UNIVERSIDAD INDUSTRIAL DE SANTANDER SANTANDER Facultad de Ingenierías Fisicoquímicas Escuela de Ingeniería de Petróleos
EER!I!I"S PR"PUEST"S #AS PIPELINES $%DRAULI!S
Ana Paula Villaquiran Vargas uan !amilo #on&ale& Angarita Nanc' Ale(andra Pati)o Arg*ello Samuel Francisco +artíne& $ern,nde&
Ing- Adol.o Polo Rodrigue&
Primer Semestre Acad/mico 0ucaramanga 1234
EER!I!I"S PR"PUEST"S #AS PIPELINES $%DRAULI!S
thickness, transports natural gas (specific grait! 3- A gas pipeline, NPS 18 with 0.375 in. wall thickness, " 0.#$ at a flow rate of 1#0 %%S&' at an inlet te)perature of #0*'. Assu)ing isother)al flow, calculate calculate the elocit! of gas at the inlet an+ outlet of the pipe if the inlet pressure pressure is 100 psig an+ the outlet pressure is 700 psig. -he ase pressure an+ ase te)perature are 1/.7 psia an+ #0*', respectiel!. Assu)e the co)pressiilit! factor Z " 0.5. hat is the pipe length for these pressures, if eleations are neglecte+2 PART A5
sing the general gas flow e4uation Q 1 =Q b
( )( ) Pb
T 1
T b P1
Z 1
Q=V ∗ A
6nowing that
V =Qb
V =
[( )( ) ] P b
T 1
T b P 1
[( )( ) ]
0,2122∗Q b Pb
D
2
Z 1 / A
T b
NPS 6all T7ic8ness ID 9 γ
18 0,375 in 17,5 in 1#0 %%S&'P
T T: P: P3 P1 ;
#0*' 50* (9sother)al flow$ #0*' 50* 1/,7 psia 100 psig 11/,7 psia 700 psig 71/,7 psia 0,5
0,#
T 1 Z T 1 1
V 1=
0,2122∗160∗10 2
17,25
6
[( )( 14,7 520
) ]
520 0,95 1214,7
V 1 " 13,1 ft
sing the relation etween : 1 an+ : P1 V 2=V 1∗ P2
( )
(
V 2=13,118∗
1214,7 714,7
)
: " , );s
PART 05 NPS 6all T7ic8ness ID 9 γ
18 0,375 in 17,5 in 1#0 %%S&'P
T T: P: P3 P1 ;
#0*' 50* (9sother)al flow$ #0*' 50* 1/,7 psia 100 psig 11/,7 psia 700 psig 71/,7 psia 0,5
0,#
'irst it is necessar! calculate e!nol+s nu)er
( )( )
ℜ= 0,0004778
P b γ ∗Q T b μ∗ D
ℜ=0,0004778
( )( 14,7 520
6
0,6∗160∗10 −6
7,23∗10 ∗17,25
)
Re = 10396902,54
Now we calculate the critical e!nol+s Nu)er using Nikura+se an+ :on 6ar)an e4uation. -he friction factor is
[ ]
1 D =4∗log 3,7 ε √ f
[
1 17,25 =4∗log 3,7 714,7 √ f 1
√ f eplacing
1 √ f
]
=19,8396
in :on 6ar)an e4uation we fin+ the critical e!nol+s nu)er.
[ ]
ℜc 1 =4∗log 1 / √ f √ f
< 0,#
(19,8386 + 0,6 )
ℜc =19,8396∗10
4
ℜc =25550087,179
-he regi)en flow is full! turulent ecause e = e c. Now using
Pan7andle 0 Equation5
( ) (
T b Q=737∗ E∗ Pb
1,02
∗
s
2
P1 − e P2 0,961
γ
1,60∗10
L=
[(
8
)
0,51
2
¿ T f ¿ Le∗Z
∗ D2,53
( ) (
=737∗
520 14,7
1,60∗10
1,02
∗
8 2
737,02∗17,25
2
( ) ) ∗(
∗
14,7 520
1,02
1 0,510
L < 3=>?11 mi-
6e'mout7 Equation5 2 2 T b P1 − P2 − E Q= 432,7 ∗ Pb γ ¿ T f ¿ Le∗Z
( )(
[
( ) ( ) ∗(
1 Q Pb L= ∗ ∗ 432.7 Tb D
2,667
)
0,5
∗ D2,667
γ ∗T e∗Z 2
P 1 − P 2
2
)]
0,5 2
L < 33>?>>@ mi A#A Equation5
2
1214,7 −714,7 0,961 0,6 ¿ 520∗ Le∗0,95
0,961
)
0,51
∗17,252,53
∗520∗0,95 2 2 1214,7 −714,7
0,6
−1
)]
( )(
2
2
T b P − P2 − E ∗ 1 Q=38,774 P b γ ¿ T f ¿ Le∗Z
[
( ) ( ) ∗(
1 Q Pb L= ∗ ∗ 38,774 Tb D
2,5
) ( [ ]) 0,5
3,7 D ε
∗ 4 log
γ ∗T e∗Z 2
P 1 − P2
2
)
0,5
(
∗
∗ D2,5
[ ])
1 3,7 D 4 log ε
]
−2
L < 3@4?> mi
1- A natural gas pipeline, N /00 with 10 )) wall thickness, transports 3. %)3;+a!. -he specific grait! of gas is 0.# an+ iscosit! is 0.0001 Poise. &alculate the alue of the e!nol+s nu)er. Assu)e the ase te)perature an+ ase pressure are 15 & an+ 101 kPa, respectiel!.
