UNIVERSIDAD INDUSTRIAL DE SANTANDER
Facultad de Ingenieras Fisicoqumicas
Escuela de Ingeniera de Petrleos
EJERCICIOS PROPUESTOS GAS PIPELINES HYDRAULICS
Ana Paula Villaquiran VargasJuan Camilo Gonzalez AngaritaNancy
Alexandra Patio ArgelloSamuel Francisco Martnez Hernndez
Ing. Adolfo Polo Rodriguez
Primer Semestre AcadmicoBucaramanga2013
EJERCICIOS PROPUESTOS GAS PIPELINES HYDRAULICS
1. A gas pipeline, NPS 18 with 0.375 in. wall thickness,
transports natural gas (specific gravity = 0.6) at a flow rate of
160 MMSCFD at an inlet temperature of 60F. Assuming isothermal
flow, calculate the velocity of gas at the inlet and outlet of the
pipe if the inlet pressure is 1200 psig and the outlet pressure is
700 psig. The base pressure and base temperature are 14.7 psia and
60F, respectively. Assume the compressibility factor Z = 0.95. What
is the pipe length for these pressures, if elevations are
neglected?
PART A:
Using the general gas flow equation:
Knowing that
NPS18Wall Thickness0,375 inID17,25 inQ160 MMSCFPD0,6T60F 520R
(Isothermal flow)Tb60F 520RPb14,7 psiaP11200 psig 1214,7 psia P2700
psig 714,7 psiaZ0,95
V1 = 13,12 ft
Using the relation between V1 and V2
V2 = 22,29 m/s
PART B:
NPS18Wall Thickness0,375 inID17,25 inQ160 MMSCFPD0,6T60F 520R
(Isothermal flow)Tb60F 520RPb14,7 psiaP11200 psig 1214,7 psia P2700
psig 714,7 psiaZ0,95
First it is necessary calculate Reynolds number:
Re = 10396902,54
Now we calculate the critical Reynolds Number using Nikuradse
and Von Karman equation.
The friction factor is:
Replacing in Von Karman equation we find the critical Reynolds
number.
- 0,6
The regimen flow is fully turbulent because Re > Rec.
Now using:
Panhandle B Equation:
L = 187,22 mi.
Weymouth Equation:
L = 117,774 mi
AGA Equation:
L = 143,79 mi
2. A natural gas pipeline, DN 400 with 10 mm wall thickness,
transports 3.2 Mm3/day. The specific gravity of gas is 0.6 and
viscosity is 0.00012 Poise. Calculate the value of the Reynolds
number. Assume the base temperature and base pressure are 15 C and
101 kPa, respectively.
NPS400Wall thickness10 mmID380 inQ 3,2*106 0,6 0,00012 PTb 15 C
288 KPb 101 kPa
To calculate Reynolds Number:
Re = 7580906,4
3. A natural gas pipeline, NPS 20 with 0,500 in. Wall thickness,
50 miles long, transports 220 MMSCFD. The specific gravity of gas
is 0,6 and viscosity is 0,000008 lb/ft-s. Calculate the friction
factor using Colebrook equation. Assume absolute pipe
roughness=750in. The base temperature and base pressure are 60F and
14,7 psia, respectively. What is the upstream pressure for an
outlet pressure of 800 psig?
Using Colebrook-white equation:
NPS 20Wall thickness0,5 inID19 inL50 miQ220*106 SCFD0,60,000008
lb/ft.s750 inTb60F 520RPb14,7 psiaP2800 psig 814,7 psia
Assuming 60F for gas flowing temperature and a compressibility
factor Zb=1
First its necessary to calculate the Reynolds number:
Re= 11729796,56
Using Colebrook equation the friction factor is:
The equation is:
Therefore:
4. For a gas pipeline flowing 3.5 Mm3/day gas of specific
gravity 0.6 and viscosity of 0.000119 Poise, calculate the friction
factor and transmission factor, assuming a DN 400 pipeline, 10 mm
wall thickness, and internal roughness of 0.015 mm. The base
temperature and base pressure are 15C and 101 kPa, respectively. If
the flow rate is increased by 50%, what is the impact on the
friction factor and transmission factor? If the pipe length is 48
km, what is the outlet pressure for an inlet pressure of 9000
kPa?
