244452165-Solucionario-Cap-4-Pares-Recuperacion-Termica-del-Petroleo-Douglas-Alvarado.pdf

ROBLEMAS ( C P ROBLEMAS C APÍTULO 4 )

F REDDY REDDYC ACHUMBA ING. JORGEV ELASQUEZ ELASQUEZ



**

** En un pozo de la Faja Petrolífera del Orinoco se está inyectando vapor de 90% de calidad a razón

de 10.000 lb/h. El pozo está situado a 400 m de la planta y la línea tiene un diámetro interno de 2,50 pulgadas y diámetro externo de 2,75 pulgadas. La temperatura ambiente en el sitio del pozo es de 80°F y generalmente no hay viento. Si la presión del vapor a la salida de la planta es de 700 Ipca, calcular las pérdidas de calor y la calidad del vapor en el cabezal del pozo.

     í   ° 30 30   

 =115,1700, =502.59°  =  =502.59°+460°=962.59° P ROBLEMAS C APÍTULO 4 ) ROBLEMAS -P ARES ( C

REDDYC ACHUMBA F REDDY ELASQUEZ ING. JORGEV ELASQUEZ



 = 502.59+60 2 =271.3°  =0,01328+2,471∗10− ∗4,247∗10− ∗  =0,01328+2,471∗10− ∗271.34,247∗10− ∗271.3 = 0.019671228 ℎ  =8,55865∗10− 1,5531∗10−+1,65602∗10− 6,92225∗10−

 =0.054257533    =0,04+6,155∗10−+1,027∗10−   =0.057454425 ℎ  =0,2382+1,39∗10−+1,027∗10−   =024272698   =2,15844∗10− 3,89367∗10−+4,12773∗10− 1,71867∗10−  =0.001371584 −

. + .  ℎ= ℎ=. + . 962.59 + 520 ℎ = 0,1714∗10− ∗ 1 ∗ 962.59 520 ∗ 962.59+520 962.59+520  ℎ=3.0417 ℎ 

P ROBLEMAS C APÍTULO 4 ) ROBLEMAS -P ARES ( C


F REDDY REDDYC ACHUMBA


ING. JORGEV ELASQUEZ ELASQUEZ

 hc d e     - 0.0757  0.3082 (log R e )  0.0379 (log R e ) 2  k ha 

log 

  Re  

5.280

d e a  a   a

 

=5.280 0.2291×30×0.054257533 =34.27 0.057454425  ] ℎ =0.0279998[ℎ 

− 1  = [ℎ + ℎ ] − 1  = [0.02079998+3.0417]  ]  =3.06249998[ℎ 

 ̇ =2rtoUtoTs  Ta∆L  ̇ =2(2.75 24 ) 3.06249998502.5960 ∗1312.33

̇ =.[ ] 





X st wh




X st gen

qL 

wLv

   ̇ = 1280 1280625 625.2.288 ℎ ⇒  = ̇ = 1280625.28 =975.84 1312.33  ℎ  ×350  =122500  =5104.15  =350    ℎ

501.4503 = 501.4502.59 712.21710.014 712.21  =710.564   975.84×1312.33 = , =0.9 710.564×5104.15

, =%








Lv @ 600F = 549.67 BTU/lbm

⁄ℎ 2132261.67  =0.75 20000  ×549.67 ⁄ =0.556  

⟦  =%⟧

0.0286×480 0.29=2.41  =  2 × √ 0.0286×480 0.5 

=0.87 P ROBLEMAS C APÍTULO 4 ) ROBLEMAS -P ARES ( C





 = 340  = 800.0 °  = 600+340 2 =470°  Khan = μan = cpan = ρan = βan =

0,023701738 0,06623352 0,247001643 0,041985394 0,001061793

BTU/h-pie-ºF lb/pie-h BTU/lb-ºF lb/pie3 ºF-1

ℎ ℎ =4.86 [ℎ ] 



 =423318.8

 =0.6902 P ROBLEMAS C APÍTULO 4 ) ROBLEMAS -P ARES ( C





ℎ

ℎ =0.576 [ℎ  ] 

 =2.884 [ℎ  ] 

