TPG 4140 Natural Gas
SUGGESTED SOLUTIONS
Examination, December 15, 2014
Professor Jon Steinar Gudmundsson
PROBLEM 1 (15 %)
(a)
Sour gases are removed first because an amine-based solution contains water. It means that
the natural gas becomes saturated with water vapour during the removal of acid gases.
Therefore, drying is carried after the removal of sour gases.
(b)
An expander gives lower temperature because the thermodynamic process is isentropic while
the Joule-Thomson valve is isenthalpic.
(c)
Used to remove sour gases; they are based on ammoniac.
(d)
Dew point is when liquid condenses out from gas. The two most important dew points are
water dew point and hydrocarbon dew point.
PROBLEM 2 (15 %)
(a)
(b)
(c)
Also
(d)
(e)
Cascade (important to name) process to produce LNG.
PROBLEM 3 (15 %)
(a)
The problem can also be solved by using the p/z equation.
(b)
Typical recovery percentage is 60 % (not an exact percentage). From the plot the value will be
between 51 and 54 GSm3.
PROBLEM 4 (20 %)
(a)
Russia, Algeria and Norway.
(b)
A gas sales agreement states that the buyer pays for a fixed volume, irrespective of whether
the buyer takes/uses the volume agreed.
(c)
Equation can be derived from the real gas law; that is, the ideal gas law with the z-factor
added. Important that to show that z-factor at standard conditions is 1.
(d)
The natural gas volume is now larger.
(e)
Reserves divided by (annual) production. This can be for a field, country and/or the world.
(f)
Increased.
(g)
Decreased, because USA uses more natural gas (shale gas) for power generation, instead of
coal.
(h)
Higher heating value, gross heating value, of natural gas. Uses in statistics for gas production
in Norway.
(i)
Atmospheric pressure and approximately -162 ºC.
(j)
The effect of pipe roughness.
PROBLEM 5 (20 %)
360 kg/h = 0.1 kg/s
k 1


 k   p2  k 
m
   1
Pideal  RT1 

M
 k 1   p1 


 
M 
   gas    gas 
  air  s.c.  M air 
M air = 28.97 [kg/kmol]
Mgas = 0.7×28.97 = 20.3 kg/kmol
k
Cp
Cv
= 2.3/2.0 = 1.15
Kompressorarbeid (adiabatisk)
m (kg/s)
0,1
M (kg/kmol)
20,43
R (J/kmolK)
8314,34
T innløp C
15
k (Cp/Cv)
1,15
p innløp bara
2
p utløp bara
8
P (MW)
0,018
T utløp C
72
Kompressorarbeid (adiabatisk)
m (kg/s)
0,1
M (kg/kmol)
20,43
R (J/kmolK)
8314,34
T innløp C
15
k (Cp/Cv)
1,15
p innløp bara
8
p utløp bara
32
P (MW)
0,018
T utløp C
72
Total 0.036 MW divided by 0.8 = 0.045 MW
Kompressorarbeid (adiabatisk)
m (kg/s)
0,1
M (kg/kmol)
20,43
R (J/kmolK)
8314,34
T innløp C
15
k (Cp/Cv)
1,15
p innløp bara
2
p utløp bara
32
P (MW)
0,039
T utløp C
141
Single stage will be a
little bit larger.
(b)
  U d 
T2  T  T1  T  exp 
L
 m C p 
T1 = 72 ºC
6” = 0.1524 m
Temperatur i rørledning
T omgivelse C
T innløp C
U W/m2C
Rate kg/s
Cp J/kgC (ikke kJ)
d (m)
L (m)
T utløp C
5
72
2
0,1
2300
0,1524
1000
6,04
Natural gas hydrate will form in the bulk of the pipeline (6 ºC < 20 ºC).
PROBLEM 6 (15 %)
(a)
LNG
(b)
Inflow and outflow performance of a gas well. The curves cross is the flow rate of the well.
(c)
The extra pressure drop in the near wellbore region due to damage. But, negative skin
indicates stimulation and less pressure drop. (s < 1 should be s < 0)
(d)
(e)