/00 10 )) 380 in 3,>10#
NPS 6all t7ic8ness ID 9 γ
0,#
B T: P:
0,0001 P 15 *& 88 6 101 kPa
-o calculate e!nol+s Nu)er
ℜ=0,5134
∗ ( ( )∗) P b T b
γ Q μ D
ℜ=0,5134
( )( 101 288
6
0,6∗3,2∗10 0,00012 ∗380
) Re = 7580906,4
4- A natural gas pipeline, NPS 0 with 0,500 in. all thickness, 50 )iles long, transports 0 %%S&'. -he specific grait! of gas is 0,# an+ iscosit! is 0,000008 l;ft
sing &olerook
( )[ Tb Pb
2
2
P 1− P2− E γ ∗ L∗T m∗ Z m
]
0,5
∗ F ∗ D2,5
NPS 6all t7ic8ness ID L 9 γ
0 0,5 in 1 in 50 )i 0>10 # S&'
?
0,000008 l;ft.s
0,#
C
750 ?in #0*' 50* 1/,7 psia 800 psig 81/,7 psia
T: P: P1
Assu)ing #0*' for gas flowing te)perature an+ a co)pressiilit! factor @ "1 'irst its necessar! to calculate the e!nol+s nu)er
( )( )
ℜ=0,0004778∗
P b T b
∗
γ ∗Q μ∗ D
( )(
6
14,7 0,6∗220∗10 ℜ=0,0004778∗ ∗ 520 0,000008∗19
) Re= 11729796,56
sing &olerook e4uation the friction factor is
√
[ ( √
√ )]
1 1,426 1 ε =−4 log + ∗ 3,7 D f ℜ f 1 = 19,541 f −3
f =2,62∗10
-he e4uation is
(
Q Pb 38,774 Tb
)( )(
(
220∗10 38,774
)(
γ ∗ L ¿ T m∗ Z m 1 2,5 1 F ∗ D
)( )( 6
14,7 520
1 19,576179
[
650,3489379= P21 −(814,7 )2
[
]
19
2,5
=[ P21− P22 ]
0,5
0,6∗50 mi∗520∗1 1
)
0,5
=[ P21− P 22 ]
0,5
0,5
650,3489379=√ P1−( 814,7 ) 2
) ( )( 1
)
0,5
2
]
( 650,3489379 )2+( 814,7 )2= P21
-herefore
P1=1042,4 psia
@- 'or a gas pipeline flowing 3.5 %)3;+a! gas of specific grait! 0.# an+ iscosit! of 0.00011 Poise, calculate the friction factor an+ trans)ission factor, assu)ing a N /00 pipeline, 10 )) wall thickness, an+ internal roughness of 0.015 )). -he ase te)perature an+ ase pressure are 15*& an+ 101 kPa, respectiel!. 9f the flow rate is increase+ ! 50B, what is the i)pact on the friction factor an+ trans)ission factor2 9f the pipe length is /8 k), what is the outlet pressure for an inlet pressure of 000 kPa2
/00C all thickness 10 )) 380 )) 48 km 3,5*10 6 m3 /day
NPS ID L Q
γ
0,6
B
0,00011 P
C
0,015 )) 15 *& 88 6 101 kPa 000 kPa
T: P: P3
D"
1∈
3
(
)
3
3
1 day 1 ft m ++++¿+ ft 3,5∗10 ∗ ∗( ) =5150055,564 0,3048 m day 24 hours hours 6
¿ 25,4 mm
¿ ¿
D =380 mm∗¿
9nitiall! we calculate the e!nol+s Nu)er
ℜ=45
ℜ=45
Q b∗γ D
(
5150055,546∗0,6 14,96
) ℜ=¿ 301103,#/
9ts necessar! to calculate the &ritical e!nol+s Nu)er using :on 6ar)an an+ Nikura+se e4uations
[ ]
D 1 =4∗log 3,7 ε √ f
[
1 380 =4∗log 3,7 0,015 √ f 1
√ f eplacing
1 √ f
]
=19,88
in :on 6ar)an e4uation we fin+ the critical e!nol+s nu)er.
