Page 193 - Fundamentals of Enhanced Oil and Gas Recovery

181
Thermal Recovery Processes

H wT 5 361.91 Btu/lbm at 387.9 F

[5 0.30
H wA 5 38 Btu/lbm at 70 F

ΔS o 5 0.31
M R 5 35 Btu/ft 3 F
M s 5 42 Btu/ft 3 F
α 5 1.2(Btu/h ft F)/42 (Btu/ft 3 F)

2
5 0.0286 ft /h
2
5 0.6857 ft /day

K h 5 1.2 Btu/h ft F.
Solution: When time is in days,
2
2
2
2
T 5 4(M S /M R ) (α/h )t 5 4(42/35) [0.6857/(32 ft) ]t 5 3.857 3 10 23 t
When time is in years,
t D 5 1.408 t, so that at t 5 4.5 years, t D 5 6.335,
H s 5 361.91 1 (0.7) (837.4) 277.94 5 870.15 Btu/Ibm
ð ð0:7Þð837:4ÞÞ
f h;v 5 5 0:674
870:15
As we know α 5 K h =M,

E h;s 5 0:33
The volume of oil displaced from the steam zone:

" #
m s H s t
V s 5 E h;s ðt D Þ
M R T s 2 T r Þ
ð
5 ½ð850 B=DÞð350:4 lbm=bblÞð870:15 Btu=lbmÞ
3 ð4:5yearsÞð365 D=yearފ=½ð35 Btu=ft 3 FÞ
2
3 ð387:9 2 110 FÞð43; 560 ft =ac ftŠð0:33Þ
5 331:5n ac ft:

N Ps 5 (7758 bbl/ac ft) [ h n S oi 2 S ors V s 5 (7758 bbl/ac ft) (0.30) (1.0) (0.31)
h t B oi B ors
(331.5 ac ft) 5 239,197 stb.
The equivalent volume of water injected is determined.
The energy content of the steam relative to the feed water temperature and the
steam leaving the boiler is computed below.
H wA 5 H ws 2 H wA 1 f sb L vs 5 361.9 2 38 1 0.8(837.4) 5 993.83 Btu/lbm.
3 3
W t 5 (850 B/D) (5.615 ft /bbl) (62.4 lbm/ft ) 3 (4.5 years) (365 D/year) 5
6
489.17 3 10 lbm.