Uploaded by Alvaro Alonso

Introduccion a la termodinamica - Michael M. Abbott H. C. Van Nes

advertisement
Subido por:
Libros de Ingeniería Química y más
https://www.facebook.com/pages/InterfaseIQ/146073555478947?ref=bookmarks
Si te gusta este libro y tienes la posibilidad,
cómpralo para apoyar al autor.
IntroduccIón a la
termodInámIca en
IngenIería químIca
00-SmithVanNess Preliminares.ind1 1
8/1/07 12:46:17
00-SmithVanNess Preliminares.ind2 2
8/1/07 12:46:17
IntroduccIón a la
termodInámIca en
IngenIería químIca
SÉPtIma edIcIón
J. m. Smith
University of California, Davis
H. c. Van ness
m. m. abbott
Rensselaer Polytechnic Institute
reVISIón tÉcnIca
misael Flores rojas
Profesor de Termodinámica
ESIQIE, Instituto Politécnico Nacional
MÉXICO • AUCKLAND • BOGOTÁ • BUENOS AIRES • CARACAS • GUATEMALA
LISBOA • LONDRES • MADRID • MILÁN • MONTREAL • NUEVA DELHI • NUEVA YORK
SAN FRANCISCO • SAN JUAN • SAN LUIS • SANTIAGO
SÃO PAULO • SIDNEY • SINGAPUR • TORONTO
00-SmithVanNess Preliminares.ind3 3
8/1/07 12:46:17
Director Higher Education: Miguel Ángel Toledo Castellanos
Director editorial: Ricardo A. del Bosque Alayón
Editor sponsor: Pablo E. Roig Vázquez
Editora de desarrollo: Lorena Campa Rojas
Supervisor de producción: Zeferino García García
Traducción: Efrén Alatorre Miguel
Esther Fernández Alvarado
Emilio Sordo Zabay
IntroduccIón a la termodInámIca en IngenIería químIca
Séptima edición
Prohibida la reproducción total o parcial de esta obra,
por cualquier medio, sin la autorización escrita del editor.
DERECHOS RESERVADOS © 2007, respecto a la séptima edición en español por
McGRAW-HILL/INTERAMERICANA EDITORES, S.A. DE C.V.
A Subsidiary of The McGraw-Hill Companies, Inc.
Edificio Punta Santa Fe
Prolongación Paseo de la Reforma 1015, Torre A
Piso 17, Col. Desarrollo Santa Fe,
Delegación Álvaro Obregón
C.P. 01376, México, D. F.
Miembro de la Cámara Nacional de la Industria Editorial Mexicana, Reg. Núm. 736
ISBN-13: 978-970-10-6147-3
ISBN-10: 970-10-6147-0
(ISBN edición anterior: 970-10-3647-6)
Traducido de la séptima edición en inglés de la obra: INTRODUCTION TO CHEMICAL ENGINEERING
THERMODYNAMICS, by J.M. Smith, H.C. Van Ness, M.M. Abbott. Copyright © 2005, 2001, 1996, 1987,
1975, 1959, 1949 by The McGraw-Hill Companies, Inc. All rights reserved.
ISBN 10: 0-07-310445-0
ISBN 13: 978-0-07-310445-4
1234567890
09865432107
Impreso en México
Printed in Mexico
00-SmithVanNess Preliminares.ind4 4
8/1/07 12:46:19
Contenido
lista de símbolos
Ix
Prefacio
xV
1
IntroduccIón
1.1 El panorama de la termodinámica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Dimensiones y unidades . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3 Medidas de cantidad o tamaño . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.4 Fuerza . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.5 Temperatura . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.6 Presión . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.7 Trabajo . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.8 Energía . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.9 Calor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1
2
2
3
5
6
8
10
15
2
la PrImera ley y otroS concePtoS báSIcoS
2.1 Experimentos de Joule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Energía interna . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3 La primera ley de la termodinámica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4 Balance de energía para sistemas cerrados . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.5 Estado termodinámico y funciones de estado . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6 Equilibrio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.7 Regla de las fases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.8 El proceso reersible . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.9 Procesos con V y P constantes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.10 Entalpía . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.11 Capacidad calorífica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.12 Balances de masa y energía para sistemas abiertos . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
21
21
21
22
23
26
29
29
31
37
38
40
44
3
ProPIedadeS VolumÉtrIcaS de FluIdoS PuroS
3.1 Comportamiento PVT de sustancias puras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 Ecuaciones de estado iriales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3 El gas ideal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
64
64
70
73
00-SmithVanNess Preliminares.ind5 5
8/1/07 12:46:19
i
3.4
3.5
3.6
3.7
Aplicación de las ecuaciones iriales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Ecuaciones cúbicas de estado . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Correlaciones generalizadas para gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Correlaciones generalizadas para líquidos . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
87
90
99
109
4
eFectoS tÉrmIcoS
4.1 Efectos del calor sensible . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2 Calores latentes de sustancias puras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.3 Calor estándar de reacción . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4 Calor estándar de formación . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.5 Calor estándar de combustión . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.6 Dependencia con la temperatura de ΔH ° . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.7 Efectos térmicos de las reacciones industriales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
125
125
133
135
136
139
140
143
5
la Segunda ley de la termodInámIca
5.1 Presentación de la segunda ley . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2 Máquinas térmicas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.3 Escalas de temperatura termodinámica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.4 Entropía . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.5 Cambios de entropía de un gas ideal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.6 Planteamiento matemático de la segunda ley . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.7 Balance de entropía para sistemas abiertos . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.8 Cálculo del trabajo ideal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.9 Trabajo perdido . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.10 La tercera ley de la termodinámica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.11 Entropía desde el punto de ista microscópico . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
159
160
161
163
167
170
173
176
181
185
188
188
6
ProPIedadeS termodInámIcaS de loS FluIdoS
6.1 Expresiones para la ealuación de una propiedad en fases homogéneas . . . . . . . . . . . . . . . .
6.2 Propiedades residuales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.3 Las propiedades residuales a partir de ecuaciones de estado . . . . . . . . . . . . . . . . . . . . . . . . .
6.4 Sistemas de dos fases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.5 Diagramas termodinámicos . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.6 Tablas de propiedades termodinámicas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.7 Correlaciones generalizadas para la ealuación de una propiedad para gases . . . . . . . . . . .
199
199
208
215
220
225
226
230
7
aPlIcacIoneS de la termodInámIca a loS ProceSoS de FluJo
7.1 Flujo en conductos de fluidos compresibles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.2 Turbinas (expansores) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.3 Procesos de compresión . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
254
255
268
273
8
generacIón de PotencIa a PartIr del calor
8.1 Planta de energía de apor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.2 Motores de combustión interna . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.3 Motores de reacción: motor de propulsión . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
290
291
302
310
9
reFrIgeracIón y lIcueFaccIón
9.1 El refrigerador de Carnot . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
317
317
00-SmithVanNess Preliminares.ind6 6
8/1/07 12:46:19
ii
9.2
9.3
9.4
9.5
9.6
Ciclo de compresión de apor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
La elección del refrigerante . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Refrigeración por absorción . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
La bomba de calor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Proceso de licuefacción . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
318
321
323
326
327
10 equIlIbrIo VaPor/líquIdo: IntroduccIón
10.1 La naturaleza del equilibrio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.2 Regla de fase . Teorema de Duhem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.3 EVL: Comportamiento cualitatio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.4 Modelos simples para el equilibrio apor/líquido . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.5 EVL mediante la ley de Raoult modificada . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.6 EVL a partir de las correlaciones del alor K . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
338
338
339
341
347
358
363
11 termodInámIca de SolucIoneS: teoría
11.1 Relación de una propiedad fundamental . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.2 Potencial químico y equilibrio de fase . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.3 Propiedades parciales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.4 Modelo de mezcla de gas ideal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.5 Fugacidad y coeficiente de fugacidad: especies puras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.6 Fugacidad y coeficiente de fugacidad: especies en solución . . . . . . . . . . . . . . . . . . . . . . . . .
11.7 Correlaciones generalizadas para el coeficiente de fugacidad . . . . . . . . . . . . . . . . . . . . . . . .
11.8 El modelo de solución ideal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.9 Propiedades de exceso . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
378
378
380
381
391
394
401
407
411
413
12 termodInámIca de SolucIoneS: aPlIcacIoneS
12.1 Propiedades de fase líquida a partir de la información de EVL . . . . . . . . . . . . . . . . . . . . . . .
12.2 Modelos para la energía de Gibbs de exceso . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12.3 Cambios en la propiedad de mezclado . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12.4 Efectos térmicos de los procesos de mezclado . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
430
430
446
449
456
13 equIlIbrIo en reaccIón químIca
13.1 Coordenada de la reacción . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13.2 Aplicación de criterios de equilibrio a las reacciones químicas . . . . . . . . . . . . . . . . . . . . . .
13.3 Cambio en la energía de Gibbs estándar y la constante de equilibrio . . . . . . . . . . . . . . . . . .
13.4 Efecto de la temperatura en la constante de equilibrio . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13.5 Ealuación de las constantes de equilibrio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13.6 Relación de las constantes de equilibrio con la composición . . . . . . . . . . . . . . . . . . . . . . . .
13.7 Conersiones de equilibrio para reacciones únicas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13.8 Regla de la fase y teorema de Duhem para sistemas con transformaciones químicas . . . . .
13.9 Equilibrios en reacciones múltiples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13.10 Celdas de combustible . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
483
484
488
489
492
496
498
502
514
518
529
14 temaS Sobre loS equIlIbrIoS de FaSe
14.1 Formulación gamma/phi del EVL . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14.2 EVL a partir de ecuaciones de estado cúbicas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14.3 Equilibrio y estabilidad . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
545
545
556
575
00-SmithVanNess Preliminares.ind7 7
8/1/07 12:46:20
viii
14.4
14.5
14.6
14.7
14.8
14.9
Equilibrio líquido/líquido (ELL) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Equilibrio vapor/líquido/líquido (EVLL) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Equilibrio sólido/líquido (ESL) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Equilibrio sólido/vapor (ESV) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Equilibrio de adsorción de gases en sólidos . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Equilibrio osmótico y presión osmótica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
581
590
597
602
606
621
15 Análisis termodinámico de procesos
15.1 Análisis termodinámico de procesos de flujo en estado estacionario . . . . . . . . . . . . . . . . . .
635
635
16 introducción A lA termodinámicA moleculAr
16.1 Teoría molecular de los fluidos . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
16.2 Segundos coeficientes viriales a partir de las funciones potenciales. . . . . . . . . . . . . . . . . . .
16.3 Energía interna de gases ideales: punto de vista microscópico . . . . . . . . . . . . . . . . . . . . . . .
16.4 Propiedades termodinámicas y mecánica estadística. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
16.5 Enlace de hidrógeno y complejo de transferencia de carga . . . . . . . . . . . . . . . . . . . . . . . . . .
16.6 Comportamiento de las propiedades de exceso . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
16.7 Bases moleculares para el comportamiento de mezclas . . . . . . . . . . . . . . . . . . . . . . . . . . . .
16.8 EVL por simulación molecular . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
647
647
654
657
660
662
665
669
672
A
Factores de conversión y valores de la constante de los gases
677
B
propiedades de especies puras
679
c
capacidades caloríficas y cambios de propiedad de formación
683
d
programas representativos para computadora
d.1 Funciones definidas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
d.2 Solución de problemas de ejemplo con Mathcad® . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
688
688
691
e
tablas de correlación generalizada de lee/Kesler
695
F
tablas de vapor
F.1 Interpolación . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
712
712
G
diagramas termodinámicos
788
H
método uniFAc
791
i
método de newton
798
Índice de autores
803
Índice analítico
807
00-SmithVanNess Preliminares.ind8 8
10/1/07 14:02:01
Lista de símbolos
A
A
A
a
a
a
a-i
B
B
B̂
B′
B0, B1
Bij
b
bi
C
C
Ĉ
C′
C 0, C1
CP
CV
CP°
ΔCP°
CPH
CPS
CP°H
Área
Energía de Helmholtz molar o específica ≡ U − TS
Parámetro, ecuaciones Empíricas, por ejemplo, en las ecuaciones (4 .4), (6 .76) y (12 .14)
Aceleración
Área molar de una fase adsorbida
Parámetro en las ecuaciones de estado cúbicas
Parámetros parciales en ecuaciones de estado cúbicas
Segundo coeficiente irial, expansión en densidades
Parámetro en ecuaciones empíricas, por ejemplo, en las ecuaciones (4 .4), (6 .76) y (12 .14)
Segundo coeficiente irial reducido, definido por la ecuación (3 .62)
Segundo coeficiente irial, expansión en presiones
Funciones de la correlación generalizada del segundo coeficiente irial
Interacción del segundo coeficiente irial
Parámetro en ecuaciones de estado cúbicas
Parámetro pracial en ecuaciones de estado cúbicas
Tercer coeficiente irial, expansión en densidades
Parámetro, en ecuaciones empíricas, por ejemplo, ecuaciones (4 .4), (6 .76) y (12 .14)
Tercer coeficiente irial reducido, definido en la página 103 .
Tercer coeficiente irial, expansión en presiones
Funciones, correlación generalizada del tercer coeficiente irial
Capacidad calorífica molar o específica, a presión constante
Capacidad calorífica molar o específica, a olumen constante
Capacidad calorífica en el estado estándar, a presión constante
Cambio en la capacidad calorífica estándar de reacción
Capacidad calorífica media para cálculos de entalpía
Capacidad calorífica media para cálculos de entropía
Capacidad calorífica estándar media para cálculos de entalpía
ix
00-SmithVanNess Preliminares.ind9 9
8/1/07 12:46:21
x
CP°S
c
D
D
D′
Ei
EK
EP
F
F
F
fi
f i°
fˆi
G
Gi°
–
Gi
GE
GR
ΔG
ΔG°
ΔGf°
g
gc
gi
H
Hi
Hi°
–
Hi
HE
HR
(HR)0, (HR)1
ΔH
∼
Δ
H
ΔH °
ΔH 0°
ΔH °f
h
I
Capacidad calorífica estándar media para cálculos de entropía
Rapidez del sonido
Cuarto coeficiente irial, expansión en densidades
Parámetro en ecuaciones empíricas, por ejemplo, en las ecuaciones (4 .4) y (6 .77)
Cuarto coeficiente irial, expansión en presiones
Niel de energía
Energía cinética
Energía potencial graitacional
Grados de libertad, regla de las fases
Fuerza
Constante de Faraday
Fugacidad, especies puras i
Fugacidad en el estado estándar
Fugacidad de la especie i en solución
Energía de Gibbs molar o específica ≡ H – TS
Energía de Gibbs en el estado estándar de la especie i
Energía de Gibbs parcial de la especie i en solución
Energía de Gibbs de exceso ≡ G – G i d
Energía de Gibbs residual ≡ G – G i g
Cambio en la energía de Gibbs por mezclado
Cambio en la energía de Gibbs estándar de reacción
Cambio en la energía de Gibbs estándar de formación
Aceleración local de la graedad
Constante dimensional = 32 .1740(lbm)(ft)(lbr)–1(s)–2
Degeneración
Entalpía molar o específica ≡ U + PV
Constante de Henry de la especie i en solución
Entalpía en el estado estándar de la especie pura i
Entalpía parcial de la especie i en solución
Entalpía de exceso ≡ H – H i d
Entalpía residual ≡ H – H i g
Funciones en la correlación de la entalpía residual generalizada
Cambio de entalpía (“calor”) por mezclado; también calor latente de transición de fase
Calor de solución
Cambio en el entalpía estándar de reacción
Calor estándar de reacción a la temperatura de referencia T0
Cambio en la entalpía estándar de formación
Constante de Planck
Representa una integral definida, por ejemplo en las ecuaciones (6 .65)
00-SmithVanNess Preliminares.ind10 10
8/1/07 12:46:22
xi
I
Kj
Ki
k
L
l
li j
m
M
M
–
Mi
ME
MR
ΔM
ΔM °
ΔM f°
m
ṁ
N
NA
n
ṅ
ñ
ni
P
P°
Pc
Pr
P 0r , P 1r
P0
pi
Pisat
Q
Q·
q
q
q
q-i
R
00-SmithVanNess Preliminares.ind11 11
Primer potencial de ionización
Constante de equilibrio para la reacción química j
Constante de equilibrio apor/líquido para la especie i ≡ yi / xi
Constante de Boltzmann
Fracción molar líquida del sistema
Longitud
Parámetro de interacción de una ecuación del estado, ecuación (14 .101)
Número de Mach
Masa molar (peso molecular)
Valor moral o específico, de una propiedad termodinámica extensia
Propiedad parcial de la especie i en solución
Propiedad de exceso ≡ M – M i d
Propiedad residual ≡ M – M i g
Cambio en una propiedad por mezclado
Cambio en una propiedad estándar de reacción
Cambio en una propiedad estándar de formación
Masa
Rapidez de flujo de masa
Número de especies químicas, regla de las fases
Número de Aogadro
Número de moles
Rapidez de flujo de moles
Moles de solente por mol de soluto
Número de moles de la especie i
Presión absoluta
Presión en el estado estándar
Presión crítica
Presión reducida
Funciones, correlación generalizada apor-presión
Presión de referencia
Presión parcial de la especie i
Presión de apor de saturación de la especie i
Calor
Rapidez de transferencia de calor
Rapidez de flujo olumétrico
Parámetro en las ecuaciones de estado cúbicas
Carga eléctrica
Parámetro parcial en las ecuaciones de estado cúbicas
Constante uniersal de los gases (tabla A .2)
8/1/07 12:46:22
xii
r
r
r
S
–
Si
SE
SR
(SR)0, (SR)1
SG
S·G
ΔS
ΔS °
ΔS °f
T
Tc
Tn
Tr
T0
Tσ
Tisat
t
t
U
U
u
V
V
–
Vi
Vc
Vr
VE
VR
ΔV
W
·
W
Wideal
·
Wideal
Wperdido
·
Wperdido
Relación de compresión
Separación intermolecular
Número de reacciones químicas independientes, regla de fases
Entropía molar o específica
Entropía parcial de la especie i en solución
Entropía de exceso ≡ S – S i d
Entropía residual ≡ S – S i g
Funciones en la correlación de entropía residual generalizada
Generación de entropía por cantidad unitaria de fluido
Rapidez de generación de entropía
Cambio de entropía por mezclado
Cambio de entropía estándar de la reacción
Cambio de entropía estándar de la formación
Temperatura absoluta en kelin o rankine
Temperatura crítica
Temperatura normal de ebullición
Temperatura reducida
Temperatura de referencia
Temperatura absoluta de los alrededores
Temperatura de saturación de la especie i
Temperatura, °C o (°F)
Tiempo
Energía interna molar o específica
Función del par potencial intermolecular
Velocidad
Volumen molar o específico
Fracción molar del sistema que es apor
Volumen parcial de la especie i en solución
Volumen crítico
Volumen reducido
Volumen de exceso ≡ V – V i d
Volumen residual ≡ V – V i g
Cambio de olumen por mezclado; también, cambio de olumen de la transición de fase
Trabajo
Rapidez de trabajo (potencia)
Trabajo ideal
Rapidez de trabajo ideal
Trabajo perdido
Rapidez de trabajo perdido
00-SmithVanNess Preliminares.ind12 12
8/1/07 12:46:23
xiii
Ws
·
Ws
xi
xv
yi
Z
Zc
Z 0, Z 1
Z
z
z
zi
Trabajo de flecha para procesos de flujo
Potencia de flecha para procesos de flujo
Fracción molar de la especie i en general o en una fase líquida
Calidad
Fracción molar de la especie i en la fase apor
Factor de compresibilidad ≡ PV / RT
Factor de compresibilidad crítica ≡ PcVc / RTc
Funciones en la correlación generalizada de factores de compresibilidad
Funcion de partición
Factor de compresibilidad de fase adsorbida definida por la ecuación (14 .108)
Eleación por encima de un niel de referencia
Fracción molar global o fracción molar en una fase sólida
Superíndices
E
av
R
s
sl
t
v
∞
Denota una propiedad termodinámica de exceso
Denota una transición de fase de la fase adsorbida a apor
Denota una propiedad termodinámica residual
Denota fase sólida
Denota transición de fase de sólido a líquido
Denota un alor total de una propiedad termodinámica extensia
Denota fase de apor
Denota un alor a dilución infinita
Letras Griegas
α
α
α, β
αβ
β
β
l
γ
γl
δ

00-SmithVanNess Preliminares.ind13 13
Función en las ecuaciones del estado cúbicas (tabla 3 .1, p .98)
Polarizabilidad
Como superíndices, identifican fases
Como superíndices, denota una transición de fase de la fase α a la fase β
Expansiidad del olumen
Parámetro en las ecuaciones de estado cúbicas
Constante de integración
Relación de las capacidades caloríficas CP / CV
Coeficiente de actiidad de la especie i en solución
Exponente politrópico
Constante en las ecuaciones de estado cúbicas
8/1/07 12:46:23
xi

0
ε
η
κ


π
µ
µ
µi
νi
ρ
ρc
ρr
σ
σ
τ
i
φi
φ̂i
φ0, φ1
, 
ω
Profundidad del pozo en la función de potencial intermolecular
Permitiidad eléctrica en el acío
Coordenada de reacción
Eficiencia
Comprensibilidad isotérmica
Presión de dispersión en fase absorida
Presión osmótica
Número de fases, regla de las fases
Coeficiente de Joule/Thompson
Momento dipolar
Potencial químico de la especie i
Número estequiométrico de la especia i
Densidad molar o específica ≡ 1/V
Densidad crítica
Densidad reducida
Constante en las ecuaciones de estado cúbicas
Diámetro de colisión molecular
Relación de temperatura ≡ T / T0 [En la ecuación (6 .77), τ ≡ 1 – Tr ]
Relación de coeficientes de fugacidad, definida por la ecuación (14 .2)
Coeficiente de fugacidad de la especie pura i
Coeficiente de fugacidad de la especie i en solución
Funciones en la correlación generalizada de coeficientes de fugacidad
constantes en las ecuaciones de estado cúbicas
Factor acéntrico
Notas
c
fs
°
–
·
ˆ
Δ
Como subíndice, denota un olumen de control
Como subíndice, denota corriente de flujo
Como superíndice, denota el estado estándar
La sobrebarra denota una propiedad parcial
El sobrepunto denota una rapidez
El circunflejo denota una propiedad en solucion
Operador de diferencia
00-SmithVanNess Preliminares.ind14 14
8/1/07 12:46:24
Prefacio
La termodinámica, uno de los temas centrales de la ciencia, está basada en leyes de aplicación universal. La
justificación para presentar el tema desde el punto de vista de la Ingeniería Química es que tenemos la convicción de que es más efectivo pensar en el contexto de la disciplina que interesa y compromete al estudiante.
Aunque es de naturaleza introductoria, el material de este texto no se considera tan simple. En realidad,
no hay manera de hacerlo sencillo, ya que un estudiante que se inicia en el tema encontrará que tiene por
delante una tarea demandante de descubrimiento. Se presentan conceptos, términos y símbolos nuevos con
una frecuencia y rapidez desconcertantes, y en este punto resulta que la memoria tiene un papel significativo.
Un reto aún más difícil es la necesidad de desarrollar la capacidad de razonamiento y aplicar los principios de
la termodinámica en la solución de problemas prácticos. Por ello, mientras se mantiene el rigor característico
del análisis termodinámico integral, hemos hecho esfuerzos para evitar la complejidad matemática innecesaria. Además, alentamos la comprensión al escribir frases simples y directas en voz activa y tiempo presente.
Existen los elementos para ofrecer la motivación requerida, pero nuestro objetivo, como ha sido en todas las
ediciones anteriores, también es proporcionar un método que pueda ser comprendido por cualquier estudiante dispuesto a ejercitarse con el esmero adecuado.
Los primeros dos capítulos del libro presentan las definiciones básicas y un desarrollo de la primera ley.
Los capítulos 3 y 4 tratan el comportamiento a través de la presión/volumen/temperatura de los fluidos y
ciertos efectos térmicos, lo cual permite la aplicación inicial de la primera ley a problemas reales. La segunda
ley y algunas de sus aplicaciones se consideran en el capítulo 5. Un tratamiento de las propiedades termodinámicas de los fluidos puros en el capítulo 6 permite la aplicación general de la primera y la segunda leyes, y
proporciona un amplio tratamiento de los procesos con flujo en el capítulo 7. Los capítulos 8 y 9 abordan
los procesos de producción de energía y de refrigeración. En el resto del libro, se hace referencia a mezclas
fluidas y se tratan temas que corresponden exclusivamente a la termodinámica de la ingeniería química. Los
capítulos 11 y 12 proporcionan una exposición detallada de la teoría y aplicación de la termodinámica de soluciones. El equilibrio de las reacciones químicas se cubre con detalle en el capítulo 13. El capítulo 14 se
ocupa de temas como el equilibrio de fase, en donde se incluye un tratamiento extenso del equilibrio vapor/
líquido, así como del equilibrio osmótico y de adsorción. El capítulo 15 considera el análisis termodinámico
de procesos reales, proporcionando una revisión de gran parte de los temas prácticos de interés para la termodinámica.
xv
00-SmithVanNess Preliminares.ind15 15
10/1/07 14:03:43
xvi
El material de estos capítulos es adecuado para un curso de un año a nivel licenciatura; no obstante, es
útil en otro tipo de cursos siempre y cuando se utilice con discreción y de manera condicionada por el contenido de estos. Los primeros 13 capítulos incluyen material que es necesario en la formación del ingeniero
químico. Por lo tanto, cuando se proporcione un curso de termodinámica de Ingeniería Química con duración
de sólo un semestre, estos 13 capítulos representan el material suficiente.
Las leyes y los principios de la termodinámica clásica no dependen de algún modelo en particular de la
estructura de la materia; están libres de cualquier consideración a nivel molecular. Sin embargo, el comportamiento exhibido por la materia (gases, líquidos y sólidos) depende de su naturaleza en particular, por lo que
en el capítulo 16 presentamos una introducción a la termodinámica molecular, a la cual se hace referencia de
manera ocasional en capítulos anteriores.
Este libro fue realizado de manera detallada a fin de convertirlo en una referencia útil para los cursos a
nivel de posgrado y la práctica profesional. No obstante, debido a consideraciones de extensión, fue necesario
hacer una selección prudente del material. De este modo, no incluimos ciertos temas dignos de atención, pero
de naturaleza especializada. Éstos incluyen aplicaciones a polímeros, electrolitos y biomateriales.
Estamos en deuda con muchas personas, estudiantes, profesores, revisores, quienes han contribuido de
diversas maneras a la calidad de esta séptima edición, tanto en forma directa como indirecta, a través de preguntas y comentarios, elogios y críticas a lo largo de 55 años y seis ediciones de evolución. A todos ellos
hacemos extensivo nuestro agradecimiento.
J. M. Smith
H. C. Van Ness
M. M. Abbott
Agradecemos en especial la valiosa contribución de los siguientes asesores técnicos para la presente
edición en español:
Alejandro J. Guzmán Gómez, Universidad Nacional Autónoma de México, FES Zaragoza
Margarita Hernández Alvarado, Universidad Tecnológica de México, campus Sur
Ma. del Carmen Doria Serrano, Universidad Iberoamericana, Ciudad de México
Luis Neri Vitela, ITESM, campus Estado de México
Rodolfo Gámez Aguilar, Instituto Tecnológico de Los Mochis
Hidelberto Hernández Frías, Instituto Tecnológico de Los Mochis
Fortunato Ramos Valenzuela, Instituto Tecnológico de Los Mochis
Pedro Rochín Angulo, Instituto Tecnológico de Culiacán
Guillermo Aguirre, Instituto Tecnológico de Mazatlán
00-SmithVanNess Preliminares.ind16 16
10/1/07 14:04:50
Capítulo 1
Introducción
1.1
EL PANORAMA DE LA TERMODINÁMICA
La ciencia de la termodinámica nació en el siglo xix, cuando surgió la necesidad de describir el funcionamiento
de las máquinas de vapor y establecer los límites de lo que éstas podían realizar. Así, como su nombre lo indica, la
termodinámica es la potencia desarrollada por el calor, con aplicaciones obvias a las máquinas térmicas, de las
cuales la máquina de vapor representa el primer ejemplo. De cualquier modo, los principios observados, válidos
para las máquinas, se han generalizado sin dificultad y ahora se conocen como la primera y la segunda leyes de
la termodinámica. Estas leyes no tienen demostración en sentido matemático; su validez estriba en la ausencia
de experiencias contradictorias. Así, la termodinámica comparte con la mecánica y el electromagnetismo la base
sobre la que se sustentan las leyes básicas, que no se deducen de algo más.
Estas leyes conducen, mediante la deducción matemática, a un sistema de ecuaciones que encuentran
aplicación en todas las ramas de la ciencia y de la ingeniería. El ingeniero químico hace frente de manera
particular a una amplia variedad de problemas; por ejemplo, el cálculo de los requerimientos de calor y de
trabajo para los procesos físicos y químicos, así como la determinación de las condiciones de equilibrio para
las reacciones químicas y para la transferencia de especies químicas entre las fases.
Las consideraciones termodinámicas no establecen la rapidez de los procesos químicos o físicos. La
rapidez depende de las fuerzas impulsoras y de la resistencia; las fuerzas impulsoras son variables termodiná­
micas, pero las resistencias no. Ninguna formulación termodinámica, es decir, de características macroscópi­
cas, revela los mecanismos microscópicos (moleculares) de los procesos físicos o químicos. Por otra parte, el
conocimiento del comportamiento microscópico de la materia puede ser útil en el cálculo de las propiedades
termodinámicas. Los valores de las propiedades son esenciales para el uso práctico de la termodinámica. El
ingeniero químico se ocupa de muchas especies químicas y a menudo no cuenta con datos experimentales,
lo cual ha conducido al desarrollo de las “correlaciones generalizadas” que proporcionan estimaciones de la
propiedad cuando se carece de información.
La aplicación de la termodinámica a cualquier problema real comienza con la identificación de un agre­
gado particular de materia como punto central de atención. Este agregado de materia se llama sistema y su
estado termodinámico está definido por algunas propiedades macroscópicas mensurables. Éstas dependen
de las dimensiones fundamentales de la ciencia, de las cuales la longitud, el tiempo, la masa, la temperatura y
la cantidad de sustancia son las de mayor interés.
Se presenta un tratamiento elemental en el capítulo 6.
01­SmithVanNess.indd 1
8/1/07 12:47:11
1.2
CAPÍTULO 1. Introducción
DIMENSIONES Y UNIDADES
Las dimensiones fundamentales son primordiales, ya que son reconocidas por nuestras percepciones sen­
soriales y no son definibles en términos de algo más simple. Sin embargo, su empleo requiere la definición
de escalas arbitrarias de medición, divididas en unidades de tamaño específico. Las unidades originales fue­
ron establecidas mediante un acuerdo internacional, y codificadas como el Sistema Internacional de Unidades
(abreviado como SI, para el Système International).
El segundo, la unidad del tiempo del SI, cuyo símbolo es s, corresponde a la duración de 9 9 63 770
ciclos de radiación asociada a una transición específica del átomo del cesio. El metro, cuyo símbolo es m, es
la unidad fundamental de longitud que se define como la distancia que la luz viaja en el vacío durante
/99 79 458 de un segundo. El kilogramo, con el símbolo kg, es la masa de un cilindro de platino/iridio que
se conserva en la Oficina Internacional de Pesas y Medidas en Sèvres, Francia. La unidad de la temperatura
es el kelvin, cuyo símbolo es K, y es igual a /73.6 de la temperatura termodinámica del punto triple del
agua. En la sección .5 se hace un análisis detallado de la temperatura, que es la dimensión característica de
la termodinámica. El mol, cuyo símbolo es mol, es la cantidad de sustancia representada por tantas entida­
des elementales (por ejemplo, moléculas) como el número de átomos que hay en 0.0 kg de carbono­.
Esto es equivalente al “gramo mol” usado de manera común por los químicos.
Los múltiplos y fracciones decimales de las unidades del SI se designan mediante prefijos. Los de
uso más frecuente se enumeran en la tabla .. De esta manera, el centímetro está determinado como cm =
0 – m, y el kilogramo como kg = 0 3 g.
Tabla 1.1: Prefijos para unidades del SI
Múltiplo
Prefijo
0−5
femto
0−
0−9
0−6
0−3
0−
pico
nano
micro
mili
centi
Símbolo Múltiplo
Prefijo
Símbolo
f
0
hecto
h
p
n
µ
m
c
03
kilo
mega
giga
tera
peta
k
M
G
T
P
06
09
0
05
Otros sistemas de unidades, como el sistema inglés de ingeniería, utilizan unidades que están relaciona­
das con las unidades del SI mediante factores fijos de conversión. Así, el pie se define como 0.3048 m, la libra
masa (lbm) como 0.4535937 kg y la libra mol (lb mol) como 453.5937 mol.
1.3
MEDIDAS DE CANTIDAD O TAMAÑO
Tres medidas de cantidad o tamaño son de uso común:
• Masa, m
• Número de moles, n
• Volumen total, V t
Para un sistema específico, estas medidas están en proporción directa una de otra. La masa, una medida primordial sin definición, se puede dividir entre la masa molar M, comúnmente llamada peso molecular, para
01­SmithVanNess.indd 2
8/1/07 12:47:11
1.4.
1.4.Force
Force
1.4.
Force
1.4.
Force
1.4.
Force
1.4.1.4.
Force
Force
1.4.
1.4.
Force
1.4.Force
Force
1.4. Fuerza
weight,
to
yield
weight, to
to yield
yieldnumber
numberofof
ofmoles:
moles:
weight,
number
moles:
333
333 333 3
3
to yield
number
of
mmmoles:
weight,
totoyield
yield
number
ofofmoles:
moles:
weight,
toweight,
yield
number
of
moles:
weight,
to
number
of
to
number
of
weight,
yield
number
moles:
obtener weight,
elweight,
número
de
moles:
toyield
yield
number
ofmoles:
moles:
nnn==
oror
mm
= m
or
m==
=Mn
Mn
Mn
M
M
m
M
m
m
mm mm
= Mn
=n=
oror or m
mMn
==m
Mn
=nnn=
or oor
m=
=m
Mn
=
Mn
nnn=
or
Mn
=M
=m
Mn
Mofof
Total
the
size
isis
quantity
MM
M=
M
Totalvolume,
volume,representing
representing
the
size
ofaor
system,
isaam
adefined
defined
quantitygiven
givenasas
asthe
theproduct
product
Total
volume,
representing
size
aasystem,
system,
defined
quantity
given
the
product
Mthe
ofof
lengths.
It
may
be
divided
by
the
mass
or
number
of
moles
of
the
system
to
yield
ofthree
three
lengths.
It
may
be
divided
by
the
mass
or
number
of
moles
of
the
system
to
yield
three
lengths.
It may
be
divided
by
the
mass
of
moles
of
the
system
to
yield
Total
volume,
representing
the
size
of aor
is
aquantity
defined
quantity
given
as
the
product
Total
volume,
representing
the
size
of
system,
isis
defined
quantity
given
asasproduct
the
product
Total
volume,
representing
the
size
of
system,
issystem,
aesdefined
defined
quantity
given
as
the
product
Total
volume,
representing
the
size
of
aaasystem,
is
aaadefined
quantity
given
as
the
product
Total
volume,
representing
size
of
aaun
system,
is
anumber
given
as
the
El volumen
total,
que
representa
el the
tamaño
de
sistema,
una
cantidad
definida
como
elproduct
producto
de
specific
or
molar
volume:
Total
volume,
representing
the
size
of
system,
defined
quantity
given
the
product
Total
volume,
representing
the
size
of
a
system,
is
a
defined
quantity
given
as
the
specific
or
molar
volume:
specific
or
molar
volume:
of
three
lengths.
It
may
be
divided
by
the
mass
or
number
of
moles
of
the
system
to
yield
ofofthree
three
lengths.
It
may
be
divided
by
the
mass
or
number
of
moles
of
the
system
to
yield
of three
three
lengths.
It
may
be
divided
by
the
mass
or
number
of
moles
of
the
system
to
yield
of
lengths.
It
may
be
divided
by
the
mass
or
number
of
moles
of
the
system
to
yield
of
lengths.
It
may
be
divided
by
the
mass
or
number
of
moles
of
the
system
to
yield
tres longitudes.
Se
puede
dividir
entre
la
masa
o
el
número
de
moles
del
sistema
para
obtener
el
volumen
lengths.
It may
be divided
by the
or number
of moles
of the
system
to yield
of threethree
lengths.
It may
be divided
by tthe
massmass
or number
of moles
of the
system
to yield
specific
or
molar
volume:
specific
orormolar
molar
volume:
specific
or
molar
volume:
VV
specific
or
volume:
or
molar
volume:
V tt
molar ospecific
específico:
tt
specific
molar
volume:
specific
or
molar
volume:
=
mV
or
V
•••Specific
volume:
V
≡
t
= mV
mV
or
V =
Specific volume:
volume:
V≡
≡
V
Specific
V
mmV tt V tor
VVVtt m
t VV t
t
t
t
t
t
= mV
VmV
• volume:
Specific
volume: ≡VV ≡
==
mV
or Vor =
Specific
volume:
≡ V ≡ or
mV
or o or
Specific
volume:
• Volumen
específico:
t=
VV=
•••Specific
volume:
•••Specific
mV
VmV
Specific
volume: VVV≡
mV
or or VV t=
Specific
volume:
≡Vm
m
m
mV≡t m
t
mV
Vt m
tt
•••Molar
VV
oror
VV
Molarvolume:
volume:
V≡≡
≡
=nV
nV
or
V t==
Molar
volume:
nV
t Vt
tnn
n
t
t
V
V
VV t VV t
t
t
t
t
t
•volume:
Molar
volume: VV ≡
= nV
VnV
Molar
volume:
≡≡V ≡ or
==
nV
or VVort=
Molar
volume:
≡VVV≡
=
nV
or oor
• Volumen
molar:
t=
•••Molar
volume:
VV=
•••Molar
−1
−1
Molar
volume:
nV
VnV
Molar
volume:
V ≡
nV
Vofspecific
nor or of
. ..
Specific
or
molar
density
is
defined
as
the
reciprocal
orormolar
n
n
n
n
Specific
or
molar
density
is
defined
as
the
reciprocal
specific
molarvolume:
volume:ρρρ≡≡
≡VV
V −1
Specific or molar density is defined as
molar
volume:
n then reciprocal of specific or
These
quantities
(V
and
ρ)
are
independent
of
the
size
of
a
system,
and
are
examples
These
quantities
(V
and
ρ)
are
independent
of
the
size
of
a
system,
and
are
examples
−1
−1ρ
−1
−1
These
quantities
(V
and
ρ)
are
independent
the
size
ofmolar
a system,
examples
LaSpecific
densidad
molar
oor
específica
sedefined
define
como
el
recíproco
del
volumen
molar
oand
específico:
Specific
molar
density
isas
as
the
reciprocal
ofor
specific
or
molar
volume:
≡−1..V −1..
Specific
orormolar
molar
density
is
defined
asas
the
reciprocal
ofof
specific
or
molar
volume:
≡
Specific
or
molar
density
is
defined
asdefined
the
reciprocal
ofofspecific
specific
or
molar
volume:
ρare
≡ρρVV
Specific
or
density
is
as
the
reciprocal
of
specific
or
molar
volume:
≡
VV
..and
or
molar
density
is
defined
the
reciprocal
of
volume:
ρpressure,
≡
−1
.
Specific
molar
density
is
defined
the
reciprocal
specific
or
molar
volume:
ρ
≡
V
ofof
intensive
thermodynamic
variables.
They
are
functions
of
the
temperature,
.
Specific
or
molar
density
is
defined
as
the
reciprocal
of
specific
or
molar
volume:
ρ
≡
V
of
intensive
thermodynamic
variables.
They
are
functions
of
the
temperature,
pressure,
and
intensive
thermodynamic
variables.
They
areof
functions
of
pressure,
and
Estas cantidades
(V
y ρ)
sonand
independientes
del independent
tamaño
desize
unofsize
sistema
son
unexamples
ejemplo
de
These
quantities
(Vare
and
ρ)independent
are
theathe
size
ofademás,
a and
system,
and
are
examples
These
quantities
(V
and
ρ)
are
independent
of
the
size
of
ay,
system,
and
are
examples
These
quantities
(V
and
ρ)
are
independent
ofare
the
size
of
aof
system,
and
are
examples
These
quantities
(V
and
ρ)
are
of
the
atemperature,
system,
and
are
These
quantities
(V
ρ)
independent
the
of
system,
are
examples
composition
of
a
system,
additional
quantities
that
independent
of
system
size.
These
quantities
(V
and
ρ)
are
independent
of
the
size
of
a
system,
and
are
examples
These
quantities
(V
and
ρ)
are
independent
of
the
size
of
a
system,
and
are
examples
composition
of
a
system,
additional
quantities
that
are
independent
of
system
size.
composition
of
a
system,
additional
quantities
that
are
independent
of
system
size.
variables
termodinámicas
intensivas.
Son
funciones
de
la
temperatura,
de
la
presión
y
de
la
composición
of
intensive
thermodynamic
variables.
They
are
functions
of
the
temperature,
pressure,
ofofintensive
intensive
thermodynamic
variables.
They
are
functions
ofoftemperature,
the
temperature,
pressure,
and and
of intensive
intensive
thermodynamic
variables.
They
are functions
functions
of the
the
temperature,
pressure,
andand
thermodynamic
variables.
They
the
pressure,
of
thermodynamic
variables.
They
are
of
pressure,
and
intensive
thermodynamic
variables.
They
arefunctions
functions
thetemperature,
temperature,
pressure,
of of
intensive
thermodynamic
variables.
They
areare
functions
of of
the
temperature,
pressure,
and and
de un sistema,
son
cantidades
adicionales
e
independientes
del
tamaño
del
sistema.
composition
of
a
system,
additional
quantities
that
are
independent
of
system
size.
composition
of
system,
additional
quantities
that
are
independent
ofofsystem
system
size.
composition
of aaof
system,
additional
quantities
thatthat
are
independent
of system
system
size.size.
composition
aaasystem,
additional
quantities
are
composition
of
additional
quantities
that
of
size.
composition
of
system,
additional
quantities
that
areindependent
independent
system
composition
of asystem,
system,
additional
quantities
thatare
areindependent
independent
of of
system
size.size.
1.4
1.4
1.4 FORCE
FORCE
1.4
FORCE
FUERZA
1.4
FORCE
1.4
FORCE
1.4
FORCE
1.4
FORCE
1.4
FORCE
1.4
FORCE
1.4
FORCE
The
TheSISIunit
unitofofforce
forceisisthe
thenewton,
newton,symbol
symbolN,N,derived
derivedfrom
fromNewton’s
Newton’ssecond
secondlaw,
law,which
whichexex-
The SI unit of force is the newton, symbol N, derived from Newton’s second law, which exLa unidadpresses
del SI que
corresponde
a la fuerza
es elmnewton,
su símboloa;
es
N,
yF
seF=
deduce
de
la newton
segunda
ley
force
FF
product
ofof
thus
The
isis
presses
force
Fasas
asthe
the
product
ofmass
mass
mand
andacceleration
acceleration
a;from
thus
=ma.
ma.second
The
newton
is de
presses
force
mass
m
and
acceleration
a;
thus
F
=
ma.
The
newton
The
SI
unit
ofthe
force
isnewton,
the
newton,
symbol
N,from
derived
Newton’s
law,
which
exThe
SI
unit
of
force
isFproduct
the
newton,
symbol
N,
derived
from
Newton’s
second
law,
which
exThe
SI
unit
of
force
is
the
newton,
symbol
N,
derived
from
Newton’s
second
law,
which
exThe
SI
unit
of
force
is
the
symbol
N,
derived
from
Newton’s
second
law,
which
exSI
unit
of
force
is
the
newton,
symbol
N,
derived
Newton’s
second
law,
which
ex−2
−2
Newton,The
que
expresa
a
la
fuerza
como
el
producto
de
la
masa
m
y
la
aceleración
a;
es
decir,
F
=
ma.
defined
as
the
force
which
when
applied
to
a
mass
of
1
kg
produces
an
acceleration
of
1
m
s
;
The
SI
unit
of
force
is
the
newton,
symbol
N,
derived
from
Newton’s
second
law,
which
exThe
SI
unit
of
force
is
the
newton,
symbol
N,
derived
from
Newton’s
second
law,
which
ex−2
definedpresses
as the
the force
force which
which
when
applied
to
mass
of 11acceleration
kg produces
producesa;
anthus
acceleration
of 11The
m ssnewton
; El is
defined
as
when
applied
to
aa and
mass
of
kg
an
acceleration
of
m
;is
F
as
the
product
of
mass
m
and
F
=
ma.
presses
force
F
as
the
product
of
mass
m
and
acceleration
a;
thus
F
=
ma.
The
newton
is
presses
force
F
as
the
product
of
mass
m
and
acceleration
a;
thus
F
=
ma.
The
newton
is
−
presses
force
F
as
the
product
of
mass
m
acceleration
a;
thus
F
=
ma.
The
newton
presses
force
F
as
the
product
of
mass
m
and
acceleration
a;
thus
F
=
ma.
The
newton
is
−2
−2
newton sepresses
define
como
fuerza
que,product
alunit
aplicarse
amuna
de
kg,
de newton
m is
s −2 is
; de
thus
the
newton
isis
athe
derived
1and
kg
m
. .. produce
presses
force
F
the
of mass
mmasa
acceleration
a; thus
= ma.
force
Fla as
product
ofrepresenting
mass
and
a; thus
Funa=Faceleración
ma.
TheThe
newton
thus
the
newton
is
derived
unit
representing
1to
kg
msss−2
−2
thus
the
newton
aaas
derived
unit
representing
1acceleration
kg
m
as
the
force
which
when
a11mass
of produces
1−
kg. produces
an acceleration
of
m;; s−2 ;
defined
asasforce
the
force
which
when
applied
to
mass
of
kg
produces
an
acceleration
of
m
defined
asdefined
the
force
which
when
applied
toapplied
ato
mass
of
kg
produces
an acceleration
acceleration
of 11of
m
defined
as
the
force
which
when
applied
aaamass
of
11is
kg
an
acceleration
11ss1−2
m
defined
as
the
which
when
applied
to
arepresenta
mass
of
kg
produces
an
of
m
;;ss1;−2
−2
−2
esta forma,
el
newton
es
una
unidad
deducida
que
kg
m
s
defined
the
force
which
when
applied
to
mass
of
1
kg
produces
an
acceleration
of
m
s
In
the
English
engineering
system
of
units,
force
treated
as
an
additional
independent
defined
as
the
force
which
when
applied
to
a
mass
of
1
kg
produces
an
acceleration
of
1
m
s
In
English
engineering
system
of units,
units,
force
is
treated
as an
an additional
additional independent
independent ;
−2
−2
−2
−2
In
the
English
system
of
force
treated
thus
the
is aunit
derived
unit representing
1m
kg
m
. as
thus
the
newton
is
aengineering
derived
unit
representing
kg
. .s−2como
thus
the
newton
isnewton
derived
unit
representing
kgfuerza
m
thus
the
newton
is
aunidades
derived
unit
representing
11m
kg
m
.force
the
newton
is
aawith
derived
representing
11la
kg
sspound
..ssis
−2
−2
Enthus
eldimension
sistema
inglés
de
de
ingeniería,
se
trata
una
dimensión
independiente
thus
the
newton
is
a
derived
unit
representing
1
kg
m
s
dimension
along
length,
time,
and
mass.
The
(lb
)
is
defined
as
thus
the
newton
is
a
derived
unit
representing
1
kg
m
s
.
f
along
with
length,
time, and
andsystem
mass. of
The
pound
force
(lbff)) is
isasdefined
defined
asthat
thatforce
force
dimension
along
length,
time,
mass.
The
pound
force
(lb
as
that
force
InEnglish
thewith
English
engineering
units,
force
isas
treated
an additional
independent
InInEnglish
the
English
engineering
system
of
units,
force
is(lb
treated
asas
an
additional
independent
In accelerates
the
English
engineering
system
of feet
units,
force
is treated
treated
as
an
additional
independent
In
the
engineering
system
of
units,
force
is
treated
as
an
additional
independent
In
the
engineering
system
of
units,
force
is
an
additional
independent
adicional,which
junto
con
la
longitud,
el
tiempo
y
la
masa.
La
libra
fuerza
),
se
define
como
la
fuerza
que
acele­
1
pound
mass
32.1740
per
second
per
second.
Newton’s
law
must
here
the
English
engineering
system
of
units,
force
is
treated
an
additional
independent
f
In
the
English
engineering
system
of
units,
force
is
treated
as
an
additional
independent
whichdimension
acceleratesalong
1 pound
pound
mass
32.1740
feet
per
second
per
second.
Newton’s
law
must
here
which
accelerates
1
mass
32.1740
feet
per
second
per
second.
Newton’s
law
must
with
length,
time,
and
mass.
The
pound
(lb
isas
defined
ashere
that force
dimension
along
with
length,
time,
and
mass.
The
pound
force
(lb
)defined
isisdefined
defined
asasthat
that
force
dimension
along
with
length,
time,
and
mass.
The
pound
force
(lbffforce
isff)defined
as
that
force
f )caso
dimension
along
with
length,
time,
and
mass.
The
pound
force
is
as
force
dimension
along
with
length,
time,
and
mass.
The
pound
force
(lb
))(lb
is
that
force
ra libra
masa
a
3.740
pies
por
segundo.
La
ley
de
Newton
debe
incluir
en
este
una
constante
de
include
a
dimensional
proportionality
constant
for
consistency
with
this
definition:
dimension
along
with
length,
time,
and
mass.
The
pound
force
(lb
)
defined
that
force
dimension
along
with
length,
time,
and
mass.
The
pound
force
(lb
)
is
defined
as
that
force
f
includewhich
dimensional
proportionality
constant
for
consistency
with
this definition:
definition:
include
aaaccelerates
dimensional
proportionality
constant
for
consistency
with
this
accelerates
1ser
pound
mass
32.1740
feet
perper
second
perfNewton’s
second.
Newton’s
law
here
which
pound
mass
32.1740
feet
per
second
per
second.
Newton’s
law
must
here
which
accelerates
pound
mass
32.1740
feet
per
second
per
second.
Newton’s
law law
must
heremust
which
accelerates
11para
mass
32.1740
feet
per
second
per
Newton’s
must
here
which
accelerates
111pound
mass
32.1740
feet
per
second
law
here
proporcionalidad
dimensional
consistente
con
esta
definición:
which
accelerates
1pound
pound
mass
32.1740
feet
per
second
persecond.
second.
Newton’s
law
must
here
which
accelerates
pound
mass
32.1740
feet
per
second
persecond.
second.
Newton’s
lawmust
must
here
a dimensional
proportionality
constant
for consistency
with
this definition:
include
dimensional
proportionality
constant
for
consistency
with
this
definition:
include
dimensional
proportionality
constant
for
consistency
withwith
thisthis
definition:
include
aaadimensional
proportionality
constant
for
definition:
include
aainclude
proportionality
constant
for
with
11consistency
include
dimensional
proportionality
forconsistency
consistency
with
this
definition:
include
adimensional
dimensional
proportionality
constant
for
consistency
withthis
thisdefinition:
definition:
Fconstant
ma
F==
= 1ma
ma
F
g
c
g
111gcc111 1
ma
F=
ma
==
ma
=FFF=
ma
ma
FFF=
ma
ma
=
gc
g
g
g
g
1
c
c
c
c
1
2
−2
g32.1740(ft)(s)
g)c )××
c 32.1740(ft)(s)
−2
1(lb
De donde,Whence,
−2
mm
f )f))==
1(lb
= 1 ××
×1(lb
1(lb
Whence,22
1(lb
Whence,
m ) × 32.1740(ft)(s)
f
ggcgc1 1(lb
1
1
1 1 c11
2
−2 −2
−2 −2
22
22Whence,
−2
=×
1(lb
) × 32.1740(ft)(s)
==×
××1(lb
1(lb
) )×
××m32.1740(ft)(s)
32.1740(ft)(s)
Whence,
1(lb1(lb
=ff))1(lb
×
1(lb
×
Whence,
f ) 1(lb
)32.1740(ft)(s)
Whence,
1(lb
=
))×
×
32.1740(ft)(s)
Whence,
2
m
m1(lb
f))1(lb
2
−2 −2
m
m
f1(lb
)=
32.1740(ft)(s)
Whence,
1(lb
=fgg32.1740(lb
×
1(lb
)
×
32.1740(ft)(s)
Whence,
−1
−2
g
m
m
fc) =
g
−1
−2
c
g
c
c
and
g
)(ft)(lb
)
(s)
c
mm)(ft)(lb
f f))−1 (s)
and
=cg32.1740(lb
32.1740(lb
)(ft)(lb
(s)−2
y
and
ggcc =
c gc
m
f
−1
−1
−2(s)−2
−1
−2
−2
and force is equivalent
)(ft)(lb
and
==g32.1740(lb
32.1740(lb
)(ft)(lb
(s)
and
=
32.1740(lb
)(ft)(lb
c = 32.1740(lb
f )−2
and
gg=4.4482216
=
))−1
and
ggcgc to
=
)(ft)(lb
))−1
c32.1740(lb
m)(ft)(lb
f(s)
mN.
−1
−2
The
pound
N.
cg
m
f(s)
m
ffm
The
pound
force is
is equivalent
equivalent
to
4.4482216
and
32.1740(lb
)−1(s)
(s)−2
and
c32.1740(lb
m )(ft)(lb
The
pound
force
4.4482216
N.
c to
m )(ft)(lb
f ) f(s)
La libra fuerza
es equivalente
a 4.4486
N.
Because
force
and
mass
are
different
concepts,
aaapound
force
and
aaapound
mass
are
Because
force
and
mass
are4.4482216
different
concepts,
pound
force
and
poundson
mass
are
Because
and
mass
are
different
concepts,
pound
pound
mass
are
Puesto
quepound
laforce
fuerza
yequivalent
la
son
conceptos
distintos,
una libra
fuerzaforce
y unaand
libra
masa
cantidades
The
pound
force
is equivalent
to 4.4482216
N.
The
pound
force
isismasa
equivalent
to
N.
TheThe
pound
force
isforce
equivalent
to 4.4482216
4.4482216
N. N.
force
is
equivalent
to
4.4482216
The
pound
is
to
N.
The
pound
force
equivalent
to
4.4482216
N.
different
quantities,
and
their
units
do
not
cancel
one
another.
When
an
equation
contains
both
The
pound
force
is
equivalent
to
4.4482216
N.
different
quantities,
and
their
units
do
not
cancel
one
another.
When
an
equation
contains
both
quantities,
and
their
units
dodifferent
not
one
When
an
equation
contains
both
diferentes,different
y estas
unidades
no
se
cancelan
entre
sí. cancel
Cuando
una
ecuación
contiene
ambas
unidades,
(lb
Because
force
and
mass
are
different
concepts,
aforce
pound
force
amass
pound
mass
are
Because
force
and
mass
are
different
pound
force
and
pound
mass
are
f) y
Because
force
and
mass
are
different
concepts,
a pound
pound
force
and
pound
mass
are
Because
force
and
mass
concepts,
aaapound
force
and
aaand
mass
Because
force
and
mass
are
different
concepts,
aanother.
and
aaand
pound
are
units,
(lb
)f))and
(lb
),
dimensional
constant
gconcepts,
appear
in
the
to
force
and
mass
are
different
concepts,
pound
force
apound
pound
mass
and
mass
areare
different
concepts,
aalso
pound
force
aequation
pound
mass
areitare
m
cgcmust
fBecause
units,Because
(lb
andforce
(lb
),the
the
dimensional
constant
must
also
appear
inand
theequation
equation
tomake
make
itareboth
m
units,
(lb
and
(lb
),
the
dimensional
constant
g
must
also
appear
in
the
to
make
it
(lb m ), la
constante
dimensional
g
también
debe
aparecer
en
la
ecuación
para
hacerla
dimensionalmente
different
quantities,
and
their
units
do
not
cancel
one
another.
When
an
equation
contains
m
c
f
different
quantities,
and
their
units
do
not
cancel
one
another.
When
an
equation
contains
both
different
quantities,
and
their
units
do
not
cancel
one
another.
When
an
equation
contains
both
c
different
quantities,
their
units
do
not
another.
When
an
contains
both
different
quantities,
and
their
units
dodonot
one
another.
When
ananequation
contains
both
dimensionally
correct.
different
quantities,
and
their
units
docancel
notcancel
cancel
one
another.
When
anequation
equation
contains
both
different
quantities,
andand
their
units
not
cancel
oneone
another.
When
equation
contains
both
dimensionally
correct.
dimensionally
correct.
correcta.units,
units,
(lb
)(lb
and
(lb
the dimensional
constant
gc must
alsoin
appear
inequation
the equation
to make
units,
(lb
)
and
(lb
),
the
dimensional
constant
g
must
also
appear
in
the
equation
to
make
it it
units,
(lb
)
and
(lb
),
the
dimensional
constant
g
must
also
appear
in
the
equation
to
make
it
m ),dimensional
fm
units,
(lb
)
and
),
the
constant
g
must
also
appear
in
the
to
make
(lb
)
and
),
the
dimensional
constant
g
must
also
appear
the
equation
to
make
it
m
c
f
m
c
f
m
c
f
c
f
units,
the dimensional
constant
gc must
appear
in equation
the equation
to make
units,
(lbf )(lb
and
(lbm(lb
), the
constant
gc must
alsoalso
appear
in the
to make
it itit
m ), dimensional
f ) and
dimensionally
correct.
dimensionally
correct.
dimensionally
correct.
dimensionally
correct.
dimensionally
correct.
dimensionally
correct.
dimensionally
correct.
2 Where
22Where
Wherenon-SI
non-SIunits
units(e.g.,
(e.g.,English
Englishunits)
units)are
areemployed,
employed,parentheses
parenthesesenclose
enclosethe
theabbreviations
abbreviationsofof
ofallall
allunits.
units.
non-SI
units
(e.g.,
English
units)
are
employed,
parentheses
enclose
the
abbreviations
units.
22
22
2
Where
units
(e.g.,
English
units)
areunidades
employed,
parentheses
enclose
the
abbreviations
ofabreviaturas
all units. de
Where
non-SI
units
(e.g.,
English
units)
are employed,
employed,
parentheses
enclose
the abbreviations
abbreviations
of
alllas
units.
Where
non-SI
unitsnon-SI
(e.g.,no
English
units)
are employed,
employed,
parentheses
enclose
the
abbreviations
of all
allof
units.
Where
non-SI
units
(e.g.,
English
are
parentheses
enclose
the
all
units.
non-SI
units
(e.g.,
English
units)
are
parentheses
enclose
abbreviations
of
units.
2 Where
2 Where
Donde seWhere
empleen
unidades
sean
del
SIunits)
(por
inglesas),
sethe
utilizarán
paréntesis
con
non-SI
units
English
units)
are employed,
parentheses
enclose
the abbreviations
ofunits.
all units.
non-SI
unitsque
(e.g.,(e.g.,
English
units)
are ejemplo,
employed,
parentheses
enclose
the
abbreviations
of all
todas las unidades.
01­SmithVanNess.indd 3
8/1/07 12:47:16
4
CHAPTER 1. Introduction
CHAPTER 1. Introduction
4
4
CAPÍTULO 1. Introducción
Weight properly refers to the force of gravity on a body, and is correctly expressed in
or in properly
pounds
force.
ofamass
areand
often
called “weights,”
refersaUnfortunately,
to
force
ofstandards
gravitysobre
on
body,
in
Elnewtons
peso Weight
se refiere
propiamente
la the
fuerza
de gravedad
un
cuerpo
y is
se correctly
expresa
deexpressed
manera and
correcta
en
a balance
toDesafortunadamente,
compare
masses is los
called
“weighing.”
Thus,
one must
discern
from
they al uso
newtons
or fuerza.
in pounds
force. Unfortunately,
standards
ofmasa
mass
often
called
“weights,”
and
newtonsuse
o enoflibras
estándares
de
aare
menudo
se
denominan
“pesos”
context
whether
force
or mass
is llama
meant
when
“weight”
incuando
a one
casual
informal
way. the
use of apara
balance
to compare
is“para
called
“weighing.”
Thus,
must
discern
from
de una balanza
comparar
masas
semasses
le
pesar”.
Por isloused
tanto,
se or
emplea
la palabra
“peso” de
context
whether
force oresmass
is meant
whendentro
“weight”
is used in
a casual
informal
una manera
ocasional
o informal
preciso
distinguir
del contexto
si se
trata deoruna
fuerza oway.
de una masa.
Example 1.1
Example
1.1 730 N in Houston, Texas, where the local acceleration of gravity
Ejemplo
1.1 weighs
An
astronaut
−2
is
=pesa
9.792730
m
What
the donde
astronaut’s
mass
on local
the moon,
where
Angastronaut
weighs
730 Nare
in
Houston,
Texas,
whereand
thelaweight
local
acceleration
Un astronauta
Ns en.Houston,
Texas,
la aceleración
de
gravedad
es deofggravity
= 9.79 m s−2.
−2 ? −2
g
=
1.67
m
s
is
g
=
9.792
m
s
.
What
are
the
astronaut’s
mass
and
weight
on
the
moon,
where
−
¿Cuáles son la masa −2
y el peso del astronauta en la Luna, donde g = .67 m s ?
g = 1.67 m s
?
Solución
1.1
Solution
1.1
1.1
Con aSolution
=With
g, laaley
Newton es:
F is:
= mg.
donde,
= de
g,
Newton’s
law
F =De
mg.
Whence,
With a = g, Newton’s law
is:
F
=
mg.
F
730 N Whence,
2
m=
= 74.55 N m−1 s
=
F
730mNs−2
2
g
9.792
m=
= 74.55 N m−1 s
=
−2
g
9.792
m
s
−,
Como elBecause
newton the
N tiene
unidades
m skg
newton
N has de
thekg
units
m s−2 ,
Because the newton N has the units kg m s−2 ,
m
= 74.55
m=
74.55 kg
kg
m = 74.55 kg
Esta masa
delmass
astronauta
independiente
de la ubicación,
perobut
el peso
depende
de on
la aceleración
This
of the es
astronaut
is independent
of location,
weight
depends
the
local delocal
la gravedad.
forma,isel
pesoondel
labut
Luna
es:
acceleration
of gravity.
Thus
theastronauta
moon
the en
astronaut’s
weight
is: on the
This
mass
of De
the esta
astronaut
independent
of location,
weight
depends
local acceleration of gravity. Thus on the moon the astronaut’s −
weight is:
F(moon)
74.55kg
kg ×× .67
1.67 m
m ss−2
F(Luna) =
= mg(moon)
mg(Luna) == 74.55
F(moon) = mg(moon) = 74.55
kg × 1.67 m s−2
or
F(moon) = 124.5 kg m s−2
= 124.5 N
−
o
F(Luna) = 4.5 kg m s = 4.5 N
or
F(moon) = 124.5 kg m s−2 = 124.5 N
Usesistema
of the English
engineering
system
of units
requiresdeconversion
of thedelas-peso del
El uso del
inglés de
unidades de
ingeniería
requiere
la conversión
−2 . With 1 N equivalent to
−of
tronaut’s
tovalores
(lbf )engineering
and
values
goftounits
(ft)(s)
English
system
requires
conversion
of the as- f) y m
astronauta
aUse
(lbf)ofweight
y the
de los
de gthe
a (pie)(s
). Puesto
que
N equivale
a 0.4809(lb
0.224809(lb
m fto
3.28084(ft):
tronaut’s weight
) and
the values of g to (ft)(s)−2 . With 1 N equivalent to
f ) andto1 (lb
a 3.8084(pie):
0.224809(lbf ) and 1 m to 3.28084(ft):
Weight of astronaut in Houston = 164.1(lbf )
El peso del astronauta en Houston = 64.(lb
f)
Weight of astronaut in Houston = 164.1(lbf )
g(Houston) = 32.13
and
g(moon) = 5.48(ft)(s)−2
−
g(Houston)
= 3.3
g(Luna)
= 5.48(pie)(s)
−2
g(Houston)
= 32.13
and y
g(moon) =
5.48(ft)(s)
Newton’s
law then gives:
then da:
gives:
Por esto,Newton’s
la ley de law
Newton
164.1(lbf ) × 32.1740(lbm )(ft)(lbf )−1 (s)−2
Fgc
=
m=
)(ft)(lbf )−1 (s)−2
164.1(lbf ) × 32.1740(lb
Fg
m −2
gc
32.13(ft)(s)
=
m=
−2
g
32.13(ft)(s)
or
m = 164.3(lb
m)
or
m = 164.3(lbm )
Thus the astronaut’s mass in (lbmm) and
weight in) (lbf ) in Houston are numerically
o
= 64.3(lb
m
almost
theastronaut’s
same, but mass
on theinmoon
not theincase:
Thus the
(lbm )this
andisweight
(lbf ) in Houston are numerically
almost
the
same,
but
on
the
moon
this
is
not
the
case:
De esta forma, la masa del astronauta
en (lbm) y (164.3)(5.48)
su peso en (lbf) en Houston son casi numéricamg(moon)
= 28.0(lbf )
=
= no es el caso:
mente iguales, pero enF(moon)
la Luna éste
(164.3)(5.48)
mg(moon)
gc
32.1740
= 28.0(lbf )
=
F(moon) =
gc
32.1740
mg.Luna/
.164:3/.5:48/
H 28:0(lbf )
F.Luna/ H
H
gc
32:1740
01­SmithVanNess.indd 4
8/1/07 12:47:19
5
1.5. Temperatura
1.5
TEMPERATURA
La temperatura se mide por lo regular con termómetros de líquidos en capilares de vidrio, donde el fluido se
expande cuando se calienta. De este modo, un tubo uniforme que está parcialmente lleno de mercurio, alcohol
o algún otro fluido, indica el grado de “calentamiento” mediante la longitud de la columna del fluido. De
cualquier modo, se asignan valores numéricos a los diversos grados de calentamiento por medio de una defi­
nición arbitraria.
Para la escala Celsius,3 el punto de hielo (punto de congelación del agua saturada con aire a la presión
atmosférica estándar) es cero, y el punto de vapor (punto de ebullición del agua pura a la presión atmosférica
estándar) es 00. Es posible asignar a un termómetro una escala numérica sumergiéndolo en un baño de hielo
y haciendo una marca correspondiente al cero en el nivel donde se encuentra el fluido, para después sumer­
girlo en agua hirviendo y hacer una marca para el valor de 00 en este nuevo nivel del fluido. La distancia
entre las dos marcas se divide en 00 espacios equidistantes denominados grados. Para extender la escala del
termómetro se marcan otros espacios de igual tamaño por debajo del cero y por encima del 00.
Todos los termómetros, sin importar el fluido que contengan, proporcionan la misma lectura en cero y
en 00 si se calibran con este método, pero generalmente en otros puntos no corresponden las lecturas porque
los fluidos varían en sus características de expansión. De esta manera se requiere una elección arbitraria del
fluido, y la escala de temperatura en el sistema SI cuya unidad es el kelvin y su símbolo es K, se apoya en
el gas ideal como fluido termométrico. Puesto que la definición de la escala Kelvin depende de las propieda­
des de los gases, se hará un análisis detallado hasta el capítulo 3. Sin embargo, se observa que la escala abso­
luta depende del concepto de un límite inferior de la temperatura.
Las temperaturas Kelvin se indican con el símbolo T, en tanto que las temperaturas Celsius se designan
con el símbolo t, y se definen en relación a las temperaturas Kelvin mediante:
t °C = T K − 73.5
La unidad de temperatura Celsius es el grado Celsius, °C, que en magnitud equivale al Kelvin.4 No obstante, las
temperaturas en la escala Celsius son 73.5 grados menores que en la escala Kelvin. De esta forma el límite infe­
rior de la temperatura, conocido como el cero absoluto en la escala Kelvin, se presenta en −73.5 °C.
En la práctica, la Escala Internacional de Temperatura de 1990 (ITS­90) se utiliza para la calibración
de instrumentos científicos e industriales.5 La escala ITS­90 se define de modo que sus valores difieren de las
temperaturas del gas ideal, pero que están en los límites de exactitud con los que actualmente se hacen las me­
diciones. Se basa en valores asignados de la temperatura para un número de estados de equilibrio de fase
reproducibles de sustancias puras (puntos fijos) y en los instrumentos estándar calibrados a estas tempera­
turas. La interpolación entre las temperaturas de punto fijo se proporciona mediante fórmulas que establecen
la relación entre las lecturas de los instrumentos estándar y los valores de la ITS­90. El termómetro de resis­
tencia de platino es un ejemplo de un instrumento estándar, que se utiliza para temperaturas que van desde
−59.35 °C (punto triple del hidrógeno) hasta 96.78 °C (el punto de congelación de la plata).
Además de las escalas Kelvin y Celsius existen otras dos que aún son utilizadas por los ingenieros de
Estados Unidos: las escalas Rankine y Fahrenheit.6 La escala Rankine es una escala absoluta que se relaciona
3 Anders
Celsius, astrónomo sueco (70­744).
Adviértase que la palabra grado no se emplea en las temperaturas dadas en kelvins, y que la palabra kelvin entendida como una
unidad no se escribe con mayúsculas.
5 El texto en lengua inglesa de la definición de ITS­90 está dado por H. Preston­Thomas, Metrologia, vol. 7, pp. 3­0, 990.
6 Gabriel Daniel Fahrenheit, físico alemán (686­736).
4
01­SmithVanNess.indd 5
8/1/07 12:47:19
6
CAPÍTULO 1. Introducción
directamente con la escala Kelvin mediante:
T(R) = .8 T K
La escala Fahrenheit se relaciona con la de Rankine por una ecuación semejante a la relación entre las escalas
Celsius y Kelvin:
t(°F) = T(R) − 459.67
De este modo, el límite inferior de la temperatura en la escala Fahrenheit es −459.67(°F). La relación entre
las escalas Celsius y Fahrenheit está dada por:
t(°F) = .8 t °C + 3
Por consiguiente, el punto de congelación del agua es 3(°F) y el punto de ebullición normal del agua es
(°F).
El grado Celsius y el kelvin representan el mismo intervalo de temperatura, al igual que el grado Fah­
renheit y el Rankine. Las relaciones entre las cuatro escalas de temperatura se ilustran en la figura .. En ter­
modinámica, se da a entender una temperatura absoluta mediante una referencia informal de la temperatura.
Celsius
Kelvin
Fahrenheit
Rankine
100(C)
373.15 K
212(F)
671.67(R)
Punto de vapor
0(C)
273.15 K
32(F)
491.67(R)
Punto de hielo
273.15(C)
0K
459.67(F)
0(R)
Cero absoluto
Figura 1.1: Relaciones entre las escalas de temperatura.
1.6
PRESIÓN
La presión P ejercida por un fluido sobre una superficie se define como la fuerza normal ejercida por el fluido
por unidad de área de la superficie. Si la fuerza se mide en N y el área en m, la unidad es el newton por metro
cuadrado o N m−, llamado pascal y representado con el símbolo Pa, y es la unidad básica de la presión para
el SI. En el sistema inglés de ingeniería una unidad común es la libra fuerza por pulgada cuadrada (psi, por
sus siglas en inglés).
El estándar primario para mediciones de presión es la balanza de peso muerto, en la cual una fuerza
conocida se equilibra con una presión del fluido que actúa sobre un área conocida, donde P ≡ F/A. En la figu­
ra . se muestra un diseño sencillo. El pistón se ajusta cuidadosamente al cilindro dejando una separación
01­SmithVanNess.indd 6
8/1/07 12:47:20
1.6. Pressure
1.6. Presión
1.6. Pressure
7
7
7
muy pequeña.
Se small.
ponen Weights
pesas en are
la bandeja
hasta
presión
del aceite,
queoil,
tiende
a hacer
el pistón se
clearance
placed on
the que
pan la
until
the pressure
of the
which
tends que
to make
eleve, se
equilibre
precisamente
por
la
fuerza
de
gravedad
en
el
pistón
y
todo
lo
que
éste
soporta.
the pistonsmall.
rise, isWeights
just balanced
by on
thethe
force
gravity
on the piston
andwhich
all that
it supports.
clearance
are placed
panof
until
the pressure
of the oil,
tends
to makeCon esta
fuerza dada
por
la
ley
de
Newton
la
presión
del
aceite
es:
Withpiston
this force
Newton’s
the pressure
of the
the
rise, given
is justby
balanced
by law,
the force
of gravity
onoil
theis:piston and all that it supports.
With this force given by Newton’s law, the pressure
of the oil is:
mg
F
=
P=
mg
FA
A
=
P=
where
m
is
the
mass
of
the
piston,
pan,
and
weights;
is la
theaceleración
local acceleration
andy A es el
A
A
donde m es la masa del pistón, de la bandeja y de las pesas; gges
local deof
la gravity;
gravedad,
A
is
the
cross-sectional
area
of
the
piston.
Gauges
in
common
use,
such
as
Bourdon
gauges,
área dewhere
sección
pistón.
Los
instrumentos
degmedida
de uso
común, como
el manómetro
de
m transversal
is the mass del
of the
piston,
pan,
and weights;
is the local
acceleration
of gravity;
and
are
by comparison
with
gauges.
Bourdon,
secalibrated
calibran
por
comparación
con
las balanzas
de peso
muerto.use, such as Bourdon gauges,
A is
the
cross-sectional
area of
thedead-weight
piston.
Gauges
in common
are calibrated by comparison with dead-weight gauges.
Pesa
Bandeja
Pistón
Cilindro
Aceite
Weight
Weight
Pan
Piston
Pan
Piston
Cylinder
Cylinder
Oil
Oil
Figure
1.2:1.2: Balanza
Figura
Dead-weight
gauge.
Figure
1.2: muerto.
de peso
Dead-weight gauge.
To pressure
A la fuentesource
To pressure
de presión
source
Because a vertical column of a given fluid under the influence of gravity exerts a pressure
at
its
base
in direct
proportion
height,
pressure
also
expressed
as la
theinfluencia
equivalent
Because
a vertical
columntode
ofitsaun
given
fluid
underisthe
influence
gravity
exerts a pressure
Puesto que una
columna
vertical
fluido
determinado
que
está of
bajo
deheight
la gravedad,
of
a
fluid
column.
This
is
the
basis
for
the
use
of
manometers
for
pressure
measurement.
at
its
base
in
direct
proportion
to
its
height,
pressure
is
also
expressed
as
the
equivalent
heightcomo la
ejerce una presión en su base que va en proporción directa con su altura, la presión también se expresa
Conversion
height
to force
perbasis
unitÉste
areathe
Newton’s
lawpressure
applied
to
the force
of
of
a fluid column.
This
isdethefluido.
for
offrom
manometers
measurement.
altura equivalente
deofuna
columna
esfollows
eluse
fundamento
para el for
uso
de
manómetros
en mediciones
gravity
acting
on
the
mass
of
fluid
in
the
column.
The
mass
is
given
by:
m
=
Ahρ,
where
A
is
Conversion
of
height
to
force
per
unit
area
follows
from
Newton’s
law
applied
to
the
force
of
de presión. La conversión de la altura a fuerza por unidad de área surge a partir de la ley de Newton aplicada
a la
the
cross-sectional
area
of
the
column,
h
is
its
height,
and
ρ
is
the
fluid
density.
Therefore,
gravity
acting
on
the
mass
of
fluid
in
the
column.
The
mass
is
given
by:
m
=
Ahρ,
where
A
is
fuerza de la gravedad que actúa sobre la masa del fluido en la columna. La masa está dada por: m = Ahρ, donde
the cross-sectional
area of the
column,
its
and
is la
thedensidad
fluid density.
Therefore,
A es el área
de sección transversal
de la
columna,
es height,
su altura
y ρρ es
del fluido.
Por lo tanto,
F h ishmg
Ahρg
P=
=
=
= hρg
FA
mg
Ahρg
A
A
P=
=
=
= hρg
The pressure to which a fluid height
A corresponds
A
Ais determined by the density of the fluid
(which
depends
itsa identity
and
temperature)
andisthe
local
acceleration
of gravity.
Thusfluid
the
The
pressure
to on
which
a fluid
height
corresponds
determined
by the del
density
of (que
the
La presión
que
corresponde
una
altura
del
fluido
se determina
por la
densidad
mismo
depende
de
◦ C in a standard gravitational
(torr)
is
the
pressure
equivalent
of
1
millimeter
of
mercury
at
0
(whichy depends
on itsyidentity
and temperature)
local acceleration
ofes
gravity.
Thusequivalente
the
su identidad
temperatura)
de la aceleración
local deand
la the
gravedad.
Así, el (torr)
la presión
field, and
ismercurio
equal to a133.322
a standard
gravitational
(torr)
is de
the
pressure
equivalent
1 millimeter
of mercury
at 0◦yCesinigual
de milímetro
0 °C enPa.
unofcampo
gravitacional
estándar,
a 33.3
Pa.
Another
unit
of
pressure
is
the
standard
atmosphere
(atm),
the approximate
averagepor
presfield,
and isdeequal
to 133.322
Pa.
Otra
unidad
presión
es la atmósfera
estándar (atm), que es la presión
promedio ejercida
la atmós­
sure
exerted
by
the
earth’s
atmosphere
at
sea
level,
defined
as
101,325
Pa,
101.325
kPa,
or El bar,
Another
unit
pressure
the standard
atmosphere
(atm), thekPa,
approximate
average MPa.
presfera de la Tierra
a nivel
delofmar,
y se is
define
como 0
35
Pa,
0.35
o
bien,
0.035
5 Pa, is equal to 0.986923(atm).
0.101325
The
bar,
an0
SI5 unit
asa 10
sure
exerted
the earth’s
atmosphere
at sea
level,
defined
una unidad
del
SI,MPa.
seby
define
como
Pa ydefined
es igual
0.98693
(atm).as 101,325 Pa, 101.325 kPa, or
Most
gauges
readings
which
areque
difference
between
the pressure
of
0.101325
MPa.
bar,
an SIgive
unit
defined
as
105 Pa,
isthe
equal
to 0.986923(atm).
La
mayoría
depressure
losThe
manómetros
de
presión
dan
lecturas
representan
la diferencia
entre la presión
de
interest
and
the
pressure
of
the
surrounding
atmosphere.
These
readings
are
known
as
gauge
Most pressure
gauges
give
readings which
are the difference
the pressure
of
interés y la presión
atmosférica
de los
alrededores.
Estas lecturas
se conocenbetween
como presiones
manométricas
pressures,
be converted
to absolute
pressures
by These
addition
of theEn
barometric
pressure.
interest
andand
theacan
pressure
ofabsolutas
the surrounding
atmosphere.
readings
are
as
gauge
y se pueden
convertir
presiones
al sumarles
la presión
barométrica.
losknown
cálculos
termodinámi­
Absolute
pressures
must
be
used
in
thermodynamic
calculations.
pressures,
andpresiones
can be converted
cos se deben
utilizar
absolutas.to absolute pressures by addition of the barometric pressure.
Absolute pressures must be used in thermodynamic calculations.
01­SmithVanNess.indd 7
8/1/07 12:47:23
CHAPTER 1. Introduction
8
8
CAPÍTULO 1. Introducción
Example
1.2
Ejemplo
1.2
A dead-weight gauge with a 1-cm-diameter piston is used to measure pressures very
Se utiliza una balanza de peso muerto con un pistón de 1 cm de diámetro para medir presiones con
accurately. In a particular instance a mass of 6.14 kg (including piston and pan)
mucha precisión. En un caso particular, una masa de 6.14 kg (incluyendo el pistón−2
y la bandeja) alcan­
brings it into balance. If the local acceleration of gravity is 9.82 m s , what is the
za el equilibrio. Si la aceleración local de la gravedad es 9.82 m s−, ¿cuál es la presión manométrica
gauge pressure being measured? If the barometric pressure is 748(torr), what is the
medida? Si la presión barométrica es 748(torr), ¿cuál es la presión absoluta?
absolute pressure?
Solución 1.2
Solution
1.2
La fuerza
ejercida por
la gravedad sobre el pistón, la bandeja y las pesas es:
The force exerted by gravity on the piston, pan, and weights is:
F = mg = (6.4)(9.8) = 60.95 N
F = mg = (6.14)(9.82) = 60.295 N
Gauge
pressure =
Presión
manométrica
60.295
F
= 76.77 N cm−2
=
A
(1/4)(π )(1)2
The
absoluteabsoluta
pressurees:
is therefore:
Por lo tanto
la presión
P = 76.77 + (748)(0.013332) = 86.74 N cm−2
P = 76.77 + (748)(0.0333) = 86.74 N cm–
P = 867.4 kPa
P = 867.4 kPa
or
o
Example 1.3
At 27◦ C the reading on a manometer filled with mercury is 60.5 cm. The local ac−2
celeration
Ejemplo
1.3 of gravity is 9.784 m s . To what pressure does this height of mercury
correspond?
A 27 °C la lectura en un manómetro lleno de mercurio es de 60.5 cm. La aceleración local de la grave­
dad es 9.784 m s −. ¿A qué presión le corresponde esta altura del mercurio?
Solution 1.3
Solución 1.3
Recall the equation in the preceding text, P = hρg. At 27◦ C the density of
is 13.53
cm−3anterior,
. Then, P = hρg. A 7 °C la densidad del mercurio es
Recuerde mercury
la ecuación
del gtexto
−3
3.53 g cm . Por esto,
P = 60.5 cm × 13.53 g cm−3 × 9.784 m s−2 = 8,009 g m s−2 cm−2
or
o
1.7
P = 60.5 cm × 3.53
g cm−3 × 9.784 m s− = 8 009 g m s− cm−
P = 8.009 kg m s−2 cm−2 = 8.009 N cm−2 = 80.09 kPa = 0.8009 bar
P = 8.009 kg m s− cm− = 8.009 N cm− = 80.09 kPa = 0.8009 bar
1.7 WORK
TRABAJO
Se realiza Work
trabajo
quewhenever
una fuerza
actúeacts
a través
de auna
distancia.
Por definición,
la cantidad
WW
is siempre
performed
a force
through
distance.
By definition,
the quantity
of de
trabajo está
dadaispor
la ecuación:
work
given
by the equation:
d W = F dl
(1.1)
dW = F dl
(.)
01­SmithVanNess.indd 8
8/1/07 12:47:24
1.7. Work
1.7. Work
Work
1.7.
1.7. Trabajo
9
99
9
where F is the component of force acting along the line of the displacement dl. When intedonde F
es la componente
de la fuerza
que acting
actúa aalong
lo largo
la of
línea
de
desplazamiento
dl.
Cuando
se inte­
where
the
component
of
force
along
thede
line
of
the
displacement
dl.
When
intewhere
isis the
component
force
the
line
displacement
dl.
integrated,FFthis
equation
yieldsof
the
workacting
of a finite
process.
Bythe
convention,
work
is When
regarded
as
gra, esta
ecuación
proporciona
el
trabajo
para
un
proceso
finito.
Por
convención,
el
trabajo
se
considera
grated,
this
equation
yields
the
work
of
a
finite
process.
By
convention,
work
is
regarded
as
grated,
equation
yields the work
of asame
finitedirection
process.asBy
work
regarded
as como
positivethis
when
the displacement
is in the
theconvention,
applied force
and is
negative
when
positivopositive
cuandowhen
el desplazamiento
está
en
la
misma
dirección
que
la
fuerza
aplicada,
y
negativo
cuando se
positive
when
the
displacement
is
in
the
same
direction
as
the
applied
force
and
negative
when
the displacement
they are in opposite
directions. is in the same direction as the applied force and negative when
encuentra
en
direcciones
opuestas.
they
are
in
opposite
directions.
they areThe
in opposite
directions.
work which
accompanies a change in volume of a fluid is often encountered in
Con frecuencia,
enwhich
termodinámica
el trabajo
se acompaña
cambio
en
el volumen
de un
The
work
which
accompanies
change
in volume
volumede
ofun
fluid isisof
often
encountered
inlíquido.
The
work
accompanies
change
in
of
aa fluid
often
encountered
in
thermodynamics. A common
exampleaais
the compression
or
expansion
a fluid
in a cylinder
Un ejemplo
común
es
la
compresión
o
expansión
de
un
fluido
en
un
cilindro,
resultado
del
movimiento
de un
thermodynamics.
A
common
example
is
the
compression
or
expansion
of
a
fluid
in
a
cylinder
thermodynamics.
common example
is the
compression
or expansion
of aonfluid
a cylinder
resulting from theAmovement
of a piston.
The
force exerted
by the piston
the in
fluid
is equal
pistón. resulting
La
fuerzafrom
ejercida
por
el
pistón
sobre
el
fluido
es
igual
al
producto
del
área
del
pistón
y
la
presión
del
resulting
from
the
movement
of
a
piston.
The
force
exerted
by
the
piston
on
the
fluid
is
equal
movement
of aand
piston.
The force
exerted
thedisplacement
piston on theoffluid
equalis
to the product the
of the
piston area
the pressure
of the
fluid.byThe
the is
piston
fluido. to
El
desplazamiento
del
pistón
es
igual
al
cambio
de
volumen
total
del
fluido
dividido
entre
el
área
to
the
product
of
the
piston
area
and
the
pressure
of
the
fluid.
The
displacement
of
the
piston
is
the product
of the
pistonchange
area and
of theby
fluid.
the piston
is
equal
to the total
volume
ofthe
thepressure
fluid divided
the The
area displacement
of the piston. ofEquation
(1.1)
del pistón.
Por
lo
tanto,
la
ecuación
(.)of
sethe
convierte
en: by
equal
to
the
total
volume
change
of
the
fluid
divided
by
the
area
of
the
piston.
Equation
(1.1)
equal
to
the
total
volume
change
fluid
divided
the
area
of
the
piston.
Equation
(1.1)
therefore becomes:
thereforebecomes:
becomes:
therefore
Vt
d W = −P A dVVt t
W=
=−P
−PAAdd A
ddW
AA
or, because A is constant,
d W = −P d V tt
(1.2)
or,abecause
because
constant,
W=
=−P
−PddVVt
(1.2) (.)
o, debido
que A es
or,
AAconstante,
isisconstant,
ddW
(1.2)
� t
� V2t
Integrating,
W = −� VV2t2 P d Vt tt
(1.3)
Integrando,
Integrating,
W=
=−
− V1t PPddVV
(1.3) (.3)
Integrating,
W
(1.3)
t
VV1t1
The“negativos”
minus signseninestas
theseecuaciones
equationssearehacen
madenecesarios
necessarypor
bylathe
sign convention
adopted
for para el
Los signos
convención
de signos
adoptada
The
minus
signs
in these
these
equations
arecylinder
made necessary
necessary
by the
the sign
sign
convention
adopted
for
The
minus
signs
in
equations
are
made
by
convention
adopted
for
work.
When
the
piston
moves
into
the
so
as
to
compress
the
fluid,
the
applied
force
trabajo. Cuando el pistón se mueve en el cilindro para comprimir el fluido, la fuerza aplicada y su desplaza­
work.
When
thepiston
piston
moves
into
the
cylinderthe
soas
as tocompress
compress
thepositive.
fluid,the
the
applied
force
the
into
cylinder
so
the
fluid,
applied
and
itsWhen
displacement
aremoves
in the
samethe
direction;
work
is
therefore
The
minus
sign menos
mientowork.
se encuentran
en la misma
dirección;
por
lo tanto,
el to
trabajo
es positivo.
Se requiere
el force
signo
and
its
displacement
are
in
the
same
direction;
the
work
is
therefore
positive.
The
minus
sign
and
its
displacement
are
in
the
same
direction;
the
work
is
therefore
positive.
The
minus
sign
is
required
because
the
volume
change
is
negative.
For
an
expansion
process,
the
applied
force
porque el cambio de volumen es negativo. Para un proceso de expansión, la fuerza aplicada y su desplaza­
isrequired
requiredbecause
becausethe
the
volume
change
negative.
For
an expansion
process,
theisapplied
appliedforce
force
volume
change
isisnegative.
an
process,
the
are
in opposite
directions.
TheFor
volume
change
in
this
case
andel signo
mientoisand
estánitsendisplacement
direcciones
opuestas.
El cambio
de volumen
en expansion
este
caso es
positivo,
y positive,
se requiere
and
its
displacement
are
in
opposite
directions.
The
volume
change
in
this
case
is
positive,
and
and
its
displacement
are
in
opposite
directions.
The
volume
change
in
this
case
is
positive,
and
the
minus
sign
is
required
to
make
the
work
negative.
menos para hacer negativo el trabajo.
theminus
minussign
signisisrequired
requiredto
tomake
makethe
thework
worknegative.
negative.
the
P
P2
PP22
2
22
P
PP
P
P
0
P1
PP10
1
t 00
V
t path.
Figura 1.3: Diagrama
que muestra
trayectoria
función de V t .
Figure 1.3:
Diagramlashowing
a P de
vs.PVen
Figure1.3:
1.3:Diagram
Diagramshowing
showingaa PPvs.
vs.VVt t path.
path.
Figure
V2tt
VV2t2
V
t
t
Vt
t
1
11
Vt
V1tt
VV1t1
VV
7
Equation (1.3) expresses the work done by a finite compression or expansion process.
777La figu­
Equation
(1.3)
expresses
the
work
done
by
a
finite
compression
or
expansion
process.
LaFigure
ecuación
(.3)
expresa
el
trabajo
hecho
por
un
proceso
de
compresión
o
expansión
finitas.
t
Equation
(1.3)
expresses
the
work
done
by
a
finite
compression
or
expansion
process.
1.3 shows a path for compression of a gas from point 1 with initial volume t Vt1t at
Figure
1.3Pshows
shows aa path
path
forla
compression
of un
gas P
from
point 1con
1 with
with
initialinicial
volume
ra .3 muestra
una
trayectoria
compresión
del
punto
volumen
V VVa presión
P
Figure
1.3
for
compression
of
aa gas
from
point
initial
volume
atat
pressure
2 para
with
volume
Vt2tt at de
pressure
1 to point
2 . This path relates the pressure at11any
t 2
pressure
P
to
point
with
volume
V
at
pressure
P
.
This
path
relates
the
pressure
at
any
al punto
con
volumen
V
a
presión
P
.
Esta
trayectoria
relaciona
la
presión
en
cualquier
punto
del
proceso
1
2
pressure
P
to
point
2
with
volume
V
at
pressure
P
.
This
path
relates
the
pressure
at
any
1
2
2
point of the process to the volume. The
2 work required is given by Eq. (1.3) and is proportional
point
ofarea
the
processthe
tothe
thevolume.
volume.
The
work
required
isofgiven
given
by
Eq.
(1.3)
and
isproportional
proportional
con el volumen.
El trabajo
requerido
está
dado
por
la
ecuación
(.3)
y esby
proporcional
al is
área
bajo
curva de la
point
the
process
to
The
work
required
and
to theof
under
curve
of Fig.
1.3.
The
SI unitis
work
isEq.
the(1.3)
newton-meter
or la
joule,
to
the
area
under
the
curve
of
Fig.
1.3.
The
SI
unit
of
work
is
the
newton-meter
or(ft
joule,
figura .3.
La area
unidad
delEnglish
trabajo
delof
SI Fig.
es el1.3.
newton­metro
o often
joule,
cuyois
es J. Enforce
el sistema
de
tosymbol
the
the curve
Thethe
SIunit
unit
of work
the foot-pound
newton-meter
or
joule,
J. Inunder
the
engineering
system
used
issímbolo
the
lbfinglés
).
symbol
Inthe
the
English
engineering
system
theunit
unitoften
oftenused
used
thefoot-pound
foot-poundforce
force(ft
(ftlb
lbff).).
ingeniería
la unidad
pie­libra
fuerza
(pie lbf)system
se emplea
con
mucha
frecuencia.
symbol
J.J.In
English
engineering
the
isisthe
7
7 However, as explained in Sec. 2.8, it may be applied only in special circumstances.
77However,
However,asasexplained
explainedininSec.
Sec.2.8,
2.8,ititmay
maybe
beapplied
appliedonly
onlyininspecial
specialcircumstances.
circumstances.
Sin embargo, como se explica en la sección .8, sólo se aplica en circunstancias especiales.
01­SmithVanNess.indd 9
8/1/07 12:47:27
0
1.8
10
10
10
CAPÍTULO 1. Introducción
CHAPTER 1. Introduction
CHAPTER 1.
1. Introduction
Introduction
CHAPTER
ENERGÍA
1.8
ENERGY
1.8 ENERGY
ENERGY
1.8
El principio general de conservación de energía se estableció alrededor de 850. El germen de este principio,
of
conservation
of
energy
established
about
1850.
The
germ
of
this
tal comoThe
se general
aplica aprinciple
la mecánica,
está implícito
en los was
trabajos
de Galileo
(564­64)
y de Isaac
The
general
principle
of conservation
conservation
of energy
energy
was
established
about
1850. The
The germ
germ
of this
thisNewton
The
general
principle
of
of
was
established
about
1850.
of
principle
as
it
applies
to
mechanics
was
implicit
in
the
work
of
Galileo
(1564–1642)
and
Isaac
(64­76).
De as
hecho,
es una
consecuencia
directa
de in
la the
segunda
ley
de Newton
sobre el movimiento,
una
principle
it
applies
to
mechanics
was
implicit
work
of
Galileo
(1564–1642)
and
Isaac
asesitdefinido
applies to
mechanics
was
in yfrom
the
of Galileo
(1564–1642)
and Isaac
Newton
(1642–1726).
Indeed,
it
follows
Newton’s
second
law
of
motion
once
vez queprinciple
el trabajo
como
producto
deimplicit
ladirectly
fuerza
el work
desplazamiento.
Newton
(1642–1726).
Indeed,
it
follows
directly
from
Newton’s
second
law
of
motion
once
Newton
(1642–1726).
Indeed,ofit force
follows
directly
from Newton’s second law of motion once
work
is
defined
as
the
product
and
displacement.
work is
is defined
defined as
as the
the product
product of
of force
force and
and displacement.
displacement.
work
Energía
cinética
Kinetic
Energy
Kinetic Energy
Energy
Kinetic
a body
mass
m,
acted
upon
by
a force
F,
is
a distance
dl
during
a differential
CuandoWhen
un cuerpo
deof
masa
el que
actúa
fuerza
F,displaced
se desplaza
una distancia
dl, durante
un intervalo
When
body
of
massm,m,
m,enacted
acted
upon
byuna
force
F, is
is
displaced
distance
dl during
during
differential
When
aatiempo
body
of
mass
upon
by
aa force
F,
displaced
aa distance
dl
aa differential
interval
of
time
dt,
the
work
done
is
given
by
Eq.
(1.1).
In
combination
with
Newton’s
second
diferencial
de
dt,
el
trabajo
que
se
realiza
está
dado
por
la
ecuación
(.).
De
manera
conjunta
interval of
of time
time dt,
dt, the
the work
work done
done is
is given
given by
by Eq.
Eq. (1.1).
(1.1). In
In combination
combination with
with Newton’s
Newton’s second
second con la
law
this
equation
becomes:
segundainterval
ley
de
Newton,
esta
ecuación
se
convierte
en:
law this
this equation
equation becomes:
becomes:
law
dW
=
ma
dl
W ==
= ma
ma dl
dl
dW
dl
dd W
ma
By
definition
the
acceleration
is
a
≡
du/dt,
where
u is
the
velocity
of
body.
Thus,
By
definition
the
acceleration
is
a
≡
du/dt,
where
is the
the velocity
velocity of
of the
the body.
body. Thus,
Thus,
By definition
the acceleration
is a ≡donde
du/dt,
where
uu is
Por definición,
la aceleración
es a ≡ du/dt,
u es
la velocidad
del cuerpo.the
De este modo,
dl
du
dl du
du dl = m dl
d
W
=
m
du
dl
=
m
du
d
W
=
m
dt
dl
=
m
d W = m dt
dt
dt du
dt
dt
Because
the
definition
of
velocity
is
≡
dl/dt,
the
expression
for
work
becomes:
Because
the definition
definition
of velocity
velocityesis
isuuu
u≡≡
≡dl/dt,
dl/dt,
the
expression
for
work becomes:
becomes:
Puesto que
la definición
de la velocidad
la the
expresión
para for
el trabajo
es:
Because
the
of
dl/dt,
expression
work
dW
=
mu
du
W ==
= mu
mu du
du
dW
dd W
mu
This
equation
may
now
be
integrated
for
a finite
change
in
velocity
from
u to
u :
This equation
equation
may
nowintegrar
be integrated
integrated
for
finitefinito
change
invelocidad
velocity from
from
to
Esta ecuación
ahora semay
puede
para unfor
cambio
en la
de uuu111a to
u:uu222::
This
now
be
aa finite
change
in
velocity
�
�
�
� u2
�
� uu 22
�
uu 221 �
22
�
u
2
2
u 2 u du = m u 2 − u 1
W
=
m
du =
=m
m 22 −
− 21
W=
=m
m u uu du
W
u1
22
22
u 11
�
�
� 2�
�
2
2
�
mu
mu
mu
2
2
2
2
1
mu
mu
2
2
mu
2
or
W
=
−
=
�
(1.4)
2
1
o
2 − mu
or
W=
= mu
=�
� mu
(1.4) (.4)
2
22
or
W
− 22 1 =
(1.4)
2
2
2
2
1
2 in Eq. (1.4) is a kinetic energy, a term introduced by Lord
Each
of
the
quantities
1 mu
2 in Eq. (1.4) is a kinetic energy, a term introduced by Lord
en
Each
of the
the
quantities
122 mu
2 in
Cada una
de las
cantidades
−¹ mu
la Eq.
ecuación
es unaenergy,
energíaa cinética,
un término
introducido
mu
(1.4) is(.4)
a kinetic
term introduced
by Lord
Each
of
quantities
8
2
²
Kelvin
in
1856.
Thus,
by
definition,
8
Kelvin
inen1856.
1856.
Thus,
by definition,
definition,
por lordKelvin
Kelvin8 8in
856.Thus,
En estos
términos, por definición,
by
1 2
(1.5)
E
K ≡
11 mu
≡
(1.5) (.5)
mu22
E
K
(1.5)
E K ≡ 22 mu
2
La ecuación
(l.4) muestra
que elthat
trabajo
hechodone
sobreonuna cuerpo
que
se acelera aitpartir
de
una
velocidad
inicial u Equation
(1.4)
shows
the
work
body
in
accelerating
from
an
initial
velocity
Equation
(1.4)
shows
that
the
work
done
on
a
body
in accelerating
accelerating
itAfrom
from
an initial
initial
velocity
hasta una
velocidad
final
u
es
igual
al
cambio
de
la
energía
cinética
del
cuerpo.
la
inversa,
si
el
cuerpo
en movi­
Equation
(1.4)
shows
that
the
work
done
on
a
body
in
it
an
velocity
u 1 to
final
velocity
u
is
equal
to
the
change
in
kinetic
energy
of
the
body.
Conversely,
if
to aaa final
final por
velocity
u 22 is
is
equal
to the
theque
change
in kinetic
kinetic
energy
ofpor
theelbody.
body.
Conversely,
ifcambio
miento se
desacelera
la
acción
de
una
fuerza
se
opone,
el
trabajo
hecho
cuerpo
es
igual
a
su
uau11moving
to
velocity
u
equal
to
change
in
energy
of
the
Conversely,
if
2
body
is
decelerated
by
the
action
of
aa resisting
force,
the
work
done
by
the
body
is
a moving
moving
body
issistema
decelerated
by the
the action
action
of
resisting
force,
the work
workendone
done
by, the
the
body is
iscinética
en energía
cinética.
En
elis
de unidades
SI
con
laSIamasa
en kg
yunits
la velocidad
m kg
s −
la energía
aequal
body
decelerated
by
of
resisting
force,
the
by
body
to
its
change
in
kinetic
energy.
In
the
system
of
with
mass
in
and
velocity
−
−
equal
to
its
change
in
kinetic
energy.
In
the
SI
system
of
units
with
mass
in
kg
and
velocity
E K tiene
unidades
de
kgenergy
min kinetic
s E. Puesto
que
newton
la of
unidad
kgiskg
mthe
s composite
, velocity
EK se mide en
−1 ,its
2ess−2
equal
to
change
energy.
Inelthe
SI m
system
units compuesta
with
mass in
and
in
m
has
the
units
of
kg
the
newton
−1 , kinetic
2 s−2 .. Because
in m
m sss−1
kinetic
energy
EK
has la
theecuación
units of
of(.4),
kg m
m2ésta
Because
the
newton
is the
the composite
composite
−2es
K con
newton­metros
o
joules.
De
acuerdo
la
unidad
del
trabajo.
−2
in
,
kinetic
energy
E
has
the
units
kg
s
.
Because
the
newton
is
K
unit
kg
m
is measured
in newton-meters or joules. In accord with Eq. (1.4), this is the
−2 ,, E
unit kg
kg m
m sss−2
EK
K is measured in newton-meters or joules. In accord with Eq. (1.4), this is the
unit
,E
K is measured in newton-meters or joules. In accord with Eq. (1.4), this is the
unit
of
work.
unit of
of work.
work.
unit
8 Lord Kelvin
8 LordoKelvin,
William Thomson
(84­907)
fue un físico
que, junto
con who,
el físico
alemán
Rudolf
Clausius
(8­888),
William Thomson
(1824–1907),
wasinglés
an English
physicist
along
with the
German
physicist
8 Lord Kelvin, or
or
William
Thomson
(1824–1907),
was an English physicist who, along with the German physicist
establecióRudolf
los8 Lord
fundamentos
para
la ciencia
moderna
de la termodinámica.
Kelvin,(1822–1888),
or
William
Thomson
(1824–1907),
English
physicist
who, along with the German physicist
Clausius
laid
the foundations
forwas
the an
modern
science
of thermodynamics.
Rudolf Clausius (1822–1888), laid the foundations for the modern science of thermodynamics.
Rudolf Clausius (1822–1888), laid the foundations for the modern science of thermodynamics.
01­SmithVanNess.indd 10
8/1/07 12:47:30
1.8. Energy
Energy
1.8. Energía
1.8.
1.8. Energy
11
11
11
2 /g , where
En el sistema
de engineering
ingeniería, lasystem,
energíakinetic
cinéticaenergy
se expresa
como −¹ mu
gc tiene
In the
the inglés
English
is expressed
expressed
as 2112/g
mu
gc el valor
c ,2donde
c
In
English
engineering −
system,
kinetic energy
is
as
mu
²
c , where gc
1
−
2
2/g
−1
−2
3.740has
y las
unidades
(lb
)(pie)(lb
)
(s)
.
De
este
modo,
la
unidad
de
la
energía
cinética
en
este
siste­
In
the
English
engineering
system,
kinetic
energy
is
expressed
as
mu
/g
,
where
m and the funits (lbm )(ft)(lbf )−1 (s)−2 . Thus the unit of
the value
value 32.1740
32.1740
kinetic
c energy gin
2 kinetic
has the
and the units (lbm )(ft)(lbf ) −1(s) −2. Thus the unit of
energy inc
ma es: has
this system
system
is:32.1740 and the units (lbm )(ft)(lbf ) (s) . Thus the unit of kinetic energy in
the
value
this
is:
−2
this system is:
)(ft)22(s)
(s)−2
mu22
(lbm)(ft)
mu
(lb
m
E
=
=
= (ft lb )
2
K
2
−1−2
−2 = (ft lbff)
E K = mu
(lb
m )(ft)f )(s)
2gc = (lbm
)(ft)(lb
(s)
−1
−2
2g
E K = c = (lbm )(ft)(lbf ) −1(s) −2 = (ft lbf )
2gc la inclusión
(lbminclusion
)(ft)(lb
Dimensional
consistency
here
requires
the
of
Aquí, laDimensional
consistenciaconsistency
dimensionalhere
requiere
defg) cof
. (s)
requires
the
inclusion
ggcc..
Dimensional consistency here requires the inclusion of gc .
Potential
Energy
Energía
potencial
Potential
Energy
Potential Energy
If aa body
body of
of mass
mass m
m is
is raised
raised from
from an initial
initial elevation
elevation zz 1 to
to aa final
final elevation
elevation zz 2,, an
an upward
upward
If
Si un cuerpo
de masa
m se mueve
desde unaanaltura
inicial z a una
final z, se debe
una fuerza
1 altura
2 ejercer
force
at
least
equal
to
the
weight
of
the
body
must
be
exerted
on
it,
and
this
force
must
move
If
a
body
of
mass
m
is
raised
from
an
initial
elevation
z
to
a
final
elevation
z
,
an
upward
1
forcealatmenos
least equal
of the ybody
bedebe
exerted
on it, aand
this force
move
ascendente
igual to
al the
pesoweight
del cuerpo,
esta must
fuerza
moverse
lo largo
de2lamust
distancia
z − z.
through
the distance
distance
− weight
z 1.. Because
Because
the
weight
ofbethe
the
body is
is the
forcethis
of gravity
gravity
on it,
it,
the force at the
least
equal tozz 2the
of thethe
body
mustof
exerted
on
it, force
and
force
must
move
through
weight
of
on
the
Puesto que
el peso del
cuerpo 2es−lazfuerza
de la gravedad
sobre
el body
mismo, the
la fuerza
mínima
requerida
está dada
1
minimum
force
required
iszgiven
given
by Newton’s
Newton’s
law:of the body is the force of gravity on it, the
through the
distance
z 2 −is
the weight
1 . Because
minimum
force
required
by
law:
por la ley
de Newton:
minimum force required is given by Newton’s law:
F=
= ma = mg
F
F = ma
ma = mg
F = ma = mg
where
g
is
the
local
acceleration
of
gravity.
The minimum
minimum work
work required
required to
to raise
raise the
the body
body is
is
is the locallocal
acceleration
of gravity.
The
donde gwhere
es product
la gaceleración
de
lathe
gravedad.
El elevation:
trabajo
mínimo requerido
para to
elevar
althe
cuerpo
es
the
of
this
force
and
change
in
where
g
is
the
local
acceleration
of
gravity.
The
minimum
work
required
raise
body
is el pro­
theesta
product
ofpor
thiselforce
anden
thelachange
in elevation:
ducto de
fuerza
cambio
elevación:
the product of this force and the change in elevation:
− z ) = mg(z
mg(z 2 −
− z1)
W=
= F(z 2 −
W
W =F(z
F(z2 − zz11) = mg(z
2− zz1))
W = F(z 2 − z 1 ) = mg(z 2 − z 1 )
− mz
mz 1gg =
= �(mzg)
�(mzg)
(1.6)
or
W=
= mz
mz 2gg −
(1.6)
or
W
2
1
o
(1.6) (.6)
or
W = mz 2 g − mz 1 g = �(mzg)
We
see
from
Eq.
(1.6)
that
work
done
on
a
body
in
raising
it
is
equal
to
the
change
in
the
We see from
Eq.
(1.6)que
thatelwork
done
on asobre
body el
incuerpo,
raising it
islevantarlo,
equal to the
change
the
En
la ecuación
(.6)
vemos
trabajo
hecho
al
esits
igual
al in
cambio
quantity
mzg.
Conversely,
if
a
body
is
lowered
against
a
resisting
force
equal
to
weight,
We
see
from
Eq.
(1.6)
that
work
done
on
a
body
in
raising
it
is
equal
to
the
change
in
the en la
quantity
mzg.
Conversely,
ifsiaelbody
is lowered
against
a resisting
force
equalcontrario,
to its weight,
the
cantidad
mzg.
De
manera
inversa,
cuerpo
desciende
con
una
fuerza
en
sentido
pero
igual a su
work
done
by the
the
body is
is equal
equal
to the
the
change
inagainst
the quantity
quantity
mzg.force
Eachequal
of the
thetoquantities
quantities
mzg
quantity
mzg.
Conversely,
if a body
ischange
loweredin
a resisting
its weight,mzg
the
done
by
body
to
the
mzg.
Each
of
9
peso, elwork
trabajo
hecho
por
el
cuerpo
es
igual
al
cambio
en
la
cantidad
mzg.
Cada
una
de
las
cantidades
in
Eq.done
(1.6)by
is athe
a potential
potential
energy.
Thus,
by definition,
definition,
work
body isenergy.
equal to
the
change
in the quantity mzg. Each of the quantities mzg mzg
9 Thus,
in
Eq.
(1.6)
is
by
9 De
9
en la ecuación
(.6)
es
una
energía
potencial.
esta
forma, por definición,
in Eq. (1.6) is a potential energy. Thus, by definition,
≡
mzg
(1.7)
E
P
(1.7)
E P ≡ mzg
≡ mzg
mzg
(1.7) (.7)
EEPP ≡
In
the
SI
system
of
units
with
mass
in
kg,
elevation
in
m,
and
the
acceleration
of
gravity
in
In the
system of units with mass in kg, elevation
in m, and the acceleration of gravity in
−2 ,SI
2 s−2 . This
En el sistema
de
unidades
SIunits
con
la the
masa
en in
kg,kg,
la m
elevación
en m,
m,
y lanewton-meter
aceleración
de
lajoule,
gravedad
en
potential
energy
has
units
of
kg
is
the
or
the
unit
m
s
In
the
SI
system
of
with
mass
elevation
in
and
the
acceleration
of
gravity
in m s−,
−2
2
−2
kg m 2 s −2. This is the newton-meter or joule, the unit
m s −2, potential energy has the units of−
la energía
potencial
tiene
unidades
de kg
m of
s kg
. Éste
newton­metro
o joule, laorunidad
del unit
trabajo, de
of work,
in agreement
agreement
with
Eq.
(1.6).