ℜc 1 =4∗log 1 / √ f √ f
−0,6
(19,88 + 0,6)
ℜc =19,88∗10
4
ℜc =2633155,765
-he flow regi)en is full! turulent ecause e = e c.
'irst calculate the friction factor an+ trans)ission factor (9nitial &on+itions$ using Panhan+le E e4uation
( )
1 = F =19,08∗ Q∗γ D √ f F =22,592 −3
f =1,959∗10
0,01961
Now calculate the friction factor an+ trans)ission factor (ith D " D o>1,5$ using Panhan+le E e4uation
( )
1 Q∗γ = F =19,08∗ D √ f
0,01961
F =22,77 −3
f =1,928∗ 10
-hen the i)pact in the friction an+ trans)issions factor is
'or the 'riction 'actor ! f =
! f =
|f − f | ∗100 f
|1,959∗10−3−1,928∗10−3| −3
1,959 ∗10
! f = 1,61
'or the -rans)ission 'actor
| F − F | ∗100
! F
F
|22,592 −22,77| ∗100
! F
22,592
! f = 0,78
∗100
'inall!, assu)ing stan+ar+ con+itions - )"15*& which )eans 886, @ "1 the co)pressiilit! factor +i)ensionless.
ith PAN$ANDLE 0 E9UATI"N for S9 units
( ) (
5
Q=1,002 ∗10 ∗
T b
1,02
∗
P b
s
2
P 1 − e P2 0,961
γ
2
¿ T m∗ L∗Z
)
0,51
∗ D2,53
-he outlet pressure will e P
(
¿ (¿ 1 ¿ ¿ 2− P 22 )0,51 ¿
)( ) ( ) Pb
Q −2
1,002∗10
1,02
T b
1
D
2,53
( γ
0,961
0,51
¿ T m L∗Z ) =¿
9000 (¿¿ 2− P22)0,51
(
3,5∗10
6
−2
1,002∗10
)( ) ( 101 288
1,02
1 2,53
380
9000 (¿¿ 2 − P22)0,51 3590,6096=¿
(3590,6096 )
1 0,51
=90002− P22
)
( 0,60,961∗288∗48∗1 ) 0,51=¿
2
9352291,437 −9000
=− P22
−7147708,56 =− P 22 8464,49 [ "Pa ] = P2
- A gas pipeline Fows 110 %%S&' gas of speciGc grait! 0.#5 an+ iscosit! of 0.000008 l;ft
γ
0, all thickness 0,375 in 1,5 in 100 km 110 MMSCFD 0,65
B
0,000008 l;ft
C
700 ?in #0*' 50* 1/,7 psia ε D = 3,64 x
T: P:
−5
10
( ) ( ) P b
e" 0,0004778*
e" 0,000/778>
T b
∗
( )∗¿ ( 14,7 520
6,5∗Q # $ %
6
0,65∗(110∗10 ) 0,000008∗19,25
)
e"#7115 &olerook<hite e4uation
(
1 2,51 ε =−2∗log ¿ + 3,7 D ℜ √ f √ f
) f= 0,011
2
'"
√ f
2
'"
√ 0,011
'" 1,07
- sing the AHA )etho+, calculate the trans)ission factor an+ friction factor for gas Fow in an NPS 0 pipeline with 0.375 in. wall thickness. -he Fow rate is 50 %%S&', gas grait! " 0.#, an+ iscosit! " 0.000008 l;ft
γ
0, all thickness 0,375 in 1,5 in 100 km 250 MMSCFD 0,#
B
0,000008 l;ft
C
#00 ?in #0*' 50*
T:
( ) ¿(
Pb ¿ e" 0,000/778 Tb
e" 0,000/778
¿
( )∗¿ ( 14,73 520
6,5∗Q #∗ D
6
6,5∗(250∗10 ) 0,000008∗19,25
Re= 13183055,69 log
'" /
log
'" /
(
( ) 3,7 D e
3,7∗19,25 −6
600∗10
)
F= 20,298
-he trans)ission factor for s)oth pipelines log
'" /
(
ℜ 1,4125∗ F t
)
Df = 0,96 F t
F t
"/
log
"/
log
F t
'" / (0,#$
(
( ) ℜ
F t
)−
6
= 22,47
(
13183055,69 1,4125∗ 22,47
F= 21,57
orking with '"0,8
−6
13183055,99 F t
log
)
)
)
2
'"
√ f
4
4
f" F 2
f"
( 20,298 )2
−3
10
f= 9,81*
PART B
9
500 %%S&'
( ) (
6,5∗Q #∗ D
)
(
6,5∗Q #∗ℜ
)
Pb ¿ e" 0,000/778 Tb
¿
So
( ) Pb
¿ ∗¿ D" 0,000/778 Tb
( ) (
14,73 ¿ D" 0,000/778 520
6
6,5∗(500∗10 ) ¿ 0,000008 ∗13183055,69
)
D= 3850 !" #$S= 42 !"
