows from a reservoir
through a mild steel pipe of 10 mm diameter and 30 m long into a second reservoir at a
pressureP 2. Plot the mass rate of flow of the air as a function of the pressureP 2. Neglect
any effects attributable to differences in level and assume an adiabatic expansion of the
air.D 0 .018 mN s/m^2 ,D 1 .36.

Solution

G/Ais required as a function ofP 2 .v 2 cannot be found directly since the downstream
temperatureT 2 is unknown and varies as a function of the flowrate. For adiabatic flow,