 = 341. 341.53 53 °° 

 =462.81 ° P ROBLEMAS C APÍTULO 4 ) ROBLEMAS -P ARES ( C




 = 340 °

423318,8021 223363,2128 200742,6137 198308,8832 198049,1267 198021,4275

340 462,811585 476,705029 478,199812 478,359353 478,376366




341,5332434 351,025112 352,0463339 352,1553308 352,1669537 352,168193

462,8115854 476,7050287 478,1998121 478,3593531 478,3763658 478,3781798

4,85767977 5,799682824 5,916711421 5,929433831 5,930793231 5,930938209

0,575507399 0,465148359 0,448899987 0,447080324 0,44688523 0,446864416

2,883815094 3,102409995 3,126925597 3,129554125 3,129834555 3,129864457

 = 2132 213226 261.1.67 67 ⁄ℎ  = 2132 213226 261.1.67 67 ⁄ℎ

 = .  ⁄ P ROBLEMAS C APÍTULO 4 ) ROBLEMAS -P ARES ( C






**

**Un pozo en Tía Juana, completado con un hoyo de 12 pulgadas de diámetro hasta una

profundidad de 3.000 pies, tiene un revestidor de producción de 8,52 pulgadas de diámetro interno y 9,60 pulgadas de diámetro externo. La temperatura promedio en la superficie es de 80 F y el gradiente geotérmico de la zona es de 0,0120 F/pie. El pozo se somete a inyección de vapor a través de una tubería de 3,25 pulgadas de diámetro interno y 3,50 de diámetro externo y recibe 20.000 lb/h de un vapor de 75 % de calidad en el cabezal y 600 F de temperatura. Calcular las pérdidas de calor y la calidad del vapor en el fondo del pozo, al c abo de 20 días de iniciada la inyección.

0.0286×1200 0.29=3.0873  =  2 × √ 0.0286×1200 0.4 P ROBLEMAS C APÍTULO 4 ) ROBLEMAS -P ARES ( C







=0.855   = 340 340 °

 = 239. 239.7 °°   = 340+239.7 2 =289.85 ° Khan = μan = cpan = ρan = βan =

0.019928871 BTU/h-pie-ºF 0.056815309 0.243091729 0.052796972 0.001334791

lb/pie-h BTU/lb-ºF lb/pie3 ºF-1

ℎ ′ ℎ ′ =2.46 ℎ   P ROBLEMAS C APÍTULO 4 ) ROBLEMAS -P ARES ( C









 = 2258 225814. 14.24 24

 =0.693 ℎ ′

ℎ ′ =0.4261 ℎ   

 =0.9975 ℎ   P ROBLEMAS C APÍTULO 4 ) ROBLEMAS -P ARES ( C







 =262.37 °   = 340°

280090.8796 232105.4087 236446.0241 236040.195 236078.0154 236074.4897





340 216.001093 326.052203 2.370999237 326.052203 223.296922 327.313877 2.33405341 327.313877 222.636964 327.195916 2.337341753 327.195916 222.698668 327.206909 2.337033856 327.206909 222.692917 327.205884 2.337062546 327.205884 222.693453 327.20598 2.337059871



0.457883665 0.430108263 0.432770207 0.432522715 0.432545791 0.43254364

0.992120498 0.985824164 0.986411911 0.986357129 0.986362236 0.98636176

 = 997948. 48.03 ⁄ℎ

 = 975050. 50.71 ⁄ℎ P ROBLEMAS C APÍTULO 4 ) ROBLEMAS -P ARES ( C





=. ⁄

Donde: Lv @ 600F = 549.67 BTU/lbm

⁄ℎ 997948.03  =0.75 20000  ×549.67 ⁄ =0.6604  

⟦  =%⟧

0.0286×2400 0.29=3.4338   =  2 × √ 0.0286×2400 0.4 






  = 340 340 °°

 = 257.6 257.626 26 °   = 340+239.7 298.813 13 ° 2 = 298.8 Khan =

0.020123099 0.057302612 0.243270501 0.052117415 0.001317667

μan =

cpan = ρan = βan =


ℎ ′ ℎ ′ =2.566 ℎ   



 =175951.6 P ROBLEMAS C APÍTULO 4 ) ROBLEMAS -P ARES ( C





 =0.6927 ℎ ′

ℎ ′ =0.3823 ℎ   

 =1.0032 ℎ   

 =345.47 ° P ROBLEMAS C APÍTULO 4 ) ROBLEMAS -P ARES ( C




  = 340°

264862.7809 219743.9 223817.8857 223437.8293 223473.1711 223469.8836








340 216.001093 326.131413 2.370999237 0.453749214 326.131413 223.255489 327.38367 2.334259545 0.426391185 327.38367 222.600457 327.266849 2.337523967 0.429007494 327.266849 222.661564 327.277712 2.337218991 0.428764773 327.277712 222.655882 327.276702 2.337247347 0.428787355 327.276702 222.65641 327.276796 2.337244709 0.428785255