, potential
energy
hasEq.
the
units
m ses .elThis
is the newton-meter
joule, the
m
swork,
of
in
with
(1.6).
acuerdoofcon
la
ecuación
(.6).
In the
the
English engineering
engineering
system, potential
potential energy
energy is
is expressed
expressed as
as mzg/g
mzg/gc.. Thus
Thus the
the
work,
in agreement
with Eq. (1.6).
In
English
system,
c
En
el of
sistema
inglés
deengineering
ingeniería,
la energía
potencial
se expresa
como mzg/g
la unidad
unit
potential
energy
in
this
system
is:
In
the
English
system,
potential
energy
is
expressed
as
mzg/g
.
Thus
the de la
c . Así,
c
unit of potential energy in this system is:
energía unit
potencial
en
este
sistema
es:
of potential energy in this system is:
−2
mzg
(lbm)(ft)(ft)(s)
)(ft)(ft)(s)−2
mzg
(lb
m
E
=
=
= (ft
(ft lb
lbf))
P
−2−2 =
−1
E P = mzg
f
m )(ft)(ft)(s)
g = (lb(lb
)(ft)(lb
(s)−2
m)(ft)(lb
f))−1 (s)
E P = gcc = (lb
=
(ft
lbf )
m
f
gc
(lbm )(ft)(lbf )−1 (s)−2
Again,
must
be
included
for dimensional
dimensional
consistency.
De nueva
cuenta,
debe be
incluirse
gc for
para
asegurar la consistency.
consistencia dimensional.
Again,
ggcc must
included
Again, gc must be included for dimensional consistency.
Energy Conservation
Conservation
Conservación
de la energía
Energy
Energy Conservation
In any
any examination
examination of
of physical
physical processes,
processes, an
an attempt
attempt is
is made
made to
to find
find or
or to
to define
define quantities
quantities
In
En cualquier
examen
de procesos
físicos of
se the
intenta
encontrar
o occur.
definirOne
cantidades
que permanezcan
which
remain
constant
regardless
changes
which
such
quantity,
early
recog-constan­
In
any
examination
of
physical
processes,
an
attempt
is
made
to
find
or
to
define
quantities
remain
constant regardless
of the
changes
which
occur. One
such quantity,
early
recogtes, sinwhich
importar
los
cambios
que
ocurran.
Una
de
estas
cantidades,
reconocida
al
principio
del
desarrollo
nized
in
the development
development
of mechanics,
mechanics,
is mass.which
The great
great
utility
of
thequantity,
law of
of conservation
conservation
whichin
remain
constant regardless
of the changes
occur.
One of
such
early recognized
the
of
utility
the
law
de la mecánica,
es la masa. La gran
utilidad deislamass.
ley deThe
conservación
de masa
sugiere
que existen otros
nized in the development of mechanics, is mass. The great utility of the law of conservation
9
9 This term was proposed in 1853 by the Scottish engineer William Rankine (1820–1872).
9 This
termpropuesto
was proposed
in 1853
byingeniero
the Scottish
engineer
William
Rankine
(1820–1872).
Este término
en 853
por el
escocés
William
Rankine
(80­87).
9 This fue
term was proposed in 1853 by the Scottish engineer William Rankine (1820–1872).
01­SmithVanNess.indd 11
8/1/07 12:47:32
CHAPTER 1. Introduction
12
CAPÍTULO 1. Introducción
of mass suggests that other conservation principles could be of comparable value. With reprincipios de conservación que podrían tener un valor comparable. Con respecto a la energía observamos que
spect to energy we observe that Eqs. (1.4) and (1.6) both show that work done on a body is
las ecuaciones (.4) y (.6) muestran que el trabajo realizado sobre un cuerpo es igual al cambio en una can­
equal to the change in a quantity which describes the condition of the body in relation to its
tidad que describe la condición del cuerpo en relación con sus alrededores. En cada caso, el trabajo efectuado
surroundings. In each case the work performed can be recovered by carrying out the reverse
puede recuperarse realizando el proceso inverso y regresando el cuerpo a su condición inicial. Esta observa­
process and returning the body to its initial condition. This observation leads naturally to the
ción conduce naturalmente a la idea de que, si el trabajo se efectúa sobre un cuerpo al acelerarlo o elevarlo,
thought that, if the work done on a body in accelerating it or in elevating it can be subsequently
éste se puede recuperar posteriormente; de esta manera, el cuerpo tiene la habilidad o capacidad de realizar
recovered, then the body by virtue of its velocity or elevation contains the ability or capacity
trabajo en virtud de su velocidad o elevación. Este concepto ha demostrado ser tan útil en la mecánica del
to do the work. This concept proved so useful in rigid-body mechanics that the capacity of a
cuerpo rígido, que a la capacidad que tiene un cuerpo para realizar trabajo se le dio el nombre de energía,
body for doing work was given the name energy, a word derived from the Greek and meaning
palabra derivada del griego, que significa “en trabajo”. Por lo tanto, se dice que el trabajo para acelerar un
“in work.” Hence the work of accelerating a body produces a change in its kinetic energy:
cuerpo produce un cambio en su energía cinética:
�
�
mu 22
W = �E KK = �
2
y el trabajo
realizado
para on
elevar
un cuerpo
produce
un cambio
en su energía
potencial:
and the
work done
a body
in elevating
it produces
a change
in its potential
energy:
W
W=
=E
EPPP =
= Δ(mzg)
�(mzg)
Si un cuerpo
recibe
cuandowhen
se eleva,
retiene
energía
hasta or
queretains
realizathis
el trabajo
If a body
is energía
given energy
it is conserva
elevated, othen
the esa
body
conserves
del cualenergy
es capaz.
Cuando
un cuerpo
se eleva
después
le permite
libremente,
en energía
until
it performs
the work
of ywhich
it issecapable.
Ancaer
elevated
body, gana
allowed
to fall cinéti­
ca lo que
pierde
en energía
potencial,
de manera
que
capacidad
parasorealizar
no for
cambia.
freely,
gains
in kinetic
energy what
it loses
in su
potential
energy
that itstrabajo
capacity
doingPara un
cuerpo que
libremente
esto significa
que: falling body this means that:
workcae
remains
unchanged.
For a freely
ΔEKK +
+�E
E P P==00
�E
K
P
o
or
mu 222 mu 212
2 −
1 + mz g − mz g = 0
22
11
2
2
La validez
esta ecuación
fue confirmada
innumerables
experimentos.
De este
modo,
desarrollo del
Thede
validity
of this equation
has beenpor
confirmed
by countless
experiments.
Thus
the el
developconcepto
de energía
condujoof
deenergy
maneraled
lógica
al principio
de conservación
de energía
para todosfor
losall
procesos
ment
of the concept
logically
to the the
principle of energy
conservation
estrictamente
amplia evidencia
experimentalevidence
que justifica
esta generalización
fue obtenida
sin
purely mecánicos.
mechanicalLa
processes.
Ample experimental
to justify
this generalization
was
dificultad.
readily obtained.
Son posibles
de energía
mecánica,
la gravitational
energía cinética
y de laenergy
potencial
Other otras
formsformas
of mechanical
energy
besidesademás
kineticde
and
potential
are gravi­
tacional.possible.
La más evidente
la energía
potencial
de configuración.
Cuando
se acomprime
un resorte, el traba­
The mostes
obvious
is potential
energy
of configuration.
When
spring is compressed,
jo es hecho
una fuerza
el resorte
puede realizar
tarde this
cierto
trabajo
contra una
workpor
is done
by anexterna.
externalPuesto
force. que
Because
the spring
can latermás
perform
work
against
fuerza de
resistencia,
posee
capacidad
para efectuarlo.
Ésta
es lawork.
energíaThis
potencial
de configuración.
a resisting
force,
thelaspring
possesses
capacity for
doing
is potential
energy of Exis­
te energía
del mismo Energy
tipo tanto
en una
estirada
en una barra
metal
deformada
en la región
configuration.
of the
sameliga
form
exists como
in a stretched
rubberdeband
or in
a bar of metal
elástica.deformed in the elastic region.
La generalidad
del principio
de conservación
de la energía
en mecánica
aumenta
si consideramos
al
The generality
of the principle
of conservation
of energy
in mechanics
is increased
if we
trabajo mismo
como
unaitself
forma
energía.
Esto esThis
claramente
ya que
los cambios
tantoand
en la ener­
look upon
work
as adeform
of energy.
is clearlyaceptable,
permissible,
because
both kineticgía cinética
como en la changes
energía potencial
iguales
al trabajo
hecho para
producirlos
potential-energy
are equal son
to the
work done
in producing
them
[Eqs. (1.4)[ecuaciones
and (1.6)]. (l.4) y
(.6)]. Sin
embargo,
el trabajo
es energía
enand
tránsito
y nunca
se considera
como
en unwork
cuerpo.
However,
work
is energy
in transit
is never
regarded
as residing
in residente
a body. When
is Cuan­
do se realiza
trabajo
no aparece
simultáneamente
trabajo en otra
se convierte
en otra
forma de
done and
doesy not
appear simultaneously
as como
work elsewhere,
it is parte,
converted
into another
form
energía.of energy.
Al agregado
o conjunto
sobre el que
se concentra
la atención
seislecalled
nombra
todo
lo demás
se
The body
or assemblage
on which
attention
is focused
thesistema.
system.AAll
else
is
le llamacalled
alrededores.
Cuando se When
realizawork
trabajo,
éste esit hecho
sobre
el sistema,
the surroundings.
is done,
is donepor
bylos
thealrededores
surroundings
on the
system, oorvicever­
sa; en tanto,
la energía
se transfiere
de los alrededores
al sistema, otodethe
manera
inversa.
formaItdeisenergía
vice versa,
and energy
is transferred
from the surroundings
system,
or the La
reverse.
conocida como trabajo sólo existe durante esta transferencia. En contraste, las energías cinética y potencial
01­SmithVanNess.indd 12
8/1/07 12:47:34
1.8. Energy
13
only during this transfer that the form of energy known as work exists. In contrast, kinetic and
1.8. Energía
3
potential energy reside with the system. Their values, however, are measured with reference
to the
surroundings;
i.e., kinetic
energy depends
velocity
with
respect
to the surroundings,
residen en
el sistema.
Sin embargo,
sus valores
se midenon
con
respecto
a los
alrededores;
es decir, la energía
and
potential
energy
depends
on
elevation
with
respect
to
a
datum
level.
Changes
in kinetic
cinética depende de la velocidad con respecto a los alrededores, y la energía potencial
depende
de la altura
and
potential
energy
do
not
depend
on
these
reference
conditions,
provided
they
are
fixed.
con respecto a un nivel de referencia. Los cambios en las energías cinética y potencial, una vez fijados, no
dependen de estas condiciones de referencia.
Example 1.4
An elevator
Ejemplo
1.4 with a mass of 2,500 kg rests at a level 10 m above the base of an elevator shaft. It is raised to 100 m above the base of the shaft, where the cable holding it
Un elevador
con The
una masa
de falls
2 500freely
kg descansa
en un
a 10 m
sobre
la base
del pozo
de un ele­
breaks.
elevator
to the base
ofnivel
the shaft
and
strikes
a strong
spring.
vador. Cuando
se
eleva
a
100
m
sobre
la
base
del
pozo,
se
rompe
el
cable
que
lo
sostiene.
El
elevador
The spring is designed to bring the elevator to rest and, by means of a catch arrangecae libremente
hacia
la
base
del
pozo
y
golpea
contra
un
fuerte
resorte.
El
resorte
está
diseñado
ment, to hold the elevator at the position of maximum spring compression. Assuming para
poner althe
elevador
en reposo
medio deand
un diseño
entire process
to y,
bepor
frictionless,
taking gde= retención,
9.8 m s−2 ,mantener
calculate:al elevador en la po­
sición de máxima compresión del resorte. Si se supone que todo el proceso es sin fricción, y tomando
− , calcule:
The potential energy of the elevator in its initial position relative to the base of
g = 9.8 m s(a)
the shaft.
a) La energía potencial del elevador en su posición inicial con respecto a la base del pozo.
(b) The work done in raising the elevator.
b) El trabajo hecho al subir el elevador.
(c) The
potentialdelenergy
of the
elevator
in its
highest
position
relative
to the
base
c) La energía
potencial
elevador
en su
posición
más
alta con
respecto
a la base
del
pozo.
of the shaft.
d) La velocidad y la energía cinética del elevador justo antes de que golpee en el resorte.
(d) The velocity and kinetic energy of the elevator just before it strikes the spring.
e) La energía potencial del resorte comprimido.
(e) The
energy
of the compressed
f) La energía
delpotential
sistema que
se compone
del elevadorspring.
y el resorte: 1) al inicio del proceso, 2) cuan­
do el(felevador
alcanza
su
altura
máxima,
3)
justo
antes
de queand
el elevador
incida
enstart
el resorte,
) The energy of the system consisting of the elevator
spring (1)
at the
4) después
de
que
el
elevador
ha
llegado
al
reposo.
of the process, (2) when the elevator reaches its maximum height, (3) just
before the elevator strikes the spring, (4) after the elevator has come to rest.
Solución 1.4
Suponga
que el subíndice
Solution
1.4 denota las condiciones iniciales; el subíndice , las condiciones cuan­
do el elevador está en su posición más alta, y el subíndice 3, las condiciones justo antes de que el
subscript
1 designate the initial conditions; subscript 2, conditions when the
elevadorLet
llegue
al resorte.
elevator is at its highest position; and subscript 3, conditions just before the elevator strikes the spring.
a) Por la ecuación (.7),
E P = mz g = ( 500)(0)(9.8) = 45 000 J
(a) By Eq. (1.7), E P1 = mz 1 g = (2,500)(10)(9.8) = 245,000 J
� z2
� z2
By Eq. (.),
(1.1),
W =
F dl =
mg dl = mg(z 2 − z 1 )
b) Por la(b)
ecuación
z1
whence
de donde
(c) By Eq. (1.7),
W = (2,500)(9.8)(100 − 10) = 2,205,000 J
W = ( 500)(9.8)(00 − 0) = 05 000 J
E P2 = mz 2 g = (2,500)(100)(9.8) = 2,450,000 J
Note that
c) Por la ecuación
(.7),
P2 −
1 . 500)(00)(9.8) = 450 000 J
EWP2==Emz
gE
=P(
Observe que
01­SmithVanNess.indd 13
z1
W = E P − E P .
8/1/07 12:47:35
4
14
14
14
14
CHAPTER
CHAPTER
1.1. Introduction
Introduction
CAPÍTULO
1. Introducción
CHAPTER
1.
Introduction
CHAPTER
1. Introduction
CHAPTER 1. Introduction
14
d) Del principio
de the
conservación
de
energía
mecánica
se puede energy,
escribir
que
lamay
suma
dethat
los
(d)
(d)
From
From
the
principle
principle
of
of
conservation
conservation
of
of
mechanical
mechanical
energy,
one
onemay
write
write
thatcam­
(d) From(d)the
principle
of conservation
of mechanical
energy, energy,
one mayone
write
that
From
the principle
of conservation
of mechanical
may
write that
bios de lasthe
energías
cinética
y potencial
durante los procesos
desde
lasthe
condiciones
a con3cones cero;
the
sum
sum
of
of
the
the
kinetickineticand
and
potential-energy
potential-energy
changes
changes
during
during
the
process
process
from
from
the sum the
of the
kineticand potential-energy
changes
during the
process
from
consum
ofthe
theprinciple
kineticand
potential-energy
changes
during
theone
process
from
From
of
conservation
of mechanical
energy,
may
writeconthat
es decir,(d)ditions
ditions
2
2
to
to
3
3
is
is
zero;
zero;
that
that
is,
is,
ditions 2ditions
to sum
3 is2zero;
that
is, that
to the
3 is
zero;
the
of
kineticandis,potential-energy changes during the process from con++
�E
==00 E oror
EE+K3 3E
−−EEKK2 E
++EEP3P30−−EEP2P2==00
�E
o
ditions�E
2�E
to
3KKis
zero;
that
P2→3
Pis,
2→3
2→3
2→3
=�E
0P2→3
�E
+
�E
=or0
orK 3 − EEKK23 K
− EPK3 2−
+2EPP2 3 =
− E P2 = 0
K 2→3 + �E
K 2→3P2→3
and
and
EPE2→3
are
zero.Therefore,
Therefore,
However,
EEP3
Sin embargo,
E
cero.
Por
�E
= zero.
0consiguiente,
or
E K 3 − E K 2 + E P3 − E P2 = 0
�E
KK
P3PTherefore,
K KyE
2+
2son
3are
2→3
are
zero.
However,
EHowever,
However,
K 2 andE
KP
2 3and E P3 are zero. Therefore,
==
EEP=2P2=
2,450,000
EEK=K3Therefore,
E P3 are
However, E K 2 and E
EEzero.
E
=450
000J JJ J
K3
J2,450,000
=32,450,000
EP
K3 = E
PK23 =
P2 = 2,450,000
=2E
EKPK23 3= 2,450,000
J
2E
(2)(2,450,000)
(2)(2,450,000)
1 1 2 2 2E KE2K232E
(2)(2,450,000)
K3
,3 , 2u3u3=
==(2)(2,450,000)
=
=
With
With
1 EEK2K
2 3=
1=2 2mu
2mu
3, 3=
3
mu
u
With
Con EEK3
=
=
mu
,
u
=
=
With
E
K3
mm
2,500
2,500
2 K 3 33 2
33
m3
m 2,500 2,500
1
2,
2 = 2E K 3 = (2)(2,450,000)
=
mu
u
With
E
K3
Whence,
Whence,
uu3m
44.27
44.27m
ms−1
s−1
3
3
2
3==
Whence,Whence,
u3 =
44.27
s−1
u 3m=
44.27
m 2,500
s−1
−
De donde,
= 44.27
44.7 m
m ss−1
Whence,
uupotential
=
33potential
(e)
(e)
Because
Because
the
the
changes
ininthe
the
energy
energy
ofofthe
thespring
spring
and
andthe
thekinetic
kineticenergy
energy
(e) Because
the
changes
inchanges
the
potential
energy
of
the
spring
the
kinetic
(e) Because
the changes
in the potential
energy
of theand
spring
and theenergy
kinetic energy
of
of
the
the
elevator
elevator
must
must
sum
sum
to
to
zero,
zero,
of
elevator
must
sum
toen
zero,
of
the
elevator
must
sum
tothe
zero,
e) the
Puesto
los
cambios
la in
energía
potencial
del resorte
y en laand
energía
cinética
del elevador
(e)que
Because
the
changes
potential
energy
of the spring
the kinetic
energy
deben ser
(spring)
(spring)
+
+
�E
�E
(elevator)
(elevator)
=
=
0
0
�E
�E
of cero,
the elevator�E
must(spring)
sum
to
zero,
PP �E (elevator)
KK= 0
+
=0
P �E P (spring)
K + �E K (elevator)
The
Theinitial
initialenergy
potential
potential
energy
energy
ofofthe
the+spring
spring
and
the
thefinal
final
kinetic
energy
energy
ofofthe
theeleele(elevador)
(spring)
�Eand
(elevator)
0kinetic
�E
(resorte)
Pspring
K and
The initial
of
the
the
final
kinetic
energy
of
the
eleThepotential
initial potential
energy
of theand
spring
the
final =
kinetic
energy
of the elevator
vator
are
are
zero;
zero;
therefore,
therefore,
the
the
final
final
potential
potential
energy
energy
of
of
the
the
spring
spring
must
must
equal
equal
the
the
vator arevator
zero;
theenergy
finalthe
potential
energy
of
the final
spring
must
equal
aretherefore,
zero;
therefore,
final
potential
energy
of the
spring
mustthe
equal
the
The
initial
potential
of
the
spring
and
the
kinetic
energy
of
the
elekinetic
kinetic
energy
energy
of
of
the
the
elevator
elevator
just
just
before
before
it
it
strikes
strikes
the
the
spring.
spring.
Thus
Thus
the
the
final
final
popoLa energía
potencial
inicial
lastrikes
energía
cinética
final
del
elevador
son cero;
por lo
kinetic
energy
of
the
elevator
justresorte
before
spring.
Thus
theThus
final
po-equal
kinetic
energy
oftherefore,
the del
elevator
justyitbefore
it the
strikes
the
spring.
the
final
povator
are
zero;
the
potential
energy
of energía
the spring
must
the justo
tential
tential
energy
energy
ofofthe
the
spring
spring
isfinal
is2,450,000
2,450,000
J.J.igual
tanto, energy
latential
energía
potencial
final
del
resorte
debe
ser
a
la
cinética
del
elevador
tential
of
the
spring
is
2,450,000
J.
energy
of the spring
is 2,450,000
energy
elevator
beforeJ. it
strikes the
Thusesthe 450
final000
po- J.
Así,just
la energía
potencial
finalspring.
del resorte
antes dekinetic
que incida
conofelthe
resorte.
tential
energy
of theand
spring
isspring
2,450,000
J. are
(f
(f
)
)
If
If
the
the
elevator
elevator
and
the
the
spring
together
together
are
taken
taken
as
as
the
the
system,
system,
the
the
initial
initial
energy
energy
(f ) If the(felevator
and the spring
are takenare
astaken
the system,
the initial
) If the elevator
and thetogether
spring together
as the system,
theenergy
initial energy
ofofthe
thethe
system
isisthe
the
potential
potential
energy
energy
ofofthe
the
elevator,
elevator,
orJ.
or245,000
245,000
J.J.energy
The
The
total
total
energy
energy
f) the
Si elsystem
elevador
ysystem
elpotential
resorte
juntos
seof
consideran
como
el sistema,
la energía
inicial
del
sistema es
of
is
energy
the
elevator,
or
245,000
The
total
of
the
system
is
the
potential
energy
of
the
elevator,
or
245,000
J.
The
total energy
(f
)
If
the
elevator
and
the
spring
together
are
taken
as
the
system,
the
initial
energy
of
of
the
the
system
system
can
can
change
change
only
only
if
if
work
work
is
is
transferred
transferred
between
between
it
it
and
and
the
the
surroundsurroundla the
energía
potencial
del
elevador,
o 45if
000
J. La
energía
total
del
sistema
sólo
cambia
si se trans­
of
system
can
change
only
if
work
is
transferred
between
it
and
the
surroundof
the
system
can
change
only
work
is
transferred
between
it
and
the
surroundofings.
the
system
the
potential
energy
of the
elevator,
245,000
J.by
The
total
energysobre el
ings.
As
Asélthe
the
elevator
elevator
isisraised,
raised,
work
work
isis
done
on
onor
the
the
system
system
by
the
surroundings
surroundings
fiere trabajo
entre
yis
alrededores.
Conforme
sedone
levanta
el
elevador
sethe
hace
trabajo
ings.
Asings.
the
elevator
islos
raised,
isifwork
done
on
the
system
by
the
surroundings
As
the elevator
is work
raised,
is
done
on the between
system
by
the
surroundings
of
the
system
can
change
only
work
is
transferred
it
and
the
surroundin
in
the
the
amount
amount
of
of
2,205,000
2,205,000
J.
J.
Thus
Thus
the
the
energy
energy
of
of
the
the
system
system
when
when
the
the
elevator
elevator
sistema
por
de of
los2,205,000
alrededores
en
unathe
cantidad
de
the
05when
000 the
J.when
De
esta
la energía
in
the amount
of 2,205,000
J. isThus
energy
of
theon
system
elevator
in
theacción
amount
J.the
Thus
energy
ofthe
system
theforma,
elevator
ings.
As
the
elevator
raised,
work
is
done
system
by
the
surroundings
reaches
reaches
its
its
maximum
maximum
height
height
is
is
245,000
245,000
+
+
2,205,000
2,205,000
=
=
2,450,000
2,450,000
J.
J.
Subsequent
Subsequent
del sistema
cuando
el height
elevador
alcanza
su
altura
máxima
es 45
000
+
05
000 = 450 000 J.
reaches
its
maximum
is
245,000
+
2,205,000
=
2,450,000
J.
Subsequent
reaches
its
maximum
height
is
245,000
+
2,205,000
=
2,450,000
J.
Subsequent
in changes
the
amount
ofocurren
2,205,000
J.
Thus
the
energy
of
the
system
when
the elevator
changes
occur
occur
entirely
entirely
within
within
the
the
system,
system,
with
with
no
no
work
work
transfer
transfer
between
between
the
the
syssys- en­
Los cambios
siguientes
por
completo
dentro
del
sistema,
sin
transferencia
trabajo
changes
occur
entirely
within
the
system,
with
no+with
work
transfer
between
the
syschanges
occur
entirely
within
the245,000
system,
no
work
transfer
between
thedesysreaches
its
maximum
height
is
2,205,000
=
2,450,000
J.
Subsequent
tem
tem
and
and
surroundings.
surroundings.
Hence
Hence
the
the
total
total
energy
energy
of
of
the
the
system
system
remains
remains
constant
constant
atat en
tre eland
sistema
y los
alrededores.
Por
lothe
tanto,
energía
total
del
sistema
permanece
tem
surroundings.
Hence the
total
energy
the
system
remains
constant
at theconstante
tem
and
surroundings.
Hence
totallaof
energy
of
the
system
remains
constant
at
changes
occur
entirely
within
the
system,
with
no
work
transfer
between
sys2,450,000
2,450,000
J.
J.
It
It
merely
merely
changes
changes
from
from
potential
potential
energy
energy
of
of
position
position
(elevation)
(elevation)
of
of
the
the
450 000
sólo
cambia changes
de
energía
potencial
deof
posición
(elevación)
del
elevador
a energía
2,450,000
J.J.and
ItÉsta
merely
from
potential
energy
position
(elevation)
of the
2,450,000
J. It changes
merely
from
potential
energy
position
(elevation)
of the
tem
surroundings.
Hence
total
of
theofsystem
remains
constant
atofof
elevator
elevator
to
tokinetic
energy
energy
ofofthe
the
the
elevator
elevator
totoenergy
potential
potential
energy
energy
ofofconfiguration
configuration
cinéticaelevator
del
elevador,
ykinetic
a energía
potencial
de
laenergy
configuración
en
el resorte.
elevator
to
kinetic
energy
of
the
elevator
to
potential
of
configuration
of
to
kinetic
energy
of
the
elevator
to
potential
energy
of
configuration
of
2,450,000
J. It merely changes from potential energy of position (elevation) of the
the
thespring.
spring.
the spring.
the
spring.
elevator
to
kinetic
energy
thedeelevator
to potential
energy of
configuration
of
Este ejemplo ilustra
el uso
de laofley
conservación
de la energía
mecánica.
Sin embargo,
se
the
spring.
This
This
example
illustrates
illustrates
application
of
ofthe
thelaw
lawofofconservation
conservation
ofofexactos
mechanical
mechanical
supone
todo
elexample
proceso
ocurre application
sinapplication
fricción;
resultados
obtenidos
son
sólo para
This que
example
illustrates
application
of
the law
of
conservation
of mechanical
This
example
illustrates
oflos
the
law of conservation
of
mechanical
energy.
energy.
However,
However,
the
the
entire
entire
processisisto
assumed
assumed
totooccur
occur
without
withoutthe
friction;
friction; the
the
procesos
tan
idealizados
como
éste.
energy.
However,
the
entire
is process
assumed
occur
friction;
energy.
theprocess
entire
process
isofassumed
occur
without
the
This However,
example
illustrates
application
theidealized
law to
ofwithout
conservation
offriction;
mechanical
results
results
obtained
obtained
are
are
exact
exact
only
only
for
for
such
such
an
an
idealized
process.
process.
results obtained
are
exactare
only
for
suchprocess
an such
idealized
process.
results
exact
only
for
idealized
energy.obtained
However,
the
entire
isanassumed
toprocess.
occur without friction; the
results obtained are exact only for such an idealized process.
During
the
the
period
period
ofofdevelopment
development
of
ofthe
thelaw
lawofofconservation
conservation
ofofmechanical
mechanical
energy,
energy,
heat
heat
During
period
development
ofdethe
of
ofdemechanical
energy,
heat
Durantethe
elDuring
periodo
del
desarrollo
la law
ley
conservación
la energía
mecánica,
en
general
During
theofperiod
of development
of de
theconservation
law of conservation
of mechanical
energy,
heat no se
was
was
not
not
generally
generally
recognized
recognized
as
as
a
a
form
form
of
of
energy,
energy,
but
but
was
was
considered
considered
an
an
indestructible
indestructible
fluid
fluid
was
notwas
generally
recognized
as of
a form
of
energy,
but
wasseof
considered
an indestructible
fluid
reconocía
alnot
calor
como
forma
deasenergía;
más
bien,
leconservation
consideraba
como
fluido
indestructible
lla­
generally
recognized
a form
of
energy,
but
was considered
an un
indestructible
fluid
During
the una
period
development
of the
law
of
mechanical
energy,
heat
called
called
caloric.
caloric.
This
This
concept
concept
was
was
firmly
firmly
entrenched,
entrenched,
and
andmuchos
for
formany
many
years
years
no
nohizo
connection
connection
was
was
called
This
concept
was
firmly
entrenched,
and
for
many
years
no
connection
was
mado caloric.
calórico.
Este
concepto
estaba
firmemente
establecido,
y
por
años
no
se
conexión
entre
el
called
caloric.
This
concept
was
firmly
entrenched,
and
for
many
years
no
connection
was
was
not between
generally
recognized
asfrom
afrom
form
of energy,
but
was
considered
anofof
indestructible
made
made
between
heat
heat
resulting
resulting
friction
friction
and
andthe
theestablished
established
forms
forms
energy.
energy.
The
Thefluid
law
lawofof
made
heat
resulting
from
friction
and
the
established
forms
of
energy.
The
law
of
calor between
resultante
de
la
fricción
y
las
formas
establecidas
de
la
energía.
Por
lo
tanto,
la
ley
de
la
conservación
made
between
heat
resulting
from
friction
and
the
established
forms
of
energy.
The
law
of
called
caloric. of
This
concept
was
firmly limited
entrenched,
and for many
years no connection
was
conservation
conservation
ofenergy
energy
was
was
therefore
therefore
limitedininapplication
application
totofrictionless
frictionless
mechanical
mechanical
proproconservation
of
energy
was
therefore
limited
in application
to frictionless
mechanical
prode la energía
estaba
limitada
a was
procesos
mecánicos
sininthe
fricción.
No to
es forms
necesaria
talmechanical
limitación;
hoy
conservation
of
energy
therefore
limited
application
frictionless
promade
between
heat
resulting
from
friction
and
established
of
energy.
The
law
of en día,
cesses.
cesses.
No
No
such
such
limitation
limitation
is
is
necessary;
necessary;
heat
heat
like
like
work
work
is
is
now
now
regarded
regarded
as
as
energy
energy
in
in
transit,
transit,
cesses.
No
such
limitation
is necessary;
heat like
work
now
as concepto
energy
transit,
tanto alcesses.
calor
como
aloftrabajo
sewas
lesisconsidera
como
energía
en regarded
tránsito,
un
que
ganó
aceptación
en
No
such
limitation
necessary;
heat
like
work
is nowtoregarded
as in
energy
in transit,
conservation
energy
therefore
limited
inisapplication
frictionless
mechanical
processes. No such limitation is necessary; heat like work is now regarded as energy in transit,
01­SmithVanNess.indd 14
8/1/07 12:47:38
Problemas
5
los años posteriores a 850, en gran parte por los experimentos clásicos de J. P. Joule. Estos experimentos
se consideran de manera detallada en el capítulo , pero primero examinaremos algunas de las caracterís­
ticas del calor.
1.9
CALOR
Sabemos por experiencia que un objeto caliente que se pone en contacto con otro objeto frío tiende a enfriar­
se, al mismo tiempo que este último se calienta. Una visión razonable es que algo se transfiere del objeto ca­
liente al frío, y a ese algo le llamamos calor Q.0 En estos términos se dice que el calor siempre fluye de una
temperatura más alta a una más baja, lo cual nos lleva al concepto de temperatura como la fuerza impulsora
para la transferencia de energía como calor. Con más precisión, la rapidez de transferencia de calor de un
cuerpo a otro es proporcional a la diferencia de temperatura entre los dos cuerpos; cuando no existe diferen­
cia de temperatura, no hay transferencia neta de calor. En sentido termodinámico, el calor nunca se considera
como algo que está almacenado dentro de un cuerpo. Al igual que el trabajo, solamente existe como energía
en tránsito desde un cuerpo hacia otro; o, en términos termodinámicos, entre un sistema y sus alrededores.
Cuando se agrega energía en forma de calor a un cuerpo, se almacena no como calor sino como energías ci­
nética y potencial de los átomos y de las moléculas que constituyen el sistema.
A pesar de la naturaleza transitoria del calor, con frecuencia se le ve sólo desde el punto de vista de su
efecto sobre el cuerpo del cual o al cual se transfiere. De hecho, alrededor de 930 las definiciones de unida­
des de calor estaban basadas en los cambios de temperatura de una masa de una unidad de agua. De esta
forma, la caloría se definió por mucho tiempo como la cantidad de calor que, al transferirse a un gramo de
agua, elevaba su temperatura un grado Celsius. Del mismo modo, la unidad térmica inglesa (o Btu, British
termal unit) se definió como la cantidad de calor que, al transferirse a una libra masa de agua, elevaba su
temperatura en un grado Fahrenheit. Aunque estas definiciones proporcionan una “sensación” de la magnitud
de estas unidades de calor, dependen de experimentos realizados con agua y, por lo tanto, están sujetas a cam­
bios en la medida en que las mediciones se vuelven más precisas. En la actualidad se reconoce a la caloría y
al (Btu) como unidades de energía, y se definen con respecto al joule, la unidad de energía del SI, que es
igual a N m. Éste es el trabajo mecánico realizado por la fuerza de un newton que actúa a través de una
distancia de un metro. Todas las demás unidades de energía se definen como múltiplos del joule. El pie­libra
fuerza, por ejemplo, es equivalente a .355879 J, la caloría es igual a 4.840 J, mientras que el (Btu) es igual a
055.04 J. La unidad SI para la potencia es el watt, cuyo símbolo es W y se define como una relación de
energía de un joule por segundo.
La tabla A. en el apéndice A proporciona una lista amplia de factores de conversión para la energía, así
como para otras unidades.
PROBLEMAS
1.1. ¿Cuál es el valor de gc y cuáles son las unidades en un sistema donde el segundo, el pie y la libra
masa están definidos como en la sección ., y el poundal es la unidad de fuerza requerida para que
a (lbm) se le dé una aceleración de (pie)(s)−?
0
Un punto de vista igualmente razonable sería considerar que se transfiere “frío” desde el objeto frío hacia el objeto caliente.
01­SmithVanNess.indd 15
8/1/07 12:47:39
CHAPTER
1.
CHAPTERCAPÍTULO
1. Introduction
Introduction
1. Introducción
16
6 16
1.2.
current
is
electrical
dimension,
ampere
(A)
1.2.
corriente
eléctrica
es fundamental
la dimensiónSI
eléctrica
fundamental
el SI,
y su unidad
el ampere (A).
1.2.LaElectric
Electric
current
is the
the
fundamental
SI
electrical
dimension,enwith
with
ampere
(A) as
asesunit.
unit.
Determine
units
for
the
following
quantities,
as
combinations
of
fundamental
SI
units.
Determine
las
unidades
de
las
siguientes
cantidades,
como
combinaciones
de
las
unidades
fundaDetermine units for the following quantities, as combinations of fundamental SI units.
mentales del SI.
(a)
(a) Electric
Electric power;
power; (b)
(b) Electric
Electric charge;
charge; (c)
(c) Electric
Electric potential
potential difference;
difference;
a)(d)
Potencia
eléctrica;
b)
carga
eléctrica;
c)
diferencia
de potencial eléctrico;
Electric
resistance;
(e)
Electric
capacitance.
(d) Electric resistance; (e) Electric capacitance.
d) resistencia eléctrica; e) capacitancia eléctrica.
sat
represented as
of
1.3.
saturation
pressure
P
is often
often
as aa function
function
of temperature
temperature
1.3.LaLiquid/vapor
Liquid/vapor
saturationde
pressure
P sat is
1.3.
presión de saturación
un líquido/vapor
Psatrepresented
con frecuencia
se representa
como una función de
by
an
equation
of
the
form:
an equation
ofmedio
the form:
laby
temperatura
por
de una ecuación de la forma siguiente:
bb
sat
sat /torr
log10 P
log
/torr =
= aa −
− t/◦◦ C + c
10 P
t/ C + c
Here,
parameters
a,
b,
are
substance-specific
constants.
Suppose
required
to
Aquí,
parámetros
y cccson
específicas
de las sustancias.
Suponga
Here,los
parameters
a, a,
b, band
and
are constantes
substance-specific
constants.
Suppose it
it is
is
requiredque
to se requiere
sat
sat
sat
represent
P
by
the
equivalent
equation:
represent
P
by
the
equivalent
equation:
representar P mediante la ecuación equivalente:
B
B
sat
ln
/kPa =
=A
A−
− T /K + C
ln P
P sat /kPa
T /K + C
Muestre
cómo
relacionados
en las
dos ecuaciones.
Show
the
parameters
in
two
equations
related.
Show how
how
theestán
parameters
in the
the los
twoparámetros
equations are
are
related.
1.4.
what
absolute
the
Fahrenheit
temperature
scales
give
1.4.
temperatura
absoluta las do
escalas
Celsiusand
y Fahrenheit
el mismo valor
¿Cuál
1.4.¿AAt
Atqué
what
absolute temperature
temperature
do
the Celsius
Celsius
and
Fahrenheitdan
temperature
scalesnumérico?
give
numerical
esthe
esesame
valor?
the
same
numerical value?
value? What
What is
is the
the value?
value?
1.5.
Pressures
up
3,000
measured
aa dead-weight
gauge.
piston
diameter
1.5.
presiones
superiores
3 000
bar se with
miden
con una balanza
peso
muerto.
El diámetro del
1.5.Las
Pressures
up to
to
3,000 bar
bara are
are
measured
with
dead-weight
gauge.deThe
The
piston
diameter
is
4
mm.
What
is
the
approximate
mass
in
kg
of
the
weights
required?
pistón
es deWhat
4 mm.
¿Cuál
es la masamass
aproximada
de las pesas
requeridas?
is 4 mm.
is the
approximate
in kg ofen
thekgweights
required?
1.6.
Pressures
up
to
measured
with
dead-weight
gauge.
The
piston
1.6.
presiones
a 3 are
000(atm)
se miden
una balanza
de peso
El diámetro del
1.6.Las
Pressures
up superiores
to 3,000(atm)
3,000(atm)
are
measured
with aacon
dead-weight
gauge.
The muerto.
piston diamdiam)
of
the
weights
required?
eter
is
0.17(in).
What
is
the
approximate
mass
in
(lb
pistón
es
de
0.7
(pulg).
¿Cuál
es
la
masa
aproximada
(lb
)
de
las
pesas
requeridas?
m
m the weights required?
eter is 0.17(in). What is the approximate mass in (lbm ) of
◦ C (open to the atmosphere at one end)
1.7.
reading
on
aa mercury
1.7.
lectura
de un
de mercurioat
5◦°C
(abierto
la atmosphere
atmósfera enatuno
sus extremos)
C (open
to athe
onede
end)
1.7.LaThe
The
reading
onmanómetro
mercury manometer
manometer
ata 25
25
−2 . Atmospheric
local
is
m
pressure
esis
56.38cm.
cm.The
La aceleración
local deof
gravedad
es de 9.83
s−. La presión
atmosférica es de
isde56.38
56.38
cm.
The
local acceleration
acceleration
oflagravity
gravity
is 9.832
9.832
m ss−2m
. Atmospheric
pressure
is
the
in
being
measured?
The
of
0.78
kPa.kPa.
¿CuálWhat
es la is
presión
absolutapressure
en kPa que
debe
medirse?
La densidad
del mercurio
a 5 °C
is 101.78
101.78
kPa.
What
is
the absolute
absolute
pressure
in kPa
kPa
being
measured?
The density
density
of
◦◦ C−3
−3
at
esmercury
de 3.534
g cm
mercury
at 25
25
C is
is.13.534
13.534 g
g cm
cm−3 ..
◦◦ F) (open to the atmosphere at one end) is
1.8.
reading
on
aa mercury
at
l.8.
lectura
en un
de mercurio
70(°F)
(abierto
a laatmosphere
atmósfera at
enone
unoend)
de sus
F) (open
to the
is extremos)
1.8.LaThe
The
reading
on manómetro
mercury manometer
manometer
at a70(
70(
−2
−. La
−2 .. Atmospheric
pressure
The
acceleration
of
is
es25.62(in).
de 5.6 (pulg).
La aceleración
de la gravedad
es de 3.43(pie)(s)
presión atmosfé­
Atmospheric
pressure
25.62(in).
The local
local
accelerationlocal
of gravity
gravity
is 32.243(ft)(s)
32.243(ft)(s)
is
Hg).
What
the
absolute
(psia)
measured?
density
rica
es de 9.86
(pulg
de is
Hg).
es lapressure
presiónin
en (psia)
que debeThe
medirse?
is 29.86(in
29.86(in
Hg).
What
is
the¿Cuál
absolute
pressure
inabsoluta
(psia) being
being
measured?
The
densityLa densidad
◦◦ F) is 13.543 g cm−3
−3
−3
of
mercury
at
70(
.
del
3.543
g cm. .
ofmercurio
mercury aat70(°F)
70( F)esis de
13.543
g cm
1.9.
Liquids
that
boil
at
low
are
stored
as
their
1.9.
líquidos
a temperaturas
relativamente
bajas
son almacenados
como
líquidos bajo
1.9.Los
Liquids
thatque
boilhierven
at relatively
relatively
low temperatures
temperatures
are often
often
stored
as liquids
liquids under
under
their
vapor
pressures,
which
at
ambient
temperature
can
be
quite
large.
Thus,
n-butane
sus
presiones
de vapor,
lasatcuales
serían
más grandes
temperatura
ambiente.
Así, el n­butano al­
vapor
pressures,
which
ambient
temperature
canabe
quite large.
Thus, n-butane
stored
aa liquid/vapor
system
is
2.581
bar
temperature
300
K.
macenado
un sistema
líquido/vapor
está aof
presión
de aa.58
bar paraof
de
stored as
as como
liquid/vapor
system
is at
at aa pressure
pressure
ofuna
2.581
bar for
for
temperature
ofuna
300temperatura
K.
33 ) of this kind is sometimes
3
done
in
spherical
tanks.
SugLarge-scale
storage
(>50
m
300
K. El almacenaje
de grandes
cantidades
(>50 m ) de
estainclase
de sustancias
en ocasiones
kind is sometimes
done
spherical
tanks. SugLarge-scale
storage (>50
m ) of this
two
reasons
why.
segest
realiza
tanques
esféricos. Sugiera dos posibles razones del por qué.
gest
twoen
reasons
why.
1.10.
first
accurate
the
of
gases
were
made
1.10.
primer
preciso paraof
las propiedades
de los gases
a alta
fue creado en
1.10.ElThe
The
first instrumento
accurate measurements
measurements
ofmedir
the properties
properties
of high-pressure
high-pressure
gases
werepresión
made by
by
E.
France
1869
and
Before
developing
the
Francia
por E. H.in
entre 869
y 893.
Antes de
desarrollar
la balanza
de peso muerto, Ama­
E. H.
H. Amagat
Amagat
inAmagat,
France between
between
1869
and 1893.
1893.
Before
developing
the dead-weight
dead-weight
gauge,
he
worked
aa mine
shaft,
and
mercury
measurements
gat
trabajó
el pozoin
una mina
y utilizó
un aamanómetro
de mercuriofor
medir presiones mayo­
gauge,
heen
worked
inde
mine
shaft,
and used
used
mercury manometer
manometer
forpara
measurements
more
bar.
Estimate
the
of
resof
400 bar.to
Determine
altura
manómetro
requerido.
ofapressure
pressure
to
more than
thanla400
400
bar.del
Estimate
the height
height
of manometer
manometer required.
required.
01­SmithVanNess.indd 16
8/1/07 12:47:41
Problems
Problemas
17
7
1.11. An instrument to measure the acceleration of gravity on Mars is constructed of a spring
1.11. Un
instrumento
para mediralamass
aceleración
de la
Marte
está the
construido
con un resorte de
from
which is suspended
of 0.40 kg.
Atgravedad
a place onenearth
where
local accelerdonde
se
suspende
una
masa
de
0.40
kg.
En
un
lugar
de
la
Tierra
donde
la
aceleración
−2
ation of gravity is 9.81
m s , the spring extends 1.08 cm. When the instrument pack- local de la
−, el resorte se extiende .08 cm. Cuando el aparato desciende en Marte, la radio
gravedad
es
9.8
m
s
age is landed on Mars, it radios the information that the spring is extended 0.40 cm.
informa
resorte acceleration
se extiende 0.40
cm. ¿Cuál es la aceleración de la gravedad marciana?
What isque
theelMartian
of gravity?
1.12.
variación
deof
la fluid
presión
de un with
fluidoheight
con laisaltura
está by
descrita
por la ecuación
diferencial:
1.12.LaThe
variation
pressure
described
the differential
equation:
dP
= −ρg
dz
Here,ρ ρes is
and yggisesthe
local acceleration
anun
ideal
Aquí,
la specific
densidaddensity
específica
la aceleración
local deofla gravity.
gravedad.For
Para
gas ideal, ρ =
gas,
ρ
=
MP/RT
,
where
M
is
molar
mass
and
R
is
the
universal
gas
constant.
M P / RT, donde M es la masa molar y R es la constante universal de los
gases.
Modele
la atmós­
Modeling
thecolumna
atmosphere
as an isothermal
column
of°C
ideal
gasdeatcalcular
10◦ C, estimate
fera
como una
isotérmica
de un gas ideal
a 0
a fin
la presiónthe
ambiental en
ambientdonde
pressure
Denver,
z =al 1(mile)
to sea
level.
ForMair,
Denver,
z −1
= in(milla)
conwhere
respecto
nivel delrelative
mar. Para
el aire,
tome
= take
9 g mol−; los
;
values
of
R
are
given
in
App.
A.
M
=
29
g
mol
valores de R se dan en el apéndice A.
1.13. A group of engineers has landed on the moon, and they wish to determine the mass
1.13. Un
ingenieros
ha aterrizado
la Luna
y desea
masaatdea algunas
ofgrupo
some de
rocks.
They have
a spring en
scale
calibrated
to determinar
read poundslamass
locationrocas. Tiene
una
escala
de
resorte
calibrada
para
leer
libras
masa
en
una
posición
donde
la
aceleración
de la
−2
where the acceleration of
gravity is 32.186(ft)(s) . One of the moon rocks gives
a
−). Una de las rocas lunares da una lectura de 8.76 en esta escala. ¿Cuál
gravedad
es
3.86(pie)(s
reading of 18.76 on this scale. What is its mass? What is its weight
on the moon?
−
esTake
la masa?
¿Cuál=es
el peso en
−2la
g(moon)
5.32(ft)(s)
. Luna? Tome g(Luna) = 5.3(pie)(s ).
1.14.
luz externa
de seguridad
de 70
wattsonseaverage,
utiliza en
0 A
horas
día.costs
Un foco nuevo
1.14.Una
A 70-watt
outdoor
security light
burns,
10 promedio
hours a day.
newalbulb
cuesta
$5.00,
y
su
tiempo
de
vida
es
de
casi
000
horas.
Si
la
electricidad
cuesta
$0.0
por kW­hora,
$5.00, and the lifetime is about 1,000 hours. If electricity costs $0.10 per kW-hour,
¿cuál
es
el
costo
anual
de
“seguridad”
por
luz?
what is the yearly price of “security,” per light?
1.15.
confinado
en un cilindro
de .5by
(pie)
de diámetro
por medio
un pistón, sobre el
1.15.Un
Agas
gasseis encuentra
confined in
a 1.25(ft)-diameter
cylinder
a piston,
on which
rests adeweight.
cual
descansa
una
pesa.
La
masa
del
pistón
y
la
pesa
en
conjunto
es
de
50(lb
).
La
aceleración
local de
m
The mass of the piston and weight together is 250(lbm ). The local acceleration
of
− y la presión atmosférica es de 30.(pulg de Hg).
−2
lagravity
gravedad
es
de
3.69(pie)(s)
is 32.169(ft)(s) , and atmospheric pressure is 30.12(in Hg).
Whatesislathe
forceen
in(lb
(lbf))ejercida
exerted on
thegas
gaspor
by la
theatmósfera,
atmosphere,
the piston,
and the
a) (a)¿Cuál
fuerza
en el
el pistón
y la pesa?
Suponga que
f
weight,
assuming
no
friction
between
the
piston
and
cylinder?
no existe fricción entre el pistón y el cilindro.
Whatesislathe
pressure
in (psia)?
b) (b)¿Cuál
presión
del of
gasthe
engas
(psia)?
(c)
If
the
gas
in
the
cylinder
is
heated,
it expands,
pushing
the piston
and weight
c) Si el gas en el cilindro se calienta, se expande
y empuja
el pistón
y la pesa
hacia arriba. Si el
upward.
If
the
piston
and
weight
are
raised
1.7(ft),
what
is
the
work
done
by(pie
the lb )? ¿Cuál
pistón y la pesa se levantan .7(pie), ¿cuál es el trabajo realizado por el gas en
f
)?
What
is
the
change
in
potential
energy
of
the
piston
and
weight?
gas
in
(ft
lb
f
es el cambio en la energía potencial del pistón y la pesa?
1.16. A gas is confined in a 0.47-m-diameter cylinder by a piston, on which rests a weight.
1.16. Un gas se encuentra confinado en un cilindro de 0.47 m de diámetro por medio de un pistón, sobre
The mass of the piston and weight together is 150 kg. The local acceleration of gravity
el cual descansa
una pesa. La masa del pistón y la pesa en conjunto es de 50 kg. La aceleración
is 9.813 m s−2 , and atmospheric pressure is 101.57 kPa.
local de la gravedad es de 9.83 m s−, y la presión atmosférica es de 0.57 kPa.
(a) What is the force in newtons exerted on the gas by the atmosphere, the piston, and
the weight,
assuming
no friction
between
piston
cylinder? el pistón y la pesa? Supon­
a) ¿Cuál
es la fuerza
en newtons
ejercida
sobrethe
el gas
porand
la atmósfera,
(b)gaWhat
is
the
pressure
of
the
gas
in
kPa?
que no hay fricción entre el pistón y el cilindro.
If the
in the del
cylinder
heated, it expands, pushing the piston and weight
b) (c)¿Cuál
es gas
la presión
gas eniskPa?
thecilindro
piston and
weight are
0.83
what el
is pistón
the work
done
the arriba. Si
c) Siupward.
el gas enIf el
se calienta,
se raised
expande
y m,
empuja
y la
pesabyhacia
in kJ?
is the
change
in potential
of therealizado
piston and
weight?
el gas
pistón
y laWhat
pesa se
levantan
0.83
m, ¿cuál energy
es el trabajo
por
el gas en kJ? ¿Cuál es el
cambio en la energía potencial del pistón y la pesa?
01­SmithVanNess.indd 17
8/1/07 12:47:43
8
CAPÍTULO 1. Introducción
1.17. Verifique si el joule es la unidad en el sistema internacional para las energías cinética y potencial.
1.18. Un automóvil que tiene una masa de 50 kg se encuentra viajando a 40 m s−. ¿Cuál es la energía
cinética en kJ? ¿Cuánto trabajo debe efectuarse para detenerlo?
1.19. Las turbinas de una planta hidroeléctrica son alimentadas por una caída de agua de 50 m de altura.
Al suponer una eficiencia de conversión de la energía potencial en eléctrica de 9%, y 8% de pérdi­
da de potencia resultante en la transmisión, ¿cuál es la rapidez de flujo de masa de agua requerida
para encender un foco de 00 watts?
1.20. A continuación se encuentra una lista de factores de conversión aproximados, que se emplean para
cálculos “rápidos”. Ninguno de ellos es exacto, pero en la mayoría hay un margen de error de ± 0%.
Haga uso de la tabla A. (apéndice A) para establecer los factores de conversión exactos.
• (atm) ≈ bar
• (Btu) ≈ kJ
• (hp) ≈ 0.75 kW
• (pulgada) ≈ .5 cm
• (lbm) ≈ 0.5 kg
• (milla) ≈ .6 km
• (cuarto) ≈ litro
• (yarda) ≈ m
Agregue sus propios términos a la lista. La idea es mantener los factores de conversión en una forma
simple que sea fácil de recordar.
1.21. Considere la siguiente propuesta para un calendario decimal. La unidad fundamental es el año de­
cimal (año), igual al número convencional de segundos (SI) requeridos para que la Tierra complete
una órbita alrededor del Sol. Otras unidades son definidas en la tabla siguiente. Desarrolle, hasta
donde sea posible, factores para convertir unidades del calendario decimal a unidades del calendario
convencional. Analice los puntos a favor y en contra de esta propuesta.
01­SmithVanNess.indd 18
Unidades del calendario decimal
Símbolo
Definición
Segundo
Seg
0–6 Año
Minuto
Hora
Día
Semana
Mes
Min
H
Día
Sem
Mes
0–5 Año
0–4 Año
0–3 Año
0– Año
0– Año
Año
Año
8/1/07 12:47:43
Problemas
9
19
Problems
Problems
19
1.22. Los costos de la energía varían en gran medida con la fuente de energía: el carbón cuesta $5.00/to­
1.22. Energy
coststiene
vary un
greatly
with
energyde
source:
coal @y$25.00/ton,
gasoline
@ a pump
nelada,
la gasolina
precio
creciente
$.00/galón
la electricidad
vale $0.000/kW­hora.
1.22. Energy costs vary greatly with energy source: coal @ $25.00/ton, gasoline
@ a pump
price of es
$2.00/gal,
and electricity
Conventional
is to put
En la práctica
conveniente
expresarlos@en$0.1000/kWhr.
una base común
en $ GJ−. practice
[Un gigajoule
equivale
price of $2.00/gal, and
electricity @ $0.1000/kWhr. Conventional
practice is to put
these on a common
basis by
expressing
them insuponga
$−1GJ−1 . valores
[A gigajoule
is approximately
aproximadamente
a 0 6 (Btu).]
Para
este propósito
de calentamiento
totales (sin
these on
a common basis
by expressing them in $ GJ −3
. [A gigajoule is approximately
106 (Btu).]
purpose,
assumeygross
ofgasolina.
29 MJ kg−1 for coal and
corrección),
de 9For
MJthis
kg−
para el carbón
de 37heating
GJ m values
para la
106 (Btu).] For
this purpose, assume gross heating values of 29 MJ kg−1 for coal and
−3
37 GJ m for gasoline.
37 GJ m−3 for gasoline.
−
a) Clasifique en orden las tres fuentes de energía con respecto a los costos de energía
−1 . en $ GJ .
(a)
Rank
order
the
three
energy
sources
with
respect
to
energy
cost
in
$
GJ
b) (a)Explique
la notable
discrepancia
que hay
enrespect
los resultados
numéricos
inciso
a). Discuta las
Rank order
the three
energy sources
with
to energy
cost in $ del
GJ−1
.
(b) Explain
the largededisparity
in the numerical
results of Part (a). Discuss the advanventajas
y
desventajas
las
tres
fuentes
de
energía.
(b) Explain the large disparity in the numerical results of Part (a). Discuss the advantages and disadvantages of the three energy sources.
tages and disadvantages of the three energy sources.
1.23. Los costos del equipo de una fábrica de especies químicas rara vez varían en proporción al tamaño.
1.23.
Chemical-plant
equipment
costs
vary in proportion
to size.
In the simplest
case,
el caso
más sencillo,
el costo
varíararely
con
S conforme
a la
1.23.EnChemical-plant
equipment
costsCrarely
varyelintamaño
proportion
to size. In
theecuación
simplestalométrica:
case,
cost C varies with size S according to the allometric equation
cost C varies with size S according to the allometric
equation
C = αSβ β
C = αS
C = αS β
La magnitud del exponente β por lo regular se encuentra entre 0 y . Para una amplia variedad de
The
size
β is typically between 0 and 1. For a wide varity of equipment types
equipos
su exponent
valorexponent
es de
aproximadamente
The size
β is
typically between0.6.
0 and 1. For a wide varity of equipment types
it is approximately 0.6.
it is approximately 0.6.
a) Para 0 < β < muestre que el costo por tamaño unitario disminuye con el incremento de tama­
(a) For 0 < β < 1, show that cost per unit size decreases with increasing size. (”Econ(a)ñoFor
0 < β < de
1, show
that cost per unit size decreases with increasing size. (”Econ(“economía
escala”).
omy of scale.”)
omy
of
scale.”)
b) Considere el caso de un tanque de almacenamiento esférico. El tamaño se mide regularmente
(b) Consider the case of a spherical
storage tank. The size is commonly measured by
(b)mediante
Consider
case ofinterno
a tspherical
storage tank.
commonly
measured
by
el the
volumen
V it. Demuestre
que The
β = size
/3.is¿De
qué parámetros
o propiedades
se
internal volume
V
.
Show
that
β
= 2/3. On what parameters or properties would
t . Show
i
internal
thatα?β = 2/3. On what parameters or properties would
espera
quevolume
dependaVila
cantidad
you expect quantity α to depend?
you expect quantity α to depend?
1.24. Un laboratorio elabora un informe con la siguiente información
de presión de vapor (P sat) para una
1.24. A laboratory reports the following vapor-pressuresat(P sat ) data for a particular organic
1.24.especie
A laboratory
the following vapor-pressure (P ) data for a particular organic
orgánicareports
particular:
chemical:
chemical:
t/◦ C P sat /kPa t/◦ C P sat /kPa
t/◦ C P sat /kPa t/◦ C P sat /kPa
−18.5
3.18
32.7
41.9
−18.5
3.18
32.7
41.9
−9.5
5.48
44.4
66.6
−9.5
5.48
44.4
66.6
0.2
9.45
52.1
89.5
0.2
9.45
52.1
89.5
11.8
16.9
63.3
129.
11.8
16.9
63.3
129.
23.1
28.2
75.5
187.
23.1
28.2
75.5
187.
Correlate the data by fitting them to the Antoine equation:
Correlate
the data de
by la
fitting
them to the
Antoinelosequation:
Haga
la correlación
información
ajustando
datos a la ecuación de Antoine:
B
ln P sat /kPa = A − B
ln P sat /kPa = A −
T /K + C
T /K + C
That is, find numerical values of parameters A, B, and C by an appropriate regression
is,encuentre
find numerical
valuesnuméricos
of parameters
B, and C by
EsThat
decir,
los valores
de losA,parámetros
A, an
B yappropriate
C medianteregression
un procedimiento de
procedure. Discuss the comparison of correlated with experimental values. What is
procedure.
DiscussTambién
the comparison
correlated
with experimental
What is
regresión
apropiado.
compareoflos
valores correlacionados
convalues.
los experimentales.
¿Cuál
the predicted normal boiling point of this chemical?
predicted
normal boiling
point
of this chemical?
esthe
el punto
de ebullición
normal
pronosticado
para esta especie química?
−1 . Between
1.25. (a) In summer 1970, the pump price of gasoline was about $0.35(gal)
−1 . Between
−. Entre
1.25.
1970,
theelpump
gasolineera
was
$0.35(gal)
1.25.
a) (a)EnInelsummer
verano de
970,
precioprice
de laofgasolina
de about
$0.35(gal)
970 y 000, la tasa
1970 and 2000, the average rate of inflation was about 5% per year. What might
1970 and
the average
rate of inflation was
5%¿Cuál
per year.
promedio
de2000,
inflación
fue de aproximadamente
5% about
por año.
es el What
preciomight
esperado para el
be the expected pump price in summer 2000? What conclusion might one reach
be thedelexpected
pump
in summerse2000?
one reach
verano
año 000?
¿A price
qué conclusión
puedeWhat
llegarconclusion
a partir de might
estos cálculos?
from this calculation?
from this calculation?
01­SmithVanNess.indd 19
8/1/07 12:47:45
0
CAPÍTULO 1. Introducción
CHAPTER 1. Introduction
20
b) Un doctor en ingeniería inició su carrera en 970, con un salario de $6 000(año)−, se retiró en
−
año
000engineer
con un salario
$80career
000(año)
tanto le
mantenerse
−1 , retiredcon este sa­
(b)el A
Ph.D.
startingdehis
in 1970. ¿Qué
at a salary
ofconvendrá
$16,000(yr)
lario
dada
una
tasa
de
inflación
de
5%
anual?
−1
in 2000 at a salary of $80,000(yr) . How well did his salary keep up with an
c) Los
incrementos
matrículas
inflation
rate ofen
5%lasper
year? de las principales universidades privadas de Estados Unidos
presentan una tasa de inflación de aproximadamente 3% anual. Haga uso de esta observación
(c) Tuition increases at major private universities in the United States have led inflapara sugerir estrategias de pago para la futura matrícula de un niño en una universidad privada.
tion rates by about 3% per year. Use this observation to suggest strategies for
Suponga la ausencia de ayuda económica, una tasa de inflación anual de 5% y una matrícula ac­
paying the future tuition for a child at a private university. Assume no financial
tual de $5 000(año)−.
aid, an annual inflation rate of 5% per year, and a current tuition of $25,000(yr)−1 .
Recall the la
compound
interest
formula:
Recuérdese
fórmula de
interés
compuesto:
C(t2 )
= (1 + i)t2 −t1
C(t1 )
whereC C
can simbolizar
be cost, salary,
etc.,
t1 andetc.,
t2 indicate
times, and itiempos,
is a ratey (inflation,
donde
puede
costos,
salarios,
t y t representan
la i es una tasa (infla­
interest,
etc.)etc.)
expressed
as acomo
decimal.
ción,
interés,
expresada
un número decimal.
01­SmithVanNess.indd 20
8/1/07 12:47:46
Capítulo 2
La primera ley y otros conceptos
básicos
2.1
EXPERIMENTOS DE JOULE
El concepto moderno de calor se desarrolló después de experimentos decisivos llevados a cabo por
James P. Joule1 (1818-1889), en el sótano de su casa cerca de Manchester, Inglaterra, durante la década
de 1840.
Los experimentos de Joule se caracterizaron por ser bastante simples en los aspectos esenciales, pero
tomó precauciones minuciosas para asegurar su exactitud. En su más famosa serie de mediciones colocó en
un recipiente aislado cantidades conocidas de agua, aceite y mercurio, y después mezcló el fluido con un agitador giratorio. La cantidad de trabajo hecha sobre el fluido con el agitador fue medida con precisión y se
anotó cuidadosamente el cambio de temperatura del fluido. Joule encontró que para cada fluido se requiere
una cantidad fija de trabajo por unidad de masa, por cada grado de aumento en la temperatura causado por el
agitador, y que la temperatura original del fluido se puede restituir por la transferencia de calor a través
del simple contacto con un objeto más frío. De esta manera, Joule demostró finalmente que existe una relación cuantitativa entre el trabajo y el calor y que, por lo tanto, el calor es una forma de energía.
2.2
ENERGÍA INTERNA
En experimentos como los realizados por Joule, la energía que se agrega a un fluido en forma de trabajo se
transfiere posteriormente desde el fluido como calor. ¿Dónde se encuentra esta energía entre su adición al
fluido y la transferencia desde el mismo? Pensando de modo lógico se diría que está contenida en el fluido,
pero en otra forma llamada energía interna.
La energía interna de una sustancia no incluye la energía que ésta puede poseer como resultado de su
posición macroscópica o su movimiento. Más bien, se refiere a la energía de las moléculas internas de la sustancia. Debido a su movimiento incesante, todas las moléculas poseen energía cinética de traslación y, exceptuando a las moléculas monoatómicas, también poseen energía cinética de rotación y vibración interna. La
1 Estos experimentos y su influencia en el desarrollo de la termodinámica fueron descritos por H. J. Steffens, James Prescott Joule
and the Concept of Energy, Neale Watson Academic Publications, Inc., Nueva York, 1979.
21
02-SmithVanNess.indd 21
8/1/07 12:48:40
22
CAPÍTULO 2. La primera ley y otros conceptos básicos
adición de calor a una sustancia aumenta su actividad molecular provocando un aumento en su energía interna. El trabajo hecho sobre la sustancia puede tener el mismo efecto, como lo demostró Joule.
La energía interna de una sustancia también incluye la energía potencial que resulta de las fuerzas intermoleculares (véase la sección 16.1). En una escala submolecular la energía se asocia con los electrones y
los núcleos de los átomos, así como con la energía de enlace que resulta de las fuerzas que mantienen unidos
a los átomos como moléculas. Esta forma de energía se llama interna para distinguirla de las energías cinética y potencial, las cuales se asocian con la sustancia debido a su movimiento o posición macroscópica y se
consideran formas externas de energía.
La energía interna no tiene una definición termodinámica concisa. Es un concepto básico en termodinámica. No se puede medir de manera directa ni existen medidores de energía interna. Como resultado, los
valores absolutos son desconocidos; sin embargo, esto no es una desventaja en el análisis termodinámico, ya
que sólo se requieren cambios en la energía interna.
2.3
LA PRIMERA LEY DE LA TERMODINÁMICA
El reconocimiento del calor y la energía interna como formas de energía hace posible generalizar la ley de la
conservación de energía mecánica (sección 1.8) que incluye el calor y la energía interna además del trabajo y
las energías externa, potencial y cinética. De hecho, la generalización se puede extender a otras formas, tales
como las energías superficial, eléctrica y magnética. La abrumadora evidencia de la validez de esta generalización ha permitido elevar su categoría al nivel de una ley de la naturaleza, conocida como la primera ley de
la termodinámica. Un enunciado formal es:
Aunque la energía adopta muchas formas, la cantidad total de energía es constante, y
cuando la energía desaparece de una forma, aparecerá simultáneamente en otras formas.
Al aplicar la primera ley a un proceso dado, la esfera de influencia del proceso se divide en dos
partes: el sistema y sus alrededores. La región en la que ocurre el proceso está separada y se conoce como
sistema; todo aquello con lo que el sistema interactúa son los alrededores. Un sistema puede ser de cualquier
tamaño, y sus fronteras pueden ser reales o imaginarias, rígidas o flexibles. Con frecuencia, un sistema se
compone de una sola sustancia; en otros casos puede ser complejo. En cualquier situación, las ecuaciones
termodinámicas se escriben con referencia a un sistema bien definido. Enfocamos nuestra atención al proceso de interés particular, así como al equipo y material directamente involucrados en el proceso. De cualquier
modo, la primera ley se aplica al sistema y a sus alrededores, y no sólo al sistema. Para cualquier proceso,
la primera ley requiere:
Δ(Energía del sistema) + Δ (Energía de los alrededores) = 0
(2.1)
donde el operador diferencia “Δ” indica cambios finitos en las cantidades encerradas entre paréntesis. El sistema puede cambiar en su energía interna, en su energía potencial o cinética, así como en las energías potencial o cinética de sus partes finitas.
En el contexto de la termodinámica, el calor y el trabajo representan energía en tránsito a través de las
fronteras que separan al sistema de sus alrededores, que nunca se encuentra almacenada o contenida en el
sistema. Por otra parte, la energía potencial, cinética e interna reside y se almacena en la materia. En la práctica, la ecuación (2.1) adopta formas especiales que son adecuadas para aplicaciones prácticas específicas. El
desarrollo de estas formas y sus posteriores aplicaciones son el tema del resto de este capítulo.
02-SmithVanNess.indd 22
8/1/07 12:48:40
2.4. Balance de energía para sistemas cerrados
2.4
23
BALANCE DE ENERGÍA PARA SISTEMAS CERRADOS
Si la frontera de un sistema impide la transferencia de materia entre éste y sus alrededores, se dice que es un
sistema cerrado, y su masa es necesariamente constante. El desarrollo de los conceptos básicos en termodinámica se facilita con un examen cuidadoso de los sistemas cerrados, y por esta razón serán tratados con todo
detalle en las secciones siguientes de este capítulo. En la práctica industrial son más importantes los procesos
en los que la materia cruza las fronteras del sistema como corrientes que entran y salen del equipo de proceso. A estos sistemas se les llama abiertos, y se abordarán en este capítulo una vez presentado el material
necesario.
Dado que en un sistema cerrado no entran ni salen corrientes, no se transporta energía asociada con la
materia a través de la frontera que divide el sistema de sus alrededores. En ese caso todo intercambio de energía entre un sistema cerrado y sus alrededores aparecerá como calor y trabajo, y el cambio de energía total de
los alrededores equivaldrá a la energía neta transferida hacia o desde ellos como calor o trabajo. Por lo tanto
el segundo término de la ecuación (2.1) se puede sustituir por
Δ(Energía de los alrededores) = ± Q ± W
El calor Q y el trabajo W siempre se refieren al sistema, y la elección de los signos empleados con estas cantidades depende de la dirección de transferencia de energía con respecto al sistema, las cuales se consideran
positivas. La convención moderna de los signos hace que los valores numéricos de ambas cantidades sean positivos para la transferencia al interior del sistema desde los alrededores. Las cantidades correspondientes
tomadas con referencia a los alrededores, Qalrededores y Walrededores, tienen el signo opuesto, por ejemplo,
Qalrededores = –Q y Walrededores = –W. Con esto entendemos:
Δ(Energía de los alrededores) = Qalrededores + Walrededores = –Q –W
(2.2)
La ecuación (2.1) ahora será:2
Δ(Energía del sistema) = Q + W
(2.3)
Esta ecuación significa que el cambio en la energía total de un sistema cerrado es igual a la energía neta transferida como calor y trabajo hacia el sistema.
Los sistemas cerrados a menudo se someten a procesos durante los cuales cambia únicamente la energía
interna del sistema. Para tales procesos, la ecuación (2.2) se reduce a:
ΔU t = Q + W
(2.3)
donde Ut es la energía interna total del sistema. La ecuación (2.3) se aplica a los procesos que involucran
cambios finitos en la energía interna del sistema. Para cambios diferenciales:
dU t = dQ + dW
(2.4)
2 La convención del signo utilizado aquí es recomendada por la Unión Internacional de Química Pura y Aplicada. De cualquier modo,
la elección original del signo para el trabajo y el que se usa en las primeras cuatro ediciones de este texto es la opuesta, y el lado derecho
de la ecuación (2.2) se escribía entonces como Q – W.
02-SmithVanNess.indd 23
8/1/07 12:48:41
24
CAPÍTULO 2. La primera ley y otros conceptos básicos
En las ecuaciones (2.3) y (2.4) los símbolos Q, W y U t están relacionados con el sistema entero, que
puede ser de cualquier tamaño y debe estar definido con claridad. Todos los términos requieren expresiones
en las mismas unidades de energía. En el sistema SI la unidad es el joule. Otras unidades en uso son la caloría,
el (pie lbf) y la (Btu).
El volumen total V t y la energía interna total U t dependen de la cantidad de material en un sistema, y se
denominan propiedades extensivas. En contraste, la temperatura y la presión, las principales coordenadas
termodinámicas para fluidos homogéneos puros, son independientes de la cantidad de materia, y se conocen
como propiedades intensivas. Para un sistema homogéneo, un medio alternativo para expresar las propiedades extensivas, como V t y U t, es:
V t = mV
o
V t = nV
y
U t = mU
o
U t = nU
donde los símbolos simples V y U representan el volumen y la energía interna de una cantidad unitaria de
materia, ya sea una unidad de masa o un mol. Éstas se denominan respectivamente propiedades específicas o
molares y son intensivas, independientes de la cantidad de materia que en realidad está presente.
Aunque V t y U t para un sistema homogéneo de tamaño arbitrario son propiedades extensivas, el volumen específico y molar V (o densidad) y la energía interna específica y molar
U son intensivas.
Observe que las coordenadas intensivas T y P carecen de contrapartes extensivas.
Para un sistema cerrado de n moles, las ecuaciones (2.3) y (2.4) se escriben ahora como:
Δ(nU) = n ΔU = Q + W
(2.5)
d(nU) = n dU = dQ + dW
(2.6)
En esta forma, dichas ecuaciones muestran de manera explícita la cantidad de sustancia contenida en el sistema.
Las ecuaciones termodinámicas con frecuencia se escriben para una unidad representativa de la cantidad de materia, ya sea unidad de masa o mol. Así para n = 1, las ecuaciones (2.5) y (2.6) serán:
ΔU = Q + W
y
dU = dQ + dW
La base para Q y W se da a entender siempre por la masa o número de moles asociado con el lado izquierdo
de la ecuación de energía.
La ecuación (2.6) es la fuente fundamental de todas las relaciones de una propiedad que une a la energía interna con las cantidades mensurables. No representa una definición de la energía interna, ya que ésta no
existe. Tampoco conduce a los valores absolutos para la energía interna. Lo que proporciona es la manera de
calcular los cambios en esta propiedad; sin ésta, la primera ley de la termodinámica no se podría formular. De
hecho, esa ley requiere de una afirmación previa de la existencia de la energía interna, la naturaleza esencial
que se resume en el siguiente axioma:
Existe una forma de energía, conocida como energía interna U, que es una propiedad intrínseca del sistema y que se relaciona mediante una función con las coordenadas mensurables que caracterizan a dicho sistema. Para sistemas cerrados, que no están en
movimiento, los cambios en esta propiedad se dan por las ecuaciones (2.5) y (2.6).
02-SmithVanNess.indd 24
8/1/07 12:48:42
25
2.4. Balance de energía para sistemas cerrados
Ejemplo 2.1
Fluye agua en una cascada de 100 m de altura. Considérese a 1 kg de agua como el sistema y suponga que no hay un intercambio de energía con los alrededores.
a) ¿Cuál es la energía potencial del agua en la parte superior de la cascada con respecto a la base
de la misma?
b) ¿Cuál es la energía cinética del agua justo antes de llegar al fondo?
c) Después de que el kg de agua entra al río que corre bajo la cascada, ¿qué cambio ocurre en su
estado?
Solución 2.1
El kg de agua no intercambia energía con los alrededores. Por lo tanto, para cada parte del proceso la ecuación (2.1) se reduce a:
Δ(Energía del sistema) = ΔU + ΔE K + ΔE P = 0
a) De la ecuación (1.7), con g igual a su valor estándar,
E P = mzg = 1 kg × 100 m × 9.8066 m s –2
kg m 2
= 980.66
= 980.66 N m = 980.66 J
s2
b) Durante la caída libre del agua no existe un mecanismo para convertir energía potencial o cinética en energía interna. Así, ΔU debe ser cero:
ΔE K + ΔE P = E K 2 – E K 1 + E P 2 – E P 1 = 0
Como una excelente aproximación, sea EK1 = EP2 = 0. Entonces,
E K 2 = E P 1 = 980.66 J
c) Conforme el kg de agua choca con el fondo y se mezcla con el agua que forma parte del río, la
turbulencia resultante tiene el efecto de convertir la energía cinética en energía interna. Durante
este proceso, ΔEP es esencialmente cero, y la ecuación (2.1) será:
ΔU + ΔE K = 0
o
ΔU = E K 2 – E K 3
No obstante, suponiendo que la velocidad del río es baja, podemos despreciar a EK3. De este modo,
ΔU = E K 2 = 980.66 J
El resultado global del proceso es la conversión de la energía potencial del agua en energía
interna de la misma. Este cambio en la energía interna se manifiesta por un aumento de la temperatura del agua. Ya que se requiere una cantidad de energía de 4 184 J kg–1 para aumentar en 1 °C
la temperatura del agua, el incremento en la temperatura es de 980.66/4 184 = 0.234 °C, suponiendo que no hay transferencia de calor con los alrededores.
02-SmithVanNess.indd 25
8/1/07 12:48:42
CHAPTER
CHAPTER2.
TheFirst
FirstLaw
Lawand
andOther
OtherBasic
BasicConcepts
Concepts
CHAPTER
2.2.The
The
First
Law
and
Other
Basic
Concepts
26
26
26
26
CAPÍTULO 2. La primera ley y otros conceptos básicos
2.5
2.5 THERMODYNAMIC
THERMODYNAMICSTATE
STATEAND
ANDSTATE
STATEFUNCTIONS
FUNCTIONS
2.5
THERMODYNAMIC
STATE
AND
STATE
FUNCTIONS
2.5 ESTADO TERMODINÁMICO Y FUNCIONES DE ESTADO
The
Thenotation
notationof
Eqs.(2.3)
(2.3)through
through(2.6)
(2.6)suggests
suggeststhat
thatthe
theinternal-energy
internal-energyterms
termson
onthe
theleft
left
The
notation
ofofEqs.
Eqs.
(2.3)
through
(2.6)
suggests
that
the
internal-energy
terms
on
the
left
are
are
different
different
in
in
kind
kind
from
from
the
the
quantities
quantities
on
on
the
the
right.
right.
Those
Those
on
on
the
the
left
left
reflect
reflect
changes
changes
in
in
the
are
different
in
kind
from
the
quantities
on
the
right.
Those
on
the
left
reflect
changes
in
the
La notación de las ecuaciones (2.3) a la (2.6) sugiere que los términos de energía interna en el ladothe
izquierdo
thermodynamic
state
state
of
thesystem
system
as
reflected
byits
thermodynamic
properties,
properties,
amongwhich
which
thermodynamic
state
ofofthe
the
system
asasreflected
reflected
by
itsitsthermodynamic
thermodynamic
properties,
among
which
son dethermodynamic
una
clase diferente
de
los
de
la derecha.
Los by
términos
de la izquierda
reflejan among
cambios
en el estado
are
aretemperature,
temperature,
pressure,
pressure,
anddensity.
density.
For
Formediante
aaahomogeneous
homogeneous
pure
puresubstance
substance
we
weknow
knowfrom
fromla temare
temperature,
pressure,
and
density.
For
homogeneous
pure
substance
we
know
from
termodinámico
del sistema.
Dichoand
estado
se refleja
sus propiedades
termodinámicas
como
experience
experience
that
that
fixing
fixing
two
two
of
of
these
these
properties
properties
also
also
fixes
fixes
all
all
the
the
others,
others,
and
and
thus
thus
determines
determines
experience
that
fixing
two
of
these
properties
also
fixes
all
the
others,
and
thus
determines
its
peratura, la presión y la densidad. Se sabe por experiencia que para una sustancia pura homogénea,its
alitsfijar dos
thermodynamic
thermodynamic
state.
state.
For
For
example,
example,
nitrogen
nitrogen
gas
gas
at
at
a
a
temperature
temperature
of
of
300
300
K
K
and
and
a
a
pressure
pressure
thermodynamic
state.
For
example,
nitrogen
gas
at
a
temperature
of
300
K
and
a
pressure
de sus propiedades termodinámicas, automáticamente se establecen las demás y así se determina su estado
55 5kPa
of
10
kPa
(1
bar)
bar)has
hasel
aaanitrógeno
fixed
fixedspecific
specific
volume
volume
and
andaaafixed
fixed
molar
molar
internal
energy.
energy.105 kPa
ofof10
10
kPa
(1(1
bar)
has
fixed
specific
volume
orordensity
density
and
fixed
molar
internal
energy.
termodinámico.
Por
ejemplo,
en forma
de or
gas
adensity
una temperatura
de
300 internal
K
y una presión
Indeed,
Indeed,
it
it
has
has
a
a
complete
complete
set
set
of
of
intensive
intensive
thermodynamic
thermodynamic
properties.
properties.
If
If
this
this
gas
gas
is
is
heated
heated
or
Indeed,
it
has
a
complete
set
of
intensive
thermodynamic
properties.
If
this
gas
is
heated
ororconjun(1 bar) tiene volumen específico o densidad fijos y una energía interna molar fija. En efecto, tiene un
cooled,
cooled,
compressed
compressed
or
or
expanded,
expanded,
and
and
then
then
returned
returned
to
to
its
its
initial
initial
temperature
temperature
and
and
pressure,
pressure,
its
cooled,
compressed
or
expanded,
and
then
returned
to
its
initial
temperature
and
pressure,
its
to completo de propiedades termodinámicas intensivas. Si este gas se calienta o enfría, se comprimeits
o expanintensiveproperties
properties
are
are
restored
to
their
initialvalues.
values.
They
They
do
do
not
notdepend
dependon
onthe
thepast
pastrecuperarán
history
history
intensive
properties
are
restored
tototheir
their
initial
values.
They
do
not
depend
on
the
past
history
de, y aintensive
continuación
regresa
a restored
su
temperatura
yinitial
presión
iniciales,
sus
propiedades
intensivas
sus
the
thesubstance
substance
nor
noron
onthe
themeans
means
by
bywhich
which
reachesaaanterior
agiven
givenstate.
state.
They
Theydepend
depend
onlymedios
on
on por
ofof
the
substance
nor
on
the
means
by
which
ititreaches
reaches
given
state.
They
depend
only
on
valoresof
iniciales.
Tales propiedades
no
dependen
de itla
historia
de
la sustancia
ni deonly
los
present
present
conditions,
conditions,
however
however
reached.
reached.
Such
Such
quantities
are
areknown
known
as
state
functions.
functions.
When
When
two
two
present
conditions,
however
reached.
Such
quantities
are
known
asasstate
state
functions.
When
two
los cuales
alcanzan
un estado
determinado,
sólo
dequantities
las
condiciones
actuales
sin importar
cómo
éstas
se hayan
33 3the
of
of
them
them
are
are
held
held
at
at
fixed
fixed
values
values
for
for
a
a
homogeneous
homogeneous
pure
pure
substance,
substance,
the
thermodynamic
thermodynamic
state
state
of
them
are
held
at
fixed
values
for
a
homogeneous
pure
substance,
the
thermodynamic
state
alcanzado. Dichas cantidades son conocidas como funciones de estado. Cuando dos de ellas se mantienen en
3 se determina
the
thesubstance
substance
is
fullydetermined.
determined.
This
Thismeans
means
that
thataaastate
state
function,
function,such
such
as
specific
internal
internal de la
ofof
the
substance
isisfully
fully
determined.
This
means
that
state
function,
such
asasspecific
specific
internal
valoresof
fijos
para
una sustancia
pura homogénea,
totalmente
el
estado
termodinámico
energy,
energy,
is
is
a
a
property
property
that
that
always
always
has
has
a
a
value;
value;
it
it
may
may
therefore
therefore
be
be
expressed
expressed
mathematically
mathematically
as
energy,
is
a
property
that
always
has
a
value;
it
may
therefore
be
expressed
mathematically
as
sustancia. Esto significa que una función de estado, tal como la energía interna específica, es una as
propiedad
a
a
function
function
of
of
such
such
coordinates
coordinates
as
as
temperature
temperature
and
and
pressure,
pressure,
or
or
temperature
temperature
and
and
density,
density,
and
and
its
its
a
function
of
such
coordinates
as
temperature
and
pressure,
or
temperature
and
density,
and
its
que siempre tiene un valor; por lo tanto, se puede expresar matemáticamente como una función de coordenavalues
values
may
may
be
identified
withpoints
points
on
onaaagraph.
graph.y densidad, mientras que sus valores se pueden idenvalues
may
bebeidentified
identified
with
points
on
graph.
das tales
como
la temperatura
ywith
presión,
o temperatura
On
On
the
the
other
other
hand,
hand,
the
the
terms
terms
on
on
the
the
right
rightsides
sidesof
Eqs.(2.3)
(2.3)through
through(2.6),
(2.6),representing
representing
On
the
other
hand,
the
terms
on
the
right
sides
ofofEqs.
Eqs.
(2.3)
through
(2.6),
representing
tificar como puntos en una gráfica.
heat
heat
and
and
work
work
quantities,
quantities,
are
are
not
not
properties;
properties;
they
they
account
account
for
for
the
the
energy
energy
changes
changes
that
thatoccur
occur
heat
and
work
quantities,
are
not
properties;
they
account
for
the
energy
changes
that
occur
Por otra parte, los términos en el lado derecho de las ecuaciones (2.3) a la (2.6) representan
cantidades
in
in
the
the
surroundings.
surroundings.
They
They
depend
depend
on
on
the
the
nature
nature
of
of
the
the
process,
process,
and
and
may
may
be
be
associated
associated
with
with
in
the
surroundings.
They
depend
on
the
nature
of
the
process,
and
may
be
associated
with
de calor y trabajo, que no son propiedades; consideran los cambios de energía que ocurren en los alrededores.
areas
areas
rather
than
thanpoints
points
onaaagraph,
graph,
as
suggested
by
byFig.
Fig.1.3.
1.3.Although
Although
time
isnot
not
aaathermothermoareas
rather
than
points
on
graph,
asassuggested
suggested
by
Fig.
1.3.
Although
time
is
not
thermoDependen
derather
la naturaleza
del on
proceso
y pueden
estar asociados
con
áreas
antestime
queis
con
puntos
en la gráfica
dynamic
dynamic
coordinate,
coordinate,
the
the
passage
passage
of
of
time
time
is
is
inevitable
inevitable
whenever
whenever
heat
heat
is
is
transferred
transferred
or
or
work
workis
is tiemdynamic
coordinate,
the
passage
of
time
is
inevitable
whenever
heat
is
transferred
or
work
isdel
como lo sugiere la figura 1.3. Aunque el tiempo no es una coordenada termodinámica, el transcurso
accomplished.
accomplished.
accomplished.
po es inevitable cada vez que se transfiere calor o se realiza un trabajo.
The
Thedifferential
differential
of
state
function
represents
represents
an
infinitesimal
infinitesimal
change
change
in
its
value.
value.
InteInteThe
differential
ofofaaastate
state
function
represents
anan
infinitesimal
change
ininsu
itsits
value.
InteLa diferencial
de una función
defunction
estado
representa
un
cambio
infinitesimal
en
valor.
La
integración
gration
gration
of
of
such
such
a
a
differential
differential
results
results
in
in
a
a
finite
finite
difference
difference
between
between
two
two
of
of
its
its
values,
values,
e.g.:
e.g.:
gration
of
such
a
differential
results
in
a
finite
difference
between
two
of
its
values,
e.g.:
de esta diferencial da como resultado una diferencia finita entre dos de sus valores, por ejemplo:
���PP2P2
���VV2V2
22
22
dddPPP=
−
−
P
P
=
=
�P
�P
and
and
dddVVV=
�V
y
==PP2P
−
P
=
�P
and
==VV2V2 2−
−−VV1V1 1=
==�V
�V
2
1
1
2
1
PP11P11
VV11V11
The
Thedifferentials
differentials
of
heatand
andno
work
work
arenot
notchanges,
changes,
but
butare
areinfinitesimal
infinitesimal
amounts.
amounts.
When
When
inteinte- estas
Las diferenciales
de calor
trabajo
sonare
cambios
sino cantidades
infinitesimales.
Cuando
se integran,
The
differentials
ofofyheat
heat
and
work
are
not
changes,
but
are
infinitesimal
amounts.
When
integrated,
grated,
these
these
differentials
differentials
give
give
not
notfinite
finite
changes,but
but
finite
finiteamounts.
amounts.Thus,
Thus,
diferenciales
no
presentan
cambios
finitos,
sinochanges,
cantidades
finitas.
Así,
grated,
these
differentials
give
not
finite
changes,
but
finite
amounts.
Thus,
���
���
dddQ
and
and
ddW
dWW=
y
QQ=
==Q
QQ
and
==W
WW
For
For
aaaclosed
closed
system
system
undergoing
undergoing
the
thesame
same
change
change
state
by
byseveral
severalde
processes,
processes,
experiexperiPara un
sistema
cerrado
sometido
al mismo
cambio
de in
estado
por
medio
diferentes
procesos, los
For
closed
system
undergoing
the
same
change
ininstate
state
by
several
processes,
experiment
ment
shows
shows
that
that
the
the
amounts
amounts
of
of
heat
heat
and
and
work
work
required
required
differ
differ
for
for
different
different
processes,
processes,
but
but
that
experimentos
muestran
queamounts
las cantidades
calor
y trabajo
difieren para
los distintos
procesos,
ment shows
that the
of heatde
and
work
requiredrequeridas
differ for different
processes,
but that
that
the
the
sum
sum
Q
Q
+
+
W
W
is
is
the
the
same
same
for
for
all
all
processes.
processes.
This
This
is
is
the
the
basis
basis
for
for
identification
identification
of
of
internal
internal
enen- de la
pero que
suma
QW
+W
es la
misma
para
todo proceso.
es el fundamento
para of
la internal
identificación
thelasum
Q+
is the
same
for all
processes.
This isÉste
the basis
for identification
entt is
t isgiven
t by
ergy
as
state
statefunction.
function.
The
The
same
sameEl
value
value
of
�U
given
Eq.
Eq.
(2.3)
(2.3)
regardless
regardless
of
the
process,
process,
energíaergy
interna
función
de
estado.
mismo
valor
ΔUby
está
dado
por
la ecuación
(2.3)
independienergy
asasaaacomo
state
function.
The
same
value
ofof�U
�U
isde
given
by
Eq.
(2.3)
regardless
ofofthe
the
process,
provided
provided
only
only
that
that
the
the
change
change
in
in
the
the
system
system
is
is
between
between
the
the
same
same
initial
initial
and
and
final
final
states.
states.
only that
the change
in the system
is between
theentre
samelos
initial
and final
states.
tementeprovided
del proceso,
siempre
que el cambio
en el sistema
ocurra
mismos
estados
inicial y final.
33For
3 For
systems
systems
more
more
complex
complex
than
than
aa simple
asimple
simple
homogeneous
homogeneous
pure
pure
substance,
substance,
the
the
number
number
of
state
state
functions
functions
that
that
must
must
be
For
systems
more
complex
than
homogeneous
pure
substance,
the
number
ofof
state
functions
that
must
bebe
arbitrarily
arbitrarilyspecified
specifiedin
orderto
definethe
thestate
stateof
thesystem
systemmay
maybe
differentfrom
fromtwo.
two.The
Themethod
methodof
determining
arbitrarily
specified
ininorder
order
totodefine
define
the
state
ofofthe
the
system
may
bebedifferent
different
from
two.
The
method
ofofdetermining
determining
this
this
number
isisis
the
the
subject
subject
of
Sec.
Sec.
2.7.simple sustancia pura homogénea, el número de funciones de estado que deben especifithis
number
the
subject
ofof
Sec.
2.7.
Para
losnumber
sistemas
más
complejos
que2.7.
una
carse en forma arbitraria con la finalidad de definir el estado del sistema puede ser distinto de dos. El método para determinar este número
es el tema de la sección 2.7.
3
02-SmithVanNess.indd 26
8/1/07 12:48:44
2.5. Thermodynamic State and State Functions
27
2.5. Estado termodinámico y funciones de estado
Example 2.2
27
A gas is confined in a cylinder by a piston. The initial pressure of the gas is 7 bar,
Ejemplo
and the2.2
volume is 0.10 m3 . The piston is held in place by latches in the cylinder wall.
apparatus
is placed
total vacuum.
What
is the
change
the
Un gasThe
estáwhole
confinado
en un cilindro
porinuna pistón.
La presión
inicial
del energy
gas es de
7 bar, of
y el
volumen
3
apparatus
if
the
restraining
latches
are
removed
so
that
the
gas
suddenly
expands
es 0.10 m . El pistón se mantiene en su lugar mediante seguros en las paredes del cilindro.toTodo el
its initialenvolume,
piston
striking
latches
at the end
the process?
aparatodouble
se encuentra
un vacíothe
total.
¿Cuál
es el other
cambio
en la energía
del of
aparato
si se eliminan los
seguros restrictivos, permitiendo así que el gas se expanda de manera repentina al doble de su volumen original,
y al final 2.2
del proceso el pistón choca con otros seguros?
Solution
SoluciónBecause
2.2 the question concerns the entire apparatus, the system is taken as the
gas, piston, and cylinder. No work is done during the process, because no force
Dado que
la pregunta
quemoves,
ver conand
todo
aparato
se tomaráthrough
como sistema
el gas,surel pistón y
external
to thetiene
system
noelheat
is transferred
the vacuum
el cilindro.
Durante
el
proceso
no
se
realiza
trabajo,
ya
que
no
se
aplica
una
fuerza
externa al
rounding the apparatus. Hence Q and W are zero, and the total energy of the
sistema,system
ni se transfiere
calor
del
vacío
hacia
los
alrededores
del
aparato.
Por
lo
tanto,
Q
does not change. Without further information we can say nothing abouty W son
cero y lathe
energía
total del
cambia.
Sin mayor
es posible
nada más
distribution
ofsistema
energy no
among
the parts
of theinformación
system. Thisnomay
well bedecir
differacerca de
la
distribución
de
la
energía
entre
las
partes
del
sistema.
Esto
puede
ser
diferente
de la
ent than the initial distribution.
distribución inicial.
Example 2.3
If the process
Ejemplo
2.3 described in Ex. 2.2 is repeated, not in a vacuum but in air at atmo-
spheric pressure of 101.3 kPa, what is the energy change of the apparatus? Assume
Si se repite
el proceso
en elbetween
ejemplo 2.2,
en el vacío
el aire a unaair
presión
atmosféthe rate
of heatdescrito
exchange
the no
apparatus
andsino
theensurrounding
is slow
rica de compared
101.3 kPa,with
¿cuál
esrate
el cambio
enthe
la energía
aparato? Suponga que la rapidez de intercamthe
at which
processdel
occurs.
bio de calor entre el aparato y el aire de los alrededores es lenta en comparación con la rapidez con la
que ocurre el proceso.
Solution 2.3
SoluciónThe
2.3system is chosen as before, but here work is done by the system in pushing
the atmosphere.
It is evaluated
product
of the force
ofsistema
atmospheric
Se eligeback
el mismo
sistema que antes,
pero ahoraaselthe
trabajo
es realizado
por el
al empujar el
pressure
on
the
back
side
of
the
piston
F
=
P
A
and
the
displacement
of the por la
atm
pistón de regreso a la atmósfera. Este trabajo se evalúa como el producto de la fuerza ejercida
t /A. Here, A is the area of the piston and �V t is the volume
�l =en�V
presión piston
atmosférica
el lado
opuesto del pistón F = Patm A y el desplazamiento del mismo, Δ l =
gas.del
This
is work
the system
on thedelsurroundings,
is a realizaΔV t / A. change
Aquí, A of
es the
el área
pistón
y ΔV tdone
es el by
cambio
de volumen
gas. Esto es eland
trabajo
negative
quantity;
thus,
do por el sistema sobre los alrededores, y es una cantidad negativa; en estos términos,
kN
W = −F �l = −Patm �V t = −(101.3)(0.2 − 0.1) kPa m3 = −10.13 2 m3
m
W = −10.13 kN m = −10.13 kJ
W = –10.13 kN m = –10.13 kJ
Heat transfer between the system and surroundings is also possible in this case,
En este but
casothe
también
esisposible
transferencia
de calor
entre el has
sistema
y susand
alrededores,
problem
workedlafor
the instant after
the process
occurred
before pero
el problema
está planeado
para elhas
instante
después
que elThus
proceso
ocurrido
y antes
appreciable
heat transfer
had time
to takedeplace.
Q is ha
assumed
to be
zero de que
se tengainelEq.
tiempo
(2.2),suficiente
giving para que se presente una transferencia de calor apreciable. Así, suponiendo que Q es cero en la ecuación (2.2), obtenemos
�(Energy of the system) = Q + W = 0 − 10.13 = −10.13 kJ
Δ(Energía del sistema) = Q + W = 0 – 10.13 = –10.13 kJ
The total energy of the system has decreased by an amount equal to the work done
on the
surroundings.
La energía
total
del sistema ha disminuido en una cantidad igual al trabajo realizado sobre los
alrededores.
o
02-SmithVanNess.indd 27
or
8/1/07 12:48:46
28
CAPÍTULO 2. La primera ley y otros conceptos básicos
Ejemplo 2.4
En la figura 2.1, cuando se lleva un sistema del estado a al b a lo largo de la trayectoria acb, fluyen
100 J de calor hacia el sistema y éste realiza 40 J de trabajo.
a) ¿Cuánto calor fluye hacia el sistema a lo largo de la trayectoria aeb, si el trabajo hecho por el
sistema es de 20 J?
b) El sistema regresa de b a a siguiendo la trayectoria bda. Si el trabajo hecho sobre el sistema es
de 30 J, ¿el sistema absorbe o libera calor? ¿Cuánto?
b
c
d
P
Figura 2.1: Diagrama para el ejemplo 2.4.
e
a
V
Solución 2.4
Supóngase que el sistema se modifica sólo en su energía interna y que se puede aplicar la ecuación (2.3). Para la trayectoria acb, y desde luego para cualquier trayectoria que conduzca de a
hacia b,
t = Q
ΔUba
acb + Wacb = 100 – 40 = 60 J
a) Para la trayectoria aeb,
t = 60 = Q
ΔUba
aeb + Waeb = Q aeb – 20
de donde
Q aeb = 80 J
b) Para la trayectoria bda,
t = – ΔU t = – 60 = Q
ΔUba
bda + W bda = Q bda + 30
ba
y
Q bda = –60 – 30 = –90 J
En consecuencia, se transfiere calor desde el sistema hacia los alrededores.
02-SmithVanNess.indd 28
8/1/07 12:48:47
29
2.6. Equilibrio
2.6
EQUILIBRIO
Equilibrio es una palabra que denota una condición estática, es decir, la ausencia de un cambio. En termodinámica significa no sólo la ausencia de un cambio sino de cualquier tendencia hacia el cambio en una escala
macroscópica. De esta manera existe un sistema en equilibrio bajo la condición de que no puede ocurrir en él
ningún cambio de estado. Puesto que cualquier tendencia hacia el cambio es causada por una fuerza impulsora de uno u otro tipo, la ausencia de estas tendencias también indica la carencia de cualquier fuerza impulsora.
Por tanto, en un sistema en equilibrio todas las fuerzas están en un balance exacto. Si en realidad ocurre un
cambio en un sistema que no está en equilibrio, éste depende tanto de la resistencia como de la fuerza impulsora. Muchos sistemas experimentan cambios no mensurables aun bajo la influencia de grandes fuerzas impulsoras, porque su resistencia al cambio es muy grande.
Las diferentes clases de fuerzas impulsoras tienden a producir distintos tipos de cambios. Por ejemplo,
las fuerzas mecánicas no equilibradas como la presión en un pistón llegan a ocasionar transferencia de energía
en forma de trabajo; las diferencias de temperatura suelen producir flujo de calor; los gradientes en el potencial químico tienden a originar que las sustancias sean transferidas de una fase a otra. En equilibrio, tales
fuerzas están balanceadas.
En muchas aplicaciones de la termodinámica no interesan las reacciones químicas. Por ejemplo, una
mezcla de hidrógeno y oxígeno en condiciones ordinarias no está en equilibrio químico, debido a la gran
fuerza impulsora para la formación de agua. Sin embargo, si la reacción química no se ha iniciado, el sistema
puede estar en equilibrio térmico y mecánico a largo plazo, y sólo se analizarán los procesos puramente físicos sin considerar las posibles reacciones químicas.
2.7
REGLA DE LAS FASES
Como se ha indicado con anterioridad, el estado de un fluido homogéneo puro se determina cada vez que se
dan valores definidos a dos propiedades termodinámicas intensivas. En contraste, cuando dos fases están en
equilibrio, el estado del sistema se establece al especificar sólo una propiedad. Por ejemplo, una mezcla de
vapor y agua líquida en equilibrio a 101.33 kPa sólo puede existir a 100 °C. Es imposible cambiar la temperatura sin modificar también la presión, si se quiere que el vapor y el agua líquida continúen existiendo en
equilibrio.
Para sistemas de fases múltiples en equilibrio, el número de variables independientes que deben fijarse
en forma arbitraria a fin de establecer su estado intensivo se proporciona por la célebre regla de las fases de J.
Willard Gibbs,4 quien la dedujo por medio de un razonamiento teórico en 1875. Aquí se presenta sin comprobación, en la forma que se aplica para sistemas que no tienen reacciones químicas:5
F=2–p +N
(2.7)
donde p es el número de fases, N es el número de especies químicas y F representa los grados de libertad del
sistema.
4
Josiah Willard Gibbs (1839-1903), físico matemático estadounidense.
La justificación de la regla de las fases para un sistema que no contiene reacciones químicas está dada en la sección 10.2, mientras
que la regla de las fases para sistemas con reacciones químicas se considera en la sección 13.8.
5
02-SmithVanNess.indd 29
8/1/07 12:48:48
30
CAPÍTULO 2. La primera ley y otros conceptos básicos
El estado intensivo de un sistema en equilibrio se establece cuando su temperatura, presión y composiciones de todas las fases se encuentran fijas. Debido a eso, éstas son variables de la regla de las fases, pero no
todas son independientes. La regla de las fases proporciona el número de variables de este conjunto, las cuales
deben especificarse de manera arbitraria para fijar todas las demás variables restantes de la regla de las fases,
y con ello el estado intensivo del sistema.
Una fase es una región homogénea de materia. Un gas o mezcla de gases, un líquido o una solución líquida y un sólido cristalino son ejemplos de fases. Una fase no necesita ser continua; son ejemplos de fases
discontinuas un gas disperso como burbujas en un líquido, un líquido disperso como gotas en otro líquido con
el cual no es miscible, y cristales de un sólido dispersos ya sea en un gas o en un líquido. En cada caso, una
fase dispersa está distribuida en todas las partes de una fase continua. Siempre ocurre un cambio repentino de
las propiedades en las fronteras entre las fases. Pueden coexistir varias fases, pero deben estar en equilibrio
para aplicar la regla de las fases. Un ejemplo de un sistema de tres fases en equilibrio es una solución salina
acuosa que está saturada en su punto de ebullición con un exceso de cristales de sal. Las tres fases (π = 3) son
la sal cristalina, la solución acuosa saturada y el vapor generado en el punto de ebullición. Las dos especies
químicas (N = 2) son agua y sal. Para este sistema, F = 1.
Las variables de la regla de las fases son propiedades intensivas, las cuales son independientes de la
extensión del sistema y de las fases individuales. De este modo, la regla de las fases proporciona la misma
información tanto para sistemas grandes como para sistemas pequeños y para cantidades diferentes de fases presentes. Además, sólo las composiciones de fases individuales son variables de las reglas de las fases.
Las composiciones totales o globales no son variables de la regla de las fases cuando ésta presenta más de
una fase.
El número mínimo de grados de libertad para cualquier sistema es cero. Cuando F = 0 el sistema es
invariante; la ecuación (2.7) se transforma en p = 2 + N. Este valor de p es el número máximo de fases que
pueden coexistir en equilibrio para un sistema que contiene N especies químicas. Cuando N = 1, este número
es 3, característico de un punto triple (véase la sección 3.1). Por ejemplo, el punto triple del agua, donde el
líquido, el vapor y la forma común de hielo existen juntos en equilibrio, se presenta a 0.01 °C y 0.0061 bar.
Cualquier cambio de estas condiciones provoca la desaparición de al menos una de estas fases.
Ejemplo 2.5
¿Cuántos grados de libertad tienen cada uno de los siguientes sistemas?
a) Agua líquida en equilibrio con su vapor.
b) Agua líquida en equilibrio con una mezcla de vapor de agua y nitrógeno.
c) Una solución líquida de alcohol en agua en equilibrio con su vapor.
Solución 2.5
a) El sistema contiene una sola especie química que existe en dos fases (una líquida y otra gaseosa). Así,
F = 2 – π+ N = 2 – 2 + 1 = 1
02-SmithVanNess.indd 30
8/1/07 12:48:48
31
2.8. El proceso reversible
Este resultado está de acuerdo con el hecho de que para una presión de agua dada sólo se tiene un
punto de ebullición. Es posible especificar la temperatura o la presión, no ambas, para un sistema
que contiene agua en equilibrio con su vapor.
b) En este caso están presentes dos especies químicas. Hay otra vez dos fases.
Ahora,
F = 2 – π+ N = 2 – 2 + 2 = 2
La adición de un gas inerte a un sistema de agua en equilibrio con su vapor cambia las características del sistema. Ahora la temperatura y la presión pueden variar en forma independiente, pero
una vez que se fijan el sistema descrito puede existir en equilibrio sólo para una composición
particular de la fase vapor. (Si se considera que la solubilidad del nitrógeno en agua es despreciable, la fase líquida es agua pura.)
c) Aquí N = 2, y π = 2, y
F = 2 – π+ N = 2 – 2 + 2 = 2
Las variables de la regla de las fases son la temperatura, la presión y las composiciones de la
fase. Las variables de la composición son las fracciones de masa o de moles de las especies en
una fase, y deben sumar la unidad para cada fase. De esa forma, si se fija la fracción molar de
agua en la fase líquida, la fracción molar de alcohol se fija automáticamente. Estas dos composiciones no se especifican en forma arbitraria.
2.8
EL PROCESO REVERSIBLE
El desarrollo de la termodinámica se facilita por la introducción de una clase especial de proceso en sistemas
cerrados que se caracteriza como reversible:
Un proceso es reversible cuando la dirección puede ser invertida en cualquier punto por
un cambio infinitesimal en las condiciones externas.
Expansión reversible de un gas
La naturaleza de los procesos reversibles se ilustra con el ejemplo de una expansión simple de gas en una
combinación pistón/cilindro. El aparato que se muestra en la figura 2.2 se supone que está en un espacio vacío. Se elige como sistema el gas retenido en el interior del cilindro; todo lo demás son los alrededores. Los
procesos de expansión ocurren cuando se retira masa del pistón. Con la finalidad de simplificarlo, se supone
que el pistón se desliza dentro del cilindro sin presentar fricción, y que ni el pistón ni el cilindro absorben o
transmiten calor. Además, ya que la densidad del gas en el cilindro es baja y que la masa del gas es pequeña,
se ignoran los efectos de la gravedad sobre el contenido del cilindro. Esto significa que los gradientes de presión inducidos por la gravedad en el gas son muy pequeños en comparación con su presión, y que los cambios
en la energía potencial del gas son despreciables en comparación con los cambios de la energía potencial del
montaje del pistón.
02-SmithVanNess.indd 31
8/1/07 12:48:48
32
CAPÍTULO 2. La primera ley y otros conceptos básicos
m
Figura 2.2: Expansión de un gas.
Δl
El pistón de la figura 2.2 confina el gas a una presión suficiente para equilibrar el peso del pistón y todo
lo que éste soporta. Tal condición de equilibrio para el sistema no debe tender al cambio. Se debe eliminar
masa del pistón para que se eleve. Primero suponga que una masa m se desliza repentinamente del pistón a
una repisa (al mismo nivel). El pistón se acelera hacia arriba y alcanza su velocidad máxima en el punto en
que la fuerza ascendente sobre el pistón se equilibra con su peso. A continuación el momentum del pistón lo
lleva a su nivel máximo donde invierte la dirección. Si el pistón se mantuviera en esta posición de elevación
máxima, su energía potencial aumentaría a un valor muy cercano al trabajo realizado por el gas durante el
recorrido inicial. Sin embargo, cuando no existe restricción, el pistón oscila con la disminución de la amplitud, hasta alcanzar finalmente una nueva posición de equilibrio a un nivel arriba de su posición inicial.
Las oscilaciones del pistón se amortiguan, ya que la naturaleza viscosa del gas de manera gradual convierte el movimiento ordenado global de las moléculas en un movimiento molecular caótico. Este proceso
disipativo transforma parte del trabajo hecho inicialmente por el gas para elevar el pistón, de nuevo en energía
interna del gas. Una vez que se inicia el proceso, ningún cambio infinitesimal en las condiciones externas
puede invertir su dirección; el proceso es irreversible.
Todos los procesos que se llevan a cabo en un tiempo finito con sustancias reales se acompañan en
cierto grado por efectos disipativos de una u otra clase y, por lo tanto, todos son irreversibles. No obstante, se
pueden imaginar procesos que están libres de efectos disipativos. Para el proceso de expansión de la figura
2.2, tales efectos tienen su origen en el retiro repentino de una masa finita del pistón. El desequilibrio resultante de fuerzas que actúan en el pistón causa su aceleración y conduce a su subsecuente oscilación. El retiro
repentino de pequeñas cantidades de masa reduce pero no elimina este efecto disipativo. Incluso el retiro de
una masa infinitesimal lleva a que el pistón oscile con una amplitud infinitesimal y, por lo tanto, a un efecto
disipativo. Sin embargo, es posible imaginar un proceso en donde se retiran pequeñas cantidades de masa una
tras otra con una rapidez tal que la elevación del pistón es continua, con pequeñas oscilaciones sólo al final
del proceso.
02-SmithVanNess.indd 32
8/1/07 12:48:49
2.8. El proceso reversible
33
El caso límite del retiro de una sucesión de masas infinitesimales del pistón se aproxima cuando las
masas m de la figura 2.2 son sustituidas por una pila de polvo, soplando en una línea de corriente muy fina
desde el pistón. Durante este proceso, el pistón se eleva con una rapidez uniforme pero muy lenta, y el polvo
se almacena siempre a niveles más altos. El sistema nunca se desplaza más que de manera diferencial, ya sea
desde su equilibrio interno o del equilibrio con sus alrededores. Si se detiene el traslado de polvo desde el
pistón y se invierte la dirección de la transferencia del polvo, el proceso invierte su dirección y continúa a la
inversa a lo largo de su trayectoria original. Al final, tanto el sistema como sus alrededores regresan virtualmente a sus condiciones iniciales. El proceso original se aproxima a la reversibilidad.
Sin la suposición de un pistón sin fricción no podemos suponer un proceso reversible. Si el pistón se
pega debido a la fricción, se debe eliminar una masa finita antes de que se libere el pistón. De esta forma, no
se mantiene la condición de equilibrio que es necesaria para la reversibilidad. Por otro lado, la fricción entre
las dos partes deslizantes es un mecanismo para la disipación de la energía mecánica en energía interna.