>- A natural gas trans)ission line transports / )illion ) 3;+a! of gas fro) a processing plant to a co)pressor station site 100 k) awa!. -he pipeline can e assu)e+ to e along a flat terrain.
&alculate the )ini)u) pipe +ia)eter re4uire+ such that the )aIi)u) pipe operating pressure is li)ite+ to 8500 kPa. -he +elier! pressure +esire+ at the en+ of the pipeline is a )ini)u) of 5500 kPa. Assu)e a pipeline efficienc! of 0,. -he gas grait! is 0,#0, an+ the gas te)perature is 18*&. se the e!)outh e4uation, consi+ering a ase te)perature"15*& an+ ase pressure"101 kPa. -he gas co)pressiilit! factor @"0,0. 100 k) /000000 ) 3;+a!
L 9 γ e
0,#
s
1
T b
0, 15 &C
288 '
T f
18 &C
291 '
E
101 kPa 8500 kPa 5500 kPa
P: P3 P1
e!)outh e4uation
( )(
s
2
2
T b P −e P2 Q=3,7435∗10 ∗ E∗ ∗ 1 Pb ( ¿ T f ¿ Le∗Z −3
−3
4000000 =3,7435∗10
D
D
2,667
(
=
2,667
( )(
∗0,92∗
288 101
∗
)
0,5
∗ D2,667 2
2
8500 −5500 0,6∗291∗100∗0,9
4000000∗101
(
2
2
8500 −5500 3,7435 ∗10 ∗0,92∗288∗ 0,6∗291∗100∗0,9 −3
)
0,5
)
)
=7878471,018
D=385,3385846 mm DN = 400 mm
0,5
∗ D2,667
=- sing the Panhan+le E e4uation, calculate the outlet pressure in a natural gas pipeline, NPS 1# with 0,50 in. all thickness, 5 )iles long. -he gas flow rate is 10 %%S&' at 100 psia inlet pressure. -he gas grait!"0,# an+ iscosit!"0,000008l;ft
1# 0,50 in 0, 15,5 in 60 ℉
520 & )
L
5 )i 80 ℉
540 & )
T f
9 P: γ
10 %%S&' 1/,73 psia
P3 s e
100 psia
0,# 1
Panhan+le E e4uation 1,02 2 s 2 T b P1 − e P2 Q=737∗ E∗ ∗ 0,961 Pb ( ¿ T f ¿ Le∗Z
( ) (
( ) (
120000000=737∗0,95∗
2,53
(
737∗15,5
2
1200
2
1200
)
0,51
14,73
120000000 2,53
− e0 P22 ∗ 0,961 0,6 ∗540∗25∗0,9
1,02
( ( )
∗0,95∗
∗ D2,53
−e0 P22 = 1,02 0,961 520 0,6 ∗540∗25∗0,9
120000000 737∗15,5
520 14,73
)
0,51
( )
∗0,95∗
520 14,73
1,02
)
1 0,51
− e0 P22 = 0,961 0,6 ∗540∗25∗0,9 1200
2
)
0,51
∗15,5 2,53
(
120000000 2,53
737∗15,5
2
1200
−
[(
( ) 520 14,73
∗0,95∗
)
∗0,6 0,961∗540∗25∗0,9 =12002− P22
120000000
737∗15,5
1304004,227 = P2
1,02
1 0,51
2,53
( )
∗0,95∗
2
P2=1141,930045 psia
520 14,73
1,02
)
1 0,51
]
∗0,6 0,961∗540∗25∗0,9 = P22