NPS400; Wall thickness: 10 mmID380 mmL48
kmQ3,5*106m3/day0,60,000119 P0,015 mmTb15 C 288 KPb101 kPaP19000
kPa
Q =
Initially we calculate the Reynolds Number:
9301103,694
Its necessary to calculate the Critical Reynolds Number using
Von Karman and Nikuradse equations:
Replacing in Von Karman equation we find the critical Reynolds
number.
The flow regimen is fully turbulent because Re > Rec.
First calculate the friction factor and transmission factor
(Initial Conditions) using Panhandle B equation:
Now calculate the friction factor and transmission factor (With
Q = Qo*1,5) using Panhandle B equation:
Then the impact in the friction and transmissions factor is:
For the Friction Factor:
For the Transmission Factor:
Finally, assuming standard conditions Tm=15C which means 288K,
Zb=1 the compressibility factor dimensionless.
With PANHANDLE B EQUATION for SI units
The outlet pressure will be
5. A gas pipeline ows 110 MMSCFD gas of specic gravity 0.65 and
viscosity of 0.000008 lb/ft-s. Calculate, using the modied
Colebrook-White equation, the friction factor and transmission
factor, assuming an NPS 20 pipeline, 0.375 in. wall thickness, and
internal roughness of 700 in. The base temperature and base
pressure are 60F and 14.7 psia, respectively.NPS20, Wall thickness:
0,375 inID19,25 inL100 kmQ110 MMSCFD0,650,000008 lb/ft-s700 inTb60F
520RPb14,7 psia = 3,64 x
Re= 0,0004778*Re= 0,0004778* Re=6271125Colebrook-White
equation
f= 0,011
F= F= F= 19,07
6. Using the AGA method, calculate the transmission factor and
friction factor for gas ow in an NPS 20 pipeline with 0.375 in.
wall thickness. The ow rate is 250 MMSCFD, gas gravity = 0.6, and
viscosity = 0.000008 lb/ft-sec. The absolute pipe roughness is 600
in. Assume a bend index of 60, base pressure =14.73 psia, and base
temperature = 60F. If the ow rate is doubled, what pipe size is
needed to keep both inlet and outlet pressures the same as that at
the original ow rate?NPS20, Wall thickness: 0,375 inID19,25 inL100
kmQ250 MMSCFD0,60,000008 lb/ft-s600 inTb60F 520R
Re= 0,0004778 Re= 0,0004778 Re= 13183055,69F= 4 F= 4F= 20,298The
transmission factor for smoth pipelinesF= 4Df = 0,96= 4 = 4 =
22,47F= 4 (0,96) F= 21,57Working with F=20,298F= f= f= f= 9,81*
PART BQ500 MMSCFDRe= 0,0004778 SoD= 0,0004778 D= 0,0004778 D=
38.50 inNPS= 42 in
7. A natural gas transmission line transports 4 million m3/day
of gas from a processing plant to a compressor station site 100 km
away. The pipeline can be assumed to be along a flat terrain.
Calculate the minimum pipe diameter required such that the maximum
pipe operating pressure is limited to 8500 kPa. The delivery
pressure desired at the end of the pipeline is a minimum of 5500
kPa. Assume a pipeline efficiency of 0,92. The gas gravity is 0,60,
and the gas temperature is 18C. Use the Weymouth equation,
considering a base temperature=15C and base pressure=101 kPa. The
gas compressibility factor Z=0,90.
L100 kmQ4000000 m3/day0,61 E0,92 Pb101 kPaP18500 kPaP25500
kPa
Weymouth equation
DN = 400 mm
8. Using the Panhandle B equation, calculate the outlet pressure
in a natural gas pipeline, NPS 16 with 0,250 in. Wall thickness, 25
miles long. The gas flow rate is 120 MMSCFD at 1200 psia inlet
pressure. The gas gravity=0,6 and viscosity=0,000008lb/ft-sec. The
average gas temperature is 80F. Assume the base pressure=14,73 psia
and base temperature=60F. The compressibility factor Z=0,90 and
pipeline efficiency is 0,95.
NPS16Wall thickness0,250 inE0,92So the diameter is15,5 in L25 mi
Q120 MMSCFDPb14,73 psia0,6P11200 psia 1
Panhandle B equation