Este análisis se ha centrado en un simple proceso de sistema cerrado, la expansión de un gas en un cilindro. El proceso opuesto, la compresión de un gas en un cilindro, se describe de la misma manera. De cualquier modo, hay muchos procesos que son manejados por el desequilibrio de fuerzas distintas de las
mecánicas. Por ejemplo, ocurre flujo de calor cuando existe una diferencia de temperatura, la electricidad
fluye bajo la influencia de una fuerza electromotriz, y las reacciones químicas suceden porque existe un potencial químico. En general, un proceso es reversible cuando la fuerza impulsora neta sólo es de tamaño diferencial. Así, el calor se transfiere en forma reversible cuando fluye desde un objeto finito a temperatura T
hacia otro objeto a la temperatura T – dT.
Reacción química reversible
El concepto de una reacción química reversible se ilustra con la descomposición del carbonato de calcio que,
al ser calentado, forma óxido de calcio y dióxido de carbono gaseoso. En equilibrio, para una temperatura
dada el sistema ejerce una presión de descomposición específica de CO2. Cuando la presión cae por abajo de
este valor, se descompone el CaCO3. Supongamos que a un cilindro se le adapta un pistón sin fricción que
contiene CaCO3, CaO y CO2 en equilibrio. Se le sumerge en un baño a temperatura constante, como se muestra en la figura 2.3, y el equilibrio térmico asegura la igualdad de la temperatura del sistema con la del baño.
La temperatura se ajusta a un valor tal que la presión de descomposición es suficiente para equilibrar el peso
sobre el pistón, una condición de equilibrio mecánico. La reacción química se mantiene balanceada (en equilibrio) por la presión del CO2. Cualquier cambio de las condiciones, incluso de manera ligera, altera el equilibrio y provoca que la reacción avance en una u otra dirección.
Si el peso se aumenta en forma diferencial, la presión del CO2 se incrementa de manera diferencial, y
el CO2 se combina con el CaO para formar CaCO3, permitiendo que la pesa baje lentamente. El calor generado por esta reacción incrementa la temperatura en el cilindro, y el calor fluye hacia el baño. La disminución
diferencial del peso sobre el pistón provoca una cadena opuesta de eventos. Se obtienen los mismos resultados si la temperatura del baño aumenta o disminuye. Si la temperatura del baño aumenta en forma diferencial,
fluye calor hacia el interior del cilindro y se descompone el carbonato de calcio. El CO2 generado ocasiona
que la presión aumente en forma diferencial, lo cual origina que se eleven el pistón y el peso. Esto continúa
hasta que se descompone totalmente el CaCO3. El proceso es reversible, ya que el sistema nunca se desplaza
del equilibrio más que de forma diferencial, y sólo una disminución diferencial de la temperatura del baño
hace que el sistema regrese a su estado inicial.
02-SmithVanNess.indd 33
8/1/07 12:48:50
34
34
CAPÍTULO
LaLaw
primera
ley y otros
CHAPTER
2. The 2.
First
and Other
Basicconceptos
Concepts básicos
w
Figura 2.3: Reversibilidad de una reacción química.
T
T
CO2
w
CO2
Figure 2.3: Reversibility of a chemical reaction.
CaCO3 �CaCO
CaO 3  CaO
Termostato
Thermostat
Some chemical reactions can be carried out in an electrolytic cell, and in this case they
Algunas
químicas
seapplied
realizanpotential
en una celda
electrolítica,
y en
esteconsists
caso pueden
mantenerse
en
may bereacciones
held in balance
by an
difference.
If such
a cell
of two
elecequilibrio
al
aplicar
una
diferencia
de
potencial.
Si
esta
celda
se
compone
de
dos
electrodos,
uno
de
zinc
y
otro
trodes, one of zinc and the other of platinum, immersed in an aqueous solution of hydrochloric
de platino,
enthat
unaoccurs
solución
acid,sumergidos
the reaction
is: acuosa de ácido clorhídrico, la reacción que ocurre es:
Zn + 2HCl � H2 + ZnCl2
cell is heldbajo
under
fixed conditions
temperature
and pressure,
and the electrodes
are con- exterLa celdaThe
se mantiene
condiciones
fijas deof
temperatura
y presión,
y los electrodos
están conectados
externally to a potentiometer.
If the electromotive
force
byequilibra
the cell isenexactly
namentenected
a un potenciómetro.
Si la fuerza electromotriz
producida
porproduced
la celda se
forma exacta
balanced by
the potential
of thelapotentiometer,
the reaction
is held inEs
equilibrium.
por la diferencia
de potencial
del difference
potenciómetro,
reacción se mantiene
en equilibrio.
posible hacer que
The proceda
reaction en
may
be made
to proceed
in the forward
by a slight
decrease
in the opla reacción
sentido
directo
disminuyendo
un pocodirection
la diferencia
de potencial
opuesta,
y se puede
potential
difference,
and it may be
by a corresponding
in la
thecelda.
potential
revertir posing
mediante
un aumento
correspondiente
enreversed
la diferencia
de potencial deincrease
la fem de
difference above the emf of the cell.
Resumen
de lasRemarks
observaciones
en los procesos
Summary
on Reversible
Processesreversibles
A reversible
process:
Un proceso
reversible:
Is frictionless.
• Es sin• fricción.
• se
Is sale
neverdel
more
than differentially
removed
from
equilibrium.
• Nunca
equilibrio
más que de una
manera
diferencial.
• Recorre
una sucesión
de estados
equilibrio.states.
• Traverses
a succession
ofde
equilibrium
• Está controlado
fuerzaswhose
cuyo desequilibrio
tiene magnitud
diferencial.
• Is drivenpor
by forces
imbalance is differential
in magnitude.
• Se puede
invertir
en cualquier
mediante
un cambio
diferencial
en conditions.
las condiciones externas.
• Can
be reversed
at any punto
point by
a differential
change
in external
• Cuando se invierte, vuelve a trazar su trayectoria y restaura el estado inicial del sistema y de sus alre• When reversed, retraces its forward path, and restores the initial state of system and
dedores.
surroundings.
Una ecuación
deducida
en lainsección
1.7gives
proporciona
el of
trabajo
de compresión
o expansión
de un gas
An equation
derived
Sec. 1.7
the work
compression
or expansion
of a gas
que es causado
porthe
el differential
desplazamiento
diferencial
un pistón
un cilindro:
caused by
displacement
of de
a piston
in a en
cylinder:
ddW
W =
Vt
= –−P
P ddV
02-SmithVanNess.indd 34
(1.2)
(1.2)
8/1/07 12:48:52
2.8.
2.8. The
The Reversible
Reversible Process
Process
2.8. El proceso reversible
35
35
35
The
The work
work done
done on
on the
the system
system is
is given
given by
by this
this equation
equation only
only when
when certain
certain characteristics
characteristics of
of
El trabajo
hecho
sobre
el
sistema
se
da
por
esta
ecuación
sólo
cuando
se
efectúan
ciertas
del
the
reversible
process
are
realized.
The
first
requirement
is
that
the
system
more
the reversible process are realized. The first requirement is that the system be
be no
no características
more than
than
procesoinfinitesimally
reversible.
El
primer
requisito
es
que
el
sistema
se
desplace
sólo
en
forma
infinitesimal
de
un
estado
displaced
from
a
state
of
internal
equilibrium,
characterized
by
uniformity
of
infinitesimally displaced from a state of internal equilibrium, characterized by uniformity of
de equilibrio
internoand
caracterizado
porsystem
la uniformidad
de lahas
temperatura
y la presión.
El sistemaincludsiempre tiene
temperature
pressure.
then
an
set
temperature
and
pressure. The
The
system
then always
always
has
an identifiable
identifiable
set of
of properties,
properties,
includun conjunto
identificable
de
propiedades,
incluyendo
la
presión
P.
El
segundo
requisito
es
que
el sistema no
ing
pressure
P.
The
second
requirement
is
that
the
system
be
no
more
than
infinitesimally
ing pressure P. The second requirement is that the system be no more than infinitesimally
se desplace
más
que
en
forma
infinitesimal
del
equilibrio
mecánico
con
sus
alrededores.
En
este
caso, la predisplaced
from
mechanical
equilibrium
with
its
surroundings.
In
this
event,
the
internal
displaced from mechanical equilibrium with its surroundings. In this event, the internal prespressión interna
P
nunca
está
más
que
un
poco
fuera
de
equilibrio
con
la
fuerza
externa,
y
se
puede
hacer
sure
P
is
never
more
than
minutely
out
of
balance
with
the
external
force,
and
we
may
make
sure P is never more than minutely out of balance with the external force, and we may make la sustituciónthe
F=
PA, que transforma
la ecuación
(1.1)
la (1.2).
Los(1.2).
procesos
para los
se reúnen
substitution
F
transforms
Eq.
into
Processes
for
which
these
the
substitution
F =
= P
PA
A that
that
transforms
Eq.en(1.1)
(1.1)
into Eq.
Eq.
(1.2).
Processes
forque
which
these estos
requisitos
se
conocen
como
mecánicamente
reversibles
y
se
puede
integrar
la
ecuación
(1.2):
requirements
are
met
are
said
to
be
mechanically
reversible,
and
Eq.
(1.2)
may
be
integrated:
requirements are met are said to be mechanically reversible, and Eq. (1.2) may be integrated:
�� V tt
V22
t
(1.3)
W
=
−
P dd V
Vt
(1.3)
W =− t P
(1.3)
V
V11t
The
reversible
process
is ideal
in
itit produces
the
possible
result.
It
aa
The reversible
reversible es
process
in that
thatde
produces
the best
best
possible
result.posible.
It represents
represents
El proceso
ideal is
enideal
el sentido
que produce
el
mejor
resultado
Representa
un
limit
to
the
performance
of
actual
processes,
but
is
never
fully
realized.
An
initial
calculation
limit
to
the
performance
of
actual
processes,
but
is
never
fully
realized.
An
initial
calculation
límite en el desempeño de los procesos reales, ya que nunca se puede realizar por completo. Con frecuencia
of work
is
made
for aa reversible
process, because
the
such
is often
often
made
reversible
the choice
choice is
islabetween
between
such aa calculation
calculation
se hace of
unwork
cálculo
inicial
del for
trabajo
para unprocess,
procesobecause
reversible,
porque
opción estriba
entre realizar o no
and
no
calculation
at
all.
The
reversible
work
as
the
limiting
value
may
be
combined
with
an
and
no
calculation
at
all.
The
reversible
work
as
the
limiting
value
may
be
combined
withapropiadas
an
estos cálculos. El trabajo reversible con el valor limitante puede combinarse con las eficiencias
appropriate
efficiency
to
yield
aa reasonable
approximation
to
the
work
of
an
actual
process.
appropriate
efficiency
to
yield
reasonable
approximation
to
the
work
of
an
actual
process.
para producir aproximaciones razonables del trabajo de los procesos reales.
Example 2.6
Ejemplo
2.6 piston/cylinder
A
A horizontal
horizontal
piston/cylinder arrangement
arrangement is
is placed
placed in
in aa constant-temperature
constant-temperature bath.
bath.
The
slides
cylinder
friction,
and
external
holds
itit pistón
The piston
pistonhorizontal
slides in
in the
the
cylinder with
with negligible
negligible
friction,
anda an
an
external force
force
holds El
Una combinación
de pistón/cilindro
se coloca en
un baño
temperatura
constante.
33. The
in
place
against
an
initial
gas
pressure
of
14
bar.
The
initial
gas
volume
is
0.03
m
in
place
against
an
initial
gas
pressure
of
14
bar.
The
initial
gas
volume
is
0.03
m
.
The
se desliza en el cilindro con una fricción insignificante, y una fuerza externa lo mantiene en su lugar en
force
on
is
and
gas
expands
external
forceinicial
on the
the
piston
is reduced
reduced
gradually,
and the
the
gas del
expands
isothermally
contra external
de
la presión
delpiston
gas que
es de 14gradually,
bar. El volumen
inicial
gas esisothermally
0.03 m3. La fuerza
as
its
volume
doubles.
If
the
volume
of
the
gas
is
related
to
its
pressure
so
as
its
volume
doubles.
If
the
volume
of
the
gas
is
related
to
its
pressure
so that
that the
theal doble
externa sobre el pistón
se reduce de manera gradual y el gas se expande de forma isotérmica
tt is constant, what is the work done by the gas in moving the external
product
PV
product
PV
is
constant,
what
is
the
work
done
by
the
gas
in
moving
the
external
de su volumen. Si el volumen del gas está relacionado con su presión, de modo que el producto PV t es
force?
force?
constante,
¿cuál es el trabajo hecho por el gas al cambiar la fuerza externa?
How
much
would
be
the
were
suddenly
How
much
work
would
be done
done
the external
external
force
were repentinamente
suddenly reduced
reduced
to
¿Cuánto
trabajo
sework
hubiera
realizado
si laififfuerza
externaforce
se reduce
a la to
mitad de
half
its
initial
value
instead
of
being
gradually
reduced?
half
its
initial
value
instead
of
being
gradually
reduced?
su valor inicial en vez de hacer la reducción de manera gradual?
Solution
2.6
Solución
2.6
El proceso
como se
en un principio,
es mecánicamente
y se puede
The
process,
out
as
is
reversible,
and
Therealizado,
process, carried
carried
outdescribió
as first
first described,
described,
is mechanically
mechanically
reversible,reversible
and Eq.
Eq. (1.3)
(1.3)
t = k, entonces P =tt k / V t y
ttPV
aplicar la
ecuación
(1.3).
Si
=
k,
then
P
=
k/V
,
and
is
applicable.
If
P
V
is applicable. If P V = k, then P = k/V , and
W
W=
=−
−
With
With
Con
and
and
y
02-SmithVanNess.indd 35
��
t
V
V22t
t
V
V11t
d V tt
P
P dV =
= −k
−k
��
t
V
V22t
t
V
V11t
t
t
dd V
V22t
V t = −k ln V
=
−k
ln
t
V
V
V tt
V1t
1
t
t
3
3
V
V
V11t =
= 0.03
0.03 m
m3
V22t =
= 0.06
0.06 m
m3
t
3
t
V 1 = 0.03 m
V 2 = 0.06 m 3
t
t
5
t
t
kk =
=P
PV
V =
=P
P11V
V11 =
= (14
(14 ×
× 10
105)(0.03)
)(0.03) =
= 42,000
42,000 JJ
t = (14 × 10 5 )(0.03) = 42 000 J
k = PV t =WP 1=V −42,000
1
W = −42,000 ln
ln 22 =
= −29,112
−29,112 JJ
W = –42 000 ln 2 = –29 112 J
8/1/07 12:48:53
36
CHAPTER 2. The First Law and Other Basic Concepts
36
CAPÍTULO 2. La primera ley y otros conceptos básicos
La presiónThe
finalfinal
es pressure is
k
42,000
= 700,000 Pa
P2 = t =
o or
7 bar7 bar
0.06
V2
En el segundo
después
haber reducido
a la mitad
el gas
In thecaso,
second
case,de
reduction
of the initial
forcelabyfuerza
half isinicial,
followed
by experimenta
sudden
una súbitaexpansion
expansiónofcontra
unaagainst
fuerzaaconstante
que equivale
a una
de of
7 bar.
Con el
the gas
constant force
equivalent
to apresión
pressure
7 bar.
tiempo, la Eventually,
transferencia
de transfer
calor regresa
al the
sistema
a una
condición
de equilibrio
idéntica
al estaheat
returns
system
to the
same final
equilibrium
state as
t es el mismo que antes, pero el
do final alcanzado
en el proceso
reversible.
Det is
esta
themanera,
same asΔV
before,
but the work accomin the reversible
process.
Thus �V
trabajo realizado
estágiven
dado by
porEq.
la ecuación
(1.3).the
Enwork
su lugar,
trabajothe
realizado
contra de
plishedno
is not
(1.3). Rather,
doneelagainst
externalenforce
la fuerza externa
es igual
a la presión
externa
equivalente
el cambio
de volumen:
equals the
equivalent
external
pressure
times thepor
volume
change:
5)(0.06
5
W
××
1010
– 0.03)
= –21
000 J J
W ==–(7
−(7
)(0.06
− 0.03)
= −21,000
Thisesprocess
is clearly
irreversible,
and compared
with thereversible
reversibleseprocess
is tiene
Este proceso
claramente
irreversible,
y comparado
con el proceso
dice que
said de:
to have an efficiency of:
una eficiencia
0.721
2121,000/29,112
000/29 112 = =
0.721
o or
72.1%
72.1%
Example
2.7
Ejemplo
2.7
The piston/cylinder arrangement shown in Fig. 2.4 contains nitrogen gas trapped be-
La combinación
que se muestra
en laThe
figura
2.4 contiene
gas
nitrógeno
atrapado
debajo
low thepistón/cilindro
piston at a pressure
of 7 bar.
piston
is held in
place
by latches.
The
del pistónspace
a una above
presiónthe
de piston
7 bar. El
pistón
se
mantiene
en
su
lugar
mediante
unos
seguros.
El
espacio
is evacuated. A pan is attached to the piston rod and a mass
sobre el pistón
es kg
evacuado.
Se une
un plato
vástago
delpiston
pistón,rod,
y una
masa
de 45 kg have
se sujeta
m of 45
is fastened
to the
pan. alThe
piston,
and
panmtogether
a al
plato. El mass
pistón,ofel23
vástago
y
el
plato
juntos
tienen
una
masa
de
23
kg.
Los
seguros
que
sostienen
kg. The latches holding the piston are released, allowing the piston to
mm
Masa
Mass
Plato
Pan
Evacuated
Espacio
space
evacuado
Figura 2.4: Diagrama
para
el ejemplo
2.7.2.7.
Figure 2.4:
Diagram
for Ex.
0.5m
m
0.5
Cylinder
Cilindro
Piston
Pistón
Latch
Seguro
Gas under
Gas
bajo
pressure
presión
rise rapidly until it strikes the top of the cylinder. The distance moved by the piston is
−2 . Discuss the energy changes that
0.5liberan
m. Thepermitiendo
local acceleration
gravity
9.8 m shasta
al pistón se
que ésteofsuba
conisrapidez
golpear la parte superior del cilindro.
of thisde
process.
El pistón occur
recorrebecause
una distancia
0.5 m. La aceleración local de la gravedad es 9.8 m s–2. Analice los
cambios de energía que ocurren debido a este proceso.
02-SmithVanNess.indd 36
8/1/07 12:48:56
37
2.9. Procesos con V y P constantes
Solución 2.7
Este ejemplo sirve para ilustrar algunas de las dificultades encontradas al analizar los procesos
irreversibles donde no hay flujo. Considere el gas sólo como sistema. De acuerdo con la definición básica, el trabajo hecho por el gas en los alrededores es igual a ∫ P ′ dV ′, donde P ′ es la
presión ejercida por el gas sobre la cara del pistón. Ya que la expansión es muy rápida, existen
gradientes de la presión en el gas, y no es posible evaluar P ′ ni la integral. Sin embargo, al regresar a la ecuación (2.1) se evita el cálculo del trabajo. El cambio en la energía total del sistema (el
t
gas) es igual al cambio en la energía interna, ΔU sistema
. Para Q = 0, los cambios de energía en los
alrededores consisten en los cambios en la energía potencial del pistón, del vástago, del plato y de
la masa m, así como de los cambios en la energía interna del pistón, del vástago y del cilindro. Por
lo tanto, la ecuación (2.1) se escribe como:
t
t
+ (ΔU alrededores
+ ΔE P alrededores) = 0
ΔU sistema
donde
Por lo tanto
ΔE P alrededores = (45 + 23)(9.8)(0.5) = 333.2 N m
t
t
ΔU sistema
+ ΔU alrededores
= –333.2 N m = –333.2 J
Los valores para ΔU tsistema y ΔU talrededores no pueden ser determinados.
2.9
PROCESOS CON V y P CONSTANTES
Para n moles de un fluido homogéneo contenido en un sistema cerrado el balance de energía es:
d(nU) = dQ + dW
(2.6)
donde Q y W representan siempre el calor y el trabajo total para cualquier valor de n. El trabajo de un proceso
mecánicamente reversible en un sistema cerrado está dado por la ecuación (1.2), que se escribe a continuación:
dW = –P d(nV)
estas dos ecuaciones se combinan:
d(nU) = dQ – P d(nV)
(2.8)
Este balance de energía es general para n moles de fluido homogéneo en un sistema cerrado que experimenta
un proceso mecánicamente reversible.
Proceso a volumen constante
Si el proceso ocurre a un volumen total constante, el trabajo es cero. Por otra parte, para sistemas cerrados el
último término de la ecuación (2.8) también es cero, porque tanto n como V son constantes. De este modo,
dQ = d(nU)(V constante)
Integrando se obtiene:
Q = n ΔU
(V constante)
(2.9)
(2.10)
De esta manera, el calor transferido en un proceso mecánicamente reversible para un sistema cerrado a volumen constante, es igual al cambio en la energía interna del sistema.
02-SmithVanNess.indd 37
8/1/07 12:48:56
38
CAPÍTULO 2. La primera ley y otros conceptos básicos
Proceso a presión constante
Resolviendo para dQ, de la ecuación (2.8) se obtiene que:
dQ = d(nU) + P d(nV)
Para un cambio de estado a presión constante:
dQ = d(nU) + d(n PV) = d[n(U + PV)]
La aparición del grupo U + PV, tanto aquí como en otras aplicaciones, sugiere la definición conveniente de
una nueva propiedad termodinámica. De esta forma, la definición matemática (y única) de la entalpía6 es:
H ≡ U + PV
(2.11)
donde H, U y V son valores molares o por unidad de masa. La ecuación anterior ahora se puede escribir
como:
La integración produce:
dQ = d(nH) (P constante)
(2.12)
Q = n ΔH (P constante)
(2.13)
De este modo, el calor transferido en un proceso mecánicamente reversible de un sistema cerrado a presión
constante, es igual al cambio en la entalpía del sistema. Al comparar las ecuaciones (2.12) y (2.13) con las
ecuaciones (2.9) y (2.10) se muestra que en los procesos a presión constante la entalpía desempeña un papel
semejante a la energía interna en los procesos a volumen constante.
2.10
ENTALPÍA
La utilidad de la entalpía se sugiere por las ecuaciones (2.12) y (2.13). También aparece en los balances energéticos para los procesos con flujo en relación con los intercambiadores de calor, evaporadores, columnas de
destilación, bombas, compresores, turbinas, máquinas, etc., para el cálculo del calor y del trabajo.
La tabulación de los valores de Q y de W para la combinación infinita de los posibles procesos es
imposible. Sin embargo, las funciones de estado intensivas, tales como el volumen, la energía interna y la
entalpía específicos, son propiedades intrínsecas de la materia. Una vez que son determinadas para una sustancia particular, sus valores para fases líquida y vapor se pueden tabular como funciones de T y P para su
uso futuro en el cálculo de Q y de W para cualquier proceso que involucre dicha sustancia. La determinación
de los valores numéricos para estas funciones de estado, así como su correlación y uso se tratarán en los
últimos capítulos.
Todos los términos de la ecuación (2.11) deben expresarse en las mismas unidades. El producto PV
tiene unidades de energía por mol o por unidad de masa, al igual que U; por lo tanto, H también tiene unidades
de energía por mol o por unidad de masa. En el sistema SI la unidad básica de la presión es el pascal o el
6 Una palabra propuesta por el físico holandés H. Kamerlingh Onnes, quien fue el primero en licuar el helio en 1908. Descubrió
la superconductividad en 1911 y ganó el premio Nobel de Física en 1913. (Véase: Comunications from the Physical Laboratory of the
University of Leiden, núm. 109, p. 3, pie de página 2, 1909.)
02-SmithVanNess.indd 38
8/1/07 12:48:56
39
2.10. Entalpía
N m–2 y, para el volumen molar, m3 mol–1. El producto PV tiene por consiguiente las unidades N m mol–1 o
J mol–1. En el sistema inglés de ingeniería una unidad común para el producto PV es el (pie lbf)(lbm)–1, que
surge cuando la presión se encuentra en (lbf)(pie)–2 y con el volumen en (pie)3(lbm)–1. Este resultado generalmente se convierte a (Btu)(lbm)–1 mediante la división entre 778.16 y haciendo uso de la ecuación (2.11),
porque la unidad inglesa de ingeniería común para U y H es el (Btu)(lbm)–1.
Dado que U, P y V son funciones de estado, la H definida por la ecuación (2.11) también es una función de estado. Como U y V, H es una propiedad intensiva del sistema. La forma diferencial de la ecuación
(2.11) es:
dH = dU + d(PV)
(2.14)
Esta ecuación se aplica siempre que ocurre un cambio diferencial en el sistema. Al efectuar la integración se
convierte en una ecuación para un cambio finito en el sistema:
ΔH = ΔU + Δ(PV)
(2.15)
Las ecuaciones (2.11), (2.14) y (2.15) se aplican a la unidad de masa de una sustancia o a un mol.
Ejemplo 2.8
Calcule ΔU y ΔH para 1 kg de agua cuando se evapora a la temperatura constante de 100 °C y a una
presión constante de 101.33 kPa. Los volúmenes específicos del agua líquida y de su vapor en estas
condiciones son 0.00104 y 1.673 m3 kg–1. Para este cambio se agrega al agua una cantidad de calor
de 2 256.9 kJ.
Solución 2.8
Se toma 1 kg de agua como el sistema, ya que es lo que nos interesa. Imaginemos que se encuentra contenido en los cilindros mediante un pistón sin fricción que ejerce una presión constante de
101.33 kPa. A medida que se agrega calor, el agua se evapora y se expande desde su volumen
inicial hasta su volumen final. La ecuación (2.13) escrita para el sistema de 1 kg se tiene a continuación:
ΔH = Q = 2 256.9 kJ
Por la ecuación (2.15),
ΔU = ΔH – Δ(PV) = ΔH – P ΔV
Evaluando el término final:
P ΔV = 101.33 kPa × (1.673 – 0.001) m3
= 169.4 kPa m3 = 169.4 kN m–2 m3 = 169.4 kJ
Entonces
02-SmithVanNess.indd 39
ΔU = 2 256.9 – 169.4 = 2 087.5 kJ
8/1/07 12:48:57
40
2.11
40
40
40
CAPÍTULO 2. La primera ley y otros conceptos básicos
CHAPTER 2. The First Law and Other Basic Concepts
CHAPTER
CHAPTER 2.
2. The
The First
First Law
Law and
and Other
Other Basic
Basic Concepts
Concepts
CAPACIDAD CALORÍFICA
2.11 HEAT
HEATCAPACITY
CAPACITY
2.11
2.11
CAPACITY
El punto
de vistaHEAT
moderno
con respecto al calor como energía en tránsito fue precedido por la idea de que un
cuerpo tiene una capacidad para el calor. Entre más pequeño fuera el cambio de temperatura en un cuerpo
The modern view of heat as energy in transit was preceded by the idea that a body has a
The
modern
view
energy
in
was
preceded
the
that
provocado
la transferencia
deas
una
cantidad
dada de
calor,
mayorby
su
capacidad.
En has
realidad,
una
The por
modern
view of
of heat
heat
as
energy
in transit
transit
was
preceded
bysería
the idea
idea
that aa body
body
has aa
capacity for heat. The smaller the temperature change in a body caused by the transfer of a
capacity
for
The
smaller
the
in
caused
by
transfer
of
capacidad
calorífica
puede
definirse
C ≡ dQ/dT. change
La
dificultad
con esta
expresión
que hace
capacity
for heat.
heat.
The
smallercomo
the temperature
temperature
change
in aa body
body
caused
by the
thees
transfer
of aque
a C, al
given quantity of heat, the greater its capacity. Indeed, a heat capacity might be defined as
given
quantity
of
the
greater
Indeed,
aa heat
capacity
be
defined
as
igual que
Q, sean
cantidades
del proceso
que de
una
estado.
obstante,
given
quantity
of heat,
heat,más
thedependientes
greater its
its capacity.
capacity.
Indeed,
heatfunción
capacitydemight
might
beNo
defined
as sugieC ≡ d Q/dT . The difficulty with this is that it makes C, like Q, a process-dependent quantity
CC ≡
.. The
difficulty
with
this
isis that
itit makes
like
aa process-dependent
re la definición
de dos
cantidades
este
nombre
hecho
son funciones dequantity
estado relacio≡ ddQ/dT
Q/dT
The
difficultycon
with
this
thatanticuado,
makes C,
C,que
likedeQ,
Q,
process-dependent
quantity
rather than a state function. However, it does suggest the definition of two quantities with this
rather
than
function.
However,
does
nadas sin
ambigüedad
con
otras funciones
deititestado.
rather
than aa state
state
function.
However,
does suggest
suggest the
the definition
definition of
of two
two quantities
quantities with
with this
this
outmoded name that are in fact state functions, unambiguously related to other state functions.
outmoded
name
that
are
in
fact
state
functions,
unambiguously
related
to
other
state
functions.
outmoded name that are in fact state functions, unambiguously related to other state functions.
Capacidad
caloríficaataConstant
volumen constante
Heat Capacity
Constant
Volume
Heat
Heat Capacity
Capacity at
at Constant Volume
Volume
The constant-volume heat capacity of a substance is defined as:
La capacidad
calorífica a volumen
constante
una sustancia
se define
The
heat
of
is defined
as:
The constant-volume
constant-volume
heat capacity
capacity
of ade
a substance
substance
defined
as: como:
� is �
�� ∂U��
∂U
(2.16) (2.16)
C ≡ ∂U
(2.16)
CCVVV≡
≡ ∂T V
(2.16)
∂∂TT VV
Esta definición
se acomoda
a ambas capacidades,
tanto
la capacidad
calorífica
como a la
capacidad
This definition
accommodates
both the molar
heata capacity
and the
specificmolar
heat capacity
(usuThis
definition
accommodates
both
the
heat
and
specific
heat
capacity
(usuThis
definition
accommodates
bothcalor
themolar
molar
heatcapacity
capacity
andthe
theen
specific
heatde
capacity
(usucalorífica
específica
(usualmente
llamada
específico),
dependiendo
todo
caso
que
U
sea
la energía
ally called specific heat), depending on whether U is the molar or specific internal energy.
called
specific
heat),
on
U
isis the
or
internal
energy.
ally
called
specific Aunque
heat), depending
depending
on whether
whether
Ureferencia
the molar
molar
or specific
specific
internal
energy. de una
internaally
molar
o
específica.
esta
definición
no
hace
a
cualquier
proceso,
se
relaciona
Although this definition makes no reference to any process, it relates in an especially simple
this
makes
no
to
process,
relates
in
especially
Although
this definition
definition
makes
no reference
reference
to any
any
process,deititun
relates
in an
an
especially simple
simple
maneraAlthough
especialmente
simple con
un
proceso
volumen
constante
sistema
cerrado,
el que la ecuaway
to a constant-volume
process
in aaclosed
system,
for which Eq.
(2.16)
may bepara
written:
way
process
way to
to aescribe
a constant-volume
constant-volume
process in
in aa closed
closed system,
system, for
for which
which Eq.
Eq. (2.16)
(2.16) may
may be
be written:
written:
ción (2.16)
se
como:
dU = C V dT
(const V )
(2.17)
dU
CCVVdT
dT
(const
(2.17)
dU
=C
(const VV))
(2.17) (2.17)
dU =
=
V dT (V constante)
�
�� T T2
Integration yields:
�U = T22 C V dT
(const V )
(2.18)
Integration
�U
(const
VV))
(2.18)
Integrando:
(V
constante)
Integration yields:
yields:
�U =
= T1 CCVV dT
dT
(const
(2.18) (2.18)
TT11
La combinación
dewith
este Eq.
resultado
ecuación (2.10)
para unconstant-volume
proceso mecánicamente
a voluThis result
(2.10) con
for alamechanically
reversible,
process777reversible,
gives:
This
result
with
Eq.
(2.10)
for
aa mechanically
reversible,
constant-volume
process
gives:
This
result
with
Eq.
(2.10)
for
mechanically
reversible,
constant-volume
process
gives:
7
men constante, nos da:
�
�� T T2
Q = n �U = n T22 C V dT
(const V )
(2.19)
Q
VV))
(2.19)
constante)
Q=
= nn �U
�U =
= nn T1 CCVV dT
dT (V(const
(const
(2.19) (2.19)
TT11
If the volume
varieselduring
the pero
process
but returns
the end
ofvalor
the process
Si el volumen
varía durante
proceso,
al final
de éste at
regresa
a su
inicial, to
noits
esinitial
correcto que
IfIf the
the volume
volume varies
varies during
during the
the process
process but
but returns
returns at
at the
the end
end of
of the
the process
process to
to its
its initial
initial
=
V1 and en las
value,
the
process
cannot
rightly
be
called
one
of
constant
volume,
even
though
V
=
V
y
ΔV
=
0.
De
cualquier
modo,
los
el proceso
se
llame
volumen
constante,
aunque
V
2 cambios
2
1
value,
= VV11 and
and
value, the
the process
process cannot
cannot rightly
rightly be
be called
called one
one of
of constant
constant volume,
volume, even
even though
though VV22 =
�Vde =
0. However,
changes indestate
functions y,
arepor
independent
of path,
and may
therefore
funciones
estado
son independientes
la trayectoria
lo tanto, pueden
calcularse
mediante
ecuacio�V
�V =
= 0.
0. However,
However, changes
changes in
in state
state functions
functions are
are independent
independent of
of path,
path, and
and may
may therefore
therefore
be
calculated
by
equations
for a truly
constant-volume
process withinicial
the same
initial
and
final (2.18)
nes para
un
proceso
a
volumen
constante
real
con
las
mismas
condiciones
y
final.
La
ecuación
�
be
process
be calculated
calculated by
by equations
equations for
for aa truly
truly constant-volume
constant-volume
process with
with the
the same
same initial
initial and
and final
final
�� C V dT
conditions.
(2.18)
, because
, C V , and T Por
are otra
all state
da entonces
ΔU = ∫ CEquation
U,then
CV gives
ygives
T son�U
todas=funciones
de
estado oU, propiedades.
parte, Q y
V dT, porque
conditions.
(2.18)
conditions. Equation
Equation
(2.18) then
then
gives �U
�U =
= CCVV dT
dT,, because
because U
U , CCVV,, and
and TT are
are all
all state
state
functions.
On the other
hand,
Q and
W depend
on
path, and
Eq. para
(2.19)
isy aWvalid
expression
W dependen
de
la
trayectoria,
y
la
ecuación
(2.19)
es
una
expresión
válida
Q,
en
general
es
cero sólo
functions.
functions. On
On the
the other
other hand,
hand, Q
Q and
and W
W depend
depend on
on path,
path, and
and Eq.
Eq. (2.19)
(2.19) isis aa valid
valid expression
expression
for
Q, and
W is inconstante.
general zero,
only
for
a constant-volume
process.
This
is the de
reason
for Q y W.
para unfor
proceso
a
volumen
Ésta
es
la
razón
para
la
distinción
entre
las
funciones
estado,
for Q,
Q, and
and W
W isis in
in general
general zero,
zero, only
only for
for aa constant-volume
constant-volume process.
process. This
This isis the
the reason
reason for
for
the distinction
functions
and Q and W . del
Theproceso
principle
thatconcepto
state functions
are y útil.
El principio
de que las between
funcionesstate
de estado
son independientes
es un
importante
the
the distinction
distinction between
between state
state functions
functions and
and Q
Q and
and W
W.. The
The principle
principle that
that state
state functions
functions are
are
independent of the process is an important and useful concept.
independent
of
process
isis an
and
independent
of the
the los
process
an important
important
and useful
useful concept.
concept.
Para
el cálculo
de
cambios
en
las propiedades,
un proceso real se puede sustituir por
For
the
calculation
of
property
changes,
an actual process may be
otro cualquiera
que
consiga
el
mismo
cambio
en
el
estado.
For
For the
the calculation
calculation of
of property
property changes,
changes, an
an actual
actual process
process may
may be
be
replaced by any other process which accomplishes the same change
replaced
by
any
other
process
which
accomplishes
the
same
change
replaced
by
any
other
process
which
accomplishes
the
same
change
De este modo,in
puede
seleccionarse un proceso alternativo, por ejemplo, debido a su simplicidad.
state.
in
in state.
state.
Such an alternative process may be selected, for example, because of its simplicity.
Such
Such an
an alternative
alternative process
process may
may be
be selected,
selected, for
for example,
example, because
because of
of its
its simplicity.
simplicity.
7
Estas restricciones sirven para excluir el trabajo de agitación, que es intrínsecamente irreversible.
7 These restrictions serve to rule out work of stirring, which is inherently irreversible.
77These
Theserestrictions
restrictionsserve
serveto
torule
ruleout
outwork
workof
ofstirring,
stirring,which
whichisisinherently
inherentlyirreversible.
irreversible.
02-SmithVanNess.indd 40
8/1/07 12:48:59
2.11. Heat Capacity
2.11. Capacidad
2.11. Heat calorífica
Capacity
2.11. Heat Capacity
41
41
41
41
HeatCapacity
Capacity
ataConstant
Constant
Pressure
Capacidad
calorífica
presión constante
Heat
at
Pressure
Heat Capacity at Constant Pressure
The constant-pressure
heat capacity se
is defined
as:
La capacidad
calorífica a presión
define como:
The constant-pressure
heat constante
capacity is defined
as:
The constant-pressure heat capacity is defined �
as: �
� ∂ H�
(2.20) (2.20)
C P ≡� ∂ H �
(2.20)
C P ≡ ∂ ∂HT P
∂
T
P
(2.20)
CP ≡
∂caloríficas
T P
De nuevo,
la
definición
se
acomoda
a
las
capacidades
molar
y
específica,
dependiendo
Again, the definition accommodates both molar and specific heat capacities, depending de
on si H es
Again,
theodefinition
accommodates
both
molarseand
specific
heat
capacities,
depending simple
on
la entalpía
molar
específica.
Esta
capacidad
calorífica
relaciona
de
una
manera
especialmente
para
whether
Hdefinition
is the molar
or specific enthalpy.
Thisand
heatspecific
capacityheat
relates
in an especially
simple
Again,
the
both
molar
capacities,
depending
onestá bien
whether
H is
the
molaraccommodates
or
specific
enthalpy.
This
heat
capacity
relates
in an especially
simple
un proceso
a
presión
constante
en
un
sistema
cerrado,
para
el
que
la
ecuación
(2.20)
de
igual
manera
way to aHconstant-pressure,
closed-system
process,
for capacity
which Eq.
(2.20) isan
equally
well written:
is the molar or specific
enthalpy.
This heat
especially
simple
way to a constant-pressure,
closed-system
process,
for which
Eq.relates
(2.20) in
is equally
well written:
escrita:whether
way to a constant-pressure, closed-system process, for which Eq. (2.20) is equally well written:
d H ==CC PdTdT (P constante)
(const P)
(2.21)
ddH
H = CPP dT
(const P)
(2.21) (2.21)
d H =�C P dT
(const P)
(2.21)
� T2T2
whence
�H =
C dT
(const P)
(2.22)
de donde
constante)
whence
�H = � TT12 C PPdT (P(const
P)
(2.22) (2.22)
whence
�H = T1 C P dT
(const P)
(2.22)
T1 process, this result may be combined with Eq. (2.13):
For a mechanically reversible, constant-P
a mechanically
reversible,
constant-P
process,
this result
be combined
with Eq. (2.13):
Para unFor
proceso
mecánicamente
reversible
a presión
constante,
estemay
resultado
puede combinarse
con la ecuaFor a mechanically reversible, constant-P process,
ción (2.13):
� T2 this result may be combined with Eq. (2.13):
� T
Q = n �H = n 2 C dT
(const P)
(2.23)
Q = n �H = n � TT12 C PPdT
(const P)
(2.23)
T
P)
(2.23) (2.23)
Q = n �H = n 1 C P dT
(P(const
constante)
T1
Because H , C , and T are state functions, Eq. (2.22) applies to any process for which P = P
Because H , C PP, and T are state functions, Eq. (2.22) applies to any process for which P2 2= P1 1
whether
is Tactually
carried
out
at constant
pressure.
However,
only
for thePmechanDado que
H, CPor
y,not
Tnot
funciones
deout
estado,
la
ecuación
(2.22)
seprocess
aplica
afor
cualquier
proceso
, it
and
are
state
functions,
(2.22)
applies
toHowever,
any
which
Because
Hor
C
Pson
2 = P1 para el
whether
it
istodas
actually
carried
at Eq.
constant
pressure.
only
for
the mechanically
reversible,
constant-pressure
process
can
heat
and
work
be
calculated
by
the
equations
cual P2whether
=
P
,
se
esté
o
no
realizando
realmente
a
presión
constante.
Sin
embargo,
sólo
para
un
proceso mecáor not it constant-pressure
is actually
carried out at constant
only by
forthe
theequations
mechan�
ically1 reversible,
can heatpressure.
and workHowever,
be calculated
� C P dT , and process
Q
=
n
�H
,
Q
=
n
W
=
−Pn
�V
.
nicamente
reversible
a
presión
constante
se
puede
calcular
el
calor
y
el
trabajo
por
medio
de
las ecuaciones
ically
can�V
heat
Q = nreversible,
�H , Q =constant-pressure
n � C P dT , and Wprocess
= −Pn
. and work be calculated by the equations
Q = n ΔH,
Q
=
n∫C
dT
y
W
=
–Pn
ΔV.
Q = n �H , PQ = n C P dT , and W = −Pn �V .
Example2.9
2.9
Example
Ejemplo
2.9
Example
2.9
Air at 1 bar and
298.15 K (25◦ ◦ C) is compressed to 5 bar and 298.15 K by two different
Air at 1 bar and 298.15 K (25 C) is compressed to 5 bar and 298.15 K by two different
mechanically
reversible
processes:
Se comprime
a1
bar
y 298.15
(25is°C),
hasta 5 bar
298.15
mediante
Air
at 1 aire
bar and
298.15
Kprocesses:
(25K◦ C)
compressed
to 5y bar
andK298.15
K bydos
twoprocesos
differentmecánimechanically
reversible
camente
reversibles
diferentes:
mechanically reversible processes:
(a) Cooling at constant pressure followed by heating at constant volume.
(a) Cooling at constant pressure followed by heating at constant volume.
(a) Cooling at constant pressure followed by heating at constant volume.
(b) Heating at constant
volumeseguido
followeddebyuncooling
at constant
pressure.
b) Calentamiento
constante
enfriamiento
a presión
constante.
(b) Heatingaatvolumen
constant
volume followed
by cooling
at constant
pressure.
(b) Heating at constant volume followed by cooling at constant pressure.
CalculeCalculate
los requerimientos
de calor
de trabajo, yand
ΔU y�U
ΔHand
del aire
unaeach
de las
trayectothe heat and
work yrequirements
�H para
of thecada
air for
path.
Calculate
the heat
and workcaloríficas
requirements
and
�U se
and
�H ofque
the son
air for
each path. de la
rias. Las
siguientes
capacidades
para
el
aire
supone
independientes
The following
heatand
capacities
for air may be assumed
independent
of temperature:
Calculate
the heat
work requirements
�U and
�H of theofair
for each path.
The following
capacities
for air may beand
assumed
independent
temperature:
temperatura:
The following heat capacities for air may be assumed independent of temperature:
−1 K−1
and
C = 29.10 J mol−1
C = 20.78
−1
–1KK
–1
C PPC=P 29.10
J mol
C VV= 20.78
CV = 20.78 and y
= 29.10
J mol
and
C P = 29.10 J mol−1 K−1
C V = 20.78
Assume also for air that P V /T is a constant, regardless of the changes it undergoes.
Assume
also for
airelthat
V /T
is aesconstant,
regardless
of the
changes
it undergoes.
Supóngase
también
para
aire P
que
PV/T
una constante,
sin considerar
los cambios
que experimen3 mol
−1
At 298.15
K and
1 bar
the
molar
of air
is 0.02479
m
3 mol
−1 .
Assume
forelair
thatthe
P Vmolar
/T isdel
avolume
constant,
the
it undergoes.
3m
–1changes
At 298.15
and
1volumen
bar
volume
of air
is
0.02479
ta. A 298.15
K yalso
1Kbar
molar
aire es
deregardless
0.02479
mof
mol
. .
At 298.15 K and 1 bar the molar volume of air is 0.02479 m3 mol−1 .
a) Enfriamiento a presión constante seguido de un calentamiento a volumen constante.
02-SmithVanNess.indd 41
8/1/07 12:49:02
42
42
42
42
42
CHAPTER 2.
CHAPTER 2.
CHAPTER 2.
CHAPTER 2.
The First Law and Other Basic Concepts
The First
First Law
Law and
and Other
Other Basic
Basic Concepts
Concepts
The
The First Law and Other Basic Concepts
42
CHAPTER
2. The 2.
First
and Other
Basicconceptos
Conceptsbásicos
CAPÍTULO
LaLaw
primera
ley y otros
Solution 2.9
2.9
Solution
Solution
2.9
Solución
Solution
2.9
In2.9
each case
take the system as 1 mol of air contained in an imaginary pisIn each case take the system as 1 mol of air contained in an imaginary pisIn
each
case arrangement.
take
the system
as 1 mol
of
aircontenido
contained
in
anare
imaginary
pisEn cadaIn
caso
considere
al sistema
como
deprocesses
aireair
enin
una
combinación
imaginaria
ton/cylinder
Because
the
considered
mechanically
each case
take
the system
as1 1mol
mol
of
contained
anare
imaginary
pisSolution
2.9arrangement.
ton/cylinder
Because
the processes
considered
mechanically
ton/cylinder
arrangement.
Because
processes
considered
are mechanically
reversible, Debido
the
piston
is estos
imagined
to the
move
in the cylinder
without
friction.
The suponde pistón/cilindro.
a que
procesos
seprocesses
consideran
mecánicamente
reversibles,
ton/cylinder
arrangement.
Because
considered
are mechanically
reversible, the
piston is imagined
to the
move
in the cylinder
without
friction. The
reversible,
the
piston
issystem
imagined
to
moveofinEl
the
cylinder
without
friction. The
In
each
case
take
theel
as
1fricción.
mol
air
contained
in es:
an imaginary
pisfinal
volume
is:
ga que el
pistón
se
mueve
en
cilindro
sin
volumen
final
reversible,
theis:piston is imagined to move in the cylinder without friction. The
final volume
final
volume is:
ton/cylinder
arrangement. Because the processes
considered are mechanically
� �
final volume is:
� 1�
P1
� 1in
�the
reversible, the piston
is Vimagined
to
move
cylinder
without
friction. The
P
1
=
0.02479
=
0.004958
m33
V2 =
1 P1 = 0.02479 � 1 � = 0.004958 m
=
V
V
2
1
P
5
2
P
1
final volume is: V2 = V1 P12 = 0.02479 5 = 0.004958 m33
V2 = V1 P2 = 0.02479 � 5 � = 0.004958 m
51 constant pressure of 1 bar until the
2 is cooled at the
(a) During the first step thePPair
1 is cooled at the
3 of 1 bar until the
(a) During
the etapa,
firstVstep
the
air
constant
pressure
=
V
=
0.02479
= 0.004958
3
a) Durante
la
primera
el
aire
se
enfría
a
presión
constante
deof1mthe
barof
hasta
que
sethe
alcanza
el
2
1
(a)
During
the
first
step
the
air
is
cooled
at
the
constant
pressure
1 at
barthe
until
final
volume
offirst
0.004958
mPair
isisreached.
The
temperature
air
end
of
32 is
5
(a)
During
theof
step
them
cooled
at
the
constant
pressure
ofair
1 at
barthe
until
final
volume
0.004958
reached.
The
temperature
of
the
endthe
of
3
3 is reached.del
.
La
temperatura
aire
al
final
de
esta
etapa
de
enfriamiento
es:
volumenfinal
final
de
0.004958
m
volume
of
0.004958
m
The
temperature
of
the
air
at
the
end
of
this cooling
step
is:
final
volumestep
of 0.004958
m3 is reached. The temperature of the air at the end of
this cooling
is:
this
coolingthe
step
is: step the air is cooled�at the constant
(a) During
first
� pressure of 1 bar until the
this cooling step is:
�
�
3 is reached.
V
0.004958
2
final volume of 0.004958
m
The
temperature
of the
� 0.004958 � = 59.63
V2 = 298.15�
T� = T V
K air at the end of
� = 59.63 K
2 = 298.15 0.004958
T �� = T11V
V
0.02479
this cooling step is:
T = T V211 = 298.15 0.004958
0.02479 = 59.63 K
T � = T11 V
1 = 298.15 � 0.02479 � = 59.63 K
V
0.02479
Whence,
V12
0.004958
De donde,
Whence,
T � = T1
= 298.15
= 59.63 K
Whence,
V
0.02479
Whence,
Q = �H = C P1�T = (29.10)(59.63 − 298.15) = −6,941 J
Q = �H = C P �T = (29.10)(59.63 − 298.15) = −6,941 J
= (29.10)(59.63
− 298.15) = −6,941 J
Q =
C
P �T
Whence,�U
= �H
�H =
−C
�(P
V)=
�H − P �V −
(29.10)(59.63
298.15) = −6,941 J
Q
P �T
�U =
= �H
�H =
− �(P
V) =
�H
− P �V
�U =
�H
−
�(P
V
)
=
�H
−
P
�V
5
= �H
−6,941
− (1V×
10�H
)(0.004958
−−
0.02479)
=
−4,958
�U
−
)=
− P �V−
5(29.10)(59.63
=
298.15)=
=−4,958
−6,941JJJ
Q=
= �(P
C
= �H
−6,941
− (1�T
×
10
)(0.004958
0.02479)
= −6,941 −P(1 × 1055 )(0.004958 − 0.02479) = −4,958 J
=
−6,941
− (1V×) 10
)(0.004958
− 0.02479) = −4,958 J
�Uthe
=second
�H −step
�(P
= �H
P �V
During
the volume
is−held
constant at V while the air is heated
During the second step the volume
is held constant at V22 while the air is heated
5
the air is heated
During
the second
step
the
volume
is held constant
at V while
to its
final
state.
By Eq.
(2.19),
−6,941
−
(1 ×
10se)(0.004958
− 0.02479)
−4,958
the airJ que
is heated
During
the=second
step
the
volume
is held constant
at Va22 while
Durante
lafinal
segunda
etapa,
el(2.19),
volumen
mantiene
constante
V=2 mientras
se calienta a
to its
state.
By
Eq.
to its final state. By Eq. (2.19),
to final.
its final
By Eq. (2.19),
su estado
Porstate.
la ecuación
(2.19),
�T = (20.78)(298.15
− 59.63)
= 4,958
J is heated
�U
= Q = C V the
the Jair
During�U
the second
is held constant
at V2 while
= (20.78)(298.15
− 59.63)
= 4,958
= Q =step
C �Tvolume
�U =ByQEq.
= C VV �T = (20.78)(298.15 − 59.63) = 4,958 J
to its final �U
state.
�T
=
(20.78)(298.15
−
59.63)
=
4,958
J
= Q = C(2.19),
V
The complete process represents
the sum of its steps. Hence,
The complete process represents the sum of its steps. Hence,
The complete
process
represents
the
sum of its steps.
Hence,= 4,958 J
�T
=
−
�U
=
Q
=
C
V
El proceso
la suma de
sus
lo 59.63)
tanto,
Thecompleto
completerepresenta
process represents
the(20.78)(298.15
sumetapas.
of its Por
steps.
Hence,
Q = −6,941 + 4,958 = −1,983 J
Q = −6,941 + 4,958 = −1,983 J
Q = −6,941
+ 4,958
= −1,983
J
The complete process represents
the sum
of its
–6 941
+ 44,958
958
==steps.
–1
983Hence,
JJ
QQ==−6,941
+
−1,983
and
�U = −4,958 + 4,958 = 0
and
�U = −4,958 + 4,958 = 0
and
�U
= −4,958
+ 4,958
=0 J
Q �U
= −6,941
+ 4,958
= −1,983
and
−4,958
+
4,958
=
0 = Q + W , and therefore,
Because the first law applies
to=the
entire
process,
�U
Because the first law appliesΔU
to the
entire
process,
�U
= Q + W , and therefore,
y
=
–4
958
+
4
958
=
0 =
Because the first law applies to the entire process, �U
Q + W , and therefore,
Because
the first law applies
to =
the−4,958
entire process,
�U
= Q + W , and therefore,
and
�U
+
4,958
=
0
0 = −1,983 + W
whence
W = 1,983 J
0 = −1,983
+ Wa todo el whence
W
=W,
1,983
Debido a que la primera
se aplica
proceso, ΔU = W
Q +=
y porJ lo tanto,
0 =ley
−1,983
+ to
W the entire
whence
1,983
Because the first
applies
process, �UW= =
Q 1,983
+ W , JJand therefore,
0 =law
−1,983
+W
whence
Equation (2.15), �H = �U + �(P V ), also applies to the entire process. But
Equation (2.15),
�H
=
�U + �(P
V ), also appliesWto=the
entire
process. But
1and
983
J process.
0 = –1�H
983P=
+VW
Equation
�U
Vdonde
), also
applies
to0,
the
entire
But
T , and(2.15),
therefore,
=+
P �(P
V de
. whence
Hence
�(P
V ) =to
T1 =
0 = −1,983
+�U
=the
1,983
J process. But
1V
1W
Equation
�HP
=
V ), also
applies
entire
T22, and (2.15),
therefore,
=+
P22�(P
V22. Hence
�(P
V) W
= 0,
and
T1 =
1
1
T1 = T2 , and therefore, P1 V1 = P2 V2 . Hence �(P V ) = 0, and
and therefore,
P1 V+1 Δ(PV),
= P2 V2también
. Hence se
�(P V ) =
0, andel proceso. Pero T = T , y
T1 = T2 , (2.15),
La ecuación
ΔH �H
= ΔU
todo
2
�H
=V
�U
= 0aplica
Equation (2.15),
= �U �H
+ �(P
), also
appliesa to
the entire process. But1
=
�U
=
0
por lo tanto,
= Ptherefore,
= �H
0,
yV2=
�U =�(P
0 V ) = 0, and
2V2. Así, Δ(PV)
P1 V1 =
P
.
Hence
T1 =PT12V, 1and
2
�H = �U = 0
(b) Two different steps are used in this case to reach the same final state of the air.
(b) Two different steps are used inΔH
this=case
to
reach the same final state of the air.
ΔU
==
0reach
(b)
Two
different
steps
are
used
in
case
to
the
same
final
statevalue
of theuntil
air.
In the
first
step the air is
heated
at�H
athis
constant
volume
equal
to final
its initial
=case
�Uto
0 the
(b)
Two
different
used at
in
reach
same
statevalue
of theuntil
air.
In the
first
step thesteps
air isare
heated
athis
constant
volume
equal
to its
initial
In
the
first
step
the
air
is
heated
at
a
constant
volume
equal
to
its
initial
value
until
the final
pressure
ofair5 is
barheated
is reached.
The air volume
temperature
attothe
end
of this
step
is:
In
first
step
theof
at a constant
equalat
itsend
initial
value
until
thethe
final
pressure
5 bar
is reached.
The para
air temperature
the
of this
step
is:
b) En este
emplean
dos
etapas
diferentes
alcanzar
mismo
estado
final
the
final
pressure
of
5 bar
isused
reached.
air
at
thefinal
end
of this
step
is:aire. En
(b)caso
Twose
different
steps
are
in thisThe
case
totemperature
reach theelsame
state
of
thedel
air.
�
�
the etapa
final pressure
5 bar is areached.
The air
at the
end ofinicial,
this step is: que se
�
� 5temperature
la primera
airethe
seofcalienta
igual
suitsvalor
Pun volumen
In the firstelstep
air
a298.15
constant
equalatoK
initial value hasta
until
� is heated
� = 1,490.75
�constante
P22 at
5volume
=
T
=
T
�
1
�
�
P
5
alcanza the
la presión
final deof5T5bar.
temperatura
del
aire
al
final
de
esta
etapa
es:
2 = 298.15
=
1,490.75
K
= La
Tis1 reached.
� bar
P
1
final pressure
The
air
temperature
at
the
end
of
this
step
is:
= 298.15 51 = 1,490.75 K
T =T P
P211 =
298.15 � 1 � = 1,490.75 K
T � = T11 P
1
PP1
15
2
= 1,490.75 K
= 298.15
T � = T1
P1
1
02-SmithVanNess.indd 42
8/1/07 12:49:07
2.11.
Capacity
2.11. Capacidad
2.11. Heat
Heatcalorífica
Capacity
43
43
43
Para esta
etapa
volumen
es constante,
y and
For
this
step
the
isis constant,
For
thisel
step
the volume
volume
constant,
and
Q
Q=
= �U
�U =
= CCVVV �T
�T =
= (20.78)(1,490.75
(20.78)(1,490.75 −
− 298.15)
298.15) =
= 24,788
24,788 JJ
In
at
its
final
En la segunda
etapa elstep
airethe
se air
enfría
a una presión
= 5to
bar
estado final:
In the
the second
second
step
the
air isis cooled
cooled
at PP =
= 5P
5 bar
bar
to
itspara
finalsustate:
state:
Q
Q=
= �H
�H =
= CCPPP �T
�T =
= (29.10)(298.15
(29.10)(298.15 −
− 1,490.75)
1,490.75) =
= −34,703
−34,703 JJ
�U
�U =
= �H
�H −
− �(P
�(PVV)) =
= �H
�H −
− PP �V
�V
55
=
= −34,703
−34,703 −
− (5
(5 ×
× 10
105)(0.004958
)(0.004958 −
− 0.02479)
0.02479) =
= −24,788
−24,788 JJ
two
steps
Para lasFor
dosthe
etapas
For
the
twocombinadas,
steps combined,
combined,
as
y como and
antes
and
as before
before
Q
Q=
= 24,788
24,788 −
− 34,703
34,703 =
= −9,915
−9,915 JJ
�U
�U =
= 24,788
24,788 −
− 24,788
24,788 =
= 00
W
W=
= �U
�U −
−Q
Q=
= 00 −
− (−9,915)
(−9,915) =
= 9,915
9,915 JJ
�H
ΔH=
= �U
ΔU
�H
=
�U ==
=000
The
changes
and
calculated
for
in
Los cambios
en las propiedades
y �H
ΔH
son calculados
para change
el
cambio
dado
en el estaThe property
property
changes �U
�UΔU
and
�Hque
calculated
for the
the given
given
change
in state
state are
are
the
same
for
both
paths.
On
the
other
hand
the
answers
to
parts
(a)
and
(b)
do son iguales
para
trayectorias.
otrahand
parte,
respuestas
a los(a)
incisos
a) show
yshow
b) demuesthe same
forambas
both paths.
On thePor
other
thelas
answers
to parts
and (b)
W
the
tran quethat
Q yQ
dependen
de laon
trayectoria.
that
QWand
and
W depend
depend
on
the path.
path.
Example
2.10
Example
2.10
Ejemplo
2.10
Calculate the internal-energy and enthalpy changes that occur when air is changed
Calculate
Calculate the
the internal-energy
internal-energy and
and enthalpy
enthalpy changes
changes that
that occur
occur when
when air
air is
is changed
changed
◦◦◦F)laand
333(lb de un
Calculefrom
los cambios
ocurren
en
energía
interna
y en its
la
cuandois
aire cambia
an
state
of
10(atm),
where
molar
from
an initial
initialque
state
of 40(
40(
F) and
10(atm),
where
its entalpía
molar volume
volume
isel36.49(ft)
36.49(ft)
(lb
−1
◦
3
–1
◦◦F)su
−1
estado mole)
inicial−1
de
40(°F)
y 10(atm),
donde
volumen
molar
es 36.49(pie)
(lbPPmol)
,isaconstant
un estado final
,, to
aa final
state
1(atm).
Assume
for
VV/T
mole)
to
final
state of
of 140(
140(
F) and
and
1(atm).
Assume
for air
air that
that
/T is
constant
−1
◦◦◦F)−1
−1
−1((PV/T
−1.es
de 140(°F)
y 1C
(atm).
Suponga
para el aire
que
mole)
and
V
= 55 and
and CCPPP== 7(Btu)(lb
7(Btu)(lb
mole)
F)−1
. constante y que CV = 5 y CP = 7 (Btu)
and that
that
C
VV=
–1
–1
(lb mol) (°F) .
Solution 2.10
Solution
Solución
2.10 2.10
Because
property
changes
independent of
process that
brings them
about,
Because
property
changes are
are
of the
the
themlos
about,
Ya que los
cambios
en la propiedad
sonindependent
independientes
delprocess
procesothat
quebrings
los causa,
cálculos se
calculations
may
be
based
on
aa two-step,
mechanically
reversible
process
in
which
calculations
may
be
based
on
two-step,
mechanically
reversible
process
in
pueden basar en un proceso de dos etapas y mecánicamente reversible, en el cual 1(lbwhich
mol) de aire
1(lb
of
(a)
at constant volume
to
the
and (b)
1(lb mole)
mole)
of air
air isis
(a) cooled
cooled
to calentado
the final
final pressure,
pressure,
(b) para
es: a) enfriado
a volumen
constante
paraatlaconstant
presión volume
final, y b)
a presión and
constante
heated
at
constant
pressure
to
the
final
temperature.
The
absolute
temperatures
heated
at
constant
pressure
to
the
final
temperature.
The
absolute
temperatures
la temperatura final. Las temperaturas absolutas están dadas en la escala Rankine:
here
here are
are on
on the
the Rankine
Rankine scale:
scale:
T1 = 40 + 459.67 = 499.67(R)
T2 = 140 + 459.67 = 599.67(R)
TT111 =
TT222 =
= 40
40 +
+ 459.67
459.67 =
= 499.67(R)
499.67(R)
= 140
140 +
+ 459.67
459.67 =
= 599.67(R)
599.67(R)
Dado que
PV = kT, la relación
T/P ratio
es constanteispara
la etapa a).
Por lo tanto,
la temperatura interBecause
Because PPVV =
= kT
kT,, the
the ratio TT/P
/P is constant
constant for
for step
step (a).
(a). The
The intermediate
intermediate
media entre
los dos pasos
es:the two steps is therefore:
temperature
between
temperature between the two steps is therefore:
= (499.67)(1/10) ==49.97(R)
TTT��� ′=
= (499.67)(1/10)
(499.67)(1/10) = 49.97(R)
49.97(R)
y los cambios
temperatura
para lasfor
dos
son: are:
and
temperature
changes
the
two
and the
thede
temperature
changes
for
theetapas
two steps
steps
are:
�T
49.97
499.67
−449.70(R)
ΔT
= 49.97
–449.70(R)
aaaa=
�T
=
49.97–−
−499.67
499.67==
=
−449.70(R)
�T
=
599.67
−
49.97
=
549.70(R)
b
�T
599.67–−49.97
49.97==549.70(R)
549.70(R)
ΔTbbb =
= 599.67
02-SmithVanNess.indd 43
8/1/07 12:49:10
44
CHAPTER
CHAPTER2.
2. The
TheFirst
FirstLaw
Lawand
andOther
OtherBasic
BasicConcepts
Concepts
CHAPTER
2.
The
First
Law
and
Other
Basic
Concepts
44
44
44
CAPÍTULO 2. La primera ley y otros conceptos básicos
For
Forstep
step(a),
(a),by
byEqs.
Eqs.(2.18)
(2.18)and
and(2.15),
(2.15),
For
Eqs.
(2.18)
and
Para la etapa
a),step
por(a),
las by
ecuaciones
(2.18)
y(2.15),
(2.15)
�U
= CC �T
�T =
= (5)(−449.70)
(5)(−449.70) =
= −2,248.5(Btu)
−2,248.5(Btu)
�Uaa =
�U
a = CVVV �Taaa = (5)(−449.70) = −2,248.5(Btu)
�H
�Haa =
= �U
�U +
+VV �P
�P
�H
a = �Uaaa + V �Paaa
=
−2,248.5
+
=
−2,248.5
+
(36.49)(1−
−10)(2.7195)
10)(2.7195) =
= −3,141.6(Btu)
−3,141.6(Btu)
= −2,248.5 + (36.49)(1
(36.49)(1
−
10)(2.7195)
=
−3,141.6(Btu)
3(atm)(ft)
which
The
factor
2.7195
the
VVV product
from
which
is an
an energy
energy
The
factorconvierte
2.7195 converts
converts
the PPPPV
product
from (atm)(ft)
El factor
2.7195
el producto
de (atm)(pie)
, el cual333,,,es
una is
unidad
de energía a
which
is
an
energy
The
factor
2.7195
converts
the
product
from
(atm)(ft)
into
(Btu).
unit,
into
(Btu).
(Btu). unit,
unit, into (Btu).
the
volume
the
air
Forstep
step
(b),
thefinal
finalfinal
volume
of
the
airis:
is:
Para la For
etapa
b),(b),
el volumen
delof
aire
es:
For
step
(b),
the
final
volume
of
the
air
is:
�
�
�
��
�
�
�
�10
PPP11TTT22
10� �599.67
599.67�
333 3
10
599.67
1
2
(pies)
VVV22 =
=
=
= VV
= 36.49
36.49
= 437.93(ft)
437.93(ft)
=
437.93(ft)
2 = V111PP22TT11 = 36.49 11
499.67
499.67
P2 T1
1
499.67
Por las By
ecuaciones
(2.22)
(2.15),
and
ByEqs.
Eqs.(2.22)
(2.22)
andy(2.15),
(2.15),
By
Eqs.
(2.22)
and
(2.15),
�H
= CC �T
�T =
= (7)(549.70)
(7)(549.70) =
= 3,847.9(Btu)
3,847.9(Btu)
�Hbb =
�H
b = CPPP �Tbbb = (7)(549.70) = 3,847.9(Btu)
�U
=
�H
−
P
�V
�U
=
�H
−
P
�V
�Ubb = �Hbb − P �Vbb
b
b
b
=
= 3,847.9
3,847.9−
−(1)(437.93
(1)(437.93−
−36.49)(2.7195)
36.49)(2.7195) =
= 2,756.2(Btu)
2,756.2(Btu)
=
3,847.9
−
(1)(437.93
−
36.49)(2.7195)
=
2,756.2(Btu)
Para lasFor
dosthe
etapas
juntas,
two
For
the
twosteps
stepstogether,
together,
For
the
two
steps
together,
ΔU=
= −2,248.5
–2
248.5 ++
756.2 ==
507.7
(Btu)
�U
507.7(Btu)
�U
=
−2,248.5
+22,756.2
2,756.2
=
507.7(Btu)
�U
=
−2,248.5
+
2,756.2
=
507.7(Btu)
ΔH
=
–3
141.6
+
3
847.9
=
706.3
(Btu)
�H
�H =
= −3,141.6
−3,141.6+
+3,847.9
3,847.9 =
= 706.3(Btu)
706.3(Btu)
�H
=
−3,141.6
+
3,847.9
=
706.3(Btu)
2.12
BALANCES DE MASA Y ENERGÍA PARA SISTEMAS ABIERTOS
2.12
2.12 MASS
MASS AND
AND ENERGY
ENERGY BALANCES
BALANCES FOR
FOR OPEN
OPEN SYSTEMS
SYSTEMS
2.12
MASS
AND
ENERGY
BALANCES
FOR
OPEN
SYSTEMS
Aunque en las secciones anteriores nos hemos concentrado en sistemas cerrados, los conceptos mencionados
Although
focus
preceding
sections
closed
systems,
the
preAlthough
the
focus of
of the
the
preceding
sections has
has
been on
on
closed
systems,
the concepts
pretienen un
uso másthe
extenso.
Las
leyes
de conservación
de been
masa
y de
energía
se aplican
aconcepts
todos los
procesos,
Although
the
focus
of
the
preceding
sections
has
been
on
closed
systems,
the
concepts
presented
find
far
more
extensive
application.
The
laws
of
mass
and
energy
conservation
apply
to
sented
find
far
more
extensive
application.
The
laws
of
mass
and
energy
conservation
apply
to espeya seansented
sistemas
o cerrados.
hecho, el sistema
incluye
el sistema
cerrado como
findabiertos
far more
extensiveDe
application.
The lawsabierto
of mass
and energy
conservation
applycaso
to
all
processes,
to
open
as
well
as
to
closed
systems.
Indeed,
the
open
system
includes
the
closed
all
processes,
to
open
as
well
as
to
closed
systems.
Indeed,
the
open
system
includes
the
closed
cial. Por
tanto, el to
resto
deaseste
se dedica
al tratamiento
sistemas
al closed
desarrollo de
allloprocesses,
open
wellcapítulo
as to closed
systems.
Indeed, the de
open
system abiertos
includes ythe
system
case.
The
system
asaaaspecial
specialmás
case.
Theremainder
remainderof
ofthis
thischapter
chapteris
istherefore
thereforedevoted
devotedto
tothe
thetreatment
treatmentof
of
ecuaciones
de as
aplicación
extensa.
system
as
special
case.
The
remainder
of
this
chapter
is
therefore
devoted
to
the
treatment
of
open
systems
and
thus
to
the
development
of
equations
of
wide
applicability.
open
systems
and
thus
to
the
development
of
equations
of
wide
applicability.
open systems and thus to the development of equations of wide applicability.
Medidas
de flujo
Measures
of
Measures
of Flow
Flow
Measures
of
Flow
Open
characterized
by
flowing
there
four
common
measures
of
Opensystems
systemsare
arecaracterizan
characterized
bycorrientes
flowingstreams;
streams;
thereare
are
fourcuatro
common
measures
offlow:
flow:
Los sistemas
abiertos
se
por
que circulan;
existen
medidas
de flujo
comunes:
Open
systems
are
characterized
by
flowing
streams;
there
are
four
common
measures
of
flow:
...
...
•••Mass
flowrate,
m
Molar
nnn
qqq
••• Velocity,
uuu
Mass
flowrate,
m ṁ ••••Rapidez
Molarflowrate,
flowrate,
Volumetric
flowrate,
Velocity,
• Rapidez
de flujo
de masa,
de flujo molar,
ṅ••• •Volumetric
Rapidez deflowrate,
flujo volumétrico,
q • Velocidad,
u
Mass
flowrate,
m
Molar
flowrate,
Volumetric
flowrate,
Velocity,
The
are
Themeasures
measures
offlow
flow
areinterrelated:
interrelated:
Las medidas
de flujo of
están
correlacionadas:
The
measures
of
flow
are
interrelated:
...
... ·
ṁ
=MnMn
m
=
m
=M
m
=
M
nn
y
and
and
and
qqqq==
uAuuuAAA
=
=
donde where
M
es laM
molar.
En Importantly,
forma
importante,
la rapidez
de flujo relate
de
masa
y molar se relaciona con la
is
mass.
mass
molar
to
where
Mmasa
ismolar
molar
mass.
Importantly,
massand
and
molarflowrates
flowrates
relate
tovelocity:
velocity:
where
M
is
molar
mass.
Importantly,
mass
and
molar
flowrates
relate
to
velocity:
velocidad:
02-SmithVanNess.indd 44
8/1/07 12:49:13
4545
2.12.
and
Balances
forforOpen
Systems
2.12.Mass
Mass
andEnergy
Energy
Balances
Open
Systems
2.12. Balances
de masa
y energía
para sistemas
abiertos
45
2.12.
Mass
and
Energy
Balances
for Open
Systems
45
2.12.
45
45
2.12. Mass
Massand
andEnergy
EnergyBalances
Balancesfor
forOpen
OpenSystems
Systems
. .
. .
mm
Aρ
(2.24a)
n .n=
u Aρ
(2.24b)
.ṁ===uuAr
u
Aρ
(2.24a)
=
uAρ
Aρ
(2.24b)
(2.24a)
= uuAr
(2.24b)
a
b
m.. = u Aρ
(2.24a)
n.ṅ. =
(2.24b)
mm =
(2.24a)
(2.24b)
= uuAρ
Aρ
(2.24a) nn =
= uuAρ
Aρ
(2.24b)
The
area
forforflow
A is isthethecross-sectional
area
and
ρ is isspecific
orormolar
area
flow
cross-sectional
areaofde
ofa aun
aconduit,
conduit,
and
specific
molar
area
flow
isis the
cross-sectional
area
of
conduit,
and
specific
orasmolar
El
área The
AThe
para
el for
flujo
es elAAárea
de
sección
transversal
conducto,
yuρrρisis
esused
la
densidad
específica
o
density.
Although
velocity
a
vector
quantity,
its
scalar
magnitude
here
thethe
The
area
for
flow
A
is
the
cross-sectional
area
of
a
conduit,
and
ρ
is
specific
or
molar
density.
Although
velocity
is
a
vector
quantity,
its
scalar
magnitude
u
used
here
as
The
area
for
flow
A
is
the
cross-sectional
area
of
a
conduit,
and
ρ
is
specific
or
molar
.
.
density.speed
velocity
is
a
vector
quantity,
its
scalar
magnitude
u
is
used
here
as
the
molar.average
Aunque
laAlthough
velocidad
es
una
cantidad
vectorial,
su
magnitud
escalar
u
se
utiliza
en
este
caso
como
la
.
.
ofofa astream
direction
normal
toto
A.A.Flowrates
m,m,
q here
represent
.uun,isis
.n,and
density.
Although
isisthe
athe
quantity,
its
scalar
magnitude
used
average speed
speed
streaminen
in
direction
normal
Flowrates
andflujo
represent
density.
Although
velocity
a vector
vector
quantity,
itsrespecto
scalar
magnitude
asṁ,the
the
average
of corriente
a velocity
stream
in
the
direction
normal
to
A. different
Flowrates
m,
and
qqhere
represent
rapidezmeasures
promedio
de
una
dirección
normal
con
a A. Lainrapidez
de
deas
ṅ, y q re.. n,
.. used
of
quantity
per
unit
of
time.
Velocity
u
is
quite
nature,
as
it
does
not
average
speed
of
a
stream
in
the
direction
normal
to
A.
Flowrates
m,
n,
and
q
represent
measures
of
quantity
per
unit
of
time.
Velocity
u
is
quite
different
in
nature,
as
it
does
not
average
speed
of
a
stream
in
the
direction
normal
to
A.
Flowrates
m,
n,
and
q
represent
measures
of
quantity
per
unit
of
time.
Velocity
u
is
quite
different
in
nature,
as
it
does
not
presenta
medidas
de
la
cantidad
por
unidad
de
tiempo.
La
velocidad
u
es
de
una
naturaleza
muy
diferente
suggest
thethe
magnitude
ofofflow.
Nevertheless,
it itis isan
important
design
parameter.
measures
quantity
unit
of
time.
Velocity
quite
in
nature,
as
suggest
magnitude
flow.
Nevertheless,
important
design
parameter.
measures
of
quantity per
per
unit
of
time.
Velocity
uanisis
quite different
different
in
nature,
as itit does
does not
not
theof
of
Nevertheless,
it es
is uan
important
design
parameter.
porque suggest
no
sugiere
lamagnitude
magnitud
delflow.
flujo.
Sin
embargo,
un
importante
parámetro
de diseño.
suggest
suggestthe
themagnitude
magnitudeof
offlow.
flow. Nevertheless,
Nevertheless,ititisisan
animportant
importantdesign
designparameter.
parameter.
Example
2.11
Example
2.11
Example
2.11
. .
Ejemplo
2.11
−1 in a pipe with inside diameter
Example
2.11
Liquid
n-hexane
flows
atata arate
ofofmm
−1 in a pipe with inside diameter
Example
2.11
. ==0.75
Liquid
n-hexane
flows
rate
0.75kg
kgsss−1
.
.
Liquid n-hexane flows at a rate of m =
0.75
kg
in a pipe with inside diameter
. · =would
.and
DLiquid
are
n,.n,
u?u?
What
these
be
the
same
mm
−1quantities
El n-hexano
circula
aq,una
de
0.75 kg
ss–1
una
tubería
con
diámetro
interno
de
..if if
n-hexane
flows
arelación
rate
of
mm. m=
0.75
in
aa pipe
with
inside
diameter
D==
=55líquido
5cm.
cm.What
What
are
q,at
and
What
would
these
quantities
befor
for
the
same
Liquid
n-hexane
flows
at
aand
rateu?
of
= would
0.75 kg
kg
s−1en
in−3
pipe
with
inside
diameter
D
cm.
What
are
q,
n,
What
these
quantities
be
for
the
same
m
·
.
.. if m· si
.
=
2
cm?
Assume
for
liquid
n-hexane
that
ρ
=
659
kg
m
.
−3
D = 5D
cm.
¿Cuáles
son
los
valores
de
q,
n
y
u?
¿Cuáles
serían
estas
cantidades
para
la
misma
DD =
=
cm.
What
q,
n,
u?
would
quantities
be
=2525cm?
cm?
Assume
for
liquid
n-hexane
that
=these
659kg
kg
m−3 .
cm. Assume
What are
are
q,liquid
n, and
and
u? What
What
would
these
quantities
be for
for the
the same
same mm ifif
D
for
n-hexane
that
ρρ =
659
–3 m−3 .
D = 2 cm?
que parafor
elliquid
n-hexano
líquidothat
r =ρρ659
kg
DD =
22 cm?
n-hexane
=
=Suponga
cm? Assume
Assume
for
liquid
n-hexane
that
= 659
659mkg
kg.m
m−3..
Solution 2.11
. .−1
Solution
2.11
Solution
2.11 q q==mρ
We
have
and
−1
. −1
We
have
mρ
and
We have
q = mρ
and
· r –1
..= −1
Tenemos
que
q
m
y
−1
We
qq =
−1 and
Wehave
have
= mρ
mρ
and
Solution
2.11
Solución
2.11 2.11
Solution
. . . . −1
n .n=
−1
.MM−1
=mm
n=
M −1
· Mm.–1
.
n· =n.n.m=
= mmM
M −1
0.75
kgkgs s−1
−1
3
−1
0.75
whence
q q== 0.75 kg −3
−1
s−1==0.00114
whence
0.00114mm3s3 s−1
−1
kgkg
mm
whence
q =659
−3 = 0.00114 m3 s−1
0.75
kg
ss−3
0.75
kg
659
3
de donde
whence
qq =
=
whence
= 659 kg m−3
= 0.00114
0.00114m
m ss−1
−3 −1
659
kg
m
−1
3
659
kg
m
)(10
))
−1
−1
. . (0.75
−1
)(103g3 ggkg
kg−1
(0.75kg
kgs ss−1
n .n=
−1
)(10
kg
)==8.703
kg
8.703mol
mols ss−1
= (0.75
−1
3−1
−1
−1
3
−1
86.177
g
mol
=
8.703
mol
n.. =
−1
)(10
g
kg
)
(0.75
kg
s
g kg )
(0.7586.177
kg s )(10
g
mol
−1
−1
=
nn =
86.177 g mol−1
= 8.703
8.703mol
molss−1
=
. .
86.177
ggmol
Given
m,m,
quantities
are
independent
velocity,
86.177
mol−1 ofofD.D. The
. these
Given
these
quantities
are
independent
The
velocity,however,
however,dedeGivenonm,
these quantities
are
independent
D.a circular
The velocity,
however,
−1 , where,offor
..diameter
pends
through
u
=
q
A
cross-section,
A Ade==
−1
Given
m,
these
quantities
are
independent
of
D.
The
velocity,
however,
de·
pends
on
diameter
through
u
=
q
A
,
where,
for
a
circular
cross-section,
−1
Given
these son
quantities
D.a circular
Thela velocity,
however,
deDada m(π/4)D
,pends
estas cantidades
independientes
de, D.
Noofobstante,
velocidad
dependeAdel
on
diameter
uare= independent
q A−1
where,
for
cross-section,
= diáme2m,
. 2For
DD==5through
pends
uu =
qqAA−1,transversal
aa circular
.diameter
For
5cm,
cm, una
(π/4)D
–1,D
pendsuon
on
diameter
=
, where,
where, for
forcircular,
circular
cross-section,
== 5 cm,
tro mediante
=22 .qA
donde
para
sección
A =cross-section,
(p/4)D2. ParaAAD=
For
= through
5through
cm,
(π/4)D
(π/4)D
�2�
For DD =
= 55cm,
cm,π π� �
(π/4)D 2.. For
2
−2
2 0.00196 m 2
A A== π 5� ×
1010−2
mm�2=
−2
×
=0.00196
0.00196m
m2
��55×
��22 =
A =4ππ
10−2 m
4
2
−2
4
AA =
5
×
10
m
=
0.00196
m
=
5 × 10 m = 0.00196 m2
4
3
−1
4
0.00114
−1
−1
0.00114mm
m3s3 ss−1
whence
u u== 0.00114
==0.582
−1
de donde
whence
0.582mm
ms ss−1
2
3
−1
3
−1 =
0.00196
m
whence
u = 0.00114
0.582
2
s
m
s
0.00114
m
0.00196
m
−1
2
whence
uu =
whence for D = 2 cm,
= 0.00196 m22 =
= 0.582
0.582m
mss−1
Similarly,
0.00196
m
0.00196
m
Similarly,
for
D
=
2
cm,
De manera
similar,for
para
D=
2 cm,
Similarly,
D=
2 cm,
Similarly,
Similarly,for
for DD =
= 22cm,
cm,
0.00114
22
0.00114= 3.63 m s−1−1
AA
===0.000314
mmm
and
u u== 0.00114
0.000314
yand
2
A
0.000314
= 3.63 m s−1
2
and
u =0.000314
A = 0.000314 m2
0.00114
0.00114 = 3.63 m s−1
0.000314
and
uu =
AA =
= 0.000314
0.000314m
m2
and
= 0.000314 =
= 3.63
3.63m
mss−1
0.000314
0.000314
Balance de masa para sistemas abiertos
Mass
MassBalance
Balancefor
forOpen
OpenSystems
Systems
Mass
Balance
for
Open
Systems
La región
del
espacio
que
es
identificada
para
el análisis de systems
sistemas abiertos se
llama volumen
de control; se
Mass
Balance
for
Open
Systems
Mass
Balance
for
Open Systems
The
Theregion
regionofofspace
spaceidentified
identifiedforforanalysis
analysisofofopen
open systemsis iscalled
calleda acontrol
controlvolume;
volume;it itis is
Theseparada
region
of
identified for
ofsurface.
open systems
is called
control
volume;
it is
encuentra
despace
sussurroundings
alrededores
mediante
una
superficie
de control.
El afluido
dentro
del volumen
de
separated
from
its
bybyanalysis
a acontrol
The
fluid
within
thethecontrol
volume
The
of
space
identified
for
analysis
open
is
aa control
volume;
itit is
separated
from
its surroundings
surroundings
control
surface.
The
fluid
within
control
volume
The
region
oftermodinámico
space
identified
for
analysis
ofsurface.
open systems
systems
is called
called
control
volume;
is
from
its
byel
aque
control
fluid
within
the
control
volume
control separated
es
elregion
sistema
para
se of
escriben
losThe
balances
de
masa
y de
energía.
El
volumen
separated
separated from
from its
its surroundings
surroundings by
by aa control
control surface.
surface. The
The fluid
fluid within
within the
the control
control volume
volume
02-SmithVanNess.indd 45
8/1/07 12:49:20
46
46
46 46
46
46
CHAPTER
2. 2.
TheThe
First
Law andand
Other
Basic
Concepts
CHAPTER
First
Other
Basic
Concepts
CHAPTER
2. The
First
LawLaw
and Other
Basic
Concepts
CHAPTER 2. The First Law and Other Basic Concepts
CAPÍTULO
LaLaw
primera
ley y otros
CHAPTER
2. The 2.
First
and Other
Basicconceptos
Conceptsbásicos
ṁ1
ṁ1 ṁ1
ṁ1
ṁ1
ṁ 3
ṁ 3 ṁ 3
ṁ 3
ṁ1
ṁ 3
ṁ 3
Control
volume
Control
volume
Control
volume
dmcvdm
/dt /dt
Control
volume
cv
dmVolumen
/dt
de control
dmcv
/dt
cvvolume
Control
dmcv /dt
dmcv /dt
Figure
2.5: Schematic
representation
of a
Figure
Schematic
representation
Figure
2.5:2.5:
Schematic
representation
of aof a
control
volume.
Figure
2.5:
Schematic
representation
a
control
volume.
Figura 2.5:
Representación
esquemática de unofvolumen
control
volume.
control
Figure volume.
2.5: Schematic representation of a
de control.
control volume.
ṁ 2
ṁ 2 ṁ 2
ṁ 2
ṁ 2
ṁ 2
Control
surface
Control
surface
Control
surface
Control surface
Superficie de control
Control surface
is the
thermodynamic
system
for which
mass
and energy
balances
are written.
The control
is the
thermodynamic
system
which
mass
energy
balances
written.
control
is the
thermodynamic
system
for for
which
mass
andand
energy
balances
are are
written.
TheThe
control
volume
shown
schematically
in
Fig.
2.5
is
separated
from
its
surroundings
by
an
extensible
is the
thermodynamic
system
for
which
mass
and
energy
balances
are
written.
The
controlpor una
de control
que
se
muestra
en
forma
esquemática
en
la
figura
2.5
está
separado
de
sus
alrededores
volume
shown
schematically
in
Fig.
2.5
is
separated
from
its
surroundings
by
an
extensible
. . from
. . its surroundings
volume shown schematically in Fig. 2.5 is separated
by an extensible
· ym
· están
control
surface.
Two streams
with
flow
rates
m
m
are
directed
into
the
control
volume
shown
schematically
in
2.5
is rates
separated
from
itsshown
surroundings
by
anThe
extensible
.and
. flujo
is the
thermodynamic
system
forFig.
which
mass
energy
balances
are
written.
superficie
de control
extensiva.
Dos
corrientes
con
relaciones
de
m
dirigidas
hacia
el volu1 and
1
2 directed
control
surface.
streams
with
shown
control
1 and
2 are
. rates
control
surface.
TwoTwo
streams
with
flowflow
m
m. 22mare
shown
directed
intointo
the the
control
. 1 mand
·
.
and
one
stream
with
flow
rate
m
is
directed
out.
Because
mass
is
conserved,
the
rate
control
surface.
Two
streams
with
flow
rates
m
and
m
are
shown
directed
into
the
control
.
volume
shown
schematically
in
Fig.
2.5
is
separated
from
its
surroundings
by
an
extensible
men de volume,
control,
y
una
corriente
con
relación
de
flujo
m
está
dirigida
hacia
afuera.
Dado
que
la
masa
se con1
2
volume,
stream
with
is .3directed
Because
mass
is conserved,
volume,
andand
oneone
stream
with
flowflow
raterate
m. 33m
is3directed
out.
Because
mass
is conserved,
the the
raterate
. out.
change
of
mass
within
the
control
volume,
dm
/dt,
equals
the
net
rate
of
flow
of
mass
volume,
and
one
stream
with
flow
rate
m
is
directed
out.
Because
mass
is
conserved,
the
rate
control
surface.
Two
streams
with
flow
rates
m
and
m
are
shown
directed
into
the
control
serva, laof
rapidez
de
cambio
de
la
masa
dentro
del
volumen
de
control,
dm
/
dt,
es
igual
a
la
rapidez
neta
cv cv /dt,
3
1 dm
2
cvthe
of change
of mass
within
control
equals
of flow
of mass de
of change
of mass
within
the the
control
volume,
dm
equals
the
net net
raterate
of flow
of mass
. volume,
cv /dt,
the
control
volume.
The
convention
is
that
flow
is
positive
when
directed
into
the
control
of
change
of
mass
within
the
control
volume,
dm
/dt,
equals
the
net
rate
of
flow
of
mass
volume,
and
one
stream
with
flow
rate
m
is
directed
out.
Because
mass
is
conserved,
the
rate está
flujo deinto
masa
en
el
interior
del
volumen
de
control.
La
convención
es
que
el
flujo
es
positivo
cuando
3 is that flow
control
volume.
convention
is positive
when
directed
control
intointo
the the
control
volume.
TheThe
convention
is
that flowcvis positive
when
directed
intointo
the the
control
and
negative
when
directed
out.
The
mass
balance
is
expressed
mathematically
by:
into
the
control
volume.
The
convention
is
that
flow
is
positive
when
directed
into
the
control
ofhacia
change
of
mass
within
the
control
volume,
dm
/dt,
equals
the
net
rate
of
flow
of
mass
dirigidovolume
el
volumen
de
control,
y
negativo
cuando
se
dirige
hacia
afuera.
El
balance
de
masa
se expresa
cv balance is expressed mathematically by:
volume
negative
when
directed
mass
volume
andand
negative
when
directed
out.out.
TheThe
mass
balance
is expressed mathematically by:
volume
negative
when
directed
out. The
mass
balance
is expressed
mathematically
by:
into
the and
control
volume.
The
convention
is that
flow
is positive
when directed
into the control
en forma
matemática
por:
dm cvThe
volume and negative when directed out.
mass. balance
by:
. = 0 is expressed mathematically (2.25)
cv �(m)
dmdm
. fs
cv +
(2.25)
=
0
(2.25)
+
�(
m)
fs
dm
(2.25)
. fs = 0
dtcv + �(m)
(2.25)
dt dt+ �(m)fs = 0
dm
.
dtcv
(2.25)
+ �(m)fs = 0
thethe
second
term
for
thethe
control
of
Fig.
is: en la figura 2.5 es:
donde elwhere
segundo
término
para
el for
volumen
devolume
control
que
se 2.5
muestra
dt
where
second
term
control
volume
of
where the second term for the control volume of Fig.Fig.
2.52.5
is: is:
where the second term for the control .volume. of Fig.
is:
. 2.5
.
.=m
. m
. m
�(
m)
−
−
.
.
.
.is:2m. 2
fs
3
1
�(
m)
=
m
−
m
−
where the second term for the control
of
Fig.
2.5
fs
3
1
�(m)
=
m
−
m
−
m
.volume
.
.
.
�(m)fsfs = m33 − m11 − m22
.
.
.
. entreexit
El operador
dedifference
diferencia
“Δ” en
caso
representa
la diferencia
los flujos
de
salida
yflows
entrada,
The difference
operator
“�”este
here
signifies
the difference
between
and entrance
flows
and y el
operator
entrance
�(signifies
m)fs =
m1 − m
3 −difference
2 between
TheThe
difference
operator
“�”“�”
herehere
signifies
themthe
difference
between
exitexit
andand
entrance
flows
andand
subíndice
indica
que
el
término
se
aplica
a
todas
las
corrientes
que
circulan.
the
subscript
“fs”
indicates
that
the
term
applies
to
all
flowing
streams.
The“fs”
difference
operator
“�”
here
signifies
the
difference
between
exit
and
entrance
flows
and
subscript
indicates
term
applies
to flowing
all flowing
streams.
.the.the
the the
subscript
“fs”“fs”
indicates
thatthat
term
applies
to all
streams.
Cuando
laWhen
relación
de
flujo
de
masa
está
dada
por
la
ecuación
(2.24a),
la
ecuación
(2.25)
se
convierte
When
the
mass
flowrate
m
is
given
by
Eq.
(2.24a),
Eq.
(2.25)
becomes:
the
“fs”
indicates
that
the
term
applies
to
all
flowing
streams.
.
Thesubscript
difference
operator
“�”
here
signifies
the
difference
between
exit
and
entrance
flows
and
mass
flowrate
is given
by Eq.
(2.24a),
(2.25)
becomes:
When the the
mass
flowrate
m. ismgiven
by Eq.
(2.24a),
Eq.Eq.
(2.25)
becomes:
en:
When the
flowrate
is given
by Eq. to
(2.24a),
Eq. (2.25)
becomes:
the subscript
“fs”mass
indicates
thatmthe
term applies
all flowing
streams.
. dmdm
cv cv
When the mass flowrate m isdm
given
by�(ρu
Eq.
(2.24a),
0 (2.25) becomes:
(2.26)
A) = Eq.
cv +
+
�(ρu
(2.26)
dm
(2.26)
A)fsfsA)=fs 0= 0
(2.26)
dtcvdt+ �(ρu
(2.26)
+ �(ρu A)fs = 0
dt
dm
dtcv
In
this
form
the mass-balance
equation
is often
called
equation.
= 0continuity
(2.26)
+
�(ρu
A)
fsthe
In this
mass-balance
isllama
often
called
equation.
De estaIn
forma,
la form
ecuación
del balance
deequation
masa
ecuación
decontinuity
continuidad.
this
form
the the
mass-balance
equation
isse
often
called
the the
continuity
equation.
dtas
flow
process
characterized
steady
state
is an
important
special
case
for
which
thisThe
form
the
mass-balance
equation
is as
often
called
the
continuity
equation.
The
flow
process
characterized
steady
state
is
an
important
special
case
for
whichen el
ElInproceso
de
flujo
caracterizado
como
un
estado
estacionario
es
un
caso
especial
e
importante,
The flow process characterized as steady state is an important special case for which
conditions
within
thedel
control
volume
do
not
change
time.
TheThe
control
volume
then
conprocess
characterized
asisdo
steady
statewith
iscon
an
important
special
case
for
which
In
thisThe
formflow
the
mass-balance
equation
often
called
the
continuity
equation.
conditions
within
the
control
volume
not
change
with
time.
control
volume
then
concual lasconditions
condiciones
dentro
volumen
de
control
no
cambian
el
tiempo.
Por
lo
tanto
el
volumen
de conwithin the control volume do not change with time. The control volume then contains
auna
constant
mass
of
fluid,
andand
the
first
or accumulation
term
of Eq.
(2.25)
is
zero,
reducing
conditions
within
the
control
volume
do
not
change
time.
The
control
volume
then
con-es cero,
The
flow
process
characterized
asprimer
steady
statewith
isdean
important
special
case
for
which
tains
a
constant
mass
of
fluid,
the
first
or
accumulation
term
of
Eq.
(2.25)
is
zero,
reducing
trol contiene
masa
constante
de
fluido
y
el
término
acumulación
de
la
ecuación
(2.25)
tains a constant mass of fluid, and the first or accumulation term of Eq. (2.25) is zero, reducing
Eq.
(2.26)
to:
tains
aecuación
constant
mass
fluid, volume
and the first
or accumulation
term The
of Eq.
(2.25)volume
is zero,then
reducing
conditions
within
the of
control
do not
change with time.
control
con(2.26)
con lo que
la
Eq.Eq.
(2.26)
to: to:(2.26) se reduce a:
�(ρu
A)fsA)=fs 0= 0 term of Eq. (2.25) is zero, reducing
Eq.
to: mass of fluid, and the first
tains(2.26)
a constant
or
accumulation
�(ρu
�(ρu A)fs = 0
�(ρu A)fs = 0
Eq. (2.26)
to:
The
term
“steady
state”
does
not
necessarily
imply
thatthat
flowrates
are constant,
merely
that thethe
term
“steady
state”
does
necessarily
flowrates
constant,
merely
TheThe
term
“steady
state”
does
notnot
necessarily
imply
that
are are
constant,
merely
thatthat
the
�(ρu
A)imply
=
0 flowrates
fsof
inflow
of
mass
is
exactly
matched
by
the
outflow
mass.
The
term
“steady
state”
does
not
necessarily
imply
that
flowrates
are
constant,
merely
thatconstantes,
the
inflow
of mass
is exactly
matched
by the
outflow
of mass.
Elinflow
término
“estado
estacionario”
noby
implica
necesariamente
que las relaciones de flujo sean
. .
of mass
is exactly
matched
the
outflow
of mass.
When
there
is
but
a
single
entrance
and
a
single
exit
stream,
the
mass
flowrate
m
is
the
inflow
of
mass
is
exactly
matched
by
the
outflow
of
mass.
.
The
term
“steady
state”
does
not
necessarily
imply
that
flowrates
are
constant,
merely
that
When
is hacia
but
a dentro
single
entrance
a single
exit
stream,
mass
flowrate
is the
sino sólo queWhen
el flujo
dethere
masa
es exactamente
igual
derrame
dethe
masa.
there
is but
a single
entrance
andand
a single
exital
stream,
the
mass
flowrate
m. ismthe
same
for
both
streams;
When
there
is corriente
butthen,
amatched
single
entrance
and
acorriente
single
exit
stream,
the
mass flowrate
mdeismasa
the es la
inflow
of
mass
isuna
exactly
by
theyoutflow
of mass.
same
for
both
streams;
then,
Cuando
sólo
hay
que
entra
una
que
sale,
la
relación
de
flujo
same for both streams; then,
.
ρ u A and
− ρa single
u A =
0 stream, the mass flowrate m is the
sameambas
for
both
streams;
then,
When
there
is butpor
a single
exit
misma para
corrientes;
esto, entrance
ρ22uρ222Au222 A
−2ρ−11uρ111Au111 A
=1 0= 0
ρ 2 u 2 A 2 − ρ1 u 1 A 1 = 0
same for both streams; then,
ρ 2 u 2 A 2 − ρ1 u 1 A 1 = 0
02-SmithVanNess.indd 46
8/1/07 12:49:29
2.12.
and
Balances
for
Systems
2.12.Mass
Mass
andEnergy
Balances
forOpen
Open
Systems
2.12. Balances
de masa
yEnergy
energía
para sistemas
abiertos
2.12.
Mass
and
Energy
Balances
for
Open
Systems
Mass
and
Energy
Balances
forfor
Open
Systems
2.12.
Mass
and
Energy
Balances
Open
Systems
2.12.2.12.
Mass
and
Energy
Balances
for Open
Systems
4747
47
47 4747
47
..
oror
const==ρρ
2u
2 2AA
2 2==ρρ
1u
1 1AA
11
o
2u
1u
.m==const
. .m
.
or
m
==const
const
=2ρ
u2ρ22A1A
=1ρ
u111AA1 1
or
m
=
=
ρA
or
m
or
m = const=const
ρ2=
u 2=ρA
2A
1A
2u
2=
2u21=
11u
2uρ
1uρ
Because
specific
volume
reciprocal
ofof
density,
Because
specific
volumeis
isthe
reciprocal
density,
Dado
que
el volumen
específico
es
elthe
recíproco
de
ladensity,
densidad,
Because
specific
volume
is
the
reciprocal
ofdensity,
Because
specific
volume
the
reciprocal
ofof
density,
Because
specific
volume
is
the
reciprocal
Because
specific
volume
is theisreciprocal
of density,
. . uu1 1AA1 1 uu2 2AA2 2 uuAA
==
=uAu 1uAA1 A
uAu222uAAA
uuAA
A.m
1 2u=
.
.u 1.m
2=
22 u A
1u 1=
1=1=== VV
2=2== VV
m=
==11V2V
m = m m=
V VV
V1 V1VV
1 2 V2VV
1V
2V2
Esta forma
de
la
ecuación
de
continuidad
es
de
uso
frecuente.
This
form
of
the
continuity
equation
finds
frequent
use.
This form of the continuity equation finds frequent use.
This
form
ofthe
the
continuity
equation
finds
frequent
use.
form
the
continuity
equation
frequent
This
of
continuity
equation
finds
frequent
use.
This This
form
ofform
theofcontinuity
equation
findsfinds
frequent
use. use.
(2.27)
(2.27)(2.27)
(2.27)
(2.27)
(2.27)
(2.27)
The
Energy
Balance
TheGeneral
General
Energy
Balance
El balance
energético
general
The
General
Energy
Balance
The
General
Energy
Balance
The
General
Energy
Balance
The
General
Energy
Balance
Because
Becauseenergy,
energy,like
likemass,
mass,isisconserved,
conserved,the
therate
rateofofchange
changeofofenergy
energywithin
withinthe
thecontrol
controlvolume
volume
Because
energy,
like
mass,
isconserved,
conserved,
the
rate
of
change
ofcambio
energy
within
the
control
volume
Ya que
la
energía
senet
conserva,
alis
igual
que
lathe
masa,
lacontrol
rapidez
de
energía
dentro
del
volumen
de
Because
energy,
like
mass,
is
conserved,
rate
of
change
ofde
energy
within
the
control
volume
Because
energy,
like
the
rate
of
change
of
energy
within
the
control
volume
Because
energy,
like
mass,
ismass,
conserved,
the rate
of
change
of
energy
within
the
control
volume
equals
the
net
rate
of
energy
transfer
into
the
control
volume.
Streams
flowing
into
and
out
ofof
equals
the
rate
of
energy
transfer
into
the
volume.
Streams
flowing
into
and
out
equals
the
net
rate
of
energy
transfer
into
the
control
volume.
Streams
flowing
into
and
out
of
control
es
igual
a
la
rapidez
neta
de
transferencia
de
energía
en
el
volumen
de
control.
Las
corrientes
que
equals
the
net
rate
of
energy
transfer
into
the
control
volume.
Streams
flowing
into
and
out
of
equals
the
net
rate
of
energy
transfer
into
the
control
volume.
Streams
flowing
into
and
out
of
equals
the
net
rate
of
energy
transfer
into
the
control
volume.
Streams
flowing
into
and
out
of
the
thecontrol
controlvolume
volumehave
haveassociated
associatedwith
withthem
themenergy
energyininitsitsinternal,
internal,potential,
potential,and
andkinetic
kinetic
the
control
volume
have
associated
with
them
energy
insystem.
internal,
potential,
and
circulan
hacia
dentro
ycontribute
hacia
fuera
del
volumen
de
control
se
asocian
con
energía
en
sus
formas
interna, pothethe
control
volume
have
associated
with
them
energy
in
internal,
potential,
and
kinetic
control
volume
have
associated
with
them
energy
inits
itsits
internal,
potential,
and
the
control
volume
have
associated
with
them
energy
in
internal,
potential,
and
kinetic
forms,
and
all
toto
the
energy
change
ofofits
the
system.
Each
unit
mass
of
akinetic
stream
forms,
and
all
contribute
the
energy
change
the
Each
unit
mass
of
akinetic
stream
1 1 2 change
forms,
and
all
contribute
to
the
energy
change
of
the
system.
Each
unit
mass
of
a
stream
tencial
ycarries
cinética,
y
todas
contribuyen
al
cambio
de
energía
del
sistema.
Cada
unidad
de
masa
de
una
forms,
and
all
contribute
to
the
energy
of
the
system.
Each
unit
mass
of
a
stream
forms,
and
all
contribute
to
the
energy
change
of
the
system.
Each
unit
mass
of
a
stream
forms,
and
all
contribute
to
the
energy
change
of
the
system.
Each
unit
mass
of
a
stream
2
UU++12 uu ++zg,
carrieswith
withitita atotal
totalenergy
energy
zg,where
whereuuisisthe
theaverage
averagevelocity
velocityofofthe
thestream,
stream,z ziscorrienis
2 + zg,
1+12 u
1 +
2 where
2+
carries
with
it
a
total
energy
U
where
u
is
the
average
velocity
of
the
stream,
carries
with
it
a
total
energy
U
u
+
zg,
where
u
is
the
average
velocity
of
the
stream,
zstream
is
te lleva
consigo
una
energía
total
donde
u
es
la
velocidad
promedio
de
la
corriente,
carries
with
it
a
total
energy
U
u
+
zg,
where
u
is
the
average
velocity
of
the
stream,
z zisis z es su
carries
with
it
a
total
energy
U
+
u
+
zg,
u
is
the
average
velocity
of
the
stream,
z
is
itsitselevation
acceleration
ofofgravity.
Thus,
each
2 2 2and
2 level,
elevationabove
abovea adatum
datum
level,
andggisisthe
thelocal
local
acceleration
gravity.
Thus,
each
stream
. .acceleration
2is
its
elevation
above
datum
level,
local
acceleration
ofgravity.
gravity.
Thus,
each
stream
itsits
elevation
above
a de
datum
and
g1gis
the
local
gravity.
Thus,
each
stream
elevación
sobre
un
nivel
referencia
yis
g+and
es
aceleración
local
deenergy
laofgravedad.
Así,
cada
corriente
transporelevation
aataat
datum
level,
and
local
acceleration
of
Thus,
each
stream
its
elevation
above
aabove
datum
level,
and
g(U
the
acceleration
of
gravity.
Thus,
each
stream
2is
transports
energy
the
rate
(U
u1laglocal
+the
zg)
net
transported
into
the
system
transports
energy
thelevel,
rate
+
+the
The
net
energy
transported
into
the
system
.m.The
.zg)
.m.
12 2 2u.��
�energy
� �transported
1
1
1
2
2
��
�
transports
energy
atthe
the
rate
+
u
+
zg)
m.
The
net
energy
transported
into
the
system
transports
energy
rate
++
u
+
zg)
m.
The
net
energy
into
the
system
ta energía
en proporción
de
Por
lo
tanto,
la
energía
neta
transportada
hacia
adentro
del
transports
energy
atthe
rate
u
+
zg)
m.
The
net
transported
into
the
system
transports
energy
at
theatrate
(U
+(U2(U
u(U
+
zg)
m.
The
net
energy
transported
into
the
system
.
.
2����
2 2−�
� �m
� , wherethe
�zg
�zg
��
�+
bybythe
UU++112 �u1 u2 .2+
theflowing
flowingstreams
streamsisistherefore
therefore
−���
theeffect
effectofofthe
theminus
minus
. fsfs, where
. �.m
2
1
1
1
2
2
2
sistema
por
las
corrientes
que
circulan
es−�
donde
el
efecto
del
signo
menos con
by
the
flowing
streams
is
therefore
U+
+u2zg
u+
+zg
zg
where
the
effect
ofthe
the
minus
byby
the
flowing
streams
is is
therefore
−�
,fswhere
thethe
effect
the
minus
the
flowing
streams
therefore
mm
, ,,where
effect
of
minus
by
the
flowing
streams
is therefore
−�
U−�
+ UU
u+
+
mzg
,mwhere
the
effect
of
theofminus
2u+
fsof
fsrate
2 out.
sign
The
within
signwith
with“�”
“�”isistotomake
makethe
theterm
termread
read2inin−−
out.
The
rate
offsenergy
energyaccumulation
accumulation
withinthe
the
..
sign
with
“�”
is
to
make
the
term
read
in
−
out.
The
rate
of
energy
accumulation
within
the
“Δ” sign
essign
hacer
que
término
se
lea
entrada-salida.
La
rapidez
de
acumulación
laand
energía
del voluwith
“�”
the
term
read
−addition
out.
The
rate
of
energy
within
thethe
with
“�”
istoincludes
tomake
make
the
term
−
out.rate
The
rate
of
energy
accumulation
within
sign
with
“�”
isvolume
toelis
make
the
term
read
inread
− in
out.
The
energy
accumulation
within
thedentro
control
this
quantity
inin
totoof
the
heat
transfer
rate
work
rate:
. de
control
volume
includes
this
quantity
in
addition
the
heat
transfer
rate
and
work
rate:
.Q.Q
.accumulation
control
volume
includes
this
quantity
in
addition
tothe
the
heat
transfer
rate
Qand
and
work
rate:del trabajo:
mencontrol
de
control
incluye
esta
además
de
la
de
transferencia
de
yrate:
la
rapidez
volume
includes
this
quantity
inin
addition
the
heat
transfer
rate
Qcalor
and
work
rate:
control
volume
includes
this
quantity
addition
to
heat
transfer
rate
Q
work
rate:
control
volume
includes
this cantidad
quantity
in
addition
to rapidez
thetoheat
transfer
rate Q
and
work
�� � �
��
��
..
d(mU
d(mU)cv
)cv
1�
1 2 2+
��
�
�. +
�
�m
��
�
�..m
��
�
��
u
zg
=
−�
U
+
u
+
zg
workrate
rate
=
−�
U
+
.Q.Q++work
d(mU
)
d(mU
)
d(mU
)
d(mU )cv dt
fsfs. +
222 .
.
.
cv cvcv
1
2
1
1
2
1
=−�
U+
+u2zg
u+
+zg
zgm
+Q+
Q+
+
work
rate
Q
work
rate
==−�
u+
work
rate
razón
de
trabajo
u+
+
mzg
+mm
Qfs+
+
work
rate
= dt
−�
U−�
+ U2 U
fs+
2
fs
fs
2
dt dtdtdt
Q̇
Q̇Q̇
Q̇
Q̇ Q̇Q̇
u2
PP
V,V,U,U,HH
P
P PP
V,U,
U,HH
H
V,
V, V,
U,
HU,
V, U, H
u1
u1u1
u1 u1uu11
u 2u 2
u 2u 2uu22
Actual
Actual
Actual
Perfil
de
Actual
Actual
Actual
velocity
velocity
velocity
velocidad
velocity
velocity
velocity
profile
profile
profile
real
profile
profile
profile
Control
Control
Volumen
Control
Control
Control
Control
volume
volume
de
control
volume
volume
volume
volume
P
u 2u 2
u 2u 2uuu222
Ẇs
ẆẆ
ss
Ẇ
ẆsẆẆ
s ss
Figure
Figure2.6:
2.6:Control
Controlvolume
volumewith
withone
oneentrance
entranceand
andone
oneexit.
exit.
Figura
2.6:
Volumen
de control
con
una
entrada
y one
una
salida.
Figure
2.6:
Control
volume
with
one
entrance
and
one
exit.
Figure
2.6:
Control
volume
with
one
entrance
and
one
exit.
Figure
2.6:
Control
volume
with
one
entrance
and
exit.
Figure
2.6:
Control
volume
with
one
entrance
and
one
exit.
The
Thework
workrate
ratemay
mayinclude
includework
workofofseveral
severalforms.
forms.First,
First,work
workisisassociated
associatedwith
withmoving
moving
The
work
rate
may
include
work
of
several
forms.
First,
work
is
associated
with
moving
The
work
rate
may
include
work
of
several
forms.
First,
work
is
associated
with
moving
The
work
rate
may
include
work
of
several
forms.
First,
work
is
associated
with
moving
The
work
rate
may
include
work
of
several
forms.
First,
work
is
associated
with
moving
the
flowing
streams
through
entrances
and
exits.
The
fluid
at
any
entrance
or
exit
has
a aset
the
flowing
streams
through
entrances
and
exits.
The
fluid
at
any
entrance
or
exit
has
setofof con el
rapidez
del
trabajo
puede
incluir
el
trabajo
defluid
varias
formas.
