Page 71 - Petroleum Production Engineering, A Computer-Assisted Approach

Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach 0750682701_chap05 Final Proof page 62 21.12.2006 2:02pm

5/62 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
5.4.2 Sonic Flow Solution (a)
Under sonic flow conditions, the gas passage rate reaches k 1:3
P outlet 2 k 1 2 1:3 1
its maximum value. The gas passage rate is expressed in ¼ ¼ ¼ 0:5459
the following equation for ideal gases: P up c k þ 1 1:3 þ 1
v ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
u !
u k 2 kþ1 P dn 200
k 1
t ¼ ¼ 0:25 < 0:5459 Sonic flow exists:
Q sc ¼ 879C D Ap up (5:8) P up 800
g g T up k þ 1
d 2 1 00
The choke flow coefficient C D is not sensitive to the Rey- ¼ ¼ 0:5
nolds number for Reynolds number values greater than d 1 2 00
6
6
10 . Thus, the C D value at the Reynolds number of 10 can Assuming N Re > 10 , Fig. 5.2 gives C D ¼ 0:62.
6
be assumed for C D values at higher Reynolds numbers. v ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
Gas velocity under sonic flow conditions is equal to u k ! 2 kþ1
u
k 1
sound velocity in the gas under the in situ conditions: q sc ¼ 879C D AP up t
s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi g g T up k þ 1

z up 2 s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1:3þ1
n ¼ n 2 up þ 2g c C p T up 1 z outlet k þ 1 (5:9) q sc ¼ (879)(0:62)[p(1) =4](800) 1:3 2 1:3 1
2
(0:6)(75 þ 460) 1:3 þ 1
or q sc ¼ 12,743 Mscf=d
p ffiffiffiffiffiffiffi
n 44:76 T up (5:10)
Check N Re :
m ¼ 0:01245 cp by the Carr–Kobayashi–Burrows cor-
5.4.3 Temperature at Choke relation.
Depending on the upstream-to-downstream pressure ratio,
the temperature at choke can be much lower than expected. N Re ¼ 20q sc g g ¼ (20)(12,743)(0:6) ¼ 1:23 10 > 10 6
7
This low temperature is due to the Joule–Thomson cooling md 2 (0:01245)(1)
effect, that is, a sudden gas expansion below the nozzle (b)
causes a significant temperature drop. The temperature
can easily drop to below ice point, resulting in ice-plugging z up P outlet k 1 k 1:3 1
if water exists. Even though the temperature still can be T dn ¼ T up ¼ (75 þ 460)(1)(0:5459) 1:3
z outlet P up
above ice point, hydrates can form and cause plugging

problems. Assuming an isentropic process for an ideal gas ¼ 465 R ¼ 5 F < 32 F
flowing through chokes, the temperature at the choke Therefore, heating is needed to prevent icing.
downstream can be predicted using the following equation:
k 1 (c)
k
z up p outlet
T dn ¼ T up (5:11)
P outlet
z outlet p up P outlet ¼ P up ¼ (800)(0:5459) ¼ 437 psia
P up
The outlet pressure is equal to the downstream pressure in
subsonic flow conditions.
Example Problem 5.2 A 0.65 specific gravity natural gas
5.4.4 Applications flows from a 2-in. pipe through a 1.5-in. nozzle-type
Equations (5.5) through (5.11) can be used for estimating choke. The upstream pressure and temperature are
100 psia and 70 8F, respectively. The downstream
pressure is 80 psia (measured 2 ft from the nozzle). The
. Downstream temperature
. Gas passage rate at given upstream and downstream gas-specific heat ratio is 1.25. (a) What is the expected
pressures daily flow rate? (b) Is icing a potential problem? (c) What
. Upstream pressure at given downstream pressure and is the expected pressure at the nozzle outlet?
gas passage
. Downstream pressure at given upstream pressure and Solution (a)
gas passage k 1:25
P outlet 2 k 1 2 1:25 1
To estimate the gas passage rate at given upstream and ¼ ¼ ¼ 0:5549
downstream pressures, the following procedure can be P up c k þ 1 1:25 þ 1
taken:
P dn ¼ 80 ¼ 0:8 > 0:5549 Subsonic flow exists:
Step 1: Calculate the critical pressure ratio with Eq. (5.1). 100
P up
Step 2: Calculate the downstream-to-upstream pressure
ratio. d 2 1:5 00
Step 3: If the downstream-to-upstream pressure ratio is ¼ 00 ¼ 0:75
d 1 2
greater than the critical pressure ratio, use Eq.
(5.5) to calculate gas passage. Otherwise, use Eq. Assuming N Re > 10 , Fig. 5.1 gives C D ¼ 1:2.
6
(5.8) to calculate gas passage.
v ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
#
"
u
Example Problem 5.1 A0.6 specific gravitygas flowsfrom u k P dn 2 P dn kþ1
k
k
t
a 2-in. pipe through a 1-in. orifice-type choke. The upstream q sc ¼ 1,248C D AP up
(k 1)g g T up P up P up
pressure and temperature are 800 psia and 75 8F,
respectively. The downstream pressure is 200 psia q sc ¼ (1,248)(1:2)[p(1:5) =4](100)
2
(measured 2 ft from the orifice). The gas-specific heat ratio v ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi #
"
u
is 1.3. (a) What is the expected daily flow rate? (b) Does u 1:25 80 2 80 1:25þ1
1:25
1:25
t
heating need to be applied to ensure that the frost does not (1:25 1)(0:65)(530) 100 100
clog the orifice? (c) What is the expected pressure at the
orifice outlet? q sc ¼ 5,572 Mscf=d