0.991724554 0.985476508 0.986058462 0.986004337 0.986009372 0.986008904

=944227 ⁄ℎ

 = 956569. 69.19 ⁄ℎ

=. ⁄ P ROBLEMAS C APÍTULO 4 ) ROBLEMAS -P ARES ( C





Lv @ 600F = 549.67 BTU/lbm

⁄ℎ 956569.19  =0.75 20000  ×549.67 ⁄ =0.6629  

⟦  =.%⟧

0.0286×12000 0.29=4.2385  =  2 × √ 0.0286×12000 0.4 

=0.855   = 340 340 °° P ROBLEMAS C APÍTULO 4 ) ROBLEMAS -P ARES ( C





 =299.25 °   = 340+299.25 319.626 26 ° 2 = 319.6 Khan =

0.020571484 0.058426619 0.243691993 0.050603076 0.001279494

μan =



ℎ ′ ℎ ′ =2.371 ℎ   



 = 7664 76640.0 0.038 38

 =0.6921 P ROBLEMAS C APÍTULO 4 ) ROBLEMAS -P ARES ( C





ℎ ′

ℎ ′ =0.2969 ℎ   

 =1.0142 ℎ   

 =362.38 °   = 340 °











Gr

233221.6515 194004.5856 197531.2731 197203.9197 197234.2115 197231.4077

Tci ºF

Tco ºF


Tins ºF

340 216.001093 326.310076 326.310076 223.162034 327.541174 327.541174 222.518069 327.426902 327.426902 222.577843 327.437476 327.437476 222.572312 327.436497 327.436497 222.572824 327.436588

hr BTU/h-pie2ºF

hc BTU/h-pie2ºF

Uto BTU/h-pie2ºF

2.370999237 2.334724647 2.337935296 2.337636852 2.337664465 2.337661909

0.444459279 0.418028931 0.420544239 0.420312031 0.420333529 0.420331539

0.990831803 0.984692186 0.985261194 0.985208533 0.985213408 0.985212956

 = 8384 838479 79.6.617 17 ⁄ℎ

=895967.9 ⁄ℎ

=. ⁄   =      Lv @ 600F = 549.67 BTU/lbm P ROBLEMAS C APÍTULO 4 ) ROBLEMAS -P ARES ( C





⁄ℎ 895967.9  =0.75 20000  ×549.67 ⁄ =0.6684  

⟦  =.%⟧

0.0286×24000 0.29=4.5851  =  2 × √ 0.0286×24000 0.4 

=0.855   = 340 340 °°

 = 317.1 317.179 79 °






 = 340+317.179 = 328. 328.6 °° 2 Khan = μan =


0.02076368 0.058907999 0.243876474 0.049976878 0.001263703




ℎ ′ ℎ ′ =2.8731 ℎ   



 =40676.1

 =0.6919 ℎ ′

ℎ ′ =0.2431 ℎ   P ROBLEMAS C APÍTULO 4 ) ROBLEMAS -P ARES ( C







 =1.0177 ℎ   

 =369.16 °   = 340°

Gr

221020.728 184057.7705 187376.0748 187068.71 187097.0938 187094.4719

Tci ºF



Tco ºF

Tins ºF



hr BTU/h-pie2ºF



hc BTU/h-pie2ºF

Uto BTU/h-pie2ºF

340 216.001093 326.384659 2.370999237 0.440595644 0.990459251 326.384659 223.123021 327.606959 2.334918869 0.414546967 0.984364699 327.606959 222.483658 327.493741 2.338107146 0.417020895 0.984928352 327.493741 222.542881 327.504196 2.337811405 0.416792976 0.984876294 327.504196 222.537412 327.503231 2.337838712 0.416814034 0.984881102 327.503231 222.537917 327.50332 2.33783619 0.416812089 0.984880658





 = 7912 791217 17.7.789 8922 ⁄ℎ

 = 8716 871694 94.7.711 ⁄ℎ

 = 8656 865657 57.1.150 5011 ⁄ℎ   =     

Donde:

⁄ℎ 871694.71  =0.75 20000  ×549.67 ⁄ =0.6707=67.07%   Lv @ 600F = 549.67 BTU/lbm






0.0286×480 0.29=2.63   =  2 × √ 0.0286×480 0.4 

=0.855  Suponiendo que

 = 340 340 °°

 =59.076 °   = 340+59.076 199.53 53  2 = 199. Khan = μan = cpan = ρan = βan =


0.017933748 0.051795843 0.241382491 0.060639768 0.001532197






ℎ ′ ℎ ′ =1.7566 ℎ   



 =1156132.609

 =0.6971 ℎ ′

ℎ ′ =0.6357 ℎ   






 =0.945 ℎ   

 =188.13 °   = 340°





Gr

Tci ºF

Tco ºF

Tins ºF

510314.7556

340 325.173984 326.541568 326.410554 326.423057 326.421863

216.001093 223.756302 223.040945 223.109476 223.102935 223.10356

325.173984 326.541568 326.410554

417382.2259 425954.5207 425133.2972 425211.6715 425204.189

hr BTU/h-pie2ºF

326.423057 326.421863 326.421977



hc BTU/h-pie2ºF

Uto BTU/h-pie2ºF

2.370999237 0.504382228 0.996516252 2.331770755 0.471723471 0.989675726 2.335327595 0.474927848 0.990329922 2.334986309 0.474622743 0.990267478 2.335018875 0.474651878 0.99027344 2.335015766 0.474649096 0.99027287

 = 6259 625906 06.9.939 3966 ⁄ℎ



.




 = 1 150 5000 006.6.7361 7361 ⁄ℎ

 = 4195 419527 27.7.799 9922 ⁄ℎ

  =     

Donde:

⁄ℎ 625906.9396  =0.75 20000  ×910.1 ⁄ =0.71.56=72%   Lv @ 300F = 910.1 BTU/lbm



=0.855  Suponiendo que

 = 340 340 °°

 =163.69 ° P ROBLEMAS C APÍTULO 4 ) ROBLEMAS -P ARES ( C





 = 340+163.69 2 =251.84 Khan = μan = cpan = ρan = βan =

0.019097748 0.054727325 0.242352051 0.05587009 0.001412188




ℎ ′ ℎ ′ =2.1452 ℎ   



 = 5085 508504. 04.52 52

 =0.6944 ℎ ′

ℎ ′ =0.5155 ℎ   P ROBLEMAS C APÍTULO 4 ) ROBLEMAS -P ARES ( C







 =0.9753 ℎ   

 =281.46 °   = 340°



Tci

Tco

Tins

hr

hc

Uto

ºF

ºF

ºF

BTU/h-pie2ºF

BTU/h-pie2ºF

BTU/h-pie2ºF

290

189.847037

285.055514

2.018720358

0.443704298

0.953417068

267139.4401 285.055514 192.433404 285.502224

2.00708469

0.432303323

0.950713338

269101.7738 285.502224

2.008129538

0.433358212

0.95096125

Gr

288859.8277

192.199738



285.461236



268921.7183 285.461236 192.221178 285.464992 2.008033614 0.433261634 0.950938534 268938.2156 285.464992 192.219214 285.464647 2.008042403 0.433270484 0.950940615 268936.7038 285.464647 192.219394 285.464679 2.008041597 0.433269673 0.950940424 P ROBLEMAS C APÍTULO 4 ) ROBLEMAS -P ARES ( C





 = 9016 901644 44.3.355 ⁄ℎ

 = 6000 600026 26.9.944 ⁄ℎ

 = 8195 819552 52.8.811 ⁄ℎ

Donde:

  =      ⁄ℎ 901644.35  =0.75 20000  ×713.9 ⁄ =0.6868=69%  

Lv @ 500F = 713.9 BTU/lbm








 Suponiendo que

 = 340 340 °°

 = 268. 268.3 °°   = 340+268.3 2 =304.15 ° Khan =

0.020238508 0.057592047 0.24337781 0.051720419 0.001307662

μan =



ℎ ′ ℎ ′ =2.6194 ℎ   



 = 1481 148166. 66.02 02 P ROBLEMAS C APÍTULO 4 ) ROBLEMAS -P ARES ( C





 =0.6925 ℎ ′

ℎ ′ =0.3632 ℎ   

 =1.006 ℎ   

 = 370 370 ° P ROBLEMAS C APÍTULO 4 ) ROBLEMAS -P ARES ( C




Gr

305549.1436 226683.7445 233766.1222 233099.0672 233161.5875 233155.7251


  = 340°



Tci ºF

Tins ºF

Tco ºF



390 242.155149 364.945335 364.945335 255.260772 367.195328 367.195328 254.083843 366.983412 366.983412 254.194692 367.003274 367.003274 254.184302 367.001412 367.001412 254.185277 367.001586