Primero,
el
trabajo
está
asociado
the
flowing
streams
through
entrances
and
exits.
The
fluid
at
any
entrance
or
exit
has
a
set
theLa
flowing
streams
through
entrances
and
exits.
The
fluid
at
any
entrance
or
exit
has
a
set
of
the
flowing
streams
through
entrances
and
exits.
The
fluid
at
any
entrance
or
exit
has
a
set
ofof
the flowing
streams
through
entrances
and
exits.
The
at
any
entrance
or
exit
has
a
set
of
average
properties,
P,
V
,
U
,
H
,
etc.
Imagine
that
a
unit
mass
of
fluid
with
these
properties
average properties, P, V , U , H , etc. Imagine that a unit mass of fluid with these properties
movimiento
deproperties,
las
corrientes
que
circulan
aImagine
través
las
ymass
las
salidas.
El
fluido
enproperties
cualquier entrada
average
properties,
UH
etc.
Imagine
that
unit
of
fluid
with
these
properties
average
properties,
V, ,V
U
,etc.
,H
etc.
aentradas
unit
mass
ofof
fluid
with
these
properties
average
P,P,
,,,U
, ,H
, ,etc.
Imagine
that
a aunit
fluid
with
these
average
properties,
P, VP,
,U
HV
Imagine
thatde
athat
unit
mass
ofmass
fluid
with
these
properties
02-SmithVanNess.indd 47
8/1/07 12:49:41
48
CHAPTER 2. The First Law and Other Basic Concepts
CHAPTER 2.
2. The
The First
First Law and
and Other Basic
Basic Concepts
Concepts
48
CHAPTER
48
CAPÍTULO
2. LaLaw
primeraOther
ley y otros conceptos
básicos
CHAPTER 2. The First Law and Other Basic Concepts
48
exists at an entrance or exit, as shown in Fig. 2.6 (at the entrance). This unit mass of fluid is
exists
at an
an
entrance
or
exit, as
as shown
shown
in Fig.
Fig.P,2.6
2.6
(at H,
theetc.
entrance).
This
unit
mass of
of fluid
fluid is
is
o salidaexists
tiene at
un
conjunto
deor
propiedades
promedio,
V, a(at
U,
Suponga
quethe
una
masa
unitaria
de fluido
entrance
exit,
in
the
entrance).
This
unit
mass
acted
upon
by
additional
fluid,
here
replaced
by
piston
which
exerts
constant
pressure
exists
at an by
entrance
or una
exit,
as here
shown
in Fig.
2.6
the
entrance).
This
unit
mass
ofpressure
fluid
is Esta
acted
upon
by
additional
fluid,
hereoreplaced
replaced
bycomo
a(at
piston
which exerts
exerts
the
constant
pressure
con estas
propiedades
existe
en
entrada
una
salida,
se
muestra
en
la
figura
2.6
(en
la
entrada).
acted
upon
additional
fluid,
by
a
piston
which
the
constant
P. Theupon
workbydone
by this fluid,
pistonhere
in moving
theby
unit
mass through
the entrance
is P V ,pressure
and the
acted
additional
replaced
aque
piston
which
exerts
the
constant
.by
P. masa
The work
work
done
by this
thissobre
piston
in
moving
the unit
unit
mass
through
the entrance
entrance
ispistón
V,, and
and the
the preunidad P.
de
del
fluido
actúa
el in
fluido
adicional,
aquí
es sustituido
por
unis
que
ejerce
The
done
piston
moving
the
mass
through
the
PPVquantities,
work
rate
is
(P
V
)
m.
Because
“�”
denotes
the
difference
between
exit
and
entrance
.
. by
P. The
work
done
this piston
indenotes
moving
the
unit massbetween
through
the and
entrance
isa P
V
, and
the
work
rate
is (P
(P
V))m.
m.
Because
“�”
denotes
the
difference
between
exit
and
entrance
quantities,
sión constante
P.
El
trabajo
hecho
por
este
pistón
en
movimiento,
de
la
unidad
de
masa
través
de
work
rate
is
V
Because
“�”
the
difference
exit
entrance
quantities,
. the system when all entrance and exit sections are taken into accountlais entrada
the
netrate
work
done
on
work
is
(P
V
)
m.
Because
“�”
denotes
the
difference
between
exit
and
entrance
quantities,
·
.dedone
the
net work
work
done
on the
the(PV)m
system
whenque
all entrance
entrance
andlaexit
exit
sectionsentre
are taken
taken
into account
account
is y de
es PV, ythe
la net
rapidez
trabajo
es
. Dado
“Δ” denota
diferencia
las
cantidades
de salida
system
when
all
and
sections
are
into
is
−�[(P
V )m]
. on
.. fs
the
net V
work
done
on the system
when cuando
all entrance
and exit sections
are
taken.into
account
is
−�[(P
V))m]
m]
.
fs
entrada,−�[(P
el trabajo
neto
realizado
en
el
sistema
se
consideran
todas
las
secciones
de
entrada
y
.
. s . In addition de sali. fs form of work is the shaft work indicated in Fig. 2.6 by rate W
Another
.
−�[(P
V
)
m]
.
·
Another
form of
of work
work is
is the
the shaft
shaft work
work indicated
indicated in
in Fig.
Fig. 2.6
2.6 by
by rate
rate W
W.ss.. In
In addition
addition
da es –Δ[(PV)m
]fs.be fsassociated
form
work Another
may
with isexpansion
or
contraction
ofinthe
control
volume
and
there
. may
Another
form es
of el
work
thepor
shaft
work
indicated
Fig.
2.6laby
rate WẆ
In
addition
s .. Además,
work
may
betrabajo
associated
with
expansion
or
contraction
of the
the
control
volume
and
there
may
Otra
forma
de
señalado
la
flecha
en
la
figura
2.6
por
rapidez
el trabajo
work
may
be
associated
with
expansion
or
contraction
of
control
volume
and
there
s by W
. . may
be
stirring
work.
These
forms
of
work
are
all
included
in
a
rate
term
represented
The
.
work
maycon
be la
associated
with
expansion
orall
contraction
of
volume
andby
there
may
be
stirring
work.
These forms
forms
of
work are
are
all
included
in
athe
ratecontrol
term
represented
by
Wtrabajo
. The
The de agi.
se puede
asociar
expansión
o
la
contracción
del
volumen
de
control
y
ahí
se
considera
be
stirring
work.
These
of
work
included
in
a
rate
term
represented
W
.
preceding
equation
may now
beofwritten:
be
stirring
work.
These
forms
work
are all included
in a rate
represented
by W . The
preceding
equation
may
now be
beestán
written:
tación. preceding
Todas
estas
formas
de trabajo
incluidas
en un término
determ
la rapidez
representado
por Ẇ. La
equation
may
now
written:
��
�
�
preceding
equation
may
now
be
written:
.
.
d(mU
)cv escribir
ecuación anterior ahora
se puede
como:
.
.
��
� �� + Q. − �[(P V )m] + W.
1 2
d(mU))cv
cv = −� �� U + 21 u 2 + zg� m
. − �[(P V )m]
.
d(mU
.. �fs + Q
.. fs
= −�
−� ��U
U+
+ 12uu2 +
+ zg
zg �m
m
+W
W
dt )cv =
.
�[(P V )m]
d(mU
. fsfs + Q. −
. fsfs +
1 2
2
dt
dt
= −� U + 2 u + zg m fs + Q − �[(P V )m]fs + W
dt in accord with the definition of enthalpy, H = U + P V , leads to:
Combination of terms
Combination of
of terms
terms in
in accord
accord with
with the
the definition
definition of
of enthalpy,
enthalpy, H
H=
=U
U+
+ PPVV,, leads
leads to:
to:
Combination
La combinación
de términos
de
con the
la��definición
UU
+ .PV,
lleva a:
� � HH=.=
Combination
of termsd(mU
in acuerdo
accord
definitionde
of entalpía,
enthalpy,
+ Pnos
V , leads
to:
)cv with
.
� �� + Q. + W.
��
1 2
d(mU))cv
.
cv = −� �� H + 21 u 2 + zg� m
.
.
d(mU
fs
. � +Q+W
+ zg
zg �m
m
= −�
−� ��H
H+
+ 12uu2 +
dt )cv =
d(mU
. fs + Q. + W.
dt
dt
= −� H + 212 u 2 + zg m fsfs + Q + W
which is usually written: dt
lo que por
lo regular
se escribe
which
is usually
usually
written:como:
which
is
written:
which is usually written:
� �
��
d(mU )cv
��H + 1 u 2 + zg�� m. �� = Q.. + W..
��
d(mU
)
(2.28) (2.28)
+
�
cv
. +W
.
d(mU
.. �fs = Q
+ zg � m
(2.28)
+ � ��H
H+
+ 1212uu22 +
dt ))cv
fs = Q. +
W.
(2.28)
d(mU
cv + �
1 2 zg m. fs
2
dt
dt
(2.28)
+ � H + 2 u + zg m fs = Q + W
dt de energía cinética de los balances de energía es la magnitud media de
La velocidad
u
en
los
términos
The velocity u in the kinetic-energy terms
energy balances is the bulk-mean velocity
· /rA. of
. u=m
The velocity
velocity
in
the
kinetic-energy
terms
of
energy
balances
is the
the en
bulk-mean
velocity un perla velocidad,
como
sethe
define
por
lakinetic-energy
ecuación,
Los
fluidos
que circulan
tuberíasprofile,
exhiben
The
uu in
the
terms
of
energy
is
velocity
as defined
by
equation,
u
=
m/ρ
A.
Fluids
flowing
in balances
pipes
exhibit
abulk-mean
velocity
as
.
.
The
velocity
u
in
the
kinetic-energy
terms
of
energy
balances
is
the
bulk-mean
as
defined
by
the
equation,
u
=
m/ρ
A.
Fluids
flowing
in
pipes
exhibit
a
velocity
profile,
as
fil de velocidad,
como
se equation,
muestra
en
figura
2.6,
elthe
cual
se incrementa
elavalor
cero
envelocity
las paredes
(la
as
defined
by the
u la
=from
m/ρ
A.
Fluids
flowing
inno-slip
pipes desde
exhibit
velocity
profile,
as
. zero
shown
in
Fig.
2.6,
which
rises
at
wall
(the
condition)
to
a
maximum
at
as
defined
by
the
equation,
u
=
m/ρ
A.
Fluids
flowing
in
pipes
exhibit
a
velocity
profile,
as
shown
in
Fig.
2.6,
which
rises
from
zero
at
the
wall
(the
no-slip
condition)
to
a
maximum
at de un
condición
de
no
deslizamiento)
hasta
llegar
a
un
máximo
en
el
centro
de
la
tubería.
La
energía
cinética
shown
in
Fig.
2.6,
which
rises
from
zero
at
the
wall
(the
no-slip
condition)
to
a
maximum
at
the
center
of the
pipe.
Therises
kinetic
energy
ofthe
a fluid
in
a pipe
depends
on itstovelocity
profile.
shown
in Fig.
2.6,
which
from
zero
at
wall
(the
no-slip
condition)
a maximum
the
center
of the
the
pipe.
The
kinetic
energy
of aa fluid
fluid
in
pipe
depends
on its
its velocity
velocity
profile.
fluido en
una
tubería
depende
de
sukinetic
perfil
de
velocidad.
Parain
eland
del
flujo
laminar,
el
perfil
es deattipo pathe
center
of
pipe.
The
energy
of
aacaso
pipe
depends
on
profile.
For
the
case
of
laminar
flow,
the
profile
is
parabolic,
integration
across
the
pipe
shows
the
center
of
the
pipe.
The
kinetic
energy
of
a
fluid
in
a
pipe
depends
on
its
velocity
profile.
For
the
case
of
laminar
flow,
the
profile
is
parabolic,
and
integration
across
the
pipe
shows
rabólico,
y the
la integración
a través
dethe
la tubería
muestra
el término
de energía
For
case
of laminar
flow,should
profile
is
parabolic,
and
integration
across cinética
the
pipe
shows
In fully
developed
turbulent
flow,
thedebería
more ser en
that
the
kinetic-energy
term
properly
be
u 22. que
For
case
ofun
laminar
flow,
the completamente
profile
is be
parabolic,
anddeveloped
integration
across
the
pipe
2. En
In fully
fully
developed
turbulent
flow,
theshows
more
that
the kinetic-energy
kinetic-energy
term
should
properly
be
u2 ..desarrollado,
sentidothat
estricto
u
flujo
turbulento
que
es
el
caso
más
común
en la prácIn
turbulent
flow,
the
more
the
term
should
properly
u
common
case in practice,
the
velocity
across
the
portion
of the
pipe isflow,
not the
far more
from
2 . major
In
fully
developed
turbulent
that
the
kinetic-energy
term
should
properly
be
u
common
case
in
practice,
the
velocity
across
the
major
portion
of
the
pipe
is
not
far
from
2
tica, la common
velocidad
a
través
de
la
parte
principal
de
la
tubería
no
se
aleja
mucho
de
lo
uniforme,
y
la
expresión
case
in
practice,
the
velocity
across
the
major
portion
of
the
pipe
is
not
far
from
uniform,
and
the
expression
u 2/2,
as used
in thethe
energy
equations,
is more
nearly
correct.
common
casethe
inexpression
practice,
the
velocity
across
major
portion
of
the pipe
is not
far from
uniform,
and
the
expression
u2 /2,
/2,
asenergía,
used
ines
themucho
energy
equations,
is more
more
nearly
correct.
u2/2, como
seAlthough
emplea
en
las
ecuaciones
de
más
correcta.
uniform,
and
u
as
used
in
the
energy
equations,
is
nearly
correct.
Eq.
(2.28) isu 2an
energy
balance
of reasonable
generality,
it has correct.
limitations.
uniform,
theEq.
expression
/2,
as usedbalance
inenergía
the energy
equations,
is more tiene
nearly
Although
Eq.
(2.28)es
isun
an
energy
balance
of reasonable
reasonable
generality,
it has
hassus
limitations.
Aunque
la and
ecuación
(2.28)
balance
de
en
general
razonable,
limitaciones.
En
Although
(2.28)
is
an
energy
of
generality,
itcontrol
limitations.
In particular,
it
reflects
the
tacit
assumption
that
the
center
of
mass
of
the
volume
is
Although
Eq.
(2.28)
is
an
energy
balance
of
reasonable
generality,
it
has
limitations.
In
particular,
it
reflects
the
tacit
assumption
that
the
center
of
mass
of
the
control
volume is
isDe este
particular,
refleja
la
suposición
tácita
de
que
el
centro
de
masa
del
volumen
de
control
es
estacionario.
In
particular,
it
reflects
the
tacit
assumption
that
the
center
of
mass
of
the
control
volume
stationary.
Thus
no terms
for
kineticand potential-energy
changes
ofofthe
fluid
in the
control
In se
particular,
ittérminos
reflects
thefor
tacit
assumption
that the center
ofymass
thedel
control
volume
is
stationary.
Thus
no terms
terms
for
kineticand
potential-energy
changes
of the
the
fluid
in the
the
control
modo, no
incluyen
para
elkineticcambio
de las
energías
cinética
potencial
fluido
encontrol
el
volumen
de
stationary.
Thus
no
and
potential-energy
changes
of
fluid
in
volume
are
included.
For
virtually
all
applications
of
interest
to
chemical
engineers,
Eq.
(2.28)
stationary.
Thus
no
terms
for
kineticand
potential-energy
changes
of
the
fluid
in
the
control
volume
are
included.
For
virtually
all
applications
of
interest
to
chemical
engineers,
Eq.
(2.28)
control.volume
En
realidad,
paramany
todas
las
aplicaciones
de interéskineticpara
losand
ingenieros
químicos,
la ecuación
(2.28) es
are included.
For
virtually
allapplications,
applications
of
interest
topotential-energy
chemical
engineers,
Eq. (2.28)
is
adequate.
For
(but
not
all)
changes
in
the
volume
are included.
virtually
all
applicationskineticoflos
interest
chemical
engineers,
Eq.yin
(2.28)
is
adequate.
For(aunque
manyFor
(but
not all)
all)
applications,
kineticandtopotential-energy
potential-energy
changes
inpotencial
the
adecuada.
Para muchas
no
todas
las
aplicaciones),
ento:
las energías
cinética
en
is
adequate.
For
many
not
applications,
and
changes
the
flowing
streams
are
also(but
negligible,
and
Eq. (2.28)kineticthencambios
simplifies
is
adequate.
For
many
(but
not
all)
applications,
and
potential-energy
changes
in
the
flowing
streams
are
also
negligible,
and
Eq.
(2.28)
then
simplifies
to:
las corrientes
flujo también
despreciables,
por esto
la ecuación
flowingdestreams
are alsoson
negligible,
and Eq.y (2.28)
then
simplifies(2.28)
to: se simplifica a:
flowing streams are also negligible,
.
. to:
d(mUand
)cv Eq. (2.28)
. then simplifies
. +W
.
d(mU))cv
(2.29)
. fs = Q
cv + �(H m)
.
.
d(mU
.
=Q
Q. +
+W
W.
(2.29) (2.29)
+ �(H
�(Hm)
m)
dt )cv +
fs =
(2.29)
d(mU
.
fs
dt
dt
(2.29)
+ �(H m)fs = Q + W
dt
48
Example
2.12
Example
2.12
Example
2.12
Ejemplo
2.12
Show
that
Eq. 2.12
(2.29) reduces to Eq. (2.3) for the case of a closed system.
Example
Show that Eq. (2.29) reduces to Eq. (2.3) for the case of a closed system.
Show that Eq. (2.29) reduces to Eq. (2.3) for the case of a closed system.
Demuestre
que
la Eq.
ecuación
se reduce
a la ecuación
(2.3)ofena el
caso system.
de un sistema cerrado.
Show
that
(2.29)(2.29)
reduces
to Eq. (2.3)
for the case
closed
02-SmithVanNess.indd 48
8/1/07 12:49:48
2.12.
2.12. Mass
Mass and
and Energy
Energy Balances
Balances for
for Open
Open Systems
Systems
49
49
49
2.12. Balances de masa y energía para sistemas abiertos
Solution
Solution
2.12
Solución
2.12 2.12
The
second
Eq.
is
streams,
and
El segundo
de laof
ecuación
(2.29)
se omitein
la absence
ausenciaof
deflowing
corrientes
de flujo,
The término
second term
term
of
Eq. (2.29)
(2.29)
is omitted
omitted
inenthe
the
absence
of
flowing
streams,
andy por lo
equation
is
then
multiplied
by
dt:
tanto la the
ecuación
se
multiplica
por
dt:
the equation is then multiplied by dt:
..
..
d(mU
= Q dt + W dt
d(mU))cv
cv = Q dt + W dt
Integration
time
Integration
over
time gives:
gives:
Integrando
a travésover
del tiempo
dado:
=
�(mU
�(mU))cv
cv =
��
tt22
tt11
��
..
Q
dt
+
Q dt +
tt22
tt11
..
W
W dt
dt
tt = Q + W
�U
�U
Q+
ΔU t =
= Q
+W
W
The
Q
and
W
terms
are
defined
by
the
integrals
The Q and W terms are defined by the integrals of
of the
the preceding
preceding equation.
equation.
Los términos Q y W se definen mediante integrales de la ecuación anterior.
o
or
or
Equation
Equation (2.29)
(2.29) may
may be
be applied
applied to
to aa variety
variety of
of processes
processes of
of aa transient
transient nature,
nature, as
as illusillusLa
ecuación
(2.29) se puede
aplicar a una variedad de procesos de naturaleza transiente (fenómeno
trated
trated in
in the
the following
following examples.
examples.
momentáneo que sucede en un sistema previo a lograr una condición de estado estacionario), como se muestra en los siguientes ejemplos.
Example 2.13
An
An evacuated
evacuated
tank is
is filled
filled with
with gas
gas from
from aa constant-pressure
constant-pressure line.
line. What
What is
is the
the relation
relation
Ejemplo
2.13 tank
between the enthalpy of the gas in the entrance line and the internal energy of the gas
between the enthalpy of the gas in the entrance line and the internal energy of the gas
in
the
between
gas
the
Se llena
gas
unNeglect
tanque heat
vacío
desde una
línea the
a
presión
constante.
incon
the tank?
tank?
Neglect
heat transfer
transfer
between
the
gas and
and
the tank.
tank.¿Cuál es la relación entre la
entalpía del gas en la línea de entrada y la energía interna del gas en el tanque? Ignore la transferencia
de calor entre el gas y el tanque.
Solution
Solution 2.13
2.13
The
The tank
tank with
with its
its single
single entrance
entrance serves
serves as
as the
the control
control volume.
volume. Because
Because there
there is
is
..
Soluciónno
2.13
no expansion
expansion work,
work, stirring
stirring work,
work, or
or shaft
shaft work,
work, W
W=
= 0.
0. If
If kinetickinetic- and
and potentialpotential-
energy
changes
are
Eq.
becomes:
energy
changes
are negligible,
negligible,
Eq. (2.29)
(2.29)
becomes:
El tanque
con una
sola entrada
sirve como
volumen
de control. Debido a que no hay trabajo de
·
expansión, de agitación o en la flecha,
W
=
0.
Si
los
cambios
en las energías cinética y potencial
d(mU
)
.. ��
d(mU )tank
tank − H ��m
=
00
−
H
=
m
son despreciables, la ecuación (2.29) será:
dt
dt
d (mU ) tanque
where
stream
−H
′ m ′ = and
0 the
where the
the prime
prime (�)
(�) identifies
identifies the
the entrance
entrance
stream
and
the minus
minus sign
sign is
is required
required
dt mass balance is:
because
it
is
an
entrance
stream.
The
because it is an entrance stream. The mass balance is:
donde la prima (′) identifica la corriente de entrada
y se requiere el signo menos porque se tiene
dm
.. �
dmtank
m
una corriente de entrada. El balance de masa
es: tank
=
m� =
dt
dt
dmtanque

m ′ = yields:
Combining
Combining these
these two
two balance
balance equations
equations
yields:
dt
d(mU
)
tank
tank
d(mU
)tank − Hde
dm
� dm
Al combinar los resultados de las dos
ecuaciones
balance:
tank
− H � dt =
= 00
dt
dt
dt
d (mU ) tanque
dmtanque
− H′
=0
dt
dt
02-SmithVanNess.indd 49
8/1/07 12:49:54
50
CHAPTER
2. The First
Law
and ley
Other
Basic Conceptsbásicos
50
CAPÍTULO
LaLaw
primera
y otros
CHAPTER
2. The2.
First
and Other
Basicconceptos
Concepts
50
5050
CHAPTER
CHAPTER2.2.The
TheFirst
FirstLaw
Lawand
andOther
OtherBasic
BasicConcepts
Concepts
CHAPTER 2. The First Law and Other Basic Concepts
50
Multiplicando
por dtby
e integrando
a través over
del tiempo
(observe
es constante)
se obtiene:
Multiplying
dt and integrating
time (noting
thatque
H � H′
is constant)
gives:
Multiplying by dt and integrating over time (noting that H� �� is constant) gives:
Multiplying
Multiplyingbybydtdtand
andintegrating
integratingover
overtime
time(noting
(notingthat
thatHH isisconstant)
constant)gives:
gives:
Δ(mU)
–time
H′Δm
=00 H � is constant) gives:
Multiplying by dt and integrating
over−
(noting
that
H � �m
�(mUtanque
)tank
tanque
tank =
− H� ���m tank
=0
�(mU )tank
�mtank
�(mU
�(mU)tank
)tank−−HH��m
tank==00
� (m −
HH
�m
Whence
m�(mU
tank
tank
2 U2 −)m
1 U−
1 =
2 =m01 )
�
Whence
m 22UU22−– m 11U11 ==HH′� (m
De donde
–−mm1)1)))
� (m22−
Whence
Whence
mmm
2U
2U
2 2−−mm
1U
1U
1 1==HH(m
2 2 −mm
11
� (m conditions
where subscripts 1 and 2mdenote
initial
and
final
in the tank.
U
−
m
U
=
H
Whence
2 2 initial
1 1and final conditions
2 − m1)
where
subscripts
1
and
2
denote
in the tank.
donde los
subíndices
1 ymass
denotan
condiciones
inicial
yconditions
final
Debido a que la masa
where
where
subscripts
subscripts
121and
andin
22denote
denote
initial
and
and
final
conditions
intanque.
inthe
thetank.
tank.
Because
the
the
tankinitial
initially
isfinal
zero,
m 1 =en0;elthen,
Because
the mass
in2the
tank
initially
isfinal
zero,conditions
m 1 = 0; then,
where
subscripts
1
and
denote
initial
and
in
the
tank.
en el tanque
es
inicialmente
cero,
m
=
0;
entonces,
Because
Becausethe
themass
massininthe
thetank
initiallyisiszero,
zero,mm1 1==0;0;then,
then,
1tankinitially
�
Because the mass in the tank initially
m 1 = 0; then,
U2 =isHzero,
�
=
H
U
H� � �
UUU2222===HH′
=H
U2heat
a result showing that in the absence of
transfer the energy of the gas contained
a result
showing
that
inenthe
absencede
of heat transfer de
thecalor,
energy
of
the gas
contained
un resultado
que
muestra
que,
ausencia
laof
energía
en
el added.
gas contenido
awithin
aresult
result
showing
showing
that
that
the
the
absence
absence
ofoftransferencia
heat
heat
transfer
theenergy
energy
ofthe
thegas
gas
contained
contained
the
tank at
theinin
end
of
the process
istransfer
equal
tothe
enthalpy
of
the
gas
within
the
tank
at
the
end
of
the
process
is
equal
to
the
enthalpy
of
the
gas
added.
awithin
result
showing
that
the
absence
ofentalpía
heat
transfer
the
energy
ofof
the
gasgas
contained
en el tanque
althe
final
delatat
proceso
es
igual
a la
deltoto
gas
agregado.
within
the
tank
tank
the
thein
end
end
ofof
the
theprocess
process
isisequal
equal
the
the
enthalpy
enthalpy
ofthe
the
gas
added.
added.
within the tank at the end of the process is equal to the enthalpy of the gas added.
Example 2.14
2.14
Example
Example
Example
2.14
2.14 heated tank for hot water contains 190 kg of liquid water at
An insulated,
electrically
Ejemplo
2.14 electrically
Example
2.14 heated
An◦insulated,
tank for hot water contains 190 kg of liquid water at
An
Aninsulated,
insulated,
electrically
heated
heated
tank
tankfor
hot
hotwater
water
contains
containsfrom
190
190kg
kgoftank
ofliquid
liquid
atat
60
C
when aelectrically
power
outage
occurs.
Iffor
water
is withdrawn
the
at water
awater
steady
60◦ C when
a power outage occurs. If water is withdrawn from the tank at a steady
.
.
◦C
◦insulated,
−1
An
electrically
heated
tankitIffor
hot
water
contains
190
kg tank
of
liquid
water
Un tanque
aislado
calienta
agua
eléctricamente
190 kg
deofagua
líquida
60
°Catcuando
60
60
Cof
when
when
a0.2
apower
power
outage
outage
occurs.
occurs.
If
water
water
isthe
iswithdrawn
withdrawn
from
from
the
tank
atat
steady
steady
, how
long
will
take
forcontiene
temperature
the
water
inaaa
the
tank
rate
m. = que
kg
s−1
◦ Cofwhen
,
how
long
will
it
take
for
the
temperature
of
the
water
in
the
tank
rate
m
=
0.2
kg
s
.
.
−1
−1
◦
◦
60
a
power
outage
occurs.
If
water
is
withdrawn
from
the
tank
at
a
steady
ocurre rate
una
interrupción
del
suministro
eléctrico.
Siwater
el
agua
se
retira
del tanque
aC,
una
, ,how
howAssume
long
long
will
willcold
itittake
take
for
for
the
thetemperature
temperature
of
ofthe
the
water
inin
the
thetank
tankestable
rate
of
ofmm
==0.2
0.2
kg
kg
to
drop
from
60
toss35
C?
enters
the tank
at
10
andproporción
negligible
◦water
.–1
−1◦◦,◦C?
to drop
from
60kg
tos35
Assume
cold
water
enters
the
tank at
10
andinen
negligible
◦ C,
◦C,
how
longliquid
willcold
itwater
take
for
the
temperature
of
the
water
the
tank
rate
oflosses
=
0.2
de m· = to
0.2
kg
sm
, ¿cuánto
tiempo
debe
transcurrir
para
que
la
temperatura
del
agua
el
tanque
todrop
drop
from
from
60
60toto
35
35tank.
C?
C?Assume
Assume
cold
water
water
enters
the
the
tank
tank
at
at
10
10
C,
and
and
negligible
negligible
heat
from
the
For
let enters
C
=
C
=
C,
independent
of
T
and
P. disV
◦ C? Assume
◦ C, and
heat
losses
from
the
tank.
For el
liquid
water
let
Centers
C PPtanque
=tank
C, independent
of Tlasand
P.
V = el
to
60 to
35tank.
cold
water
10°C,
minuyaheat
dedrop
60
afrom
35from
°C?
Suponga
que
agua
fría
entra
aat10
y que
pérdidas
de
heat
losses
losses
from
the
the
tank.For
Forliquid
liquid
water
water
let
letCC
==CCPthe
C,independent
independent
ofofnegligible
TTand
and
P.P.
V Ven
P==C,
heat losses
from
the
tank. For liquid
water
let Clíquida
= supone
C, independent
and
P. indeV = C Pse
calor desde
el tanque
son
despreciables.
Para
el agua
que C V =ofCTP =
C son
Solution
2.14
pendientesSolution
de T y P . 2.14
. 2.14
.
Solution
Solution
2.14
. = 0. Additionally, assume perfect mixing of the contents of the
Here, .Q.. =
W
Solution
2.14
.. =
Here, Q
=
W
0. Additionally, assume perfect mixing of the contents of the
Solución 2.14
Here,
Here,Q
Q
=
WW
0.
0.Additionally,
Additionally,
assume
perfect
mixing
mixing
ofoftank
the
thecontents
contents
ofthe
the
tank;
this
implies
that
the propertiesassume
of the perfect
water
leaving
the
are
thoseofof
the
. =implies
. ==that
tank; this
the propertiesassume
of the water
leaving
theoftank
are
those of
of the
the
·W
Here,
Q· the
=implies
=that
0. the
Additionally,
perfect
mixing
the
contents
tank;
tank;
this
this
implies
that
the
properties
properties
of
of
the
the
water
water
leaving
leaving
the
the
tank
tank
are
are
those
those
of
of
the
the
water
in
tank.
With
the
mass
flowrate
into
the
tank
equal
to
the
mass
flowrate
En este water
caso, Q
Wtank.
= 0. Además,
se supone
unainto
mezcla
perfecta
detolos
contenidos
del tanque;
in =the
Withthe
theproperties
mass
flowrate
theleaving
tank
equal
theare
mass
flowrate
tank;
this
implies
that
of theinto
water
theand
tank
those
of and
the
water
water
in
inthe
the
tank.
tank.With
With
the
the
mass
flowrate
flowrate
the
thetank
tank
equal
to
tothe
themass
mass
flowrate
flowrate
out,
m
is
constant;
moreover,
the
differences
between
inlet
outlet
kinetic
cv
esto implica
que
las
propiedades
delmass
agua
que
deja into
el tanque
sonequal
las mismas
que
las
deland
agua en el
out,
m
is
constant;
moreover,
the
differences
between
inlet
and
outlet
kinetic
cv the tank. With the mass flowrate into the tank equal to the mass flowrate
water
in
out,mmcvla
is
constant;
constant;
moreover,
moreover,
the
thedifferences
differences
between
between
inlet
inletand
and
outlet
outlet
kinetic
and
andde flujo
potential
energies
can
be neglected.
Equation
(2.29)
istanque
therefore
written:
cvis
tanque.out,
Como
relación
de
flujo
de masa
hacia
el interior
delis
es igual
akinetic
la
relación
potential
energies
can
be
neglected.
Equation
(2.29)
therefore
written:
out,
m is
constant;
moreover,
the differences
between
inlet
and written:
outlet
kinetic and
potential
energies
energies
bebeneglected.
neglected.
Equation
Equation
isistherefore
therefore
written:
de masapotential
hacia cv
afuera,
mcv can
escan
constante;
por otra
parte, (2.29)
se(2.29)
pueden
despreciar
las diferencias entre las
dU Equation
potential energies can be neglected.
(2.29)
is
therefore
written:
.
energías cinética y potencial de entrada
y de+salida.
PorHlo) tanto,
.
= 0 la ecuación (2.29) se escribe:
m(H
−
m dU
dU
mdU
. . − H11) = 0
dt + m(H
m(H
−−HH
mmdU
dt ++m(H
1 )1 )==00
.
+ m(H
−H
m dtdtrefer
1 ) = 0 of the tank and H is the
where unsubscripted quantities
to the
contents
dt refer to
where unsubscripted quantities
the contents of the tank and H11 is the
the
where
whereunsubscripted
unsubscripted
quantities
quantities
refer
refertoto
the
the
contents
contents
ofofthe
theCtank
tank
andHH
specific
enthalpy
ofcarecen
the
water
entering
the
tank.
With C
=
C,
1 1isisthe
P =and
donde las
cantidades
queof
de entering
subíndices
se
refieren
y H1 es la
specific
enthalpy
the
water
thethe
tank.
With al
CofVVcontenido
= Ctank
= del
C, tanque
where
quantities
refer the
to
contents
the
and
specific
specificunsubscripted
enthalpy
enthalpyofofthe
the
water
waterentering
entering
thetank.
tank.
With
WithCC
=
=CCPPP==C,
C, H1 is the
V
V
entalpíaspecific
específica
deldU
agua quedT
entra entering
en el tanque.
DadoWith
que C
CV ==CCP ==
C,C,
enthalpy
water
the tank.
V
P
dT
dU of=the
=
C(T
−
T
)
C
and
H
−
H
1
dT
dU
dU
CdT
and
H − H1 = C(T − T11)
dt ==
dt
y
=
=
C(T
C(T
−
−
T
T
)
=
C
C
and
and
H
H
−
−
H
H
dt
dt
11
1 1)
dT
dU
dtdt = C dtdt
=
C(T
−
T
and
H
−
H
1
1)
The energy balance
thendtbecomes, on rearrangement,
dt
Theenergético
energy balance
then
becomes,
on rearrangement,
El balance
será
en
ese
caso,
después
de
una
reordenación,
The
Theenergy
energybalance
balancethen
thenbecomes,
becomes,ononrearrangement,
rearrangement,
m dT
The energy balance then becomes, on rearrangement,
dt = −mm
. dT
dTT
m
dt = − m
. TdT
−
dtdt==−− m
− T11
. . T dT
T−−TT
11
dt = − mm
. Tarbitrary
Integration
from
t = 0 (where
T =
Ttiempo
to
time
yields:
0) m
T
− T1 time
Integrando
desde t =from
0 (donde
= T0) en
unT
arbitrario
t sett yields:
obtiene:
Integration
t = 0 T(where
T =
)
to
arbitrary
0
Integration
Integrationfrom
fromt t==00(where
(whereTT==TT
arbitrarytime
yields:
0 )0 )toto
�timet tyields:
�arbitrary
�
�
m0 ) to
T − T1��
Integration from t = 0 (where T = T
arbitrary
time t yields:
�
�
m
T
−
T
t =− m
1
. ln
−TT
T
t = −mm
1 1�
. ln �TTT−−
t t==−−m
− TT111
. .lnln T00 −
0 0−−TT
11
t = − mm
. ln TT
T0 − T1
m
02-SmithVanNess.indd 50
8/1/07 12:50:02
51
2.12.Mass
Massand
andEnergy
EnergyBalances
Balancesfor
forOpen
OpenSystems
Systems
51
2.12.
Mass
and
Energy
Balances
for
Open
Systems
51
2.12.
2.12.
Mass
and
Energy
Balances
for
Open
Systems
51
2.12. Balances
de masa
y energía
para sistemas
51
2.12. Mass
and Energy
Balances
for Openabiertos
Systems
Substitution
of
numerical
values
into
this
equation
gives,
for
the
conditions
of
this
Substitution
of
numerical
values
into
this
equation
gives,
for
the
conditions
of
this
Substitution of numerical values into this equation gives, for the conditions of this
problem,
problem,
Substitution
of numerical
values
intoecuación
gives, para
for the
of de
thiseste proSustituyendo
los valores
numéricos
en esta
se��
obtiene,
lasconditions
condiciones
� equation
problem,
�this
Substitution of numerical values
into�
this
equation
190
35−
−10
10� gives, for the conditions of this
190
35
−
10
blema, problem,
190
35
658.5s s
==−
−−190 ln
lnln� 35 − 10 � =
==658.5
problem,
tt t=
0.2 �60
60−
10� 658.5 s
0.2
60
−−10
0.2
t = − 190
ln 35
− 10
10 = 658.5 s
60
− 10temperature
t minutes
=
− 0.2for
lnthe
= 658.5 in
s the
Thus,ititittakes
takesabout
about11
11
minutes
for
the
water
temperature
thetank
tankto
dropfrom
from
Thus,
takes
about
11
water
tank
totodrop
drop
from
Thus,
minutes
for
the
water
temperature
ininthe
0.2
60
− 10
◦
◦
C.
60
to
35
◦
C.
60
to
35
Thus,
it
takes
about
11
minutes
for
the
water
temperature
in
the
tank
to
drop
from
C. about 11 minutes for the water temperature in the tank to drop from
60
to 35
Thus,
it ◦takes
Así, le toma
C.de 11 minutos a la temperatura del agua en el tanque disminuir de 60 a 35 °C.
60
tocerca
35
◦
60 to 35 C.
51
Energy
Balancespara
forSteady-State
Steady-State
Flow
Processes
Energy
Balances
for
Steady-State
Flow
Processes
Balances
energéticos
los procesosFlow
de flujo
en estado estacionario
Energy
Balances
for
Processes
Energy Balances for Steady-State Flow Processes
Flowprocesses
processes
forwhich
which
theaccumulation
accumulationterm
termof
Eq.(2.28),
(2.28),d(mU
d(mU))cv
) /dt,
/dt,is
zeroare
aresaid
saidtoto
Flow
processes
for
which
the
accumulation
term
ofofEq.
Eq.
(2.28),
d(mU
/dt,
isiszero
zero
are
said
Energy
Balances
for
Steady-State
Flow
Processes
Flow
for
the
cvcv
Los procesos
de flujo
para
los cuales
el
término de acumulación
de la ecuación
(2.28),
d(mU)
/dt estocero, se
cv
occurprocesses
steadystate.
state.
Asdiscussed
discussed
withrespect
respect
themass
massbalance,
balance,
this
means
that
the
mass
occur
atatsteady
steady
state.
As
discussed
with
respect
to
the
mass
balance,
this
means
that
the
mass
Flow
for
which
the
accumulation
term to
oftothe
Eq.
(2.28),
d(mU )this
/dt,
is zero
are
said
to
cv
at
As
means
that
the
mass
dice queoccur
ocurren
en estado
estacionario.
Alwith
igual que
con
el
análisis
del
balance
de ismasa,
esto
significa
Flow
processes
for
which
the
accumulation
term
of
Eq.
(2.28),
d(mU
)
/dt,
zero
are
said
to que
cv
of
the
system
within
the
control
volume
is
constant;
it
also
means
that
no
changes
occur
with
of
the
system
within
the
control
volume
is
constant;
it
also
means
that
no
changes
occur
with
occur
at
steady
state.
As
discussed
with
respect
to
the
mass
balance,
this
means
that
the
mass
the
system
within
the
control
volume
is
constant;
it
also
means
that
no
changes
occur
with
la masaof
del
sistema
dentro
del
volumen
de
control
es
constante;
también
significa
que
no
ocurre
ningún
camoccur
steady
state.the
As
discussed
with respect
to the
mass
balance,
means
that
the
mass
time
in
theproperties
properties
thefluid
fluid
within
the
control
norat
entrances
and
exits.
No
time
inat
the
properties
ofofthe
the
fluid
within
the
control
volume
nor
atatthat
itsitsthis
entrances
and
exits.
No
of
thein
system
within
control
volume
is
constant;
itvolume
also means
no
changes
occur
with
the
of
within
the
control
volume
nor
its
entrances
and
exits.
No
bio contime
el
tiempo
en
las
propiedades
del
fluido
dentro
del
volumen
de
control
ni
en
sus
entradas
ni
salidas.
No
of
theinsystem
within
the
control
is
it also
means
noThe
changes
occur
with
expansion
ofthe
thecontrol
control
volume
ispossible
possible
underthese
these
circumstances.
The
only
work
the
expansion
of
the
control
volume
iswithin
possible
under
these
circumstances.
only
work
ofof
the
time
theof
properties
of the
fluidvolume
theconstant;
control
volume
nor at that
its entrances
and
exits.
No
expansion
volume
is
under
circumstances.
The
only
work
of
the
hay expansión
del
volumen
de
control
bajo
estas
circunstancias.
El
único
trabajo
del
proceso
es
el
de
la
flecha
time
in the
properties
of
the
fluid
the under
control
volume
nor atbecomes:
its entrances
exits.
No
process
shaft
work,
and
the
general
energy
balance,
Eq.
(2.28),
becomes:
process
isisshaft
shaft
work,
and
the
general
energy
balance,
Eq.
(2.28),
expansion
of
thework,
control
volume
iswithin
possible
these
circumstances.
The onlyand
work
of the
process
is
and
the
energy
balance,
Eq.
(2.28),
becomes:
y el balance
energético
general,
de
la general
ecuación
(2.28),
será:these
expansion
of
the
control
volume
is
possible
under
circumstances.
The
only
work
of
the
process is shaft work, and the general energy balance, Eq. (2.28), becomes:
��
��
�� ��� Eq. (2.28),
process is shaft work, and the ��
general
energy balance,
becomes:
.
.
.
.
�
.W
..m�. ==Q.Q++W
(2.30)(2.30)
����H
HH+
++112u12u2u2 2+
++zg
zgzg� m
m
(2.30)
�
(2.30)
. �fsfsfs= Q. + W. ss s
21 2
��
�
� H + 21 u 2 + zg m. fs = Q. + W. s
(2.30)
� H + 2 u + zg m fs = Q + Ws
(2.30)
Although
“steady
state”
does
not
necessarily
imply
“steady
flow,”
the
usual
application
Although
“steady
state”
does
not
necessarily
imply
“steady
flow,”
the
usual
application
ofof ecuastate” does
not necessarily
imply“flujo
“steady
flow,” la
theaplicación
usual application
of
AunqueAlthough
el “estado“steady
estacionario”
no implica
necesariamente
estable”,
usual de esta
this
equation
is
to
steady-state,
steady-flow
processes,
because
such
processes
represent
the
this
equation
is
to
steady-state,
steady-flow
processes,
because
such
processes
represent
the
Although
“steady
state”
does
not
necessarily
imply
“steady
flow,”
the
usual
application
of
this
equation
is
to
steady-state,
steady-flow
processes,
because
such
processes
represent
the
ción es Although
para procesos
en8 8estado
estacionario
y de flujoimply
estable,
ya queflow,”
tales procesos
elof
modelo
“steady
state”
does not
necessarily
“steady
the
usualrepresentan
application
industrial
norm.
8
industrial
norm.
this
equation
is
to
steady-state,
steady-flow
processes,
because
such
processes
represent
the
8
industrial
norm.
industrial.
this equation
is 8specialization
tospecialization
steady-state,results
steady-flow
processes,
because
such
processes
represent
the
further
results
when
thecontrol
control
volume
hasbut
but
oneentrance
entrance
and
one y una
AAfurther
further
when
the
control
volume
has
but
one
entrance
and
one
industrial
norm.
A
when
the
volume
one
and
one
Un
resultado
másspecialization
cuando
el volumen
dehas
control
tiene
sólo una
entrada
8especializadoresults
. obtiene
.se
industrial
norm.
.
exit.
The
same
mass
flowrate
m
then
applies
to
both
streams,
and
Eq.
(2.30)
reduces
to:
exit.
The
same
mass
flowrate
m
then
applies
to
both
streams,
and
Eq.
(2.30)
reduces
to:
A
further
specialization
results
when
the
control
volume
has
but
one
entrance
and
one
exit.
The
same
mass
flowrate
m
then
applies
to
both
streams,
and
Eq.
(2.30)
reduces
to:
salida. Por lo Atanto
se aplica
la misma. relación
de the
flujo
de masa
a ambas
corrientes,
y la ecuación
specialization
when
control
volume
has
one reduces
entrance
and one (2.30)
exit.
Thefurther
same mass
flowrate m.results
then
and
Eq.but
(2.30)
to:
�� streams,
�� applies to both
se reduce
a:
.
.
.
.
�
�
.
.
exit. The same mass flowrate m
then
applies
tozgzg
both
and
.W Eq. (2.30) reduces to: (2.31)
.mstreams,
(2.31)
+112u12u2u2 2+
++
m
==Q.Q++W
���
HH+
+
zg � m
(2.31)
�
H
. = Q. + W. ss s
21 2
�
�
(2.31) (2.31)
� H + 21 u 2 + zg m. = Q. + W. s
+ zg
m =case
Q +and
Ws “�”
(2.31)
�omitted
H + 2inuinthis
wheresubscript
subscript“fs”
“fs”has
hasbeen
been
omitted
this
simple
case
and
“�”denotes
denotesthe
thechange
change
from
where
subscript
“fs”
has
been
simple
denotes
the
change
from
where
this
simple
case
and
“�”
from
. en esteincaso
. omitted
donde elwhere
subíndice
“fs”
ha
sido
omitido
simple
y
“Δ”
denota
el
cambio
de
la
entrada
a la salida.
.
entrance
to
exit.
Division
by
m
gives:
entrance
to
exit.
Division
by
m
gives:
subscript
“fs”
has
been
omitted
in
this
simple
case
and
“�”
denotes
the
change
from
entrance
to· exit. Division
by m. omitted
gives: in this simple case and “�” denotes the change from
where
subscript
“fs”
has
been
Dividiendo
entre
m
se
obtiene:
entrance to exit. Division by m. gives:
..
..W.
entrance to exit. Division by
���m gives:
��� Q
Q..Q W
W
ss
2
1
2
1
���H
HH+
++12u2u2u +
++zg
zgzg� =
==Q.. .+
++ ..s. =
==Q
QQ+
++W
WW
�
ss s
.m W
.m
s
21 2
m
m
�
�
m. s = Q + Ws
� H + 21 u 2 + zg = m
Q. + W
� H + 2 u + zg = m. + m. = Q + Ws
m
m
�u
22 2
�u
�u
or
�H
+
(2.32a)
+
g
�z
=
or
�H
+
(2.32a)
+
g
�z
=
Q
++W
WW
o
(2.32a)
2
or
�H + �u
(2.32a)
+ g �z = QQ+
ss s
2222 + g �z = Q + Ws
or
�H + �u
(2.32a)
or
�H + 2 + g �z = Q + Ws
(2.32a)
2
Esta ecuación
es la expresión
matemática
deexpression
la
primera
ley
para
un law
proceso
en
estado
estacionario
y flujo estaThis
equation
is
the
mathematical
expression
of
the
first
law
for
a
steady-state,
steady-flow
This
equation
is
the
mathematical
of
the
first
for
a
steady-state,
steady-flow
This equation is the mathematical expression of the first law for a steady-state, steady-flow
ble entreThis
unaequation
entrada
yisuna
salida.
Todos
los
términos
representan
energía
por
unidad
demass
masa
del
fluido.
process
between
one
entrance
and
one
exit.
All
terms
represent
energy
per
unit
mass
of
fluid.
process
between
one
entrance
and
one
exit.
All
terms
represent
energy
per
unit
of
fluid.
the
mathematical
expression
of
the
first
law
for
a
steady-state,
steady-flow
process
between
one
entrance
and one
exit. All of
terms
represent
energy
per unit mass
of fluid.
This
equation
isthe
the
mathematical
expression
the
firstthe
law
for
steady-state,
Enprocess
todas
las
ecuaciones
de balance
energético
escritas
hasta
ahora
sea supone
que
lasteady-flow
unidad
de
In
all
of
energy-balance
equations
so
far
written,
the
energy
unit
is
presumed
to
be
theenergía
between
one
entrance
and
one
exit.
All
terms
represent
energy
per
unit
mass
of
fluid.
In
all
of
the
energy-balance
equations
so
far
written,
energy
unit
is
presumed
to
be
the
In between
all of theone
energy-balance
equations
so
farterms
written,
the energy
unitper
is unit
presumed
to be
the
process
and
one
exit.
All
represent
energy
mass
of
fluid.
es el joule,
deIn
acuerdo
con
elentrance
sistema
deequations
unidades
SI.
Para
el
sistema
inglés
deisunidades
de
ingeniería,
joule,
in
accord
with
the
SI
system
of
units.
For
the
English
engineering
system
of
units,
the los
joule,
in
accord
with
the
SI
system
of
units.
For
the
English
engineering
system
of
units,
all
of
the
energy-balance
so
far
written,
the
energy
unit
presumed
to
be
the
joule, In
in accord
the SI system
of units.soFor
English
of units,
of thewith
energy-balance
equations
far the
written,
the engineering
energy
unit issystem
presumed
to be the
the
joule, in all
accord
with
the SI system
of units. For
the
English
engineering
system
of units,
the
joule,
in accord with the SI system of units. For the English engineering system of units, the
88
Anexample
exampleof
steady-stateprocess
processthat
thatis
notsteady
steadyflow
flowis
waterheater
heaterin
whichvariations
variationsin
flowrate
rateare
are
An
example
ofofaa asteady-state
steady-state
process
that
isisnot
not
steady
flow
isisaa awater
water
heater
ininwhich
which
variations
ininflow
flow
rate
are
8 An
8 An compensated
exactly
compensated
bychanges
changes
inthe
therate
rate
ofheat
transfer
sothat
that
throughout
remain
constant.
exactly
by
changes
inprocess
the
rate
ofque
heat
transfer
so
that
temperatures
throughout
remain
constant.
Un ejemplo
de
un proceso
de
estado
estacionario
notransfer
tiene
flujo
estable
es unheater
calentador
de agua
en el
cual
las variaciones
en el
example
of a by
steady-state
that
isheat
not
steady
flow
istemperatures
atemperatures
water
in which
variations
in flow
rate are
exactly
compensated
in
of
so
throughout
remain
constant.
8 An example of a steady-state process that is not steady flow is a water heater in which variations in flow rate are
caudal sonexactly
compensadas
exactamente
con los
cambios
la rapidez
de calor,
de modo remain
que las constant.
temperaturas permanezcan
compensated
by changes
in the
rate ofenheat
transferdesotransferencia
that temperatures
throughout
compensated
by changes in the rate of heat transfer so that temperatures throughout remain constant.
constantesexactly
en todas
partes.
8
02-SmithVanNess.indd 51
8/1/07 12:50:10
52
CHAPTER
CHAPTER 2.
2. The
The First
First Law
Law and
and Other
Other Basic
Basic Concepts
Concepts
52
52
CAPÍTULO 2. La primera ley y otros conceptos básicos
kinetickinetic- and
and potential-energy
potential-energy terms,
terms, wherever
wherever they
they appear,
appear, require
require division
division by
by the
the dimensional
dimensional
términosconstant
de las energías
cinética
y1.8).
potencial,
dondequiera
que aparezcan,
requieren
la división entre la consg
(Secs.
1.4
and
In
this
event
Eq.
(2.32a),
for
example,
is
written:
c
constant gc (Secs. 1.4 and 1.8). In this event Eq. (2.32a), for example, is written:
tante dimensional gc (secciones 1.4 y 1.8). En este caso, por ejemplo, la ecuación (2.32a) se escribe:
2
�u
�u 2 + gg �z = Q + Ws
�H
+
�H + 2gc + gc �z = Q + Ws
2gc
gc
(2.32b)
(2.32b)(2.32b)
the
usual
for
�H
and
the
kinetic
energy,
and
are
Aquí, laHere,
unidad
para
la energía
cinética,
energía energy,
potencial
y work
el trabajo
Here,
thehabitual
usual unit
unit
forΔH
�Hy Q
andesQ
Qelis
is(Btu);
the (Btu);
(Btu);
kinetic
energy,lapotential
potential
energy,
and
work
are se ex−1 –1
usually
expressed
as
(ft
lb
).
Therefore
the
factor
778.16(ft
lb
)(Btu)
must
be
used
with
the
−1
f
f
presan generalmente
como
(pie
lb
).
Por
lo
tanto,
el
factor
778.16(pie
lb
)(Btu)
se
debe
utilizar
con
f
usually expressed as (ft lbff). Therefore the factor 778.16(ft lbf )(Btu)
must be used with the los térappropriate
terms
to
put
them
all
in
consistent
units
of
either
(ft
lb
)
or
(Btu).
f
minos apropiados
para
poner
a
todos
en
unidades
convenientes
en
(pie
lb
)
o
en
(Btu).
appropriate terms to put them all in consistent units of either (ft lbff ) or (Btu).
In
many
applications,
kineticand
potential-energy
terms
are
omitted,
they
are
En muchas
aplicaciones
se
omiten
los
términos
de
las
energías
cinética
y
potencial
porque
In many applications, kinetic- and
potential-energy terms are omitted, because
because
theyson
aredespre9 For such cases, Eqs. (2.32a) and (2.32b) reduce to:
9 En
compared
with
other
terms.
9
ciables negligible
en
comparación
con
otros
términos.
estos
casos
las
ecuaciones
(2.32a)
y
(2.32b)
se
reducen
a:
negligible compared with other terms. For such cases, Eqs. (2.32a) and (2.32b) reduce to:
(2.33)
(2.33) (2.33)
�H
�H =
=Q
Q+
+W
Wss
Esta expresión
de la primera
ley para
un proceso
en estado estacionarioprocess
y flujo estable
es análoga
a la ecuación
This
This expression
expression of
of the
the first
first law
law for
for aa steady-state,
steady-state, steady-flow
steady-flow process is
is analogous
analogous to
to Eq.
Eq. (2.3)
(2.3)
(2.3) para
un
proceso sin
flujo. Sin
embargo,
más que
la energía
interna,energy
la entalpía
esthermodynamic
una propiedad termofor
a
nonflow
process.
However,
enthalpy
rather
than
internal
is
the
for a nonflow process. However, enthalpy rather than internal energy is the thermodynamic
dinámica
de gran importancia.
property
property of
of importance.
importance.
A
Calorimeter
Enthalpy
Measurements
Calorímetro
flujo parafor
mediciones
la entalpía
A Flow
Flow de
Calorimeter
for
Enthalpy de
Measurements
The
of
(2.32)
(2.33)
to the
solution
of
problems
requires
enThe application
application
of Eqs.
Eqs. (2.32)
(2.32)yand
and
(2.33)
the
solution
of practical
practicalprácticos
problems
requires
enLa aplicación
de
las ecuaciones
(2.33)
paratolaits
solución
de problemas
requiere
de
valores de
thalpy
values.
Because
H
is
a
state
function,
values
depend
only
on
point
conditions;
once
thalpy
values.
Because
H
is
a
state
function,
its
values
depend
only
on
point
conditions;
once
entalpía.determined,
Porque H es
una
función
de
estado,
sus
valores
dependen
sólo
de
condiciones
puntuales;
una vez
they
may
be
for
subsequent
use
for
the
same
sets
of
conditions.
To
determined,
they
may
be tabulated
tabulated
for
subsequent
use
forvez
the que
same
sets
of el
conditions.
To this
thisde condeterminados
pueden
ser
tabulados
para
el
uso
consecutivo
cada
se
tenga
mismo
conjunto
end,
Eq.
(2.33)
may
be
to
laboratory
processes
designed for
enthalpy measurements.
end,
Eq.
(2.33)
may
be applied
applied
to se
laboratory
processes
measurements.
diciones.
Con
este
fin, la
ecuación
(2.33)
puede aplicar
a losdesigned
procesosfor
de enthalpy
laboratorio
diseñados específicamente para medir la entalpía.
Section 1
Section 1
Sección 1Heater
Heater
T2
T2
T2
P2
P2
P2
Calentador
Constant
Constant
temperature
temperature
Baño a
bath
bath
temperatura
constante
Supply
Supply
Suministro
Valve
Valve
Válvula
Discharge
Discharge
Descarga
Applied
Applied
emf
emf
fem
aplicada
Section 2
Section 2
Sección 2
Figure
Figure 2.7:
2.7: Flow
Flow calorimeter.
calorimeter.
Figura 2.7: Calorímetro de flujo.
A
A simple
simple flow
flow calorimeter
calorimeter is
is illustrated
illustrated schematically
schematically in
in Fig.
Fig. 2.7.
2.7. Its
Its essential
essential feature
feature
is
an
electric
resistance
heater
immersed
in
a
flowing
fluid.
The
design
provides
for
minimal
is
an
electric
resistance
heater
immersed
in
a
flowing
fluid.
The
design
provides
for
minimal
Un calorímetro de flujo simple se ilustra en forma esquemática en la figura 2.7. Su característica esencial es un calentador de resistencia eléctrica sumergido en un líquido que circula. El diseño proporciona para
9 Exceptions are applications to nozzles, metering devices, wind tunnels, and hydroelectric power stations.
9 Exceptions are applications to nozzles, metering devices, wind tunnels, and hydroelectric power stations.
9 Las excepciones son las aplicaciones a las toberas, a los aparatos de medición, a los túneles de viento y a las estaciones hidroeléctricas.
02-SmithVanNess.indd 52
8/1/07 12:50:13
53
2.12. Balances de masa y energía para sistemas abiertos
velocidad y cambios de elevación mínimos de la sección 1 a la sección 2, haciendo despreciables los cambios en
las energías cinética y potencial del fluido. Además, como no se hace trabajo en la flecha del sistema, la ecuación
(2.33) se reduce a ΔH = H2 – H1 = Q. La rapidez de transferencia de calor al fluido se determina por la resistencia del calentador y por la corriente que pasa a través de éste. En la práctica se necesita poner atención a un gran
número de detalles, pero en principio la operación del calorímetro de flujo es simple. Las medidas del rendimiento térmico y la relación de flujo permiten el cálculo del cambio ΔH entre las secciones 1 y 2.
Como ejemplo, se determinan fácilmente las entalpías del H2O como líquido y como vapor. El baño a
temperatura constante se llena de una mezcla de hielo molido y agua para mantener una temperatura de 0 °C.
Se suministra agua líquida al aparato y el serpentín que lleva el agua a través del baño a temperatura constante es lo suficientemente extenso, de modo que el líquido emerge a la temperatura del baño que es esencialmente de 0 °C. La temperatura y la presión en la sección 2 se miden con instrumentos adecuados. Los valores
de la entalpía del H2O para diversas condiciones en la sección 2 están dados por:
H2 = H1 + Q
donde Q es el calor agregado por unidad de masa del flujo de agua.
La presión varía en cada una de las experimentaciones, pero en el intervalo aquí encontrado tiene un
efecto despreciable sobre la entalpía del agua que entra, y para propósitos prácticos H1 es constante. Los valores absolutos de la entalpía, al igual que los valores absolutos de la energía interna son desconocidos. Por lo
tanto, se le puede asignar a H1 un valor arbitrario como base para los demás valores de la entalpía. Estableciendo H1 = 0 para el agua líquida a 0 °C obtenemos:
H2 = H1 + Q = 0 + Q = Q
Los valores de la entalpía se pueden tabular para las temperaturas y presiones que existen en la sección
2, para una gran cantidad de mediciones. Además, las mediciones del volumen específico realizadas para
estas mismas condiciones se pueden agregar a la tabla, junto con los valores correspondientes de la energía
interna calculados por medio de la ecuación (2.11), U = H – PV. De esta manera, las tablas de propiedades
termodinámicas se compilan sobre toda la gama útil de condiciones. El uso más extenso posible para esta
tabulación es para el H2O y se conoce como las tablas de vapor.10
La entalpía se puede tomar igual a cero para algunos otros estados del líquido a 0 °C. La elección es
arbitraria. Las ecuaciones termodinámicas, tales como las ecuaciones (2.32) y (2.33) se aplican a los cambios
de estado, para los cuales las diferencias de entalpía son independientes de la localización del punto cero. Sin
embargo, una vez que se elige un punto cero arbitrario para la entalpía, no se puede hacer una elección arbitraria para la energía interna, debido a que los valores de la energía interna pueden calcularse a partir de los
de la entalpía mediante la ecuación (2.11).
Ejemplo 2.15
Para el calorímetro de flujo recién analizado se toman los siguientes datos con el agua como líquido de
prueba:
Relación de flujo = 4.15 g s –1
t1 = 0 °C
t 2 = 300 °C
P2 = 3 bar
10 Las tablas de vapor se dan en el apéndice F. Las tablas para otras sustancias se encuentran en la literatura. Un análisis de compilaciones de las propiedades termodinámicas aparece en el capítulo 6.
02-SmithVanNess.indd 53
8/1/07 12:50:14
54
CHAPTER
2.
The First
Law
Other Basic
Concepts
CHAPTER
2.
First
Law and
and
Basic
Concepts
CHAPTER
2. The
The2.
First
and Other
Other
Basicconceptos
Conceptsbásicos
CAPÍTULO
La Law
primera
ley
y otros
54
54
54
Rate of heat addition from resistance heater = 12,740 W
of
from
=
Rapidez Rate
de adición
deaddition
calor desde
calentadorheater
de resistencia
12 740 W
Rate
of heat
heat
addition
fromelresistance
resistance
heater
= 12,740
12,740=W
W
The
The water
water is
is completely
completely vaporized
vaporized in
in the
the process.
process. Calculate
Calculate the
the enthalpy
enthalpy of
of steam
steam at
at
The
water is
completely
vaporized
in liquid
the
process.
the enthalpy
El agua300
se ◦evapora
el proceso.
Calcule
la entalpía
a 300 °Cofysteam
3 bar, at
conside3totalmente
bar
based en
on
0 for
water
atCalculate
0◦◦C. del vapor
C and
and
bar
on H
H=
=
300◦◦C
andel33agua
bar based
based
= 00 for
for liquid
liquid water
water at
at 00◦C.
C.
rando H300
= 0C
para
líquidaon
a 0H°C.
Solution
Solution
2.15
Solución
2.15 2.15
Solution
2.15
2 son
If
�z
anddespreciables,
�u 222 are
negligible
are zero,
H
and
s and
2y = Q,
Si Δz y Δu
y si W and
y Hif1 W
son
cero,H1entonces
Hthen
If
H
2 = Q,
If �z
�z and
and �u
�u are
are negligible
negligible Sand
and if
if W
Wss and
and H
H11 are
are zero,
zero, then
then
H22 =
= Q,
Q, and
and
−1
12,740
J
s
−1
12,740 J ss−1 = 3,070 J g−1
H
−1
2 = 12,740 J −1
H
=
gg ss−1
H22 =
= 4.15
= 3,070
3,070 JJ gg−1
4.15
−1
4.15 g s
Example
2.16
Example
2.16
Example
2.16
Ejemplo
2.16
Air at 1 bar and 25◦◦C enters a compressor at low velocity, discharges at 3 bar, and
Air
at
Air at
at 11 bar
bar and
and 25
25◦C
C enters
enters aa compressor
compressor at
at low
low velocity,
velocity, discharges
discharges
at 33 bar,
bar, and
and
−1
enters
a
to
a
velocity
of
600
ss−1
at
the
Se introduce
aire
a 1 barin
ywhich
25 °C it
un compresor
a baja
velocidad,
sem
descarga
a 3initial
bar ycondientra en una
enters
aa nozzle
nozzle
in
which
ititaexpands
expands
to
aa final
final
velocity
of
m
at
initial
condienters
nozzle
inand
which
expands
tothe
final
velocity
of a600
600
m
s−1240
at the
the
initial
condi–1
tions
of
pressure
temperature.
If
work
of
compression
is
kJ
per
kilogram
tobera en
la
cual
se
expande
a
una
velocidad
final
de
600
m
s
las
condiciones
iniciales
de presión
tions
of
pressure
and
IfIf the
work
tions
of
pressure
and temperature.
temperature.
theduring
work of
of compression
compression is
is 240
240 kJ
kJ per
per kilogram
kilogram
of
air,
how
much
heat
must
be
removed
compression?
y temperatura.
Si
el
trabajo
de
compresión
es
de
240
kJ
por
kg
de
aire,
¿cuánto
calor
debe
eliminarse
of
of air,
air, how
how much
much heat
heat must
must be
be removed
removed during
during compression?
compression?
durante la compresión?
Solution 2.16
Solution
Solución
2.16 2.16
Solution
2.16
Because
the air
returns to
its initial
conditions
of T
and
P,
the overall
process
Because
air
to
conditions
TT el
and
P,
overall
process
Dado que
el aire the
regresa
a sus condiciones
iniciales
de Tof
proceso
no produce
ningún
Because
the
air returns
returns
to its
its initial
initial
conditions
ofy P,
and
P, the
thetotal
overall
process
produces
no
change
in
enthalpy
of
the
air.
Moreover,
the
potential-energy
change
produces
no
change
in
enthalpy
of
the
air.
Moreover,
the
potential-energy
cambio en
la entalpía
del aire.
Por
otra parte,
elair.
cambio
en la energía
potencial del change
aire se supone
produces
no
change
in
enthalpy
of
the
Moreover,
the
potential-energy
change
of
the air
is
presumed negligible.
Neglecting also
the initial
kinetic
energy of
the
of
air
is
negligible.
also
initial
kinetic
of
que es insignificante.
Ignorando
también Neglecting
la
energía cinética
aire,energy
escribimos
of the
the
air
is presumed
presumed
negligible.
Neglecting
also the
theinicial
initial del
kinetic
energy
of the
thela ecuaair,
we
write
Eq.
(2.32a)
as:
air,
write
ción (2.32a)
como:
air, we
we
write Eq.
Eq. (2.32a)
(2.32a) as:
as:
2
uu222
Q
=
u2 − W
s
Q
=
Q = 222 −
−W
Wss
2
The
kinetic-energy
term
is
as
The
kinetic-energy
term
is evaluated
evaluated
as follows:
follows:
Thede
kinetic-energy
term
evaluated
follows:forma:
El término
energía cinética
seisevalúa
de laassiguiente
�
2
2
��
�
11 �
m
m
11 2
2
2
m
m
180,000
1 600
m22
m �2 =
1uu222 =
=
=
180,000
600
22 u22 = 22 600 ss
= 180,000 ss22
2
2
s 2 kg
s
m
2 kg
−1 = 180 kJ kg−1
m
2
=
180,000
·
=
180,000
m
kg
−1 = 180 kJ kg−1
m kg = 180,000 N
=
N
m
kg
−1 = 180 kJ kg−1
= 180,000
180,000 ss222 ·· kg
=
180,000
N
m
kg
kg
kg
s
−1
–1
Then
Q
==180
240
−60
kJ
kg
−1
En tal caso
180−
240 =
= –60
Then
Q
−
=
−60
kJ
kg
Then
QQ=
= 180
180
−– 240
240
=
−60kJ
kJkg
kg−1
Heat
in
the amount
of 60
kJ
must be
removed per
kilogram
of air
compressed.
Heat
in
amount
kJ
be
kilogram
Se deben
eliminar
kJ de of
calor
kilogramo
de aireper
comprimido.
Heat
in the
the60
amount
of 60
60 por
kJ must
must
be removed
removed
per
kilogram of
of air
air compressed.
compressed.
Example
2.17
Example
2.17
Example
2.17
Ejemplo
2.17
◦
Water at 200( ◦F) is pumped from a storage tank at the rate of 50(gal)(min)−1
−1.
Water at 200( ◦F) is pumped from a storage tank at the rate of 50(gal)(min) −1. The
The
Water
at 200(
F) °F)
is pumped
from
aatstorage
tank
atlathe
rate
of
50(gal)(min)
. –1The
motor
for
the
pump
supplies
work
the
rate
of
2(hp).
The
water
goes
through
a
Se bombea
agua
a
200(0
de un tanque
de
almacenaje
en
proporción
de 50(gal)(min)
. El
motor
for
the
pump
supplies
work
at
the
rate
of
2(hp).
The
water
goes
through
aa motor
motor
for
the
pump
supplies
work
at
the
rate
of
2(hp).
The
water
goes
through
−1
heat
exchanger,
giving
up
heat
at
the
rate
of
40,000(Btu)(min)
,
and
is
delivered
to
a
de la bomba
suministra
trabajo
con
un
valor
de
2(hp).
El
agua
pasa
a
través
de
un
intercambiador
de
−1
heat
giving
up
heat
rate
of
40,000(Btu)(min)
is
to
−1,, and
heat exchanger,
exchanger,
giving
upproporción
heat at
at the
the
rate
of
40,000(Btu)(min)
and
is adelivered
delivered
to aatanque
–1, y se
second
storage
tank
at
an
elevation
50(ft)
above
the
first
tank.
What
is
the
temperature
calor, proporcionando
calor
a
una
de
40
000(Btu)(min)
entrega
un
segundo
second
storage
tank
elevation
50(ft)
above
the
first
What
is the temperature
second
storage
tank at
attoan
an
elevation
50(ft)
above
thetanque.
first tank.
tank.
What
temperature
of
the
water
delivered
the
second
tank?
de almacenaje
a
una
elevación
de
50(pies)
sobre
el primer
¿Cuál
esislathe
temperatura
del agua
of
the
water
delivered
to
the
second
tank?
of
the
water
delivered
to
the
second
tank?
entregada al segundo tanque?
02-SmithVanNess.indd 54
8/1/07 12:50:18
2.12.
2.12.Mass
Massand
andEnergy
EnergyBalances
Balancesfor
forOpen
OpenSystems
Systems
5555
55
2.12. Balances de masa y energía para sistemas abiertos
Solución
2.17 2.17
Solution
Solution
2.17
Éste esThis
un proceso
en estado estacionario yprocess
flujo establewhich
para el que(2.32b)
se aplica la ecuación
(2.32b).
Thisisisa asteady-state,
steady-state,steady-flow
steady-flow processfor
for whichEq.
Eq. (2.32b)applies.
applies. The
The
Las velocidades
inicial
y final delofagua
eninlosthetanques
detanks
almacenaje
son insignificantes,
y se
initial
and
final
velocities
water
storage
are
negligible,
and
the
initial and final velocities
of water in the storage tanks are negligible, and the
2/2g . Los
puede omitir
el 2término
Δu
términos
restantes
se
expresan
en
unidades
de
(Btu)(lb
)–1
c
m
2
term
�u
/2g
may
be
omitted.
The
remaining
terms
are
expressed
in
units
of
c c may be omitted. The remaining terms are expressed in units of
term �u /2g
−1
◦
mediante
el uso
factoresuse
de conversión apropiados.
A 200(°F)
la densidad
del agua es
−1through
◦ F)the
) de
(Btu)(lb
mm
through
useofofappropriate
appropriateconversion
conversionfactors.
factors.AtAt200(
200(F)
thedenden(Btu)(lb
–3, )y 1(pie)
3 es equivalente
−3
3 3is equivalent
60.1 (lbsity
)(pie)
a
7.48(gal);
así,
la
rapidez
de
flujo
de
masa
es:
−3
m
of
water
is
60.1(lb
)(ft)
,
and
1(ft)
to
7.48(gal);
thus
the
mass
mm )(ft) , and 1(ft) is equivalent to 7.48(gal); thus the mass
sity of water is 60.1(lb
flowrate
is:
flowrate is:
–1
(50)(60.1/7.48) = 402(lbm)(min)−1
(50)(60.1/7.48)
)(min)
mm
(50)(60.1/7.48)==402(lb
402(lb
)(min)−1
De donde
Q −40,000/402
= –40 000/402 = –99.50(Btu)(lbm))–1
−1
Whence
QQ=
mm )−1
Whence
= −40,000/402==−99.50(Btu)(lb
−99.50(Btu)(lb
−1 , the shaft work is:
Because
isisequivalent
toto42.41(Btu)(min)
Dado que
1 (hp)1(hp)
es
equivalente
a 42.41
(Btu)(min)–1, el−1
trabajo
en lawork
flecha
Because
1(hp)
equivalent
42.41(Btu)(min)
, the shaft
is:es:
−1
= =(2)(42.41)/(402)
)–1
WW
(2)(42.41)/(402) =
==0.21(Btu)(lb
sW
mm
(2)(42.41)/(402)
0.21(Btu)(lb
)−1
s s=
–2−2
−2
IfIfthe
local
taken
value,
32.174(ft)(s)
, the
Si el valor
local
devalue
g seoftoma
elasthe
valor
estándar,
32.174(pie)(s)
, el
término
de la energía
the
local
value
ofg gisiscomo
takenas
thestandard
standard
value,
32.174(ft)(s)
, thepotentialpotentialenergy
term
potencial
se convierte
en:
energy
termbecomes:
becomes:
−2
−2
32
.174(pie)(s)−2
50
(pie)
32.174(ft)(s)
50(ft)
ggg
32.174(ft)(s)
50(ft)
Δ
=
z
�z
=
−1 −2
−2 ⋅· ·
−1
�z
=
−1
32
.32.174(lb
174(lbmm)()(ft)(lb
pie
)(
lb
)
(
s
)
778
.
16
(
pie
( Btu−1
)−1
−1
−2
ggcgc
32.174(lb
)
(s)
778.16(ft
f))(Btu)
f f ) (s)
f )(Btu)
)(ft)(lb
778.16(ftlblb
lb
c
m
f
f
==0.06(Btu)(lb
)−1
mm
0.06(Btu)(lb
)−1
Equation
(2.32b)
now yields
�H
Equation
(2.32b)
yieldsΔH:
�H: :
La ecuación
(2.32b)
ahoranow
produce
gg
= −99.50 + 0.21 − 0.06 = −99.35(Btu)(lbm )−1−1
�H
s s−− �z
�H==QQ++WW
gcgc �z = −99.50 + 0.21 − 0.06 = −99.35(Btu)(lbm )
◦ ◦ is:
value
for
enthalpy
ofofagua
liquid
water
200(
The
value
forthe
theentalpía
enthalpy
liquid
waterat
200(F)
F)
El valorThe
en steam-table
lasteam-table
tabla de vapor
para
la
del
líquida
aat200(°F)
es:is:
−1
)−1
HH
1 11==
168.09
(Btu)(lbmmm
))–1
H
=168.09(Btu)(lb
168.09(Btu)(lb
Thus,
Thus,
De esta forma,
and
and
y
�H
2 2−−HH
1 1==HH
2 2−−168.09
�H==HH
168.09==−99.35
−99.35
ΔH = H2 – H1 = H2 – 168.09 = –99.35
HH
)−1
mm
2 2==168.09
168.09−−99.35
99.35==68.74(Btu)(lb
68.74(Btu)(lb
)−1
–1
H
168.09
–this
99.35
= 68.74
(Btu)(lb
The
enthalpy
isisfound
2 = having
m) the
Thetemperature
temperatureofofwater
water
havingthis
enthalpy
foundfrom
from
thesteam
steamtables:
tables:
La temperatura que el agua tiene a esta entalpía
se encuentra
en las tablas de vapor:
◦◦
t t==100.74(
F)
100.74(F)
t = 100.74(°F)
are small compared with Q, and for practical
InInthis
s sand
c )�z
and(g/g
(g/g
thisexample
exampleWW
c )�z are small compared with Q, and for practical
purposes
they
could
be
neglected.
theyy could
be neglected.
En este purposes
ejemplo W
(g/gc)Δz
son pequeños en comparación con Q, y para fines prácticos se
s
pueden despreciar.
02-SmithVanNess.indd 55
8/1/07 12:50:22
56
CAPÍTULO 2. La primera ley y otros conceptos básicos
PROBLEMAS
2.1. Un recipiente no conductor lleno con 25 kg de agua a 20 °C tiene un agitador, el cual se mueve
debido a la acción de la gravedad sobre una pesa con masa de 35 kg. La pesa cae lentamente a una
distancia de 5 m accionando el agitador. Suponiendo que todo el trabajo hecho en la pesa se transfiere al agua y que la aceleración local de la gravedad es 9.8 m s–2, determine:
a)
b)
c)
d)
e)
La cantidad de trabajo realizado en el agua.
El cambio en la energía interna del agua.
La temperatura final del agua, para la cual CP = 4.18 kJ kg–1 °C–1.
La cantidad de calor que debe extraerse del agua para que regrese a su temperatura inicial.
El cambio en la energía total del universo debido a: 1) el proceso de bajar el peso, 2) el proceso
de enfriamiento del agua para que regrese a su temperatura inicial, y 3) ambos procesos juntos.
2.2. Trabaje de nuevo con el problema 2.1 para un recipiente aislado que cambia de temperatura junto con
el agua, y que tiene una capacidad térmica equivalente a 5 kg de agua. Trabaje el problema con:
a) El agua y el recipiente como el sistema; b) considere sólo al agua como el sistema.
2.3. Un huevo, inicialmente en reposo, se deja caer sobre una superficie de concreto y se rompe. Si se
considera al huevo como el sistema,
a)
b)
c)
d)
e)
¿Cuál es el signo de W?
¿Cuál es el signo de ΔEP?
¿A qué es igual ΔEK?
¿A qué es igual ΔU t?
¿Cuál es el signo de Q?
Al configurar este proceso, suponga que pasa el tiempo suficiente para que el huevo roto regrese a
su temperatura inicial. ¿Cuál es el origen de la transferencia de calor del inciso e)?
2.4. Un motor eléctrico bajo carga fija toma 9.7 amper a 110 volts, liberando 1.25(hp) de energía mecánica. ¿Cuál es la intensidad de transferencia de calor desde el motor en kW?
2.5. Un mol de gas en un sistema cerrado se somete a un ciclo termodinámico de cuatro etapas. Use la
información que se proporciona en la tabla siguiente para determinar los valores numéricos de las
cantidades faltantes, es decir, “complete los espacios en blanco”.
02-SmithVanNess.indd 56
Paso
ΔU t/J
12
23
Q/J
W/J
–200
?
–6 000
?
–3 800
?
34
?
–800
300
41
4 700
?
?
12 341
?
?
–1 400
8/1/07 12:50:23
Problemas
57
2.6. Explique la factibilidad de enfriar su cocina durante el verano abriendo la puerta de un refrigerador
accionado eléctricamente.
2.7. Un laboratorio de renombre reporta coordenadas del punto cuádruple de 10.2 Mbar y 24.1 ºC para
el equilibrio en las cuatro fases de las formas alotrópicas sólidas del exótico producto químico bmiasmone. Evalúe la afirmación.
2.8. Un sistema cerrado sin reacción química contiene a las especies 1 y 2 en equilibrio vapor/líquido.
La especie 2 es un gas muy ligero, esencialmente insoluble en la fase líquida. La fase vapor contiene ambas especies. Se agregan algunas moles adicionales de la especie 2 al sistema, el cual regresa
a su T y P iniciales. Como resultado del proceso, ¿el número total de moles del líquido se incrementa, disminuye o no cambia?
2.9. Un sistema compuesto por cloroformo, 1,4-dioxano y etanol existe como un sistema de dos fases
vapor/líquido a 50 °C y 55 kPa. Se encuentra que después de agregar cierta cantidad de etanol puro,
el sistema regresa al equilibrio de dos fases a su T y P iniciales. ¿En qué aspecto el sistema cambia?
¿En qué aspecto no cambia?
2.10. Para el sistema descrito en el problema 2.9:
a) ¿Cuántas variables de la regla de la fase, además de T y P, se deben seleccionar para fijar las
composiciones de ambas fases?
b) Si la temperatura y la presión siguen siendo las mismas, ¿puede la composición global del sistema cambiar (por la adición o eliminación de materia) sin afectar las composiciones de las fases líquida y vapor?
2.11. Un tanque que contiene 20 kg de agua a 20 °C cuenta con un agitador que proporciona trabajo al
agua a la intensidad de 0.25 kW. ¿Cuánto tiempo transcurre para incrementar la temperatura del
agua a 30 °C si no hay pérdida de calor desde el agua? Para el agua, CP = 4.18 kJ kg–1 °C–1.
2.12. Se agrega a un sistema cerrado 7.5 kJ de calor a medida que su energía interna disminuye 12 kJ.
¿Cuánta energía se transfiere como trabajo? Para un proceso que ocasiona el mismo cambio de estado pero cuyo trabajo es cero, ¿cuánto calor se transfiere?
2.13. Un bloque de acero fundido de 2 kg tiene una temperatura inicial de 500 °C. En un tanque de 5 kg
de acero perfectamente aislado se tienen 40 kg de agua a 25 °C. Se sumerge el bloque en el agua y
se permite al sistema llegar al equilibrio. ¿Cuál es la temperatura final? Ignore cualquier efecto de
expansión o contracción y suponga constantes el calor específico del agua que es de 4.18 kJ kg–1
K–1 y del acero 0.50 kJ kg–1 K–1.
2.14. Un fluido incompresible (r = constante) está contenido en un cilindro aislado con un pistón sin
fricción hecho a la medida. ¿Puede transferirse energía al fluido en forma de trabajo? ¿Cuál es el
cambio en la energía interna del fluido cuando la presión aumenta de P1 a P2?
2.15. Un kg de agua líquida a 25 °C:
a) Experimenta un aumento en la temperatura de 1 K. ¿Cuánto es ΔU t, en kJ?
02-SmithVanNess.indd 57
8/1/07 12:50:23
58
CAPÍTULO 2. La primera ley y otros conceptos básicos
b) Experimenta un cambio en la elevación Δz. El cambio en la energía potencial ΔEP es igual a la
que presenta ΔU t para el inciso a). ¿Cuánto es Δz en metros?
c) Se acelera desde el reposo hasta la velocidad final u. El cambio en la energía cinética ΔEK es
igual al que presenta ΔU t en el inciso a). ¿Cuánto es u, en m s–1?
Compare y analice los resultados de estos tres incisos.
2.16. Un motor eléctrico funciona “caliente” en servicio, debido a sus irreversibilidades internas. Se sugiere que la pérdida de energía asociada sea minimizada por el aislamiento térmico del armazón del
motor. Realice un comentario crítico con respecto a esta sugerencia.
2.17. Una hidroturbina funciona con una carga de 50 m de agua. Los conductos de entrada y salida tienen
2 m de diámetro. Estime la potencia mecánica desarrollada por la turbina para una velocidad de
salida de 5 m s–1.
2.18. El agua líquida a 180 °C y 1 002.7 kPa tiene una energía interna (en una escala arbitraria) de 762.0
kJ kg–1 y un volumen específico de 1.128 cm3 g–1.
a) ¿Cuál es su entalpía?
b) El agua se hace llegar al estado de vapor a 300 °C y 1 500 kPa, donde su energía interna es
2 784.4 kJ kg–1 y su volumen específico es 169.7 cm3 g–1. Calcule ΔU y ΔH para el proceso.
2.19. Un cuerpo sólido con temperatura inicial T0 se sumerge en. un baño de agua a la temperatura inicial
Tw0. Se transfiere calor del sólido al agua en proporción Q = K · (Tw – T), donde K es una constante Tw y T son los valores instantáneos de las temperaturas del agua y del sólido. Desarrolle una expresión para T en función del tiempo t. Compruebe su resultado para los casos límites, t = 0 y t =
∞. Ignore los efectos de la expansión o contracción y suponga constante el calor específico tanto
para el agua como para el sólido.
2.20. Observe una lista de operaciones unitarias comunes:
a) Intercambiador de calor de una sola tubería; b) intercambiador de calor de doble tubería; c) bomba; d) compresor de gas; e) turbina de gas; f) válvula reguladora, y g) tobera.
Desarrolle una forma simplificada del balance energético general en estado estacionario más apropiado para cada operación. Indique y justifique de manera cuidadosa cualquier suposición que haga.
2.21. El número de Reynolds, Re, es un grupo adimensional que caracteriza la intensidad de un flujo. Para
Re grandes, un flujo es turbulento; para Re pequeños, es laminar. Para el flujo en la tubería, Re ≡
urD/m, donde D es el diámetro de la tubería y m es la viscosidad dinámica.
.
a) Si D y m están fijos, ¿cuál es el efecto al aumentar la relación de flujo de masa m en Re?
.
b) Si m y m están fijos, ¿cuál es el efecto sobre Re al aumentar D?
2.22. Un líquido incompresible (r = constante) fluye de manera estable a través de un conducto de sección transversal circular con diámetro en aumento. En el punto 1, el diámetro es de 2.5 cm y la velocidad es 2 m s–1; en el punto 2, el diámetro es de 5 cm.
02-SmithVanNess.indd 58
8/1/07 12:50:24
Problems
Problemas
Problems
59
59
59
(a) Whatvelocidad
is the velocity
at location
2?
a) (a)¿Cuál
punto
2?
Whatesislathe
velocityen
at el
location
2?
−1 ) of the fluid between locations 1 and 2?
–1
(b) es
What
is the kinetic-energy
change −1
b) (b)¿Cuál
el
cambio
en
la
energía
cinética
) delfluid
fluido
entre locations
los puntos11and
y 2?
What is the kinetic-energy change (J kg(J kg
) of the
between
2?
2.23. A stream
of warmde
water iscaliente
produced
in aproceso
steady-flow
mixing
by combining
2.23.
corriente
un
de
mezcla
de process
flujo
al combinar 1.0
2.23.SeAproduce
stream una
of warm
water isagua
produced
in–1en
a steady-flow
mixing
process
byestable
combining
kg s−1
ofacool
water
at0.8
25◦kg
C with
0.8agua
kg s−1
of hotawater
at Durante
75◦ C. During
mixing,
kg1.0
s–1kg1.0
des−1
agua
fría
25
°C
con
s
de
caliente
75
°C.
la
mezcla
◦
−1
◦
of cool water at 25 C with 0.8 kg s of –1
hot water
at 75 C. During mixing, se pierde
heat is
lost
to the surroundings
at thederate
of
30
kJ
s−1 . What
is
the temperature
of the
calor
hacia
los
alrededores
en
proporción
30
kJ
s
.
¿Cuál
es
la
temperatura
de
la
de
−1
heat is lost to the surroundings at the rate of 30 kJ s . What is the temperature of −1
thecorriente
−1 . –1
K–1
warm-water
stream?
Assume
the specificdel
heat
of water
constant
at 4.18
kJ
kgkJ
agua
caliente?
Suponga
que
el
calor
específico
agua
es
constante
e
igual
a
4.18
kg
K
.
−1
−1
warm-water stream? Assume the specific heat of water constant at 4.18 kJ kg K .
2.24. Gas
bled from a tank.
heat
transfer
between the
gas
and the el
tank,
2.24.
queisestá
deNeglecting
un tanque.
Ignore
la
transferencia
calor
gasshow
y el tanque
2.24.Imagine
Gas is bled
fromsaliendo
a tank. gas
Neglecting
heat transfer
between
the gasde
and
the entre
tank, show
that mass
and
energy balances
produce
the
differential
equation:
y that
demuestre
que
los
balances
energéticos
y
de
masa
producen
la
ecuación
diferencial:
mass and energy balances produce the differential equation:
dU
dm
dU
dm=
�
=
H −U
m
H� − U
m
� is the specific enthalpy of the
Here, U and m refer to the gas remaining in the tank;
� is H
the
of the del gas que
Here,
U
and
m
refer
to
the
gas
remaining
in
the
tank;
H
� enthalpy
En estegas
caso,
U y mthe
setank.
refieren
al gas
restante
en el can
tanque;
′ esspecific
la entalpía
específica
leaving
Under
what
conditions
one H
assume
� =HH ?= H ?
gas
leaving
the
tank.
Under
what
conditions
can
one
assume
H
abandona el tanque. ¿Bajo qué condiciones se supone que H ′ = H?
◦ C flows in a straight horizontal pipe in which there is no exchange of
2.25. Water at
◦ C28
2.25.
Water
at
28
flows
in atubería
straight horizontal
pipeeninlawhich
there
is no −1
exchange calor
of o de tra2.25. Fluye agua
28 °Corenwork
una
recta,
cual no
eithera heat
with thehorizontal
surroundings.
Its velocity
ishay
14intercambio
m s in ade
pipe with
−1
either
heat
or
work
with
the
surroundings.
Its
velocity
is
14
m
s
in
a
pipe
with
–1
bajo con
alrededores.
Su of
velocidad
de 14
m s into
en una
tubería where
con diámetro
interno
de 2.5 cm
anlos
internal
diameter
2.5 cm es
until
it flows
a section
the pipe
diameter
an internal
diameter
of 2.5 cm
until
it flows into
section aumenta
where the
pipe diameter ¿Cuál es el
hasta
que
fluye
aincreases.
una sección
donde
el diámetro
de laachange
tubería
repentinamente.
abruptly
What
is the
temperature
of the water
if the downstream
abruptly
increases.
What isagua
the si
temperature
change of abajo
the water
thecm?
downstream
cambio
en la
temperatura
elcm]?
diámetro
estemperature
deif3.8
¿Si
el diámetro
diameter
is 3.8 cm?del
If it is 7.5
Whatcorriente
is the maximum
change
for an es de
diameter
is
3.8
cm?
If
it
is
7.5
cm]?
What
is
the
maximum
temperature
change
for an
7.5 cm?enlargement
¿Cuál es el in
máximo
cambio de temperatura para un alargamiento de la tubería?
the pipe?
enlargement in the pipe?
2.26. Se
comprimen
cincuenta(50)
kmol
porishora
de aire defrom
P1 = P1.2
bar
aP
compresor
2.26.
Fifty (50)
kmol per hour
of air
compressed
1.2
bar
P2bar
= en
6.0unbar
in
2 =to6.0
1 =
2.26.con
Fifty
(50)
kmol La
perpotencia
hour of mecánica
air is compressed
from
P1 98.8
= 1.2
barLas
to temperaturas
P2 = 6.0
baryinlas velocidaflujo
estable.
transmitida
es de
a steady-flow
compressor.
Delivered
mechanical
powerkW.
is 98.8
kW.
Temperatures
and
a steady-flow
compressor. Delivered mechanical power is 98.8 kW. Temperatures and
des
son:
velocities are:
velocities are:
T1 T=1300
K K
T2T2= =
520
KK
= 300
520
T1 = 300
K –1
T2 = 520
K
−1
u1 u=110=m10s−1m s
uu2 2= =
3.53.5
mm
s–1s−1
u 1 = 10
ms
u 2 = 3.5
m s−1
CalculeEstimate
la rapidez
transferencia
de calor
delthe
compresor.
Suponga
paraforelair
airethat
queCC =
= 7 R,
thederate
of heat transfer
from
compressor.
Assume
R y que
Estimate the rate of heat transfer from the compressor. Assume for air that C P = P72 PR 2
la entalpía
de la presión.
and es
thatindependiente
enthalpy is independent
of pressure.
and that enthalpy is independent of pressure.
2.27. Nitrogen
flows
at steady
state through
a horizontal,
insulated
pipe with
insidecon
diameter
2.27. Circula
nitrógeno
en estado
estacionario
a través
de una tubería
horizontal,
aislada
un diámetro
2.27. Nitrogen flows at steady state through a horizontal, insulated pipe with inside diameter
A pressure
results
fromválvula
flow through
a partially
opened
valve.unJust
interiorofde1.5(in).
1.5 (pulg.).
Debido drop
al flujo
por una
parcialmente
abierta
se produce
descenso
of 1.5(in). A pressure drop results from flow through a partially opened valve. Just
◦ F), and
upstream
from
valve the
pressureesisde100(psia),
temperatureesisde
120(
en la presión.
Antes
dethe
la válvula
la presión
100(psia),the
la temperatura
120(°F)
y la velo◦
upstream from the valve the pressure
is 100(psia), the temperature is 120( F), and
–1. Si la−1
the average
velocity
is 20(ft)(s)
. If thea la
pressure
justla downstream
the valve
cidad promedio
es de
20(pie)(s)
presión
salida de
válvula es defrom
20(psia),
¿cuál es la
the average velocity is 20(ft)(s)−1 . If the pressure just downstream from the valve
is 20(psia),
whatque
is the
Assume
for nitrogen
P V /Ty C
isPconstant,
temperatura?
Suponga
paratemperature?
el nitrógeno PV/T
es constante,
CVthat
= (5/2)R
= (7/2)R. (Los
is 20(psia), what is the temperature? Assume for nitrogen that P V /T is constant,
= están
(5/2)R,
anden
C el
(7/2)R.A.)
(Values for R are given in App. A.)
valoresCde
dados
apéndice
V R
P =
C V = (5/2)R,
and C P = (7/2)R.
(Values for R are given in App. A.)
2.28. Water
flows
through
horizontalhorizontal
coil heated
from
the outsidedesde
by high-temperature
flue gases
2.28.
través
de
unaserpentín
que
esoutside
calentado
el exterior mediante
2.28.Circula
Water agua
flowsathrough
a horizontal
coil heated from
the
by high-temperature
flue
gases.
As
it
passes
through
the
coil
the
water
changes
state
from
liquid
at
200
kPaestado
degases.
combustión
de alta through
temperatura.
Mientras
pasachanges
a travésstate
del serpentín
el agua
cambia
As it◦ passes
the coil
the water
from liquid
at−1
200
kPa del –1
◦ C. Its
and
80
C
to
vapor
at
100
kPa
and
125
entering
velocity
is
3
m
s
and
its
exit
líquido
y 80
°C akPa
vapor
a 125
100◦ C.
kPaIts
y 125
°C. La
velocidad
desentrada
es de
−1 and its
and 80a◦ 200
C to kPa
vapor
at 100
and
entering
velocity
is 3 m
exit3 m s y su
–1. Determine
velocity
is 200
mdes−1
. Determine
the heat el
transferred
through por
the coil
per de
unitmasa
mass of
velocidad
de
salida
es
200
m
s
calor
transferido
unidad
−1
velocity is 200 m s . Determine the heat transferred through the coil per unit mass of de agua a
water.
Enthalpies
of
the
inlet
and
outlet
streams
are:
través
serpentín.ofLas
las corrientes
de entrada y salida son:
water.delEnthalpies
theentalpías
inlet andde
outlet
streams are:
−1 ; Outlet: 2,726.5 kJ kg−1
Inlet: 334.9 kJ
kg
−1
−1
Inlet: 334.9
kJ kg
kg–1
Entrada:
334.9kJ
kJkg
kg–1; Outlet:
salida: 22,726.5
726.5 kJ
02-SmithVanNess.indd 59
8/1/07 12:50:28
60
CHAPTER 2. The First Law and Other Basic Concepts
60
CAPÍTULO 2. La primera ley y otros conceptos básicos
2.29. Steam flows at steady state through a converging, insulated nozzle, 25 cm long and
2.29. Circula
estado estacionario
través
de entrance
una tobera
convergente
aislada, de and
25 cm de largo
with anvapor
inleten
diameter
of 5 cm. At athe
nozzle
(state
1), the temperature
y pressure
un diámetro
de
entrada
de
5
cm.
En
la
entrada
de
la
tobera
(estado
1),
la
temperatura
are 325◦ C and 700 kPa, and the velocity–1is 30 m s−1 . At the nozzle exit (statey la presión
son
325
°C
y 700
kPa, y la velocidad
es de
m◦ C
s and
. En350
la salida
de la tobera
(estado
2), the steam
temperature
and pressure
are30
240
kPa. Property
values
are: 2), la temperatura y la presión del−1
vapor son 240 °C y 3503kPa.
Los
valores
de
sus
propiedades
son:
H1 = 3,112.5 kJ kg
V1 = 388.61 cm g−1
−13 –1
2,945.7
kg−1
V2 = V
667.75
cm3 gcm
H1H=2 =
3 112.5
kJ kJ
kg–1
g
1 = 388.61
is the kJ
velocity
nozzle cm
exit,
3 gand
–1 what is the exit diameter?
HWhat
= 2 945.7
kg–1 of the steam at
V the
= 667.75
2
2
= 20.8
C P =de29.1
J mol−1y ◦¿cuál
C−1 for
nitrogen
gas:
2.30.¿Cuál
In theesfollowing
takedel
C Vvapor
la velocidad
en and
la salida
la tobera?,
es el
diámetro
de salida?
–1 °C–1 para
C, contained
in aelrigid
vessel,gaseoso:
is heated to 250◦ C.
(a) Threea moles
of nitrogen
at 30J◦ mol
2.30. Considere
CV = 20.8
y CP = 29.1
nitrógeno
How much heat is required if the vessel has a negligible heat capacity? If the
vessel weighs 100 kg and has a heat capacity of 0.5 kJ kg−1 ◦ C−1 , how much heat
a) Tres moles de nitrógeno a 30 °C, contenidas en un recipiente rígido, se calientan a 250 °C.
is required?
¿Cuánto calor se requiere si el recipiente
tiene una capacidad calorífica insignificante? Si el
◦ is contained in a piston/cylinder arrangement.
(b)recipiente
Four moles
nitrogen
at 200
pesaof100
kg y tiene
unaC capacidad
calorífica de 0.5 kJ kg–1 °C–1, ¿cuánto calor se
How
much
heat
must
be
extracted
from
this
system, which is kept at constant
requiere?
◦ C if the heat capacity of the piston and cylinder is nepressure,
to
cool
it
to
40
b) Cuatro moles de nitrógeno a 200 °C están contenidas en una combinación de pistón/cilindro.
glected?
¿Cuánto
calor debe extraerse de este sistema, que se mantiene a presión constante, para enfriar-
lo a 40 °C si la capacidad calorífica del pistón y del cilindro se desprecia?
2.31. In the following take C V = 5 and C P = 7(Btu)(lb mole)−1 (◦ F)−1 for nitrogen gas:
2.31. Considere a CV = 5 y CP = 7(Btu)(lb mol)–1◦(°F)–1 para el nitrógeno gaseoso:
(a) Three pound moles of nitrogen at 70( F), contained in a rigid vessel, is heated to
350(◦ F). How much heat is required if the vessel has a negligible heat capacity? If
a) Tres libras mol de nitrógeno a 70(°F), contenidas en un recipiente
rígido, se calientan a 350(°F).
it weighs 200(lbm ) and has a heat capacity of 0.12(Btu)(lbm )−1 (◦ F)−1 , how much
¿Cuánto calor se requiere
si el recipiente tiene una capacidad calorífica insignificante? Si el
heat is required?
recipiente pesa 200(lbm) y tiene una capacidad
calorífica de 0.12(Btu)(lbm)–1(°F)–1, ¿cuánto
◦ F) is contained in a piston/cylinder arrange(b)calor
Foursepound
moles
of
nitrogen
at
400(
necesita?
ment.libras
How mol
much
must be
extracted
from
this system,
which
is kept at conb) Cuatro
deheat
nitrógeno
a 400(°F)
están
contenidas
en una
combinación
de pistón/cilin◦ F) if the heat capacity of the piston and cylinder
stant
pressure,
to
cool
it
to
150(
dro. ¿Cuánto calor debe extraerse de este sistema, que se mantiene a presión constante, para
is neglected?
enfriarlo
a 150(°F) si se desprecia la capacidad calorífica del pistón y del cilindro?
2.32.Encuentre
Find the equation
for para
the work
of a reversible,
isothermalisotérmica
compression
of 1 moldeof1gas
2.32.
la ecuación
el trabajo
de una compresión
reversible
mol de gas en
in combinación
a piston/cylinder
assembly if thesimolar
volumemolar
of thedel
gasgas
is está
givendado
by por
una
de pistón/cilindro
el volumen
V =
RT
+b
P
whereb by and
R constantes
are positivepositivas.
constants.
donde
R son
◦ F) [state
2.33.
a 200(psia)
y 600(°F)
[estado
1] entra
en unaaturbina
través deauna
tubería de 3 pulgadas de
1] enters
turbinea through
3-inch-diameter
2.33.Vapor
Steam
at 200(psia)
and 600(
−1 . The–1exhaust
diámetro
con
una velocidad
de 10(pie)(s)
. La descarga
desde
la turbina
se efectúa
a través
de una
pipe with
a velocity
of 10(ft)(s)
from the
turbine
is carried
through
a
◦ F) y
tubería
de 10 pulgadas
y está
5(psia)
200(°F)
[estado
2].power
¿Cuáloutput
es la potencia de
10-inch-diameter
pipe de
anddiámetro
is at 5(psia)
anda200(
[state
2]. What
is the
salida
la turbina?
of thedeturbine?
H1 = 1,322.6(Btu)(lbm )−1
H1 = 1 322.6(Btu)(lbm)–1 −1
H2 = 1,148.6(Btu)(lbm )
H2 = 1 148.6(Btu)(lbm)–1
02-SmithVanNess.indd 60
V1 = 3.058(ft)3 (lbm )−1
V = 3.058(pie)3(lb )–1
V21= 78.14(ft)3 (lbmm)−1
V2 = 78.14(pie)3(lbm)–1
8/1/07 12:50:30
Problems
Problemas
61
61
61
Problems
2.34. Carbon dioxide gas enters a water-cooled compressor at conditions P11 = 15(psia) and
2.34. Dióxido de carbono
entra
a
un
compresor
que
se
enfría
con
agua
en
las
condiciones
P
1 = 15(psia) y
T1 = 50(◦◦F), and is discharged at conditions P22 = 520(psia) and T22 = 200(◦◦F).
T1 =1 50(°F), y descarga en las condiciones P2 = 520(psia)
y T2 = 200(°F). El CO2 que−1
circula
−1entra,
through
a 4-inch-diameter
pipe
a velocity
of15(psia)
20(ft)(s)
,
The entering
COgas
22 flows
2.34.a Carbon
enters
apulgadas
water-cooled
compressor
at with
conditions
Pde
–1and
1 =
través dedioxide
una
tubería
de
4
de
diámetro
con
una
velocidad
20(pie)(s)
,
y
se
descarga
◦ F), and through
and
is 50(
discharged
a 1-inch-diameter
pipe. =The
shaft work T
supplied
to◦the
at
520(psia)
200(
F). al compre1 = de
2 =
aT
través
una
tuberíaisdedischarged
1 pulgada
de conditions
diámetro. ElP2trabajo
de flechaand
que se
proporciona
−1
−1
−1 ,
compressor
isCO
5,360(Btu)(mol)
.
What
is
the
heat-transfer
rate
from
the
compressor
flows
through
a
4-inch-diameter
pipe
with
a
velocity
of
20(ft)(s)
The
entering
–1
2
sor es de 5 360(Btu)(mol)
. ¿Cuál es la rapidez de transferencia de calor desde el compresor en
−1
−1?
in (Btu)(hr)
and
is discharged
through a 1-inch-diameter pipe. The shaft work supplied to the
–1?
(Btu)(hr)
−1
−1
−1
H11 = 307(Btu)(lb
V11−1=. What
9.25(ft)
(lbm
)−1
compressor
is 5,360(Btu)(mol)
is 33the
heat-transfer
rate from the compressor
m
m)
m
−1
−1
3
−1
in
(Btu)(hr)
?
–1
3
–1
−1
3(lbm )(lb
−1 )
330(Btu)(lb
V22 =V0.28(ft)
H1H=22 =
307(Btu)(lb
m) m
1 = 9.25(pie)
m
m)
m
−1
H1 = 307(Btu)(lb–1
V1 = 9.25(ft)3 (lbm3)−1 –1
m)
V2mechanically
= 0.28(pie)
(lb
2 = 330(Btu)(lb
m)Q for
m)
2.35.HShow
W and
reversible
nonflow process are given
−1 an arbitrary
3 (lb )−1
H2 = that
330(Btu)(lb
V2 = 0.28(ft)
m)
m
by:
�
2.35. Demuestre que W y Q�para un proceso arbitrario sin flujo, mecánicamente
reversible, están dados
2.35.por:
Show that W and
Q forVandarbitrary
reversible
are given
W =
P − �(Pmechanically
V)
Q = �H nonflow
− V dprocess
P
by:
�
�
d P −reversibly
�(P V ) at constant
Q =pressure
�H − from
V danP initial state of
2.36. One kilogram W
of =
air isVheated
300 K and 1 bar until its volume triples. Calculate W , Q, �U , and �H for the
33 molconstante
−1
−1 K−1
−1 de un
2.36.
kilogramo
deofaire
calienta
enV forma
presión
inicial
de 300
process.
Assume
forseisair
that Preversibly
/T = reversible
83.14
baracm
C PPestado
= 29of
J
2.36.Un
One
kilogram
air
heated
at constant
pressure
from anand
initial
state
−1
−1
−1
−1
K300
y1K
barK
hasta
el
triple
de
su
volumen.
Calcule
W,
Q,
ΔU
y
ΔH
para
el
proceso.
Suponga
para
el
mol
.
and 1 bar until its volume
triples. Calculate W–1, Q,
�U , and �H for the
3 mol–1 K–1 y C = 29 J mol
–1
3
−1
−1
aire
que
PV/T
=
83.14
bar
cm
K
.
P
process. Assume for air that P V /T = 83.14 bar cm mol K ◦ and C P = 29 J
2.37. mol
The−1
conditions
of a gas change in a steady-flow process from 20 ◦C and 1,000 kPa
K−1 .
◦
◦
to condiciones
60 C and 100
Devise
a reversible
nonflow
(any number
2.37. Las
dekPa.
un gas
cambian
en un proceso
deprocess
flujo estable
de 20 °Cofysteps)
1 000 for
kPa a 60 °C y
accomplishing
of state,
calculate
�U
andnúmero
�H
process
the este camkPa.
Diseñe this
un
reversible
flujo (cualquier
de◦the
etapas)
paraonlograr
2.37.100
The
conditions
of proceso
achange
gas change
in and
a sin
steady-flow
process
from for
20
C
and
1,000
kPa
basis
of
mol
of kPa.
gas.
gascalcule
that
PΔU
V /T
is constant,
C VV = of
(5/2)R,
bio
de
y,100
con
baseAssume
en 1 mol
dethe
gas,
y ΔH
para
proceso.
Asuma
para
to
60◦estado
C 1and
Devise
afor
reversible
nonflow
process
(anyelnumber
steps) and
for el gas que
C PP =
PV/T
es (7/2)R.
constante,
CVchange
= (5/2)R
y CP =and
(7/2)R.
accomplishing
this
of state,
calculate �U and �H for the process on the
basis of 1 mol of gas. Assume for the gas that P V /T is constant, C V = (5/2)R, and
2.38.a)C
(a)Un=
An
incompressible
fluid
(ρ r==constant)
through
a pipe
of tubería
constantcon
cross2.38.
fluido
incompresible
(con
constante)flows
circula
a través
de una
un área de sec(7/2)R.
P
sectional
area.
If
the
flow
is
steady,
show
that
velocity
u
and
volumetric
flowrate
ción transversal constante. Si el flujo es estable, demuestre que la velocidad u y la relación
qflujo
are
constant.
volumétrico qfluid
son constantes.
2.38. (a)deAn
incompressible
(ρ = constant) flows through a pipe of constant crossA corriente
chemically
gas is
stream
flows
steadily
through
pipe
of aconstant
crosssectional
area.
If the
flow
steady,
show
that velocity
u aand
volumetric
b)(b)Una
dereactive
gas
químicamente
reactivo
fluye
de
manera
estable
travésflowrate
de
una tubería con
area.transversal
Temperature
andtemperatura
pressure vary
pipevarían
length.
Which
of the
qsectional
are
constant.
área
de sección
fija. La
y.lawith
presión
con
la
longitud
de la tubería.
.
following
are
necessarily
constant:
m,through
n, q,constantes:
u? a pipe ofm·constant
de las quantities
siguientes
cantidades
necesariamente
, n·, q, u? cross(b)¿Cuál
A
chemically
reactive
gas
streamson
flows
steadily
sectional area. Temperature and pressure vary with pipe length. Which of the
. .estimating pressure drop owing
2.39.ElThe
mechanical-energy
balance
provides
a basis
for
2.39.
balance
de laquantities
energía mecánica
proporciona
un fundamento
following
are necessarily
constant:
m,
n, q, u? para calcular la caída de presión que
to
friction
in
fluid
flow.
For
steady
flow
of
an
incompressible
in afluido
horizontal
pipe
resulta de la fricción en el flujo del fluido. Para un flujo establefluid
de un
incompresible
en una
of
constant
cross-sectional
area,
it
may
be
written,
2.39.tubería
The mechanical-energy
provides
a basis
estimating
pressure
drop owing
horizontal de área balance
de sección
transversal
fija,for
puede
establecerse
que:
to friction in fluid flow. For steady
fluid in a horizontal pipe
�Pflow of
2 an incompressible
+ be written,
f FF ρu 22 = 0
of constant cross-sectional area, it may
�L
D
�P
2
2 11
11 011 proporciona
wherefF fes
Fanning
friction
the following
expression
for
donde
el the
factor
de fricción
defactor.
Fanning.
la siguiente
expresión
para fF
F
+ Churchill
fChurchill
F is
F ρu = gives
�L
D
f Fcorresponde
for turbulentalflow:
que
flujo con turbulencia:
F
��
� � Churchill11
where f F is the Fanning friction factor.
the
following
expression for
−2
� gives
�0.9
−2
0.9
�
7
f F for turbulent flow:f = 0.3305 ln 0.27 +
F
F
D
Re
� �
� �0.9 ��−2
�
7
f F = 0.3305 ln 0.27 +
11
D
Re
11AIChE
AIChE J.,
J., vol.
vol. 19,
19, pp.
pp. 375–376,
375–376, 1973
1973
11
AIChE J., vol. 19, pp. 375-376, 1973.
11 AIChE J., vol. 19, pp. 375–376, 1973
02-SmithVanNess.indd 61
8/1/07 12:50:37
62
CHAPTER 2. The First Law and Other Basic Concepts
62
CAPÍTULO 2. La primera ley y otros conceptos básicos
Here,Re
Reesiselthe
Reynolds
number (see
Pb.el2.21),
and 2.21)
�/D isy /D
the dimensionless
pipe
Aquí,
número
de Reynolds
(véase
problema
es la aspereza adimensional
de
flowturbulento
obtains forpara
Re >
laroughness.
tubería. SeTurbulent
obtiene flujo
Re3,000.
