Do problems 8.1, 8.2, & 8.4 on page 183.
CBEN 408 Spring 2016 - 1 - February 20, 2016

Solutions
Problem #8.1 (10 points)
A 6 in. pipe (internal area = 0.2006 ft² [0.01864 m²] gas line transmits 20 MMscfd (0.536 × 10 6
Nm³/d). An 8 mile (12.9 km) segment of the line is pigged. Assuming that the average pipeline temperature is 80°F (27°C) and average operating pressure is 250 psig (17.2 barg), what would be the average pig speed (mph, km/h) in the line and the minimum time it would take to traverse the pigged segment? Assume that there is minimal gas slippage around the pig & ideal gas behavior.
Solution
The pig will be pushed ahead of the gas flow at a velocity established by the gas’s average linear velocity. The following must be considered when calculating the flows:
 The standard volumetric flow must be corrected for the actual pressure & temperature to get the actual volumetric flow at pipeline conditions.
 Because we are not given any information about the gas we will assume it is at ideal gas conditions & no additional correction to the actual volumetric flowrate need be made for its compressibility factor.
 The average linear velocity of the gas will be the ratio of the actual volumetric flowrate & the pipeline’s cross-sectional area.
 Remember that the standard temperature when using US customary units is 60°F whereas the temperature for normal conditions using metric units is 0°C (32°F).
Using the ideal gas law to correct the standard volumetric flowrate to the actual flowrate:
V act
 V std



T
T act std



P
P std act

 
  20 10 6 ft³ day














14.7 psia

 6 ft³ day
 48,060 ft³ hr



The average linear velocity is: u 
V act
A cross

48,060 ft³ ft hr
 45.4
mi hr
So, with no slippage, the time for the pig to traverse the pipeline is: t
8 mi
45.4
mi hr
 0.18 hr  11 min
CBEN 408 Spring 2016 - 2 - February 20, 2016

Problem #8.2 (5 points)
Assume that the gas in the line in Exercise 8.1 condenses 0.01 GPM (1.41 m³/10 6 m³) over the 8 mile (12.9 km) distance and the line is pigged once a day. How many gallons (m³) of condensate would the pig remove from the line?
Solution
The value for the liquid content takes into account the pipeline length so it does not have to be further used in any calculation. The total liquids that drop out are:
V liq




0.01
gal





6 ft³ day




So, in a day’s time 200 gal will accumulate.
200 gal day
.
CBEN 408 Spring 2016 - 3 - February 20, 2016

Problem #8.4 (10 points)
Compute the concentration of methanol needed to provide 18°F (10°C) of subcooling into the hydrate region using the Hammerschmidt and Nielsen & Bucklin equations. Assume the specific gravity of methanol is 0.79.
Solution
The following are the Hammerschmidt equations to estimate the mass fraction inhibitor required to provide a specified hydrate subcooling:
X i

 T
 
 
F°
M
F°

M
2335
for US customary units & X i

 T
 
 
C°
M
C°

M
1297
for metric units.
Since methanol has a molecular weight of 32.042 then:
X
MeOH

 
 
18 32.042


 0.198
Note that you get the same answer using the metric units:
X
MeOH

 
 
10 32.042


 0.198
.
The following are the Nielsen and Bucklin equations to estimate the water’s mole fraction required to have a specified hydrate subcooling:
 
F°
  129.6 ln
 
for US customary units &
 
C°
  72.0ln
 
for metric units.
Using the US customary unit equation: x w
 exp



  
F°
 129.6




 exp



18
 129.6



 0.8703
Now converting to mass fraction methanol:
X
MeOH
 x
MeOH
M
MeOH x M w
 x
MeOH
M
MeOH


 

0.8703 18.015
 

 
 0.209
.
Specifying the mass fraction of methanol is sufficient to report the concentration. If you’d like to report as the volume fraction (based on ideal liquid volumes) then the volume fraction methanol predicted by the Hammerschmidt equation is:
CBEN 408 Spring 2016 - 4 - February 20, 2016

v
MeOH


1 
X
MeOH


MeOH
X
MeOH

 w
X
MeOH

MeOH



 and for the Nielsen and Bucklin equation:
1



0.198
0.79



0.198
0.79


 v
MeOH





0.209
0.79
  





0.209
0.79



 0.25
.
 0.24
CBEN 408 Spring 2016 - 5 - February 20, 2016