hr BTU/h-pie2ºF

hc BTU/h-pie2ºF

Uto BTU/h-pie2ºF

2.762771233 2.688750436 2.695216807 2.694606257 2.694663468 2.694658103

0.478545476 0.433257614 0.437719084 0.437302758 0.437341812 0.437338151

1.027793839 1.018197402 1.019099083 1.019014553 1.019022479 1.019021736

 = 1194 119411 115.5.84 84 ⁄ℎ

 = 1350 135006 060.0.62 62 ⁄ℎ

 = 1237 123753 531.1.85 85 ⁄ℎ

Donde:

  =      P ROBLEMAS C APÍTULO 4 ) ROBLEMAS -P ARES ( C




⁄ℎ 1350060.62  =0.75 20000  ×172.1 ⁄ =0.3577=36%   Lv @ 600F = 172.1 BTU/lbm

Con la tasa de inyección únicamente cambia xst al final debido a que solo se lo utiliza en la ecuación para xst. Entonces:

  =     

⁄ℎ 1023391.98  =0.75 1000  ×549.67 ⁄ =1.11=111%   Lv @ 600F = 549.67 BTU/lbm

  =     

⁄ℎ 1023391.98  =0.75 2500  ×549.67 ⁄ =0.0052=0.52%   Lv @ 600F = 549.67 BTU/lbm P ROBLEMAS C APÍTULO 4 ) ROBLEMAS -P ARES ( C





  =     

⁄ℎ 1023391.98  =0.75 5000  ×549.67 ⁄ =0.3776=38%   Lv @ 600F = 549.67 BTU/lbm

  =     

⁄ℎ 1023391.98  =0.75 10000  ×549.67 ⁄ =0.5638=56.38%   Lv @ 600F = 549.67 BTU/lbm

Δ

 = 3411 341130 30.6.666 ⁄ℎ   =     

Donde:

⁄ℎ 341130.66  =0.75 20000  ×549.67 ⁄ =0.6879=69%   Lv @ 600F = 549.67 BTU/lbm






Δ

=511696 ⁄ℎ

Donde:

  =     

⁄ℎ 511696  =0.75 20000  ×549.67 ⁄ =0.6569=66%  

Lv @ 600F = 549.67 BTU/lbm

Δ

 = 6822 682261 61.3.322 ⁄ℎ   =     

Donde:

⁄ℎ 682261.32  =0.75 20000  ×549.67 ⁄ =0.6258=63%   Lv @ 600F = 549.67 BTU/lbm

Δ

 = 8528 852826 26.6.655 ⁄ℎ   =      P ROBLEMAS C APÍTULO 4 ) ROBLEMAS -P ARES ( C





Donde:

⁄ℎ 852826.65  =0.75 20000  ×549.67 ⁄ =0.5948=60%   Lv @ 600F = 549.67 BTU/lbm

1000000 980000 R O L A C E D S A D I D R É P

960000 940000 920000 900000 880000 860000 0

200

400

60 0

800

1000

1200

TIEMPO

67,2 67 R 66,8 O P A V66,6 E D D A66,4 D I L A C 66,2

66 65,8 0

200

400

600

800

1000

1200

TÍIEMPO






1600000 1400000 1200000 R O L 1000000 A C E D S 800000 A D I D R 600000 É P

400000 200000 0 0

100

200

300

400

500

600

700

800

TEMPERATURA

80

70

60 R O50 P A V E D40 D A D I L A30 C

20

10

0 0

100

200

300

400

500

600

700

800

TEMPERATURA






1023392

R O L A C E D S A D I D R É P

1023392 0

2000

4000

6000

8000

10000

12000

TASA DE INYECCIÓN

80 60 40 20 R O P A V E D D A D I L A C

0 0

2000

4000

6000

8000

10000

12000

-20 -40 -60 -80

-100 -120

TASA DE INYECCIÓN






900000 800000 700000 R O600000 L A C E 500000 D S A 400000 D I D R É 300000 P

200000 100000 0 0

500

1000

1500

2000

2500

3000

PROFUNDIDAD



244452165-Solucionario-Cap-4-Pares-Recuperacion-Termica-del-Petroleo-Douglas-Alvarado.pdf

Recommend Documents