> 3 000.
Consider elthe
flow
liquid
water at
25◦ C.a 25
For°C.
one
of cada
the sets
given
Considere
flujo
delofagua
en estado
líquido
Para
uno of
de conditions
los conjuntos
de condiciones
.
−1
−1 ). Assume
·
below,
determine m (in
kg s ) m
and
kPa m(en
�/D = 0.0001.
dados
a continuación,
determine
(en�P/�L
kg s–1), y(in∆P/∆L
kPa m–1). Suponga
que /D = 0.0001.
◦ C, ρ = 996 kg m−3
–1 sm
–1−1
For el
liquid
at a2525
= 9.0
s−1 . Verify
Para
aguawater
líquida
°C, r = 996 kg m–3, yand
m =µ9.0
× 10×–410
kg−4mkg
. Verifique
si el flujo tiene
that the flow is turbulent.
turbulencia.
(a) D = 2 cm, u = 1 m s−1
a) D = 2 cm, u = 1 m s–1 −1
(b) D = 5 cm, u = 1 m s
b) D = 5 cm, u = 1 m s–1 −1
2 cm,
ms
c) (c)D D
= 2=cm,
u =u5 =
m 5s–1
−1
(d)
D
=
5
cm,
u
=
5
–1
d) D = 5 cm, u = 5 m s m s
2.40.Un
A sistema
system de
of propano
propane yand
n-butane
exists
in equilibrio
two-phasedevapor/liquid
equilibrium aat10 bar y 323
2.40.
n-butano
existe
en un
dos fases vapor/líquido
10
bar
and
323
K.
The
mole
fraction
of
propane
is
about
0.67
in
the
phase de 0.40 en
K. La fracción molar del propano es aproximadamente 0.67 en la fase vaporvapor
y alrededor
and
about
0.40
in
the
liquid
phase.
Additional
pure
propane
is
added
to
the
system,
la fase líquida. Se agrega propano puro adicional al sistema, que de nuevo lleva al equilibrio a las
which is
equilibrium
andpresentes.
P, with both
liquid
vapor
mismas
T brought
y P con again
ambastofases,
líquidaaty the
de same
vaporTaún
¿Cuál
es eland
efecto
de agregar el
phases
still
present.
What
is
the
effect
of
the
addition
of
propane
on
the
mole
fractions
propano sobre las fracciones molares de propano en las fases líquida y de vapor?
of propane in the vapor and liquid phases?
2.41. Seis especies químicas se encuentran presentes en cantidades significativas en un sistema de frac2.41.cionamiento
Six chemical
aremetano,
present etano,
in significant
amounts
in a light-ends
despecies
petróleo:
propano,
isobutano,
n-butano ypetroleum
n-pentano.fracUna mezcla de
tionation system: methane, ethane, propane, isobutane, n-butane, and n-pentane. A
estas especies existe en equilibrio vapor/líquido en un recipiente cerrado. ¿De cuántas variables
mixture of these species exists in vapor/liquid equilibrium in a closed vessel. On how
de la regla de fase dependen las composiciones de las fases además de T y P?
many phase-rule variables in addition to T and P do the compositions of the phases
Sidepend?
T y P permanecen iguales, ¿existe alguna manera de que la composición del contenido total del
recipiente pueda modificarse (agregando o eliminando material) sin afectar las composiciones de
If fases?
T and P are to remain the same, is there any way the composition of the total conlas
tents of the vessel can be changed (by adding or removing material) without affecting
compositions
2.42. Sethe
introduce
etilenoofa the
unaphases?
turbina a 10 bar y 450 K, y se vacía a 1 (atm) y 325 K. Para m· = 4.5 kg s–1,
determine el costo C de la turbina. Establezca las suposiciones que considere necesarias.
2.42. Ethylene enters a turbine at 10 bar and 450 K, and exhausts at 1(atm) and 325 K. For
.
you make.
m = 4.5 kg s−1 , determine the cost C of the–1turbine. State any assumptions
·
Datos: H1 = 761.1
H2 = 536.9 kJ kg−1
C/$ = (15 200)(|W.|/kW)0.573
Data: H1 = 761.1
H2 = 536.9 kJ kg
C/$ = (15, 200)(|W |/ kW)0.573
2.43. Para incrementar la temperatura de una casa, la calefacción debe modelarse como un sistema abier2.43.to,The
of a home to
its temperaature
must
be modeled
as anuna
open
system,
ya heating
que la expansión
delincrease
aire doméstico
a presión
constante
produce
fuga
de aire hacia los
because
expansion
of
the
household
air
at
constant
pressure
results
in
leakage
air las mismas
exteriores. Suponiendo que las propiedades molares del aire que abandona la casaofsean
to
the
outdoors.
Assuming
that
the
molar
properties
of
air
leaving
the
home
are
the
que las que corresponden al aire en el hogar, demuestre que los balances molar y de energía produsame
as those of
the air diferencial:
in the home, show that eneregy and mole balances yield the
cen
la siguiente
ecuación
following differential equation:
.
dU
dn
+n
Q = −P V
dt
dt
.
Here, ·Q is the rate of heat transfer to the air in the home, and t is time. Quantities P,
Aquí,
es la
de calor para el aire en la casa, y t es el tiempo. Las cantidades
V , n,Qand
Urapidez
refer tode
thetransferencia
air in the home.
P, V, n y U se refieren al aire dentro de la casa.
.
2.44. (a) Water flows through the nozzle of a garden hose. Find an expression for m in terms
2.44. a) Elofagua
fluye
a
través
de
la
boquilla
de
una
manguera
de
jardín.
Encuentre
expresión para
nozzle
line pressure P1 , ambient pressure P2 , inside hose diameter D1 , and una
m· en términos de la presión de la línea P1, la presión ambiental P2, el diámetro interno de la
02-SmithVanNess.indd 62
10/1/07 14:06:09
Problemas
63
manguera D1 y el diámetro de la boquilla de salida D2. Suponga que se tiene un flujo estable,
así como operación isotérmica y adiabática. Para el agua líquida modelada como un fluido incompresible, H2 – H1 = (P2 – P1)/r para una temperatura constante.
b) De hecho, el flujo no puede ser realmente isotérmico: esperamos que T2 > T1, obedeciendo a la
fricción del fluido. Por lo tanto, H2 – H1 = C(T2 – T1) + (P2 – P1)/r, donde C es el calor específico del agua. De modo direccional, ¿de qué forma la incorporación del cambio de temperatura
podría afectar el valor de m· como fue calculado en el inciso a)?
02-SmithVanNess.indd 63
8/1/07 12:50:39
Capítulo 3
Propiedades volumétricas
de fluidos puros
Las cantidades de trabajo y calor que se necesitan para llevar a cabo procesos industriales se calculan a partir
del conocimiento de propiedades termodinámicas, tales como la energía interna y la entalpía. Para los fluidos,
esas propiedades con frecuencia son evaluadas a partir de medidas del volumen molar como una función de
la temperatura y la presión, y proporcionan relaciones de presión/volumen/temperatura (PVT), las cuales se
expresan en forma matemática como ecuaciones de estado. La ecuación menos compleja, PV = RT, proporciona el modelo realista más simple del comportamiento de un fluido. Además las ecuaciones de estado sirven
para la medición de los fluidos y la dimensión de recipientes y tuberías.
En este capítulo se describe, en primer lugar, la naturaleza general del comportamiento PVT de fluidos
puros; de ahí, se continúa con un tratamiento detallado del gas ideal; a continuación, la atención se enfoca en
ecuaciones de estado más realistas, que proporcionan los fundamentos para la descripción cuantitativa del
comportamiento de los fluidos reales; por último, se presentan las correlaciones generalizadas que permiten
la predicción del comportamiento PVT de fluidos, de los cuales no se tiene información experimental.
3.1
COMPORTAMIENTO PVT DE SUSTANCIAS PURAS
Las líneas 1-2 y 2-C de la figura 3.1 representan las condiciones de presión y temperatura de una sustancia
pura en donde existen las fases líquida y sólida en equilibrio con una fase de vapor. Estas líneas de presión de
vapor en función de temperatura caracterizan las relaciones de equilibrio sólido/vapor (línea 1-2) y líquido/
vapor (línea 2-C). La relación de equilibrio sólido/líquido está representada en la línea 2-3. Las tres líneas
despliegan las condiciones de P y T, en las cuales es posible coexistir en dos fases y separa en el diagrama las
regiones de una sola fase. Así, la línea 1-2, la curva de sublimación, separa las regiones de sólido y de gas; la
línea 2-3, la curva de fusión, separa las regiones de sólido y líquido; la línea 2-C, la curva de vaporización,
separa las regiones de líquido y de gas. El punto C se conoce como el punto crítico; sus coordenadas Pc y Tc
corresponden a la presión y temperatura más altas a las cuales se observa que una especie química pura existe en equilibrio vapor/líquido. Las tres líneas se encuentran en el punto triple, donde las tres fases coexisten
en equilibrio. De acuerdo con la regla de la fase, ecuación (2.7), el punto triple es invariante (F = 0). Si el
sistema existe a lo largo de cualquiera de las líneas de dos fases de la figura 3.1, éste es univariante (F = 1),
mientras que en las regiones de una sola fase es divariante (F = 2).
64
03-SmithVanNess.indd 64
8/1/07 12:51:57
65
3.1. Comportamiento pvt de sustancias puras
Es posible representar los cambios de estado mediante líneas en el diagrama PT: un cambio isotérmico
por una línea vertical y un cambio isobárico por una línea horizontal. Cuando una línea así cruza una frontera
de fase, se presenta un cambio súbito en las propiedades del fluido a T y P constantes; por ejemplo, la vaporización para la transición de líquido a vapor.
3
Presión
Pc
Curva de fusión
A
Región de fluido
C
Región líquida
Curva de
vaporización
B
Región sólida
Región gaseosa
2
1
Punto
triple
Figura 3.1: Diagrama PT
para una sustancia pura.
Región
de vapor
Curva de
sublimación
Tc
Temperatura
Es evidente que el agua en un matraz abierto es un líquido que está en contacto con el aire a través de
un menisco. Si el matraz se encuentra sellado y el aire se extrae, el agua se vaporiza para reemplazar el aire,
y el H2O llena el matraz. Aunque la presión en el matraz se reduce, todo parece igual. El agua líquida reside
en la parte inferior del matraz porque su densidad es mucho mayor que la del vapor de agua, y las dos fases
están en equilibrio en las condiciones representadas por un punto de la curva 2-C de la figura 3.1. Las propiedades del líquido y del vapor son muy diferentes. No obstante, si la temperatura aumenta de tal forma que el
estado de equilibrio avanza de manera ascendente a lo largo de la curva 2-C, las propiedades de las dos fases
se hacen cada vez más parecidas; así, en el punto C se hacen idénticas y el menisco desaparece. Una consecuencia es que las transiciones de líquido a vapor es posible que sucedan a lo largo de trayectorias que no
cruzan la curva de vaporización 2-C, es decir, de A hasta B. Por lo tanto, la transición de líquido a gas es gradual y no incluye etapa de vaporización.
La región que existe a temperaturas y presiones mayores que Tc y Pc se marca por las líneas discontinuas en la figura 3.1; éstas no representan fronteras de fase, sino más bien los límites establecidos por los
significados concedidos con las palabras líquido y gas. En general, se considera una fase líquida si se produce vaporización a partir de la reducción de presión a temperatura constante. Se considera gas una fase si se
produce una condensación a partir de la reducción de la temperatura a presión constante. Debido a que ninguno de estos procesos puede iniciarse más allá de la región de las líneas punteadas, se le llama región de
fluido.
La región gaseosa algunas veces está dividida en dos partes, como se indica mediante la línea vertical
discontinua de la figura 3.1. A un gas a la izquierda de esta línea, que puede ser condensado por compresión
a temperatura constante o por enfriamiento a presión constante, se le llama vapor. Un fluido que existe a una
temperatura mayor que Tc se conoce como supercrítico. Un ejemplo es el aire atmosférico.
03-SmithVanNess.indd 65
8/1/07 12:51:58
66
CAPÍTULO 3. Propiedades volumétricas de fluidos puros
Diagrama PV
La figura 3.1 no proporciona información alguna acerca del volumen, tan sólo despliega las fronteras de fase
en un diagrama PT. En un diagrama PV [figura 3.2a)] estas fronteras, a su vez, serán regiones donde dos fases
(sólido/líquido, sólido/vapor y líquido/vapor) coexisten en equilibrio. Estas regiones están separadas por
curvas limítrofes que representan fases únicas, cuyas cantidades relativas determinan los volúmenes molares
(o específicos) en puntos intermedios. Aquí, el punto triple de la figura 3.1 es una línea horizontal, donde
coexisten las tres fases a una sola temperatura y presión.
Sólido/líquido
C
Fluido
P
Gas
Líquido/vapor
Vapor
Sólido/vapor
Vc
C
Pc
Pc
Q
Líquido
P
Sólido
Pc
Líquido
Tc
N
Va
p
or
Tc
T  Tc
Tc
J
K
Líquido/vapor
B
D
T1  Tc
T2  Tc
Vc
V
a)
V
b)
Figura 3.2: Diagramas PV para una sustancia pura. a) Se muestran las regiones sólida, líquida y gaseosa. b) Se muestran
las regiones de líquido, de líquido/vapor y de vapor con isotermas.
La figura 3.2b) muestra las regiones de líquido, de líquido/vapor y de vapor en un diagrama PV, con
cuatro isotermas sobrepuestas. Las isotermas en la figura 3.1 son líneas verticales, y a temperaturas mayores
que Tc no cruzan una frontera de fase. En la figura 3.2b) la isoterma marcada con T > Tc es, por lo tanto,
uniforme.
Las líneas señaladas con T1 y T2 corresponden a temperaturas subcríticas y se componen de tres segmentos. El segmento horizontal de cada isoterma representa todas las posibles combinaciones de líquido y de
vapor en equilibrio, variando desde 100% líquido en el extremo izquierdo hasta 100% vapor en el derecho. El
lugar de estos puntos extremos es la curva en forma de domo marcada con BCD, donde la mitad izquierda (de
B a C) representa líquidos de una sola fase en sus temperaturas de vaporización (ebullición) y la mitad derecha
(de C a D), vapores de una sola fase a sus temperaturas de condensación. Los líquidos y vapores representados
por BCD se conocen como saturados, y las fases coexistentes se conectan mediante el segmento horizontal de
la isoterma a la presión de saturación específica para ésta. También conocida como la presión de vapor, está
dada por un punto en la figura 3.1, donde una isoterma (línea vertical) cruza la curva de vaporización.
La región de dos fases líquido/vapor se encuentra debajo del domo BCD, la región de líquido subenfriado se localiza a la izquierda de la curva de líquido saturado BC y la región de vapor sobrecalentado está si-
03-SmithVanNess.indd 66
8/1/07 12:51:59
3.1. Comportamiento pvt de sustancias puras
67
tuada a la derecha de la curva de vapor saturado CD. El líquido subenfriado existe para temperaturas por
abajo y, el vapor sobrecalentado en temperaturas por arriba del punto de ebullición para una presión determinada. Las isotermas en la región de líquido subenfriado son de pendiente muy pronunciada, porque el volumen de los líquidos cambia muy poco con grandes variaciones en la presión.
Los segmentos horizontales de las isotermas en la región de dos fases se vuelven progresivamente más
cortos a temperaturas altas, y al final se reducen a un punto en C. Por esto, la isoterma crítica, marcada como
Tc, exhibe una inflexión horizontal en el punto crítico C en la parte superior del domo, donde las fases líquida
y de vapor no pueden distinguirse entre sí.
Comportamiento crítico
Se obtiene cierta idea acerca de la naturaleza del punto crítico a partir de una descripción de los cambios que
ocurren cuando se calienta una sustancia pura en un tubo recto, sellado y de volumen constante. Estos procesos se indican en la figura 3.2b) mediante las líneas discontinuas verticales. También es posible trazar en el
diagrama PT de la figura 3.3, donde la curva de vaporización (figura 3.1) aparece como una línea continua, y
las líneas discontinuas son trayectorias a volumen constante en las regiones de una sola fase. Si el tubo se
llena, ya sea con líquido o vapor, el proceso de calentamiento produce cambios que se sitúan a lo largo de las
líneas discontinuas, por ejemplo, el cambio de E a F (líquido subenfriado) y el de G a H (vapor sobrecalentado). Las líneas verticales correspondientes de la figura 3.2b) no se muestran, pero se encuentran a la izquierda
y a la derecha de BCD, respectivamente.
V2l
Vc
Líquido
C
l
V1
P
F
Q
N
(J, K)
G
E
V1v
H
V2v
Figura 3.3: Diagrama PT para un fluido puro donde se
muestra la curva de presión de vapor y las líneas a volumen
constante en las regiones de una sola fase.
Vapor
T
Si el tubo se llena sólo de manera parcial con líquido (el resto es vapor en equilibrio con el líquido), el
calentamiento provoca, al principio, los cambios descritos por la curva de la presión de vapor (línea continua)
de la figura 3.3. Para el proceso indicado con la línea JQ de la figura 3.2b), de manera inicial, el menisco está
cerca de la parte superior del tubo (punto J), y el líquido se expande lo suficiente debido al calentamiento
hasta que llena completamente el tubo (punto Q). En la figura 3.3 el proceso describe una trayectoria que va
de (J, K) hasta Q, y con calentamiento adicional se desvía de la curva de presión de vapor siguiendo la línea
del volumen molar constante V2l .
El proceso indicado por la línea KN de la figura 3.2b) comienza con un nivel inferior del menisco en el
tubo (punto K), y el calentamiento origina que el líquido se vaporice hasta el punto en que el menisco retro-
03-SmithVanNess.indd 67
8/1/07 12:52:00
68
68
68
68
68
CAPÍTULO
Propiedades
volumétricas
deFluids
fluidos puros
CHAPTER
3.
Properties
of
CHAPTER
3.3.Volumetric
Volumetric
Properties
of Pure
Pure
CHAPTER
3.3. Volumetric
Properties
ofofFluids
Pure
CHAPTER
Volumetric
Properties
PureFluids
Fluids
cede hacia el fondo del tubo (punto N). En la figura 3.3 el proceso se indica mediante una trayectoria que va
bottom
of
tube
N
3.3
the
process
traces
atraces
(J,
K
))(J,
to
With
bottombottom
of the
theof
tube
(point
N).). On
On
Fig.
3.3
the
process
tracestraces
a path
pathaafrom
from
(J,
K(J,
to KN
N
With
the
tube
NN).Fig.
Fig.
3.3
the
NN. . With
of
the(point
tube(point
(point
).elOn
On
Fig.
3.3
theprocess
process
path
from
K)..)to
to
With
de (J, K) a N. bottom
Con calentamiento
adicional,
proceso
continúa
a lo largo
de lapath
líneafrom
del
molar
consvv. volumen
vv.
heating
the
path
continues
along
the
line
of
constant
molar
volume
V
furtherfurther
heating
the
path
continues
along
the
line
of
constant
molar
volume
V
.
vfurther
heating
the
path
continues
along
the
line
of
constant
molar
volume
V
2
2
further heating the path continues along the line of constant molar volume
V22 .
tante V 2.
For aa unique
filling
of
tube,
particular
intermediate
meniscus
level,
heating
unique
fillingfilling
of the
theof
tube,
with
particular
intermediate
meniscus
level, the
the
heating
For
aaunique
tube,
aaparticular
intermediate
meniscus
level,
the
For
unique
filling
ofthe
thewith
tube,aawith
with
particular
intermediate
meniscus
level,
the
heating
Para un For
llenado
único
del tubo,
con
una
altura
intermedia
particular
del
menisco,
el proceso
deheating
calentaprocess
follows
a
vertical
line
on
Fig.
3.2(b)
that
passes
through
the
critical
point
C.
Physically,
process
follows
a
vertical
line
on
Fig.
3.2(b)
that
passes
through
the
critical
point
C.
Physically,
process
follows
a
vertical
line
on
Fig.
3.2(b)
that
passes
through
the
critical
point
C.
Physically,
process
follows
a vertical
on 3.2b)
Fig. 3.2(b)
that por
passes
through
the critical
point C. Physically,
miento describe
una línea
vertical
en la line
figura
que pasa
el punto
crítico
C. Físicamente,
el proceso
heating
does
produce
much
in
level
the
meniscus.
As
critical
point
is
heatingheating
does not
not
produce
much change
change
in the
thein
level
of
theof
meniscus.
As the
theAs
critical
point point
ispointisis
does
not
change
the
level
the
meniscus.
the
does
not
produce
much
change
theof
level
the
meniscus.
As
the
critical
no produce unheating
gran cambio
enproduce
el nivel much
del
menisco.
Ainmedida
queofel
proceso
se acerca
alcritical
punto
crítico,
el
approached,
the
meniscus
becomes
indistinct,
then
hazy,
and
finally
disappears.
On
Fig.
3.3
approached,
the
meniscus
becomes
indistinct,
then
hazy,
and
finally
disappears.
On
Fig.
3.3
approached,
the
meniscus
becomes
indistinct,
then
hazy,
and
finally
disappears.
On
Fig.
approached,
the meniscus
indistinct,
thendesaparece.
hazy, and finally
disappears.
Fig.3.3
3.3
menisco se torna
indistinguible,
despuésbecomes
es nebuloso
y al final
En la figura
3.3, laOn
trayectoria
the
first
the
curve,
proceeding
from
(J,
K
))(J,
to
the path
path
first
follows
the vapor-pressure
vapor-pressure
curve, curve,
proceeding
from point
point
(J,
K(J,
toKthe
the
critical
path
first
follows
vapor-pressure
proceeding
))tocritical
the
the
pathfollows
first
follows
the
vapor-pressure
proceeding
from
point
to
the
critical
primero siguethe
la curva
de
presión
dethe
vapor,
procediendocurve,
del punto
(J, K) alfrom
puntopoint
crítico
C,Kdonde
se critical
introdupoint
C,
where
it
enters
the
single-phase
fluid
region,
and
follows
V
,
the
line
of
constant
point point
C,
where
it
enters
the
single-phase
fluid
region,
and
follows
V
,
the
line
of
constant
c
c
C,
where
it
enters
the
single-phase
fluid
region,
and
follows
V
,
the
line
of
point C,
where
enters
the single-phase
region, and
follows
Vcc , igual
the line
of constant
constant
ce a la región fluida
de una
solait fase
y sigue
a Vc, la líneafluid
de volumen
molar
constante
al volumen
crímolar
equal
to
the
critical
volume
of
fluid.
molar volume
volume
equal equal
to
theto
critical
volume
of the
theof
fluid.
volume
molarvolume
volume
equal
tothe
thecritical
critical
volume
ofthe
thefluid.
fluid.
tico del fluido.molar
Single-Phase
Region
Single-Phase
Region
Region
Regiones Single-Phase
de una sola fase
For
regions
of
diagram
where
aa single
phase
exists,
Fig.
implies
aa relation
conFor the
the
regions
of the
theof
diagram
wherewhere
single
phasephase
exists,
Fig. 3.2(b)
3.2(b)
implies
relation
con- conFor
the
the
aasingle
Fig.
implies
aarelation
For
theregions
regions
of
thediagram
diagram
where
single
phaseexists,
exists,
Fig.3.2(b)
3.2(b)
implies
relation
connecting
P,
V
,
and
T
.
Expressed
analytically,
as
f
(P,
V,
T
)
=
0,
such
a
relation
is
known
as
P,
V
,
and
T
.
Expressed
analytically,
as
f
(P,
V,
T
)
=
0,
such
a
relation
is
known
as aas
necting
P,
V
,
and
T
.
Expressed
analytically,
as
f
(P,
V,
T
)
=
0,
such
a
relation
is
Para lasnecting
regiones
del
diagrama
donde
sólo
existe
una
fase,
la
figura
3.2b)
implica
una
relación
que
conecta
P,
necting P, V , and T . Expressed analytically, as f (P, V, T ) = 0, such a relation isknown
known
as
PVT
of
state.
It
relates
pressure,
molar
or
specific
volume,
and
temperature
for
a
PVTse
equation
of
state.
It
relates
pressure,
molar
or
specific
volume,
and
temperature
for
a
aaequation
PVT
equation
of
state.
It
relates
pressure,
molar
or
specific
volume,
and
temperature
for
a
V y T, laaacual
expresa
de
manera
analítica
mediante
f
(P,
V,
T)
=
0;
una
relación
así
se
conoce
como
ecuación
PVT equation of state. It relates pressure, molar or specific volume, and temperature for a
pure
homogeneous
in
equilibrium
states.
The
ideal-gas
of
P
V
,,homogéneo
pure
homogeneous
fluid
in
equilibrium
states.
The
ideal-gas
equation
of state,
state,
Pstate,
V=
=
RT
has
pure
homogeneous
fluid
The
equation
of
PPRT
VV =
RT
de estado
PVT.
Relaciona
lafluid
presión,
elin
volumen
molarstates.
o específico
y equation
la
temperatura
para
un
fluido
pure
homogeneous
fluid
inequilibrium
equilibrium
states.
Theideal-gas
ideal-gas
equation
ofstate,
=has
RT, ,has
has
approximate
validity
for
the
low-pressure
gas
region
of
Fig.
3.2(b),
and
is
discussed
in
detail
validity
for
the
low-pressure
gas
region
of
Fig.
3.2(b),
and
is
discussed
in
detail
approximate
validity
for
the
low-pressure
gas
region
of
Fig.
3.2(b),
and
is
discussed
in
puro en approximate
estados
de
equilibrio.
La
ecuación
de
estado
de
gas
ideal,
PV
=
RT,
tiene
una
validez
aproximada
para
approximate validity for the low-pressure gas region of Fig. 3.2(b), and is discussed indetail
detail
Sec.
3.3.
indel
Sec.
3.3.
inin
Sec.
3.3.
la regiónin
gas
de baja
Sec.
3.3.presión de la figura 3.2b), y se discutirá con detalle en la sección 3.3.
An
of
solved
for
any
one
of
the
three
quantities
,,P,
or
An equation
equation
of state
state
may
be
solved
for
any
one
ofone
thetres
three
quantities
P,
or
as
An
equation
of
state
may
be
solved
for
any
of
the
three
VTT
or
as
Una ecuación
de
puede
resolverse
para
cualquiera
de
las
cantidades,
P,P,
V oVVT,
como
una
Anestado
equation
ofmay
statebe
may
be
solved
for
any
one
of
the
threequantities
quantities
P,
V, ,as
or TTfunas
a
function
of
the
other
two.
For
example,
if
V
is
considered
a
function
of
T
and
P,
then
of
the
other
two.
For
example,
if
V
is
considered
a
function
of
T
and
P,
then
aa function
of
the
other
two.
For
example,
if
V
is
considered
a
function
of
T
and
P,
then
ción de alasfunction
otras
dos.
Por
ejemplo,
si
se
considera
a
V
como
una
función
de
T
y
P,
entonces
V
=
V(T,
P),
y
function of the other two. For example, if V is considered a function of T and P, then
V
P),
and
V=
=V
VVV(T,
(T,
P),
and
==V
P),
�
V(T,
(T,
P),and
and
� �
� �
�
�
� �
�
∂∂V
V �∂∂VV � ∂∂V
V �∂∂VV �
ddV
d
P
(3.1)
(3.1)
V=
=ddVV == dT
dT +
+dT
d
P
(3.1) (3.1)
dP
dT
(3.1)
∂∂TT PP∂∂TT
∂∂+
P
P+ TT∂∂PP T d P
PP
T
The
partial
derivatives
in
this
equation
have
definite
physical
meanings,
and
related
to
The
partial
derivatives
in
this
equation
have
definite
physical
meanings,
and yare
are
related
to two
twotototwo
The
partial
derivatives
in
this
equation
have
definite
physical
meanings,
and
are
related
The
partial
derivatives
in
this
equation
have
definite
physical
meanings,
and
are
related
two
Las derivadas parciales en esta ecuación tienen significados físicos bien definidos
están
relacionadas
con
properties,
commonly
tabulated
for
liquids,
and
defined
as
follows:
properties,
commonly
tabulated
for
liquids,
and
defined
as
follows:
properties,
commonly
tabulated
for
liquids,
and
defined
as
follows:
properties,
commonly
tabulated
for
liquids,
and
defined
as
follows:
dos propiedades que comúnmente se tabulan para los líquidos, y que se definen como:
�
� �
�� ��
11 ∂∂V
1V1 ∂∂VV
•
Volume
expansivity:
β
≡
(3.2)
• Volume
expansivity:
(3.2) (3.2)
••Volume
expansivity:
• Coeficiente
de
expansión
volumétrica:β ≡ Vββ≡≡∂ T
(3.2)
(3.2)
Volume
expansivity:
V ∂VTV PP∂∂TT P
P
�
� �
�� ��
11 ∂∂V
1V1 ∂∂VV
•• Isothermal
compressibility:
κ
≡
−
(3.3)
Isothermal
compressibility:
κ
≡
−
(3.3) (3.3)
•
Isothermal
compressibility:
κ
≡
−
• Isothermal
compressibility: κV
(3.3)
• Compresibilidad
isotérmica:
(3.3)
∂∂V
P
V≡ −
PV TT∂∂PP T
T
In
Equations
(3.1)
(3.3)
In combination
combination
Equations
(3.1)
through
(3.3) yield:
yield:
In
Equations
(3.1)
through
(3.3)
yield:
La combinación
de
las ecuaciones
(3.1)through
a(3.1)
(3.3)
produce
la ecuación:
Incombination
combination
Equations
through
(3.3)
yield:
ddV
V ddVV
=
κκdT
dd −
P
(3.4)
(3.4)
= ββ dT
dT
−
P−κκddPP
(3.4) (3.4)
==−
β
βdT
(3.4)
V
V
VV
Las isotermas
para la fase
de laphase
partephase
izquierda
deside
la
figura
3.2b)
son are
de
muy
pronunThe
for
the
liquid
on
the
left
Fig.
very
and
The isotherms
isotherms
forlíquida
the
liquid
phase
on
theon
left
side
of
Fig.
3.2(b)
arependiente
very
steep
and
The
for
the
the
left
side
of
Fig.
are
very
The isotherms
isotherms
for
the liquid
liquid
phase
on
the
leftof
side
of3.2(b)
Fig. 3.2(b)
3.2(b)
aresteep
very steep
steep and
and
ciada y closely
muy cercanas
entre
sí.
De
esta
manera,
(∂V/∂T)
y
(∂V/∂P)
y,
por
tanto,
b
y
k
son
pequeños.
Este
and
(∂
V
/∂
P)
and
hence
both
β
and
κ
are
small.
spaced.
Thus
both
(∂
V
/∂
T
)
P
T
and
(∂
V
/∂
P)
and
hence
both
β
and
κ
are
small.
closely
spaced.
Thus
both
(∂
V
/∂
T
)
P
T
closely
and (∂
(∂VVT/∂
/∂P)
P)TT and
and hence
hence both
both ββ and
and κκ are
are small.
small.
closely spaced.
spaced. Thus
Thus both
both (∂
(∂VVP/∂
/∂TT))PP and
comportamiento
característico
debehavior
losof
(lejos
de
la
región
delregion)
puntoregion)
crítico)
sugiere
una
idealización,
This
behavior
liquids
(outside
the
critical
suggests
an
This characteristic
characteristic
behavior
oflíquidos
liquids
(outside
the
critical
region)
suggests
an idealization,
idealization,
This
of
(outside
the
suggests
an
This characteristic
characteristic
behavior
of liquids
liquids
(outside
the critical
critical
region)
suggests
an idealization,
idealization,
comúnmente
utilizada
en
mecánica
de
fluidos,
que
se
conoce
como
el
fluido
incompresible,
para
el
queboth
b yk
commonly
employed
in
fluid
mechanics
and
known
as
the
incompressible
fluid,
for
which
both
commonly
employed
in
fluid
mechanics
and
known
as
the
incompressible
fluid,
for
which
both
commonly
commonlyemployed
employedininfluid
fluidmechanics
mechanicsand
andknown
knownas
asthe
theincompressible
incompressiblefluid,
fluid,for
forwhich
which
both
son cero.
ningún
fluido
real
es
incompresible,
pero
la
idealización
es
muy
útil
porque
con
frecuenββ De
and
κ
are
zero.
No
real
fluid
is
truly
incompressible,
but
the
idealization
is
useful,
because
andhecho,
κ
are
zero.
No
real
fluid
is
truly
incompressible,
but
the
idealization
is
useful,
because
ββand
andκκ are
arezero.
zero. No
Noreal
realfluid
fluidisistruly
trulyincompressible,
incompressible,but
butthe
theidealization
idealizationisisuseful,
useful,because
because
cia proporciona
unaamodelo
bastante
realista
delmodel
comportamiento
del
para
finespurposes.
prácticos.
No existe
itit provides
sufficiently
realistic
model
of
behavior
for
many
practical
No
provides
sufficiently
realistic
model
of liquid
liquid
behavior
forlíquido
many
practical
purposes.
No
ititprovides
aasufficiently
realistic
of
behavior
for
purposes.
No
provides
sufficiently
realistic
model
ofliquid
liquid
behavior
formany
manypractical
practical
purposes.
No
ningunaequation
ecuación
de
estado
para
un
fluido
incompresible,
porque
V
es
independiente
de
T
y
P.
of
state
exists
for
an
incompressible
fluid,
because
V
is
independent
of
T
and
P.
equation
of
state
exists
for
an
incompressible
fluid,
because
V
is
independent
of
T
and
P.
equation
independent
of
TT and
P.
equationof
ofstate
stateexists
existsfor
foran
anincompressible
incompressiblefluid,
fluid,because
becauseVV isis
independent
of
and
P.
Para líquidos,
b casi
esalways
positiva
(es positive
una(liquid
excepción
agua líquida
entre
0C
°C)
k es
nece◦an
For
ββ siempre
is
positive
water
between
00◦◦C
◦CCexcepFor liquids
liquids
is almost
almost
always
positive
(liquid
waterelwater
between
C and
and
4◦◦and
Cyis
is444an
excepFor
ββisisalmost
always
(liquid
00◦◦C4C
isisyan
Forliquids
liquids
almost
always
positive
(liquid
waterbetween
between
and
anexcepexcepsariamente
positiva.
En
condiciones
no
cercanas
al
punto
crítico
b
y
k
son
funciones
débiles
de
la
temperatution),
and
κ
is
necessarily
positive.
At
conditions
not
close
to
the
critical
point,
β
and
κ
are
tion),
and
κ
is
necessarily
positive.
At
conditions
not
close
to
the
critical
point,
β
and
κ
are
tion),
tion),and
andκκ isisnecessarily
necessarilypositive.
positive. At
Atconditions
conditionsnot
notclose
closetotothe
thecritical
criticalpoint,
point,ββ and
andκκ are
are
03-SmithVanNess.indd 68
8/1/07 12:52:03
3.1.
Behavior
of of
Pure
Substances
6969
3.1.PVT
PVT
Behavior
Pure
Substances
69
69
3.1. PVT
PVT Behavior
Behavior of
of Pure
Pure Substances
Substances
69
3.1.
69
3.1. PVT Behavior of Pure Substances
3.1. PVT Behavior
of Pure
Substances
3.1. Comportamiento
de
sustancias
puras Thus for small changes in T and P little error is69
69
weak
functions
ofpvt
temperature
and
pressure.
weak
functions
of
temperature
and
pressure. Thus
Thus for
for small
small changes
changes in
in TT and
and P
P little
little error
error69
is
3.1.
PVT
Behavior
of Pure Substances
weak
functions
of
temperature
and
pressure.
is
weak
functions
of
temperature
and
pressure.
Thus
for
small
changes
in
T
and
P
little
error
is
weak
functions
of
temperature
and
pressure.
Thus
for
small
changes
in
T
and
P
little
error
is
introduced
if they
areare
assumed
constant.
Integration
Eq.
(3.4)
then
yields:
introduced
if they
they
are
assumed
constant.
Integration
of
Eq.
(3.4)
then
yields:
weak
functions
of
temperature
and
pressure.
Thusof
for
small
changes
in
T and P little error is
introduced
if
assumed
constant.
Integration
of
Eq.
(3.4)
then
yields:
introduced
ifeste
they
are
assumed
constant.
Integration
of
Eq.
(3.4)
then yields:
yields:
if
they
are
assumed
constant.
Integration
of
Eq.
(3.4)
then
ra y la introduced
presión.
De
modo,
para
cambios
pequeños
de
T
y
P
se
introduce
error is
si se les
weak
functions
of
temperature
and
pressure.
Thus
for
small
changes
in
T un
andpequeño
P little error
introduced if they are assumedVconstant.
Integration
of
Eq.
(3.4)
then
yields:
2V
222 pressure. Thus for small changes in T and P little error is
V
weak
functions
temperature
and
=
β(T
−
T
)
−
κ(P
−
P
)
(3.5)
ln ln
V
considera
constantes.
Laofintegración
de
la
ecuación
(3.4)
produce:
V
2
1
2
1
=
β(T
−
T
)
−
κ(P
−
P
)
(3.5)
introduced
if they
are
assumed
constant.
Integration
of
Eq.
(3.4)
then
yields:
22 = β(T222 − T111) − κ(P222 − P111)
(3.5)
ln
Vconstant.
= β(T
β(T
− TT11)) −
− of
κ(P
−(3.4)
(3.5)
ln
κ(P
PP11))then yields:
(3.5)
ln
1V
introduced if they are assumed
Integration
Eq.
22 −
22 −
1121 =
V
V
=
β(T
−
T
)
−
κ(P
−
P
)
(3.5)
ln V
1
2
1
2
1
1
V
2
V21 =than
This
is is
a less
restrictive
approximation
the
assumption
an
incompressible
fluid.
This
a less
restrictive
approximation
than
the
fluid.
β(T
T1assumption
) − κ(P2of−of
P1an
) incompressible
(3.5) (3.5)
ln V
This
is
less
restrictive
approximation
than
the
assumption
of
an
incompressible
fluid.
2−
This is
is aaa less
less restrictive
restrictive approximation
approximation
than
theTassumption
ofPan
incompressible fluid.
fluid. (3.5)
This
than
the
) − κ(P2 −of
) incompressible
ln V1 = β(T
2−
1assumption
1an
This is a less restrictive approximation
than the assumption of an incompressible fluid.
V1
Thisaproximación
is a less restrictive
the assumption
an incompressible
Ésta es una
menosapproximation
restrictiva quethan
la suposición
de unof
fluido
incompresible.fluid.
This is a less restrictive approximation than the assumption of an incompressible fluid.
Example
3.1
Example
3.1
Example 3.1
3.1 ◦ ◦◦
Example
For
liquid
acetone
at at
2020
C◦ C
and
1 bar,
Example
3.1
For
liquid
acetone
and
1 bar,
For
liquid
acetone
at
20
C
and
bar,
For liquid
liquid acetone
acetone
at 20
20◦◦◦C
C and
and 111 bar,
bar,
For
at
Example
3.1
−3
◦
−1
−6−6
−1−1
Ejemplo
3.1
For liquid
acetone
at
20
C
and
1
bar,
−3
◦
−1
−6
−1
−3
◦
−1
Example
3.1
C
κ
=
6262
××
1010
bar
ββ
=
1.487
×
10
−3
◦
−1
−6
−1
C
κκ =
=
1.487
×
10
◦
=
62
×
10−6
bar−1
β = 1.487 × 10−3
−3 ◦◦C−1
−1
−6 bar
−1
3 333−1−1
−1
VV
==
1.287
cmcm
g33 g
−1
1.287
V
=
1.287
cm
gg−1
−1
For liquid
at
20◦ C
and 1 bar,
C
κ
=
62
×
10
bar
V
=
1.287
cm
=acetone
1.487 ×
× 10
10−3
κ
=
62
×
10
bar
V
=
1.287
cm
g
ββ =
1.487
3
−6 bar−1
−1
liquid
acetone
at
20 y◦C
and
Para For
la For
acetona
líquida
a×20
°C
1−1
bar, 1 bar,
C
κ
=
62
×
10
V
=
1.287
cm
g
β
=
1.487
10
acetone,
find:
For
acetone,
find:
−3 ◦ C−1
−6 bar−1
−1
For
acetone,
find:
κ = 62 × 10−6
V = 1.287 cm33 g−1
β = 1.487
× 10−3
For acetone,
acetone,
find:
For
find:
◦ C−1
−1
◦
κ
=
62
×
10
bar
V
=
1.287
cm
g
β
=
1.487
×
10
◦
◦
For
acetone,
find:
(a)(a)
The
value
of of
(∂ (∂
P/∂
T )TV ) at
2020
C◦ C
and
1 bar.
The
value
P/∂
and
1 bar.
V at
(a)
The
value
of
(∂
P/∂
at
20
C
and
bar.
(a) The
The value
value
of (∂
(∂P/∂
P/∂TTT)))VVVV at
at 20
20◦◦◦C
C and
and 111 bar.
bar.
For (a)
acetone,
find: of
Para la For
acetona,
encuentre:
◦ C◦◦◦ and 1 bar to
(a)
The
value
of
(∂
P/∂
T
)
at
20
C
and
1
bar.
acetone,
find:
V byby
(b)(b)
The
pressure
generated
heating
at at
constant
VV
from
2020
C
and 1
bar
to
The
pressure
generated
heating
constant
from
◦
C
bar
to
(b)
The
pressure
generated
by
heating
at
constant
V
from
20
◦◦C
(a)
value
of
(∂
P/∂
T
)
at
20
C
and
1
bar.
(b)
The
pressure
generated
by
heating
at
constant
V
from
20
C and
and 111 bar
bar to
to
V
◦
(b)
The
pressure
generated
by
heating
at
constant
V
from
20
◦◦◦C.
◦ C and
30
a) El valor
deC.
(∂P/∂T)
20
°CTy)1
(a)
The
value
P/∂
at by
20◦heating
C and 1at
bar.
(b)3030
pressure
generated
constant
V
from
20
and
1
bar
to
Vofa(∂
30
C.
V bar.
◦
◦
30◦C.
C.
◦ C and 1 bar to
(b) 30
The
pressure
generated
by heating
at constant
V from
20
b) La(c)
presión
generada
por
calentamiento
a V from
constante
°C
1
hasta
30 °C.
◦ C◦◦◦desde
◦ C◦bar
C.
◦◦◦C
and
1V20
bar
to yto
020
and
1010
The
change
in in
volume
forfor
a by
change
20
C
and
1
bar
0
and
bar.
(c)
The
change
volume
a
change
from
20
1bar.
bar
to
(b)
pressure
generated
heating
at
constant
from
◦
C
and
1
bar
to
0
C
and
10
bar.
(c)
The
change
in
volume
for
a
change
from
20
◦
◦C
◦
◦10
30
C.
c) El cambio
de
volumen
para
una
modificación
de
20
°C
y
1
bar
a
0
°C
y
bar.
C
and
1
bar
to
0
C
and
10
bar.
(c) The
The
change
in
volume
for
a
change
from
20
C
and
1
bar
to
0
C
and
10
bar.
(c)
change
in
volume
for
a
change
from
20
◦ C.
◦
◦
30
(c) The change in volume for a change from 20 C and 1 bar to 0 C and 10 bar.
(c) The change in volume for a change from 20◦◦ C and 1 bar to 0◦◦ C and 10 bar.
Solución
3.1
(c)
The change
Solution
3.1
Solution
3.1 in volume for a change from 20 C and 1 bar to 0 C and 10 bar.
Solution
3.1
Solution
3.1
a) LaSolution
derivada
(∂P/∂T)
determina
con
la aplicación
de la ecuación
(3.4)
para
elthe
caso
en el que
(a)
The
derivative
P/∂
T )T
determined
byby
application
of of
Eq.Eq.
(3.4)
to to
the
case
3.1 V(∂se(∂
V )is
(a)
The
derivative
P/∂
determined
application
(3.4)
case
V is
V
(a)
The
derivative
(∂
P/∂
is
determined
by
application
of
Eq.
(3.4)
to
the
case
V
(a)
Theyderivative
derivative
(∂P/∂
P/∂
is determined
determined by
by application
application of
of Eq.
Eq. (3.4)
(3.4) to
to the
the case
case
The
(∂
TTTd)))V
V es constante
dV
=
0:
V is
V
Solution
3.1
for(a)
which
V
is
constant
and
=
0:
for
which
V
is
constant
and
V
=
0:
(a)
The
derivative
(∂ P/∂and
T )VddddV
is
determined
by application of Eq. (3.4) to the case
for
which
V
is
constant
and
V
== 0:
0:
Solution
3.1
for which
which
V is
is constant
constant
and
V=
0:
for
V
(a) The
derivative
(∂ P/∂and
T )Vd V
is determined
by(V
application
of Eq. (3.4) to the case
for
which
V is constant
=
0:
constante)
β dT
κ determined
dκκPdd P
==
0 (const
V )V
β
dT
−
00 by
(const
)) of Eq. (3.4) to the case
(a)
derivative
(∂ P/∂
)V−
is
application
ββTdT
dT
−
P
=
(const
V
for The
which
V is�constant
and
d
V
=
0:
dT
−
κ
d
P
=
0
(const
V
)
β
−
κ
d
P
=
0
(const
V
)
��
�
for which V is constant
andβ d V− 1.487
= 0:
−3−3
�
�
dP ×
=×
0 10
−3(const V )
∂�
10
−3
�P∂∂ P
� β dT
�
β κ 1.487
P
1.487
×
10
−3
−1
−3
−1
or
=−=
=
2424
bar
−1
β=
dTβββ
κ 1.487
d1.487
P =×
0−6
(const
Vbar
)◦ C◦◦◦◦◦−1
P �=
×
10
10
�∂∂P
or
=
C
or
=
=
=
24
bar
C
o
−1
−3
−6 (const
−6
∂�T∂∂ T
κ
β
dT
−
κ
d
P
=
0
V
)
62
×
10
−6
or
=
=
=
24
bar
C−1
or
=
=
=
24
bar
C
P
1.487
×
10
β
κ
62
×
10
V�V
T
κ
V
62
×
10
◦
−1
−6
−6
V = κκ = 62
−3
∂
T
or
=
24
bar
C
∂
T
62
×
10
×
10
P �VV
1.487
×1010
β
� ∂∂ T
−6−3
−1
κ = 1.487
62 ××
orIf β and κ are assumed
= 24 bar ◦◦ C−1
∂P V =
10
β
◦ C◦◦◦ temperature
−6
(b)(b)
constant
in
the
10
interval,
then
thethe
b) Si se
supone
que
b
y
k
son
constantes
en
el
intervalo
de
temperatura
de 10 °C,
por
lo tanto la
If
β
and
κ
are
assumed
constant
in
the
10
C
temperature
interval,
then
∂
T
κ
or
=
=
=
24
bar
C
62
×
10
(b)
If
β
and
κ
are
assumed
constant
in
the
10
C
temperature
interval,
then
the
◦
◦
V
−6
(b) If
If ββderived
and κκ in
are
assumed
in
the
10◦C
C temperature
temperature interval,
interval, then
then the
the
(b)
and
are
constant
in
the
10
∂assumed
Tmay
κconstant
62
×
10
equation
(a)
be
written
(V
=
const):
V
ecuación
deducida
en
a)
se
puede
escribir
como
(V
=
constante):
equation
derived
in
(a)
may
be
written
(V
=
const):
(b)
If β and
κ are
constant
10 C temperature interval, then the
equation
derived
in
(a)
may
be
written
(V
const):
equation
derived
in assumed
(a) may
may be
be
writtenin
(Vthe
const):
equation
derived
in
(a)
written
(V
=== const):
(b) If β and
κ are
constant
in(Vthe
10◦◦ C temperature interval, then the
equation
derived
in assumed
(a) may βbe
written
=
const):
β
β
(b)
If
β
and
κ
are
assumed
constant
in
the
10
temperature
interval, then the
�P
==
==
(24)(10)
=C
240
barbar
ββwritten
�P
�T
(24)(10)
=
240
equation derived in (a)
may
be�T
(V
= const):
�P
=
�T
=
(24)(10)
=
240
bar
κbe
�P
=
�T =
= (V
(24)(10)
= 240
240 bar
bar
�T
(24)(10)
=
κβ
equation derived in (a)�P
may=
written
= const):
κ
�P = κβκ �T = (24)(10) = 240 bar
=
P1=
+1 κ+
�P
===
1(24)(10)
+
240
==
241
bar
and
P2P
y
P
�P
1+
240
=
241
bar
and
222 =
=
P
+
�P
=
+
240
=
241
bar
and
P
�P
�T
240
bar
==
P1111β
+�T
�P=
=(24)(10)
+ 240
240=
=240
241bar
bar
and
P
+
�P
=
111 +
=
241
bar
and
PP�P
22 =
κ
and
P2 = P1 κ+ �P = 1 + 240 = 241 bar
Direct
substitution
into
Eq.Eq.
gives:
c) La (c)
sustitución
directa
en la
(3.5)
da:
P(3.5)
�P
= 1 + 240 = 241 bar
and
Pecuación
(c)
Direct
substitution
into
(3.5)
gives:
2 =
1+
(c)
Direct
substitution
into
Eq.
(3.5)
gives:
(c) Direct
Direct substitution
substitutionPinto
into
Eq.
(3.5)
gives:
(c)
(3.5)
P1 +
�Pgives:
= 1 + 240 = 241 bar
and
2 = Eq.
(c) Direct V
substitution
into Eq.
(3.5) gives:
2V
222
−3−3
−6−6
V
−3
−6
(1.487
××
10
)(−20)
−−
(62(62
××
1010
)(9)
==
−0.0303
ln ln
−3
−6
V2=
V
(1.487
10
)(−20)
−0.0303
(c) Direct
substitution
into
Eq.
(3.5)
gives:
2 =
=
(1.487
×
10
)(−20)
−
(62
×
10
)(9)
=
−0.0303
ln
−3
−6)(9)
−3
V1V
= (1.487
(1.487
×Eq.
10−3
)(−20)
− (62
(62 ×
× 10
10−6
)(9) =
= −0.0303
−0.0303
ln
10
)(−20)
−
)(9)
(c) Directln
substitution
into×
(3.5)
gives:
1121 =
V
−6
V121 = (1.487 × 10 )(−20) − (62 × 10 )(9) = −0.0303
ln V
V
−3
−6
V
V2V
3 333−1−1
222ln V21 = (1.487 × 10−3 )(−20) − (62 × 10−6 )(9) = −0.0303
V
−1
0.9702
and
=
(0.9702)(1.287)
1.249
cmcm
g33 gg−1
−1
V2=
V
=
V
(0.9702)(1.287)
=
1.249
2ln
V0.9702
× 10 V2)(−20)
− (62 × 10 =
)(9)
= −0.0303
222 =
=
0.9702
and
V
=
(0.9702)(1.287)
=
1.249
cm
1 = (1.487and
−1
V1V
=
0.9702
and
V
=
(0.9702)(1.287)
=
1.249
cm
y =
0.9702
and
V
=
(0.9702)(1.287)
=
1.249
cm
gg−1
22
1121 V1
V
3
V
=
0.9702
and
V
=
(0.9702)(1.287)
=
1.249
cm
g
V
1
2
1
V2 �V = V − V = 1.249 − 1.287 = −0.038 cm3 g333−1−1
−1 3 −1
Then,
−1
V
2V
1V
Then,
�V
=
1.249
1.287
=
−0.038
gg−1
2−
1=
0.9702
and
V2 =−
(0.9702)(1.287)
=cm
1.249
cm g
Then,
�V
=
V
−
V
=
1.249
−
1.287
=
−0.038
cm
−1
33 g
21 = �V
Then, V
�V =
=V
V2222 −
−and
V1111 =
= 1.249
1.249
−
1.287
=
−0.038
cm
g
Then,
V
−
1.287
=
−0.038
cm
V1 = 0.9702
V
=
(0.9702)(1.287)
=
1.249
cm3 g−1
2 − 1.287 = −0.038 cm3 g−1
Then, V1
�V = V2 − V1 = 1.249
En tal caso,
Then,
�V = V2 − V1 = 1.249 − 1.287 = −0.038 cm3 g−1
Then,
�V = V2 − V1 = 1.249 − 1.287 = −0.038 cm3 g−1
03-SmithVanNess.indd 69
8/1/07 12:52:12
70
70
70
3.2
ECUACIONES
DE ESTADO
VIRIALES
3.2 VIRIAL EQUATIONS
OF
STATE
3.2 VIRIAL EQUATIONS OF STATE
CAPÍTULO
Propiedades
volumétricas
deFluids
fluidos puros
CHAPTER 3.3.Volumetric
Properties
of Pure
CHAPTER 3. Volumetric Properties of Pure Fluids
Las isotermas para gases y vapores, que se localizan arriba y a la derecha de CD en la figura 3.2b), son curvas
Isotherms for gases and vapors, lying above and to the right of C D in Fig. 3.2(b), are relarelativamente simples para las que V disminuye a medida que P se incrementa. Aquí, el producto PV para una
Isotherms
forcurves
gases for
andwhich
vapors,
aboveasand
to the right
of Cthe
D product
in Fig. 3.2(b),
relatively
simple
V lying
decreases
P increases.
Here,
P V forare
a given
T determinada es mucho más constante que cualquiera de sus integrantes, y por tanto más fácil de representar
tively
simple
curvesmore
for which
decreases
as P
increases.
Here, the product
P Vmore
for a easily
given
T
should
be much
nearlyVconstant
than
either
of its members,
and hence
de manera analítica como una función de P. Esto sugiere expresar a PV para una isoterma, por medio de
T should beanalytically
much moreasnearly
constant
of its
members,Pand
more easily
represented
a function
of P.than
Thiseither
suggests
expressing
V forhence
an isotherm
by a
un desarrollo en serie de potencias de P:
represented
as a function of P. This suggests expressing P V for an isotherm by a
power
seriesanalytically
in P:
power series in P:
P V = a + b P + c P 22 + · · ·
2
If b′, ≡
B ��, ′,c etc.,
≡ aCen��, tal
etc.,caso,
then, P V = a + b P + c P + · · ·
Si b ≡ aB
c ≡a aC
If b ≡ a B � , c ≡ aC � , etc., then,
P V = a(1 + B �� P + C �� P 22 + D �� P 33 + · · ·)
(3.6) (3.6)
�
� 2
� 3
P
V
=
a(1
+
B
P
+
C
P
+
D
P
+
·
·
·)
(3.6)
��, Cson
��, etc.,
donde a,where
B ′, Ca,′, B
etc.,
constantes
para una
y especie
química
are constants
for atemperatura
given temperature
and
a givendeterminadas.
chemical species.
� , etc.,
Enwhere
principio,
derecho
de
ecuación
(3.6)
es infinite
una serie
infinita.
Dechemical
cualquier
modo,
en la práctiB � ,elClado
are
constants
for
given
and
a given
species.
Ina,principle,
the
right
sidela of
Eq.a(3.6)
istemperature
an
series.
However,
in practice
a finite
ca se emplea
un
número
finito
de In
términos.
hecho,
la información
muestra
a presiones
In of
principle,
the
right
side
an
infinite
However,
inque
practice
a finite
number
terms is
used.
fact,ofPEq.
VDe
T(3.6)
dataisshow
that atseries.
low PVT
pressures
truncation
after
twobajas el
truncamiento
después
de is
dosused.
términos
con
proporciona
satisfactorios.
number
of terms
In fact,
Pfrecuencia
V T data show
that at resultados
low pressures
truncation after two
terms
often
provides
satisfactory
results.
terms often provides satisfactory results.
Ideal-Gasde
Temperatures;
Universal
Gas
Constant
Temperaturas
gases ideales;
constante
universal
de los gases
Ideal-Gas�� Temperatures;
Universal
Gas
Constant
��
Parameters B , C , etc., in Eq. (3.6) are species dependent and functions of temperature, but
� , etc.,
Los parámetros
B ′,a BC
en
ecuación
son
de and
laof
temperatura
y dependen
de la
Parameters
Cetc.,
in
Eq. (3.6)(3.6),
are
species
dependent
functions
of
temperature,
butespecie,
parameter
is� ,′,found
byla
experiment
to be
the funciones
same
function
temperature
for all chemical
pero, enspecies.
forma experimental,
seby
demuestra
que
elof
parámetro
a es
la misma
de la
parameter
a isisfound
by
experiment
to be
the
same function
of temperature
for
alltemperatura
chemical
This
shown
measurements
volume
as
a function
offunción
P for various
gases at para
todas lasconstant
especiestemperature.
químicas.
Esto
muestra
mediciones
del
como
una
función
species.
This
is shown
bysemeasurements
of de
volume
as a of
function
forfour
various
gases
atde P paFigure
3.4, fora través
example,
is a plot
P Vvolumen
vs.ofPPfor
gases
at the
ra diferentes
gases
atemperature
temperatura
Porlimiting
ejemplo,
desame
PV
en
constant
temperature.
Figure
3.4,The
for
example,
islaafigura
plotP of
Pes
VPuna
vs.
P
forthe
four
gases
at
triple-point
ofconstante.
water.
value
of
V3.4as
→ gráfica
0 is
forfunción
allthe
of de P,
para cuatro
gases en
la temperatura
del punto
del agua.
Todos
el mismo
valor
límite de
triple-point
of water.
limiting
value
of Pbecomes:
Vlos
as gases
P →tienen
0 is the
same for
all of
the
gases.
Intemperature
this
limit (denoted
by The
thetriple
asterisk),
Eq. (3.6)
PV conforme
P →In0.this
En limit
este límite
(denotado
por∗el asterisco),
labecomes:
ecuación 3.6 será:
the gases.
(denoted
by the asterisk),
Eq.
(3.6)
(P V )∗ = a = f (T )
(P V )∗ = a = f (T )
Figura 3.4: (PV)*, el valor límite
limiting
value
ofVes)P∗V
as
de PV conforme
P
→(P0,
Figure 3.4:
, the
P → 0,
independent
independiente
delis
gas.
limiting
value
of P V asof the
gas.
P → 0, is independent of the
gas.
0
0
H2
PV/cm3 bar mol1
Figure 3.4: (P V )∗∗, the
3 bar
3 bar
�1 �1
PV/cm
PV/cm
molmol
H22
H
N22
N
Air
Air
N
2
�1
� 22,711.8
22,711.8
cm33 bar
bar mol
mol3�1
(PV)*t*t �
(PV)
22 711.8
cm bar mol1
(PV)*  cm
t
(PV)*t � 22,711.8 cm3 bar mol�1
Air
O22
O
O2
H2
N2
Aire
O2
� 273.16
273.16 K
K�
� triple
triple point
point of
of water
water
TT �
T  273.16
K  punto
triple del agua
T � 273.16 K � triple point of water
0
P
P
P
This property
gases
basis for establishing
an absolute
temperature
scale. All
that Todo
Esta propiedad
de losof
gases
esiselthe
fundamento
para establecer
una escala
de temperatura
absoluta.
This es
property
of gases
is the basis
establishing
an
absolute
scale.
thatespecíis requiere
required
is la
theasignación
arbitrary
assignment
functional
relationship
) and the
assignment
lo que se
arbitraria
deoffor
lathe
relación
funcional
f (T) y temperature
laf (T
asignación
de
unAll
valor
issolo
required
thelaarbitrary
themás
functional
f (T )internacional
andthe
theone
assignment
a specific
value
to a single
point onofthe
scale.
The relationship
simplest procedure,
adopted
fico a unof
puntoisde
escala.
Elassignment
procedimiento
simple,
adoptado
a nivel
para definir la
of a specific
value
toes:a the
single
pointscale
on the
scale.
internationally
to
define
Kelvin
(Sec.
1.5): The simplest procedure, the one adopted
escala Kelvin
(sección
1.5),
internationally to define the Kelvin scale (Sec. 1.5):
03-SmithVanNess.indd 70
8/1/07 12:52:17
3.2. Virial
VirialEquations
EquationsofofState
State
71
71
3.2.
71 71
Virial
Equations
State
3.2.
Virial
of State
3.2. 3.2.
Ecuaciones
deEquations
estadoofviriales
71
3.2. Virial Equations of State
71
Makes(P
(PVV)∗)∗∗∗directly
directlyproportional
proportionaltotoTT, ,with
withRRasasthe
theproportionality
proportionalityconstant:
constant:
•• Makes
• Makes
(P V(P)∗de
proportional
to Tto, with
R asRthe
proportionality
constant:
• Makes
Vdirectly
)manera
directly
T , with
as the
constant:
• Establecer
(PV)*
queproportional
sea directamente
proporcional
aproportionality
T, con R como
la constante de propor• Makes (P V )∗ directly proportional to T , with
R
as
the
proportionality
constant:
∗
cionalidad:
(PVV)∗)∗∗ ==aa≡≡RT
RT
(3.7)
(P
(3.7)
(P V(P)∗V=
(3.7)
) a=≡a RT
≡ RT
(3.7)
∗
(P V ) = a ≡ RT
(3.7)
(3.7)
Assignsthe
thevalue
value273.16
273.16KKtotothe
thetemperature
temperatureofofthe
thetriple
triplepoint
pointofofwater
water(denoted
(denotedby
by
•• Assigns
• Assigns
the the
value
273.16
K toKthe
temperature
of the
triple
point
of water
(denoted
by by
• Assigns
value
273.16
to the
temperature
of the
triple
point
of water
(denoted
subscript
t):273.16
subscript
t):
• Asignar
el valor
de
K a la temperatura del punto
triple
delpoint
agua of
(denotado
por el subíndice
t):
• subscript
Assigns
the
of the
triple
water (denoted
by
t): value
subscript
t): 273.16 K to the temperature
∗
∗
∗
(PVV)t)∗t ==RR××273.16
273.16KK
(3.8)
(P
(3.8)
subscript t):
(P V(P)∗tV=
K K
(3.8)
)t R=×R273.16
× 273.16
(3.8) (3.8)
∗ t
(P V )t = R × 273.16 K
(3.8)
DivisionofofEq.
Eq.(3.7)
(3.7)by
byEq.
Eq.(3.8)
(3.8)gives:
gives:
Division
El resultado
de
la Eq.
ecuación
(3.7) por la (3.8) nos da:
Division
of la
Eq.
(3.7)
bydeEq.
(3.8)
gives:
Division
ofdivisión
Eq.
(3.7)
by
(3.8)
gives:
Division of Eq. (3.7) by Eq. (3.8) gives:
/K
(PVV)∗)∗∗
TT/K
(P
(P V(P)∗V )∗∗∗ ==T /K
T /K
(P∗VV=
273.16
K
(P
)t)∗t∗t =T273.16
/K K K
(P V
V
(P
(P))∗tV =
)t 273.16
273.16
K
∗
(P V )t
273.16 K
(PVV)∗)∗∗
(P
/K==273.16
273.16
(3.9) (3.9)
o
(P V(P)∗V )∗∗∗
oror
TT/K
(3.9)
∗VV))∗t
(P
or or
T /K
= 273.16
(3.9)
T /K
= 273.16
(3.9)
(P
(P
V
)
t
∗
∗
t
or
T /K = 273.16 (P V(P)t∗V )t
(3.9)
(P V )t
Equation
(3.9)
establishes
the
Kelvin
temperature
scale
throughout
thetemperature
temperature
range
for
Equation
establishes
the
Kelvin
temperature
throughout
the
range
for
La ecuación
(3.9)(3.9)
establece
la escala
Kelvin
de
temperaturascale
en todo
el intervalo
de
temperatura
para
el que se
Equation
(3.9)
establishes
Kelvin
temperature
scale
throughout
the the
temperature
range
for for
Equation
(3.9)
establishes
the
Kelvin
temperature
scale
throughout
temperature
range
∗)∗∗the
are
experimentally
accessible.
which
values
of
(P
V
are
experimentally
accessible.
which
values
of
(P
V
)
tieneEquation
acceso experimental
a los
valores
de temperature
(PV)*.
∗
∗
(3.9)
establishes
the
Kelvin
scale
throughout
the
temperature
range
for
experimentally
accessible.
which
values
of (P
experimentally
accessible.
which
values
of V(P) Vare
) are
The
state
thelimiting
limiting
condition
where
→
deserves
discussion.
The
∗aaare
El estado
deof
un
gas
lagas
condición
límite,accessible.
donde
P →where
0 merece
cierto
análisis.discussion.
A
medida que
disminuThe
state
ofof)en
gas
atatthe
condition
PP →
00deserves
The
experimentally
which
values
(P
V
The
state
of
a
gas
at
the
limiting
condition
where
P
→
0
deserves
discussion.
The
The
state
of
a
gas
at
the
limiting
condition
where
P
→
0
deserves
discussion.
The
molecules
making
up
a
gas
become
more
and
more
widely
separated
as
pressure
is
decreased,
ye la presión
sobre
un
gas,
las
moléculas
del
mismo
se
separan
cada
vez
más
y
el
volumen
de
las
moléculas
se
molecules
making
up
a
gas
become
more
and
more
widely
separated
as
pressure
is
decreased,
The making
state
of up
a gas
thebecome
limiting
condition
where
Pseparated
→
0 deserves
discussion.
The
molecules
aup
gas
become
more
andand
more
widely
assmaller
pressure
is decreased,
molecules
making
aat
gas
more
more
widely
separated
as
pressure
is decreased,
and
the
volume
ofthe
the
molecules
themselves
becomes
asmaller
smaller
and
fraction
ofthe
thetotal
total
convierte
en
una
fracción
cada
vez
más
pequeña
del
volumen
total
ocupado
por
el
gas.
Además,
las
fuerzas
and
the
volume
of
molecules
themselves
becomes
a
and
smaller
fraction
of
molecules
making
up
a gas
become
more and
more
widely
separated
assmaller
pressure
is of
decreased,
and
the
volume
of the
molecules
themselves
becomes
a smaller
andand
smaller
fraction
the
total
and
the
volume
of the
molecules
themselves
becomes
a smaller
fraction
of es
the
totalvez más
volume
occupied
bythe
the
gas.
Furthermore,
the
forces
attraction
between
molecules
become
de atracción
entre
lasofmoléculas
se
vuelven
más pequeñas
debido
a que
lasmaller
distancia
entre
ellas
cada
occupied
by
gas.
Furthermore,
the
forces
ofofattraction
between
molecules
become
andvolume
the volume
thethe
molecules
themselves
becomes
aofsmaller
andbetween
fraction
of
the
total
volume
occupied
by
gas.
Furthermore,
the
forces
attraction
molecules
become
volume
occupied
by
the
gas.
Furthermore,
the
forces
of
attraction
between
molecules
become
eversmaller
smaller
because
theconforme
increasing
distances
betweenthem
them
(Sec.16.1).
16.1).
thelimit,
limit,indetermigrande
(sección
16.1).
Enthe
el límite,
Pthe
→
0, las moléculas
están
separadas
por
ever
because
ofofthe
increasing
distances
between
(Sec.
InIndistancias
the
asas
volume
occupied
by
Furthermore,
forces
of attraction
between
molecules
become
ever
smaller
because
of gas.
the
increasing
distances
between
them
(Sec.
16.1).
Inbecome
the
limit,
as as
ever
smaller
because
of
the
increasing
distances
between
them
(Sec.
16.1).
In
the
limit,
P
→
0,
the
molecules
are
separated
by
infinite
distances.
Their
volumes
negligible
nadas.
Sus
volúmenes
se vuelven
despreciables
en comparación
con
el volumen
total
dellimit,
gas, yaslas fuerzas
P smaller
→
0, the
molecules
separateddistances
by infinite
distances.
Their
volumes
become
negligible
ever
because
ofare
theare
increasing
between
them
(Sec.
16.1).
In
the
P →
0, the
molecules
separated
by
distances.
Their
volumes
become
negligible
Pcompared
→
0, the
molecules
are
separated
by
infinite
distances.
Their
volumes
become
negligible
with
the
totalvolume
volume
ofthe
theinfinite
gas,
and
the
intermolecular
forces
approach
zero.
Theseestableintermoleculares
tienden
a
cero.
Estas
condiciones
definen
un
estado
de
gas
ideal,
y
la
ecuación
(3.9)
compared
with
the
total
of
gas,
and
the
intermolecular
forces
approach
zero.
These
P →
0, the
molecules
are volume
separated
by
infinite
distances.
Their forces
volumes
become
negligible
compared
with
the
total
volume
of
the
gas,
and
the
intermolecular
approach
zero.
These
compared
with
the
total
of
the
gas,
and
the
intermolecular
forces
approach
zero.
These
conditions
define
an
ideal-gas
state,
and
Eq.
(3.9)
establishes
the
ideal-gas
temperature
scale.
ce lacompared
escala
dewith
temperatura
del
gas ideal.
La
constante
deestablishes
proporcionalidad
Rapproach
de latemperature
ecuación
(3.7)
se conoce
conditions
define
an ideal-gas
and
Eq.the
(3.9)
the
ideal-gas
scale.
theantotal
volume
ofstate,
theand
gas,
and
intermolecular
forces
zero. These
conditions
define
ideal-gas
state,
Eq.
(3.9)
establishes
the the
ideal-gas
temperature
scale.
conditions
define
an
ideal-gas
state,
and
Eq.
(3.9)
establishes
ideal-gas
temperature
scale.
The
proportionality
constant
R
in
Eq.
(3.7)
is
called
the
universal
gas
constant.
Its
numerical
como
constante
universal
de
los
gases.
Su
valor
numérico
se
determina
mediante
la
ecuación
(3.8)
a
The
proportionality
constant
R
in
Eq.
(3.7)
is
called
the
universal
gas
constant.
Its
numerical
conditions
define anconstant
ideal-gas
Eq.
establishes
the ideal-gas
temperature
scale. partir de
The
proportionality
Rstate,
inREq.
(3.7)
is(3.9)
called
the the
universal
constant.
Its numerical
The
proportionality
constant
inand
Eq.
(3.7)
is called
universal
constant.
Its numerical
value
isdetermined
determined
bymeans
means
of
Eq.
(3.8)
from
experimental
Pgas
VTTgas
data:
información
PVT
experimental:
value
is
by
of
Eq.
(3.8)
from
experimental
P
V
data:
Thevalue
proportionality
constant
RofinEq.
Eq.
(3.7)
is from
called
the universal
Its numerical
value
is determined
by
means
(3.8)
from
experimental
P VPTgas
is determined
by means
of
Eq.
(3.8)
experimental
Vdata:
T constant.
data:
value is determined by means of Eq. (3.8) from experimental
P
V
T
data:
∗
∗
(PVV))∗t
(P
(P V(P)∗tV )t∗tt
RR==
273.16
K
R =R =(P273.16
V )∗tK K
273.16
K
R = 273.16
273.16
K pressure,
Because
dataen
cannot
fact
betaken
taken
zero
pressure,
datataken
taken
finite
are
Because
PPVVTTPVT
data
cannot
ininfact
be
atat
zero
data
atat
finite
pressures
are
Ya que
la
información
realidad
no taken
es
a unadata
presión
cero,
toma
apressures
presiones
Because
P VPTVdata
cannot
in fact
be
at zero
pressure,
taken
at finite
pressures
arevalue
Because
Ttodata
cannot
in fact
be posible
taken
attomarse
zero
pressure,
data
taken
atsefinite
pressures
arefinitas y
extrapolated
the
zero-pressure
state.
Determined
as
indicated
by
Fig.
3.4,
the
accepted
extrapolated
to
the
zero-pressure
state.
Determined
as
indicated
by
Fig.
3.4,
the
accepted
value
después
se extrapola
althe
estado
dein3presión
cero.
Determinado
como
se
en3.4,
la 1figura
3.4,value
elare
valor
Because
P V∗∗to
T the
data
cannot
fact
bestate.
taken
at zero as
pressure,
data
taken
at
finite
pressures
extrapolated
zero-pressure
state.
Determined
indicated
byindica
Fig.
3.4,
the
accepted
extrapolated
to
zero-pressure
Determined
asfollowing
indicated
by Fig.
the
accepted
value acepta−1
−1
−1
V)22
)∗∗t is711.8
is22,711.8
22,711.8
cm333mol
bar
mol
leading
the
value
R:
*(P
3 bar
–1mol
111
ofof(P
bar
, ,leading
totoas
the
following
value
ofofR:
do de
(PV)
es
cm
,
lo
cual
nos
lleva
al
siguiente
valor
de
∗Vis
3cm
−1
1R:
t
−1
extrapolated
to
the
zero-pressure
state.
Determined
indicated
by
Fig.
3.4,
the
accepted
value
t
t
of
(P
V
)
22,711.8
cm
bar
mol
,
leading
to
the
following
value
of
R:
of (PtV )t is 22,711.8 cm bar mol , leading to the following value of R:
of (P V )∗t is 22,711.8 cm3 bar mol−1 , 3leading to−1the following value of R:1
−1
barmol
mol−1
22,711.83cm
cm333 bar
22,711.8
3
−1 K
−1
−1 −1 = 83.1447 cm
−1
−1
barmol
mol−1
bar bar
molmol
cm cm
= 83.1447 3cm333 bar
RR22,711.8
== 22,711.8
K−1
−1 K
−1−1
3
−1
273.16
K
= 83.1447
cm cm
R =R 22,711.8
bar bar
molmol
= 83.1447
=
K−1
273.16
K
bar
mol
cm
273.16
K K
273.16
= 83.1447 cm3 bar mol−1 K−1
R=
273.16
K
Through
the
use
of
conversion
factors,
R
may
beexpressed
expressedininvarious
variousunits.
units.Commonly
Commonlyused
used
Through the use of conversion factors, R may be
Through
theare
use
of conversion
factors,
R may
be expressed
in various
units.
Commonly
usedused
Through
the
use
of
conversion
factors,
R may
be expressed
in various
units.
Commonly
values
given
in
Table
A.2
of
App.
A.
values
arede
given
in
Table
A.2factors,
of App.
A.
Mediante
elare
uso
factores
deA.2
conversión,
se
puede
expresar en
unidades.
En general,
Through
the
use
ofTable
conversion
may
be expressed
in diversas
various units.
Commonly
usedse usan los
values
given
in
of App.
A.RR A.
values
are
given
in Table
A.2
of App.
valores
dados
la tabla
A.2 del
A.
values
areen
given
in Table
A.2apéndice
of App. A.
1 http://physics.nist.gov/constants.
11http://physics.nist.gov/constants.
1 http://physics.nist.gov/constants.
1 http://physics.nist.gov/constants.
http://physics.nist.gov/constants.
http://physics.nist.gov/constants.
1 http://physics.nist.gov/constants.
03-SmithVanNess.indd 71
8/1/07 12:52:20
72
CHAPTER
3.3.Volumetric
Volumetric
Properties
of Pure
Pure
Fluids
CAPÍTULO
Propiedades
volumétricas
deFluids
fluidos puros
CHAPTER 3.
CHAPTER
Properties
of
CHAPTER
72
72
Dos formas
de laof
ecuación
Two Forms
Forms
of
the Virial
Virialvirial
Equation
Two
the
Equation
A useful
useful
auxiliarytermodinámica
thermodynamicauxiliar
property
ismediante
defined by
by
the equation:
equation:
Se define
una propiedad
útilis
la the
ecuación:
A
auxiliary
thermodynamic
property
defined
PV
Z≡
≡ PV
Z
RT
RT
RT
(3.10) (3.10)
(3.10)
This dimensionless
dimensionless
ratio
is called
called
the
compressibility
factor.
With
this definition
definition
and
with
Esta relación
adimensional ratio
se llama
factorthe
de compressibility
compresibilidad.factor.
Con esta
definición
y con aand
= RT
[ecuación
This
is
With
this
with
a
=
RT
[Eq.
(3.7)],
Eq.
(3.6)
becomes:
(3.7)], la
ecuación
(3.6)
será:
a = RT [Eq. (3.7)], Eq. (3.6) becomes:
Z=
= 11 +
+B
B���� P
P+
+C
C ���� P
P2222 +
+D
D���� P
P3333 +
+ ·· ·· ··
Z
(3.11) (3.11)
(3.11)
2
2
An
alternative
expression
for
Z is
also
in
common
use:
An alternative
alternative
expression
for
is also
also
in common
common
use:
22
An
expression
is
in
Una expresión
alternativa
tambiénfor
deZZuso
común
para Z es:use:
Z=
= 11 +
+
Z
C
B
D
C + D
B+ C
D + ···
B
+ V 222 +
3 + ···
V
V
V
V2
V 333
(3.12) (3.12)
(3.12)
� C′,��, etc.,
� etc.,
Ambas Both
ecuaciones
conocen are
como
expansiones
viriales,
y los
parámetros
B ′, CB
y B,and
C, D, etc.,
of
these
equations
known
as
virial
expansions,
and
the
parameters
D
Both
of these
theseseequations
equations
are
known
as virial
virial expansions,
expansions,
and
the parameters
parameters
B′,���,,,DC
C�� ,, D
D���,,, etc.,
etc.,
and
Both
of
are
known
as
and
the
B
and
�
se llaman
coeficientes
viriales.
Los
parámetros
B ′ Parameters
y B son los B
segundos
coeficientes
viriales;
C ′ y C son los
B,
C,
D,
etc.,
are
called
virial
coefficients.
B
are
second
virial
coefficients;
��� and
B,
C,
D,
etc.,
are
called
virial
coefficients.
Parameters
B
and
B
are
second
virial
coefficients;
B,� C, D, etc., are called virial coefficients. Parameters B and B are second virial coefficients;
tercerosC
viriales,
etc.
Para ciertoetc.
gasFor
los aacoeficientes
viriales
son funciones
C
are
third
virial
coefficients;
given
gas
the
virial
coefficients
are
functions
��� and
Ccoeficientes
and C
C are
are third
third
virial
coefficients;
etc.
For
given gas
gas the
the
virial coefficients
coefficients
areúnicamente
functions de la
C
and
virial
coefficients;
etc.
For
a given
virial
are
functions
temperatura.
of
temperature
only.
of
temperature
only.
of temperature only.
Los dosThe
conjuntos
ecuaciones
(3.12)
estánasrelacionados
de la manera sitwo sets
setsdeof
ofcoeficientes
coefficientsen
in las
Eqs.
(3.11) and
and(3.11)
(3.12)yare
are
related
follows:
The two
coefficients
in
Eqs.
(3.11)
(3.12)
related
as follows:
guiente:
B
B���� =
= B
B
RT
RT
RT
(3.13a)
(3.13a)
a)
C ���� =
=
C
C−
−B
B2222
C
(RT ))2222
(RT
(3.13b)
(3.13b)
b)
D−
− 3BC
3BC +
+ 2B
2B3333
�� = D
D
�
D� =
(RT ))3333
(RT
(3.13c)
(3.13c)
c)
To derive
derive these
these relations,
relations, we
we set
set Z
Z=
=P
P V /RT
/RT in
in Eq.
Eq. (3.12), and
and solve
solve for
for P.
P. This
This allows
allows elimelimTo
Para deducir
estas relaciones,
establecemos
ZV
= PV/RT
en la(3.12),
ecuación (3.12)
y resolvemos
para P. Esto perination
of
P
on
the
right
of
Eq.
(3.11).
The
resulting
equation
reduces
to
a
power
series
in
1/V
inationPofdePlaonderecha
the right
(3.11).(3.11).
The resulting
equation
reduces
a power
series
in de
1/V
mite eliminar
deofterm
laEq.
ecuación
La ecuación
resultante
setoreduce
a una
serie
potencias
which
may
be
compared
by
term
with
Eq.
(3.12)
to
yield
the
given
relations.
They
hold
which
may
be compared
term
by term
with Eq.
(3.12)
to yield(3.12)
the given
relations.
They
hold
en 1/V que
es
posible
compararse
término
a
término
con
la
ecuación
para
lograr
las
relaciones
que se
exactly
only
for
the
two
virial
expansions
as
infinite
series,
but
are
acceptable
approximations
exactlySe
only
for the
theprecisamente
two virial
virial expansions
expansions
as
infinite
series, but
but viriales
are acceptable
acceptable
approximations
exactly
only
for
two
as
infinite
series,
are
approximations
proporcionan.
cumplen
sólo
para
las
dos
expansiones
como
series
infinitas,
pero
son
for
the
truncated
forms
used
in
practice.
for the
the truncated
truncated
forms
used
in practice.
practice.
for
forms
in
aproximaciones
aceptables
paraused
las formas
truncadas que se emplean en la práctica.
Many other
other equations
equations of state
state have
have been
been proposed
proposed for
for gases,
gases, but
but the virial
virial equations
equations are
are
Se han Many
propuesto
muchas otrasofecuaciones
de estado
para los gases,
perothe
las ecuaciones
viriales son las
the
only
ones
firmly
based
on
statistical
mechanics,
which
provides
physical
significance
to
the only
based on
statistical
mechanics,
which provides
physical un
significance
to
únicas que
tienenones
unafirmly
base teórica
firme
en la mecánica
estadística,
que proporciona
significado
físico a
the
virial
coefficients.
Thus,
for
the
expansion
in
1/V
,
the
term
B/V
arises
on
account
of
the virialviriales.
coefficients.
Thus,
for the
in 1/V
, theel2term
B/VB/V
arises
ona account
oflas inter­
los coeficientes
De esta
manera,
paraexpansion
el desarrollo
en 1/V,
término
surge
causa
de
term, on
on account of
of three-body
three-body
interactions between
between pairs
pairs of
of molecules
molecules (Sec.
(Sec. 16.2);
16.2); the C/V
C/V 222 term,
interactions
acciones
entre pares de moléculas
(sección 16.2);
el términothe
C/V 2, con
motivoaccount
de las interacciónes
entre un
interactions;
etc.
Because
two-body
interactions
are
many
times
more
common
than
threeinteractions;
Because
two-body interactions
are many
more common
than que
threeconjunto
de tres, etc.etc.
Ya que
las interacciones
entre conjuntos
de dostimes
son mucho
más comunes
entre tres,
body interactions,
interactions, and
and three-body
three-body interactions
interactions are
are many
many times
times more
more numerous
numerous than
than four-body
y éstas body
a su vez
son más numerosas
que las correspondientes
entre conjuntos
de cuatro, etc.,four-body
las contribuciointeractions,
etc.,
the
contributions
to
Z
of
the
successively
higher-ordered
terms
decrease
etc., de
theorden
contributions
to Z of the con
successively
higher-ordered terms decrease
nes a Z interactions,
de estos términos
superior disminuyen
mucha rapidez.
rapidly.
rapidly.
2
Proposed
by
H.
Kamerlingh
Onnes,
“Expression
of
the
Equation
of
State
of
Gases
and
Liquids
by
Means
of
222Proposed
Propuesto
Proposed
by H.
H. Kamerlingh
Kamerlingh
Onnes, “Expression
“Expression
of the
the Equation
Equation
of
Stateand
of Liquids
Gases and
and
Liquidsofby
by
MeansCommunicaof
by
Onnes,
of
State
of
Gases
Liquids
Means
of
por H. Kamerlingh
Onnes, “Expression
of the Equation
of
State ofof
Gases
by Means
Series”,
Proposed
by H. Kamerlingh
Onnes,
“Expression
of
the
Equation ofof State
of no.
Gases
and
Liquids by Means of
Series,”
Communications
from
the
Physical
Laboratory
of
the
University
Leiden,
71,
1901.
Series,”
Communications
from
the
Physical Laboratory
Laboratory
of the
the
University
of Leiden,
Leiden, no.
no. 71,
71, 1901.
1901.
Communications
Physical
of
University
tions fromSeries,”
the Physical
Laboratoryfrom
of thethe
University
of
Leiden, núm.
71,
1901. of
Series,”
Communications
from
the
Physical
Laboratory
of
the
University
of
Leiden,
no.
71,
1901.
03-SmithVanNess.indd 72
8/1/07 12:52:24
3.3. The
The Ideal
Ideal Gas
Gas
3.3.
3.3. El3.3.
gas ideal
The
3.3.
The Ideal
Ideal Gas
Gas
3.3
73
73
73
73
73
3.3 GAS
THE IDEAL GAS
EL
3.3 THEIDEAL
IDEAL GAS
2, ,,etc.,
Because
the terms
terms
B/V
C/Vde22,,laetc.,
etc.,
of the
the virial
virial
expansion
[Eq.
(3.12)]
arise
on account
account
of
the
B/V
C/V
of
(3.12)]
arise
on
of
Ya que Because
los
términos
B/V, C/V
expansión
virialexpansion
[ecuación [Eq.
(3.l2)]
surgen
a causa
de las interaccio22, etc., of the virial expansion [Eq. (3.12)] arise on account of
Because
the
terms
B/V
,
C/V
Because
the
terms
B/V
,
C/V
,
etc.,
of
the
virial
expansion
[Eq.
(3.12)]
arise
on
account
molecular
interactions,
the
virial
coefficients
B,
C,
etc.,
would
be
zero
were
no
such
interacmolecular
interactions,
the
virial
coefficients
B,
C,
etc.,
would
be
zero
were
no
such
interacnes moleculares, los coeficientes viriales B, C, etc., serían cero donde no existen estas interacciones. of
Con esto,
molecular
interactions,
the
coefficients
B,
molecular
interactions,
the virial
virial
coefficients
B, C,
C, etc.,
etc.,
would be
be zero
zero were
were no
no such
such interacinteractions
to
exist,
and
the virial
virial
expansion
would reduce
reduce
to: would
tions
to
exist,
the
expansion
would
to:
la expansión
virial
seand
reduce
a:
tions
tions to
to exist,
exist, and
and the
the virial
virial expansion
expansion would
would reduce
reduce to:
to:
= 11 o PPPV
= RT
RT
Z=
=
1
or
=
RT
ZZ
or
VV =
ZZ =
or
P
= 11
or
PV
V=
= RT
RT
De hecho,
las interacciones
moleculares
existen
y influence
ejercen
influencia
sobrebehavior
el comportamiento
observaMolecular
interactions
do in
in fact
fact exist,
exist,
and
influence
the observed
observed
behavior
of real
real gases.
gases.
Molecular
interactions
do
and
the
of
do de gases
reales.
Conforme
disminuye
la
presión
a
temperatura
constante,
V
aumenta
y
las
contribuciones
de
Molecular
interactions
do
in
fact
exist,
and
influence
the
observed
behavior
of
real
gases.
Molecular
interactions
do intemperature,
fact exist, andVVinfluence
observed
behavior of
gases.
As pressure
pressure
is reduced
reduced
at constant
constant
temperature,
increasesthe
and
the contributions
contributions
of real
the terms,
terms,
As
is
at
increases
and
the
the
2, D/V
3, etc., en la ecuación (3.12) disminuyen. Para P → 0, Z se aproxima a uno, pero
2
3
2
3
los términos
B/V,
C/V
As
pressure
is
reduced
at
constant
temperature,
V
increases
and
the
contributions
of
the
terms,
As
pressure
reduced
temperature,
increases
the contributions
terms,
D/V
...atin
inconstant
Eq. (3.12),
(3.12),
decrease.VFor
For
→ 0,
0,and
approaches
unity,of
notthe
because
B/V
C/V is
,, D/V
,, ...
Eq.
decrease.
PP →
ZZ approaches
unity,
not
because
B/V
,, C/V
22, D/V
33,coeficientes
no por of
algún
viriales
sino
porque
tiende
infinito. Thus
Por
loin
en elas
...
(3.12),
decrease.
For
0,
unity,
not
B/V
,, cambio
C/V
, en
D/V
,virial
... in
in Eq.
Eq.
(3.12),
decrease.
For P
P
→
0, ZZ aapproaches
approaches
unity,
not
because
B/V
C/V
of
any
change
in los
the
virial
coefficients,
but
because
VV→
becomes
infinite.
Thus
intanto,
the because
limit
aslímite, a
any
change
in
the
coefficients,
but
because
V
becomes
infinite.
the
limit
medidaof
que
P
→
0,
la
ecuación
de
estado
se
reduce
a
la
misma
forma
simple
que
para
el
caso
hipotético
any
change
in
the
virial
coefficients,
but
because
V
becomes
infinite.
Thus
in
the
limit
as
of
any0,
in the virial
coefficients,
V becomes
Thus in the limit
as donP→
→
0,change
the equation
equation
of state
state
reduces to
tobut
the because
same simple
simple
form as
asinfinite.
for the
the hypothetical
hypothetical
case of
of
P
the
of
reduces
the
same
form
for
case
de B = C
=
·
·
·
=
0;
es
decir,
P
→
0,
the
equation
of
state
reduces
to
the
same
simple
form
as
for
the
hypothetical
case
of
P=
→C
=
C0,=
=the
= 0;
0; i.e.,
i.e.,of state reduces to the same simple form as for the hypothetical case of
BB
·· ·· ··equation
=
B
B=
=C
C=
= ·· ·· ·· =
= 0;
0; i.e.,
i.e.,
→ 11
or
→ RT
RT
ZZZ→
PPVV →
→
→ 1 or
o PV
RT
ZZ →
or
P
→ 11
or
PV
V→
→ RT
RT
We know
know from
from the
the phase
phase rule
rule that
that the
the internal
internal energy
energy of
of aa real
real gas
gas is
is aa function
function of
of
We
De
la regla
dewell
la fase
sabemos
querule
la energía
interna
de
un
gas real
unafrom
función
tanto
de la
We
know
the
phase
that
internal
energy
of
aaesreal
gas
is
aa function
of
Weas
know
from
the
phase
rule
that the
the
internal
energy
ofresults
real
gas
isforces
function
ofpresión
pressure
as
as of
of
temperature.
This
pressure
dependency
results
forces
between
pressure
well from
as
temperature.
This
pressure
dependency
from
between
como depressure
la
temperatura.
Esta
dependencia
de
la
presión
aparece
como
resultado
de
las
fuerzas
entre
las
as
well
as
of
temperature.
This
pressure
dependency
results
from
forces
between
pressure
as wellIf
of forces
temperature.
dependency
results
forces
the
molecules.
Ifas
such
forces did
did not
not This
exist,pressure
no energy
energy
would be
be required
requiredfrom
to alter
alter
the between
averagemo­lécu­
the
molecules.
such
exist,
no
would
to
the
average
las. Si estas
fuerzas
no
existieran,
no
se
necesitaría
energía
alguna
para
alterar
la
distancia
intermolecular
the
molecules.
If
such
forces
did
not
exist,
no
energy
would
be
required
to
alter
the
average
the molecules. distance,
If such forces
did not no
exist,
no energy
be required
alter volume
the average
intermolecular
distance,
and therefore
therefore
no energy
energy
would would
be required
required
to bring
bringtoabout
about
volume
and prointermolecular
and
would
be
to
and
medio y,
en
consecuencia,
no
se
requeriría
energía
para
originar
cambios
de
volumen
y
presión
en
intermolecular
distance,
and
therefore
no
energy
would
be
required
to
bring
about
volume
and
intermolecular
distance,
therefore
no energy would
be required
tothe
bring
aboutof
andun gas a
pressure
changes
in aa gas
gasand
at constant
constant
temperature.
We conclude
conclude
that in
in
the
absence
ofvolume
molecular
pressure
changes
in
at
temperature.
We
that
absence
molecular
temperatura
constante.
Concluimos
que
en
ausencia
de
interacciones
moleculares,
la
energía
interna
del gas depressure
changes
in
a
gas
at
constant
temperature.
We
conclude
that
in
the
absence
of
molecular
pressure changes
in a gas energy
at
constant
thatonly.
in theThese
absence
of molecular
interactions,
the internal
internal
energy
of aatemperature.
gas depends
dependsWe
on conclude
temperature
only.
These
considerations
interactions,
the
of
gas
on
temperature
considerations
pende exclusivamente
de
la
temperatura.
Estas
consideraciones
acerca
del
comportamiento
de
un
gas
interactions,
the
internal
energy
of
a
gas
depends
on
temperature
only.
These
considerations
interactions,
theof
energy of
gaswhich
depends
on temperatureforces
only. exist
These
considerations
of
the behavior
behavior
ofinternal
hypothetical
gasa in
in
which
no intermolecular
intermolecular
forces
exist
and
of aa real
real gas
gas hipoté­
of
the
aa hypothetical
gas
no
and
of
tico en of
el
que
no
existen
fuerzas
moleculares,
y
de
un
gas
real
en
el
límite
conforme
la
presión
tiende
the
behavior
of
a
hypothetical
gas
in
which
no
intermolecular
forces
exist
and
of
a
real
gas
of the
behavior
of a hypothetical
gaszero
in which
no
intermolecular
existgas
andas
a real
gas a cero,
in
the limit
limit as
as pressure
pressure
approaches
zero
lead to
to the
the
definition of
offorces
an ideal
ideal
gas
asofone
one
whose
in
approaches
lead
definition
an
whose
conducen
a
la
definición
de
un
gas
ideal
como
aquel
cuyo
comportamiento
macroscópico
se
caracteriza
in
the
limit
as
pressure
approaches
zero
lead
to
the
definition
of
an
ideal
gas
as
one
whose
in the limit as
pressure
zero
macroscopic
behavior
is approaches
characterized
by:lead to the definition of an ideal gas as one whose por:
macroscopic
behavior
is
characterized
by:
macroscopic
behavior
is
characterized
by:
macroscopic behavior is characterized by:
• La ecuación
de estado
The equation
equation
of state:
state:
•• The
of
•• The
equation
The equation of
of state:
state:
= RT
RT (ideal
(ideal
gas)
(3.14) (3.14)
(gas ideal)
PPVV =
gas)
(3.14)
P
(ideal
(3.14)
PV
V=
= RT
RT
(ideal gas)
gas)
(3.14)
•
An
internal
energy
that
is
a
function
of
temperature
only:
• Una •energía
internaenergy
que esthat
unaisfunción
sóloofdetemperature
la temperatura:
An internal
a function
only:
•• An
An internal
internal energy
energy that
that is
is aa function
function of
of temperature
temperature only:
only:
U=
=U
U(T
(T )) (ideal
(ideal
gas)
(3.15) (3.15)
U
gas)
(3.15)
(gas ideal)
U
(ideal
(3.15)
U=
=U
U(T
(T ))
(ideal gas)
gas)
(3.15)
Implied Property Relations for an Ideal Gas
Relaciones
una propiedad
implícitas
paraGas
un gas ideal
Impliedde
Property
Relations
for an Ideal
The definition
definition of
of heat
heat capacity
capacity at
at constant
constant volume,
volume, Eq.
Eq. (2.16),
(2.16), leads
leads for
for an
an ideal
ideal gas
gas to
to the
the
The
The
definition
of
heat
capacity
at
constant
volume,
Eq.
(2.16),
leads
for
an
ideal
gas
to
the
The
definition
of
heat
capacity
at
constant
volume,
Eq.
(2.16),
leads
for
an
ideal
gas
to
thea la conconclusion
that
C
is
a
function
of
temperature
only:
La definición
de capacidad
térmica
a volumen
constante,
ecuación (2.16), para un gas ideal conduce
conclusion
that C VV is
a function
of temperature
only:
conclusion
that
C
is
a
function
of
temperature
only:
conclusion
that
C
is
a
function
of
temperature
only:
V
clusión de que CV es una Vfunción exclusiva�
�de la�
�temperatura:
∂U �
dU(T
(T ))
∂U
dU
�
�
� =
C
≡
= dU
(T ))
(3.16)
=C
CVV (T
C VV ≡ ∂U
(3.16)
(T
∂U
dU
(T )) =
∂
T
dT
C
(3.16)
CVV ≡
≡ ∂ T VV =
= dT =
(T ))
(3.16) (3.16)
=C
CVV (T
∂∂TT VV
dT
dT
The defining
defining equation
equation for
for enthalpy,
enthalpy, Eq.
Eq. (2.11),
(2.11), applied
applied to
to an
an ideal
ideal gas,
gas, leads
leads to
to the
the conclusion
conclusion
The
The
defining
for
Eq.
(2.11),
applied
an
ideal
gas,
leads
The H
defining
equation
forof
enthalpy,
Eq.(2.11),
(2.11),
applied to
angas
ideal
gas,conduce
leads to
to athe
the
conclusion
that
Hque
also
isequation
function
ofenthalpy,
temperature
only: aplicada
La ecuación
define
a la entalpía,
ecuación
atoun
ideal,
la conclusion
conclusión de que
that
also
is
aa function
temperature
only:
that
H
also
is
of
only:
thates
Huna
alsofunción
is aa function
function
oflatemperature
temperature
only:
H también
sólo de
temperatura:
H≡
≡U
U+
+ PPVV =
=U
U(T
(T )) +
+ RT
RT =
=H
H(T
(T ))
(3.17)
H
(3.17)
H
(3.17)
H≡
≡U
U+
+P
PV
V=
=U
U(T
(T )) +
+ RT
RT =
=H
H(T
(T ))
(3.17) (3.17)
03-SmithVanNess.indd 73
8/1/07 12:52:27
74
74
74
74
CHAPTER 3. Volumetric Properties of Pure Fluids
CAPÍTULO
Propiedades
volumétricas
deFluids
fluidos puros
CHAPTER
Properties
of
CHAPTER 3.
3.3.Volumetric
Volumetric
Properties
of Pure
Pure
Fluids
The heat
capacity
at constant
pressure
byecuación
Eq. (2.20),
likecomo
CV , C
isVa, es
function
of sólo
P , defined
La capacidad
calorífica
a presión
constante
CP, C
definida
por la
(2.20)
una función
The
heat
capacity
at
constant
pressure
C
by
Eq.
(2.20),
like
C
aa function
of
The
heat
capacity
at
constant
pressure
C
defined
by
Eq.
(2.20),
like
C
is
function
of
PP,, defined
VV,, is
temperature
only:
de la temperatura:
�
�
temperature
temperature only:
only:
∂
H
d
H
(T
)
�
�
� =
C P ≡ �∂∂H
(3.18)
ddH
H
H(T
(T)) = C P (T )
C
(3.18)
CPP ≡
≡ ∂T P =
= dT =
(T))
(3.18) (3.18)
=C
CPP(T
∂∂TT PP
dT
dT
A useful relation between C P and C V for an ideal gas comes from differentiation of Eq. (3.17):
useful
C
and
C
for
ideal
from
Eq.
Arelación
useful relation
relation
CPPun
and
CVVideal
for an
an
ideal gas
gas
comes
from differentiation
differentiation
of
Eq. (3.17):
(3.17):
Una útilA
entre Cbetween
y CV para
gas
proviene
decomes
la derivada
de
la ecuación of
(3.17):
Pbetween
dH
dU
C P = ddH
(3.19)
H = dU
dU + R = C V + R
C
(3.19)
CPP =
= dT =
+ RR
(3.19) (3.19)
= dT +
+ RR =
=C
CVV +
dT
dT
dT
dT
This equation does not imply that C P and C V are themselves conEsta ecuación
no implica
que
C
C
enthat
síC
constantes
para un gas
ideal,
This
does
not
imply
that
and
C
conP ybut
V sean
This equation
equation
does
not
imply
that
Cmismas
andvary
CVV are
are
themselves
conPPthey
stant
for
an ideal
gas,
only
withthemselves
temperature
in sino
que varían
únicamente
con
la
temperatura
de
tal
manera
que
su
diferencia
es
igual
stant
for
an
ideal
gas,
but
only
that
they
vary
with
temperature
stant afor
anthat
ideal
gas,
but onlyisthat
they
with temperature in
ina R.
such
way
their
difference
equal
to Rvary
.
such
a
way
that
their
difference
is
equal
to
R
.
such a way that their difference is equal to R .
Para cualquier
gasgas
ideal
las(3.16)
ecuaciones
(3.16)lead
y (3.18)
For anycambio
change de
of estado
state ofen
anun
ideal
Eqs.
and (3.18)
to: nos conducen a:
For
For any
any change
change of
of state
state of
of an
an ideal
ideal gas
gas Eqs.
Eqs. (3.16)
(3.16) and
and (3.18)
(3.18) lead
lead to:
to:
�
dU = C V dT (3.20a)
a) �U = �� C V dT (3.20b)
b)
dU
dU =
=C
CVV dT
dT (3.20a)
(3.20a) �U
�U =
= C
CVV dT
dT (3.20b)
(3.20b)
�
d H = C P dT (3.21a)
a) �H = �� C P dT (3.21b)
b)
ddH
H=
=C
CPP dT
dT (3.21a)
(3.21a) �H
�H =
= C
CPP dT
dT (3.21b)
(3.21b)
both the
internal
energy
an son
idealfunciones
gas are functions
temperatureDU para
V of
Ya que Because
tanto la energía
interna
como
CV deand
unCgas
ideal
sólo de laof
temperatura,
Because
both
the
internal
energy
and
C
of
an
ideal
gas
are
functions
of
temperature
Because
both
the
internal
energy
and
C
of
an
ideal
gas
are
functions
of
temperature
VVEq.
only,
�U
for
an
ideal
gas
is
always
given
by
(3.20b),
regardless
of
the
kind
of process
un gas ideal siempre se da por la ecuación (3.20b), sin considerar la clase de proceso que ocasione
el cambio.
only,
�U
for
an
ideal
gas
is
always
given
by
Eq.
(3.20b),
regardless
of
the
kind
of
only,
�U
for
an
ideal
gas
is
always
given
by
Eq.
(3.20b),
regardless
of
the
kind
of process
process
causing
the
change.
This
is
demonstrated
in
Fig.
3.5,
which
shows
a
graph
of
internal
energy
Esto se prueba en la figura 3.5, la cual exhibe una gráfica de la energía interna como una función
del volumen
causing
the
This
is
in
which
a graph
of
energy
causing
the change.
change.
This
is demonstrated
demonstrated
in Fig.
Fig.
3.5,
which shows
shows
graph
of internal
internal
as
alafunction
of molar
volume
with temperature
as3.5,
parameter.
Because
U is independent
of
V,
molar con
temperatura
como
parámetro
(cantidad
que
es constante
bajoacierto
conjunto
deenergy
condiciones,
as
aa function
of
molar
volume
with
temperature
as
parameter.
Because
U
is
independent
of
VV
as
function
of
molar
volume
with
temperature
as
parameter.
Because
U
is
independent
of
a
plot
of
U
vs.
V
at
constant
temperature
is
a
horizontal
line.
For
different
temperatures,
U,, de U
pero es posible que sea diferente bajo otras condiciones). Ya que U es independiente de V, una gráfica
aa plot
of
VV at
temperature
is
horizontal
line.
different
temperatures,
U
plot
of U
U vs.
vs.
at constant
constant
temperature
is aaeach
horizontal
line. For
For
different
temperatures,
U
has
different
values,
with
a
separate
line
for
temperature.
Two
such
lines
are
shown
in
en función de V a temperatura constante es una línea horizontal. A diferentes temperaturas, U tiene valores
has
has different
different values,
values, with
with aa separate
separate line
line for
for each
each temperature.
temperature. Two
Two such
such lines
lines are
are shown
shown in
in
distintos, con una línea separada para cada temperatura. En la figura 3.5 se muestran dos de estas líneas, una
U2
U
U22
Figure 3.5: Internal energy changes for an ideal
Figure
Figure 3.5:
3.5: Internal
Internal energy
energy changes
changes for
for an
an ideal
ideal
gas.
Figura 3.5:
Cambios en la energía interna de un gas
gas.
gas.
ideal.
U
U
U U1
U
U11
b
bb
c
cc
d
dd
b
U2
c
U1
2
d
T2
T1
TT1
a
aa
U
T2
TT2
1
T1
a
V
VV
V
Fig. 3.5, one for temperature T1 and one for a higher temperature T2 . The dashed line connectFig.
3.5,
for
and
temperature
TT22..the
The
dashed
connectFig.points
3.5, one
one
for temperature
temperature
and one
one for
for aa higher
higher
temperature
Thetemperature
dashed line
lineincreases
connecting
a and
b representsTTa11 constant-volume
process
for which
ing
points
a
and
b
represents
a
constant-volume
process
for
which
the
temperature
increases
ing
points
a
and
b
represents
a
constant-volume
process
for
which
the
temperature
increases
from T1 to T2 and the internal energy changes
by �U = U2 − U1 . This change in internal
�
para la temperatura
y la the
otra
para unaenergy
temperatura
superior
. La
que une
a los puntos a
2=
from
TTT221 and
internal
by
�U
U
in
from TT1is
to
and
the(3.20b)
internal
changes
by. The
�UTdashed
=
U2línea
−U
Udiscontinua
. This
This change
change
in internal
internal
11.connecting
1 to
2−
energy
given
by Eq.
asenergy
�U =changes
C V dT
lines
points
a and c
�
�
y b representa
un
proceso
a
volumen
constante
para
el
que
la
temperatura
aumenta
de
T
a
T
y laccenergía
1
energy
is
(3.20b)
as
=
dT
.. The
lines
points
aa2and
energy
is given
given
byd Eq.
Eq.
(3.20b)other
as �U
�U
= C
CVVnot
dToccurring
The dashed
dashed
lines connecting
connecting
points
and
and
points
a andby
represent
processes
at constant
volume but
which
also
interna cambia
por
DU
=
U
–
U
.
Este
cambio
en
la
energía
interna
es
conocido
por
la
ecuación
(3.20b)
2represent
1
and
aa and
other
not
constant
but
also
and points
points
and dd represent
other
processes
not occurring
occurring
constant
volume
but which
which
also como
lead
from an
initial
temperature
T1 processes
to
a final temperature
T2at
.atThe
graphvolume
shows that
the
change
DU = ∫ lead
C
dT.
Las
líneas
discontinuas
que
unen
a
los
puntos
a
y
c,
así
como
a
los
puntos
a
y
d,
representan
V from
an
temperature
TT22.. The
shows
change
lead
an initial
initial
temperature
T11 to
to aaasfinal
final
temperature
The graph
graph
shows
that
the
change
in
U from
for these
processes
is the T
same
for temperature
the
process,
andthat
it isthe
therefore
� constant-volume
otros procesos
que
no ocurren
a volumen
constante,
pero
que también conducen
de una
temperatura
inicial T1
in
U
for
these
processes
is
the
same
as
for
the
constant-volume
process,
and
it
is
therefore
in
U
for
these
processes
is
the
same
as
for
the
constant-volume
process,
and
it
is
therefore
given by the same equation, namely, �U = �� C V dT . However, �U is not equal to Q for
a la temperatura
final
T
.
La
gráfica
muestra
que
el
cambio
en
U
para
estos
procesos
es
el
mismo
que
2 equation,
given
given by
by the
the same
same
equation, namely,
namely, �U
�U =
= C
CVV dT
dT.. However,
However, �U
�U is
is not
not equal
equal to
to Q
Q for
for para el
03-SmithVanNess.indd 74
8/1/07 12:52:32
75
75
75
proceso a volumen constante, y esto es consecuencia de la misma ecuación, a saber, DU = ∫ CV dT. No obstante, DU
no processes,
es igual a Q
para estos
procesos,
queonQTdepende
T2, sino
también
de la trathese
because
Q depends
notya
only
butsólo
alsode
onTthe
of the
process.
1 y path
1 and T2 no
these processes, because Q depends not only on T11 and T22 but also on the path of the process.
yectoriaAn
delentirely
proceso.
A la entalpía
H de un
gas ideal
seenthalpy
le aplicaHunofanálisis
analogous
discussion
applies
to the
an idealcompletamente
gas. (See Sec. análogo.
2.11.) (Véase
An entirely analogous discussion applies to the enthalpy H of an ideal gas. (See Sec. 2.11.)
la sección 2.11.)
The ideal gas is a model fluid described by simple property relations, which are freThe ideal
is a model
fluid described
by simple
property
whichsimple,
are freElquently
gas ideal
es approximations
ungas
modelo
de fluido
mediante
relaciones
de relations,
unacalculations,
propiedad
good
when descrito
applied to
actual gases.
In process
gases atque con
quently
good
approximations
when
applied
to
actual
gases.
In
process
calculations,
gases
at
frecuencia
son una
aproximación
cuando
se aplican aideal,
los gases
reales. equations
En los cálculos
de los procesos,
pressures
upbuena
to a few
bars may often
be considered
and simple
then apply.
pressures
up
to
a
few
bars
may
often
be
considered
ideal,
and
simple
equations
then
apply.
los gases con presión hasta de unos pocos bar con frecuencia se consideran ideales, y en tal caso se les aplican
ecuaciones sencillas.
3.3. El gas ideal
3.3. The Ideal Gas
3.3. The Ideal Gas
Equations for Process Calculations for Ideal Gases
Equations for Process Calculations for Ideal Gases
Ecuaciones
para cálculos
de procesos:
gases
Process calculations
provide work
and heat quantities.
Theideales
work of a mechanically reversible
Process calculations provide work and heat quantities. The work of a mechanically reversible
closed-system process is given by Eq. (1.2), here written for a unit mass or a mole:
closed-system
process
is given bycantidades
Eq. (1.2), de
here
written
for a unit
mass orpara
a mole:
Los cálculos
de procesos
proporcionan
trabajo
y calor.
El trabajo
un proceso mecánicamente reversible en un sistema cerrado se conoce
por
la
ecuación
(1.2),
y
en
este
caso
se
escribe para un mol
d W = −P d V
d W = −P d V
o una unidad de masa:
For an ideal gas in any closed-system process, the first law as given by Eq. (2.6) written for a
For an ideal gas in any closed-system process,
the first law as given by Eq. (2.6) written for a
dWEq.
