# 5.29: Untitled Page 90

## Chapter 5

6,895 Pa

MPa

2,900 psia

 20.00 MPa

6

psia

10 Pa

6,895 Pa

MPa

2,100 psia

 14.48 MPa

6

psia

10 Pa

 5 C K

90 F

32 F 

 273 15

.

K  305.4 K

9 F

C

 5 C K

45 F

32 F 

 273 15

.

K  280.4 K

9 F

C

 0.3048 m

50 ft/s 

 15.24 m/s

ft

0.0254 m

D

D

20 in 

 0.508 m

o

1

in

D

0.508 m2

2

2

A

A

 0 2027

.

m

o

1

4

4

Since the natural gas is assumed to be pure methane, we can obtain the molecular mass from Table A2 in Appendix A where we find MW

 16 . 043 g/mol . The volume per mole of methane can be CH4

determined using the ideal gas law given by Eq. 5‐3 along with the value of the gas constant found in Table 5‐1. The volume per mole at the entrance is determined as

3

m Pa

8 314

.

305 4 . K

RT

mol K

4

3

V n

 1 . 2 10 m / mol (1)

o

6

p

o

19 98

.

10 Pa

while the volume per mole at the exit is given by

RT

4

3

V n

 1 5

. 3  10 m / mol

(2)

1

p 1

The gas densities at the entrance and exit of the pipeline are given by

Two‐Phase Systems & Equilibrium Stages 161

MW CH

16 043

.

g/mol

4

(

)

CH4 o

4

3

V n

1 2

.  10

m /mol

o

(3)

3

3

 133 , 700 g/m

 133 . 7 kg/m

MW CH

4

3

(

)

 104 . 8 kg/m

(4)

CH4 1

1

V n

To perform a mass balance for the pipeline, we begin with the species mass balance given by

d

dV   v n dA

r dV

(5)

A

A A

A

dt V

A

V

Since there are no chemical reactions and there is no accumulation, the first and last terms in this result are zero and Eq. 5 simplifies to

dA  0

(6)

v n

A A

A

For a single component system, the species velocity is equal to the mass average velocity (see Chap. 4) and Eq. 6 takes the form

dA  0

(7)

v n

A

A

The control volume is constructed in the obvious manner, thus there is an entrance in Glenpool, OK and an exit in Lincoln, NE. Since the mass average velocity and the diameter of the pipeline are given, we express the mass balance as

(

) v A

 (

) v A

(8)

CH4 o o

o

CH4 1 1

1

The only unknown in this result is the velocity of the gas at the exit of the pipeline. Solving for v we obtain

1

3

A (

)

o

CH

133 7

. kg/m

4 o

v  

 v 

15 . 24 m/s

 19 . 44 m/s

(9)

1

o

3

A (

) 

1

CH

104 8

. kg/m

4 1 

The mass flow rate is a constant given by

162