= –PdV
unit mass or a mole, may be combined with
(3.20a) to give:
unit mass or a mole, may be combined with Eq. (3.20a) to give:
Para un gas ideal en cualquier proceso de sistema cerrado, la primera ley se conoce por la ecuación (2.6),
Q + d Wcombinarla
= C V dT
escrita para una masa unitaria o un mol, y esddposible
Q + d W = C VV dT con la ecuación (3.20a) para dar:
Substitution for d W and solution fordQ
d Q+ dW
yields
equation valid for an ideal gas in any
= Can
V dT
Substitution for d W and solution for d Q yields an
equation valid for an ideal gas in any
mechanically reversible closed-system process:
mechanically
reversible
closed-system
La sustitución
para dW
y la solución
para dQ process:
conducen a una ecuación que es válida para un gas ideal en
cualquier proceso mecánicamente reversible
de=sistema
+ P dV
dQ
C V dTcerrado:
d Q = C VV dT + P d V
dQ = CV dT + PdV
This equation contains the variables P, V , and T , only two of which are independent.
This equation contains the variables P, V , and T , only two of which are independent.
Working
equations
forlas
d Qvariables
and d WP,depend
which
variables is Las
selected
as
Esta
ecuación
contiene
V y T,on
pero
sólopair
dos of
sonthese
independientes.
ecuaciones
de
Working equations for d Q and d W depend on which pair of these variables is selected as
upon which
variable
is eliminated
by Eq.
Withcomo
P = independiente;
RT /V ,
trabajo independent;
para dQ y dWi.e.,
dependen
de cuál
par de
estas variables
se (3.14).
seleccione
es decir,
independent; i.e., upon which variable is eliminated by Eq. (3.14). With P = RT /V ,
de qué variable se elimine mediante la ecuación (3.14). Con P = RT/V,
dV
d Q = C VV dT + RT d V
d Q = C V dT + RT V
V
(3.22)
(3.22)
dV
d W = −RT d V
d W = −RT V
V
(3.23)
(3.23)
With V = RT /P and with C VV given by Eq. (3.19), the equations for d Q and d W become;
V y=con
RTC/P
and with C Vpor
given
by Eq. (3.19),
forpara
d Q and
W become;
Con V =With
RT/P,
la ecuación
(3.19),the
lasequations
ecuaciones
dQ yd dW
se convierten en:
V determinada
dP
d Q = C PP dT − RT d P
d Q = C P dT − RT P
P
(3.24)
(3.24)
dP
d W = −R dT + RT d P
d W = −R dT + RT P
P
(3.25)
(3.25)
T el
= P V /R,es
thesimplemente
work is simply
= −Py dcon
V , and
withvez
C V given
again
(3.19),(3.19),
Con T =With
PV/R,
dW dd=W
conoce
porby
la Eq.
ecuación
V otra
With
T = trabajo
P V /R, the
work is simply
W–P=dV,
−P d V , C
and
with C Vsegiven
again
by
Eq.
(3.19),
V
CP
CV
(3.26) (3.26)
d Q = C VV V d P + C PP P d V
(3.26)
d Q = R V d P + R P dV
R
R
Es posible aplicar estas ecuaciones para gases ideales a diversas clases de procesos, como los descritos
These
equations may
be applied
fordeducción
ideal gases
various
kindsesofcerrado
processes,
a continuación.
Las suposiciones
implícitas
en su
sonto
el sistema
y queas
el deproceso es
These
equations may
be applied
for ideal gases
toque
various
kinds of processes,
as
described
in
what
follows.
The
assumptions
implicit
in
their
derivation
are
that
the
system
is
mecánicamente
reversible.
scribed in
what follows. The assumptions implicit in their derivation are that the system is
closed and the process is mechanically reversible.
closed and the process is mechanically reversible.
03-SmithVanNess.indd 75
8/1/07 12:52:34
76
76 76
767676
76
76
76 76
76
76 76
CAPÍTULO
3. Propiedades
volumétricas
de
fluidos puros
CHAPTER
3. 3.
Volumetric
Properties
of Pure
Fluids
CHAPTER
CHAPTER
Volumetric
Volumetric
Properties
Properties
Pure
ofPure
Pure
Fluids
Fluids
CHAPTER
3.3.Volumetric
Volumetric
Properties
ofof
Pure
Fluids
CHAPTER
3.
Volumetric
Properties
of
Pure
Fluids
CHAPTER
3.
Volumetric
Properties
of
Fluids
CHAPTER
3.
Volumetric
Properties
of
Pure
Fluids
CHAPTER
3.
Properties
of
Pure
Fluids
CHAPTER
CHAPTER
3. 3.
Volumetric
Properties
of Pure
Fluids
CHAPTER
CHAPTER
3. Volumetric
Volumetric
Properties
Properties
of
of Pure
Pure
Fluids
Fluids
Proceso
isotérmico
Isothermal
Process
Isothermal
Isothermal
Process
Process
Isothermal
Process
Isothermal
Process
Isothermal
Process
Isothermal
Process
Isothermal
Process
Isothermal
Process
Isothermal
Process
Process
ByIsothermal
Eqs. (3.20b) and
(3.21b),
�U
= �H
==00=
De las
ecuaciones
(3.20b)
y (3.21b),
DU
=
DH
By
By
Eqs.
Eqs.
(3.20b)
(3.20b)
and
and
(3.21b),
(3.21b),
�U
�U
�H
=�H
�H
By
Eqs.
(3.20b)
and
(3.21b),
�U
==
�H
==
0=0000= 0
By
Eqs.
Eqs.
(3.20b)
(3.20b)
and
and
(3.21b),
(3.21b),
�U
�U
=
=
�H
By
Eqs.
(3.20b)
and
(3.21b),
�U
=
By
Eqs.
(3.20b)
and
(3.21b),
�U
=
�H
=
By
Eqs.
(3.20b)
and
(3.21b),
�U
=
�H
=
ByBy
Eqs.
(3.20b)
and
(3.21b),
�U
=
�H
=
0=
By
By Eqs.
Eqs. (3.20b)
(3.20b) and
and (3.21b),
(3.21b),
�U
�UV=
= �H
�H0�H
=
= 000 P
2V
V
V
P2P2P
2
2
2PP
2
2
By
Eqs.
(3.22)
and
(3.24),
Q
=
RT
ln
=
−RT
ln
V
VV=
V−RT
VVln
PP
2222=−RT
222 P2
2−RT
By
By
Eqs.
Eqs.
(3.22)
(3.22)
and
and
(3.24),
(3.24),
Q
=
Q
=
RT
RT
ln
ln
=
−RT
ln
ln
2V
2P
PP2ln
By
Eqs.
(3.22)
and
(3.24),
Q
=
RT
ln
ln
2
2 1P
V
V
De By
las
ecuaciones
(3.22)
y
(3.24),
By
Eqs.
(3.22)
and
(3.24),
Q
=
RT
=
ln
2
2
22
1
Eqs.
(3.22)
and
(3.24),
QQRT
=
RT
ln
=
−RT
ln
By
Eqs.
(3.22)
and
(3.24),
Q
=
RT
ln
=
−RT
Eqs.
(3.22)
and
(3.24),
Q
=
ln
=
−RT
ln
V
V
P
P
By
Eqs.
(3.22)
and
(3.24),
=
RT
ln
=
−RT
ln
V
P
ByBy
Eqs.
(3.22)
and
(3.24),
Q
=
RT
ln
=
−RT
ln
1 1V 1 1=
By
By Eqs.
Eqs. (3.22)
(3.22) and
and (3.24),
(3.24),
Q
Q=
= RT
RTVln
ln
= −RT
−RTPln
ln1 1P111 1 P1
VV11111 V1
PP111
V11 V
P11PP
V
P
V
2
2P
V
V
P
V
P
2
2
2
2
2
2
By
Eqs.
(3.23)
and
(3.25),
W
=
−RT
ln
=
RT
ln
V
VV=
PP
V=
VV2V
PP2P
222=RT
222 P2
2 ln
By
By
Eqs.
Eqs.
(3.23)
(3.23)
and
and
(3.25),
(3.25),
==
−RT
=−RT
−RT
ln
RT
RT
ln
ln
PP2ln
By
Eqs.
(3.23)
and
(3.25),
WW=
=W
−RT
ln−RT
2ln=
2 1P
V2V
22RT
22
1V
By
By
Eqs.
Eqs.
(3.23)
(3.23)
and
and
(3.25),
(3.25),
W
W
=
−RT
ln
ln
=
=
RT
RT
ln
ln
By
Eqs.
(3.23)
and
(3.25),
W
=
ln
=
RT
Eqs.
(3.23)
and
(3.25),
W
−RT
ln
ln
V
P
P
By
Eqs.
(3.23)
and
(3.25),
W
=
−RT
ln
=
RT
ln
De By
las
(3.23)
y
(3.25),
V
P
Byecuaciones
Eqs.
(3.23)
and
(3.25),
W
=
−RT
ln
=
RT
ln
1
1
1
1V
By
By Eqs.
Eqs. (3.23)
(3.23) and
and (3.25),
(3.25),
W
W=
= −RT
−RTVln
ln
=1RT
RTPln
ln1 P111 1 P1
VV1111 =
V
PP111
V1from
P11 P
1V
V11 Eq.
P
Note
that
Q
=
−W
,
a
result
that
also
follows
(2.3).
Therefore,
Note
Note
that
that
Q
=
Q
−W
=
−W
,
a
,
result
a
result
that
that
also
also
follows
follows
from
from
Eq.
Eq.
(2.3).
(2.3).
Therefore,
Therefore,
Note
that
Q
=
−W
,
a
result
that
also
follows
from
Eq.
(2.3).
Therefore,
Note
Note
that
that
Q
Q
=
=
−W
−W
,
,
a
a
result
result
that
that
also
also
follows
follows
from
from
Eq.
Eq.
(2.3).
(2.3).
Therefore,
Therefore,
Note
that
Q
=
−W
,
a
result
that
also
follows
from
Eq.
(2.3).
Therefore,
Note
that
Q
=
−W
,
a
result
that
also
follows
from
Eq.
(2.3).
Therefore,
Note
that
Q
=
−W
,
a
result
that
also
follows
from
Eq.
(2.3).
Therefore,
Note
that
Q
=
−W
,
a
result
that
also
follows
from
Eq.
(2.3).
Therefore,
Note
that
=
= −W
−W,, aaresultado
result
result that
that
also
follows
follows from
from
Eq.
(2.3).
(2.3).aTherefore,
Therefore,
ObserveNote
que that
Q
=Q
–QW,
como
dealso
la ecuación
(2.3).Eq.
Debido
eso,
V
P
2V
VV
P2P2P
2V2=
2V
2PP
2 (const T )
2V
Q
=
−W
=
RT
ln
−RT
ln
VV
PP
222=−RT
222 P(const
2 −RT
2(const
QQ
=
Q=
−W
=
−W
=
=
RT
RT
ln
ln
−RT
=
ln
ln
(const
T ) T)) ) T )
2V
2P
V
PP2ln
Q
=
−W
=
RT
ln
=
ln
T(const
2
2 1P
V
V
2
2
22 (const
1
Q
=
−W
−W
=
=
RT
RT
ln
ln
=
=
−RT
−RT
ln
ln
(const
(const
Q
−W
=
RT
ln
=
−RT
QQ==
−W
=
RT
ln
=
−RT
ln
TT
)))TT
V
V
P
P
Q
=
−W
=
RT
ln
=
−RT
ln
(const
V
P
−W
=
RT
ln
=
−RT
ln
(const
(T
constante)
1V 1 1=
1P1 1 1V
1
Q
Q=
= −W
−W =
= RT
RTVln
ln
=
−RT
−RT
ln
ln
(const
(const
TTT)))
P
V
P
PP11 P
VV11111 1
PP11111 1
V11 V
(3.27)
(3.27)
(3.27)
(3.27)
(3.27)
(3.27)
(3.27)
(3.27)
(3.27)
(3.27)
(3.27)
(3.27)
(3.27)
Isobaric
Process
Isobaric
Isobaric
Process
Process
Isobaric
Process
Isobaric
Process
Isobaric
Process
Isobaric
Process
Isobaric
Process
Proceso
isobárico
Isobaric
Process
Isobaric
Isobaric
Process
Process
By Eqs. (3.20b) and (3.21b),
By
By
Eqs.
Eqs.
(3.20b)
(3.20b)
and
and
(3.21b),
(3.21b),
By
Eqs.
(3.20b)
and
(3.21b),
By
Eqs.
(3.20b)
and
(3.21b),
By
Eqs.
(3.20b)
(3.21b),
By
Eqs.
(3.20b)
and
(3.21b),
By
Eqs.
(3.20b)
and
(3.21b),
By
Eqs.
(3.20b)
and
(3.21b),
Byecuaciones
Eqs.
(3.20b)
andand
(3.21b),
�� �
�� �
By
By Eqs.
Eqs.
(3.20b)
(3.20b)
and
and
(3.21b),
De las
(3.20b)
y (3.21b),
(3.21b),
��� �����C V�dT
��� �����C P�dT
�U
=
and
�H
=
�U
�U
=CC
CVdT
dTdT
and
andand
�H
�H
=CCPCC
CPdT
dT
�U
==�U
and
�H
==
VdT
VdT
PdT
PdT
VC
�U
=
C
and
�H
=
=
�H
=
C
�H
C
�U
==
dT
�H
==
dT
VVV dT
PPP dT
V and
�U
=
C
dT
and
�H
=
C
dT
�U�U
dT
andand
�H
dT
VC
PC
�U
�U
=
=CCV=
C
dT
and
and
�H
�H
=
=CCP=
C
dTP dT
y VV dT
PP dT
and
by by
Eqs.
(3.24)
and
(3.25),
and
and
Eqs.
Eqs.
(3.24)
(3.24)
and
and
(3.25),
(3.25),
and
bybyby
Eqs.
(3.24)
and
(3.25),
and
by
Eqs.
Eqs.
(3.24)
(3.24)
and
and
(3.25),
(3.25),
and
by
Eqs.
(3.24)
and
by
(3.24)
and
(3.25),
y deand
lasand
ecuaciones
(3.24)
yand
(3.25),
and
by
Eqs.
(3.24)
(3.25),
and
byEqs.
Eqs.
(3.24)
and
(3.25),
�� (3.25),
and
and
by
by Eqs.
Eqs.
(3.24)
(3.24)
and
and
(3.25),
(3.25),
�
�
�� �����C P�dT
Q
=
and
WW
= −R(T
TT−1 ))T )
2−−
dT and
and
and
−R(T
=−R(T
−R(T
=
Q=
=CC
CPdT
QQ
=
WWW
==
−R(T
PdT
PdT
2−
2T−
1T1−
1
PC
1))T
and
W
=
−R(T
Q
=
C
and
=
Q
y and
W
=222−
−R(T
C
dT
W
QQ
==
CCP=
PPP dT
1−
Q
=
C
dT
and
W
=
−R(T
−
TT11111))))) T1 )
dT
and
W=
=
−R(T
−22222T2T−
PC
1 )2T
dTP dTand
and
and
W
W−R(T
=
=
−R(T
−R(T
−
Q
Q=
=Q
C
PP dT
Note
that
Q
=
�H
,, aa ,result
also
given
by
Eq.
(2.13).
Therefore,
Note
Note
that
that
Qthat
=
Q=
�H
=
�H
aresult
result
also
also
given
given
by
by
Eq.
Eq.
(2.13).
(2.13).
Therefore,
Therefore,
Note
that
Q=
=
�H
,=
aresult
result
also
given
by
Eq.
(2.13).
Therefore,
Observe
que
Q
∆H
es
un
resultado
también
conocido
por
laTherefore,
ecuación
(2.13). Por lo tanto,
Note
Note
that
that
Q
Q
=
�H
�H
,�H
,result
aaresult
result
also
also
given
given
by
by
Eq.
Eq.
(2.13).
(2.13).
Therefore,
Note
Q
,
a
result
also
given
by
Eq.
(2.13).
Therefore,
Note
that
Q
=
�H
,
a
also
given
by
Eq.
(2.13).
Therefore,
Note
that
Q
=
�H
,
a
result
also
given
by
Eq.
(2.13).
Therefore,
Note
that
Q
=
�H
,
a
also
given
by
Eq.
(2.13).
Therefore,
Note
Note that
that Q
Q=
= �H
�H,, aa result
result also
also given
given �
by
by Eq.
Eq. (2.13).
(2.13). Therefore,
Therefore,
� �
��� �����C P�dT
Q
=
�H
=
(const
P)
(3.28)
Q
=
Q
�H
=
�H
=
=CC
CPdT
dTdT
(const
(const
P)
P) P)
(3.28)
(3.28)
(3.28)
(P
constante)
Q=Q
=
�H
= �H
(const
P)
(3.28)
PdT
PdT
PC
Q
=
=
�H
�H
=
=
C
dT
(const
(const
P)
P)
(3.28)
(3.28)
Q
=
=
C
(const
(3.28)
QQ
�H
=
C
dT
(const
P)
(3.28)
P
P
P
P
Q
=
�H
=
C
dT
(const
P)
(3.28)
=
�H
=
C
dT
(const
P)
(3.28)
Q
Q=
= �H
�H =
= PCCP dT
dT
(const
(const P)
P)
(3.28)
(3.28)
PP
Isochoric
(Constant-V
))Process
Process
Isochoric
Isochoric
(Constant-V
(Constant-V
)Process
Process
Isochoric
(Constant-V
Proceso
isocórico
(V constante)
Isochoric
(Constant-V
Process
Isochoric
(Constant-V
)))Process
Isochoric
(Constant-V
) Process
Isochoric
(Constant-V
)))Process
Isochoric
(Constant-V
Process
Isochoric
Isochoric
(Constant-V
(Constant-V
)
Process
Process
Equations (3.20b) and (3.21b) again apply:
Equations
Equations
(3.20b)
(3.20b)
and
and
(3.21b)
(3.21b)
again
again
apply:
apply:
Equations
(3.20b)
and
(3.21b)
again
apply:
Equations
(3.20b)
and
(3.21b)
again
Equations
(3.20b)
and
(3.21b)
again
apply:
Equations
(3.20b)
and
(3.21b)
again
apply:
Equations
(3.20b)
and
(3.21b)
again
apply:
Equations
(3.20b)
and
(3.21b)
again
apply:
Se aplican
de
nuevo
las
ecuaciones
(3.20b)
yapply:
(3.21b):
Equations
(3.20b)
and
(3.21b)
again
apply:
�
�� �
Equations
Equations
(3.20b)
(3.20b)
and
and
(3.21b)
(3.21b)
again
apply:
���again
� � apply:
�
�
��� �����C P�dT
� ���C V dT
�U
=
and
�H
=
�U
�U
=CC
CVdT
dTdT
and
andand
�H
�H
=CCPCC
CPdT
dT
�U
==�U
and
�H
==
VdT
VdT
PdT
PdT
VC
�U
=
C
and
�H
=
=
�H
=
y C
�H
C
�U
==
dT
and
�H
==
dT
VVV dT
PPP dT
V �U
=
C
dT
and
�H
=
C
dT
�U�U
dT
andand
�H
dT
VC
PC
�U
�U
=
=CCV=
C
dT
and
and
�H
�H
=
=CCP=
C
dTP dT
VV dT
PP dT
By
Eq.
(3.22)
and
the
basic
equation
for
work,
By
By
Eq.
Eq.
(3.22)
(3.22)
and
the
the
basic
basic
equation
equation
for
for
work,
work,
Eq.
(3.22)
and
the
basic
equation
for
work,
De By
laBy
ecuación
(3.22)
yand
la
ecuación
fundamental
delwork,
trabajo,
By
Eq.
Eq.
(3.22)
(3.22)
and
and
the
basic
basic
equation
equation
for
work,
work,
By
Eq.
(3.22)
and
the
basic
equation
for
Eq.
(3.22)
and
the
basic
equation
for
work,
By
Eq.
(3.22)
and
the
basic
equation
for
work,
ByBy
Eq.
(3.22)
and
thethe
basic
forfor
work,
��equation
�
By
By Eq.
Eq.
(3.22)
(3.22)
and
and
the
the
basic
basic
equation
equation
for
for
work,
work,
�
�
����� �
�
�
�
�
�
��P d V = 0
�
�
�
Q=
dTdT and
=
−
�P
��C
VdT
y and
andandWW
WW
==
−
=−
−
PdPPdVPdV
dVVP=
V0d=0=
=
Q=
=
C
QQ
=
C
and
=W
−
=
VCVdT
VdT
VC
Q
=
C
dT
and
W
=
−
and
000
Q
Q
=
C
dT
W
=
dT
and
W
=
−
P
QQ
==
C
V
V
V
V
dT
and
W
=
−
PP
VV00=
=
Q
=
C
dT
=
dV
Vddd=
VV dT
dT andand
and W W
W−
=
=−
−Pd−
P
d=
V
=
=V000= 0
Q
Q=
=C V C
C
V
Note
that
Q
=
�U
,, aa ,result
also
given
by
Eq.
(2.10).
Therefore,
Observe
que
Qthat
∆U
es,�U
también
un
resultado
conocido
porTherefore,
la ecuación
(2.10). Por lo tanto,
Note
Note
that
Qthat
=
Q=
�U
=
aresult
result
also
also
given
given
by
by
Eq.
Eq.
(2.10).
(2.10).
Therefore,
Therefore,
Note
that
Q=
=
�U
aresult
result
also
given
by
Eq.
(2.10).
Note
Note
that
that
Q
Q
=
�U
�U
,,result
aaresult
result
also
also
given
given
by
by
Eq.
Eq.
(2.10).
(2.10).
Therefore,
Therefore,
Note
Q
=
�U
,
a
result
also
given
by
Eq.
(2.10).
Therefore,
Note
that
Q
=
�U
,
a
also
given
by
Eq.
(2.10).
Therefore,
Note
that
Q
=
�U
,
a
result
also
given
by
Eq.
(2.10).
Therefore,
Note
that
Q
=
�U
,
a
also
given
by
Eq.
(2.10).
Therefore,
Note
Note that
that Q
Q=
= �U
�U,, aa result
result also
also given
given by
by
Eq.
Eq.
(2.10).
(2.10).
Therefore,
Therefore,
�� �
��� �����C V�dT
(3.29)
(V
constante)
Q
=
�U
=
(const
V
))V )
(3.29)
=
Q=
�U
=
�U
=CC
CVdT
dTdT
(const
(const
V)V
(3.29)
(3.29)
QQ
=
�U
==�U
(const
V
(3.29)
VdT
VdT
VC
Q
Q
=
�U
�U
=
=
C
dT
(const
(const
V
)
)
(3.29)
(3.29)
Q
=
=
C
(const
V
)
(3.29)
QQ
==
�U
=
C
dT
(const
V
)
(3.29)
V
V
V
V
Q
=
�U
=
C
dT
(const
V
)
(3.29)
�U
=
C
dT
(const
V
)
(3.29)
Q
Q=
= �U
�U =
= V CCV dT
dT
(const
(const VV))
(3.29)
(3.29)
VV
03-SmithVanNess.indd 76
8/1/07 12:52:52
3.3.
The
Ideal
Gas
3.3.
The
Ideal
Gas
3.3.
The
Ideal
Gas
3.3.
The
Ideal
Gas
3.3.
The
Ideal
Gas
3.3.
The
Ideal
Gas
3.3.
The
Ideal
Gas
3.3.
The
Ideal
Gas
3.3.The
TheIdeal
IdealGas
Gas
3.3. El3.3.
gas
ideal
77
7777
7777
77
77
77
7777
77
Adiabatic Process; Constant Heat Capacities
Adiabatic
Process;
Constant
Heat
Capacities
Adiabatic
Process;
Constant
Heat
Capacities
Adiabatic
Process;
Constant
Heat
Capacities
Adiabatic
Process;
Constant
Heat
Capacities
Adiabatic
Process;
Constant
Heat
Capacities
Adiabatic
Process;
Constant
Heat
Capacities
Proceso
adiabático:
capacidades
caloríficas
constantes
Adiabatic
Adiabatic
Process;
Process;
Constant
Constant
Heat
Heat
Capacities
Capacities
An
An adiabatic
adiabatic process
process is
is one
one for
for which
which there
there is
is no
no heat
heat transfer
transfer between
between the
the system
system and
and its
its
AnAn
adiabatic
process
is is
one
for
which
there
is is
no
transfer
between
thethe
system
andand
its
An
adiabatic
process
isisone
one
for
which
there
isisno
no
heat
transfer
between
the
system
and
itsitsits
An
adiabatic
process
isone
one
for
which
there
isheat
heat
transfer
between
the
system
and
adiabatic
process
for
which
there
heat
transfer
between
system
its
An
adiabatic
process
for
which
there
nono
heat
transfer
between
the
system
and
surroundings;
i.e.,
ddQ
0.
Each
of
Eqs.
(3.22),
(3.24),
and
(3.26)
may
therefore
be
set
equal
An
adiabatic
process
is=
one
for
which
there
isisno
no
heat
transfer
between
the
system
and
its
surroundings;
i.e.,
Qis
=one
0.0.
Each
ofof
Eqs.
(3.22),
(3.24),
and
(3.26)
may
therefore
bebe
set
equal
An
An
adiabatic
adiabatic
process
process
is
one
for
for
which
which
there
there
is
no
heat
heat
transfer
transfer
between
between
the
the
system
system
and
and
its
its
surroundings;
i.e.,
d
Q
=
0.
Each
of
Eqs.
(3.22),
(3.24),
and
(3.26)
may
therefore
be
set
equal
surroundings;
i.e.,
d
Q
=
0.
Each
of
Eqs.
(3.22),
(3.24),
and
(3.26)
may
therefore
be
set
equal
surroundings;
i.e.,
d
Q
=
Each
Eqs.
(3.22),
(3.24),
and
(3.26)
may
therefore
set
equal
surroundings;
i.e.,
d
Q
=
0.
Each
of
Eqs.
(3.22),
(3.24),
and
(3.26)
may
therefore
be
set
equal
Un proceso
adiabático
esddaquel
en
el
que
no
hay(3.22),
transferencia
de
calor
entre
sistemathe
y set
sus
alrededores;
surroundings;
i.e.,
Q=
=C0.
0.
Each
of
Eqs.
(3.22),
(3.24),
and
(3.26)
mayel
therefore
be
setequal
equal
to
zero.
Integration
with
and
C
constant
then
yields
simple
relations
among
variables
surroundings;
i.e.,
Q
Each
of
Eqs.
(3.24),
and
(3.26)
may
therefore
be
V
P
to
zero.
Integration
with
C
and
C
constant
then
yields
simple
relations
among
the
variables
V
P
surroundings;
surroundings;
i.e.,
i.e.,
d
d
Q
Q
=
=
0.
0.
Each
Each
of
of
Eqs.
Eqs.
(3.22),
(3.22),
(3.24),
(3.24),
and
and
(3.26)
(3.26)
may
may
therefore
therefore
be
be
set
set
equal
equal
to
zero.
Integration
with
C
and
C
constant
then
yields
simple
relations
among
the
variables
to
zero.
Integration
with
C
and
C
constant
then
yields
simple
relations
among
the
variables
to
zero.
Integration
with
C
and
C
constant
then
yields
simple
relations
among
the
variables
to
zero.
Integration
with
C
and
C
constant
then
yields
simple
relations
among
the
variables
esto es,to
dQ
= 0.
Por
tanto,
es
cada then
una
de
las ecuaciones
(3.22),
(3.24)the
y (3.26)
a cero. La
V mechanically
P igualar
to
zero.
Integration
with
C
and
constant
then
yields
simple
relations
among
the
variables
VV V Vand
PP P P
TT
,,zero.
P,
and
VV ,lo
valid
for
reversible
adiabatic
compression
or
expansion
of
ideal
Integration
with
Cposible
CC
constant
yields
simple
relations
among
variables
V andC
and
,valid
for
mechanically
reversible
adiabatic
compression
or
expansion
ofof
ideal
to
zero.
Integration
Integration
with
with
C
C
CPPreversible
constant
then
then
yields
yields
simple
simple
relations
relations
among
among
the
the
variables
V Vand
Pconstant
T ,TTto
P,
and
V
,VVvalid
for
mechanically
adiabatic
compression
or
expansion
ofvariables
ideal
, Tzero.
P,
and
,V
for
mechanically
reversible
adiabatic
compression
or
expansion
of
ideal
,P,
P,
and
,valid
valid
for
mechanically
reversible
adiabatic
compression
or
expansion
ideal para
,gases.
P,
and
,
valid
for
mechanically
reversible
adiabatic
compression
or
expansion
of
ideal
integración
con
C
y
C
constantes
produce
relaciones
sencillas
para
las
variables
T,
P
y
V,
válida
T
,
P,
and
V
,
valid
for
mechanically
reversible
adiabatic
compression
or
expansion
of
ideal
V
P
For
example,
Eq.
(3.22)
becomes:
P,
and
V
valid
for
mechanically
reversible
adiabatic
compression
ororexpansion
expansion
ofofideal
ideal
For
example,
Eq.
(3.22)
becomes:
TTgases.
T,,gases.
,P,
P,
and
and
V
Vexample,
,, ,valid
valid
for
for
mechanically
mechanically
reversible
reversible
adiabatic
adiabaticde
compression
compression
or
expansion
of
ideal
gases.
For
example,
Eq.
(3.22)
becomes:
gases.
For
example,
Eq.
(3.22)
becomes:
For
Eq.
(3.22)
becomes:
gases.
example,
Eq.
(3.22)
becomes:
compresión
oFor
expansión
adiabática,
mecánicamente
reversible
gases ideales
Por
ejemplo,
la
ecuación
gases.
For
example,
Eq.
(3.22)
becomes:
gases.
For
example,
Eq.
(3.22)
becomes:
gases.
gases.
For
For
example,
example,
Eq.
Eq.
(3.22)
(3.22)
becomes:
becomes:
dT
d
V
R
(3.22) será:
dT
d
V
R
dTdT
dRVdRdV
R RR
dT
VdVV
dT=
−
dT
dT
ddV
VV
RR
=
−
==
−−
=
T
=
=−
−−C
VV d
T
C
dT
dT
R
=
−
T
C
V
T TTT ==C
VVV
CC
VdV
VC
V
V
V
V
−
−
TT
CCV V
VV
T
C
V
V
gives:
Integration
with
C VV constant
constant
gives:
Integration
with
constant
gives:
Integration
with
C
constant
gives:
Integration
with
CVC
constant
gives:
Integration
with
CVconstant
gives:
Integration
with
Vconstant
V
gives:
Integration
with
C
VVconstante
De la
integración
con
CC
se
obtiene:
constant
gives:
Integration
with
C
V
� �
constantgives:
gives:
Integration
Integrationwith
withCCV Vconstant
��
R/C
R/CVV
��
TT22� �
VVR/C
1�
�
���
R/C
R/C
R/C
V
VR/C
1�
VV V V
�
�
T2 TT22TT
V
V
V
=
V
R/C
1
1
2
1
V
1
V
=
�
�
�
�
2 = V11 R/C
R/C
T
VV
2
=
== VVVV
TT2T11=
2
T
2
2
1
1
=
T1 TT11TT
V2VV2V
V
1=
TT11 =
V2V22 2
T
V
1
1
2
2
Similarly,
Eqs.
(3.24)
and
(3.26)
lead
to:
Similarly,
Eqs.
(3.24)
and
(3.26)
lead
to:
Similarly,
Eqs.
(3.24)
andand
(3.26)
lead
to:
Similarly,
Eqs.
(3.24)
and
(3.26)
lead
to:
Similarly,
Eqs.
(3.24)
and
(3.26)
to: conducen a:
Similarly,
Eqs.
(3.26)
lead
to:
Similarly,
Eqs.
(3.24)
and
(3.26)
lead
to:
De manera
similar,
las(3.24)
ecuaciones
(3.24)
ylead
(3.26)
Similarly,
Eqs.
(3.24)
and
(3.26)
lead
to:
� R/C
�
� �
Similarly,
Similarly,Eqs.
Eqs.(3.24)
(3.24)and
and(3.26)
(3.26)
to:
��
C
PPleadto:
R/Clead
CPP/C
/CVV
��
�
��
��
�
�PPR/C
TT22� �
PP22� �
VVC1�
2�
�
�
��
���
R/C
C/C
/C
R/C
CPC
/C
R/C
C
/C
P
PVV/C
V
PR/C
P�
2
1V
PPP
PV
P
P
�
�
�
VV
T
P
P
T2 TT22T2 2=
P�
P
V
P
V
and
=
PP2P
P
V
R/C P
C PP/C
2�
2
1
2 2�2�
2
1
2
1
P
V
2
1
V
=
and
=
�
�
�
�
2
1
R/C
R/C
CC
/C
/C
T
P
P
V
P
P
P
P
VV
2
2
2
1
y
=
and
=
=
and
=
=
and
=
T
P
P
V
=
and
=
= PPP
and
=
112=
112
1
2
T
P
V
T
T
P
P
V
V
1
2
2
2
2
2
1
1
and
T1 TT11TT
P1PP1P
P1PP1P
V2VV2V
P
P=
V
1=
1=
and
and
1=
P1P11 1
P1P11 =
V2V22 2
TTmay
T1 1 also
P
P
V
1
1
1
1
1
2
2
These
equations
be
expressed
as:
These
equations
may
also
bebe
expressed
as:
These
equations
may
also
be
expressed
as:as:
These
equations
may
also
be
expressed
as:
These
equations
may
also
expressed
as:
These
equations
may
be
Estas
ecuaciones
también
esalso
posible
expresarlas
como:
These
equations
may
also
beexpressed
expressed
as:
These
equations
may
also
be
expressed
as:
These
Theseequations
equationsmay
mayalso
alsobebeexpressed
expressedas:
as:
γγ−1
(1−γ
)/γ
−1
(1−γ
)/γ =
T
V
=
const.
(3.30a)
T
P
const.
(3.30b)
P VVγγγγγ γ=
const.
(3.30c)
const.
(3.30a)
TPTP
const.
(3.30b)
const.
(3.30c)
)/γ
−1
(1−γ
)/γ
γ −1
(1−γ
)/γ
γγV
−1
)/γ
γ=
−1
)/γ
constante
(3.30a)
constante
(3.30b)
constante
(3.30c)
T VTTγT
const.
(3.30a)
T PTT(1−γ
=
const.
(3.30b)
P VPPγP
const.
(3.30c)
==
const.
(3.30a)
==
const.
(3.30b)
==
const.
(3.30c)
TV
=
const.
(3.30a)
P(1−γ
=
const.
(3.30b)
=
const.
(3.30c)
VV−1
=
const.
(3.30a)
P (1−γ
=
const.
(3.30b)
VVP=
(3.30c)
T
V
=
const.
(3.30a)
T
P
=
const.
(3.30b)
P
VγγV=
=const.
const.
(3.30c)
γ
−1
(1−γ
)/γ
T
V
=
const.
(3.30a)
T
P
=
const.
(3.30b)
P
V
=
const.
(3.30c)
γ
−1
γ
−1
(1−γ
(1−γ
)/γ
)/γ
γ
TTVV ==const.
const. (3.30a)
(3.30a) TTPP
==const.
const. (3.30b)
(3.30b) PPVV ==const.
const. (3.30c)
(3.30c)
C PP
CC
γγC≡
(3.31)
PC
C
PPC
(3.31)
PP
C
γ ≡
(3.31)
≡
(3.31)
γ≡
≡
(3.31)(3.31)
C
γγ γ≡
(3.31)
≡
(3.31)
VVP
C
C
CPPC
γ
≡
(3.31)
C
C
C
V
V
V
C
V
γ γ≡≡C VV
(3.31)
(3.31)
CCV V
Equations
(3.30)
are
restricted
in
application
to
ideal
with
conLas ecuaciones
(3.30)
restringen
su aplicación
a gases ideales
congases
capacidades
caloríficas
Equations
(3.30)
are
restricted
inin
application
toto
ideal
gases
with
conEquations
(3.30)
are
restricted
in in
application
toto
ideal
gases
with
conEquations
(3.30)
are
restricted
in
application
to
ideal
gases
with
conEquations
(3.30)
are
restricted
application
ideal
gases
with
conEquations
(3.30)
are
restricted
gases
with
conEquations
(3.30)
are
restricted
inapplication
application
toideal
ideal
gases
with
constant
heat
capacities
undergoing
mechanically
reversible
adiabatic
Equations
(3.30)
are
restricted
in
application
to
ideal
gases
with
conconstantes
que
son
sometidos
a
expansión
o
compresión
adiabática
mecánicamente
restant
heat
capacities
undergoing
mechanically
reversible
adiabatic
Equations
Equations
(3.30)
(3.30)
are
are
restricted
restricted
inin
application
application
toto
ideal
ideal
gases
gases
with
with
conconstant
heat
capacities
undergoing
mechanically
reversible
adiabatic
stant
heat
capacities
undergoing
mechanically
reversible
adiabatic
stant
heat
capacities
undergoing
mechanically
reversible
adiabatic
stant
heat
capacities
undergoing
mechanically
reversible
adiabatic
stant
heat
capacities
undergoing
mechanically
reversible
adiabatic
expansion
or
compression.
stant
heat
capacities
undergoing
mechanically
reversible
adiabatic
versible.
expansion
or
compression.
stant
stant
heat
heat
capacities
capacities
undergoing
undergoing
mechanically
mechanically
reversible
reversible
adiabatic
adiabatic
expansion
oror
compression.
expansion
or
compression.
expansion
compression.
expansion
compression.
expansion
oror
compression.
expansion
or
compression.
expansion
expansionor
orcompression.
compression.
Para gases ideales, el trabajo de cualquier proceso adiabático en sistema cerrado se conoce por:
For
ideal
gases,
the
work
of any
adiabatic
closed-system
process
is given
by:
For
ideal
gases,
the
work
any
adiabatic
closed-system
process
by:
ForFor
ideal
gases,
thethe
work
of of
any
adiabatic
closed-system
process
is given
by:by:
For
ideal
gases,
the
work
ofof
any
adiabatic
closed-system
process
isis
given
by:
For
ideal
gases,
the
work
of
any
adiabatic
closed-system
process
isgiven
given
by:
ideal
gases,
work
adiabatic
closed-system
process
is
For
ideal
gases,
the
work
ofany
any
adiabatic
closed-system
process
isgiven
given
by:
For
ideal
gases,
the
work
of
any
adiabatic
closed-system
process
is
given
by:
dW
=
dU
=
C
dI
V
For
Forideal
idealgases,
gases,the
thework
workofofany
any
adiabatic
adiabatic
closed-system
closed-system
process
process
is
is
given
given
by:
by:
ddW
= dU
=C
VV dT
dT
dU
CC
d WddW
=dW
dU
=dU
C
dT
==
dU
==
CdT
dT
=
=
=
=
VdT
VC
dW
WW
=dU
dU
=
C
VV VdT
dT
d
W
=
dU
=
C
V
dT
dT
d
d
W
W
=
=
dU
dU
=
=
C
C
V
V
constant
C VV ,,
W =
�U =
C
�T
(3.32)
Para CVFor
constante,
(3.32)
For
constant
CVC
�T
(3.32)
V�T
For
constant
W
=�U
=�T
�T
(3.32)
ForFor
constant
C VC
WW
=W
�U
=�U
C
(3.32)
For
constant
C,C
W
==
�U
==
C
(3.32)
constant
,V, V, ,
=
(3.32)
VC
V
For
constant
C
W=
=�U
�U
=
C
�T
(3.32)
VVC
VV V�T
For
constant
C
,
W
=
�U
=
C
�T
(3.32)
V
V
For
Forconstant
constant
C
CV V, , of
WW
=V�U
�U
==CCV V�T
�Telimina
(3.32)
(3.32)
Alternative
forms
Eq.
(3.32)
result
ifif=
C
is
eliminated
in favor
the
heat-capacity
ratio
γγ :: capaciAlternative
forms
Eq.
(3.32)
result
C
iseliminated
eliminated
of
the
heat-capacity
ratio
Veliminated
Se obtienen
formas
alternativas
de
la
ecuación
(3.32)
se
Cof
en
favor
de la ratio
relación
Vthe
Alternative
forms
of of
Eq.
(3.32)
result
if C
is
in
favor
of of
the
heat-capacity
γratio
:γγde
Alternative
forms
ofof
Eq.
(3.32)
result
Cif
is
inin
favor
of
the
heat-capacity
ratio
Alternative
forms
of
Eq.
(3.32)
result
iscuando
eliminated
infavor
favor
of
the
heat-capacity
Alternative
forms
(3.32)
result
ififVifC
is
in
heat-capacity
ratio
:γ: γ: :
Vis
Alternative
forms
ofEq.
Eq.
(3.32)
result
C
iseliminated
eliminated
infavor
favor
of
the
heat-capacity
ratio
VVC
V
Alternative
forms
of
Eq.
(3.32)
result
if
C
eliminated
in
favor
of
the
heat-capacity
ratio
γ
V isiseliminated
dades caloríficas
γforms
: formsofofEq.
Alternative
Alternative
Eq.
(3.32)
(3.32)
result
result
if
if
C
C
eliminated
in
in
favor
favor
of
of
the
the
heat-capacity
heat-capacity
ratio
ratio
γ
γ:: :
V
V
C
C
R
R
+
R
P
V
C
C
R
R
+
R
P
V
C=VC
R RRRR
R
CC
C+
+
RRR=
γγC≡
1+
or
C VV =
RRRR
PC
V
+R+
PPC
VVC
=R
PP
V+
C
+
RR=
CC
RR
R
γ ≡
=
1==
+
or or
=
≡
==
111+
+1+
oror
CC
=
γ≡
C
−R 11
γγ γ≡
1=
V=
≡≡C
==C
=
++C
oror C VC
C
==γγ R
VVP=
VV
VV
VVC
C
C+
C
CPPC
CVVCC
R
+=
R
V=
V
γ
≡
=
=
1
+
or
C
V
C
C
C
γ
−
1−
C
C
γ
C
C
γ−
C
C
C
γ
−
V
V
V
V C
V C
V V == V C
VV
VV
γ
−11−11
V
V
V
γ γ≡≡
=
=
1
1
+
+
or
or
C
C
=
=
o
V
V
C
C
C
γ
−
V
V
V
CC
CC
CC
γ γ−−111
33
Where
by
definition,
3 33 3
Where
byby
definition,
Donde
por
definición,
Where
by
definition,
Where
by
definition,
Where
definition,
Where
by
definition,
Where
by
definition,
33
Where
by
definition,
33
Where
Wherebybydefinition,
definition,
VV
VV
VV
33If C and C are constant, γ is necessarily constant. For an ideal gas, the assumption of constant γ is equivalent
If3 CVV and CPP are constant, γ is necessarily constant. For an ideal gas, the assumption of constant γ is equivalent
3 If 33CIf
C
constant,
γ isγγnecessarily
constant.
For
an
ideal
gas,
the
assumption
of constant
γ isγγequivalent
If3VIf
and
are
constant,
necessarily
constant.
For
anan
ideal
gas,
the
assumption
ofof
constant
equivalent
If
are
constant,
γis
isnecessarily
necessarily
constant.
For
anideal
ideal
gas,
the
assumption
ofconstant
constant
is
equivalent
CCand
and
CCare
are
constant,
necessarily
constant.
For
an
ideal
gas,
the
assumption
of
constant
isisγis
equivalent
Pand
VC
PC
C
constant,
γisisis
constant.
For
gas,
the
assumption
equivalent
Vand
Pare
V
3the
to
assumption
that
the
themselves
are
constant.
This
is
only
the
/C
V
P
VV ≡
to 3the
assumption
that
the heat
heat capacities
themselves
are For
constant.
This
isthe
theassumption
only way
way that
that
the ratio
ratioγγC
CisPPequivalent
/C
≡ γγ
If
C
and
CPC
are
constant,
γcapacities
necessarily
constant.
an ideal
gas,
of constant
3the
Vassumption
Pthat
If
IfCassumption
C
and
CC
are
constant,
constant,
γcapacities
γis
isnecessarily
necessarily
constant.
constant.
For
For
an
an
ideal
ideal
gas,
gas,
the
assumption
assumption
of
of
constant
constant
γCC
γconstante
is
is
equivalent
the
assumption
that
the
heat
capacities
themselves
areare
constant.
This
is
the
only
way
that
the
ratio
Csea
/C
≡
to
the
assumption
the
heat
capacities
themselves
are
constant.
This
only
way
that
the
ratio
/C
≡V≡
to
that
the
heat
capacities
themselves
are
constant.
This
isthe
the
only
way
that
the
ratio
C
to
to
the
assumption
the
heat
capacities
themselves
constant.
This
isismonotonic
the
only
way
that
the
ratio
/C
≡
γγ≡γ γ
Pequivalent
Vdifference
Vand
Pthat
Pare
Pand
V
PC
V/C
to
that
the
heat
themselves
are
constant.
This
is
the
only
way
that
the
ratio
C
/C
P
V
and
the
C
−
C
=
R
can
both
be
constant.
Except
for
the
gases,
both
C
and
actually
Si Cto
ythe
C
son
constantes,
necesariamente
es
constante.
Para
un
gas
ideal,
la
suposición
de
que
a la
P
Vγ
V the
P
P
V
P
V
and
the
difference
C
−
C
=
R
can
both
be
constant.
Except
for
the
monotonic
gases,
both
C
C
actually
assumption
thatP
the−heat
capacities
themselves
are constant.
This
is monotonic
the only way
that both
the ratio
C P /C
≡equivale
γ
V
P
V
V
and
the
difference
C
C
=
R
can
both
be
constant.
Except
for
the
gases,
C
and
C
actually
to
tothe
the
assumption
assumption
that
that
the
the
heat
heat
capacities
themselves
themselves
are
are
constant.
constant.
This
This
is
isthe
the
only
only
way
way
that
that
the
ratio
ratio
CC
/C
/C
≡
≡
andand
the
difference
C PCC
−
R=
can
both
beson
constant.
Except
for
the
monotonic
gases,
both
C the
and
C
actually
and
the
difference
−
C=
can
both
be
constant.
Except
for
the
monotonic
gases,
both
and
CC
actually
the
difference
−
C
=
RRcapacities
can
both
be
constant.
Except
for
the
monotonic
gases,
both
CC
and
actually
P
P
V
V
V
P
V
P
V
P
V
and
the
difference
C
−
C
=
R
can
both
be
constant.
Except
for
the
monotonic
gases,
both
C
and
C
actually
P
V
P
V
P
V
P
V
increase
with
temperature,
but
the
ratio
γ
is
less
sensitive
to
temperature
than
the
heat
capacities
themselves.
P
V
P
V
creencia
de
que
las
capacidades
caloríficas
mismas
constantes.
Ésta
es
la
única
manera
en
que
la
relación
C
/
C
≡
y
laγ γdiferencia
increase
with temperature,
γ is be
lessconstant.
sensitiveExcept
to temperature
than the heat
capacities
P C
VV actually
and
the difference
C P − C Vbut=the
R ratio
can both
for the monotonic
gases,
both Cthemselves.
and
P
and
and
the
thedifference
difference
CC
−Cbut
C
=
=
Rratio
Rcan
both
be
be
constant.
constant.
Except
Except
forforthan
the
the
monotonic
monotonic
gases,
gases,
both
boththemselves.
Cthemselves.
C
andCC
actually
increase
with
temperature,
but
the
ratio
γ can
isγγboth
less
sensitive
to temperature
the
heat
capacities
themselves.
increase
with
temperature,
but
the
ratio
isγis
less
sensitive
to
temperature
than
the
heat
capacities
with
temperature,
the
ratio
is
less
sensitive
totemperature
temperature
than
the
heat
capacities
temperature,
the
less
sensitive
to
temperature
the
heat
capacities
PExcepto
P−
Vbut
Vbut
Pthemselves.
Pand
V Vactually
with
temperature,
the
ratio
γis
less
sensitive
to
the
heat
capacities
C – C increase
=increase
Rincrease
seanwith
constantes.
para
los
gases
monotónicos,
en
realidad
Cthan
ythan
C
aumentan
con la themselves.
temperatura,
pero la relación
γ
γ
γ
P
increase with temperature, but the ratio γ is less sensitive to temperatureP thanVthe heat capacities themselves.
increase
increasewith
withtemperature,
temperature,but
butthe
theratio
ratioγ γisisless
lesssensitive
sensitivetototemperature
temperaturethan
thanthe
theheat
heatcapacities
capacitiesthemselves.
themselves.
es menos sensible a la temperatura que a las mismas capacidades caloríficas.
03-SmithVanNess.indd 77
V
γ
8/1/07 12:53:09
78
78
78
78
78
CHAPTER
CHAPTER 3.
3. Volumetric
Volumetric Properties
Properties of
of Pure
Pure Fluids
Fluids
CHAPTER 3.
3.3.Volumetric
Volumetric
Properties
of Pure
Pure
Fluids
CHAPTER
Properties
of
CHAPTER
CAPÍTULO
Propiedades
volumétricas
deFluids
fluidos puros
RR �T
�T
W
R �T
�T
W=
=C
CVV �T
�T =
= γR
W=
=C
CVVV �T
�T =
= γ−
W
− 11
− 11
γγ −
Because
RT
=
RT
=
expression
may
written:
1V
Como RT
V
es posible
escribirla
como:
Because
= 2PP=
VP112and
and
RT22expresión
= PP22VV22,, this
this
expression
may be
be
written:
1 = P1VRT
1 y11 RT
2, esta
Because
RT
=P
P1111V
V111 and
and
RT222 =
=P
P222V
V222,, this
this expression
expression may
may be
be written:
written:
Because
RT
RT
111 =
RT
−
RT
V
−
P
V
P
RT22 − RT11 = P22 V22 − P11 V11
W
(3.33)
RT2 −
− RT
RT11 = P
V2 −
−P
P11V
V11
P2V
W=
= RT
(3.33) (3.33)
W=
= γγ22 −
= 22 γγ22 −
(3.33)
W
(3.33)
− 11 1 =
− 111 1
− 11
− 11
γγ −
γγ −
Equations
(3.32)
and
are
compression
expansion
proEquations
(3.32)
and (3.33)
(3.33)
are general
general
for
adiabatic
compression
and
expansionadiabático
proLas ecuaciones
(3.32)
y (3.33)
son generales
parafor
unadiabatic
proceso de
compresiónand
y expansión
en
Equations
(3.32)
and whether
(3.33)
are
general
for
adiabatic
compression
and
expansion
proEquations
(3.32)
and
(3.33)
are
general
for
adiabatic
compression
and
expansion
processes
in
a
closed
system,
reversible
or
not,
because
P,
V
,
and
T
are
state
funccesses
in
a
closed
system,
whether
reversible
or
not,
because
P,
V
,
and
T
are
state
funcun sistema
cerrado,
sea
o
no
reversible,
ya
que
P,
V
y
T
son
funciones
de
estado,
independientes
de
la
trayeccessesindependent
in aa closed
closed system,
system,
whether
reversible
or2 not,
not,
becauseunknown.
P, V
V ,, and
and TT are
are state
state of
funccesses
in
whether
reversible
or
because
P,
functions,
of
However,
and
are usually
tions,
independent
of path.
path.
However,
and
unknown. deElimination
Elimination
of VV22 (3.33)
toria. De
cualquier
modo, por
lo general
T2 y VTT
se VV
conocen.
La eliminación
V2 de la ecuación
2 are usually
222no
tions,
independent
of
path.
However,
T
and
V
are
usually
unknown.
Elimination
of
V2
tions,
independent
of
path.
However,
T
and
V
are
usually
unknown.
Elimination
of
V
2
2
2
2
from
Eq.
by
(3.30c),
valid
only
for
reversible
leads
the
2
from
Eq. (3.33)
(3.33)
by Eq.
Eq.que
(3.30c),
validsólo
only2para
for mechanically
mechanically
reversible processes,
processes,
leads
to
the22
mediante
la ecuación
(3.30c),
es válida
procesos
mecánicamente
reversibles,
nosto
conduce
a la
from
Eq.
(3.33)
by
Eq.
(3.30c),
valid
only
for
mechanically
reversible
processes,
leads
to
the
from
Eq.
(3.33)
by
Eq.
(3.30c),
valid
only
for
mechanically
reversible
processes,
leads
to
the
expression:
expression:
expresión:
expression:
expression:
�
�
�
�
�
�
��
��
�(γ
�(γ
�P �
�P �
�
�
��
��
(γ−1)/γ
−1)/γ
(γ−1)/γ
−1)/γ
�
�
�
RT
PP11VV11 �
1
2
�(γ
�
� P22 �
�
�
RT
P
(γ−1)/γ
−1)/γ − 1 =
(γ−1)/γ
−1)/γ − 1
1
2 (γ
(γ
−1)/γ
(γ
−1)/γ
(3.34)
W
=
RT111
P222
P222
V111
P111V
RT
P
P
−
1
=
−
1
(3.34) (3.34)
W = γP
PP11
PP11
− 11 =
= γγ −
− 11
(3.34)
W=
= γ−
−
−
(3.34)
W
− 11
− 11
− 11
P111
− 11
P111
γγ −
P
γγ −
P
The
result
is
the
PP and
given
by
Theelsame
same
result
is obtained
obtained
when
the relation
relation
between
and VVpor
given
by Eq.
Eq. (3.30c)
(3.30c)
isseused
used
Se obtiene
mismo
resultado
cuando
la relación
entrebetween
P y V conocida
la ecuación
(3.30c)is
usa para la
��when
Thethe
same
result is
is W
obtained
when
the
relation(3.34)
between
P and
and
V for
given
bygases
Eq. (3.30c)
(3.30c)
is used
used
The
same
result
obtained
when
the
relation
between
P
V
given
by
Eq.
is
for
integration,
=
−
P
d
V
.
Equation
is
valid
only
ideal
with
constant
�
�
for the
d VLa. Equation
valid
only for
ideal
gases
with con
constant
integración
de integration,
la expresiónW
W=
= –−∫ P P
dV.
ecuación (3.34)
(3.34) is
sólo
es válida
para
gases
ideales
capacidades
for the
the
integration,adiabatic,
W=
=−
− P
P ddV
V .. Equation
Equation
(3.34) is
is valid
valid only
only for
for ideal
ideal gases
gases with
with constant
constant
for
integration,
W
(3.34)
heat
capacities
reversible,
heatconstantes,
capacities in
adiabatic,
mechanically
reversible, closed-system
closed-system
processes.
caloríficas
yinen
procesosmechanically
adiabáticos mecánicamente
reversiblesprocesses.
de sistema cerrado.
heat capacities
capacities
in
adiabatic,
mechanically
reversible,
closed-system
processes.
heat
in
adiabatic,
mechanically
reversible,
closed-system
processes.
applied
to
gases,
(3.30)
yield
approxWhen
applied
to real
real
gases,
Eqs.
(3.30) through
through
(3.34) often
oftenfrecuencia
yield satisfactory
satisfactory
approxCuandoWhen
se aplican
a gases
reales,
lasEqs.
ecuaciones
(3.30) a (3.34)
(3.34)
con
proporcionan
aproximaWhen
appliedthe
to real
real
gases, Eqs.
Eqs. (3.30)
(3.30)
through
(3.34) often
often
yield
satisfactory
approxWhen
applied
to
gases,
through
(3.34)
yield
satisfactory
approximations,
provided
deviations
from
ideality
are
relatively
small.
For
monatomic
gases,
imations, provided
they deviations
ideality are
small.
For monatomic
gases,
ciones satisfactorias,
siempre
cuando lasfrom
desviaciones
de lorelatively
ideal sean
relativamente
pequeñas.
Para gases
imations,
provided the
the deviations
deviations
from1.4
ideality
are relatively
relatively
small. For
For simple
monatomic
gases,
provided
from
ideality
are
small.
monatomic
gases,
γimations,
values
of
for
gases
and
for
polyatomic
γ=
= 1.67;
1.67;
approximate
valuesaproximados
of γγ are
are 1.4de
forγdiatomic
diatomic
gases
and 1.3
1.3son
for de
simple
polyatomic
monoatómicos,
γ =approximate
1.67; los valores
para gases
diatómicos
1.4 para
gases diatómiγ
=
1.67;
approximate
values
of
γ
are
1.4
for
diatomic
gases
and
1.3
for
simple
polyatomic
γ
=
1.67;
approximate
values
of
γ
are
1.4
for
diatomic
gases
and
1.3
for
simple
polyatomic
such
as
.. como CO2, SO2, NH3 y CH4.
2 ,, SO
gases
suchgases
as CO
CO
SO22,, NH
NH33,simples,
, and
and CH
CH4tales
cos y degases
1.3 para
poliatómicos
gases such
such as
as CO
CO2222,, SO
SO222,, NH
NH333,, and
and CH
CH4444..
gases
Whence
Whence
Whence
Whence
De donde,
Polytropic
Polytropic Process
Process
Polytropic
Process
Proceso
politrópico
Because polytropic means “turning many ways,” polytropic process suggests a model of some
Because polytropic means “turning many ways,” polytropic process suggests a model of some
Because polytropic
polytropic
means “turning
“turning
many ways,”
ways,”
polytropic
processbysuggests
suggests
model of
of some
some
Because
means
many
polytropic
process
aa model
With
itit is
as
represented
empirical
versatility.
With δδ asignifica
a constant,
constant,
is defined
defined
as aa process
process
represented
by the
thepolitrópicos
empirical equation:
equation:
Debidoversatility.
a que politrópico
“cambios
de muchas
maneras”,
los procesos
sugieren un moversatility. With
With δδ aa constant,
constant, itit is
is defined
defined as
as aa process
process represented
represented by
by the
the empirical
empirical equation:
equation:
versatility.
delo con algo de versatilidad. Con d como una constante, éste se define como un proceso representado por la
PPVV δδδ =
(3.35a)
ecuación empírica:
= constant
constant
(3.35a)
PV
V δδ =
= constant
constant
(3.35a)
P
(3.35a)
d
PV =(3.30a)
constante (3.30b) are readily derived:
(3.35a)
For
For an
an ideal
ideal gas
gas equations
equations analogous
analogous to
to Eqs.
Eqs. (3.30a) and
and (3.30b) are readily derived:
For an
an ideal
ideal gas
gas equations
equations analogous
analogous to
to Eqs.
Eqs. (3.30a)
(3.30a) and
and (3.30b)
(3.30b) are
are readily
readily derived:
derived:
For
Es fácil deducir las ecuaciones
análogas a las ecuaciones (3.30a)
y (3.30b) para un gas ideal:
δ−1 =
(1−δ)/δ =
TT VV δ−1
constant
(3.35b)
TT PP(1−δ)/δ
=
constant
(3.35b)
= constant
constant (3.35c)
(3.35c)
δ−1
(1−δ)/δ
δ−1
(1−δ)/δ
δ−1 =
Vd–1
= constant
constant
(3.35b) TT P
P(1−δ)/δ = constant
constant (3.35c)
TT V
(3.35b)
TV
= constante
(3.35b) TP(1–d)/d = =
constante (3.35c)
(3.35c)
��
When
the
relation
between
P
and
V
is
given
by
Eq.
(3.35a),
evaluation
When the relation between P and V is given by Eq. (3.35a), evaluation of
of �� PP ddVV yields
yields
When
the relation
relation
between
P dada
and V
Vpor
is given
given
by Eq.
Eq.(3.35a),
(3.35a), la
evaluation
of deP
P∫ddPV
V dV
yields
When
the
P
and
is
by
(3.35a),
evaluation
of
yields
Cuando
la relación
entre between
P
y Vδ:está
la ecuación
evaluación
produce la
Eq.
Eq. (3.34)
(3.34) with
with γγ replaced
replaced by
by δ:
Eq.
(3.34)
with
replaced by
by
δ:d:
Eq.
(3.34)
with
γγ replaced
δ:
ecuación
(3.34)
con
γ reemplazada
por
�
�
��
�
�(δ−1)/δ
�P �
�
�
(δ−1)/δ
�
�
RT
1
2
�
�
�
�
P2 (δ−1)/δ
RT1
(δ−1)/δ − 1
(δ−1)/δ
(3.36)
W
=
P222
RT111
P
−
1
(3.36)
W = δRT
PP11
− 11
(3.36) (3.36)
W=
= δ−
−
(3.36)
W
− 11
δ
−
1
P
δ−1
P111
Moreover,
Moreover, for
for constant
constant heat
heat capacities,
capacities, the
the first
first law
law solved
solved for
for Q
Q yields:
yields:
Moreover,
for constant
constant
heat capacities,
capacities,
thela
first
law solved
solved
for Q
Q yields:
yields:
Moreover,
for
heat
the
first
law
for
Además,
para capacidades
caloríficas
constantes,
primera
ley resuelta
para
Q produce:
�
�
�
��
�
(δ−1)/δ
�
�
��
�
�
(δ
�(δ−1)/δ
� PP22 �
(δ −
− γγ )RT
)RT11 �
(δ−1)/δ − 1
(δ−1)/δ
(3.37)
Q
P222 (δ−1)/δ
(δ −
− γγ )RT
)RT111
P
−1
(3.37)
Q=
= (δ(δ
−
1)(γ
−
1)
P
− 11
(3.37) (3.37)
Q=
= (δ − 1)(γ − 1)
−
(3.37)
Q
P11
(δ −
− 1)(γ
1)(γ −
− 1)
1)
P111
(δ
P
03-SmithVanNess.indd 78
8/1/07 12:53:21
79
3.3. El gas ideal
Los procesos descritos en esta sección corresponden a las cuatro trayectorias que se muestran en la figura 3.6
para valores específicos de d:
• Proceso isobárico: mediante la ecuación (3.35a), d = 0.
• Proceso isotérmico: por la ecuación (3.35b), d = 1.
• Proceso adiabático: d = γ.
• Proceso isocórico: mediante la ecuación (3.35a), dV/dP = V/Pd; para V constante, d = ±∞.
0
1
P

Figura 3.6: Trayectorias de procesos
politrópicos caracterizados por valores
específicos de d.

V
Proceso irreversible
Todas las ecuaciones desarrolladas en esta sección fueron deducidas para procesos mecánicamente reversibles en sistemas cerrados para gases ideales. Sin embargo, las ecuaciones que proporcionan cambios de propiedad (dU, dH, ∆U y ∆H) son válidas para gases ideales sin considerar el proceso. se aplican de igual modo
a procesos reversibles e irreversibles tanto en sistemas cerrados como abiertos, ya que los cambios en las
propiedades dependen sólo de los estados inicial y final del sistema. Por otra parte, una ecuación para Q o W,
a menos que sea igual al cambio en una propiedad, se somete a las restricciones en su deducción.
El trabajo de un proceso irreversible se calcula con un procedimiento de dos pasos. Primero, se determina W para un proceso mecánicamente reversible que alcanza el mismo cambio de estado que el proceso
irreversible real. Segundo, este resultado se multiplica o se divide por una eficiencia para obtener el trabajo
real. Si el proceso produce trabajo, el valor absoluto para el proceso reversible es muy grande y debe multiplicarse por una eficiencia. Si el proceso requiere trabajo, el valor para el proceso reversible es muy pequeño
y debe dividirse entre la eficiencia.
Las aplicaciones de los conceptos y de las ecuaciones de esta sección se ilustran en los siguientes ejemplos. En particular, el trabajo de los procesos irreversibles se trata en el ejemplo 3.4.
Véase el problema 3.13.
03-SmithVanNess.indd 79
8/1/07 12:53:23
80
CAPÍTULO 3. Propiedades volumétricas de fluidos puros
Ejemplo 3.2
Se comprime aire a partir de la condición inicial de 1 bar y 25 °C hasta alcanzar un estado final de 5
bar y 25 °C mediante tres procesos diferentes y mecánicamente reversibles en un sistema cerrado:
a) Calentamiento a volumen constante seguido por un enfriamiento a presión constante.
b) Compresión isotérmica.
c) Compresión adiabática seguida por enfriamiento a volumen constante.
Suponga que el aire es un gas ideal con capacidades caloríficas constantes, CV = (5/2)R y C P = (7/2)R.
Calcule el trabajo requerido, el calor transferido y los cambios en la energía interna y la entalpía del
aire para cada proceso.
10
8
Figura 3.7: Diagrama para el ejemplo 3.2.
P/bar
c
6
2
4
a
b
2
1
0
5
10
15
20
25
V  103/m3
Solución 3.2
Elija como sistema 1 mol de aire. Para R = 8.314 J mol–1 K–1,
CV = 20.785 CP = 29.099 J mol–1 K–1
Los estados inicial y final del aire son idénticos a los del ejemplo 2.9, donde:
V1 = 0.02479 V2 = 0.004958 m3
Ya que T es la misma al principio y al final del proceso, para todos los casos:
DU = DH = 0
a) El proceso en este caso es exactamente el del ejemplo 2.9b), para el cual:
Q = –9 915 J y 03-SmithVanNess.indd 80
W = 9 915 J
8/1/07 12:53:24
3.3. El 3.3.
gas ideal
The Ideal Gas
3.3.
3.3. The
The Ideal
Ideal Gas
Gas
3.3.
The
Ideal
Gas
81
81
81
81
81
b) Aplicamos
la ecuación
la compresión
isotérmica
un gas
(b) Equation
(3.27)(3.27)
for thepara
isothermal
compression
of ande
ideal
gas ideal:
applies:
(b)
Equation
(3.27)
for
the
isothermal
compression
of
an
ideal
(b) Equation
Equation (3.27)
(3.27) for
for the
the isothermal
isothermal compression
compression of
of an
an ideal
ideal gas
gas applies:
applies:
(b)
gas
applies:
P1
1
Q = −W = RT ln P
(8.314)(298.15) ln 111 = −3,990 J
PP111 =
Q
(8.314)(298.15)
2 =
Q=
= −W
−W =
= RT
RT ln
ln P
=
(8.314)(298.15)ln
ln 55 =
= −3,990
−3,990 JJJ
Q
=
−W
=
RT
ln
=
(8.314)(298.15)
ln
=
−3,990
P
PP222
55
(c) The
initial
step
of adiabaticadiabática
compression
takes
the air
it to
its final
c) La etapa
inicial
la compresión
toma
el aire
y lo
lleva
afinal
un volume
volumenof
(c)
initial
step
of
adiabatic
compression
takes
the
air
it
volume
of
3de
(c) The
The
initial
step
of (3.30a),
adiabaticthe
compression
takes
thepoint
air
to its
its final
volume
offinal de
(c)
The
initial
step
of
adiabatic
compression
takes
the
air
ititis:to
to
its
final
volume
of
0.004958
m
.
By
Eq.
temperature
at
this
3
3
0.004958
m
.
Usando
la
ecuación
(3.30a),
la
temperatura
en
este
punto
es:
3
3
0.004958
m
.
By
Eq.
(3.30a),
the
temperature
at
this
point
is:
0.004958 m
m .. By
By Eq.
Eq. (3.30a),
(3.30a), the
the temperature
temperature at
at this
this point
point is:
is:
0.004958
� �γ −1
�
�0.4
�
�
�
�
�V
�γγγ−1
� 0.02479
�0.4
�
−1 = (298.15) �
0.4 = 567.57 K
−1
VV1111�
0.02479
T �� = T1 �
0.02479 0.4
0.02479
�� = T1 V
T
=
(298.15)
=
V
0.004958
= TT11 V2
= (298.15)
(298.15) 0.004958
= 567.57
567.57 K
K
TT =
=
=
567.57
K
0.004958
VV222
0.004958
For this step, Q = 0, and by Eq. (3.32), the work of compression is:
=
0,
and
(3.32),
work
compression
For this
this
step,
Qpor
= la
0, ecuación
and by
by Eq.
Eq.
(3.32),
the
workdeof
ofcompresión
compression
is:
For
this
step,
Q
=
0,
and
by
Eq.
(3.32),
the
work
of
compression
is:
En esta For
etapa
Q step,
= 0 y,Q
(3.32),
el the
trabajo
es:is:
W = C V �T = (20.785)(567.57 − 298.15) = 5,600 J
W
�T =
= (20.785)(567.57
(20.785)(567.57 −
− 298.15)
298.15) =
= 5,600
5,600 JJJ
W=
=C
CVVV �T
�T
=
(20.785)(567.57
−
298.15)
=
5,600
W
=
C
For the constant-volume step, no work is done; the heat transfer is:
For
constant-volume
step,
no
work
done;
For the
the
constant-volume
step,
noefectúa
work is
istrabajo;
done; the
the
heat transfer
transfer is:
is:
For
the
constant-volume
step,
no
work
is
done;
the
heat
transfer
is:
Para la etapa
a volumen
constante
no se
la heat
transferencia
de calor es:
�
Q = �U = C V (T2 − T � ) = −5,600 J
Q
(T222 −
−T
= −5,600
−5,600 JJJ
Q=
= �U
�U =
=C
CVVV(T
(T
−
TT��))) =
=
−5,600
Q
=
�U
=
C
Thus for process (c),
for
process
(c),
De esta Thus
manera,
el proceso
c),
Thus
for para
process
(c),
Thus
for
process
(c),
WW
==
5,600
J J and
5 600
y QQ
==
–5−5,600
600 J J
W
and
Q
W=
= 5,600
5,600 JJJ
and
Q=
= −5,600
−5,600 JJJ
W
=
5,600
and
Q
=
−5,600
Aunque Although
los cambios
la propiedad
∆U y�U
∆H and
son cero
cada for
proceso,
y W dependen
de
theenproperty
changes
�H para
are zero
each Q
process,
Q
Although
the
changes
�U
and
�H
are
zero
for
each
process,
Q
Although
theQproperty
property
changes
�U
and
�H
are
zero
for
each
process,
Q Ya que
the
property
�U
�H
are
zero
process,
Q
la trayectoria,
aquí
= –W. Lachanges
figura
proceso
en each
un
diagrama
PV.
and Although
W pero
are path-dependent,
but
here3.7
Q muestra
=and
−W
.cada
Figure
3.7for
shows
each
process
and
W
are
path-dependent,
but
here
Q
=
−W
3.7
each
process
and
W
arediagram.
path-dependent,
but
here
Qfor
=each
−W..of
. Figure
Figure
3.7 shows
shows
each
process
and
are
path-dependent,
but
here
Q
=
−W
Figure
3.7
shows
each
process
el trabajo
para
cada
uno de estos
procesos
mecánicamente
reversibles
es conocido
por W = –∫ P
on
aW
PV
Because
the
these mechanically
reversible
� work
on
a
P
V
diagram.
Because
the
work
for
each
of
these
mechanically
reversible
on aa PPde
diagram.
Because
the
work
for
each
of
these
mechanically
reversible
on
VVcada
diagram.
for
each
of
these
reversible
��� Pwork
dV, el trabajo
proceso
es proporcional
área
total
debajo
de
las trayectorias
en el diagraprocesses
is
given
byBecause
W
= −the
d V ,al
the
work
for
eachmechanically
process
is proportional
processes
is
given
by
W
−
P
dddV
,,, the
work
for
each
is
proportional
processes
is
given
by
W=
=
− on
the
work
forfrom
each1process
process
islos
proportional
processes
by
W
=
−
PP the
VVde
work
for
each
process
proportional
ma PV que
vatotal
de 1is
a given
2.
Los
tamaños
relativos
estas
áreas
corresponden
ais
valores
numéricos
to
the
area
below
the
paths
Pthe
V
diagram
to 2. The
relative
sizes
to
area
below
paths
on
diagram
from
1 to
to the
the total
total
area
below the
theto
paths
on the
the P
diagram
from
to 2.
2. The
The relative
relative sizes
sizes
the
total
area
below
the
paths
the
PPV
VV values
diagram
2.
The
relative
sizes
de W. to
of
these
areas
correspond
the on
numerical
offrom
W . 11 to
of
of these
these areas
areas correspond
correspond to
to the
the numerical
numerical values
values of
of W
W...
of
these
areas
correspond
to
the
numerical
values
of
W
Ejemplo
3.3 3.3
Example
Example
3.3
Example
3.3la siguiente secuencia de procesos mecánicamente reversibles en un sisteExample
3.3
Un gas ideal experimenta
An ideal gas undergoes the following sequence of mechanically reversible processes
An
gas
undergoes
ma cerrado:
Anaideal
ideal
gas
undergoes the
the following
following sequence
sequence of
of mechanically
mechanically reversible
reversible processes
processes
An
ideal
gas
undergoes
the
following
sequence
of
mechanically
reversible
processes
in
closed
system:
in
in a
closed system:
system:
in
aa closed
closed
system:
a) Desde un estado inicial de 70 °C y◦ 1 bar, se comprime adiabáticamente hasta la temperatura
de
(a) From an initial state of 70◦ C and 1 bar, it is compressed adiabatically to 150◦◦ C.
◦
◦
◦
◦
From
an
initial
state
of
70
C
and
1
bar,
it
is
compressed
adiabatically
to
150
C.
150 (a)
°C.
(a) From
From an
an initial
initial state
state of
of 70
70 C
C and
and 11 bar,
bar, itit is
is compressed
compressed adiabatically
adiabatically to
to 150
150 C.
C.
(a)
◦
(b) It is then
from
150
to °C
70 C at constant
pressure.
b) A continuación
secooled
enfría de
150
a 70
constante.
(b) It is then
cooled
from
150
to 70◦◦a◦Cpresión
at constant
pressure.
(b) ItIt is
is then
then cooled
cooled from
from 150
150 to
to 70
70 C
C at
at constant
constant pressure.
pressure.
(b)
Finally,
it is expanded
isothermally
to its
state.
c) Por (c)
último,
se expande
isotérmicamente
hasta
su original
estado original.
(c)
(c) Finally,
Finally, it
is expanded
expanded isothermally
isothermally to
to its
its original
original state.
state.
(c)
Finally,
itit is
is
expanded
isothermally
to
its
original
state.
Q, para
�U , cada
and �H
of procesos
the three yprocesses
andcompleto.
for the entire
CalculeCalculate
W, Q, ∆UW
y ,∆H
uno for
de each
los tres
para el ciclo
Tomecycle.
CV = (3/2)R
Calculate
W
,,, Q,
,,, and
�H
for
Calculate
W(3/2)R
Q, �U
�U
and
�H
for each
each of
of the
the three
three processes
processes and
and for
for the
the entire
entire cycle.
cycle.
Calculate
Q,
�U
and
for
each
of
the
three
processes
and
for
the
entire
cycle.
C V =W
and
C P�H
= (5/2)R.
y C P = Take
(5/2)R.
Take
Take C
CVVV =
= (3/2)R
(3/2)R and
and C
CPPP =
= (5/2)R.
(5/2)R.
Take
C
=
(3/2)R
and
C
=
(5/2)R.
03-SmithVanNess.indd 81
8/1/07 12:53:31
82
CHAPTER
CHAPTER 3.
3. Volumetric
Volumetric Properties
Properties of
of Pure
Pure Fluids
Fluids
CHAPTER 3. Volumetric Properties of Pure Fluids
82
82
82
82
82
CHAPTER 3. Volumetric Properties of Pure Fluids
CHAPTER 3. Volumetric Properties of Pure Fluids
CAPÍTULO
3. Propiedades volumétricas de fluidos puros
3
3
70°C 3
70°C
70°C
3
70°C 3
70°C
Figure
Figure 3.8:
3.8: Diagram
Diagram for
for Ex.
Ex. 3.3.
3.3.
Figure 3.8: Diagram for Ex. 3.3.
Figure 3.8: Diagram for Ex. 3.3.
FiguraFigure
3.8: Diagrama
para el
3.8: Diagram
forejemplo
Ex. 3.3.3.3.
PP
P
P
P
b 22
b
150°C
b 2 150°C
150°C
b 2
150°C
b 23
b 2
150°C
150 °C
70 °C
a
a
a
c
c
c
P
a
a
c
c
c
a 70°C
70°C
1 70°C
1
1
70°C
1 70°C
1
VV
V
V
V
1
70 °C
V
Solution 3.3
Solution
3.3
Solución
3.3
Solution
3.3
−1
−1
–1, −1 K −1,
R
Para Solution
R =For
8 314
mol
For
R=
=J 8.314
8.314
mol
3.3–1 JJKmol
−1 K−1 ,
For R = 8.314
Solution
3.3 J mol K ,
−1 −1
−1 K−1 ,
C
12.471
C
For R = 8.314
mol
CV J=
=
12.471
CP =
= 20.785
20.785 JJ mol
mol−1 K
K−1
−1 K−1 ,
C VVJ =
12.471
C PP = 20.785 J mol−1 K−1
For R = 8.314
mol
−1 K
is
shown
on
3.8.
Take
as
aa basis
11−1
mol
12.471
C Pfigura
= 20.785
J mol
V = en
The
cycle
is C
shown
on aa P
PV
V diagram
diagram
in
Fig.
3.8.
Take
as
basis
mol
of
gas.
El cicloThe
estácycle
representado
PVin
enFig.
la
3.8. Tome
como
baseof
1 gas.
mol
−1
The
cycle
isCshown
onuna diagrama
P V diagram
in
Fig.
Take
as
a basis
mol
of
gas. de gas.
C P 3.8.
= 20.785
J mol
K1−1
V = 12.471
a) Para (a)
un
gas
ideal
que
se
somete
a
una
compresión
adiabática,
Q
=
0,
de
donde
an
gas
adiabatic
compression,
Q
(a) For
For
an ideal
ideal
gas undergoing
undergoing
adiabatic
compression,
Qa=
= 0;
0; whence
whence
The
cycle
shown
a P V diagram
in Fig.
3.8. Take asQ
1 mol of gas.
(a) For
an isideal
gas on
undergoing
adiabatic
compression,
=basis
0; whence
The cycle is shown on a P V diagram in Fig. 3.8. Take as a basis 1 mol of gas.
W
=
−
JJ
(a) For an ideal�U
gas=
adiabatic
compression,
Q ==
whence
�U
=undergoing
W=
=C
CV �T
�T
= (12.471)(150
(12.471)(150
− 70)
70)
=0;998
998
W = C VV �T
= (12.471)(150
− 70)
J
(a) For an ideal�U
gas=
undergoing
adiabatic
compression,
Q ==0;998
whence
�U
=
W
=
C
�T
=
(12.471)(150
−
70)
=
998
J
�T
=
(20.785)(150
−
70)
=
1,663
J
�H
=
C
V = (20.785)(150 − 70) = 1,663 J
�H = CPP �T
�U
==
W C=P C
= (12.471)(150
− 70)
= 998J J
�H
�T
= (20.785)(150
− 70)
= 1,663
V �T
from
Eq.
(3.30b):
Pressure
is found
found
from
Eq.
(3.30b):
Pressure
P22 is
La presión
P2 se P
puede
encontrar
a
partir
de
la
ecuación
(3.30b):
�H
=
C
�T
=
(20.785)(150
−
70)
=
1,663
J
PEq. (3.30b):
from
Pressure P2 is found
�H
=
C
�T
=
(20.785)(150
−
70)
=
1,663
J
P
�
�
�
� 2�
�γγ/(γ
/(γ
−1)
2.5
Eq.−1)
(3.30b): �150 + 273.15 �2.5
Pressure P2 is found
� TTfrom
2 �γ /(γ −1) = (1) � 150 + 273.15 �2.5 = 1.689 bar
P
=
P
2
1
+
273.15
Eq. (3.30b):
Pressure PP22 is=found
P1 TTfrom
= (1) 150
= 1.689 bar
2
P2 = P1 � T11 �γ /(γ −1) = (1) � 70
70 +
+ 273.15
273.15 �2.5 = 1.689 bar
70
+
273.15
T
150
+
273.15
21
P2constant-pressure
= P1 � T2 �γ /(γ −1)
= (1) � 150 + 273.15 �2.5 = 1.689 bar
(b)
process,
T
(b) For
For this
this
constant-pressure
process,
1
P
=
P
=
(1) 70 + 273.15
= 1.689 bar
1
(b) For this2 constant-pressure
process,
T1 constante:
70 + 273.15
b) Para este proceso a presión
=
Q
(b) For this constant-pressure
process,
�T
= (20.785)(70
(20.785)(70 −
− 150)
150) =
= −1,663
−1,663 JJ
Q=
= �H
�H =
=C
CP �T
= (20.785)(70 − 150) = −1,663 J
Q = �H = C PP �T
(b) For this constant-pressure
process,
(20.785)(70
150)==−998
−1,663
Q =�U
�H=
�T
(12.471)(70
−−
150)
J J
P �T==
�U
==C
CVC
V �T = (12.471)(70 − 150) = −998 J
�T==(12.471)(70
(20.785)(70−−150)
150)==−998
−1,663
Q =�U
�H==CC
J J
V P�T
W
=
�U
−
Q
=
−998
−
(−1,663)
=
665
J
�U
==
(12.471)(70
− 150) =
W=
= C�U
−Q
−998 − (−1,663)
= −998
665 J J
V �T
W=
=C�U
− Q = −998 − (−1,663) = 665 J
�U
V �T = (12.471)(70 − 150) = −998 J
(c)
process,
(c) For
For ideal
ideal gases
gases
undergoing
an
isothermal
process,=�U
�U
and �H
�H are
are zero;
zero;
W undergoing
= �U − Qan
= isothermal
−998
− (−1,663)
665 and
Jand
(c) (3.27)
For ideal
gases
undergoing
an
isothermal
process,=
�U
�H are zero;
Eq.
yields:
W
=
�U
−
Q
=
−998
−
(−1,663)
665
J
Eq. (3.27) yields:
Eq. For
(3.27)
yields:
(c)
ideal
gases undergoing an isothermal process, �U and �H are zero;
c) Para (c)
gases
ideales
que seundergoing
someten
a an
un isothermal
proceso
isotérmico,
∆H �H
son cero;
la ecuación
P
P
For
ideal
gases
process, ∆U
�U y1.689
and
are zero;
Eq.Q(3.27)
yields:
P2
1.689
P3
=
1,495
JJ
=
=
P22 =
P33 =
= RT
RT ln
ln P
= (8.314)(343.15)
(8.314)(343.15) ln
ln 1.689
=
1,495
Q
= −W
−W
= RT
RT ln
ln P
(3.27) produce:
Eq. Q
(3.27)
yields:
= −W = RT ln P11 = RT ln P11 = (8.314)(343.15) ln 11 = 1,495 J
1
P21
1.689
P31
Q = −W = RT ln P3 = RT ln P2 = (8.314)(343.15) ln 1.689 = 1,495 J
Q = −W = RT ln P1 = RT ln P1 = (8.314)(343.15) ln 1 = 1,495 J
P1
P1
1
03-SmithVanNess.indd 82
8/1/07 12:53:49
3.3. The
The Ideal
Ideal Gas
Gas
3.3.
3.3.
The
Ideal
Gas
The
3.3. El3.3.
gas ideal
3.3.
The Ideal
Ideal Gas
Gas
83
83
83
83
83
83
For the
the
entire cycle,
cycle,
For
entire
For
the
entire
cycle,
Para todo
ciclo,
For
entire
Forelthe
the
entire cycle,
cycle,
Q=
= 000 −
− 1,663
1,663 +
+ 1,495
1,495 =
= −168
−168 JJJ
Q
Q
=
−
1,663
+
1,495
=
−168
Q
Q=
= 00 −
− 1,663
1,663 +
+ 1,495
1,495 =
= −168
−168 JJ
W =
= 998
998 +
+ 665
665 −
− 1,495
1,495 =
= 168
168 JJJ
W
W
=
998
+
665
−
1,495
=
168
W
W =
= 998
998 +
+ 665
665 −
− 1,495
1,495 =
= 168
168 JJ
�U
=
998
−
998
+
0
=
0
�U
�U =
= 998
998 −
− 998
998 +
+ 000 =
= 000
�U
�U =
= 998
998 −
− 998
998 +
+ 0=
=0
�H
=
1,663
−
1,663
+
= 000
�H
=
1,663
−
1,663
+
�H =
= 1,663 −
− 1,663 +
+ 000 =
=
�H
�H = 1,663
1,663 − 1,663
1,663 + 00 =
= 00
The property
property changes
changes �U
�U and
and �H
�H both
both are
are zero
zero for
for the
the entire
entire cycle,
cycle, because
because the
the
The
The
property
changes
�U
and
�H
both
are
zero
for
the
entire
cycle,
because
the
The
property
changes
�U
and
both
are
zero
for
the
entire
cycle,
the
Los cambios
en
la final
propiedad
∆U
yidentical.
∆H�H
son cero
para
elthat
ciclo
completo,
yathe
quebecause
los estados
The
property
changes
�U
and
�H
both
are
zero
for
the
entire
cycle,
because
the inicial
initial
and
final
states
are
identical.
Note
also
that
Q
=
−W
for
the
cycle.
This
initial
and
states
are
Note
also
Q
=
−W
for
cycle.
This
initial
and final
final
states are
are
identical.
Note
also
thatelQ
Q=
= −W
−W
for
the cycle.
cycle.
This
initial
and
states
identical.
Note
also
that
the
This
y final son
idénticos.
Observe
también
que==Q
W para
Estefor
resultado
surge
de la priinitial
and
final
states
are
identical.
Note
that Qciclo.
= −W
for
the cycle.
This
follows
from
the
first law
law
with
�U
0.= – also
follows
from
the
first
with
�U
0.
follows
from
the
first
law
with
�U
=
0.
follows
from
the
first
law
with
�U
=
0.
mera leyfollows
con ∆Ufrom
= 0.the first law with �U = 0.
Example
3.4
Example
3.4
Example
3.4
Ejemplo
3.4
If
the
processes
of Ex.
Ex. 3.3
3.3 are
are carried
carried out
out irreversibly
irreversibly but
but so
so as
as to
to accomplish
accomplish exactly
exactly
If
the
processes
If the processes of
of
Ex.
3.3
are
carried
out
irreversibly
but
so
as
to
accomplish
exactly
IfIf the
the processes
processes of
of Ex.
Ex. 3.3
3.3 are
are carried
carried out
out irreversibly
irreversibly but
but so
so as
as to
to accomplish
accomplish exactly
exactly
the same
samedel
changes
of
state—
the same
same changes
changes
in P,
P, T
U
and H
H—
— then
then different
differentlos misthe
changes
state—
the
in
the
same
changes
of
state—
the
same
changes
in
P,
TT ,,,, U
U
and
H
—
then
different
Si los procesos
ejemploof
3.3
son irreversibles,
pero de modo
se,,,, and
consigan
the
same
changes
of
state—
the
changes
in
P,
T
U
H
then
different
the
same
changes
of
state—
the same
sameQ
changes
in
P, que
T ,step
U
, and
and
H—
— exactamente
then
different
values
of
Q
and
W
result.
Calculate
Q
and
W
if
each
step
is
carried
out
with an
an
values
of
Q
and
W
result.
Calculate
and
W
if
each
is
carried
out
with
valuesde
ofestado
Q and
and(los
W mismos
result. cambios
Calculate
QP, and
and
WH),ifif por
each
step se
is producen
carried out
out
with
an
mos cambios
enQ
U yW
lo tanto
diferentes
values
of
Q
step
with
values
of of
Q 80%.
and W
W result.
result. Calculate
Calculate
Q T,
and
W if each
each
step is
is carried
carried out
with an
anvalores
efficiency
of
80%.
efficiency
efficiency
ofQ80%.
80%.
de Q y efficiency
W. Calculeof
y W si cada etapa se lleva a cabo con una eficiencia de 80%.
efficiency of 80%.
Solution
3.4
Solución
3.4
Solution
3.4
Solution
3.4
Solution 3.4
If the
the same
same
changes
of state
statede
asestado
in Ex.
Ex.que
3.3 en
are carried
carried
out3.3
by irreversible
irreversible
processes,
If
changes
of
as
in
3.3
are
out
by
If
the
same
changes
of
state
as
in
Ex.
3.3
are
carried
out
by
irreversible
processes,
Si se realizan
los mismos
cambios
ejemplo
medio deprocesses,
procesos irreverIf
the
same
changes
of
state
as
in
3.3
are
out
by
irreversible
processes,
If
the
same
changes
offor
state
assteps
in Ex.
Ex.
3.3
areelcarried
carried
out
bypor
irreversible
processes,
the
property
changes
for
the
steps
are
identical
with
those
of
Ex.
3.3.
However,
the
property
changes
the
are
identical
with
those
of
Ex.
3.3.
However,
the
property
changes
for
the
steps
are
identical
with
those
of
Ex.
3.3.
However,
sibles, los
cambios
en
la
propiedad
para
las
etapas
son
idénticos
a
los
del
ejemplo
3.3. No obstanthe
property
changes for
the
the
property
the steps
steps are
are identical
identical with
with those
those of
of Ex.
Ex. 3.3.
3.3. However,
However,
the values
values
ofychanges
Q
and
Wfor
change.
the
of
Q
and
W
change.
the
values
of
Q
and
W
change.
te, los valores
de
Q
W
se
modifican.
the
the values
values of
of Q
Q and
and W
W change.
change.
(a)
For
mechanically
reversible,
adiabatic compression,
compression,
W==
=
998
J. If
If
the
process
(a)
For
mechanically
reversible,
adiabatic
W
998
process
(a) For
For
mechanically
reversible,
adiabatic
compression,
W
=
998
J.
If
the
process
a) Para (a)
una
compresión
adiabática,
mecánicamente
reversible, W
998
J. J.
Si
elthe
proceso
tiene 80%
mechanically
reversible,
adiabatic
compression,
W
=
998
J.
the
process
(a)
For
mechanically
reversible,
adiabatic
compression,
W
=
998
J. If
If
the
process
is
80%
efficient
compared
with
this,
W
=
998/0.80
=
1,248
J.
This
step
cannot
is
80%
efficient
compared
with
this,
W
=
998/0.80
=
1,248
J.
This
step
cannot
is 80%
80%enefficient
efficient
compared
with
this,
W
= 998/0.80
998/0.80
= J.
1,248
J. This
This
step
cannot
de eficiencia
comparación
con
esto,
W
=
998/0.80
=
1
248
En
este
caso
no
es
posible
que
is
compared
with
this,
W
=
=
1,248
J.
step
cannot
is
80%
efficient
with
this, W = 998/0.80 = 1,248 J. This step cannot
here
beadiabática.
adiabatic.compared
By
the
first
law,
here
be
adiabatic.
By
the
first
law,
here
be
adiabatic.
By
the
first
law,
esta etapa
sea
Por
la
primera
ley,
here
be
adiabatic.
By
the
first
law,
here be adiabatic. By the first law,
Q=
= �U
�U −
−W
W=
= 998
998 −
− 1,248
1,248 =
= −250
−250 JJJ
Q
Q
=
�U
−
W
=
998
−
1,248
=
−250
Q
Q=
= �U
�U −
−W
W=
= 998
998 −
− 1,248
1,248 =
= −250
−250 JJ
(b) The
The
work
for the
thede
mechanically
reversible
coolingreversible
process is
ises665
665
J. For
For
the el pro(b)
work
for
mechanically
cooling
process
J.
the
b) El trabajo
para
el proceso
enfriamientoreversible
mecánicamente
de 665
J. Para
(b)
The
work
for
the
mechanically
reversible
cooling
process
is
665
J.
For
the
(b)
The
work
for
the
mechanically
reversible
cooling
process
is
665
J.
For
the
(b)
The
work
for
the
mechanically
reversible
cooling
process
is
665
J.
For
the
irreversible
process,
W
=
665/0.80
=
831
J,
and
irreversible
process,
W
=
665/0.80
=
831
J,
and
ceso irreversible,
= 665/0.80
= 831
J, y = 831
irreversibleWprocess,
process,
W=
=
665/0.80
831 J, and
and
irreversible
irreversible process, W
W = 665/0.80
665/0.80 =
= 831 J,
J, and
Q=
= �U
�U −
−W
W=
= −998
−998 −
− 831
831 =
= −1,829
−1,829 JJJ
Q
Q
=
�U
−
W
=
−998
−
831
=
−1,829
Q
Q=
= �U
�U −
−W
W=
= −998
−998 −
− 831
831 =
= −1,829
−1,829 JJ
c) A medida
que
el trabajo
es by
realizado
por elin
en the
esta
etapa, el trabajo
(c) As
As
work
is done
done
by
the system
system
insistema
this step,
step,
the irreversible
irreversible
workirreversible
in absolute
absoluteen valor
(c)
work
is
the
this
work
in
(c)
As
work
is
done
by
the
system
in
this
step,
the
irreversible
work
in
absolute
(c)
As
work
is
by
the
system
in
this
step,
the
(c)
As
work
is done
done
by
the
system
in
this
step,J:J:
the irreversible
irreversible work
work in
in absolute
absolute
absolutovalue
es
menor
que
el
trabajo
reversible
de
–1
495
value
is
less
than
the
reversible
work
of
−1,495
J:
is
less
than
the
reversible
work
of
−1,495
value is
is less than
than the reversible
reversible work of
of −1,495 J:
J:
value
value is less
less than the
the reversible work
work of −1,495
−1,495 J:
W=
= (0.80)(−1,495)
(0.80)(−1,495) =
= −1,196
−1,196 JJJ
W
W
=
(0.80)(−1,495)
=
−1,196
W
W=
= (0.80)(−1,495)
(0.80)(−1,495) =
= −1,196
−1,196 JJ
Q=
= �U
�U −
−W
W=
= 000 +
+ 1,196
1,196 =
= 1,196
1,196 JJJ
Q
Q
=
�U
−
W
=
+
1,196
=
1,196
Q
Q=
= �U
�U −
−W
W=
= 00 +
+ 1,196
1,196 =
= 1,196
1,196 JJ
For the
the entire
entire cycle,
cycle, �U
�U and
and �H
�H are
are zero,
zero, with
with
For
For
the
entire
cycle,
�U
and
�H
are
zero,
with
entire
and
are
zero,
Para el For
ciclo
completo,
∆ U �U
y�U
∆H
son�H
cero,
For the
the
entire cycle,
cycle,
and
�H
arecon
zero, with
with
Q=
= −250
−250 −
− 1,829
1,829 +
+ 1,196
1,196 =
= −883
−883 JJJ
Q
Q
=
−250
−
1,829
+
1,196
=
−883
Q
Q=
= −250
−250 −
− 1,829
1,829 +
+ 1,196
1,196 =
= −883
−883 JJ
W=
= 1,248
1,248 +
+ 831
831 −
− 1,196
1,196 =
= 883
883 JJJ
W
W
=
1,248
+
831
−
1,196
=
883
W
W=
= 1,248
1,248 +
+ 831
831 −
− 1,196
1,196 =
= 883
883 JJ
A summary
summary of
of these
these results
results and
and those
those for
for Ex.
Ex. 3.3
3.3 is
is given
given in
in the
the following
following table;
table;
A
A
summary
of
these
results
and
those
for
Ex.
3.3
is
given
in
the
following
table;
A
summary
of
these
results
and
those
for
Ex.
3.3
is
given
in
the
following
table;
A
summary
of
these
results
and
those
for
Ex.
3.3
is
given
in
the
following
table;
En la siguiente
tabla
se
proporciona
un
resumen
de
estos
resultados
y
los
correspondientes
para
values
are
in
joules.
values
values are
are in
in joules.
joules.
values
in
values
are
in joules.
joules.
el ejemplo
3.3,are
donde
los valores se proporcionan en joules.
03-SmithVanNess.indd 83
8/1/07 12:54:02
84
CAPÍTULO 3. Propiedades volumétricas de fluidos puros
Mecánicamente reversible, ejemplo 3.3
∆U
∆H
a)
998
1 663
b)
–998
c)
Suma
Irreversible, ejemplo 3.4
W
∆U
∆H
Q
W
0
998
998
1 663
–250
1 248
–1 663
–1 663
665
–998
–1 663
–1 829
831
0
0
1 495
–1 495
0
0
1 196
–1 196
0
0
–168
168
0
0
–883
883
Q
Es un ciclo que requiere trabajo y produce una cantidad igual de calor. La característica más
sorprendente de la comparación mostrada en la tabla es que el trabajo total requerido cuando el
ciclo se compone de tres etapas irreversibles es más de cinco veces el trabajo total requerido
cuando las etapas son mecánicamente reversibles, aun cuando cada etapa irreversible se supone
con 80% de eficiencia.
Ejemplo 3.5
Se confina una cantidad de gas nitrógeno en un cilindro vertical mediante un pistón sin fricción. La
parte superior del pistón está abierta hacia la atmósfera. El peso del pistón hace que la presión del
nitrógeno sea 0.35 bar mayor que la de la atmósfera de los alrededores, que es de 1 bar y 27 °C. De
este modo, el nitrógeno se encuentra inicialmente a una presión de 1.35 bar y se encuentra en equilibrio mecánico y térmico con respecto a sus alrededores. El pistón es empujado dentro del cilindro, de
modo que comprime el nitrógeno hasta una presión de 2.7 bar. A esta presión, se permite que el nitrógeno llegue al equilibrio térmico con la atmósfera circundante a 27 °C. En este punto, el pistón se fija
en su sitio mediante seguros.
Los seguros se retiran liberando al pistón de su restricción, y con el tiempo el aparato regresa al
equilibrio mecánico y térmico con sus alrededores. Discuta la aplicación de la termodinámica a este
proceso. Suponga que el nitrógeno en estas condiciones es un gas ideal.
Solución 3.5
Cuando los seguros que sostienen el pistón sin fricción son removidos, dicho pistón se mueve
rápidamente hacia arriba, y debido a su inercia al ir más allá de su posición de equilibrio. Esta
expansión inicial se aproxima a un proceso adiabático reversible, ya que se produce poca turbulencia de un solo recorrido del pistón y porque la transferencia de calor es relativamente lenta. Sin
embargo, la oscilación consecutiva del pistón introduce la irreversibilidad como resultado de la
agitación y la turbulencia tanto del gas como de la atmósfera. Este proceso sigue durante un tiempo considerable, durante el cual se presenta transferencia de calor en una cantidad suficiente para
regresar el nitrógeno a su temperatura inicial de 27 °C a una presión de 1.35 bar.
No es posible especificar la trayectoria de un proceso irreversible, y esto hace imposible el
cálculo tanto de Q como de W. A diferencia del calor y el trabajo, es posible calcular los cambios
de la propiedad del sistema, ya que dependen únicamente del estado inicial y final, y éstos son
conocidos. Tanto ∆U como ∆H para el proceso de expansión son cero, porque las temperaturas
03-SmithVanNess.indd 84
8/1/07 12:54:03
The
Ideal
Gas Gas
3.3.
The Ideal
3.3. 3.3.
El gas
ideal
85 85
85
◦ C. The
inicial
final
son
27
°C.
primera
aplica
tanto para
procesos
reversiarey zero,
because
theLa
initial
and ley
final
temperatures
are 27
first
are zero,
because
the
initial
andsefinal
temperatures
are
27◦ C.irreversibles
The law
firstapplies
lawcomo
applies
bles, to
y nos
lleva
a
que:
irreversible
as well
as to as
reversible
processes,
and itand
becomes:
to irreversible
as well
to reversible
processes,
it becomes:
==
Q
+WW
de donde
==
–W−W
�U DU
=
+Q
Q =QQ
−W
�UQ
+==W00 =
0 whence
whence Aunque
no
es posible
calcular
Qcan
ni
W,
sus
absolutos
sonvalues
los values
mismos.
Eltheresultado
Although
neither
Q
norQ W
their
absolute
are the
Although
neither
nor
W be
cancalculated,
be valores
calculated,
their
absolute
aresame.
same. del
proceso
es
la
elevación
del
pistón
y
la
atmósfera,
así
como
una
disminución
en
la
compensación
The process
results
in elevation
of theofpiston
and the
and aand
compenThe process
results
in elevation
the piston
andatmosphere,
the atmosphere,
a compenen la sating
energía
interna
dethe
lainatmósfera
circundante.
decrease
in
internal
energy
of theofsurrounding
atmosphere.
sating
decrease
the internal
energy
the surrounding
atmosphere.
Ejemplo 3.6
Example
3.63.6
Example
En una
que
contiene
una
parcialmente
cerrada,
circulavalve.
aire con
una rapidez
Air tubería
flows
athorizontal
a at
steady
rate
through
a válvula
horizontal
pipe pipe
to a to
partly
closed
The
Air flows
a steady
rate
through
a horizontal
a partly
closed
valve.
The
uniforme.
La
tubería
que
deja
la
válvula
es
lo
suficientemente
larga
con
respecto
a
la
tubería
de entrapipe pipe
leaving
the valve
is enough
larger
thanthan
the entrance
pipe pipe
that that
the kinetic-energy
leaving
the valve
is enough
larger
the entrance
the kinetic-energy
da, de
modo
que
el
cambio
de
energía
cinética
del
aire
a
medida
que
fluye
a
través
de
la
válvula
change
of the
flows
through
the valve
is negligible.
The The
valvevalve
and and
connecting
change
ofair
theas
airit as
it flows
through
the valve
is negligible.
connecting es
despreciable.
Tanto
la
válvula
como
las
tuberías
de
conexión
se
encuentran
bien
aisladas.
◦ CLas
pipespipes
are well
insulated.
The The
conditions
of the
fromfrom
the valve
are 20
are well
insulated.
conditions
of air
theupstream
air upstream
the valve
are
20◦ Ccondiciones
del
aire
corriente
arriba
de
la
válvula
son
de
20
°C
y
6
bar,
y
la
presión
corriente
abajo
es de 3
and and
6 bar,
and and
the downstream
pressure
is 3 bar.
If airIfisair
regarded
as an
gas, gas,
6 bar,
the downstream
pressure
is 3 bar.
is regarded
asideal
an ideal
bar. what
Si el aire
se
considera
como
un
gas
ideal,
¿cuál
es
la
temperatura
del
aire
a
cierta
distancia
cois theistemperature
of the
distance
downstream
fromfrom
the valve?
what
the temperature
ofair
thesome
air some
distance
downstream
the valve?
rriente abajo de la válvula?
Solución
3.6 3.6 3.6
Solution
Solution
El flujo a través de una válvula parcialmente cerrada se conoce como un proceso de estrangulaFlowFlow
through
a partly
closed
valvevalve
is known
as a throttling
process.
The system
is is
through
a partly
closed
is known
as a throttling
process.
The system
ción. El sistema se aísla, lo cual hace que se ignore a Q; además, son despreciables los cambios
insulated,
making
Q negligible;
moreover,
the potential-energy
and kinetic-energy
insulated,
making
Q negligible;
moreover,
the potential-energy
and kinetic-energy
de energías potencial y cinética. Ya que no hay trabajo de flecha, Ws = 0. Por tanto, la ecuación
changes
are negligible.
Because
no shaft
workwork
is accomplished,
Ws =W0.
Hence,
changes
are negligible.
Because
no shaft
is accomplished,
0. Hence,
s =
(2.32) se reduce a ∆H = 0. Así, para un gas ideal,
Eq. (2.32)
reduces
to: �H
= 0. =
Thus,
for anfor
ideal
gas, gas,
Eq. (2.32)
reduces
to: �H
0. Thus,
an ideal
� T2 � T2
de donde �H �H
= = C P dT
C P=dT0 =
0 whence
whence T2 =TT21= T1
T1
T1
The result
that �H
=0=
is general
for a for
throttling
process,
because
the assumpThe result
that �H
0 is general
a throttling
process,
because
the assumpEl resultado de que ∆H = 0 es general para un proceso de estrangulación, ya que son válidas las
tionstions
of negligible
heat transfer
and potentialand kinetic-energy
changes
are usuof negligible
heat transfer
and potentialand kinetic-energy
changes
are ususuposiciones de que la transferencia de calor y los cambios en las energías potencial y cinética son
ally valid.
If theIffluid
is an is
ideal
gas, no
temperature
change
occurs.
The throttling
ally valid.
the fluid
an ideal
gas,
no temperature
change
occurs.
The throttling
despreciables. Si el fluido es un gas ideal, no ocurre cambio en la temperatura. El proceso de esprocess
is inherently
irreversible,
but this
immaterial
to thetocalculation,
because
process
is inherently
irreversible,
butisthis
is immaterial
the calculation,
because
trangulación es inherentemente irreversible, pero esto no tiene importancia en el cálculo, ya que
Eq. (3.21b)
is valid
for anfor
ideal
gas whatever
the process.
Eq. (3.21b)
is valid
an ideal
gas whatever
the process.
la ecuación (3.21b) es válida para un gas ideal sin importar el proceso.
Example
Ejemplo
3.7 3.73.7
Example
−1
−1
inejemplo
Ex.
the
flow
rate rate
of
isair
1 mol
and
if, both
upstream
and un
downstream
If in 3.6
Ex.3.6
3.6la
the
flow
ofair
the
is es
1 smol
s s–1and
if both
upstream
and
downstream
Si enIfel
rapidez
dethe
flujo
del
aire
1 mol
y si
la
tubería
tiene
diámetro
interno de
pipes
have
an
inner
diameter
of
5
cm,
what
is
the
kinetic-energy
change
of
the
air
pipes
have
an
inner
diameter
of
5
cm,
what
is
the
kinetic-energy
change
of
theand
air
and del
5 cm, tanto en la entrada como en la salida de la válvula, ¿cuál es el cambio en−1la energía
cinética
−1 .
what
is
its
temperature
change?
For
air,
C
=
(7/2)R
and
M
=
29
g
mol
.
what
is
its
temperature
change?
For
air,
C
=
(7/2)R
and
M
=
29
g
mol
–1
P
aire y cuál es su cambio en la temperatura? ParaP el aire,
C P = (7/2)R y M = 29 g mol .
03-SmithVanNess.indd 85
8/1/07 12:54:36
86
86
86
86
86
86
86
86
86
CHAPTER
CHAPTER 3.
3. Volumetric
Volumetric Properties
Properties of
of Pure
Pure Fluids
Fluids
CHAPTER
3.
Volumetric
Properties
of
Pure
Fluids
CHAPTER
3.
Volumetric
Properties
of
Fluids
CHAPTER
3.
Volumetric
Properties
of
Pure
CHAPTER 3. Volumetric Properties of Pure
Pure Fluids
Fluids
CAPÍTULO
Propiedades
volumétricas
deFluids
fluidos puros
CHAPTER 3.3.Volumetric
Properties
of Pure
CHAPTER 3. Volumetric Properties of Pure Fluids
Solución
3.7
Solution
3.7
Solution
3.7
Solution
3.7
Solution
3.7
Solution
Solution 3.7
3.7
3.7
By
Eq.
(2.24b),
Por laSolution
ecuación
(2.24b),
By Eq.
Eq. (2.24b),
(2.24b),
Solution
3.7
By
..
.
nV
nn..
.
nV
..
.
=
uu =
nV
n
.
.
=
=
.
nV
n
nV
n
= nV
u = Aρ
A
n. =
Aρ
A
=
By
Eq.
(2.24b),
u
=
.
By
Eq.
(2.24b),
u
=
Aρ
A
= nV
By Eq. (2.24b),
u = Aρ
n
.
.
Aρ
A
A
�� π ��u = Aρ
A
n = 2nV
By Eq. (2.24b),
π
2
π D 22 = �
π �u(5=×Aρ
−2
−3
=
By
Eq.
(2.24b),
A
2
−2
−3 m
where
A
=
)
=
10
π
π
�
�
�
�
2
2
where
A
=
=
)
= 1.964
1.964 ×
× 10
10−3
m22
D
(5
×
10
donde where
−2
�
�
Aρ
A
π
π
A = π4π
=
)
=
1.964
×
10
m
D
(5
×
10
4
2
2
−2
−3
π D22 =
4 � (5
where
=
= � 4π
= 1.964
1.964 ×
× 10
10−3
m222
(5 ×
× 10
10−2
−2))22 =
−3 m
where
AA
44 D
where
A=
=π
44 D2 = �π
44 � (5 × 10−2 )2 = 1.964 × 10−3 m2
4π
4π
where
A = volume
× 10 −3is:
m2
D 2 = as
(5 ×by
10the
The
upstream
molar
equation
2 = 1.964
−2))ideal-gas
The
upstream
molar
volume
as4 given
given
by
the
ideal-gas
equation
is:
1.964
× del
10 gas
mideal es:
where
A = 4arriba,
D =que
×by
10the
El volumen
molar
corriente
se(5
conoce
porideal-gas
la=ecuación
The
upstream
molar
volume
as
given
equation
is:
4
4
The
upstream
molar
volume
as
given
by
the
ideal-gas
equation
is:
The
upstream
molar
volume
as
given
by
the
ideal-gas
equation
is:
The upstream molar volume as given by the ideal-gas equation is:
RT
(83.14)(293.15)
11
The upstream
molar
volume
as given by
the
ideal-gas
equation
3
RT
(83.14)(293.15)
−6
−3
−1
−6
−3is:
−1
=
×
m
×
10
V
RT
(83.14)(293.15)
1 =
The upstream
molar
as given by
ideal-gas
is:
= RT
= volume
= 4.062
4.062equation
× 10
10−3
m33 mol
mol−1
× the
10−6
V11 =
RT
(83.14)(293.15)
(83.14)(293.15)
1
=
=
=
4.062
×
10
m
mol
×
10
V
P
6
1
3
1
−6
−3
−1
1
33 mol−1
RT
(83.14)(293.15)
−6
−3
P
6
1
1
=
=
=
4.062
×
10
m
×
10
V
−6
−3
−1
=
4.062
×
10
m
mol
×
10
VV111 =
P1 =
6
= RT
=
=
4.062
×
10
m
mol
×
10
(83.14)(293.15)
P
6
1
P
6
11
3
−6
−3
−1
1
P
6
RT
(83.14)(293.15)
× −3
10 −6 = 4.062 × 10 −3 m 3 mol −1
V1 =
1 =
−3
(1)(4.062
×
10 m mol
× −3
10)) = 4.062 ×−1
V1 = P1 =
6
(1)(4.062
× 10
10
−1
Then,
2.069
m
ss−1
)=
(1)(4.062
×
10
P1 uu11 =
6
−3
−3
Then,
=
=
2.069
m
−3
)
(1)(4.062
×
10
−3
)
(1)(4.062
×
10
Then,
u
=
=
2.069
m
s
−3
1.964
10
−1
(1)(4.062
1.964 ×
××
1010
−3−3 ) =
En tal caso,
Then,
=
= 2.069
2.069 m
m ss−1
−1
Then,
uuu1111 =
1.964
×
10
Then,
= (1)(4.062
−3
−3
) = 2.069 m s−1
×10
10
1.964
×
10
−3
−3
1.964
×
1.964
×
10
)
(1)(4.062
×
10
Then,
u
=
=
2.069
m
s
1
If
the
is
changed
from
upstream
Then,
u 1 = 1.964
= 2.069
m s−1
If
the downstream
downstream temperature
temperature
is little
little
changed
from the
the
upstream temperature,
temperature,
× 10−3
If
the
downstream
temperature
is
little
changed
from
the
upstream
temperature,
−3
1.964
×
10
then
to
a
good
approximation:
If
the
downstream
temperature
is
little
changed
from
the
upstream
temperature,
If
the
downstream
temperature
is
little
changed
from
the
upstream
Si la temperatura
corriente
abajo
cambia
un
poco
de
la
correspondiente
a la temperature,
corriente
arriba, por
then
to
a
good
approximation:
If
the
downstream
temperature
is
little
changed
from
the
upstream
temperature,
then to a good approximation:
then
to
a
good
approximation:
to
a
good
approximation:
If
the
downstream
temperature
is
little
changed
from
the
upstream
temperature,
lo tantothen
para
una
buena
aproximación:
then
a good approximation:
If thetodownstream
temperatureand
is little changed
from 4.138
the upstream
temperature,
−1
−1
uu22 =
then to a goodV
22 =
Vapproximation:
= 2V
2V11
and
= 2u
2u11 =
= 4.138
4.138 m
m sss−1
=
2V
and
u2 =
2u
=
m
then to a goodV
approximation:
2
1
1
−1
−1
=
2V
and
=
2u
=
4.138
m
VV
y uuu222 =
V222 =
= 2V
2V111 and
and
= 2u
2u111 =
= 4.138
4.138 m
m sss−1
The
rate
of
change
in
kinetic
energy
is
therefore:
V
=
2V
and
u
=
2u
=
4.138
m
s−1
1
The rate
rate of
of change
change
in2Vkinetic
kinetic
energy
is therefore:
therefore:
The
energy
V22 =in
and is
u22 = 2u11 = 4.138 m s−1
1
The
rate
of
changedein
in
kinetic
energy
is
therefore:
Debido The
aThe
esto,
laof
rapidez
de
la energía
cinética es:
rate
change
kinetic
energy
is
rate
incambio
kinetic
energy
is therefore:
therefore:
..of change
..
11 22
11 22
�(
u
)
=
n
M
�(
u
)
The ratem
of
change
in
kinetic
energy
is
therefore:
.
.
m.of
�(
M �(
�(energy
21 u 2 ) =
21 u 2 )
.. �(
nn...M
The ratem
change
in kinetic
1221 u22) =
1221 u22) is therefore:
m
�(
=
M
�(
2
2
m
m. �(
�(2212uuu 2))) =
= nnn. M
M �(
�(2212uuu 2))) −3 (4.138
(4.13822 −
− 2.069
2.06922))) = 0.186 J s−1
1×210 −3) (4.138
−1
(1
×
29
−
2.069
2
2 = 0.186 J s−1
m. �( 121u 22) =
n
M
�(
u
)
2
2
.
= (1
(1 ×
× 29
2921×
× 210
10−3 ))(4.138
−
2.069
(4.138
=
= 0.186 J s−1
22 2.069
m �( 2 u ) =
= (1
nM
�(
u 10
) −3
−3
−1
−
2.0692))) =
(4.1382 −
2×
=
(1
×
29
×
10
)
=
0.186
J
s
−3
−1
×
29
)
0.186
J
s
2
2
2
= (1 × 29 × 10 ) (4.138 2 −
22 2.069 2) = 0.186 J s−1
2
−3
−
2.069
)
(4.138
=
(1
×
29
×
10
)
=
0.186
J
s
In
work,
(2.31),
becomes:
−3 ) the
In the
the absence
absence of
of heat
heat
transfer
and
work,
the energy
energy
balance,=Eq.
Eq.
(2.31),
becomes:
= transfer
(1 × 29and
× 10
0.186
J s−1
2 balance,
In
the
absence
of
heat
transfer
and
work,
the
energy
balance,
Eq.
(2.31),
becomes:
2
In
the
absence
of
heat
transfer
and
work,
the
energy
balance,
Eq.
(2.31),
becomes:
In
the
absence
of
heat
transfer
and
work,
the
energy
balance,
Eq.
(2.31),
becomes:
In thedeabsence
of heatde
transfer
work,
the
Eq. (2.31),
becomes:
..
.. elenergy
.. balance,
En ausencia
transferencia
calor
y11and
de
balance
de11energía,
ecuación
(2.31), será:
22 trabajo,
22
+
u
)
m
=
m
�H
+
m
�(
u
)
=
0
In the absence of heat�(H
transfer
and
work,
the
energy
balance,
Eq.
(2.31),
becomes:
.
.
.
2
2
�(H
+
u
)
m
=
m
�H
+
m
�(
u
)
=
0
1
1
+ 122211and
u222)m
+ m...balance,
�( 122211u222) Eq.
= 0(2.31), becomes:
... = m
... �H
In the absence of heat�(H
transfer
work,
the
energy
�(H
m=
=
m
�H
+
m
�(
=
�(H
... �H
.. 222uuu1 )))22=
..+
.. C
�(H +
+ 222uuu11 )))m
=m
m
�H +
+m
m. �(
�(
= 000
22m
C PP �T
.)) =
1 212u
1 2112u
nC
�T
+
m
�(
)) =
+
m
�(
m
C
.
.
.+
.. C
P
P �T
2
2
+
m
�(
u
nC
�T
+
m
�(
u
=
m
�(H
u
)
m
=
m
�H
+
m
�(
u
)
=
0 000
.
.
.
P
C
.
.
.
2
2
) = nC
+
m
)=
�T�(H
+ m..+�(
m.. C
M
.m
.. �(
21 u
21 u1221 u)22=
P �T
12211u
MPPP �T
. �H
)222m
+
m
�(
0 00
2
1
u
)
=
nC
�T
+
m
�(
u
)
=
�T
+
m
�(
m
u
)
=
nC
�T
+
m
�(
u
)
=
+
m
�(
m
P
2
2
M
�T + m �(22 u ) = nC
m. C
M
. .PP �T + m. �(2122 u2 ) = 0
M
MPP �T + m.. �( 121u 22) = nC
C
�T
+
m
�(
u
)
=
m
1
.
.
.
2
. P �T
21 u 2 ) = 00
2 =
−0.186
Whence
(1)(7/2)(8.314)�T
m
=−
nC
�T + m �(22 u ) =
mM
= �(
−0.186
Whence
(1)(7/2)(8.314)�T
=
−
m... P�(
�( 2112 uuu+
2))) m
2
=
−0.186
Whence
(1)(7/2)(8.314)�T
=
−
m
�(
1
2
M
. �(121uu22)) =
−0.186
Whence
(1)(7/2)(8.314)�T
=
−
m
Whence
(1)(7/2)(8.314)�T
=
−
22 u ) =
= −0.186
−0.186
Whence
(1)(7/2)(8.314)�T
=
−m
m. �(
�(2K
and
�T
=
−0.0064
2
and
�T =
==−0.0064
−0.0064
K
Whence
(1)(7/2)(8.314)�T
−m. �( 12K
and
�T
1u 2) = −0.186
De donde
Whence
(1)(7/2)(8.314)�T
−m �(K
and
�T
=
−0.0064
K
and
�T
−0.0064
2 u ) = −0.186
and
�T =
==
−0.0064
K
andClearly,
�T = −0.0064
K
temperature
change
Clearly, the
the assumption
assumption of
of negligible
negligible
temperature
change across
across the
the valve
valve is
is
andClearly,
�T = −0.0064
K
the
assumption
of
negligible
temperature
change
across
the
valve
is
justified.
Even
for
an
upstream
pressure
of
10
bar
and
a
downstream
pressure
Clearly,
the
assumption
of
negligible
temperature
change
across
the
valve
Clearly,
the
assumption
of
negligible
temperature
change
across
the
valve
is
justified.
Even
for
an
upstream
pressure
of
10
bar
and
a
downstream
pressure
Clearly,
the for
assumption
of negligible
temperature
change
across thepressure
valve is
is
justified.
Even
an
upstream
pressure
of 10
and
a downstream
y
DT
= the
–0.0064
K bar
of
1
bar
and
for
the
same
flow
rate,
temperature
change
is
only
−0.076
K.
justified.
Even
for
an
upstream
pressure
of
10
bar
and
a
downstream
pressure
justified.
Even
for
an
upstream
pressure
of
10
bar
and
a
downstream
pressure
Clearly,
the
assumption
of
negligible
across
the
valve
is
of
1
bar
and
for
the
same
flow
rate,
the
temperature
change
is
only
−0.076
K.
justified.
Even
for
an
upstream
pressure
of 10 bar and
a downstream
pressure
of
1
bar
and
for
the
same
flow
rate,
the
temperature
change
is
only
−0.076
K.
Clearly,
the
assumption
of
negligible
temperature
change
across
the
valve
is
We
that,
except
for
very
unusual
conditions,
�H
=
is
of
bar
and
for
the
same
flow
rate,
the
temperature
change
is
only
−0.076
K.
of
111conclude
bar
for
the
same
flow
rate,
the
temperature
change
is
−0.076
K.
justified.
Even
an
upstream
pressure
of
10 bar and
a downstream
pressure
We
conclude
that,
except
for
very
unusual
conditions,
�H
=despreciable
is aaa satisfactory
satisfactory
of
bar and
and
forfor
the
same
flow
rate,
the
temperature
change
is000 only
only
−0.076
K. de la
We
conclude
that,
except
for
very
unusual
conditions,
�H
=
is
satisfactory
justified.
Even
for
an
upstream
pressure
of
10
bar
and
a
downstream
pressure
Es evidente,
se
justifica
la
suposición
del
cambio
de
temperatura
a
través
energy
for
aexcept
throttling
process.
We
conclude
that,
for
very
unusual
conditions,
�H
=
is
satisfactory
We
except
very
unusual
conditions,
�H
=
00 only
is
of
1conclude
barbalance
and that,
for
samefor
flow
rate,
the temperature
change
−0.076 K.
energy
balance
forthe
throttling
process.
We
that,
for
very
unusual
conditions,
�H
=is
is aaa satisfactory
satisfactory
balance
for
aaexcept
throttling
process.
of
1conclude
bar
and
the
same
flow
rate,
the
is0corriente
only
−0.076
válvula.energy
Aun
para
unafor
presión
corriente
arriba
detemperature
10
bar y unachange
presión
abajoK.de 1 bar
energy
balance
for
a
throttling
process.
energy
balance
for
a
throttling
process.
We
conclude
that,
except
for
very
unusual
conditions,
�H
=
0
is
a
satisfactory
energy
balancethat,
for
a throttling
process.
We conclude
very
unusual
conditions,es�H
= 0 –0.076
is a satisfactory
para la misma
relación
de except
flujo,
elfor
cambio
en la temperatura
de sólo
K. Concluimos
energy balance for a throttling process.
energy
balance
for a throttling
process.
que, excepto
para
muy pocas
condiciones
inusitadas, ∆H = 0 es un balance de energía satisfactorio para un proceso de estrangulación.
03-SmithVanNess.indd 86
8/1/07 12:54:59
3.4. Aplicación
de las ecuaciones
viriales
3.4.
of
Equations
3.4. Application
Application
of the
the Virial
Virial
Equations
3.4. Application of the Virial Equations
87
87
87
87
3.4 APLICACIÓN
DE LAS
VIRIALES
3.4
APPLICATION
OFECUACIONES
THE VIRIAL
EQUATIONS
3.4
3.4 APPLICATION
APPLICATION OF
OF THE
THE VIRIAL
VIRIAL EQUATIONS
EQUATIONS
Las dos formas de la expansión virial conocidas en las ecuaciones (3.11) y (3.12) son series infinitas. Para
The
The two
two forms
forms of
of the
the virial
virial expansion
expansion given
given by
by Eqs.
Eqs. (3.11)
(3.11) and
and (3.12)
(3.12) are
are infinite
infinite series.
series. For
For
propósitos
su of
usothe
es virial
práctico
sólo cuando
convergencia
rápida,
esto es,series.
cuandoFor
dos o tres
Thetécnicos,
two forms
expansion
givenlaby
Eqs. (3.11) es
andmuy
(3.12)
are infinite
engineering
engineering purposes
purposes their
their use
use isis practical
practical only
only where
where convergence
convergence isis very
very rapid,
rapid, that
that is,
is, where
where
términos
son
suficientes
para
una
aproximación
razonable
a
los
valores
de
las
series.
Esto
ocurre
para
gases y
engineering
purposes
their
use
is
practical
only
where
convergence
is
very
rapid,
that
is,
where
two
two or
or three
three terms
terms suffice
suffice for
for reasonably
reasonably close
close approximations
approximations to
to the
the values
values of
of the
the series.
series. This
This
vaporesistwo
de
presiones
bajas
a
moderadas.
or
three
terms
suffice
for
reasonably
close
approximations
to
the
values
of
the
series.
This
is realized
realized for
for gases
gases and
and vapors
vapors at
at low
low to
to moderate
moderate pressures.
pressures.
is realized for gases and vapors at low to moderate pressures.
1.00
1.00
1.00
0.75
Z�
�
PV/RT
PV/RT
Z Z�
PV/RT
Z  PV/RT
1.00
0.75
0.75
0.75
150
(F)
0.50
0.25
0.50
0.50
0.50
�150
�150
�150
(�F)
(�F)
(�F)
0 (
F)
400(F)
440000((��FF))
250(F)
00(�F)
2254
500((��FF))
250(�F)
100(F)
110000((��FF))
100(�F)
F)
0 (
F))
00((��FF)
0 (�
)
(�FF)
5500((��F)
�
� 50
�
)
))
( F
0
�
((�FF )
0
10
0
F
0
�

0
1 (
��1 00
1
�
5
0 0.25
0.2501 000
2 000
3 000
1000
2000
0.25 0
1000
2000
0
P(psia)
1000
2000
P(psia)
P(psia)
P(psia)
Figura 3.9: Gráfica del factor de compresibilidad
Figure
3.9:
Figure
3.9: Compressibility-factor
Compressibility-factor
para
el metano.
Figure
3.9: Compressibility-factor
graph
graph for
for methane.
methane.
graph for methane.
3000
3000
3000
La figura
3.9 muestra
una gráfica del factor de compresibilidad
para el metano.
Los valores del factor
Figure
Figure 3.9
3.9 shows
shows aa compressibility-factor
compressibility-factor graph
graph for
for methane.
methane. Values
Values of
of the
the compressibilcompressibilFigure
3.9
shows
a
compressibility-factor
graph
for
methane.
Values
of
the
compressibilde compresibilidad
Z
(como
se
calculan
a
partir
de
la
información
PVT
del
metano
mediante
la ecuación
que
ity
ity factor
factor ZZ (as
(as calculated
calculated from
from PPVVTT data
data for
for methane
methane by
by the
the defining
defining equation
equation ZZ =
= PPVV/RT
/RT))
ity
factor
Z
(as
calculated
from
P
V
T
data
for
methane
by
the
defining
equation
Z
=
P
V
/RT
)
la define
Z
=
PV/RT)
están
graficados
en
función
de
la
presión
para
varias
temperaturas
constantes.
Las
isoare
are plotted
plotted vs.
vs. pressure
pressure for
for various
various constant
constant temperatures.
temperatures. The
The resulting
resulting isotherms
isotherms show
show graphgraphare plotted vs.
pressure
various
constant
temperatures.
The resulting
show graphtermas ically
resultantes
muestran
defor
manera
gráfica
lo que
se intenta
representar
en isotherms
formaisotherms
analítica
con la expanically what
what the
the virial
virial expansion
expansion in
in PP isis intended
intended to
to represent
represent analytically.
analytically. All
All isotherms origiorigiically
what
the
virial
expansion
in
P
is
intended
to
represent
analytically.
All
isotherms
origisión virial
en
P.
Todas
las
isotermas
se
originan
en
el
valor
Z
=
1
para
P
=
0,
y
a
bajas
presiones
las
isotermas
nate
nate at
at the
the value
value ZZ =
= 11 for
for PP =
= 0,
0, and
and the
the isotherms
isotherms are
are nearly
nearly straight
straight lines
lines at
at low
low pressures.
pressures.
nate
at
the
value
Z
=
1
for
P
=
0,
and
the
isotherms
are
nearly
straight
lines
at
low
pressures.
son líneas
casi
rectas.
Así,
la
tangente
a
una
isoterma
en
P
=
0
es
una
buena
aproximación
de
la
isoterma
Thus
Thus the
the tangent
tangent to
to an
an isotherm
isotherm at
at PP =
= 00 isis aa good
good approximation
approximation of
of the
the isotherm
isotherm from
from PP →
→ 00
Thus
the
tangent
to
an
isotherm
at
P
=
0
is
a
good
approximation
of
the
isotherm
from
P
→
0
desde Pto
→
0
para
alguna
presión
finita.
Al
derivar
la
ecuación
(3.11)
para
cierta
temperatura
se
obtiene:
to some
some finite
finite pressure.
pressure. Differentiation
Differentiation of
of Eq.
Eq. (3.11)
(3.11) for
for aa given
given temperature
temperature gives:
gives:
to some finite pressure. Differentiation
of
Eq.
(3.11)
for
a
given
temperature
gives:
�
� �
�
�∂∂ZZ �
�
�
� 2
=
∂Z
= BB � +
+ 2C
2C �PP +
+ 3D
3D �PP 2 +
+ ······
∂∂PP TT = B � + 2C � P + 3D � P 2 + · · ·
∂P T �
�
� �
�∂∂ZZ �
�
=
from
which,
∂
Z
= BB ��
from which,
∂∂PP TT;P=0
=
B
a partir from
de la which,
cual,
∂ P T ;P=0
;P=00
�
equation
the
ZZ =
11también
+
result
also
truncating
Thus the
the de
equation
of
the tangent
tangent
line
=
+ BB ��P,
P,esaa un
result
also given
given
by
truncating
Así, la Thus
ecuación
la rectaof
tangente
es: Z =line
1 +isis
B ′P,
que
resultado
que seby
obtiene
al truncar la
Thus
the
equation
of
the
tangent
line
is
Z
=
1
+
B
P, a result
also given
by
truncating
Eq.
(3.11)
to
two
terms.
Eq.
(3.11)
to two
terms.
ecuación
(3.11)
a
dos
términos.
Eq. (3.11)
to two terms.form of this equation results from substitution for B �� by Eq. (3.13a):
A
A more
more
common
of thisesta
equation
results
from substitution
by Eq. (3.13a):
Una forma
máscommon
común deform
expresar
ecuación
es sustituyendo
B ′ porfor
la B
ecuación
A more
common
form
of this equation
results
from substitution
for
B � by Eq.(3.13a):
(3.13a):
PPVV
BBPP
ZZ =
(3.38)
= PV =
(3.38) (3.38)
= 11 +
+ BP
Z = RT
(3.38)
RT = 1 + RT
RT
RT
RT
This
aa direct
ZZ and
often
This equation
equation
expresses
direct proportionality
proportionality
between
and P,
P, and
andseisisaplica
often applied
applied
toa tempeEsta ecuación
expresaexpresses
una proporcionalidad
directa entre between
Z
y P, y con
vaporesto
This equation
expresses
a direct proportionality
between
Z frecuencia
and P, At
andhigher
is oftena applied
to
at
temperatures
up
pressures.
vapors
at subcritical
subcritical
temperatures
up to
to their
theirAsaturation
saturation
pressures.
At
higher temperatures
temperatures
raturas vapors
subcríticas
hasta
sus
presiones
de
saturación.
temperaturas
altas
se
proporciona
una
aproximación
vapors at subcritical temperatures up to their saturation pressures. At higher temperatures
03-SmithVanNess.indd 87
8/1/07 12:55:09
CHAPTER 3.3.Volumetric
Properties
of Pure
CAPÍTULO
Propiedades
volumétricas
deFluids
fluidos puros
88
88
it provides
a reasonable
for gases
up to a pressure
of several
with
the que
razonable
para los gases
hasta unaapproximation
presión de varios
bar, incrementando
el intervalo
de bars,
presión
a medida
range increasing as the temperature increases.
aumentapressure
la temperatura.
Equation
(3.12)
as well
may betruncarse
truncatedatodos
twotérminos
terms for
application
at lowa bajas
pressures:
La ecuación
(3.12)
también
es posible
para
su aplicación
presiones:
PV
B
=1+
(3.39) (3.38)
RT
V
De cualquier
modo,
ecuación
(3.38)
es más conveniente
y suand
aplicación
es tan
exacta
la (3.39).
De
However,
Eq.la(3.38)
is more
convenient
in application
is normally
at least
as como
accurate
as
este modo,
cuando Thus
la ecuación
virial
se trunca
a dos
términos to
setwo
prefiere
la Eq.
ecuación
Eq. (3.39).
when the
virial
equation
is truncated
terms,
(3.38)(3.38).
is preferred.
El segundo
virial
B depende
desubstance
la sustancia
y de una and
función
de la temperatura.
Los valores
The coeficiente
second virial
coefficient
B is
dependent
a function
of temperature.
5
Además, es posible
experimentales
están disponibles
para varios
calcularestimation
el segundoof
coeficiente
Experimental
values are available
forgases.
a number
of gases.5 Moreover,
second virial
cuando virial
no hay
información
disponible,
como
analiza
en la sección
3.6.
coefficients
is possible
where
no se
data
are available,
as discussed
in Sec. 3.6.
Para presiones
que están
arriba
del intervalo
de aplicabilidad
de labut
ecuación
peropresabajo de la
For pressures
above
the range
of applicability
of Eq. (3.38)
below (3.38),
the critical
presiónsure,
crítica,
ecuación
virialtruncated
truncadatoa three
tres términos
con provides
frecuencia
proporciona
excelentes
resultados.
thelavirial
equation
terms often
excellent
results.
In this case
En este Eq.
caso,
en la the
ecuación
(3.12)inla1/V
expansión
en 1/V estomuy
que en
la ecuación
Así, cuan(3.12),
expansion
, is far superior
Eq. superior
(3.11). Thus
when
the virial(3.11).
equation
do la ecuación
virialtosethree
trunca
a tres
la forma
es:
is truncated
terms,
thetérminos,
appropriate
form apropiada
is:
Z=
Z=
PV
B
C
=1+ + 2
RT
V
V2
(3.40) (3.40)
0
0
100
�200
200
0
C
B
B
B
4,000
4 000
2,000
2 000
0
0
�2,000 2 000
�100
�300
C
C
�2
C/cm
C/cm66mol
mol�2
100
B/cm3 mol1
Figura 3.10:
Coeficientes
Figure
3.10: Virialviriales B y
C de la progresión
densidad
para
coefficientsenBlaand
C
el nitrógeno.
for nitrogen.
�1
B/cm
B/cm33mol
mol�1
100
300
�4,000 4 000
0 100
100200
T/K
200300
T/K
300400
C/cm6 mol2
This equation
be solved
directly for
butpero
is cubic
in volume.
SolutionLa
forsolución
V is para
Esta ecuación
se puedecan
resolver
directamente
parapressure,
la presión,
es cúbica
en el volumen.
readily
accomplished
byun
an esquema
iterative scheme,
in Ex.se3.8.
V se obtiene
fácilmente
usando
iterativo as
deillustrated
cálculo, como
ilustra en el ejemplo 3.8.
400
Values of C, like those of B, depend on the gas and on temperature. However, much less
is known
virial
coefficients
than aboutdel
second
data forse sabe
Los
valoresabout
de C,third
al igual
que
los de B, dependen
gas yvirial
de lacoefficients,
temperatura.though
Sin embargo,
number
of gases
are found
in the literature.
virial
beyond
the third
are
mucho amenos
sobre
los terceros
coeficientes
virialesBecause
que acerca
de coefficients
los segundos
coeficientes
viriales,
aun
cuando rarely
es posible
encontrar
en la literatura
información
que rara
vez se conocen
known
and because
the virial alguna
expansion
with morepara
thanvarios
threegases.
terms Ya
becomes
unwieldy,
los coeficientes
más allá de los terceros, y ya que la expansión virial es muy difícil de manejar debido
its use isviriales
uncommon.
a que tiene más de tres términos, su uso no es muy común.
55 J. H. Dymond and E. B. Smith, The Virial Coefficients of Pure Gases and Mixtures, Clarendon Press, Oxford,
1980.
J. H. Dymond y E. B. Smith, The Virial Coefficients of Pure Gases and Mixtures, Clarendon Press, Oxford, 1980.
03-SmithVanNess.indd 88
8/1/07 12:55:13
3.4. Application of the Virial Equations
3.4. Application
of the Virial
Equations
3.4. Aplicación
de las ecuaciones
viriales
89
89
89
Figure 3.10 illustrates the effect of temperature on the virial coefficients B and C for
La figura
3.10 3.10
ilustra el efecto the
de laeffect
temperatura
sobre losoncoeficientes
viriales B y BC and
para el for
nitrógeno;
Figure
temperature
virialthe
coefficients
nitrogen;
althoughillustrates
numerical values
areofdifferent
for otherthe
gases,
trends are similar.C The
aunque nitrogen;
los
valores
numéricos
son
diferentes
para
otros
gases,
las
tendencias
son
similares.
La
curva
although
numericalthat
values
are different
for other gases,
trends are
similar. Thede la ficurve of Fig.
3.10 suggests
B increases
monotonically
with Tthe
; however,
at temperatures
gura 3.10
sugiere
que
B
aumenta
en
forma
monotónica
con
T;
no
obstante,
a
temperaturas
mucho
mayores que
curve
Fig. than
3.10 shown
suggests
B increases
monotonically
with Tdecreases.
; however,The
at temperatures
much of
higher
B that
reaches
a maximum
and then slowly
temperature
las mostradas
B
alcanza
un
máximo
y
en
seguida
disminuye
lentamente.
La
dependencia
de
la
temperatura
de
much
higherofthan
B reaches
a maximum
and then slowly
decreases.
The temperature
dependence
C isshown
more difficult
to establish
experimentally,
but its
main features
are clear: C
C es más
difícil
de
establecer
de
manera
experimental,
pero
sus
características
principales
son
claras:
C
es
dependence
C istemperatures,
more difficultpasses
to establish
experimentally,
its main features
C
is negative atoflow
through
a maximum atbut
a temperature
nearare
theclear:
critical,
negativais
a
bajas
temperaturas,
pasa
por
un
máximo
a
una
temperatura
cercana
a
la
crítica,
y
después
disminunegative
at low
temperatures,
passes
through aT maximum
at a temperature near the critical,
and
thereafter
decreases
slowly with
increasing
.
ye de forma
lenta
conforme
T aumenta.
and thereafter
decreases
slowly
withby
increasing
T .known as extended virial equations, is illusA class
of
equations
inspired
Eq. (3.12),
Hay
unaA
de
ecuacionesinspired
inspiradas
en la
(3.12),
que se
como
ecuaciones
6
class
of equations
by
Eq.
(3.12),
known
asconoce
extended
virial
equations,viriales
is illus-extenditrated
byclase
the Benedict/Webb/Rubin
equation:
6
6
das. Éstas
se
ilustran
mediante
la
ecuación
de
Benedict/Webb/Rubin:
trated by the Benedict/Webb/Rubin equation:
RT
b RT − a
B0 RT − A0 − C0 /T 2
P = RT + B00 RT − A002 − C00/T 22 + b RT 3− a
V
V
V
P=
+
+
V
V 22 �
V 33
�
γ
−γ
aα
c
+ aα6 + 3c 2 � 1 + γ 2 � exp −γ2
+ V 66 + V 33 T 22 1 + V 22 exp V 22
V
V T
V
V
where A0 , B0 , C0 , a, b, c, α, and γ are all constant for a given fluid. This equation and its
A000,,a,Bb,
C0a
a,γ b,
c,constantes
α,
and γ are
all
forpetroleum
a given Esta
fluid.
This equation
and its
00, c,
0, y
donde Awhere
son
paraare
unconstant
fluido
y susindustries
modificaciones,
modifications,
despite
their
complexity,
used
indeterminado.
the
andecuación
natural-gas
0, B0, C
modifications,
despite
their
complexity,
are
used
in
the
petroleum
and
natural-gas
industries
a pesar de
complejidad,
se utilizan
en las
industrias
del petróleo
y delgases.
gas natural para hidrocarburos ligeros
forsu
light
hydrocarbons
and a few
other
commonly
encountered
forotros
light gases
hydrocarbons
and aencontrados.
few other commonly encountered gases.
y algunos
comúnmente
Example 3.8
Example
3.8for the virial coefficients of isopropanol vapor at 200◦ C are:
Ejemplo
3.8values
Reported
Reported values for the virial coefficients of isopropanol vapor at 200◦◦C are:
Los valores reportadosBpara
los coeficientes
de isopropanol
a 200 °C son:
C vapor
= −26,000
cm6 mol−2
= −388
cm3 mol−1 viriales del
66
−1
−2
−1
−2
C = −26,000 cm
B = −388 cm33 mol
3 mol–1 C
6 molmol
–2
B
=
–388
cm
=
–26
000
cm
◦
Calculate V and Z for isopropanol vapor at 200 C and 10 bar by:
Calculate V and Z for isopropanol vapor at 200◦◦C and 10 bar by:
Calcule V y(a)
Z para
vapor deequation;
isopropanol
200 °C y 10
bar mediante:
Theelideal-gas
(b)a Equation
(3.38);
(c) Equation (3.40).
(a) The ideal-gas equation; (b) Equation (3.38); (c) Equation (3.40).
a) La ecuación del gas ideal;
b) La ecuación (3.38);
c) La ecuación (3.40).
Solution
3.8
Solución
3.8
Solution 3.8
La temperatura
absoluta
es T = 473.15
y el valor
de la constante
The absolute
temperature
is T K,
= 473.15
K, apropiado
and the appropriate
valuede
oflos
thegases
gas es R =
3 bar
−1 K−1
The
absolute
temperature
T mol
= 473.15
K,. and the appropriate value of the gas
83.14 cm
mol
.
constant
is –1
RK
=–183.14
cm3isbar
−1 K−1
−1.
constant is R = 83.14 cm33 bar mol−1
a) Para (a)
un gas
Z =gas,
1, yZ = 1, and
For ideal,
an ideal
(a) For an ideal gas, Z = 1, and
RT
(83.14)(473.15)
V = RT = (83.14)(473.15) = 3,934 cm3 mol−1
−1
P
10
V =
=
= 3,934 cm33 mol−1
P
10
(b) Solving Eq. (3.38) for V gives:
b) Resolviendo
paraEq.
V de
la ecuación
(3.38) se obtiene:
(b) Solving
(3.38)
for V gives:
RT
V = RT + B = 3,934 − 388 = 3,546 cm3 mol−1
−1
V = P + B = 3,934 − 388 = 3,546 cm33 mol−1
P
6 M. Benedict, G. B. Webb, L. C. Rubin, J. Chem. Phys., vol. 8, pp. 334–345, 1940; vol. 10, pp. 747–758, 1942.
66M. Benedict, G. B. Webb, L. C. Rubin, J. Chem. Phys., vol. 8, pp. 334–345, 1940; vol. 10, pp. 747–758, 1942.
M. Benedict, G. B. Webb y L. C. Rubin, J. Chem. Phys., vol. 8, pp. 334-345, 1940; vol. 10, pp. 747-758, 1942.
03-SmithVanNess.indd 89
8/1/07 12:55:20
90
CHAPTER 3.
3.3.Volumetric
Properties
of Pure
Pure
CAPÍTULO
Propiedades
volumétricas
deFluids
fluidos puros
CHAPTER
CHAPTER 3. Volumetric
Volumetric Properties
Properties of
of Pure Fluids
Fluids
CHAPTER 3. Volumetric Properties of Pure Fluids
CHAPTER 3. Volumetric Properties of Pure Fluids
PV
V
V
3,546
P
V
3,546 = 0.9014
Whence,
Z=
= PV =
= V =
= 3,546
0.9014
De donde,
Whence,
Z
V
V
3,546
Whence,
Z = PRT
=
=
= 0.9014
RT
RTV/P
/P
3,934 =
3,934
V = RT
Whence,
Z = PRT
RT /P = 3,546
3,934 = 0.9014
Whence,
Z = RT = RT /P = 3,934 = 0.9014
RT Eq.
/P as:(3.40)
3,934 se escribe como:
(c) To
To facilitate
facilitate
iteration,
write
(3.40)
c) Para facilitar
el
proceso
de
iteración,
laRT
ecuación
(c)
iteration,
write
(3.40)
(c) To facilitate iteration,
write Eq.
Eq.
(3.40) as:
as:
(c) To facilitate iteration, write Eq. (3.40)
�
� as:
�
�
(c) To facilitate iteration, write Eq. (3.40)
�
� as: B
RT
C�
�
B
RT
C
B
RT
C
1
+
=
+
V
i+1
= P �11 +
V
Bi +
C222 �
+V
+V
Vi+1
i+1 = RT
P
Bii + V
Ci
= RT
Vi+1
P 1+ V
V
V
Vi+1 = P 1 + Vi + Viii22
P
Vi
Vthe
i first iteration, i = 0, and
where
subscript
denotes
the iteration
iteration
number.
For
subscript
iii denotes
the
number.
the
0,
donde elwhere
subíndice
i denota
el número
de iteración.
ParaFor
la primera
iteración,
= 0,
where
subscript
denotes
the iteration
number.
For
the first
first iteration,
iteration, iii =
=
0, yand
and
where subscript i denotes the iteration�
number. For the
first iteration, i = 0, and
�
�
�
where subscript i denotes the iteration�number. For the
� first iteration, i = 0, and
RT �
C�
B
RT �11 +
C�
B +
V11 =
= RT
+ C
+ B
V
C22
B +V
V = RT
P 1+ V
V
P
V11 = RT
P 1 + VB00 + VC02
V1 = P 1 + V00 + VV00022
P
V0
V0
where V
V00 =
= 3,934,
3,934, the
the ideal-gas
ideal-gas value.
value. Numerically,
Numerically,
where
where
V
=
3,934,
the
ideal-gas
value.
Numerically,
0
donde Vwhere
=
3
934,
el
valor
del
gas
ideal.
En
forma
numérica,
0
V = 3,934, the ideal-gas
Numerically, �
�
� value.
�
where V00 = 3,934, the ideal-gas
� value.
26,000 �
388Numerically,
26,000
388
�
�
26,000
388
−
= 3,539
3,539
= 3,934 �11 −
−
V11 =
V
26,000222 � =
388 −
= 3,539
− (3,934)
3,934 1 − 3,934
V = 3,934
26,0002 = 3,539
388 − (3,934)
V11 = 3,934 1 − 3,934
3,934
(3,934)
V1 = 3,934 1 − 3,934 − (3,934) 2 = 3,539
3,934
The second
second iteration
iteration depends
depends on this
this
result: (3,934)
The
The second iteration depends on
on this result:
result:
The second
iteration
on
result:
�this
� depends
�
�
La segunda
depende
de
este
resultado:
�
�
The iteración
second iteration
depends
on
this
result:
�
�
�
�
388
RT �
C�
26,000 ��
B
RT
C
388
26,000
B
�
�
388
RT
C
26,000
B
V22 =
= 3,934 �11 +
+
= 3,495
3,495
=
+ 2� =
−
�1 +
V
388 −
C22 = 3,934
26,000222 � =
B +
V = RT
3,934 1 + 3,539
= 3,495
+V
+V
− (3,539)
P 11 +
P
388 − (3,539)
C112 = 3,934 1 + 3,539
26,0002 = 3,495
V22 = RT
P 1+ V
VB111 + V
3,539
(3,539)
V
V2 = P 1 + V1 + V112 = 3,934 1 + 3,539 − (3,539) 2 = 3,495
P
V
3,539 (3,539)
1
Iteration
continues until
until1 theVdifference
difference
Vi+1
−V
Vii is
is insignificant,
insignificant, and
and leads
leads after
after
i+1 −
Iteration
V
Iteration continues
continues until the
the difference
V
−
V
i is insignificant, and leads after
i+1
7
7
five
iterations
to
the
final
value,
Iteration
continues
until
the
difference
V
−
V
is
insignificant,
and
leads
after
i+1
7
five
to
final
value,
five iterations
iterations
to the
theuntil
finalthe
value,
Iteration
continues
difference
Vii isinsignificante,
insignificant, yand
leads de
after
La iteración
continúa
hasta
diferencia
Vi V+i+1
Vi sea
después
cinco ite7
1–−
five iterations
to
the que
finallavalue,
7
iterations
the final,
final value,
3
−1
racionesfive
se obtiene
el to
valor
V=
= 3,488 cm
cm3 mol−1
−1
V
V = 3,488
3,488 cm3 mol
mol−1
V = 3,488 cm33 mol–1
−1
=
cmwith
molthis result, the ideal-gas value is
= 33,488
488 cm
mol
from which
which Z =
= 0.8866. In
InVVcomparison
from
from which Z
Z = 0.8866.
0.8866. In comparison
comparison with
with this
this result,
result, the
the ideal-gas
ideal-gas value
value is
is
13%
too
high
and
Eq.
(3.38)
gives
a
value
1.7%
too
high.
from
which
Z
=
0.8866.
In
comparison
with
this
result,
the
ideal-gas
value
13%
too
high
and
Eq.
(3.38)
gives
a
value
1.7%
too
high.
a partir from
del cual
Z =Z0.8866.
En
comparación
con1.7%
este this
resultado,
del gas
idealis
13%
too
high
and
(3.38)
a value
too
high.
which
= Eq.
0.8866.
Ingives
comparison
with
result, el
thevalor
ideal-gas
value
is es 13%
too high(3.38)
and Eq.
(3.38) gives
value1.7%
1.7% too high.
mayor y13%
la ecuación
proporciona
unaavalor
13%
too high and Eq.
(3.38) gives
value 1.7%mayor.
too high.
90
90
90
90
90
3.5
ECUACIONES
CÚBICAS DE
3.5 CUBIC
CUBIC EQUATIONS
OF ESTADO
STATE
3.5
3.5 CUBIC EQUATIONS
EQUATIONS OF
OF STATE
STATE
3.5
CUBIC
EQUATIONS
OF
STATE
Si una ecuación
de
estado
representa
el
comportamiento
3.5 CUBIC EQUATIONS OF STATEPVT para líquidos y vapores, debe abarcar un amplio
If de
antemperaturas
equation of
of state
is to
to represent
the
PV
V T ser
behavior
of both
both liquids
liquids
and
vapors,grandes
it must
must
intervalo
y presiones.
Aun así,the
no P
debe
tan compleja
paraand
presentar
If
is
of
vapors,
If an
an equation
equation of state
state
is to represent
represent
the
PVT
T behavior
behavior
of bothcomo
liquids
and
vapors, it
it mustdificulencompass
a
wide
range
of
temperatures
and
pressures.
Yet
it
must
not
be
so
complex
asel to
to
If
an
equation
of
state
is
to
represent
the
P
V
T
behavior
of
both
liquids
and
vapors,
it
must
tades numéricas
o
analíticas
para
su
aplicación.
Las
ecuaciones
polinomiales
que
son
cúbicas
en
volumen
encompass
aa wide
range
it
not
complex
as
encompass
wide
range
of
temperatures
and
pressures.
Yet
it must
must
not be
be
sovapors,
complex
as to
If
an equation
of state
is of
to temperatures
represent
the and
P V Tpressures.
behaviorYet
of both
liquids
andso
it must
present
excessive
numerical
or
analytical
difficulties
in
application.
Polynomial
equations
that
encompass
a
wide
range
of
temperatures
and
pressures.
Yet
it
must
not
be
so
complex
as
to
molar ofrecen
un
compromiso
entre
la
generalidad
y
la
simplicidad
adecuadas
para
muchos
propósitos.
De
present
or
in
equations
present excessive
excessive
numerical
or analytical
analytical difficulties
difficulties
in application.
application.
Polynomial
equationsasthat
that
encompass
a widenumerical
range of temperatures
and pressures.
Yet it mustPolynomial
not be so complex
to
are
cubic
in
molar
volume
offer
a
compromise
between
generality
and
simplicity
that
is
suitable
present
excessive
numerical
or
analytical
difficulties
in
application.
Polynomial
equations
that
hecho, las
ecuaciones
cúbicas
son
las
más
sencillas,
pero
capaces
de
representar
el
comportamiento
tanto
de
are
in
volume
aa compromise
between
and
simplicity
is
are cubic
cubic
in molar
molarnumerical
volume offer
offer
compromise
between
generality
andPolynomial
simplicity that
that
is suitable
suitable
present
excessive
or analytical
difficulties
ingenerality
application.
equations
that
to
many
purposes.
Cubic
equations
are
in
fact
the
simplest
equations
capable
of
representing
cubic
in
molar
volume
offer
a
compromise
between
generality
and
simplicity
that
is
suitable
líquido are
como
de
vapor.
to
Cubic
equations
are
the
equations
capable
representing
to many
many
purposes.
Cubic offer
equations
are in
in fact
factbetween
the simplest
simplest
equations
capable of
of
representing
are
cubicpurposes.
in molar volume
a compromise
generality
and simplicity
that
is suitable
both
liquid
and vapor
vapor
behavior.
to
many
purposes.
Cubic
equations are in fact the simplest equations capable of representing
both
liquid
and
behavior.
both
liquid
and vapor
behavior.
to
many
purposes.
Cubic
equations are in fact the simplest equations capable of representing
both liquid and vapor behavior.
liquid and vapor behavior.
Losboth
esquemas de iteración, incorporados en los paquetes de software, llevan a cabo esta iteración de manera inadvertida y rutinaria.
777Iteration
Iteration schemes,
schemes, incorporated in
in software packages,
packages, accomplish this
this iteration routinely
routinely and unobserved.
unobserved.
Iteration schemes, incorporated
incorporated in software
software packages, accomplish
accomplish this iteration
iteration routinely and
and unobserved.
7 Iteration schemes, incorporated in software packages, accomplish this iteration routinely and unobserved.
7 Iteration schemes, incorporated in software packages, accomplish this iteration routinely and unobserved.
03-SmithVanNess.indd 90
8/1/07 12:55:29
91
3.5. Ecuaciones cúbicas de estado
3.5. Cubic Equations of State
91
La ecuación de estado de van der Waals
The van der Waals Equation of State
La primera ecuación de estado cúbica que se considera práctica fue propuesta por J.8 D. van der Waals en
1873: The first practical cubic equation of state was proposed by J. D. van der Waals in 1873:
P=
RT
a
− 2
V −b
V
(3.41) (3.41)
En este Here,
caso, aa yand
b son
constantes
positivas; when
cuando
sonare
cero,
se the
recupera
la ecuación
gas ideal.
b are
positive constants;
they
zero,
ideal-gas
equationdel
is recovered.
Conocidos
losvalues
valoresofde
a y bb para
fluido particular,
es posible
calcular
como una
Given
a and
for a un
particular
fluid, one can
calculate
P as aPfunction
of función
V for de V
para valores
diferentes
deTT.. La
figura
3.11isesa un
diagrama
PVdiagram
esquemático,
quethree
muestra
isotermas. Está
various
values of
Figure
3.11
schematic
PV
showing
suchtres
isotherms.
sobrepuesto
el “domo”isque
los estados states
de líquido
y de vapor
saturados.
Para lavapor.
isoterma
Superimposed
therepresenta
“dome” representing
of saturated
liquid
and saturated
ForT1 > Tc,
la presión
una función
que
decrece
monótonamente
conforme
aumenta
el
volumen
molar.
La
isoterma
the es
isotherm
T1 > T
,
pressure
is
a
monotonically
decreasing
function
with
increasing
molar
c
crítica (marcada
como
Tc) contiene
inflexión
horizontal
C, característica
crítico. Para la
volume. The
critical
isothermuna
(labeled
Tc ) contains
the en
horizontal
inflection del
at Cpunto
characteristic
isotermaofTthe
Tc, la presión
en forma Trápida
en la región de líquido subenfriado conforme V aumenta;
2 < critical
point. decrece
For the isotherm
2 < Tc , the pressure decreases rapidly in the subcooleddespués,liquid
cruzaregion
la línea
de increasing
líquido saturado,
a un mínimo,
se eleva a un
máximo
a continuación
with
V ; afterllega
crossing
the saturated-liquid
line,
it goesythrough
a mini- decrece, cruza
la línea
saturadoand
y continúa
hacia abajo
dentro
la región de vapor
sobrecalentado.
mum,
risesdetovapor
a maximum,
then decreases,
crossing
the de
saturated-vapor
line and
continuing
Las
isotermasinto
experimentales
no exhibenregion.
esta transición uniforme de líquido saturado a vapor saturado;
downward
the superheated-vapor
más bien, contienen
un segmento
horizontal
dethis
la región
de transition
dos fases donde
el líquidoliquid
y el vapor
Experimental
isotherms
do notdentro
exhibit
smooth
from saturated
to saturados coexisten
diferentes
presión de
saturación
o dethevapor.
Este comportamiento,
saturateden
vapor;
rather,proporciones
they containaalahorizontal
segment
within
two-phase
region where que
se muestra
mediante
la línea
discontinua
de la
figurain3.11,
no esproportions
analítico, yatesthe
aceptado
como
conducta
saturated
liquid
and saturated
vapor
coexist
varying
saturation
or una
vapor
inevitable
no realista
las ecuaciones
en laline
región
de dos
fases.
pressure.
Thisdebehavior,
shown de
by estado
the dashed
in Fig.
3.11,
is nonanalytic, and we accept
Enasrealidad,
comportamiento
PV predicho
en esta
región,
mediante
ecuaciones
inevitableelthe
unrealistic behavior
of equations
of state
in the
two-phase
region. de estado cúbicas
apropiadas, noActually,
es del todo
ficticio.
Cuando
la
presión
disminuye
en
un
líquido
saturado
sin que
hayaissitios de
the P V behavior predicted in this region by proper cubic equations
of state
formación
de
núcleos
de
vapor
en
un
experimento
controlado
de
manera
cuidadosa,
la
vaporización
no ocurre
not wholly fictitious. If pressure is decreased on a saturated liquid devoid of vapor-nucleation
y la fasesites
líquida
persiste
sólo
a
presiones
que
razonablemente
están
por
debajo
de
la
presión
de
vapor.
De
in a carefully controlled experiment, vaporization does not occur, and liquid persists alone
manera to
similar,
el
aumento
en
la
presión
en
un
vapor
saturado
en
un
experimento
apropiado
no
ocasiona
pressures well below its vapor pressure. Similarly, raising the pressure on a saturated vapor
condensación
y el vapor
persistedoes
a presiones
que,
de manera razonable,
poralone
arribatode
la presión
de vapor.
in a suitable
experiment
not cause
condensation,
and vapor están
persists
pressures
well
Estos estados
de no
equilibrio
o metaestables
para el líquido
sobrecalentado
para el vapor
subenfriado
se
above the
vapor
pressure.
These nonequilibrium
or metastable
states ofysuperheated
liquid
and
aproximan
por aquellas
de la isoterma
queportions
se encuentran
región de
dos fases
subcooled
vapor partes
are approximated
by PV
those
of the PenVlaisotherm
which
lie inadyacentes
the two- a los
estados phase
de líquido
y de
vapor to
saturados.
region
adjacent
the states of saturated liquid and saturated vapor.
Las ecuaciones
cúbicas
de
estado
tienen
raícesroots,
para el
las be
cuales
es posible
Cubic equations of state
have
threetres
volume
of volumen,
which twodemay
complex.
Phys-que dos
sean complejas.
Los
valores
de
V
físicamente
significativos
siempre
son
reales,
positivos
y
mayores
ically meaningful values of V are always real, positive, and greater than constant b. For an que la
constante
b. Para at
una
a T > Tc,tolaFig.
figura
3.11
muestra
la solución
para
de cualquier
isotherm
T isoterma
> Tc , reference
3.11
shows
that que
solution
for V at
anyV positive
valuevalor
of positivo de P
P da
sólo
una
de
estas
raíces.
Para
la
isoterma
crítica
(T
=
T
),
esto
también
es
cierto,
excepto
c
yields only one such root. For the critical isotherm (T = Tc ), this is also true, except at a la
presión the
crítica
donde
existenwhere
tres raíces
a Vroots,
lasequal
isotermas
T < isotherms
Tc, la ecuación
c. Paraall
at T puede
< Tc , exhibir
critical
pressure,
thereiguales
are three
to Vc .a For
una o tres
raíces
reales,
dependiendo
de
la
presión.
Aunque
estas
raíces
son
reales
y
positivas,
no
hay estados
the equation may exhibit one or three real roots, depending on the pressure. Although these
físicos estables
para
la
parte
de
una
isoterma
que
está
entre
líquido
y
vapor
saturados
(debajo
del
“domo”).
roots are real and positive,
they are not
physically stable states for the portion of an isotherm
saturado(líquido) y V saturado(vapor), son estados estables, conectaSólo laslying
raícesbetween
para P =saturated
P saturadoliquid
, es decir,
V
and saturated vapor (under the “dome”). Only for the vapor or
dos porsaturation
la parte horizontal
desatlaare
isoterma
real.
Para
otras
indican
mediante
las líneas
ho­
pressure P
the roots,
V sat
(liq)
and presiones
V sat (vap),(que
stablesestates,
lying
at the ends
of
saturado), la raíz más pequeña es un volumen líquido o
rizontales
de
la
figura
3.11
arriba
y
abajo
de
P
the horizontal portion of the true isotherm. For other pressures (as indicated by the horizontal
“semejante
un líquido”,
y la3.11
másabove
grande
es below
un volumen
o “semejante
vapor”.
La tercera raíz, que
linesa shown
on Fig.
and
P sat ),de
thevapor
smallest
root is a al
liquid
or “liquid-like”
está entre
los
otros
valores,
no
es
importante.
volume, and the largest is a vapor or “vapor-like” volume. The third root, lying between the
other values, is of no significance.
Johannes
Diderik van der Waals (1837-1923). Físico holandés ganador del premio Nobel de Física en 1910.
8 Johannes
Diderik van der Waals (1837–1923), Dutch physicist who won the 1910 Nobel Prize for physics.
03-SmithVanNess.indd 91
8/1/07 12:55:31
92
CAPÍTULO 3. Propiedades volumétricas de fluidos puros
CHAPTER 3.
3. Volumetric
Volumetric Properties
Properties of
of Pure
Pure Fluids
Fluids
CHAPTER
92
92
C
C
C
Figura 3.11: Isotermas
proporcionadas
por
ecuaciónas
Figure 3.11:
3.11:una
Isotherms
Figure
Isotherms
as
cúbica degiven
estado.
by
a
cubic
equation of
of
given by a cubic equation
state.
state.
T1  Tc
P
T �T
T11 � Tcc
PP
Psaturado
Tc
T
Tcc
sat
PPsat
T2  Tc
T �T
T22 � Tcc
Vsat urado(líquido)
sat (liq)
VVsat (liq)
Vsat urado(vapor)
sat (vap)
VVsat (vap)
V
VV
Una ecuación
estado
cúbica of
genérica
A Generic
GenericdeCubic
Cubic
Equation
State
A
Equation
of State
Desde laSince
introducción
de la ecuación
de van
Waals
se han propuesto
varias cubic
ecuaciones
de estado
cúbicas.
the introduction
introduction
of the
the van
van
derder
Waals
equation,
several hundred
hundred
equations
of state
state
Since the
of
der
Waals
equation,
several
cubic equations
of
Todas son
casos
la ecuación:
have
beenespeciales
proposed99de
All
are special
special cases
cases of
of the
the equation:
equation:
have
been
proposed
.. All
are
(V −
− η)
η)
RT
θθ(V
−
P=
= RT −
P
2 + κ V + λ)
V
−
b
(V
−
b)(V
2
V − b (V − b)(V + κ V + λ)
Here, b,
b, θθ,, κ,
κ, λ,
λ, and
and ηη are
are parameters
parameters which
which in
in general
general depend
depend on
on temperature
temperature and
and (for
(for mixmixHere,
Aquí, b,tures)
q, k, composition.
l y h son parámetros
que
enequation
general dependen
de
la temperatura
y de la composición
(para las
Although
this
appears
to
possess
great
flexibility,
it
has
inherent
tures) composition. Although this equation appears to possess great flexibility, it has inherent
10
mezclas).
Aunque
esta
ecuación
parece
tener
gran
flexibilidad,
posee
limitaciones
inherentes
debido
limitations because
because of
of its
its cubic
cubic form.
form.10 It
It reduces
reduces to
to the
the van
van der
der Waals
Waals equation
equation when
when ηη =
= b,
b, a su
limitations
10 Ésta se reduce
forma cúbica.
a
la
ecuación
de
van
der
Waals
cuando
h
=
b,
q
=
a
y
k
=
l
=
0.
= a,
a, and
and κκ =
= λλ =
= 0.
0.
θθ =
Una clase
importante
de ecuaciones
cúbicas results
resultafrom
de lathe
ecuación
anterior
haciendo
lasasdesignaAn
important
class
of cubic
cubic equations
equations
preceding
equation
with the
the
An important class of
results from
the preceding
equation
with
asciones: signments:
signments:
= bb
ηη =
= a(T
a(T ))
θθ =
= (�
(� +
+ σσ )b
)b
κκ =
= �σ
�σ bb22
λλ =
It is
is thus
thus transformed
transformed into
into an
an expression
expression general
general enough
enough to
to serve
serve as
as aa generic
generic cubic
cubic equation
equation of
of
It
Para una revisión, véase J. O. Valderrama, Ind. Eng. Chem. Res., vol. 42, pp. 1603-1618, 2003.
M. M. 9Abbott, AIChe J., vol. 19, pp. 596-601, 1973; Adv. in Chem. Series 182, K. C. Chao y R. L. Robinson, Jr., eds., pp. 47-70,
For aWashington,
a review,
review, see
see D.
J. O.
O.
Valderrama,
Ind. Eng.
Eng. Chem.
Chem. Res.,
Res., vol.
vol. 42,
42, pp.
pp. 1603–1618,
1603–1618, 2003.
2003.
9 For
J.
Ind.
Am. Chem. Soc.,
C.,Valderrama,
1979.
10
10 M. M. Abbott, AIChE J., vol. 19, pp. 596–601, 1973; Adv. in Chem. Series 182, K. C. Chao and R. L. Robinson,
10
M. M. Abbott, AIChE J., vol. 19, pp. 596–601, 1973; Adv. in Chem. Series 182, K. C. Chao and R. L. Robinson,
Jr., eds.,
eds., pp.
pp. 47–70,
47–70, Am.
Am. Chem.
Chem. Soc.,
Soc., Washington,
Washington, D.C.,
D.C., 1979.
1979.
Jr.,
03-SmithVanNess.indd 92
8/1/07 12:55:36
3.5.
Equations
of
3.5. Ecuaciones
cúbicas
de estado
3.5.
Cubic
Equations
of
State
3.5. Cubic
Cubic
Equations
ofState
State
3.5.
3.5. Cubic
Cubic Equations
Equations of
of State
State
93
93
93
93
93
93
Así, ésta
se transforma
en una
general
bastante
útil assignment
como
una ecuación
de estado
cúbica genérica,
state,
which
to
all
here
of
parameters:
state,
which
reduces
to
all
others
of
interest
here
upon
assignment
of
appropriate
parameters:
state,
whichreduces
reduces
toexpresión
allothers
othersof
ofinterest
interest
hereupon
upon
assignment
ofappropriate
appropriate
parameters:
state,
which
reduces
to
all
others
of
interest
here
upon
assignment
of
appropriate
parameters:
state,
which
reduces
to all ecuaciones
others of interest
here pero
uponbajo
assignment
of appropriate
parameters:
la cual se
reduce
a todas
las otras
de interés,
la designación
de parámetros
apropiados:
RT
a(T
RT
a(T
a(T)))
RT −
a(T
P
(3.42)
PP =
=
(3.42)
−
RT
a(T )) + σ b)
−(V + �b)(V
= V RT
(3.42)
P
=
(3.42)
−
−
b
VV −
−
(V
+
�b)(V
+
b)
P=V
(3.42) (3.42)
−bbb − (V
(V +
+�b)(V
�b)(V +
+σσσb)
b)
V − b (V + �b)(V + σ b)
For
For
given
equation,
and
are
pure
numbers,
the
same
for
all
substances,
whereas
paramForaaaagiven
givenequation,
equation,����and
andσσσ are
arepure
purenumbers,
numbers,the
thesame
samefor
forall
allsubstances,
substances,whereas
whereasparamparamFor
given
equation,
are
pure
numbers,
the
same
for
all
substances,
whereas
paramPara una
ecuación
conocida,
 and
y s σσson
números
puros,
iguales
para
todas
las of
sustancias,
mientras
que los
For
a
given
equation,
�
and
are
pure
numbers,
the
same
for
all
substances,
whereas
parameters
a(T
)
and
b
are
substance
dependent.
The
temperature
dependence
a(T
eters
a(T
and
are
substance
dependent.
The
temperature
dependence
of
a(T
isisspecific
specific
to
etersa(T
a(T)))and
andbbbare
aresubstance
substancedependent.
dependent. The
Thetemperature
temperaturedependence
dependenceof
ofa(T
a(T))))is
specificto
to
eters
is
specific
to
parámetros
a(T)
y
b
dependen
de
la
sustancia.
La
dependencia
de
la
temperatura
de
a(T)
se
especifica
para
eters
a(T
)
and
b
are
substance
dependent.
The
temperature
dependence
of
a(T
)
is
specific
to
each
each
equation
of
state.
For
the
van
der
Waals
equation,
a(T
=
isisaaaasubstance-dependent
substance-dependent
eachequation
equationof
ofstate.
state. For
Forthe
thevan
vander
derWaals
Waalsequation,
equation,a(T
a(T)))) =
= aaaa is
substance-dependent
each
equation
of
state.
For
the
van
der
Waals
equation,
a(T
=
is
substance-dependent
cada ecuación
de
estado.
Para
la
ecuación
de
van
der
Waals,
a(T)
=
a,
es
una
constante
que
depende
del
tipo
each
equation
of
state.
For
the
van
der
Waals
equation,
a(T
)
=
a
is
a
substance-dependent
constant,
constant,
and
=
=
0.
constant,and
and���=
=σσσ =
=0.
0.
constant,
0.
de sustancia,
y  and
= s ��= =
0. σσ =
constant,
and
=
= 0.
Determination
Determination
of
Equation-of-State
Parameters
Determinationof
ofEquation-of-State
Equation-of-StateParameters
Parameters
Determination
of
Equation-of-State
Parameters
Determinación
de los
de la ecuación
de estado
Determination
ofparámetros
Equation-of-State
Parameters
The
The
constants
in
an
equation
of
state
for
particular
substance
may
be
evaluated
by
fit
to
Theconstants
constantsin
inan
anequation
equationof
ofstate
statefor
foraaaparticular
particularsubstance
substancemay
maybe
beevaluated
evaluatedby
byaaaafit
fitto
to
The
constants
in
an
equation
of
state
particular
substance
may
be
evaluated
by
fit
to
Las constantes
enPuna
ecuación
de estado
para for
unaaasustancia
particular
es
posible
evaluarlas
mediante
The
constants
in
an
equation
of
state
for
particular
substance
may
be
evaluated
by
a
fit
toun ajusavailable
V
T
data.
For
cubic
equations
of
state,
however,
suitable
estimates
are
usually
available
PPV
VVT
TT data.
data.
For
cubic
equations
of
state,
however,
suitable
estimates
are
usually
available P
data. For
For cubic
cubic equations
equations of
of state,
state, however,
however, suitable
suitable estimates
estimates are
are usually
usually
available
te de lafound
información
disponible.
Noequations
obstante, of
para
lasc .however,
ecuacionesthe
de estado
cúbicas,
lasusually
estimaciones
available
PV
TPVT
data.
Forcritical
cubic
state,
suitable
estimates
are
found
from
values
for
the
constants
TTccccand
and
PPccc...Because
Because
the
critical
isotherm
exhibits
foundfrom
fromvalues
valuesfor
forthe
thecritical
criticalconstants
constantsTT
andP
Becausethe
thecritical
criticalisotherm
isothermexhibits
exhibitsaaaa
found
from
values
for
the
critical
constants
and
P
Because
critical
isotherm
exhibits
c
c
adecuadas
usualmente
se
encuentran
a
partir
de
los
valores
para
las
constantes
críticas
T
y
P
.
Ya
que la
found
frominflection
values foratthe
critical
constants
Tmay
andimpose
Pc . Because
the critical isotherm
a isoterc
cexhibits
cmay
horizontal
the
horizontal
the
critical
point,
we
impose
the
mathematical
conditions:
horizontalinflection
inflectionat
thecritical
criticalpoint,
point,we
wemay
impose
themathematical
mathematicalconditions:
conditions:
horizontal
inflection
atatthe
the
critical
point,
we
may
impose
the
mathematical
conditions:
ma crítica
exhibe
una
inflexión
horizontal
en
el
punto
crítico,
podemos
imponer
las
condiciones
matemáticas:
horizontal inflection at the critical point, we may impose the mathematical conditions:
�
�� 2 �
��
�
�
�
��∂ P �
��
�
�
�
�
∂∂∂∂2222P
∂
P
P
� ∂∂PP�
P
P
2
=
(3.43)
=
(3.44)
=
(3.43)
=
(3.44)
P
P2
=0000
(3.43)
=0000
(3.44)
=
(3.43)
=
(3.44)
∂∂∂∂∂V
∂∂∂∂∂V
VV TTTT;cr
=0
(3.43)
(3.44)
VV22222 T ;cr = 0
V
V
;cr
;cr
;cr
T
;cr
T
;cr
∂ V TT ;cr
∂ V TT ;cr
;cr
;cr
T ;cr
where
the
“cr”
denotes
point.
of
(3.42)
donde el
subíndice
“cr” denota
punto the
crítico.
Derivando
la ecuación (3.42)
produce
expresiones
para amwhere
the
subscript
“cr”
denotes
the
critical
point.
Differentiation
of
Eq.
(3.42)
yields
expreswhere
thesubscript
subscript
“cr”el
denotes
thecritical
critical
point.Differentiation
Differentiation
ofEq.
Eq.
(3.42)yields
yieldsexpresexpreswhere
the
subscript
“cr”
denotes
the
critical
point.
Differentiation
of
Eq.
(3.42)
yields
expreswhere
the
subscript
“cr”
denotes
the
critical
point.
Differentiation
of
Eq.
(3.42)
yields
expres,
T
=
T
,
and
V
=
V
sions
for
both
derivatives,
which
may
be
equated
to
zero
for
P
=
P
c
c
c
bas derivadas,
las
cuales
se
pueden
igualar
a
cero
para
P
=
P
,
T
=
T
y
V
=
V
.
La
ecuación
de
estado
,
T
=
T
,
and
V
=
V
sions
for
both
derivatives,
which
may
be
equated
to
zero
for
P
=
P
sions
for
both
derivatives,
which
may
be
equated
to
zero
for
P
=
P
,
T
=
T
,
and
V
=
V
c
cP = Pccc,c T = Tccc, and V = Vccc.... puede
sions
for
both
derivatives,
which
may
be
equated
to
zero
for
= Tcthree
, andequations
V, V=, VTcc,. a(T ) y
sions
forlas
both
derivatives,
which
may
be
equated
tocritical
zero
for
P =cinco
Pcc , T
The
equation
of
state
may
itself
be
written
for
the
conditions.
These
escribirse
para
condiciones
críticas.
Estas
tres
ecuaciones
contienen
constantes:
The
equation
of
state
may
itself
be
written
for
the
critical
conditions.
These
The
equation
of
state
may
itself
be
written
for
the
critical
conditions.
Thesecthree
threePequations
equations
c c c
c
The equation of state may itself be written for the critical conditions.
These
three
equations
The
equation
of
state
may
itself
be
written
for
the
critical
conditions.
These
three
equations
five
),
several
ways
to
these
ccc,,,,V
b. Entrecontain
las diferentes
manerasP
tratar
más
adecuada
eliminación
de Vc para obcontain
five
constants:
PPde
V
TTcccc,,estas
,,a(T
a(T
),),and
and
b.
Of
the
several
ways
to
treat
these
equations,
contain
fiveconstants:
constants:
Vcccc,,,,TT
a(Tcccecuaciones,
andb.
b. Of
Oflathe
the
several
wayses
tolatreat
treat
theseequations,
equations,
c
contain
five
constants:
P
V
a(T
),
and
b.
Of
the
several
ways
to
treat
these
equations,
c
cc , a(Tcc ), and b. Of the several ways to treat these equations,
contain
five
constants:
Pcc ,a(T
Vcc ,) Tyof
to
relating
.
the
is
tener expresiones
que
relacionen
bV
aV
yyield
Tyield
razón es que
Pc y Tca(T
soncccc)))conocidas
con
másTT
to
yield
expressions
relating
a(T
and
to
PPccc and
and
TTcprecisión
the
most
suitable
isiselimination
elimination
of
VVcP
to
expressions
relating
a(T
)and
andbbbbto
to P
and
themost
mostsuitable
suitable
elimination
of
c of
c. Laexpressions
cc cto
cc...
yield
expressions
relating
a(T
to
P
and
the
most
suitable
is
elimination
cc ) and
yieldaccurately
expressions
relating
a(T
and b to Pccc and Tccc .
the
most
suitable
is
elimination
of
Vccc tomore
reason
is
that
P
and
T
are
usually
known
than
V
.
c
c
c
que Vc.The
The
reason
is
that
P
and
T
are
usually
more
accurately
known
than
V
.
The
reason
is
that
P
and
T
are
usually
more
accurately
known
than
V
.
The reason is that Pcccc and Tcccc are usually more accurately known than Vcccc.
The
reason
is thatequivalente
Pc but
and
Tc are
usually
more accurately
known
than Vc .for
An
straightforward,
is
illustrated
the
der
Un
procedimiento
pero
más directo
seprocedure
ilustra para
ecuación
de
der
Puesto que
An
equivalent,
but
more
straightforward,
procedure
isisla
illustrated
for
the
van
der
Waals
Anequivalent,
equivalent,
butmore
more
straightforward,
procedure
illustrated
forvan
thevan
vanWaals.
derWaals
Waals
An
equivalent,
but
more
straightforward,
procedure
is
illustrated
for
the
van
der
Waals
An
equivalent,
but
more
straightforward,
procedure
is
illustrated
for the van der Waals
Because
V
=
V
for
each
of
the
three
roots
at
the
critical
point,
c
V = Vc equation.
para
cada
una
de
las
tres
raíces
en
el
punto
crítico,
equation.
Because
V
=
V
for
each
of
the
three
roots
at
the
critical
point,
equation.
Because
V
=
V
for
each
of
the
three
roots
at
the
critical
point,
equation. Because V = Vcccc for each of the three roots at the critical point,
equation. Because V = Vc for each of the three roots at the critical point,
3
(V
(V
−
VVccc))))3333=
=
(V −
−V
=0000
(V
−
V
=
(V − Vccc )3 = 0
3
2
2 − V 333= 0
or
V
(A)
or
VV3333−
−
3V
VV2222+
+
3V
VV −
−
VVc33 =
=
(A)
or
−3V
3VcccV
+3V
3Vc2222V
−V
=000
(A) (A)
o
or
V
−
3V
V
+
3V
V
(A)
or
V 3 − 3Vccc V 2 + 3Vcccc2c V − Vcccc3c = 0
(A)
Equation
(3.41),
written
for
(T
=
T
,
P
=
P
)
and
expanded
in
polynomial
form,
becomes:
Equation
(3.41),
written
for
(T
TTccc=c,,,P
PP)=
=
P
and
expanded
in
form,
becomes:
Equation
(3.41),
written
for=(T
(T =
=
=P
Pcccc)))and
andexpanded
expanded
in polynomial
polynomial
form,
becomes:
La ecuación
(3.41),
escrita
para for
(T
, PT
desarrollada
en forma
polinomial,
es: becomes:
Equation
(3.41),
written
P
in
form,
c =y �
�
Equation
(3.41),
written
for
(TTc=
=
P
in polynomial
polynomial
form,
becomes:
��cc ) and expanded
��Tcc , PRT
�
�
RT
ab
�
RTcccc� V 2222+ aaaa V − ab
ab = 0
3
RT
ab
V
(B)
cc V
VV33333−
VV22 +
−
+
+
VV −
−
=
(B)
− bbbb+
+ RT
+ Pa V
− ab
=000
(B) (B)
V
−
+
=
(B)
P
P
c
c
PPccc V + P
PPcccV − P
PPcccc = 0
V − b+ P
(B)
c
c
c
Pc
Pc
Pc
where
parameters
aaaaand
dependent,
but
independent
of
where
parameters
and
are
substance
dependent,
but
independent
of
temperature.
where
parameters
andbdependientes
aresubstance
substance
dependent,
but
independent
oftemperature.
temperature.
where
parameters
and
bbbare
are
substance
dependent,
but
independent
of
temperature.
donde los
parámetros
a
y
b
son
de
la
sustancia,
pero
independientes
de la temperatura.
whereTerm-by-term
parameters
a
and
b
are
substance
dependent,
but
independent
of
temperature.
comparison
of
Eqs.
(A)
and
(B)
provides
three
equations:
Term-by-term
comparison
of
Eqs.
(A)
and
(B)
provides
three
equations:
Term-by-term
comparison
of
Eqs.
(A)
and
(B)
provides
three
equations:
Term-by-term
comparison
of
Eqs.
(A)
and
(B)
provides
three
equations:
Al comparar
término
a
término
las
ecuaciones
(A)
y
(B)
se
obtienen
tres
ecuaciones:
Term-by-term comparison of Eqs. (A) and (B) provides three equations:
RT
ab
RT
ab
RTcccc (C) 3V 222= aaaa
ab
3
RT
ab
3V
(D)
V
(E)
cc
3V
=
+
(C)
3V
=
(D)
V
=
(E)
cc222c =
cc3333c3=
a
ab
3Vcccc=
=bbbb+
+ RT
(C)
3V
=
(D)
V
=
(E)
3V
=
+
(C)
3V
(D)
V
=
(E)
P
P
P
c
c
c
c
c
c
c
PPccc
PPccc
PPcccc
3Vc = b + P
(C) 3Vc = P
(D) Vc = P
(E)
Pcc
Pcc
Pcc
Solving
Eq.
for
a,
Eq.
for
Solving
Eq.
(D)
for
a,
combining
the
result
with
Eq.
(E),
and
solving
for
gives:
Solving
Eq.(D)
(D)
forpara
a,combining
combining
the
resultwith
with
Eq.
(E),and
andsolving
solving
forbbbgives:
gives:
Al resolver
la ecuación
(d)
a, combinarthe
el result
resultado
con
la(E),
ecuación
(e)
y resolver
para b se encuentra que:
Solving
Solving Eq.
Eq. (D)
(D) for
for a,
a, combining
combining the
the result
result with
with Eq.
Eq. (E),
(E), and
and solving
solving for
for bb gives:
gives:
111
2
bbbb=
aaaa=
VVccc2222c2
=
VVccc
=
3P
= 311 V
=3P
3PccccV
V
=
V
=
3P
c
b = 333Vccc
a = 3Pc Vcc
3
03-SmithVanNess.indd 93
8/1/07 12:56:00
94
CHAPTER 3.
3. Volumetric
Volumetric Properties
Properties of
of Pure
Pure Fluids
Fluids
CHAPTER
94
CHAPTER 3. Volumetric Properties of Pure Fluids
94
CHAPTER 3. Volumetric Properties of Pure Fluids
94
94
CAPÍTULO 3. Propiedades volumétricas de fluidos puros
Substitution
for
b
in
Eq.
(C)
allows
solution
for V
Vc , which
which can
can then
then be
be eliminated
eliminated from
from the
the
Substitution for
for bb in
in Eq.
Eq. (C)
(C) allows
allows solution
solution for
for
Substitution
Vcc,, which
can then
be eliminated
from the
equations
for
a
and
b:
equations
forfor
a and
and
b:Eq. (C) allows solution for Vc, which can then be eliminated from the
Substitution
b inb:
equations
for
Sustituyendo
a b en
laa ecuación
(c) se puede encontrar a V2c, la
que después se elimina de las ecuaciones para
27 R
1 RT
RTcc
R22 TTc222
RTcc
and b:33 RT
27
a y b: equations for a V
= 3 RTc
a=
= 27 R Tcc
b=
= 11 RT
c=
a
b
V
c
Pcc
64 2Pcc 2
Pccc
a = 64
b = 818 RT
Vc = 838 RT
P
P
T
27 RP
a = 64 Pc c
b = 8 Pc c
Vc = 8 Pc c
Although these
these equations
equations
may not
not yield
yield the
the best
best
possible
results, they
they provide
provide
reasonable valval8 Pcmay
64 possible
Pc
8 Pc reasonable
Although
results,
Although
these
equations
may be
notdetermined,
yield the
best
possible
results,
they provide
reasonable
values
which
can
almost
always
because
critical
temperatures
and
pressures
(inrazonaues
which
can almost
almost
always
be
determined,
because
critical
temperatures
and
pressures
(in
Aunqueues
conwhich
estasthese
ecuaciones
no
se
obtienen
los the
mejores
resultados
posibles,
proporcionan
valores
Although
equations
may
notdetermined,
yield
best
possible
results,
theysíprovide
reasonable
valcan
always
be
because
critical
temperatures
and
pressures
(in
contrast
to
extensive
P
V
T
data)
are
often
known,
or
can
be
reliably
estimated.
contrast
to extensive
extensive
PV
V Tdeterminar,
data)
are
often
known,
or can
can
be reliably
reliably
estimated.
bles quecontrast
casiwhich
siempre
pueden
ya que
lasbecause
temperaturas
ytemperatures
presiones
críticas
(en contraste
ues
canse
almost
always
beare
determined,
critical
and pressures
(in con la
to
often
known,
or
be
estimated.
Substitution
forPV
VccTin
indata)
the equation
equation
for
theconocidas
critical
compressibility
factor
reduces
it imimSubstitution
for
the
for
the
critical
compressibility
factor
reduces
it
gran cantidad
de
información
PVT)
con
frecuencia
son
o
se
calculan
de
manera
fiable.
contrast
to
extensive
P
V
T
data)
are
often
known,
or
can
be
reliably
estimated.
Substitution
Vc in the equation for the critical compressibility factor reduces it immediately
to:de V for
mediately
to:
La
sustitución
para el factor
de critical
compresibilidad
crítico la
reduce
inmediatamente
a:
c en
Substitution
forlaVcecuación
in the equation
for Pthe
compressibility
factor
reduces
it immediately
to:
Vcc
V
PccV
33
Z
≡
=
P
3
mediately to:
c
c c =
Zc ≡
≡ RT
8
Z
c = 8
c
RT
P
c Vccc
RT
83
≡
=
Z
c
A
single
value
for
Z
,
applicable
alike
to
all
substances,
results whenever
whenever the
the parameters
parameters of
of aa
A single
single value
value for
for Z
Z cc,, applicable
applicable alike
alike to
to all
all substances,
substances,
results
8 results
c manera
whenever
the parameters
a que se
ResultaA
un solo valor
para
Zcc, el of
cual
es apropiado
deRT
igual
para
todas las
sustancias,
cadaofvez
two-parameter
equation
state
are
found
by
imposition
of
the
critical
constraints.
Different
two-parameter
equation
of state
state are
are
found
bysubstances,
impositionresults
of the
the critical
critical constraints.
constraints.
Different
A
single
value
for
Z
,
applicable
alike
to
all
whenever
the
parameters
of
two-parameter
equation
of
found
by
imposition
of
Different
encuentran
los
parámetros
de
una
ecuación
de
estado
de
dos
parámetros
mediante
la
imposición
de
las
c
values are
are found
found for
for different
different equations
equations of
of state,
state, as
as indicated
indicated in
in Table
Table 3.1,
3.1, p.
p. 98.
98. Unfortunately,
Unfortunately,a restricvalues
two-parameter
equation
of
state
are
found
by
imposition
of
the
critical
constraints.
Different
values
are
found
for
different
equations
of
state,
as
indicated
in
Table
3.1,
p.
98.
Unfortunately,
ciones críticas.
Se
encuentran
valores
diferentes
para
las
distintas
ecuaciones
de
estado,
como
se
indica en la
the values
values so
so obtained
obtained do
do not
not in
in general
general agree
agree with
with those
those calculated
calculated from
from experimental
experimental
values
the
values
values
are
found
for
different
equations
of
state,
as
indicated
in
Table
3.1,
p.
98.
Unfortunately,
the
values
so
obtained
do
not
in
general
agree
with
those
calculated
from
experimental
values
tabla 3.1.
Por
desgracia,
los
valores
así
obtenidos
por
lo
general
no
están
de
acuerdo
con
los
calculados
a
of TTcc,, P
Pcc,, and
and V
Vcc;; each
each chemical
chemical species
species in
in fact
fact has
has its
its own
own value
value of
of ZZcc.. Moreover,
Moreover, the
the values
values
of
values
the
values
so
obtained
do
not
in
general
agree
with
those
calculated
from
experimental
T
,
P
,
and
V
;
each
chemical
species
in
fact
has
its
own
value
of
Z
.
Moreover,
the
values
partir deof
los
valores
experimentales
de
T
,
P
y
V
;
de
hecho,
cada
especie
química
tiene
su
propio
valor
de
Z
c
c
c
c
c various
c
c substances are almost all smaller than any of the
c.
given in
in Table
Table B.1
B.1 of
of App.
App. B
B for
for
given
various
substances
are
almost
allZsmaller
smaller
than any
any
of the
the
of
T
,
P
,
and
V
;
each
chemical
species
in
fact
has
its
own
value
of
.
Moreover,
the
values
given
in
Table
B.1
of
App.
B
for
various
substances
are
almost
all
than
of
c
c
c
c
Además,
casi
todos
los
valores
proporcionados
en
la
tabla
B.1
del
apéndice
B
para
diferentes
sustancias
son
equation values
values given
given in
in Table
Table 3.1.
3.1.
equation
given An
inque
Table
B.1
of
App.
for
various
substances
are
almost
all smaller
any
equation
values
given
in
Table
3.1.
más pequeños
cualquiera
de
los B
valores
obtenidos
a partircubic,
de las
ecuaciones
en lathan
tabla
3.1.of the
analogous
procedure
applied
to the
the
generic
Eq.
(3.42),
yields
expressions
for
An values
analogous
procedure
applied
to
generic
cubic, Eq.
Eq.
(3.42),
yields
expressions
for
equation
given
in
Table
3.1.
An
analogous
procedure
applied
to
the
generic
cubic,
(3.42),
yields
expressions
for
Un
procedimiento
análogo
que
se
puede
aplicar
a
la
cúbica
genérica,
ecuación
(3.42),
produce
expreparameters
a(T
)
and
b.
For
the
former,
c
parameters
a(Tc)) and
andprocedure
b. For
For the
theapplied
former,to the generic cubic, Eq. (3.42), yields expressions for
Anparámetros
analogous
parameters
a(T
b.
former,
siones para
los
a(T
)
y
b.
Para
la
anterior,
c
c
R222TTc222
parameters a(Tc ) and b. For the former,
R
=�
�R
a(Tcc)) =
Tcc
a(T
P
2T
a(Tc ) = � RP
c2
Pcc c
)
=
�
a(T
c
This result
result
is extended
extended
to temperatures
temperatures
other
than the
thePde
critical
by introduction
introduction
ofintroducción
a dimensionless
dimensionless
Este resultado
se puede
extender
a temperaturas
diferentes
mediante laof
de una funThis
is
to
other
than
critical
by
c la crítica
This
resultα(T
is
extended
to
temperatures
other
than
the critical
by
introduction of
aa dimensionless
function
)
that
becomes
unity
at
the
critical
temperature.
Thus
ción adimensional
a(T
), que
es igualunity
a uno
en
lacritical
temperatura
crítica. Thus
Así
rr ) rthat
function
α(T
becomes
at
the
temperature.
This result
isrextended
to temperatures
other
thantemperature.
the critical byThus
introduction of a dimensionless
function
α(T
) that becomes
unity at the
critical
function α(Tr ) that becomes unity at the critical temperature.
Thus
)R222TTc222
α(Trr )R
α(T
a(T )) =
=�
� α(T
(3.45) (3.45)
r )R Tcc
a(T
(3.45)
Pcc 2 2
a(T ) = � α(TrP
(3.45)
)R
T
Pc c
a(T ) = �
(3.45)
P
Function
α(T
)
is
an
empirical
expression,
specific
tocuna
a particular
particular
equation
of state
state (Table
(Table
3.1).3.1). El
La función
a(Tr)α(T
es una
empírica,
específica
para
ecuaciónequation
de estado
particular
(tabla
rr ) isexpresión
Function
an
empirical
expression,
specific
to
a
of
3.1).
Function
α(T
)given
is an empirical
expression, specific to a particular equation of state (Table 3.1).
r
Parameter
b
is
by:
parámetro
b se conoce
por: by:
Parameter
isr )given
given
Function α(T
is an by:
empirical expression, specific to a particular equation of state (Table 3.1).
Parameter
bb is
RTcc
RT
Parameter b is given by:
b=
=�
� RT
(3.46)
b
(3.46)
Pccc
b = � RT
(3.46)
(3.46)
P
c
b = � Pc
(3.46)
Pc
In these
these equations
equations �
� and
and �
� are
are pure
pure numbers,
numbers, independent
independent
of substance
substance and
and determined
determined for
for aa
In
of
In
these equations
�ofand
� from
are pure
numbers,
independent
of
determined for
a
particular
equation
state
the
values
assigned
to �� and
and
σsubstance
.sustancia,and
En estasparticular
ecuaciones
W y Ψofson
números
puros
e independientes
de la
que se determinan para una
equation
state
from
the
values
assigned
to
σ
.
In these
equations
�
and
�
are
pure
numbers,
independent
of
substance
and
determined
for
a
particular
equation
of
state
from
the
values
assigned
to
�
and
σ
.
The modern
modern
development
oflos
cubic
equations
of state
state
was
initiated in
in 1949
1949 by
by publication
publication
ecuación de estado
particular
a partir deof
valores
asignados
a  was
y s.initiated
The
development
cubic
equations
of
particular
equation
of
state
from
the
values
assigned
to
�
and
σ
.
The
modern
development
of
cubic
equations
of
state
was
initiated
in
1949
by
publication
11
11
ofdesarrollo
the Redlich/Kwong
Redlich/Kwong
(RK)
equation:11
Elof
moderno de
las equation:
ecuaciones
de estado cúbicas se inició en 1949 con la publicación de la
the
(RK)
The
modern development
of cubic equations of state was initiated in 1949 by publication
of the
Redlich/Kwong
(RK)
equation:
11
ecuación Redlich/Kwong (RK):
11
RT
a(T ))
of the Redlich/Kwong (RK) equation:
RT
a(T
P=
= RT
− a(T
(3.47)
)
P
−
(3.47)
VRT
−
b
V
(V
+) b)
b)
P
= V
−
(3.47)
−
b
V
(V
+
a(T
(V + b)
P = V − b − V−1/2
(3.47) (3.47)
−1/2
V
−
b
V
(V .+
)
=
T
. b)
where
a(T
)
is
given
by
Eq.
(3.45)
with
α(T
r ) = Trr−1/2
where
a(T
)
is
given
by
Eq.
(3.45)
with
α(T
r
.
where a(T ) is given by Eq. (3.45) with α(Tr ) = Tr
−1/2
–1/2 .
)
=
T
where
a(T
)
is
given
by
Eq.
(3.45)
with
α(T
r
r
donde a(T)
.
11 se conoce por la ecuación (3.45) con a(T ) = T
r pp. 233–244,
r
Otto Redlich
Redlich and
and J.
J. N.
N. S.
S. Kwong,
Kwong, Chem.
Chem. Rev.,
Rev., vol.
vol. 44,
44,
1949.
11 Otto
pp. 233–244, 1949.
11 Otto
Redlich and J. N. S. Kwong, Chem. Rev., vol. 44, pp. 233–244, 1949.
11 Otto Redlich and J. N. S. Kwong, Chem. Rev., vol. 44, pp. 233–244, 1949.
11
Otto Redlich and J. N. S. Kwong, Chem. Rev., vol. 44, pp. 233–244, 1949.
Otto Redlich y J. N. S. Wong, Chem. Rev., vol. 44, pp. 233-244, 1949.
03-SmithVanNess.indd 94
8/1/07 12:56:17
3.5. Ecuaciones cúbicas de estado
3.5.
3.5.
Cubic
Cubic
Equations
Equations
ofofState
of
State
3.5.
Cubic
Equations
State
3.5.
Cubic
Equations
of
State
959595
95
95
Teorema de estados correspondientes; factor acéntrico
Theorem
Theorem
ofofof
of
Corresponding
Corresponding
States;
States;
Acentric
Acentric
Factor
Factor
Theorem
Corresponding
States;
Acentric
Factor
Theorem
Corresponding
States;
Acentric
Factor
Las observaciones
experimentales
muestran
que
los factores
de Z
compresibilidad
Zfluids
para
diferentes
fluidos
Experimental
Experimental
observation
observation
shows
shows
that
that
compressibility
compressibility
factors
factors
Z for
different
for
different
fluids
exhibit
exhibit
simisimiExperimental
observation
shows
that
compressibility
factors
Zfor
different
fluids
exhibit
simiExperimental
observation
shows
that
compressibility
factors
Zfor
different
fluids
exhibit
simiexhiben
una
conducta
similar
cuando
se
correlacionan
con
una
función
de
temperatura
reducida
T
y una
r
larlarlar
behavior
lar
behavior
when
when
correlated
correlated
asasaas
as
afunction
function
ofofreduced
of
reduced
temperature
temperature
TrTrand
reduced
reduced
pressure
pressure
PrP; P
r and
behavior
when
correlated
afunction
reduced
temperature
reduced
pressure
; r;
behavior
when
correlated
afunction
of
reduced
temperature
TTand
r and reduced pressure rPr ;
presión
reducida
P
;
por
definición,
r
bybyby
definition,
by
definition,
definition,
definition,
TT TT
PP PP
and
yand
PrPr≡
TrTr≡
≡
P≡
≡
r≡
r≡
and
TTr≡
and
P
r
TcTcTTc
PcPcPPc
c
c
These
dimensionless
dimensionless
thermodynamic
thermodynamic
coordinates
coordinates
provide
provide
the
the
basis
basis
for
the
for
the
simplest
simplest
form
form
Estas These
coordenadas
termodinámicas
adimensionales
proporcionan
las
bases
para
lasimplest
forma
más
simple
These
dimensionless
thermodynamic
coordinates
provide
the
basis
for
the
simplest
form
These
dimensionless
thermodynamic
coordinates
provide
the
basis
for
the
form
ofofthe
of
the
theorem
theorem
of
corresponding
of
corresponding
states:
states:
del teorema
de
estados
correspondientes:
ofthe
thetheorem
theoremofofcorresponding
correspondingstates:
states:
All
fluids,
fluids,
when
when
compared
compared
atatthe
at
the
same
same
reduced
reduced
temperature
temperature
and
and
rereAll
fluids,
when
compared
the
same
reduced
temperature
and
reTodosAll
los
fluidos,
cuando
se
les compara
con
la misma
temperatura
y presión
reducidas,
All
fluids,
when
compared
at
the
same
reduced
temperature
and
reduced
duced
pressure,
pressure,
have
have
approximately
approximately
the
the
same
same
compressibility
compressibility
factor,
factor,
duced
have
approximately
the
factor,
tienen aproximadamente
mismo
factor de compresibilidad
y todos se desvían
del comducedpressure,
pressure,el
have
approximately
thesame
samecompressibility
compressibility
factor,
and
and
all
deviate
all
deviate
from
from
ideal-gas
ideal-gas
behavior
behavior
to
about
to
about
the
the
same
same
degree.
degree.
and
all
deviate
from
ideal-gas
behavior
to
about
the
same
degree.
portamiento
del
gas
ideal
casi
al
mismo
grado.
and all deviate from ideal-gas behavior to about the same degree.
Corresponding-states
Corresponding-states
correlations
correlations
ofofZof
of
based
ononon
this
on
this
theorem
theorem
are
are
called
called
two-parameter
two-parameter
correcorreCorresponding-states
correlations
Zbased
based
this
theorem
are
called
two-parameter
correCorresponding-states
correlations
ZZ
this
theorem
are
called
two-parameter
correLas correlaciones
de estados
correspondientes
debased
Ztwo
están
basadas
en este
teorema
yPAlthough
se. conocen
como
lations,
lations,
because
because
they
they
require
require
use
use
of
the
of
the
two
reducing
reducing
parameters
parameters
T
T
and
and
P
.
Although
these
thesecorrec
c
c
c
lations,
because
they
require
use
of
the
two
reducing
parameters
T
and
P
.
Although
these
cTc and cPc . Although these
lations,
because they
require
use of the
two
reducing
parameters
laciones
de
dos
parámetros,
porque
requieren
del
uso
de
los
dos
parámetros
reducidos
T
,
y
P
.
Aunque
c
c
correlations
correlations
are
are
very
very
nearly
nearly
exact
exact
for
the
for
the
simple
simple
fluids
fluids
(argon,
(argon,
krypton,
krypton,
and
and
xenon)
xenon)
systematic
systematic estas
correlations
are
very
nearly
exact
for
the
simple
fluids
(argon,
krypton,
and
xenon)
systematic
correlations
are
very
nearly
exact
for
the
simple
fluids
(argon,
krypton,
and
xenon)
systematic
correlaciones
están
cerca
de
ser
exactas
para
fluidos
simples
(argón,
kriptón
y
xenón)
se
observan
desviaciodeviations
deviations
are
observed
are
observed
for
more
for
more
complex
complex
fluids.
fluids.
Appreciable
Appreciable
improvement
improvement
results
results
from
from
introintrodeviations
are
observed
for
more
complex
fluids.
Appreciable
improvement
results
from
introdeviations
are
observed
for
more
complex
fluids.
Appreciable
improvement
results
from
intrones sistemáticas
para
fluidos
más
complejos.
Los
resultados
mejoran
de
manera
apreciable
a
partir
de
la
duction
duction
ofofaof
of
a third
third
corresponding-states
corresponding-states
parameter
parameter
(in
addition
(in
addition
totoTto
to
and
PcP),cPP
characteristic
ofofof
ofintrocTcand
c and
c),),
duction
athird
corresponding-states
parameter
(in(in
addition
),ccharacteristic
characteristic
duction
athird
corresponding-states
parameter
addition
TTcand
characteristic
ducción
de
un
tercer
parámetro
de
estados
correspondientes
(además
de
T
,
y
P
),
característico
c
c
molecular
molecular
structure;
structure;
the
the
most
most
popular
popular
such
such
parameter
parameter
isisthe
the
acentric
acentric
factor
factor
ω,ω,ω,
introduced
ω,
introduced
bybyby
byde la
molecular
structure;
the
most
popular
such
parameter
the
acentric
factor
introduced
molecular
structure;
the
most
popular
such
parameter
isis
the
acentric
factor
introduced
estructura
molecular;
el
más
popular
de
estos
parámetros
es
el
factor
acéntrico
w,
introducido
por
K.
S.
Pitzer
12
12
12
K.K.S.
K.
S.
Pitzer
andand
and
coworkers.
coworkers.12
S.Pitzer
Pitzer
coworkers.
K.
S.
Pitzer
coworkers.
12 and
y colaboradores.
The
The
acentric
acentric
factor
factor
for
for
a apure
a pure
pure
chemical
chemical
species
species
isisdefined
defined
with
with
reference
reference
totoits
to
vapor
its
vapor
The
acentric
factor
for
chemical
species
defined
with
reference
itsits
vapor
The
acentric
factor
for
apure
chemical
species
isis
defined
with
reference
to
vapor
El factor
acéntrico
para
una
especie
química
pura
estáofdefinido
confluid
respecto
a su presión
de vapor.
pressure.
pressure.
Because
Because
the
the
logarithm
logarithm
of
the
of
the
vapor
vapor
pressure
pressure
a
of
pure
a
pure
fluid
is
approximately
is
approximately
linear
linear
pressure.
Because
the
logarithm
of
the
vapor
pressure
of
a
pure
fluid
is
approximately
linear
pressure.
Because
thepresión
logarithm
of thedevapor
pressure
ofesa aproximadamente
pure fluid is approximately
linear con
Puestoinque
el
logaritmo
de
la
de
vapor
un
fluido
puro
lineal
en
relación
in
the
reciprocal
reciprocal
ofofabsolute
of
absolute
temperature,
temperature,
inthe
the
reciprocal
absolute
temperature,
in
the
of
absolute
temperature,
el recíproco
de reciprocal
la temperatura
absoluta,
satsat
d log
dlog
log
Psaturado
P sat
dlog
rPrrsat
d dlog
PrrP
= S= S
d(1/T
d(1/T
) )==S S
dd(1/T
(d(1/T
1 / rTrr ))rr)
satthe
saturado
sat
where
PrPsat
Pisla
the
reduced
reduced
vapor
vapor
pressure,
pressure,
the
reduced
reduced
temperature,
temperature,
and
Sla
the
the
slope
slope
ofofuna
of grádondewhere
Pwhere
es
presión
de vapor
reducida,
TTrrTes
temperatura
reducida,
yand
Sand
es
pendiente
de
rthe
the
reduced
vapor
pressure,
the
reduced
temperature,
SisSS
is
the
slope
rwhere
ris
isis
the
vapor
pressure,
TTrisla
isis
the
reduced
temperature,
and
isis
the
slope
of
rPrrsatissat
sat reduced
saturado
sat
a
plot
of
log
of
log
P
P
vs.
1/T
vs.
1/T
.
Note
.
Note
that
that
“log”
“log”
denotes
denotes
a
logarithm
a
logarithm
to
the
to
the
base
base
10.
10.
fica dea log
P
en
función
de
1/T
.
Observe
que
“log”
denota
un
logaritmo
de
base
10.
sat
r
r
aplot
plot
of
log
P
vs.
1/T
.
Note
that
“log”
denotes
a
logarithm
to
the
base
10.
r
r rP r vs. 1/T
r of log
r r . Note that “log” denotes a logarithm to the base 10.
a plot
r
IfIfthe
the
two-parameter
two-parameter
theorem
theorem
ofofcorresponding
of
corresponding
states
states
were
were
generally
generally
valid,
valid,
the
the
slope
slope
Si el teorema
detwo-parameter
estados correspondientes
con
dos parámetros
enwere
general
fuera
válido,
la
pendiente
S
the
two-parameter
theorem
corresponding
states
were
generally
valid,
the
slope
IfIf
the
theorem
of
corresponding
states
generally
valid,
the
slope
would
S
would
be
the
be
the
same
same
for
for
all
pure
all
pure
fluids.
fluids.
This
This
is
observed
is
observed
not
not
to
be
to
true;
be
true;
each
each
fluid
fluid
has
has
its
its
sería laSSmisma
para
todos
los
fluidos
puros.
Sin
embargo,
no
se
ha
observado
que
esto
sea
cierto;
cada
fluido
would
be
the
same
for
all
pure
fluids.
This
is
observed
not
to
be
true;
each
fluid
has
its
S would be the same for all pure fluids. This is observed not to be true; each fluid has its
own
own
characteristic
characteristic
value
value
ofofS,
of
which
S,
which
could
could
ininprinciple
in
principle
serve
serve
asasaas
as
a third
third
corresponding-states
tiene su
propio
valor característico
deS,
S,which
que
en
principio
puede
servir
como
elcorresponding-states
tercer
parámetro de estados
own
characteristic
value
S,
which
could
principle
serve
athird
corresponding-states
own
characteristic
value
of
could
in
principle
serve
athird
corresponding-states
parameter.
parameter.
However,
However,
Pitzer
Pitzer
noted
noted
that
that
all
vapor-pressure
all
vapor-pressure
data
data
for
for
the
the
simple
simple
fluids
fluids
(Ar,
(Ar,
Kr,
Kr,
correspondientes.
No
obstante,
Pitzer
observó
que
todos
los
datos
de
presión
de
vapor
para
los
fluidos
simples
parameter.
However,
Pitzer
noted
that
all
vapor-pressure
data
for
the
simple
fluids
(Ar,
Kr,
parameter. However, Pitzer noted that all vapor-pressure
data for the simple fluids
(Ar,
Kr,
satsat
sat vs.
saturado
Xe)
Xe)
lie
on
lie
the
on
the
same
same
line
line
when
when
plotted
plotted
as
log
as
log
P
P
vs.
1/T
1/T
and
and
that
that
the
the
line
line
passes
passes
through
through
sat
(Ar, Kr,
Xe)
se
encuentran
sobre
la
misma
línea
cuando
se
grafica
log
P
en
función
de
1/T
,
y
que
la línea
r
r
Xe)
lie
on
the
same
line
when
plotted
as
log
P
vs.
1/T
and
that
the
line
passes
through
r
r
r
r r rand that the line passes through
Xe)satliesaton the same line when plotted as log rPr vs. 1/T
saturado
sat
=
−1.0
=
−1.0
at
T
at
=
T
0.7.
=
0.7.
This
This
is
illustrated
is
illustrated
in
Fig.
in
Fig.
3.12.
3.12.
Data
Data
for
for
other
other
fluids
fluids
define
define
log
log
P
P
sat
pasa por
log
P
=
–1.0
en
T
=
0.7.
Esto
se
ilustra
en
la
figura
3.12.
Información
para
otros
fluidos
r
r
=
−1.0
at
T
=
0.7.
This
is
illustrated
in
Fig.
3.12.
Data
for
other
fluids
define
log
P
r
r
r
log rPrr = −1.0 at rTr = 0.7. This is illustrated in Fig. 3.12. Data for other fluids definedefine
other
other
lines
lines
whose
whose
locations
locations
can
can
bebefixed
be
fixed
ininrelation
in
relation
totopara
the
to
the
line
line
for
the
for
the
simple
simple
fluids
fluids
(SF)
(SF)
byla
the
by
the
otras líneas
cuyas
posiciones
se fijan
con
respecto
arelation
la
línea
fluidos
simples
(FS)
mediante
diferencia:
other
lines
whose
locations
can
fixed
the
line
for
the
simple
fluids
(SF)
byby
the
other
lines
whose
locations
can
be
fixed
in
relation
to
the
line
for
the
simple
fluids
(SF)
the
difference:
difference:
difference:
difference:
saturado
saturado
sat
sat
sat
loglog
P
–log
loglog
Psatrsat
PrPsat
P(SF)
(SF)
−−
−
PrP
Psat
sat(SF)
rlog
log
log
log
(SF)
−
log
rP r(SF)
rP r
r
r
The
The
acentric
acentric
factor
factor
isisdefined
is
defined
as
this
difference
difference
evaluated
evaluated
at
=
0.7:
rTr=
r0.7:
The
acentric
defined
asthis
this
difference
evaluated
atTat
0.7:
El factor
acéntrico
sefactor
define
como
laasdiferencia
evaluada
en
Tr =at0.7:
The
acentric
factor
is
defined
as
this
difference
evaluated
TTr=
=
0.7:
satsat sat )
ωω≡
≡
−1.0
−−log(P
−
log(P
T)rT=0.7
Tr =0.7
≡−1.0
−1.0
log(P
r r )rsat
ωω
≡
−1.0
−
log(P
)r T=0.7
=0.7
r
r
(3.48)
(3.48)(3.48)
(3.48)
(3.48)
Therefore
can
can
bebe
determined
be
determined
for
any
for
any
fluid
fluid
from
from
TcTa,cTT
P
P
,de
and
a aTsingle
measuremeasurePor loTherefore
tanto,
w seωωpuede
determinar
para
cualquier
fluido
,aPsingle
yvapor-pressure
devapor-pressure
una sola medición
de la precP, and
csingle
c, vapor-pressure
Therefore
can
determined
for
any
fluid
from
,partir
, cand
measureTherefore
ωω
can
be
determined
for
any
fluid
from
c , cPc , and a single vapor-pressure measure=
0.7.
=
0.7.
Values
Values
of
ω
of
and
ω
and
the
the
critical
critical
constants
constants
T
,
T
P
,
,
P
and
,
and
V
for
a
for
number
a
number
of
of se
ment
ment
made
made
at
T
at
T
sión de
vapor
hecha
a
T
=
0.7.
Los
valores
para
w
y
las
constantes
críticas
T
,
P
y
V
para
varios
fluidos
r
r
c
c
c
c
c
r
c
c
c
ment
0.7.Values
Valuesofofωωand
andthe
thecritical
criticalconstants
constantsTcT,cP
, cP,cand
, andVcVcfor
fora anumber
numberofof
mentmade
madeatatTrTr==0.7.
fluids
fluids
are
are
listed
listed
in
App.
in
App.
B.
B.
encuentran
en
el
apéndice
B.
fluids
fluidsare
arelisted
listedininApp.
App.B.B.
12
La descripción
completa
seinS.
encuentra
en K.
S. Pitzer, Thermodynamics,
3a.
ed., apéndice
3,
McGraw-Hill,
1212
12 Fully
Fully
described
described
ininK.
K.
S. Pitzer,
Thermodynamics,
Thermodynamics,
3d3ded.,
3d
App.
ed.,
App.
3,3,McGraw-Hill,
3, McGraw-Hill,
New
New
York,
York,
1995.
1995. Nueva York.
12
Fully
described
K.K.
S.Pitzer,
Pitzer,
Thermodynamics,
ed.,
App.
McGraw-Hill,
New
York,
1995.
Fully
described
in
S.
Pitzer,
Thermodynamics,
3d
ed.,
App.
3,
McGraw-Hill,
New
York,
1995.
03-SmithVanNess.indd 95
8/1/07 12:56:34
96
CAPÍTULO
Propiedades
volumétricas
deFluids
fluidos puros
CHAPTER 3.3.Volumetric
Properties
of Pure
96
log Prsat
Figura 3.12:
Dependencia
Figure
3.12: Approximate
aproximada
con
respecto
a la
temperature
dependence
of
temperatura
de
la
presión
the reduced vaporde vapor
reducida.pressure.
�1
�2
log Prsaturado
1.0
0
1.0
1.2
0
1.2
1.4
1/Tr
1.4
1.6
1/Tr
1.6
1.8
1.8
2.0
2.0
1
Pendiente  2.3
Slope � �2.3
(Ar, Kr, Xe)
(Ar, Kr, Xe)
2
1
1
Slope �Pendiente
�3.2
 3.2
1
1
 1.43

T 1.430.7
�
�
(n-Octane) (n-octano)
Tr
0.7 r
La definición de w hace que su valor sea cero para el argón, kriptón y xenón, y la información experiThefactores
definition
ω makes its value
krypton,
and correlacionados
xenon, and experimental
mental produce
de of
compresibilidad
parazero
los for
tresargon,
fluidos
que están
por las mismas
data yield
factors
for all three
that es
arelacorrelated
by thedel
same
curvesteorema
curvas cuando
Z secompressibility
representa como
una función
de Tr fluids
y Pr. Ésta
premisa básica
siguiente
when
Z is representedcon
as tres
a function
of Tr and Pr . This is the basic premise of the following
de estados
correspondientes
parámetros:
three-parameter theorem of corresponding states:
Todos los fluidos que tienen el mismo valor w, cuando se les compara con la misma Tr y Pr
Allmismo
fluidsvalor
having
the
same se
value
of ω ,del
when
compared atdethe
same
tienen el
de Z,
y todos
desvían
comportamiento
gas
idealTelr mismo
and
P
,
have
about
the
same
value
of
Z
,
and
all
deviate
from
idealr
grado.
gas behavior to about the same degree.
VaporVapor
y las &
raíces
de la ecuación
cúbica
Vapor-Like
Roots ofde
theestado
Generic
Cubicgenérica
Equation of State
may solve
explicitly
for para
its three
roots,
the generic
cubicdeequation
of state,
AunqueAlthough
es posibleone
resolver
en forma
explícita
sus tres
raíces,
la ecuación
estado cúbica
genérica,
13 Convergence
13 Los problemas
Eq.
(3.42),
practiceusualmente
far more commonly
by iterative
procedures.
ecuación
(3.42),
enislainpráctica
se resuelvesolved
mediante
procedimientos
iterativos.
problems se
areevitan
most más
likelyfácil
avoided
when
the equation
is rearranged
to a forma
form suited
to the
solude convergencia
cuando
la ecuación
se reacomoda
en una
adecuada
para
encontrar
for a particular
root.
largestesroot,
i.e.,
vapor orde
vapor-like
volume,
Eq. (3.42)
una raíztion
en particular.
Para la
raízFor
másthe
grande,
decir,
unavolumen
vapor o de
algo parecido,
la is
ecuación
multiplied
(V − b)/RT
. It can se
then
be written:
(3.42) se
multiplicathrough
por (V by
– b)/RT.
Ésta entonces
puede
escribir como:
V =
RT
a(T )
V −b
+b−
P
P (V + �b)(V + σ b)
(3.49) (3.49)
La solución
para for
V seVpuede
encontrar
con unaofrutina
de un paquete
software.
Solution
may be
by trial,mediante
iteration,ensayo,
or withiteración
the solveoroutine
a software
package.deAn
initial estimate
ideal-gas
value
RTRT/P.
/P. For
this este
valuevalor
is substituted
onde V en
Una estimación
inicial for
paraVVisesthe
el valor
del gas
ideal
Paraiteration,
la iteración,
se sustituye
the rightde
side
of Eq. (3.49).
resulting
value of
thelado
leftizquierdo
is then returned
to the
right al lado
el lado derecho
la ecuación
(3.49).The
El valor
resultante
deVV on
en el
se regresa
después
the process
the change
is suitably small.
derechoside,
y el and
proceso
continúacontinues
hasta queuntil
el cambio
en Vin
esVadecuadamente
pequeño.
Una ecuación
para Z,forequivalente
a lato(3.49)
se obtiene
a través
la sustitución
ZRT/P.
Además,
An equation
Z equivalent
Eq. (3.49)
is obtained
byde
substituting
V =VZ=RT
/P. In
la definición
de dos
adimensionales
conduce
a la simplificación.
Así,
addition,
the cantidades
definition of
two dimensionless
quantities
leads to simplification.
Thus,
13 Estos13
Such procedures
are built into
computerdentro
software
packages
With these
one
procedimientos
se encuentran
integrados
de los
paquetesfor
detechnical
software calculations.
de las computadoras
parapackages
cálculos técnicos.
Con
can solve
routinely
for V Vinmediante
equationsuna
such
as (3.42)
with little como
thought
to how
is done.
careful
estos paquetes
es posible
calcular
rutina
en ecuaciones
la as
(3.42)
conitmuy
pocaHowever,
labor mental
de thought
cómo fue hecho.
should
be tenerse
given toespecial
the question
of whether
the answers
Sin embargo,
debe
cuidado
con la pregunta
de siare
lasreasonable.
respuestas en realidad son razonables.
03-SmithVanNess.indd 96
8/1/07 12:56:39
3.5.
Cubic
Equations
of
State
3.5. Cubic
Cubiccúbicas
Equations
of State
State
3.5.
Equations
of
3.5. Ecuaciones
de estado
3.5.
3.5. Cubic
Cubic Equations
Equations of
of State
State
97
97
97
97
97
97
a(T
P
a(T)))
bbbPP
a(T
≡
≡
(3.50)
(3.51)
))
bPP
a(T
β≡
≡ bRT
≡ a(T
(3.50) qqq ≡
(3.51)
ββ
(3.50)
(3.51)
RT
ββ ≡
(3.50)
(3.51)
≡ RT
≡ bbbRT
(3.50) qq ≡
(3.51)
RT
RT
RT
bbRT
RT
RT
With
these
substitutions
Eq.
(3.49)
becomes:
With
these
substitutions
Eq.
(3.49) becomes:
becomes:
With
these
substitutions
(3.49)
Mediante
estas
sustituciones,
enEq.
la ecuación
(3.49) se obtiene:
With
With these
these substitutions
substitutions Eq.
Eq. (3.49)
(3.49) becomes:
becomes:
−
Z−
− ββ
β
ZZ
(3.52)
=
+
−
qβ
ZZ −
− ββ+ σβ)
(3.52) (3.52)
Z=
= 111 +
+ ββ
β−
− qβ
qβ (Z + �β)(Z
(3.52)
ZZ
(3.52)
ZZ =
(3.52)
= 11 +
+ ββ −
− qβ
qβ(Z
(Z +
+ �β)(Z
�β)(Z +
+ σβ)
σβ)
(Z
(Z +
+ �β)(Z
�β)(Z +
+ σβ)
σβ)
Equations
(3.50)
and
(3.51)
in
combination
with
Eqs.
(3.45)
and
(3.46)
yield:
Con lasEquations
ecuaciones
(3.50)
y (3.51)
combinación
con las
ecuaciones
(3.46) se obtiene:
Equations
(3.50)
and
(3.51)en
in combination
combination with
with
Eqs.
(3.45) and
and(3.45)
(3.46)yyield:
yield:
(3.50)
and
(3.51)
in
Eqs.
(3.45)
(3.46)
Equations
Equations (3.50)
(3.50) and
and (3.51)
(3.51) in
in combination
combination with
with Eqs.
Eqs. (3.45)
(3.45) and
and (3.46)
(3.46) yield:
yield:
�
α(T
Prrr
� α(T
α(Trrr)))
PP
�
=
�
(3.53)
qq =
=
(3.54)
�
α(T
P
�
α(Trr))
β=
=�
�P
(3.53)
=
(3.54)
ββ
(3.53)
q
(3.54)
r
r
�
ββ =
(3.53)
(3.54)
=�
�TT
(3.53) qq =
= �
(3.54)
Trrr
� TT
Trrr
TTrr
�
�TTrr
Iterative
solution
of
Eq.
(3.52)
starts
with
the
value
=
on
the
right
side.
La solución
iterativa
de la
(3.52)
el
valor
=substituted
1 sustituido
ladoside.
derecho. El
Iterative
solution
ofecuación
Eq. (3.52)
(3.52)
startsinicia
withcon
the value
value
Z de
= 1Z11 substituted
substituted
on en
theelright
right
side.
Iterative
solution
of
Eq.
starts
with
the
ZZ
=
on
the
Iterative
solution
of
Eq.
(3.52)
starts
with
the
ZZ process
=
11 substituted
on
the
right
side.
Iterative
solution
of
Eq.
(3.52)
starts
with
the value
value
=hasta
substituted
on
the
right
side.final de
The
calculated
value
of
Z
is
returned
to
the
right
side
and
the
continues
to
convergence.
valor calculado
de
Z
se
regresa
al
lado
derecho
y
el
proceso
continúa
la
convergencia.
El
valor
The
calculated
value
of
Z
is
returned
to
the
right
side
and
the
process
continues
to
convergence.
The calculated value of Z is returned to the right side and the process continues to convergence.
The
of
is
returned
to
the
side
the
process
continues
The calculated
calculated
value
of ZZmediante
isthe
returned
toZRT/P.
the right
right
side and
and
theRT
process
continues to
to convergence.
convergence.
final
value
of
yields
volume
root
through
=
/P.
Z produce
valor
del value
volumen
V=
Theelfinal
final
value
of ZZ
Z yields
yields
the volume
volume
root through
through VV
V=
= ZZ
Z RT
RT/P.
/P.
The
value
of
the
root
The
The final
final value
value of
of ZZ yields
yields the
the volume
volume root
root through
through VV =
= ZZRT
RT/P.
/P.
Liquid
&
Liquid-Like
Roots
of
the
Generic
Cubic
Equation
of
State
Liquid
&raíces
Liquid-Like
Roots of
of the
the
Genericcúbica
Cubicgenérica
Equation of
of State
State
Liquid
Liquid-Like
Roots
Generic
Cubic
Equation
Líquido
y las&
de la ecuación
de
Liquid
&
Roots
Generic
Liquid
& Liquid-Like
Liquid-Like
Roots of
of the
theestado
Generic Cubic
Cubic Equation
Equation of
of State
State
Equation
(3.49)
may
be
solved
for
the
in
the
numerator
of
the
final
fraction
to
give:
Equation (3.49)
(3.49) may
may be
be solved
solved for
for the
the VV
V in
in the
the numerator
numerator of
of the
the final
final fraction
fraction to
to give:
give:
Equation
Equation
(3.49)
may
be
for
of
the
final
fraction
to
give:
Equation(3.49)
(3.49)es
may
be solved
solved
for the
the
inenthe
the
numerator
of
the
final
fraction
to
give:
�
�
Con la ecuación
posible
resolver
paraVVVin
elnumerator
numerador
de
la
fracción
final
obteniendo:
�� RT + b P − V P ��
�
�
�
�
RT
+
b
P
−
V
P
RT + b P − V P
=
+
(V
+
�b)(V
+
b)
(3.55)
bbPP −
RT +
+a(T
− VV PP
V=
= bbb +
+ (V
(V +
+ �b)(V
�b)(V +
+ σσ
σb)
b) RT
(3.55)
VV
(3.55)
VV =
(3.55)
= bb +
+ (V
(V +
+ �b)(V
�b)(V +
+ σσb)
b)
(3.55) (3.55)
a(T)))
a(T
a(T
a(T))
This
equation
with
starting
value
of
=
on
the
right
side
converges
upon
iteration
to
This equation
equation with
with aaa starting
starting value
value of
of VV
V =
= bbb on
on the
the right
right side
side converges
converges upon
upon iteration
iteration to
to aaa
This
Esta ecuación
con
un valor
inicial
de value
V
= b of
en
el
converge
por iteración
a una
raíz de
o
This
with
starting
=
on
side
upon
iteration
to
aa
This equation
equation
with aaroot.
starting
value
of V
V lado
= bb derecho,
on the
the right
right
side converges
converges
upon
iteration
to líquido
liquid
or
liquid-like
liquid or
or liquid-like
liquid-like root.
root.
liquid
de algoliquid
parecido.
or
liquid-like
root.
liquid
or
liquid-like
root.
An
equation
for
equivalent
to
Eq.
(3.55)
is
obtained
when
Eq.
(3.52)
is
solved
for
the
An equation
equation for
for ZZ
Z equivalent
equivalent to
to Eq.
Eq. (3.55)
(3.55) is
is obtained
obtained when
when Eq.
Eq. (3.52)
(3.52) is
is solved
solved for
for the
the
An
Una
ecuación
para Zfor
quethe
a lato
se obtiene
cuandowhen
la ecuación
(3.52)
se resuelve
para Z en
An
equation
ZZequivale
equivalent
Eq.
is
Eq.
is
for
Annumerator
equation
for
equivalent
to(3.55)
Eq. (3.55)
(3.55)
is obtained
obtained
when
Eq. (3.52)
(3.52)
is solved
solved
for the
the
Z
in
the
of
final
fraction:
in the
the numerator
numerator of
of the
the final
final fraction:
fraction:
ZZ in
�
�
el numerador
denumerator
la fracciónof
ZZ in
the
in the
the
numerator
offinal:
the final
final fraction:
fraction:
�1 + β − Z�
�
�
�
�11 +
�
+ ββ −
− ZZ �
+
σβ)
(3.56)
Z
=
β
+
(Z
+
�β)(Z
1
+
β
−
Z
1
+
β
−
Z
+
σβ)
(3.56)
Z
=
β
+
(Z
+
�β)(Z
(3.56)
Z = β + (Z + �β)(Z + σβ)
qβ
(3.56)
ZZ =
+ σβ)
σβ)
(3.56) (3.56)
= ββ +
+ (Z
(Z +
+ �β)(Z
�β)(Z +
qβ
qβ
qβ
qβ
For
iteration
starting
value
of
=
is
substituted
on
the
right
side.
Once
is
known,
the
For iteration
iteration aaa starting
starting value
value of
of ZZ
Z=
= ββ
β is
is substituted
substituted on
on the
the right
right side.
side. Once
Once ZZ
Z is
is known,
known, the
the
Para la For
iteración,
unis
inicial
de of
Z
en el on
lado
UnaOnce
vez que
se conoce,
For
aavalor
starting
value
is
the
right
ZZ is
the
For iteration
iteration
starting
value
of=ZZb=
=seββsustituye
is substituted
substituted
on
thederecho.
right side.
side.
Once
isZknown,
known,
the la raíz
volume
root
V
=
Z
RT
/P.
volume root
root is
is VV =
= ZZ RT
RT/P.
/P.
volume
del volumen
es
V
=
ZRT/P.
volume
root
ZZRT
/P.
volume
root is
is VVof=
=state
RT
/P. express Z as a function of Tr and Pr are said to be generalized,
Equations
which
and PPrr are
are said
said to
to be
be generalized,
generalized,
Equations of
of state
state which
which express
express ZZ as
as aa function
function of
of TTrr and
Equations
Las
ecuaciones
deof
estado
que
expresan
enaagases
función
de
TTTrrryand
Pr Any
se
que to
deben
ser generalizadas,
PPrrdice
are
said
Equations
state
express
ZZZall
as
function
of
and
are
said
to be
be generalized,
generalized,
Equations
of
state which
which
expressato
as
function
of
because
of
their
general
applicability
and
liquids.
equation
of
state
can
be
because of
of their
their general
general applicability
applicability to
to all
all gases
gases and
and liquids.
liquids.
Any
equation of
of state
state can
can be
be
because
Any
equation
debido because
aput
su
campo
de
aplicación
general
para
todos
los
gases
y
los
líquidos.
Es
posible
presentar
cualquier
of
general
applicability
to
gases
liquids.
Any
of
can
because
of their
their
general
applicability
to all
allcorrelation
gases and
andfor
liquids.
Any equation
equation
of state
state
can be
be
into
this
form
to
provide
a
generalized
the
properties
of
fluids.
This
allows
put into
into this
this form
form to
to provide
provide aa generalized
generalized correlation
correlation for
for the
the properties
properties of
of fluids.
fluids. This
This allows
allows
put
ecuación
de
estado
en
esta
forma
para
proporcionar
una
correlación
generalizada
para
las
propiedades
de los
put
to
aa generalized
correlation
for
of
This
allows
put into
into this
this form
form
to provide
provide
generalized
correlation
for the
the properties
properties
of fluids.
fluids.
Thissuch
allows
the
estimation
of
property
values
from
very
limited
information.
Equations
of
state,
as
the estimation
estimation of
of property
property values
values from
from very
very limited
limited information.
information. Equations
Equations of
of state,
state, such
such as
as
the
fluidos.the
Esto
permite
el
cálculo
de
valores
de
la
propiedad
a
partir
de
información
muy
limitada.
Las
ecuacioof
values
limited
information.
Equations
of
such
as
the estimation
estimation
of property
property
values from
from very
very
limitedwhich
information.
Equations
of state,
state,
such
as
van
der
Waals
and
Redlich/Kwong
equations,
express
as
functions
of
and
the van
van der
der Waals
Waals and
and Redlich/Kwong
Redlich/Kwong equations,
equations, which
which express
express ZZ
Z as
as functions
functions of
of TT
Trr and
and
the
nes de estado,
la two-parameter
de van
Waals
y la de Redlich/Kwong,
expresan
a Zfunctions
sólo como
the
van
der
Waals
and
Redlich/Kwong
equations,
which
express
ZZ as
of
TTrrr and
the
vancomo
der
Waals
andder
Redlich/Kwong
equations,
whichque
express
as
functions
offunciones
and de Tr
only,
yield
corresponding
states
correlations.
The
Soave/Redlich/Kwong
P
r
only, yield
yield two-parameter
two-parameter corresponding
corresponding states
states correlations.
correlations. The
The Soave/Redlich/Kwong
Soave/Redlich/Kwong
PPrr only,
14
15
yield
corresponding
states
The
Soave/Redlich/Kwong
PPrr only,
only,
yield two-parameter
two-parameter
corresponding
states
correlations.
The
Soave/Redlich/Kwong
y Pr, producen
correlaciones
de
estado
correspondientes
decorrelations.
dos15
parámetros.
La
ecuaciónfactor
de Soave/Redlich/
14 and
15
(SRK)
equation
the
Peng/Robinson
(PR)
equation,
in
which
the
acentric
enters
14
(SRK)
equation
and
the
Peng/Robinson
(PR)
equation,
in
which
the
acentric factor
factor enters
enters
(SRK)
equation
and
the
Peng/Robinson
(PR)
equation,
in
which
the
acentric
14
15
14 and
15 in
14
15 en
equation
Peng/Robinson
(PR)
equation,
which
acentric
factor
enters
(SRK)
equation
andr the
the
Peng/Robinson
(PR)
equation,
in
which the
the
acentric
factor
enters
Kwong(SRK)
(SRK)
y
la
de
Peng/Robinson
(PR),
las
que
el
factor
acéntrico
se
introduce
a
través
de la funthrough
function
α(T
;
ω)
as
an
additional
parameter,
yield
three-parameter
correspondingthrough function
function α(T
α(Trr;; ω)
ω) as
as an
an additional
additional parameter,
parameter, yield
yield three-parameter
three-parameter correspondingcorrespondingthrough
through
function
α(T
;
ω)
as
an
additional
parameter,
yield
three-parameter
correspondingthrough
function
α(T
;
ω)
as
an
additional
parameter,
yield
three-parameter
correspondingción a(Tstates
;
w)
como
un
parámetro
adicional,
produce
correlaciones
de
estados
correspondientes
con
tres parár
r
correlations.
The
numerical
assignments
for
parameters
�,
σ
,
�,
and
�,
both
for
these
r
states correlations.
correlations. The
The numerical
numerical assignments
assignments for
for parameters
parameters �,
�, σσ,, �,
�, and
and �,
�, both
both for
for these
these
states
correlations.
The
numerical
assignments
for
�,
,, �,
and
�,
both
for
states
correlations.
Thevan
numerical
assignments
for parameters
parameters
�, σσpara
�,are
and
�,ecuaciones
both
for these
these
metros.states
Las
asignaciones
numéricas
para
los
parámetros
,
s,
W
y
Ψ,
ambas
y
equations
and
for
the
der
Waals
and
Redlich/Kwong
equations,
given
in
Table
3.1.
equations and
and for
for the
the van
van der
der Waals
Waals and
and Redlich/Kwong
Redlich/Kwong equations,
equations, are
are given
given in
in Table
Table 3.1.
3.1.para las
equations
equations
and
for
the
van
der
Waals
and
Redlich/Kwong
equations,
are
given
in
3.1.
equations
and
for
thegiven
van
derα(T
Waals
and
Redlich/Kwong
equations,
are
given
in Table
Table
3.1.
ecuaciones
de
van
der
Waals
y
de
Redlich/Kwong,
se
proporcionan
en
la
tabla
3.1.
También
se
dan
expresioExpressions
are
also
for
;
ω)
for
the
SRK
and
PR
equations.
r
Expressions are
are also
also given
given for
for α(T
α(Trr;; ω)
ω) for
for the
the SRK
SRK and
and PR
PR equations.
equations.
Expressions
Expressions
are
also
given
for
α(T
;
ω)
for
the
SRK
and
PR
equations.
Expressions
are
also
given
for
α(T
;
ω)
for
the
SRK
and
PR
equations.
nes para a(Tr; w) para las ecuaciones SRKrr y PR.
14
Soave,
Chem.
Eng.
Sci.,
vol.
27,
pp.
1197–1203,
1972.
14 G.
14
G. Soave,
Soave, Chem.
Chem. Eng.
Eng. Sci.,
Sci., vol.
vol. 27,
27, pp.
pp. 1197–1203,
1197–1203, 1972.
1972.
G.
14
14G.
Soave,
Chem.
Eng.
Sci.,
vol.
27,
pp.
1197–1203,
1972.
G.
Soave,
Chem.
Eng.
Sci.,
vol.
27,
pp.
1197–1203,
1972.
14 G. Soave,
15
Chem.
Eng.
Sci.,
vol.
27,
pp.
1197-1203,
1972.
Peng
and
D.
B.
Robinson,
Ind.
Eng.
Chem.
Fundam.,
vol.
15,
pp.
59–64,
1976.
15 D.-Y.
15
D.-Y.
Peng
and
D.
B.
Robinson,
Ind.
Eng.
Chem.
Fundam.,
vol. 15,
15, pp.
pp. 59–64,
59–64, 1976.
1976.
D.-Y.
Peng
and
D.
B.
Robinson,
Ind.
Eng.
Chem.
Fundam.,
vol.
15 D. Y.15
15
Peng
y D.
B. and
Robinson,
Ind. Eng. Chem.
Fundam.,
15, pp.vol.
59-64,
1976.
D.-Y.
Peng
Ind.
Chem.
Fundam.,
15,
59–64,
D.-Y.
Peng
and D.
D. B.
B. Robinson,
Robinson,
Ind. Eng.
Eng.
Chem. vol.
Fundam.,
vol.
15, pp.
pp.
59–64, 1976.
1976.
03-SmithVanNess.indd 97
8/1/07 12:56:53
98
98 98
98
98
98
CHAPTER
CHAPTER
3. Volumetric
3. Volumetric
Properties
Properties
of Pure
of Pure
Fluids
Fluids
CHAPTER 3. Volumetric Properties of Pure Fluids
CHAPTER
3.
Volumetric
Properties
of
Pure
Fluids
CHAPTER 3. Volumetric Properties of Pure Fluids
CAPÍTULO 3. Propiedades volumétricas de fluidos puros
Table
Table
3.1:3.1:
Parameter
Parameter
Assignments
Assignments
for for
Equations
Equations
of State
of State
Table
3.1: Parameter
Assignments
forlas
Equations
of State
TablaTable
3.1: Asignación
de parámetros
para
ecuaciones
de estado
3.1:
Parameter
Assignments
for
Equations
of
State
Table 3.1:For
Parameter
Assignments
for
Equations
For
use use
withwith
Eqs.
Eqs.
(3.49)
(3.49)
through
through
(3.56)
(3.56) of State
For
use
with
Eqs. (3.49) through
(3.56)
ParaFor
su use
usowith
con las
(3.49)(3.56)
a (3.56)
Eqs.ecuaciones
(3.49) through
through
For use
with Eqs.
(3.49)
(3.56)
Eq.Eq.
of de
State
ofestado
State α(T
α(T
�Ω �
� Ψ
�
Z c Z cZc
r )r) r )
� � �
sσ σ σ
Ecuación
a(T
Eq. of
State
α(T
�
�
Z
r)
Eq.
of
State
α(T
)
σ
�
�
�
Zccc
r
Eq.
of
State
α(T
)
σ
�
�
�
Z
r
vdWvdW
(1873)
00 0
00 0
1/81/8 27/64
27/64
vdW
(1873)
(1873)
11 1
1/8
27/64 3/8 3/83/8
vdW (1873)
1−1/2
0
0
1/8
27/64
3/8
−1/2
–1/2
vdW
(1873)
1
0
0
1/8
27/64
3/8
1
1
0
0
0.08664
0.08664
0.42748
0.42748
1/3
1/31/3
RK
RK
(1949)
(1949)
T
T
RK (1949)
10
00
0.08664
T
vdW
(1873)
1
1/8
27/64
3/8
r r −1/2
r
1
0
0.08664 0.42748
1/3
RK (1949)
Tr−1/2
−1/2 † †
1
0
0.08664
0.42748
1/3
RK
(1949)
T
SRK
(1972)
(1972)α aαSRK
(Trr r ;(T
ω)
0.086640.42748
0.42748 1/3
1/3
r;; ω)
†
11 1√ 00√
00 0√ 0.08664
0.08664
1/3
RK
(1949)
T
SRKSRK
(1972)
11√
0.08664
0.42748
SRK
SRK
(1972) SRK
αSRK
(T(T
0.08664 0.42748
0.42748
1/3 1/3
r ;rω)
‡ω)
√
√
†† ‡ 1 +11
(T
;
ω)
1
0
0.08664
0.42748
1/3
SRK
(1972) αα
αPR
†
r
PR
PR
(1976)
(1976)
α
SRK
(T
;
(T
ω)
;
+
2
2
1
−
1
−
2
2
0.07780
0.07780
0.45724
0.45724
0.30740
0.30740
PR
r
r
SRK
(1972)
(T
;
ω)
0
0.08664
0.42748
1/3
‡
w)
√2 1 − √
√2 0.07780 0.45724 0.30740
PR (1976)
αSRK
1 +√
PR (Trr; ω)‡
PR (1976)
(1976)
αPR
(Trr ;; ω)
ω)‡‡
+ 22 11 −
− 22 �0.07780
0.07780
0.45724
0.30740
PR�(T
��0.45724
�α
�
PR
11+ +
0.30740
2��
2
1
√
2
1
–
√
2
0.45724
0.30740
PR †(1976)
a
(T
;
w)
PR
��
� r
1/2 1/2
† α (T ;(Tω); =
2 )ω0.07780
21�
2
α†SRK
ω)
=
1
+
1
(0.480
+
(0.480
+
1.574
+
1.574
ω
−
ω
0.176
−
0.176
ω
)
−
1
T
−
T
��
�
�
1/2
r r
SRK
r r �� 2
2
�
�
αSRK (Tr ; ω) = 1 + (0.480 + 1.574 ω −20.176 ω12/ 2) 12 − Tr1/2
1/2 2
†α
(T
ω)
=
+ (0.480
(0.480
+ω1.574
1.574
ω−
−
0.176
ω2 )))11 −
−�TTr �
† a †α
��2��2
SRK
1�=
+
+
ω
(T
w)rr ;;=ω)
+�(0�11.480
+ 1.574
− 0.176
ω 0.176
) (1 − Tω
SRK
r; (T
r
�r
��
1/2 1/2

2
2
‡ α ‡SRK
α
(T
;
(T
ω)
;
=
ω)
=
1
+
1
(0.37464
+
(0.37464
+
1.54226
+
1.54226
ω
−
ω
0.26992
−
0.26992
ω
)
ω
1
)
−
1
T
−
T
�
�
��2
1/2
PR
PR
r
r
r
‡ α (T ; ω) = � 1 + (0.37464 + 1.54226 ω − 0.26992 ω2 ) � 1 − T r ��
1/2 22
2
PR
r
r
2
‡
2
1
/
2
1/2
‡ a ‡ (T
αPR
(T
ω)
=
+ (0.37464
(0.37464
+ω
1.54226
ω−
−ω0.26992
0.26992
ω2))) 11 −
− TTr
1=
PR
w)rr ;;=ω)
+ (011.37464
+ 1.5422
− 0.26992
) (1 − Trω
+
+
1.54226
ω
PR α
r; (T
r

Example
Example
3.9
Ejemplo
3.9 3.9
Example
3.9
Example
3.9
Example
3.9
Given
Given
thatthat
thethe
vapor
vapor
pressure
pressure
of n-butane
of n-butane
at 350
at 350
K isK 9.4573
is 9.4573
bar,bar,
findfind
thethe
molar
molar
Given that the vapor pressure of n-butane at 350 K is 9.4573 bar, find the molar
Teniendo en cuenta que la presión de vapor para el n-butano a 350 K es 9.4573 bar, encuentre los
Given
that
vapor
pressureand
ofand
n-butane
at 350
350 K
K n-butane
is 9.4573
9.4573
findconditions
the
molar
volumes
volumes
of (a)
ofthe
(a)
saturated-vapor
saturated-vapor
(b)
(b)
saturated-liquid
saturated-liquid
n-butane
atbar,
these
at these
conditions
Given
that
vapor
pressure
of
at
is
find
the
molar
volumes
of the
(a)
saturated-liquid
n-butane
at
conditions
volúmenes
molares
de saturated-vapor
a) vapor
saturadoand
y n-butane
b)(b)
líquido
saturado
de
n-butanobar,
enthese
estas
condiciones,
mevolumes
ofthe
(a)
saturated-vapor
and
(b)
saturated-liquid
n-butane
at
these
conditions
asvolumes
given
as given
byof
by
the
Redlich/Kwong
Redlich/Kwong
equation.
equation.
(a)
saturated-vapor
and (b) saturated-liquid n-butane at these conditions
asecuación
given byde
theRedlich/Kwong.
Redlich/Kwong equation.
diante la
as given
given by
by the
the Redlich/Kwong
Redlich/Kwong equation.
equation.
as
Solution
Solution
3.9
Solución
3.9 3.9
Solution
3.9
Solution
3.9
Solution
A partir
deValues
losofvalores
de
Pc para
elfrom
n-butano
del
apéndice
Values
Tof3.9
and
Tc and
PcTfor
n-butane
n-butane
from
App.
App.
B yield:
B
yield: B, se obtiene:
cPcy for
Values of cTc and
Pc for
n-butane from App. B yield:
Values
of
T
and
P
for
n-butane
from
App.
B
yield:
c
c
Values of Tc and
Pc for n-butane from App. B yield: 9.4573
350350
9.4573
350 = 0.8233
9.4573= 0.2491
Pr =
= 0.8233 andyand
Pr =9.4573
= 0.2491
Tr =
Tr = 350
= 0.8233
and
Pr =
= 0.2491
Tr =425.1
425.1
37.96
37.96=
350
9.4573
=
=
=
0.8233
and
P
0.2491
T
425.1
37.96
and
Prr = 37.96 = 0.2491
Trr = 425.1 = 0.8233
425.1
37.96
Parameter
Parameter
q
is
q
given
is
given
by
Eq.
by
Eq.
(3.54)
(3.54)
with
with
�,
�,
�,
and
�,
and
α(T
α(T
)
for
)
for
the
the
RK
RK
equation
from
from
r (Tr) para laequation
El parámetro
q se qconoce
por
ecuación
(3.54)
W, α(T
Ψr yr )a
ecuación
RK de la
Parameter
is given
bylaEq.
(3.54) with
�, con
�, and
for the RK equation
from
Parameter
q
is
given
by
Eq.
(3.54)
with
�,
�,
and
α(T
)
for
the
RK
equation
from
Table
Table
3.1:
3.1:
r
Parameter
q
is
given
by
Eq.
(3.54)
with
�,
�,
and
α(T
)
for
the
RK
equation
from
tabla 3.1:
r
Table 3.1:
Table 3.1:
3.1:
Table
−1/2−1/2
�Tr�T
� �−3/2−3/2 0.42748
0.42748
−1/2
r
−3/2−3/2
�Tr−1/2
0.42748
−1/2= =�
q =q =�T
= 6.6048
Tr T
(0.8233)
(0.8233)
−3/2
−3/2= 6.6048
r = =0.42748
�
q = �T
=
=
= 6.6048
T
(0.8233)−3/2
rr�Tr
−3/2 0.08664
�T
�
�
0.08664
r−3/2
�
0.42748
r
−3/2
q
=
=
=
=
6.6048
T
(0.8233)
�T
�
0.08664
r
q = �Tr = � Tr
= 0.08664 (0.8233)
= 6.6048
�Eq.
rr from
Parameter
Parameter
β isβfound
is �T
found
from
Eq.
(3.53):
(3.53):0.08664
Parameter
β
is found afrom
Eq.
(3.53):
El parámetro
b se β
determina
partir
de(3.53):
la ecuación (3.53):
Parameter
β
is found
found from
from
Eq.
(3.53):
Parameter
is
Pr Eq.
Pr (0.08664)(0.2491)
(0.08664)(0.2491)
P
(0.08664)(0.2491)
β =β�= �
=(0.08664)(0.2491)= 0.026214
= 0.026214
Prrr=
β = �TrP
=
= 0.026214
Tr= (0.08664)(0.2491)
0.8233
0.8233
β
=
�
=
0.026214
T
0.8233
r
β =�T =
= 0.026214
0.8233
Trrwrite
0.8233
(a) (a)
ForFor
the the
saturated
saturated
vapor,
vapor,
write
the the
RK
RK
form
form
of Eq.
of Eq.
(3.52)
(3.52)
which
which
results
results
upon
upon
(a) For the saturated vapor, write the RK form of Eq. (3.52) which results upon
(a)
For
the
saturated
vapor,
write
the
RK
form
of
Eq.
(3.52)
which
results
upon
substitution
substitution
of
appropriate
of
appropriate
values
values
for
for
�
and
�
and
σ
from
σ
from
Table
Table
3.1:
3.1:
a) Para (a)
el
vapor
saturado,
escribiendo
la forma
RKform
laofecuación
(3.52)
lo que
resulta
por sustiFor the
saturated
vapor,values
write
the �RK
Eq.
(3.52)
which
results
upon
substitution
of appropriate
for
and
σdefrom
Table
3.1:
substitution
of appropriate
appropriate
values
for
and 3.1:
σ from
from Table
Table 3.1:
3.1:
tución de
valores apropiados
para values
y s defor
la ��tabla
substitution
of
and
σ
(Z −
(Zβ)
− β)
(Z − β)
Z =Z 1=+1β+−βqβ
− qβ(Z
−
β)
Z = 1 + β − qβZ (Z
Z
+
(Z
β)
+ β)
= 11 +
+β
β−
− qβ
qβ Z(Z
(Z−+β)
β)
ZZ =
Z
(Z
+
β)
Z (Z + β)
03-SmithVanNess.indd 98
8/1/07 12:57:27
3.6.
3.6. Generalized
Generalized Correlations
Correlations for
for Gases
Gases
3.6. Correlaciones
generalizadas
para
3.6. Generalized
Correlations
forgases
Gases
99
99
99
99
Iteration
with
an
value
Z=
on
Z
Thus,
Iteration
with
an initial
initial
value
= 11 converges
converges
on
Z=
= 0.8305.
0.8305.
Thus,
La iteración
con with
un
valor
inicialvalue
Z = 1ZZconverge
en Z =on
0.8305.
Así,
Iteration
an initial
= 1 converges
Z = 0.8305.
Thus,
Z
v
Z RT
RT = (0.8305)(83.14)(350)
(0.8305)(83.14)(350) = 2,555 cm33 mol −1
V
V vv =
= Z RT
= (0.8305)(83.14)(350)
= 2,555 cm mol−1
P
9.4573
V = P =
= 2,555 cm3 mol−1
9.4573
P
9.4573
3
−1
−1..
An
An experimental
experimental value
value is
is 2,482
2,482 cm
cm33 mol
mol−1
An
experimental
value
is
2,482
cm
mol
.
3
–1
Un valor experimental es 2 482 cm mol .
(b)
For
apply
Eq.
in
its
RK
(b)líquido
For the
the saturated
saturated liquid,
liquid,
apply
Eq. (3.56)
(3.56)
in en
its su
RK form:
form:
b) Para (b)
el
aplicaapply
la ecuación
(3.56)
For thesaturado
saturatedseliquid,
Eq. (3.56)
in its
RKforma
form:RK:
�
�1 + β − Z �
�
� 1 + β − Z�
Z
=
β
+
Z
(Z
+
β)
1
+
β
−
Z
Z = β + Z (Z + β)
qβ
Z = β + Z (Z + β)
qβ
qβ
(1.026214
(1.026214 −
−Z
Z ))
o
or
Z
(1.026214 − Z )
or
Z=
= 0.026214
0.026214 +
+Z
Z (Z
(Z +
+ 0.026214)
0.026214) (6.6048)(0.026214)
or
Z = 0.026214 + Z (Z + 0.026214) (6.6048)(0.026214)
(6.6048)(0.026214)
initial
is
=
the
right
this
equation.
Iteration
La etapaThe
inicial
es step
la sustitución
de Z of
= bZ
el on
lado
deof
esta
iteración conThe
initial
step
is substitution
substitution
of
Z en
=β
β
on
thederecho
right side
side
of
thisecuación.
equation.La
Iteration
The
initial
step
is
substitution
of
Z
=
β
on
the
right
side
of
this
equation.
Iteration
leads
to
convergence
on
the
value
Z
=
0.04331.
Whence,
duce a laleads
convergencia
en el valor
= 0.04331.
De donde,
to convergence
on theZvalue
Z = 0.04331.
Whence,
leads to convergence on the value Z = 0.04331. Whence,
Z
l
Z RT
RT = (0.04331)(83.14)(350)
(0.04331)(83.14)(350) = 133.3 cm33 mol−1
V
Vll =
= Z RT
= (0.04331)(83.14)(350)
= 133.3 cm mol−1
9.4573
V = P
=
= 133.3 cm3 mol−1
P
9.4573
P
9.4573
3
−1
3
–1
An
experimental
value
is
115.0
cm
An
experimental
value
is
115.0
cm
mol−1..
Un valor experimental es 115.0 cm mol .33 mol
An experimental value is 115.0 cm mol−1 .
v
l
For
For comparison,
comparison, values
values of
of V
V vv and
and V
Vll calculated
calculated for
for the
the conditions
conditions of
of Ex.
Ex. 3.9
3.9 by
by all
all four
four
For
comparison,
values
of
V
and
V
calculated
for the conditions
of Ex. 3.9 by all four
v
l
of
the
cubic
equations
of
state
considered
here
are
summarized
as
follows:
Por
los valores
de considered
V y V calculados
las condiciones
del ejemplo 3.9 para las cuatro
of comparación,
the cubic equations
of state
here are con
summarized
as follows:
of the cubic equations vof state
considered
here are summarized as follows:
3
−1
ecuaciones de estado cúbicasV se
resumen
de la siguiente forma: V ll /cm 33 mol −1
−1
V vv/cm
/cm33 mol
mol−1
V /cm mol−1
V /cm mol
V l /cm3 mol−1
l RK
3
–1
Exp.
Exp.
V v/cm3RK
mol–1 SRK
Exp. vdW
vdW
RK
SRK PR
PR
Exp. vdW
vdWV /cm
RK molSRK
SRK PR
PR
Exp. vdW RK SRK PR
Exp. vdW RK SRK PR
Exp.
PR
Exp. 191.0
vdW 133.3
RK 127.8
SRK 112.6
PR
2,482
2,667
2,555
2,520
115.0
2,482vdW
2,667 RK
2,555 SRK
2,520 2,486
2,486
115.0
191.0
133.3
127.8
112.6
2,482 2,667 2,555 2,520 2,486 115.0 191.0 133.3 127.8 112.6
2 482 2 667 2 555 2the
520 2 486 115.0
191.0 133.3
127.8 112.6
The
The Soave/Redlich/Kwong
Soave/Redlich/Kwong and
and the Peng/Robinson
Peng/Robinson equations
equations were
were developed
developed specifically
specifically for
for
The
Soave/Redlich/Kwong
and the Peng/Robinson
equations were developed specifically for
vapor/liquid
equilibrium
calculations
(Sec.
14.2).
vapor/liquid equilibrium calculations (Sec. 14.2).
vapor/liquid
equilibrium
calculations
(Sec.
14.2). found
Las ecuaciones
de Soave/Redlich/Kwong
de
Peng/Robinson
fueron
forma such
específica
Roots
of
of
most
with
aa software
package
as
Roots
of equations
equations
of state
state yare
are
most easily
easily found
withdesarrolladas
software en
package
such
as para
�
�
R
R
Roots
ofvapor/líquido
equations
ofwhich
state iteration
are14.2).
mostis easily
foundpart
with
a software
package such
as
cálculosMathcad
de
equilibrio
(sección
R
R
�
�
or
Maple
,
in
an
integral
of
the
equation-solving
routine.
Mathcad�
or Maple�
in which iteration is an integral part of the equation-solving routine.
R or
R ,, in
Mathcad
which
iteration
isand
an integral
part
ofmás
the fácil
equation-solving
routine.
Las
raícesvalues
de lasMaple
ecuaciones
de estado
se encuentran
conparticular
un paquete
deof
software
Starting
or
bounds
may
be
appropriate
to
root
Starting
values
or
bounds
may
be required,
required,
and must
mustenbe
beforma
appropriate
to the
the
particular
root
of
R
�
Starting
values
or bounds
may
be
required,
and3.9
must
be appropriate
to the
particular
root
of
como Mathcad®
o
Maple®,
en
que
la
iteración
es
una
parte
integral
de
la
rutina
de
solución
de
la
ecuación.
R
�
interest.
A
Mathcad
program
for
solving
Ex.
is
given
in
App.
D.2.
interest. A Mathcad�
program for solving Ex. 3.9 is given in App. D.2.
R program
interest.
A de
Mathcad
forosolving
Ex. 3.9
given in App.
Se puede
requerir
los valores
iniciales
de frontera
másisapropiados
paraD.2.
la raíz que nos interese. Un programa Mathcad® para solucionar el ejemplo 3.9 está dado en el apéndice D.2.
3.6
3.6 GENERALIZED
GENERALIZED CORRELATIONS
CORRELATIONS FOR
FOR GASES
GASES
3.6 GENERALIZED CORRELATIONS FOR GASES
CORRELACIONES
GENERALIZADAS PARA GASES
Generalized correlations find widespread use. Most popular are correlations of the kind de-
3.6
Generalized
Generalized correlations
correlations find
find widespread
widespread use.
use. Most
Most popular
popular are
are correlations
correlations of
of the
the kind
kind dedeGeneralized
correlations
find widespread
use.
Most popular
are Z
correlations
of second
the kindvirial
developed
by
Pitzer
and
coworkers
for
the
compressibility
factor
and
for
the
veloped bygeneralizadas
Pitzer and coworkers
for
the
compressibility
factor
Z and for son
the las
second
virial
Las correlaciones
tienen
un
uso
muy
extendido.
Las
más
populares
desarrolladas
por
16
veloped
by B.
Pitzer
and coworkers for the compressibility factor Z and for the second virial
16
coefficient
16
coefficient
B.16
Pitzer ycoefficient
colaboradores
para
el
factor
de
compresibilidad
Z
y
para
el
segundo
coeficiente
virial
B.
B.
16
16
16See
See Pitzer,
Pitzer, op.
op. cit.
cit.
16 See
Pitzer, op. cit.
Véase Pitzer, op. cit.
03-SmithVanNess.indd 99
8/1/07 12:57:46
100
CAPÍTULO
3. Volumetric
Propiedades
volumétricas
de Fluids
fluidos puros
CHAPTER 3.
Properties
of Pure
100
1.2
1.2
4.0
4.0
1.0
1.0
0.8
0.8
Z 0 0.6
Z 0 0.6
0.4
0.4
Tr  0.7
Tr � 0.7
Gases
Gases
0.9
0.9
1.5
1.5
1.2
1.2
1.0
1.0
Región de
Two-phase
dos
fases
region
0.7
0.7
C
C
0.2
0.2
0
0 0.05
0.05
0.9
0.9
Líquidos comprimidos
Compressed
(Tr  1.0) liquids
(Tr � 1.0)
0.1
0.1
0.2
0.2
0.5
0.5
Pr
1.0
1.0
2.0
2.0
5.0
5.0
10.0
10.0
Pr
Figura 3.13: La correlación de Lee/Kesler para Z 0 = F 0 (Tr, Pr).
Figure 3.13: The Lee/Kesler correlation for Z 0 = F 0 (Tr , Pr ).
Correlaciones de Pitzer para el factor de compresibilidad
Pitzer Correlations for the Compressibility Factor
The correlation
for la
Z is:
La correlación
para Z toma
forma:
Z = Z 0 + ωZ 1
(3.57) (3.57)
0 y Z 1 son
Z 0 and
Z 1 arede
functions
ofr. both
Tr and
Whenesωel=caso
0, asdeislos
thefluidos
case for
the simple
funciones
Tr y de P
Cuando
w =P0,r . como
simples,
el segundo
donde Zwhere
0 becomes identical with Z . Thus a generalized corre0
fluids,
the
second
term
disappears,
and
Z
término desaparece, y Z es idéntico a Z. Así, una correlación generalizada para Z como función de Tr y Pr,
0 = F 0(T , P ).
forenZinformación
as a functionpara
of Treland
Pr based
on data
just argon,
krypton,laand
xenonZprovides
que estálation
basada
argón,
el kriptón
y elforxenón,
proporciona
relación
r r
0 = F 0 (T , P ). By itself, this represents a two-parameter correspondingthe
relationship
Z
r
r
Por sí misma, ésta representa una correlación
de estados correspondientes con dos parámetros para Z. Ya que
states
correlation
for Z . Because
second
term of relativamente
Eq. (3.57) is apequeña
relativelypara
small
correction
el segundo
término
de la ecuación
(3.57) the
es una
corrección
esta
correlación, su
0
0
to
this
correlation,
its
omission
does
not
introduce
large
errors,
and
a
correlation
for
Z
be
omisión no introduce grandes errores y es posible usar una correlación para Z y así obtenermay
estimaciones
used
alone
for
quick
but
less
accurate
estimates
of
Z
than
are
obtained
from
a
three-parameter
rápidas de Z, pero menos precisas que las que se consiguen a partir de una correlación de tres parámetros.
Lacorrelation.
ecuación (3.57) es una relación lineal simple entre Z y w para valores conocidos de Tr y de Pr. La
and
Equation (3.57)
a simple
linear
and
for given values
of Tr de
información experimental
paraisZ,
de fluidos
querelation
no son between
simples, Z
que
sonω graficados
en función
w a Tr y
.
Experimental
data
for
Z
for
nonsimple
fluids
plotted
vs.
ω
at
constant
T
and
P
P
r
r
r do
Pr constantes,
producen líneas aproximadamente rectas y sus pendientes proporcionan valores
para
Z 1 a par1
1 = F 1provide
approximately
straight
lines,
and their Z
slopes
tir de losindeed
cualesyield
es posible
construir la
función
generalizada
(Tr, Pr).values for Z from which
1
1
function
Z = del
F (T
bedesarrollada
constructed.por Lee y Kesler17 es la más favorecida.
r, P
r ) canla
Dethe
lasgeneralized
correlaciones
disponibles
tipo
Pitzer,
17 has
Of
the
Pitzer-type
correlations
available,
the
onededeveloped
by de
Lee
and Kesler
Aunque su desarrollo se fundamenta en una forma modificada
la ecuación
estado
de Benedict/Webb/
17
17 B. I. Lee and M. G. Kesler, AIChE J., vol. 21, pp. 510–527, 1975.
B. I. Lee
y M. G. Kesler, AIChE J., vol. 21, pp. 510-527, 1975.
03-SmithVanNess.indd 100
8/1/07 12:57:54
3.6. Generalized
Generalized Correlations
Correlations for
for Gases
Gases
101
3.6.
101
3.6.
Correlations
for
Gases
101
101 101
3.6. Generalized
Generalized
Correlations
forgases
Gases
3.6. Correlaciones
generalizadas
para
3.6. Generalized Correlations for Gases
101
found
greatest
favor.
Although
its
development
is
based
on
a
modified
form
of
the
Bene0
1
found
greatest
favor.
Although
its
development
islos
based
on
aamodified
modified
form
of
the
BeneRubin,found
ésta
adquiere
formaAlthough
de las tablas
que
presentanis
valores
Z y Z form
comoof
funciones
de T y de
greatest
its
on
ade
0Benefound
greatestlafavor.
favor.
Although
itsitdevelopment
development
onwhich
modified
ofthe
theZBenedict/Webb/Rubin
equation
of state,
state,
takes the
the form
formisbased
ofbased
tables
presentform
values
of
and r
000and
dict/Webb/Rubin
equation
of
it
takes
of
tables
which
present
values
of
Z
and
Pr, quedict/Webb/Rubin
se
proporcionan
en
el
apéndice
E
de
las
tablas
E.1
a
E.4.
El
uso
de
estas
tablas
con
frecuencia
requieequation
of
state,
it
takes
the
form
of
tables
which
present
values
of
Z
1 as functions
state,
takes
the form
tables
which
present
values
ZBeneand
found
greatest favor.
Although
itsitdevelopment
is of
based
a modified
form
of
theUse
Zdict/Webb/Rubin
ofequation
Tr and
and
Pof
These
are
given
in App.
App.
E as
asonTables
Tables
E.1
through
E.4.of
of
r .. These
111as
as
functions
of
T
P
are
given
in
E
E.1
through
E.4.
Use
of
Z
0
r
r
re de interpolación,
la
cual
se
trata
al
inicio
del
apéndice
F.
La
naturaleza
de
la
correlación
se
indica
en
Z
functions
of
T
and
P
.
These
are
given
in
App.
E
as
Tables
E.1
through
E.4.
Use
of
functions
Tr r andinterpolation,
P
These
are
given
in
App.
Ethe
as beginning
Tables
E.1ofthrough
Useand
of la fiZ astables
r r .state,
dict/Webb/Rubin
equation
of
it takes
theisform
of at
tables
which present
values
of Z
these
often of
requires
which
treated
App.
F.E.4.
The
nature
0 en función which
these
tables
often
requires
interpolation,
is
treated
at
the
beginning
of
App.
F.
The
nature
1
gura 3.13,
que
es
una
gráfica
de
Z
de
P
para
seis
isotermas.
these
tables
often
requires
interpolation,
which
is
treated
at
the
beginning
of
App.
F.
The
nature
these
tables
oftenof
isin
treated
at
the
beginning
App. F.E.4.
TheUse
nature
Z the
ascorrelation
functions
Tr andinterpolation,
Prby
. These
arewhich
given
App.
EP
asr for
Tables
E.1 of
through
of
of
isrequires
indicated
Fig. 3.13,
3.13,
a rplot
plot
of
Z0000vs.
vs.
six isotherms.
isotherms.
of
the
correlation
is
indicated
by
Fig.
of
P
for
six
La
correlación
de Lee/Kesler
proporciona
resultados
para
gases
que
sonF.noThe
polares
o ligeraof
the
correlation
is
indicated
by
Fig.
3.13,
aaaplot
of
ZZconfiables
vs.
P
six
isotherms.
of
theThe
correlation
is
indicated
by
Fig.
3.13,
plot
of
Z
vs.
P
for
six
isotherms.
rrrfor
these
tables
often
requires
interpolation,
which
is
treated
at
the
beginning
of
App.
nature
Lee/Kesler correlation
correlation provides
provides reliable
reliable results
results
for gases
gases which
which are
are nonpolar
nonpolar or
or only
only
The
Lee/Kesler
for
0
mente polares;
para
éstos,
los
errores
indicados
no
son
mayores
de
2
o
3%.
Cuando
se
aplica
a
gases
muy
Lee/Kesler
correlation
results
for
which
are
ororonly
of theThe
correlation
is indicated
Fig.
3.13,reliable
a plot
ofresults
vs.
for indicated.
six
isotherms.
The
Lee/Kesler
correlation
provides
reliable
forPgases
which
arenonpolar
nonpolar
only
rgases
slightly
polar;
for these,
these,
errorsby
ofprovides
no
more
than
2 or
or
3Z percent
percent
are
When
applied
to
slightly
polar;
for
errors
of
no
more
than
2
3
are
indicated.
When
applied
to
polaresslightly
o
asociados,
se
pueden
esperar
errores
más
grandes.
polar;
for
errors
ofofno
22oror
33percent
are
indicated.
applied
toto
slightly
polar;
forthese,
these,
errors
nomore
morethan
than
percent
are
indicated.
When
applied
The
Lee/Kesler
provides
reliable
results
for be
gases
which areWhen
nonpolar
or only
highly
polar
gases
or
tocorrelation
gases
that
associate,
larger
errors
can
expected.
highly
polar
gases
or
to
gases
that
associate,
larger
can
be
expected.
Los
gases
cuánticos
(por
ejemplo,
hidrógeno,
helio
yerrors
neón)
nobe
se
ajustan
al mismo
comportamiento
de
highly
polar
gases
or
to
gases
that
associate,
larger
errors
can
expected.
highly
polar
gases
or
to
gases
that
associate,
larger
errors
can
be
expected.
slightly
polar;
for
these,
errors
of
no
more
than
2
or
3
percent
are
indicated.
When
applied
The quantum
quantum gases
gases (e.g.,
(e.g., hydrogen,
hydrogen, helium,
helium, and
and neon)
neon) do
do not
not conform
conform to
to the
the same
sameto
The
estados highly
correspondientes
como
los
fluidos
normales.
Algunas
veces
su
análisis
mediante
las
correlaciones
The
quantum
gases
(e.g.,
hydrogen,
helium,
and
neon)
not
conform
totothe
polar
gases or
to gases
that
associate,
largerTheir
errors
can
bedo
expected.
The
quantum
gases
(e.g.,
hydrogen,
helium,
andtreatment
neon)
do
not
conform
thesame
same
corresponding-states
behavior
as
do
normal fluids.
fluids.
by
the
usual correlations
correlations
is
corresponding-states
behavior
as
do
normal
Their
treatment
by
the
usual
is
acostumbradas
se adapta
a través
del
uso
denormal
parámetros
críticos
eficaces
que
son
dependientes
desame
lais
temperacorresponding-states
behavior
as
do
fluids.
Their
treatment
by
the
usual
correlations
18
corresponding-states
behavior
as
do
normal
fluids.
Their
treatment
by
the
usual
correlations
is
The
quantum
gases
(e.g.,
hydrogen,
helium,
and
neon)
do
not
conform
to
the
For
sometimes
accommodated
by
use
of
temperature-dependent
effective
critical
parameters.
18
18
18
For
sometimes
accommodated
by
use
of
temperature-dependent
effective
critical
parameters.
18
Para
el
hidrógeno,
el
gas
cuántico
más
encontrado
en
los
procesos
químicos,
las
ecuaciones
recomentura. sometimes
For
accommodated
use
effective
critical
parameters.
For
sometimes
accommodated
by
use
oftemperature-dependent
temperature-dependent
effective
critical
parameters.
corresponding-states
asof
do
normal fluids.
Their
treatment
by
the usual
correlations
is
hydrogen,
the
quantumbehavior
gasby
most
commonly
found in
in
chemical
processing,
the
recommended
hydrogen,
the
quantum
gas
most
commonly
found
chemical
processing,
the
recommended
18 For
dadas son:
hydrogen,
the
quantum
gas
most
commonly
found
in
chemical
processing,
the
recommended
hydrogen,
the
quantum
gas
most
commonly
found
in
chemical
processing,
the
recommended
sometimes
accommodated
by
use
of
temperature-dependent
effective
critical
parameters.
equations are:
are:
equations
equations
equations
are:
hydrogen,are:
the quantum gas most commonly
found in chemical processing, the recommended
43.6
43.6
(para
el2H
(for H
H
/K =
=
) 2)
(3.58) (3.58)
Tc /K
43.6
43.6
equations are:
(for
(3.58)
T
21.8
(for
HH222)))
(3.58)
TTcc/K
(3.58)
(for
21.8
c /K== 1 + 43.6
21.8
21.8
++ 2.016
T
(for H2 )
(3.58)
Tc /K =111+
2.016
T
21.8TT
2.016
2.016
1+
2.016 T
20.5
20.5
Pc /bar
(for H
H
/bar =
=
) )
(3.59) (3.59)
20.5
20.5
el2H
P
(3.59)
(for
c
44.2 (para
PPc /bar
(3.59)
(for
(3.59)
(forHH222)))2
44.2
c /bar== 1 + 20.5
44.2
44.2
++ 2.016
T
Pc /bar =111+
)
(3.59)
(for
H
2
2.016
44.2TTT
2.016
2.016
1+
2.016
51.5T
−1 =
51.5
Vc /cm
/cm3333mol
mol−1
(3.60)
(for H
H2 ))
51.5
−1 =
51.5
V
(3.60)
(for
9.91 (para
VVcc/cm
) )2)
(3.60)
(for
elH22H
mol−1== 1 − 9.91
(3.60) (3.60)
(forH
9.91
c /cm mol
2
51.5
9.91
3
1
−
−1
2.016
T
1
−
Vc /cm mol = 1 −2.016 T
(3.60)
(for H2 )
9.91TT
2.016
2.016
1Use
− of these effective critical parameters for hywhere
T
is
absolute
temperature
in
kelvins.
where
isisabsolute
absolute
temperature
ininkelvins.
kelvins.
Use of
of
these
effective
critical
parameters
for
hy2.016
T effective
where
temperature
in
whereTTrequires
Tis
absolute
temperature
kelvins.
ofthese
these
effectivecritical
criticalparameters
parametersfor
forhyhydrogen
the further
further
specification
that Use
ωUse
= 0.
0.
drogen
requires
the
specification
that
ω
=
donde drogen
Tdrogen
es la requires
temperatura
absoluta
en kelvins.
El
uso
de
parámetros
paraforelhyhidrógeno
the
specification
that
ωω
=
thefurther
further
specification
that
=0.estos
0. these
where
Trequires
is absolute
temperature
in kelvins.
Use
of
effective críticos
critical eficaces
parameters
requieredrogen
la especificación
de que w = 0.
requires theadicional
further specification
that ω = 0.
Pitzer Correlations
Correlations for
for the
the Second
Second Virial
Virial Coefficient
Coefficient
Pitzer
Pitzer
PitzerCorrelations
Correlationsfor
forthe
theSecond
SecondVirial
VirialCoefficient
Coefficient
Correlaciones
de Pitzer
para
el compressibility-factor
segundo
coeficiente
virial
The
tabular
nature
of the
the generalized
generalized
compressibility-factor
correlation
is aa disadvantage,
disadvantage, but
but
Pitzer
Correlations
for the
Second Virial
Coefficient
The
tabular
nature
of
correlation
is
The
tabular
ofofthe
compressibility-factor
correlation
isisaadisadvantage,
but
Thecomplexity
tabularnature
nature
thegeneralized
generalized
compressibility-factor
correlation
disadvantage,
but
the
of the
the
functions
Z0000 and
and
Z1111 precludes
precludes their
their accurate
accurate
representation
by simple
simple
Z
representation
by
the
complexity
of
functions
Z
and
Z
precludes
their
accurate
representation
by
simple
the
complexity
of
the
functions
Z
and
Z
precludes
their
accurate
representation
by
simple
the
complexity
of
the
functions
Z
The
tabular
nature
of
the
generalized
compressibility-factor
correlation
is
a
disadvantage,
equations.
However,
we can
can give
give
approximate
analytical
expression
to these
these
functions
forbut
a pero la
La naturaleza
tabular
de la correlación
generalizada
delanalytical
factor deexpression
compresibilidad
es functions
una desventaja,
equations.
However,
we
approximate
to
for
0approximate
1 precludes
equations.
However,
we
can
give
approximate
analytical
expression
to
these
functions
for
equations.
However,
we
can
give
analytical
expression
to
these
functions
foraaa
0
1
and
Z
their
accurate
representation
by
simple
the
complexity
of
the
functions
Z
limited
range
of
pressures.
The
basis
for
this
is
Eq.
(3.38),
the
simplest
form
of
the
virial
complejidad
derange
las funciones
Z y The
Z impide
su this
representación
exacta
medianteform
ecuaciones
sencillas.
Sin
limited
of
pressures.
basis
for
is
Eq.
(3.38),
the
simplest
of
the
virial
limited
range
ofofpressures.
basis
for
this
Eq.
(3.38),
the
form
ofofthe
limited
range
pressures.
The
basis
for
thisisis
Eq.de
(3.38),
thesimplest
simplest
form
thevirial
virial
equations.
However,
we canThe
give
approximate
analytical
expression
topara
these
for
a
equation:
embargo,
es posible
dar
una
expresión
analítica
aproximada
estas
funciones
unfunctions
intervalo
limitado
de
equation:
equation:
equation:
limited
range
ofesto
pressures.
The basis
for Bla
this
is Eq.
(3.38),
the
simplest
form
of the virial
Pforma
P
presiones.
La base
para
es la ecuación
(3.38),
más
simple
de
la
ecuación
virial:
r
B
P
P
Z=
= 11 +
+ BBPP =
(3.61)
=1+
+ B̂
B̂ PP
rr
equation:
(3.61)
RT ==111+
Tr r
ZZZ=
(3.61)
=11++ RT
(3.61)
+B̂B̂T
r
B
P
P
RT
TTr
RT
Z =1+
(3.61) (3.61)
= 1 + B̂ r r
RT
Tr
where B̂
B̂ is
is a reduced
reduced second
second virial
virial coefficient,
coefficient, given
given by:
by:
where
isisaaareduced
second
virial
coefficient,
given
reduced
secondvirial
virialreducido,
coefficient,
givenby:
by:
donde where
B̂where
es unB̂B̂segundo
coeficiente
conocido
por:
where B̂ is a reduced second virial coefficient, B
given
by:
B
P
c
PPcc
(3.62)
B̂ =
= BBP
(3.62)
B̂
RTcc
(3.62)
B̂
(3.62) (3.62)
B̂== RT
c
B
P
RT
c
RT
(3.62)
B̂ = c c
RTc
18 J. M. Prausnitz, R. N. Lichtenthaler, and E. G. de Azevedo, Molecular Thermodynamics of Fluid-Phase Equilib18 J.
M. Prausnitz,
Prausnitz, R.
R. N.
N. Lichtenthaler,
Lichtenthaler, and
and E.
E. G.
G. de Azevedo,
Azevedo, Molecular
Molecular Thermodynamics
Thermodynamics of Fluid-Phase EquilibEquilib18
18
J.
M. Prausnitz,
R. N.
Lichtenthaler,
andUpper
E. G.de
de Azevedo,
Fluid-Phase Equilibria,
3dJ.M.
ed.,
pp. 172–173,
172–173,
Prentice
Hall PTR,
PTR,
Saddle
River, Molecular
NJ, 1999.
1999. ThermodynamicsofofFluid-Phase
ria,
3d
ed.,
pp.
Prentice
Hall
Upper
Saddle
River,
NJ,
J. ria,
M. Prausnitz,
R.
N. Lichtenthaler
y E.PTR,
G. deUpper
Azevedo,
Molecular
Thermodynamics
of Fluid-Phase Equilibria, 3a. edición, pp.
3d
ed.,
pp.
172–173,
Prentice
Hall
Saddle
River,
NJ,
1999.
18
ria,
3d
ed.,
pp.
172–173,
Prentice
Hall
PTR,
Upper
Saddle
River,
NJ,
1999.
J. M. Prausnitz,
R. N. Lichtenthaler,
and E. G. de Azevedo, Molecular Thermodynamics of Fluid-Phase Equilib172-173, Prentice-Hall
PTR, Upper
Saddle River, 1999.
ria, 3d ed., pp. 172–173, Prentice Hall PTR, Upper Saddle River, NJ, 1999.
18
03-SmithVanNess.indd 101
8/1/07 12:58:17
102
CHAPTER 3.
3. Volumetric
Volumetric
Properties
of Pure
Pure
Fluids
CHAPTER
Properties
of
CAPÍTULO
Propiedades
volumétricas
deFluids
fluidos puros
CHAPTER 3.3.Volumetric
Properties
of Pure
Fluids
102
102
102
Thus,
Pitzer and
and coworkers
coworkers
proposed
second correlation,
correlation, which
which
yields
values
for B̂:
B̂:
Thus,
Pitzer
proposed
second
values
for
Así, Pitzer
y colaboradores
recomiendan
unaaasegunda
correlación,
la cualyields
produce
valores
para B̂:
Thus, Pitzer and coworkers proposed a second correlation, which yields values for B̂:
+ ωB
ωB111
= BB000 +
B̂B̂ =
B̂ = B 0 + ωB 1
(3.63) (3.63)
(3.63)
(3.63)
Equations
(3.61)
and (3.63)
(3.63)
together
become:
(3.61)
and
together
become:
Al unir Equations
las ecuaciones
(3.61)
y (3.63)
se obtiene:
Equations
(3.61)
and (3.63)
together
become:
PPr
PPr
+ ωB
ωB111 Prrr
= 11 +
+ BB000 Prrr +
ZZ =
Z = 1 + B 0 TTrrr + ωB 1 TTrrr
Tr
Tr
Comparison
of this
this equation
equation
withse
Eq.
(3.57) provides
provides
the following
following
identifications:
Comparison
of
with
Eq.
(3.57)
the
identifications:
Al comparar
esta ecuación
con
la (3.57)
obtienen
las identificaciones
siguientes:
Comparison of this equation with Eq. (3.57) provides the following identifications:
PPr
= 11 +
+ BB000 Prrr
(3.64) (3.64)
(3.64)
ZZ0000 =
(3.64)
Z = 1 + B 0 TTrrr
Tr
PPr
and
= BB1111 Prrr
and
ZZ1111 =
and
Z = B TTrrr
y
Tr
Second virial
virial coefficients
coefficients are
are functions
functions of
of temperature
temperature only,
only, and
and similarly
similarly BB0000 and
and BB1111 are
are funcfuncSecond
Second
virial
coefficients
are
functions
of
temperature
only,
and
similarly
B
andthe
B following
are funcLos segundos
viriales son
funciones
sólo
de la temperatura,
y de manera
similar
B0 y B1 son
tions of
ofcoeficientes
reduced temperature
temperature
only.
They are
are reasonably
reasonably
well represented
represented
by
tions
reduced
only.
They
well
by
the
following
tions
of
reduced
temperature
only.
They
are
reasonably
well
represented
by
the
following
19
funciones
sólo de19
la temperatura reducida. Se representan bien mediante las ecuaciones siguientes:19
19
equations:
equations:
equations:19
0.422
0.422
= 0.083
0.083 −
− 0.422
BB0000 =
1.6
1.6
B = 0.083 − TTrr1.6
Trr1.6
(3.65)
(3.65)
(3.65)
0.172
0.172
= 0.139
0.139 −
− 0.172
BB1111 =
4.2
4.2
B = 0.139 − TTrr4.2
Trr4.2
(3.66)
(3.66)
(3.66)
The simplest
simplest form
form of
of the
the virial
virial equation
equation has
has validity
validity only
only at
at low
low to
to moderate
moderate pressures
pressures
The
La forma
más
simple form
de la of
ecuación
virial
tiene validez
sólo aonly
presiones
a moderadas,
donde Z
theThe
virial
equation
has validity
at lowdetobajas
moderate
pressures
whereThe
issimplest
linear in
in pressure.
pressure.
generalized
virial-coefficient
correlation
is therefore
therefore
useful
where
ZZ is
linear
The generalized
virial-coefficient
correlation
is
useful
es una función
lineal
de
la
presión.
Por
lo
tanto,
la
correlación
generalizada
del
coeficiente
virial
es
útil
sólo
where
Z
is
linear
in
pressure.
The
generalized
virial-coefficient
correlation
is
therefore
useful
0
1
0
1
only
where
and ZZ 11 are
are at
at least
least approximately
approximately linear
linear functions
functions of
of reduced
reduced pressure.
pressure. FigFigonly
ZZ 00 and
0 y Zwhere
1 son al
donde Zonly
menos
aproximadamente
funciones
lineales
de
la
presión
reducida.
En
la
figura
3.14
se
where
Z
and
Z
are
at
least
approximately
linear
functions
of
reduced
pressure.
Fig0
0
ure 3.14
3.14 compares
compares the
the linear
linear
relation of
of ZZ 00 to
to PPrrr as
as given
given by
by Eqs.
Eqs. (3.64)
(3.64) and
and (3.65)
(3.65) with
with values
values
ure
relation
0 en función
comparaure
la 3.14
relación
lineal
de
Z
de
P
como
se
conocen
por
las
ecuaciones
(3.64)
y
(3.65)
con
los
compares
the
linear
relation
of
Z
to
P
as
given
by
Eqs.
(3.64)
and
(3.65)
with
values
r
0
r
0
from the
the Lee/Kesler
Lee/Kesler compressibility-factor
compressibility-factor correlation,
correlation, Tables
Tables E.1
E.1 and
and E.3.
E.3. The
The two
two corcorof ZZ 000from
of
valores of
de ZZ from
de la the
correlación delcompressibility-factor
factor de compresibilidad
de Lee/Kesler,
tablas
E.1
y E.3.
Las dos
correlation,
Tables
E.1
and
E.3.
The
cor-correlarelations
differ Lee/Kesler
by less
less than
than 2%
2% in
in the
the region
region above
above
the dashed
dashed
line of
of
the
figure.
For two
reduced
relations
differ
by
the
line
the
figure.
For
reduced
ciones difieren
endiffer
menos
de
2%than
en la
región
que
está above
arriba the
de ladashed
línea punteada
defigure.
la figura.
Para
temperaturelations
by
less
2%
in
the
region
line
of
the
For
reduced
temperatures greater
greater than
than TTrrr ≈
≈ 3,
3, there
there appears
appears to
to be
be no
no limitation
limitation on
on the
the pressure.
pressure. For
For lower
lower
temperatures
ras reducidas
mayoresgreater
que Trthan
≈ 3,Tparece
ser
queappears
no haytolímite
enlimitation
la presión.
Para
valores menores
que Tr, el
temperatures
≈
3,
there
be
no
on
the
pressure.
For
lower
r
the allowable
allowable pressure
pressure range
range decreases
decreases with
with decreasing
decreasing temperature.
temperature. A
A point
point is
is
values of
of TTrrr the
values
intervalo
de presión
permisible
disminuye
conforme
se reduce
ladecreasing
temperatura.
Sin embargo,
se alcanza
un
the
allowable
pressure
range
decreases
with
temperature.
A
point
is
values
of
T
20
r
20
20
reached, however,
however, at
at TTrrr ≈
≈ 0.7
0.7 where
where the
the pressure
pressure range
range is
is limited
limited by
by the
the saturation
saturation pressure.
pressure.
reached,
20 Esto 20
punto enreached,
Tr ≈ 0.7,
donde el
intervalo
de
presión
está limitado
por
la presión
de saturation
saturación.pressure.
se
indica
however,
at
T
≈
0.7
where
the
pressure
range
is
limited
by
the
1
r
1
This is
is indicated
indicated by
by the
the left-most
left-most segment
segment of
of the
the dashed
dashed line.
line. The
The minor
minor contributions
contributions of
of ZZ 11 to
to
This
por el segmento
que estáby
enthe
el extremo
izquierdo
de the
la línea.
Seline.
ignoran
las
contribuciones
más
de
Thiscorrelations
is indicated
left-most
segment
of
dashed
The
minor
contributions
of Zpequeñas
to
the
are
here
neglected.
In
view
of
the
uncertainty
associated
with
any
generalized
the
correlations
are
here
neglected.
In
view
of
the
uncertainty
associated
with
any
generalized
Z 1 a lasthe
correlaciones.
En
vista
de
la incertidumbre
asociada
con
cualquier
correlación
generalizada,
las
descorrelations
are
here
neglected.
In
view
of
the
uncertainty
associated
with
any
generalized
0
0
correlation,
deviations of
of no
no more
more than
than 2%
2% in
in ZZ 0 are
are not
not significant.
significant.
correlation,
deviations
viaciones
en Z 0 no mayores
irrelevantes.
correlation,
deviationsdeof2%
noson
more
than 2% in Z 0 are not significant.
The relative
relative simplicity
simplicity of
of the
the generalized
generalized second-virial-coefficient
second-virial-coefficient correlation
correlation does
does much
much
The
La simplicidad
relativa
de la correlación
generalizada
del coeficiente virial
la hace muy
recomendable.
The relative
the generalized
second-virial-coefficient
correlation
does
much
to recommend
recommend
it. simplicity
Moreover,oftemperatures
temperatures
and
pressures of
of many
many chemical-processing
chemical-processing
opto
it.
Moreover,
and
pressures
opPor otratoparte,
las temperaturas
y presiones
de operación
de
varios procesos
químicos
se encuentranopdentro de
recommend
it.
Moreover,
temperatures
and
pressures
of
many
chemical-processing
erations lie
lie within
within the
the region
region where
where itit does
does not
not deviate
deviate by
by aa significant
significant amount
amount from
from the
the
erations
la regiónerations
donde no
sewithin
desvían
enregion
una cantidad
significativa
de la correlación
del factor
de compresibilidad.
Al
lie
the
where
it
does
not
deviate
by
a
significant
amount
from
the
compressibility-factor correlation.
correlation. Like
Like the
the parent
parent correlation,
correlation, itit is
is most
most accurate
accurate for
for nonpolar
nonpolar
compressibility-factor
igual que
la correlación generadora,
ésta esLike
másthe
exacta
para
especies no
polares,
y menosfor
para
moléculas alcompressibility-factor
correlation.
parent
correlation,
it
is
most
accurate
nonpolar
species and
and least
least accurate
accurate for
for highly
highly polar
polar and
and associating
associating molecules.
molecules.
species
tamentespecies
polaresand
y asociadas.
least accurate for highly polar and associating molecules.
19
19
19These
These
correlations
first
appeared
in
1975
in
the
third
edition
of
this
book,
attributed
as
personal
communication
Thesecorrelations
correlationsfirst
firstappeared
appearedin
in1975
1975in
inthe
thethird
thirdedition
editionof
ofthis
thisbook,
book,attributed
attributedas
asaaapersonal
personalcommunication
communication
19 These
correlations
first appeared
in 1975 in the third edition of this book, attributed as a personal communication
to
M.
Abbott,
who
developed
them.
to
M.
M.
who
them.
to M.
M.
M.Abbott,
Abbott,aparecieron
whodeveloped
developed
them.
19 Estas
correlaciones
por
vez
primera
en
1975
en
la
tercera
edición
de
este
libro,
como
una
comunicación
personal de
to M. M. Abbott, who developed them.
M. M. Abbott,
quien
las
desarrolló.
20
20
20Although
Although
the
Lee/Kesler
tables
of
Appendix
list
values
for
superheated
vapor
and
subcooled
liquid,
they
do
not
Althoughthe
theLee/Kesler
Lee/Keslertables
tablesof
ofAppendix
AppendixEEElist
listvalues
valuesfor
forsuperheated
superheatedvapor
vaporand
andsubcooled
subcooledliquid,
liquid,they
theydo
donot
not
20 Although
20 Aunque
Lee/Kesler
tables
of Appendix
E listvalores
values para
for superheated
vapor and subcooled
liquid,
they do not
lasvalues
tablasthe
de
Lee/Kesler
del apéndice
E exhiben
el vapor sobrecalentado
y el líquido
subenfriado,
no proporprovide
values
at
saturation
conditions.
provide
at
conditions.
provide
values
atsaturation
saturation
conditions.
provide
at saturation
conditions.
cionan valores
en values
condiciones
de saturación.
03-SmithVanNess.indd 102
8/1/07 12:58:34
3.6. Correlaciones
generalizadas
para
3.6. Generalized
Generalized
Correlations
forgases
Gases
3.6.
Correlations
for
Gases
103
103
4.0
4.0
4.0
4.0
1.0
1.0
1.0
1.0
Z0
ZZ000
Z
2.4
2.4
2.4
2.4
1.8
1.8
1.8
1.8
0.9
0.9
0.9
0.9
0.8
0.8
0.8
0.8
103
T  0.8
TTrr �
� 0.8
0.8 0.9
T
rr � 0.8
0.9
0.9
0.9
1.0
1.0
1.0
1.0
1.1
1.1
1.1
1.1
1.5
1.5
1.5
1.5
1.3
1.3
1.3
1.3
0.7
0.7
0.70.0
0.7
0.0
0.0
0.0
0.5
0.5
0.5
0.5
1.0
1.0
1.0
1.0
1.5
1.5
1.5
1.5
Pr
P
Prrr
P
2.0
2.0
2.0
2.0
2.5
2.5
2.5
2.5
0 correlación
Figura 3.14:
Comparación
de correlaciones
para for
Z 0.ZLa
del coeficiente
virial is
se represented
representa por
medio de líFigure
3.14:
Comparison
of
correlations
The
virial-coefficient
correlation
by
the
Figure 3.14:
3.14: Comparison
Comparison of
of correlations
correlations for
for Z
Z00... The
The virial-coefficient
virial-coefficient correlation
correlation is
is represented
represented by
by the
the
Figure
neas rectas;
la correlación
de Lee/Kesler
mediante
los
puntos.
En
la región
por
encima
de la
líneathe
discontinua,
las dos
straight
lines;
the
Lee/Kesler
correlation,
by
the
points.
In
the
region
above
the
dashed
line
two
straight lines;
lines; the
the Lee/Kesler correlation,
correlation, by
by the
the points.
points. In
In the
the region
region above
above the
the dashed
dashed line
line the
the two
two
straight
correlaciones
difierendiffer
por Lee/Kesler
menos
de
2%.
correlations
by
less
than
2%.
correlations differ
differ by
by less
less than
than 2%.
2%.
correlations
Correlaciones
parafor
elthe
tercer
coeficiente
virial
Correlations
for
the
Third
Virial Coefficient
Coefficient
Correlations
Third
Virial
Accurate precisa
data for
forpara
third
virial
coefficients
are viriales
far less
less es
common
than
for
second
virial
coefAccurate
data
third
coefficients
are
far
common
second
coefLa información
losvirial
terceros
coeficientes
bastantethan
másfor
escasa
quevirial
para los
segundos
ficients.
Nevertheless,
generalized
correlations
for
third
virial
coefficients
do
appear
in
the
ficients.
Nevertheless,
generalized
for correlaciones
third virial coefficients
do appear
the
coeficientes
viriales.
No obstante,
aparecencorrelations
en la literatura
generalizadas
para losinterceros
coliterature.
literature.
eficientes
viriales.
Equation
(3.40)
may
be written:
written:
Equation
(3.40)
may
be
La ecuación
(3.40)
puede
escribirse:
Z=
= 11 +
+ Bρ
Bρ +
+ Cρ
Cρ 222
Z
(3.67) (3.67)
(3.67)
where
ρes=
=la1/V
1/V
is molar
molar
density.
Rewritten
in reduced
reduced
form, reducida,
this equation
equation
becomes:
donde rwhere
= 1/Vρ
densidad
molar.
Al volverse
a escribir
en forma
esta becomes:
ecuación se convierte en:
is
density.
Rewritten
in
form,
this
�
�
�
Pr �222
Pr
P
P
rr + Ĉ
rr
Z
=
1
+
B̂
(3.68) (3.68)
Z = 1 + B̂ T Z + Ĉ T Z
(3.68)
Trrr Z
Trrr Z
donde el
segundo
coeficiente
virialvirial
reducido
está definido
por la by
ecuación
(3.62)and
y elthe
tercer
coeficiente
where
the reduced
reduced
second
coefficient
B̂ is
is defined
defined
Eq. (3.62),
(3.62),
reduced
third virial
where
the
second
virial coefficient
B̂
by Eq.
and the
reduced
third
reducido
se
define
como:
virial coefficient
coefficient is
is defined
defined as:
as:
virial
CP
Pc222
C
Ĉ
≡
Ĉ ≡ R 222Tcc222
R Tc
cc
A Pitzer-type
Pitzer-type
correlation
forseĈ
Ĉescribe:
is written:
written:
Una correlación
tipo correlation
Pitzer
para Ĉ
A
for
is
+ω
ωC
C 111
Ĉ =
=C
C 000 +
Ĉ
03-SmithVanNess.indd 103
(3.69) (3.69)
(3.69)
8/1/07 12:58:52
CHAPTER 3. Volumetric Properties of Pure Fluids
104
104
CHAPTER 3. Volumetric Properties of Pure Fluids
CAPÍTULO 3. Propiedades volumétricas de21fluidos puros
An expression for
as a function of reduced temperature is given by Orbey and Vera:
0
An expression for C as a function of reduced temperature is given by Orbey and21Vera:21
0.02432
0.00313
Una expresión para C 0 como una función
de temperatura
reducida
la dan Orbey y Vera:
C 0 = 0.01407 +
−
(3.70)
Tr
T 10.5
0.02432
0.00313
C 0 = 0.01407 +
− r10.5
(3.70) (3.70)
Tr
Tr
1
The expression for C given by Orbey and Vera is replaced here by one that is algebraically
Aquí, la expresión para C 1 conocida
por Orbey y Vera se reemplaza por una que en términos algebraicos es
simpler,
but essentially
numerically:
given
by Orbey
and Vera is replaced here by one that is algebraically
The expression
for C 1 equivalent
más simple,
pero numéricamente
es equivalente:
simpler, but essentially equivalent numerically:
0.05539 0.00242
C 1 = −0.02676 +
−
(3.71) (3.71)
Tr2.7
Tr10.5
0.05539
0.00242
1
−
(3.71)
C = −0.02676 +
2.7
La ecuación (3.68) es cúbica en Z, y no es posibleTexpresarla
la forma de la ecuación (3.57). Al esTen10.5
Equation (3.68) is cubic in Z , and cannot ber expressedr in the form of Eq. (3.57). With
pecificarse Tr y Pr, la solución para Z es por iteración. Una aproximación inicial de Z = 1 en el lado derecho
Tr andEquation
Pr specified,
solution
for ZZ is
by iteration.
initial guess
Z =of1 Eq.
on the
rightWith
side
(3.68)
is cubic
, and
cannot
beAn
expressed
in theofform
(3.57).
de la ecuación
(3.68) por
lo general
nosinlleva
a una
convergencia
rápida.
of
usually solution
leads to rapid
Tr Eq.
and(3.68)
Pr specified,
for Z convergence.
is by iteration. An initial guess of Z = 1 on the right side
of Eq. (3.68) usually leads to rapid convergence.
104
C0
Condiciones
de validez
aproximada
de laofecuación
del gas
ideal
Conditions
of Approximate
Validity
the Ideal-Gas
Equation
Conditions
of arises
Approximate
Validity
the Ideal-Gas
The
question often
as to when the
ideal-gasof
equation
may be used Equation
as a reasonable ap-
Con frecuencia surge la pregunta acerca de cuándo la ecuación del gas ideal puede utilizarse como una aproxitooften
reality.
Figure
canthe
serve
as
a guide.
The
question
arises
to3.15
when
ideal-gas
equation
be used as a reasonable apmaciónproximation
razonable
a la
realidad.
Laasfigura
3.15
puede
servir
como unamay
guía.
proximation to reality. Figure 3.15 can serve as a guide.
10
10
10
Z 0 � 1.02
1
1
1
Figure 3.15: Region where Z 0 lies
between
0.98 Region
and 1.02,
and the
Figure 3.15:
where
Z 0 lies
Figura 3.15:
Región
dondeisZa0 reasonable
reside
ideal-gas
equation
between 0.98 and 1.02, and the
entre 0.98
y 1.02, y la ecuación del gas
approximation.
ideal-gas equation is a reasonable
ideal es una aproximación razonable.
approximation.
Z 0  1.02
Z 0 � 1.02
Z 0 � 0.98
Pr
0.1
Pr
Pr 0.1
0.1
Z 0 � 0.98
Z 0  0.98
0.01
0.01
0.001
0.001
0.01
0
0
0.001
1
0
1
1
2
Tr
2
Tr
3
2
Tr
3
4
3
4
4
21 H. Orbey and J. H. Vera, AIChE J., vol. 29, pp. 107–113, 1983.
21 H. Orbey and J. H. Vera, AIChE J., vol. 29, pp. 107–113, 1983.
21
H. Orbey y J. H. Vera, AIChE J., vol. 29, 107-113, 1983.
03-SmithVanNess.indd 104
8/1/07 12:59:01
3.6. Generalized
Generalized Correlations
Correlations for
for Gases
Gases
3.6.
3.6.
Correlations
Gases
3.6. Generalized
Generalized
Correlations
for
Gases
3.6. Correlaciones
generalizadas
parafor
gases
105
105
105
105
105
Example 3.10
3.10
Example
Example
3.10
Example
3.10
Ejemplo
3.10
Determine the
the molar
molar volume
volume of
of n-butane
n-butane at
at 510
510 K
K and
and 25
25 bar
bar by
by each
each of
of the
the following:
following:
Determine
Determine the molar volume of n-butane at 510 K and 25 bar by each of the following:
Determine the molar volume of n-butane at 510 K and 25 bar by each of the following:
(a) The
The ideal-gas
ideal-gas equation.
equation.
(a)
(a)
equation.
(a) The
The ideal-gas
ideal-gas
equation.
La ecuación
del
gas ideal.
(b) The
The generalized
generalized compressibility-factor
compressibility-factor correlation.
correlation.
(b)
(b)
compressibility-factor
correlation.
(b) The
The generalized
generalized
compressibility-factor
correlation.
La correlación
generalizada
del factor de compresibilidad.
(c) Equation
Equation (3.61),
(3.61), with
with the
the generalized
generalized correlation
correlation for
for B̂.
B̂.
(c)
La ecuación
(3.61)(3.61),
con lawith
correlación
generalizada
para B̂.for
(c)
the
correlation
(c) Equation
Equation
(3.61),
with
the generalized
generalized
correlation
for B̂.
B̂.
(d) Equation
Equation (3.68),
(3.68), with
with the
the generalized
generalized correlations
correlations for
for B̂
B̂ and
and Ĉ.
Ĉ.
(d)
La ecuación
(3.68)(3.68),
con laswith
correlaciones
generalizadas
parafor
B̂ yB̂
(d)
the
correlations
and
(d) Equation
Equation
(3.68),
with
the generalized
generalized
correlations
for
B̂Ĉ.
and Ĉ.
Ĉ.
Determine
the molar
molar de
volume
of n-butane
510
K mediante:
and 25 bar by each of the following:
Determine
el volumen
n-butano
a 510 K at
y 25
bar
a)
b)
c)
d)
Solution
3.10
Solución
3.10 3.10
Solution
Solution
Solution 3.10
3.10
a) Por la(a)
ecuación
del gas ideal,
(a)
By the
the ideal-gas
ideal-gas
equation,
By
equation,
(a)
(a) By
By the
the ideal-gas
ideal-gas equation,
equation,
RT
(83.14)(510)
−1
V=
= RT
mol−1
= (83.14)(510)
= 1,696.1
1,696.1 cm
cm33 mol
RT
(83.14)(510)
RT
(83.14)(510)
V
=
=
33
−1
P
25
VV =
=
= P =
mol−1
=
= 1,696.1
1,696.1 cm
cm mol
25
PP
25
25
(b)valores
From the
the
values
of
and P
Pen
given
in Table
Table
B.1
of App.
App.B,
B,
b) De los
de Tvalues
conocidos
la tabla
B.1 del
apéndice
(b)
From
TTcc and
in
B.1
of
B,
cc given
c y Pc of
(b)
(b) From
From the
the values
values of
of TTcc and
and PPcc given
given in
in Table
Table B.1
B.1 of
of App.
App. B,
B,
510
25
510
25
= 510
= 25
= 1.200
1.200
Pr =
= 0.659
0.659
510 =
25 =
TTrr =
P
425.1
37.96
TTrr =
=
PPrrr =
=
= 425.1
= 37.96
= 1.200
1.200
= 0.659
0.659
425.1
37.96
425.1
37.96
Interpolation
in
Tables
E.1
and
E.2
then
provides:
Interpolation
in tablas
TablesE.1
E.1yand
then provides:
La interpolación
en las
E.2E.2
proporciona:
Interpolation
in
E.2
then
Interpolation
in Tables
Tables E.1
E.1 and
and
E.2
then provides:
provides:
0
1
=
0.865
Z
= 0.038
0.038
0 =0.865
11 =
ZZ 000Z=
0.865 ZZ
11 0.038
ZZ=
=
ZZ =
= 0.865
0.865
= 0.038
0.038
Thus,
by
Eq.
(3.57)
with
ω
=
0.200,
Thus, by
Eq. (3.57)(3.57)
with ωcon
=w
0.200,
Así, mediante
la ecuación
= 0.200,
Thus,
Thus, by
by Eq.
Eq. (3.57)
(3.57) with
with ω
ω=
= 0.200,
0.200,
0
1
+ ωZ
ωZ 1 =
= 0.865
0.865 +
+ (0.200)(0.038)
(0.200)(0.038) =
= 0.873
0.873
= ZZ 0 +
ZZ =
ZZ =
+ ωZ
ωZ11 =
= 0.865
0.865 +
+ (0.200)(0.038)
(0.200)(0.038) =
= 0.873
0.873
= ZZ00 +
(0.873)(83.14)(510)
Z RT
RT
−1
(0.873)(83.14)(510)
Z
mol−1
= (0.873)(83.14)(510)
= 1,480.7
1,480.7 cm
cm33 mol
and
V=
= ZZ RT
(0.873)(83.14)(510)
RT =
=
and
V
33
−1
−1
P
25
mol
=
=
1,480.7
cm
and
V
=
P
25
mol
=
=
1,480.7
cm
and
V
=
y
PP
25
25
If ZZ 11,, the
the secondary
secondary term,
term, is
is neglected,
neglected, ZZ =
= ZZ 00 =
= 0.865.
0.865. This
This two-parameter
two-parameter
If
1, Z = Z 0Z
3 con
−1
If
secondary
term,
=
=cm
This
two-parameter
If ZZ11,, the
the
secondary
term, is
is neglected,
neglected,
Z=1,467.1
= ZZ00 Así,
= 0.865.
0.865.
This
two-parameter
Si se desprecia
el
término
secundario,
Z
0.865.
la
correlación
dethan
estados co3
−1
mol
,
which
is less
less
than
corresponding-states
correlation
yields
V
=
is
corresponding-states correlation yields V = 1,467.1 cm
33 mol
−1
−1,,, which
3cm
–1, que
mol
which
is
corresponding-states
correlation
yields
VV1==467.1
1,467.1
mol
which
is less
less than
than
corresponding-states
correlation
yields
1,467.1
cm
rrespondientes
de
dos
parámetros
se
obtiene
V
=
cm
mol
es
1%
menor
que
el valor
1%
lower
than
the
value
given
by
the
three-parameter
correlation.
1% lower than the value given by the three-parameter correlation.
1%
lower
than
value
given
by
the
three-parameter
correlation.
1%
lower
than the
the
value
given
by
the
three-parameter
correlation.
conocido
por
la correlación
de
tres
parámetros.
(c) Values
Values of
of B
B00 and
and B
B11 are
are given
given by
by Eqs.
Eqs. (3.65)
(3.65) and
and (3.66):
(3.66):
(c)
00 and
0
1
(c)
Values
of
B
B
(c)
Values
of
B
and
B11 are
are given
given
by Eqs.
Eqs. (3.65)
(3.65) and
and (3.66):
(3.66):
c) Los valores de B y B se conocen
por by
las
ecuaciones
(3.65)
y (3.66):
00 = −0.232
11 = 0.059
B
B
B
=
0.059
B 00 =0 −0.232
11
1B=
=
BB B=
==−0.232
−0.232
= 0.059
0.059
–0.232 B B
0.059
Equations (3.63)
(3.63) and
and (3.61)
(3.61) then
then yield:
yield:
Equations
Equations
(3.61)
yield:
Equations (3.63)
(3.61)
then
yield:
De las ecuaciones
(3.63) and
yand
(3.61)
sethen
obtiene:
00 + ωB11 = −0.232 + (0.200)(0.059) = −0.220
B̂
=
B
B̂ = B 00 + ωB 11 = −0.232 + (0.200)(0.059) = −0.220
B̂B̂ =
+ ωB
ωB =
= −0.232
−0.232 +
+ (0.200)(0.059)
(0.200)(0.059) =
= −0.220
−0.220
= BB +
0.659
0.659
= 0.879
0.879
= 11 +
+ (−0.220)
(−0.220) 0.659
0.659 =
ZZ =
1.200
=
ZZ =
= 0.879
0.879
= 11 +
+ (−0.220)
(−0.220)1.200
1.200
1.200
3 mol−1
−1 , a value
3
from
which
V
=
1,489.1
cm
less
than
1% higher
higher than
than that
that given
given
from which V = 1,489.1 cm33 mol−1
value less than 1%
−1,,, a
from
which
VV == 1,489.1
cm
mol
aa value
less
than
1%
higher
than
that
given
from
which
1,489.1
cm
mol
value
less
than
1%
higher
than
that
given por la
3
–1
by
the
compressibility-factor
correlation.
de la cual
V = 1 489.1 cmcorrelation.
mol , un valor que es 1% mayor que el conocido
by se
theobtiene
compressibility-factor
by
the
compressibility-factor
correlation.
by
the
compressibility-factor
correlation.
correlación del factor de compresibilidad.
03-SmithVanNess.indd 105
8/1/07 12:59:21
106
CHAPTER 3.
3.
CHAPTER
CHAPTER
CHAPTER 3.
3.
CHAPTER
3.
Volumetric Properties
Properties of
of Pure
Pure Fluids
Fluids
Volumetric
Volumetric
Volumetric Properties
Properties of
of Pure
Pure Fluids
Fluids
Volumetric
Properties
of
Pure
CAPÍTULO 3. Propiedades volumétricas de Fluids
fluidos puros
106
106
106
106
106
(d) Values
Values of
of C
C000 and
and C
C111 are
are given
given by
by Eqs.
Eqs. (3.70)
(3.70) and
and (3.71):
(3.71):
(d)
(d)
Values
of
C
and
C
are
by
Eqs.
(3.70)
(3.71):
0
0
0
1
(d) Values
Values
of yC
CC and
and
C11 dados
are given
given
by
Eqs.
(3.70) and
and
(3.71):
d) Los valores
de Cof
están
por by
lasEqs.
ecuaciones
(3.70)
y (3.71):
(d)
C
are
given
(3.70)
and
(3.71):
000 = 0.0339
111 = 0.0067
C
C
C
C1 11 =
0.0067
0=
CCC
=
0.0339
0==0.0339
0.0339
C= =
= 0.0067
0.0067
0.0339 CC
0.0067
C0 =
0.0339
C
=
0.0067
Equation
(3.69)
then
yields:
Equation
(3.69)
then
yields:
Equation
(3.69)
yields:
Por lo tanto,
la ecuación
(3.69)
nos da:
Equation
(3.69) then
then
yields:
Equation
(3.69)
then
yields:
0
00 + ω C111 = 0.0339 + (0.200)(0.0067) = 0.0352
Ĉ
=
C
ωC
= 0.0339
+ (0.200)(0.0067)
= 0.0352
Ĉ
C 00 +
ĈĈ =
=
=C
C +
+ω
ωC
C11 =
= 0.0339
0.0339+
+(0.200)(0.0067)
(0.200)(0.0067) =
= 0.0352
0.0352
+
ω
C
=
0.0339
+
(0.200)(0.0067)
=
0.0352
Ĉ =
C
With
this
value
of
Ĉ
and
the
value
of
B̂
from
part
(c),
Eq.
(3.68)
becomes,
Ĉ and
value
of B̂
part (c),
Eq. (3.68)
becomes,
Con esteWith
valorthis
de value
Ĉ y elofvalor
de the
B̂ del
inciso
c),from
la ecuación
(3.68)
se convierte
en,
With
With this
this value
value of
of Ĉ
Ĉ and
and the
the value
value of
of B̂
B̂ from
from part
part (c),
(c), Eq.
Eq. (3.68)
(3.68) becomes,
becomes,
With
this
value
of
Ĉ
and
the
value
of
B̂
from
part
(c),
Eq.
(3.68)
becomes,
�
�
�
�
�
�
�� 0.659
��
�� 0.659
��222
0.659 �
0.659 �
� 0.659 �
� 0.659 �
Z
=
1
+
(−0.220)
+
(0.0352)
0.659 +
0.659 22
Z
11 +
(−0.220)
(0.0352)
0.659
0.659
ZZ =
=
+
(−0.220)
+
(0.0352)
1.200Z
1.200Z
=
1
+
(−0.220)
+
(0.0352)
1.200Z
1.200Z
Z = 1 + (−0.220) 1.200Z + (0.0352) 1.200Z
1.200Z
1.200Z
1.200Z
1.200Z
0.121
0.0106
0.121
0.0106
or
Z = 11 −
− 0.121
+ 0.0106
0.121
0.0106
or
Z
0.121
0.0106
or
ZZ =
=
11−
+
Z +
Z22
or
=
−
+
Z
o
or
Z =1− Z + Z
ZZ222
ZZ
Z
−1
Whence,
Z
=
0.876
and
V
=
1,485.8
cm333 mol
mol−1
−1
Whence,
Z
0.876
and
V
1,485.8
cm
Whence,
ZZ =
=
0.876
and
VV =
=
1,485.8
cm
3 mol
−1
3
Whence,
=
0.876
and
=
1,485.8
cm
mol
3 mol
-1−1
Whence,
ZZ
==0.876
cm
mol
De donde,
0.876 and
y VV ==1 1,485.8
485.8 cm
The value
value of
of V
V differs from
from that
that of
of part
part (c)
(c) by
by about
about 0.2%.
0.2%. An
An experimental
experimental
The
The
value
of
VV differs
differs
that
of
part
(c)
by
about
0.2%.
An
experimental
3from
−1
The
value
of
differs
from
that
of
part
(c)
by
about
0.2%.
An
experimental
El valor
de
V
difiere
del
que
corresponde
al
inciso
c)
en
aproximadamente
Un valor
value
for
V
is
1,480.7
cm
mol
.
Significantly,
the
results
of
parts
(b), 0.2%.
(c), and
and
3
−1
The
value
of
V
differs
from
that
of
part
(c)
by
about
0.2%.
An
experimental
3 mol
−1 .. Significantly,
value
for
V
cm
the
results
of
parts
(b),
(c),
value
for
VV is
isis 1,480.7
1,480.7
cm
Significantly,
the
results
of
parts
(b),
(c),
and
33mol
−1
–1
3
−1
value
for
1,480.7
cm
mol
.
Significantly,
the
results
of
parts
(b),
(c),
and
experimental
para
V
es
1
480.7
cm
mol
.
De
manera
significativa,
los
resultados
de
los
incisos
(d)
are
in
excellent
agreement.
Mutual
agreement
at
these
conditions
is
suggested
value
for
V
is
1,480.7
cm
mol
.
Significantly,
the
results
of
parts
(b),
(c),
and
(d)
are
in
excellent
agreement.
Mutual
agreement at
these conditions
is
(d)
are
in
excellent
agreement.
Mutual
at
conditions
isissuggested
suggested
(d)seFig.
are
in
excellent
agreement.
Mutual agreement
agreement
at these
these
conditions
suggested
b), c) y d)
encuentran
en
excelente
concordancia.
La
mutua
correlación
de
estas
condiciones
se
by
3.14.
(d)
are
in
excellent
agreement.
Mutual
agreement
at
these
conditions
is
suggested
by
Fig.
3.14.
by
Fig.
3.14.
byla
Fig.
3.14.3.14.
sugiere en
figura
by
Fig.
3.14.
Example 3.11
3.11
Example
Example
Ejemplo
3.11 3.11
What pressure
is generated when
when 1(lb mol)
mol) of methane
methane is stored
stored in aa volume
volume of 2(ft)
2(ft)33
What
What pressure
pressure is
is generated
generated when 1(lb
1(lb mol) of
of methane is
is stored in
in a volume of
of 2(ft)3
What
pressure
is generated
generated
when
1(lb mol)
mol)
offollowing:
methane is
is stored
stored in
in aa volume
volume of
of 2(ft)
2(ft)33
at
122(
calculations
on each
of
theof
◦◦ F)? Base
What
pressure
is
when
1(lb
methane
◦ F)?
at
122(se
Base
calculations
on
each
of
the
following:
¿Qué presión
genera
se almacena
1(lb
de metano en un volumen de 2(pie) 3 a 122(°F)?
at
Base
calculations
on
of
the
at 122(
122(◦◦F)?
F)?
Basecuando
calculations
on each
each
ofmol)
the following:
following:
at
122(
F)?
Base
calculations
on
each
of
the
following:
Apoye sus cálculos
lo siguiente:
(a) The en
ideal-gas
equation.
(a)
(a) The
The ideal-gas
ideal-gas equation.
equation.
(a) The
The del
ideal-gas
equation.
(a)
ideal-gas
equation.
a) La ecuación
gas ideal.
(b) The Redlich/Kwong equation.
(b)
The Redlich/Kwong
equation.
(b)
Redlich/Kwong
equation.
b) La ecuación
Redlich/Kwong.
(b) The
The de
Redlich/Kwong
equation.
(b)
The
Redlich/Kwong
equation.
(c) A generalized correlation.
A
correlation.
(c)
AA generalized
generalized
correlation.
c) Una(c)
correlación
generalizada.
(c)
generalized
correlation.
(c)
A generalized
correlation.
Solution
3.11
Solución
3.11 3.11
Solution
Solution
3.11
Solution 3.11
3.11
Solution
a) Por la(a)
ecuación
del gas ideal,
By the
the ideal-gas
ideal-gas
equation,
(a)
By
equation,
(a)
By
the
ideal-gas
equation,
(a)
By
the
ideal-gas
equation,
(a) By the ideal-gas equation,
(0.7302)(122 +
+ 459.67)
459.67)
RT
(0.7302)(122
+
RT
= (0.7302)(122
= 212.4(atm)
212.4(atm)
P = RT
(0.7302)(122
+459.67)
459.67) =
RT
=
P
(0.7302)(122
+
459.67)
RT
=
=
212.4(atm)
PP =
=
V
2
=
=
212.4(atm)
=
V
2
= 212.4(atm)
P= V =
22
VV
2
(b) The
The
pressure
as given
given
by the
the Redlich/Kwong
Redlich/Kwong
equation
is:
(b)
as
by
equation
b) La presión
sepressure
conoce por
la ecuación
de
Redlich/Kwong
que es:is:
(b)
equation
is:
(b) The
The pressure
pressure as
as given
given by
by the
the Redlich/Kwong
Redlich/Kwong
equation
is:
(b)
The
pressure
as
given
by
the
Redlich/Kwong
equation
is:
a(T ))
RT
a(T
))
RT
− a(T
(3.47)
P=
= RT
a(T
RT
−
(3.47)
P
a(T+) b)
(3.47)
PP =
VRT
− bb −
V (V
(V
−V
(3.47)
=V
+
b)
−
(3.47)
P=
VV −
−
b
V
(V
+
b)
− bb VV(V
(V +
+ b)
b)
V−
03-SmithVanNess.indd 106
(3.47)
8/1/07 12:59:39
107 107
3.6. Generalized
Correlations
Gases
3.6. Correlaciones
generalizadas
parafor
gases
107
3.6. Generalized
Generalized Correlations
Correlations for
for Gases
Gases
3.6.
107
107
–1/2
−1/2
3.6.
3.6. Generalized
Generalized
Correlations
Correlations
for
for
Gases
Gases
107
Values
a(T
b come from
(3.45) and (3.45)
(3.46),ywhere
Tr r) =inTr 107
Los valores
de of
a(T)
y )b and
se desarrollan
de Eqs.
las ecuaciones
(3.46),α(T
donde
en
r ) = a(T
Eq. (3.45). Con
Withlos
values
of Tde
P
from
Table
B.1
converted
to
(R)
and
(atm),
−1/2
la ecuación
valores
T
y
P
del
apéndice
B
convertidos
a
(R)
y
(atm),
c and
c
−1/2 in
Values of
of a(T
a(T )) and
and bb come
come from
fromc Eqs.
Eqs.c (3.45)
(3.45) and
and (3.46),
(3.46), where
where α(T
α(Tr )) =
= TTr−1/2
Values
in
−1/2in
Values
of
)))and
bbbcome
from
Eqs.
(3.45)
and
(3.46),
where
α(T
TTTrr−1/2
rrrr)))=
Values
Values
of
ofa(T
a(T
a(T
and
and
come
come
from
from
Eqs.
Eqs.
(3.45)
(3.45)
and
and
(3.46),
(3.46),
where
where
α(T
α(T
=
=
in
in
r
r
and
P
from
Table
B.1
converted
to
(R)
and
(atm),
Eq.
(3.45).
With
values
of
T
c andTPc from
581.67
Eq.
(3.45).
With
values
of
T
Table
B.1
converted
to
(R)
and
(atm),
from
converted
Eq.
=
=B.1
1.695
and
andPPPcccc=
from
fromTable
Table
Table
B.1
B.1
converted
convertedto
to
to(R)
(R)
(R)and
and
and(atm),
(atm),
(atm),
Eq.
Eq.(3.45).
(3.45).
(3.45).With
With
Withvalues
values
valuesof
of
ofTTTcccrcand
Tc
343.1
T
581.67
581.67
Tr =
= TTTT =
= 581.67
= 1.695
1.695
581.67
581.67
T
=
TTTrr =
=
=
343.1
−1/2
2 (343.1)
2
= TTTccc(0.7302)
=
= 343.1
=
=1.695
1.695
1.695
343.1
rr =
(1.695)
6
TTcc
343.1
343.1
a = 0.42748
= 453.94(atm)(ft)
(pie)6
−1/2
2
2
45.4
(1.695)−1/2
(0.7302)22 (343.1)
(343.1)22
−1/2 (0.7302)
(1.695)
−1/2(0.7302)
22(343.1)
a=
= 0.42748
0.42748(1.695)
= 453.94(atm)(ft)
453.94(atm)(ft)666
(1.695)
(1.695)−1/2
(0.7302)
(0.7302)
(343.1)
(343.1)22 =
=
(0.7302)(343.1)
66
aaaa=
0.42748
453.94(atm)(ft)
45.4
33
=
= 0.42748
0.42748
=
= 453.94(atm)(ft)
453.94(atm)(ft)
45.4
= 0.4781(ft)
b = 0.08664 45.4
(pie)
45.4
45.4
45.4
(0.7302)(343.1)
3
(0.7302)(343.1)
(0.7302)(343.1)
= 0.4781(ft)
0.4781(ft)
b=
= 0.08664
0.08664
3333
(0.7302)(343.1)
(0.7302)(343.1)
Substitution of numerical
values
into 45.4
the Redlich/Kwong
equation
now yields:
=
=
0.4781(ft)
bbbb=
0.08664
=
=
0.4781(ft)
0.4781(ft)
=
= 0.08664
0.08664
45.4
45.4
Sustituyendo los valores numéricos en la ecuación
de Redlich/Kwong, se obtiene:
45.4
45.4
Substitution of
of numerical
numerical
values into
into the
the Redlich/Kwong
Redlich/Kwong
equation now
now yields:
yields:
(0.7302)(581.67)
453.94
Substitution
values
equation
Substitution
now
−the
=equation
187.49(atm)
Substitution
SubstitutionPof
of
of=numerical
numerical
numericalvalues
values
valuesinto
into
into
the
theRedlich/Kwong
Redlich/Kwong
Redlich/Kwong
equation
equation
now
nowyields:
yields:
yields:
2 − 0.4781
(2)(2453.94
+ 0.4781)
(0.7302)(581.67)
(0.7302)(581.67)
453.94
453.94
P=
= (0.7302)(581.67)
−
= 187.49(atm)
187.49(atm)
(0.7302)(581.67)
(0.7302)(581.67)
453.94
453.94
P
−
=
=
−
2−
− 0.4781
0.4781
(2)(2
+ 0.4781)
0.4781) =
(c) BecausePPPthe
here is high,
the generalized
compressibility-factor
cor=
= pressure
−
−(2)(2
=
=187.49(atm)
187.49(atm)
187.49(atm)
(2)(2
+
2222−
0.4781
+
0.4781)
− 0.4781
0.4781
(2)(2
(2)(2 +
+of0.4781)
0.4781)
relation is the proper −
choice.
In the absence
a known value for Pr , an iterative
(c) Because
Because the
the pressure
pressure here
here is
is high,
high, the
the generalized
generalized compressibility-factor
compressibility-factor corcor(c)
(c)
Because
here
is
high,
the
compressibility-factor
corc) Ya que
este caso
lapressure
presión
alta,
elección
apropiada es
la correlación del factor
procedure
isthe
based
on thees
following
equation:
(c)
(c)en
Because
Because
the
the
pressure
pressure
here
here
is
isla
high,
high,
the
thegeneralized
generalized
generalized
compressibility-factor
compressibility-factor
corcor-de com,
an
iterative
relation
is
the
proper
choice.
In
the
absence
of
a
known
value
for
P
r
,
an
iterative
relation
isausencia
the proper
proper
choice.
In
the absence
absence
of
a
known
value
for
P
r
,
an
iterative
relation
is
the
choice.
In
the
of
a
known
value
for
P
presibilidad.
En
de
un
valor
conocido
para
P
,
el
procedimiento
iterativo
se
apoya
an iterative
iterative en la
relation
relation is
is is
the
thebased
proper
proper
choice.
In
In the
the
absence
absence of
ofr aa known
known value
value for
for PPrrr,, an
procedure
onchoice.
the
following
equation:
Z RT
Z (0.7302)(581.67)
procedure
is based
based on
on
the
following
equation:
procedure
is
the
following
equation:
siguiente
ecuación:
=the
=
= 212.4 Z
procedure
procedure is
is based
basedPon
on
the following
following
equation:
equation:
V
2
Z
RT
Z
(0.7302)(581.67)
RT = ZZ(0.7302)(581.67)
(0.7302)(581.67) = 212.4 Z
P=
= ZZZZRT
RT
RT =
ZZ(0.7302)(581.67)
(0.7302)(581.67)
P
=
= 212.4
212.4 ZZ
=
V
2 becomes:
=
,=this
equation
Because P = Pc P
r=
PPr =
=
= 45.4P
=
= 212.4
212.4 ZZ
V
VVV
2222
Pr =
= 45.4P
45.4P
, this
this equation
equation becomes:
becomes:
Because P
P=
=P
Pc P
45.4P
Z
Because
r rrr,, this
equation
becomes:
Ya que Because
P
= PcPr PP
=P =
45.4
esta
ecuación
Z
=
or
Pr =
=
Pr
Prrr,r =
=
=45.4P
45.4P
45.4P
this
thisserá:
equation
equation
becomes:
becomes:
Because
Because
=
= PP
PccccPPP
rr,,0.2138
212.4
0.2138
45.4Pr
Z
45.4P
Z=
= 45.4P
or
Pr =
= ZZZZ
= 0.2138
0.2138 P
Pr
rrrr =
45.4P
45.4P
Z
or
P
r
r
Z
=
or
P
=
=
0.2138
P
212.4
0.2138
One now assumes
a starting
value for
Z , say
gives Pr = 4.68, and
ZZ =
= 212.4
or
orZ = 1.
PPrrrThis
=
= 0.2138
=
= 0.2138
0.2138
PPrrr o 212.4
0.2138
212.4
0.2138
0.2138
allows a new value 212.4
of
Z to be calculated by Eq. (3.57) from
values interpolated
= 4.68,
4.68, and
and
One now
now assumes
assumes aa starting
starting value
value for
for ZZ,, say
say ZZ =
= 1.
1. This
This gives
gives P
Pr =
One
r = 4.68,
One
now
assumes
a
starting
value
for
Z
,
say
Z
=
1.
This
gives
PPPrWith
in Tables
E.3
and
E.4
at
the
reduced
temperature
of
T
=
1.695.
this and
new
r
=
=
4.68,
4.68,
and
and
One
One
now
now
assumes
assumes
a
a
starting
starting
value
value
for
for
Z
Z
,
,
say
say
Z
Z
=
=
1.
1.
This
This
gives
gives
r
r
allows
a
new
value
of
Z
to
be
calculated
by
Eq.
(3.57)
from
values
interpolated
Se puede
suponer
un valor
para
Z, digamos
Z =the
1. Ésta
da Pvalues
=continues
4.68,
y permite
rvalues
allows
new
value
of Zinicial
Z of
to P
be
calculated
by and
Eq.
(3.57)
from
interpolated
allows
aaaanew
of
to
be
calculated
by
Eq.
(3.57)
from
interpolated
calculated,
procedure
until nocalcular
value
of
Znew
, a value
new
value
r is
allows
allows
new
value
value
of
of
Z
Z
to
to
be
be
calculated
calculated
by
by
Eq.
Eq.
(3.57)
(3.57)
from
from
values
values
interpolated
interpolated
in
Tables
E.3
and
E.4
at
the
reduced
temperature
of
T
=
1.695.
With
this
new
r
con la ecuación
un valor
de
Z astep
partir
de los
valores
interpolados
lasfound
tablas E.3 y
in
Tables(3.57)
E.3
and
E.4
atnuevo
the
reduced
temperature
ofThe
=
1.695.
With
this
new
in
E.3
and
at
the
reduced
temperature
of
TTTTrr =
1.695.
new
significant
change
occurs
from
one
to the
next.
final
valueWith
of Zenthis
so
in
inTables
Tables
Tables
E.3
and
andE.4
E.4
E.4
at
atof
the
the
reduced
temperature
temperature
of
of
=
1.695.
1.695.
With
With
this
this
new
new
rr =
is calculated,
calculated,
and
the
procedure
continues
until
no recienvalue
of ZZE.3
, aareducida
new
value
Prreduced
E.4 a lavalue
temperatura
de
T
=
1.695.
Con
este
valor
nuevo
de
Z
se
calcula
un
valor
r
is
and
the
procedure
continues
until
no
value
of
,
new
value
of
P
calculated,
and
the
continues
until
no
of
ZZat
,,,aaP
new
PPPrr is
is
0.890
= value
4.14. of
This
may
be confirmed
byprocedure
substitution
into Eq.
(3.57)
rnew
is
is
calculated,
calculated,
and
and
the
the
procedure
procedure
continues
continues
until
until
no
noun paso
value
value
of
of
Z
a
new
value
value
of
of
r
r
significant
change
occurs
from
one
step
to
the
next.
The
final
value
of
Z
so
found
te de Prsignificant
,of
y el
procedimiento
continúa
queE.3
no
ocurra
cambio
de
significant
change
occurs
fromhasta
one
step
to the
the
next.
The final
final
value
of
so
found
change
occurs
one
step
to
next.
The
value
ZZZZ so
found
Z 1 from
Tables
and
E.4ningún
interpolated
atsignificativo
Pof
4.14
and
values
for
Z 0 and
r =
significant
significant
change
change
occurs
occurs
from
from
one
one
step
step
to
to
the
the
next.
next.
The
The
final
final
value
value
of
of
so
so
found
found
is
0.890
at
P
=
4.14.
This
may
be
confirmed
by
substitution
into
Eq.
(3.57)
r
al siguiente.
valor
de=Z0.012,
que
se may
encuentra
es de 0.890
Pr = 4.14.
Éste
se(3.57)
puede confirisTr0.890
0.890
at P
PWith
=
4.14.
This
may be
be confirmed
confirmed
by para
substitution
into
Eq.
(3.57)
rfinal
is
at
=
4.14.
This
by
substitution
into
Eq.
=El1.695.
ω
0=and
is
isla0.890
0.890
at
at
4.14.
4.14.
This
This (3.57)
may
may be
be
confirmed
confirmed
by
substitution
substitution
into
into
Eq.
(3.57)
(3.57)
0 y Zat
1 de
rZ0en
Z111 from
from
Tables
E.3
and
E.4 by
interpolated
P
=Eq.
4.14
and
of
values
forPPrrZ
0=
r las
mar porof
sustitución
la
ecuación
de
los
valores
para
Z
tablas
E.3 y E.4,
of
values
for
and
Z
Tables
E.3
and
E.4
interpolated
at
P
=
4.14
and
r
ZZZ11 from
Tables
E.3
and
E.4
interpolated
at
PPPr =
4.14
and
values
for
ZZZ00 and
and
and
from
from
Tables
Tables
E.3
E.3
and
and
E.4
E.4
interpolated
interpolated
at
at
=
=
4.14
4.14
and
and
of
of
values
values
for
for
r
r
=
1.695.
With
ω
=
0.012,
T
0
1
r
interpolados
en
P
=
4.14
y
T
=
1.695.
Con
w
=
0.012,
=
1.695.
With
ω
=
0.012,
T
r
r
+
ωZ
=
0.887
+
(0.012)(0.258)
=
0.890
Z
=
Z
r
TTTr =
=
=1.695.
1.695.
1.695.With
With
Withωωω===0.012,
0.012,
0.012,
rr
0
1
+ ωZ
ωZ(0.890)(0.7302)(581.67)
= 0.887
0.887 +
+ (0.012)(0.258)
(0.012)(0.258) =
= 0.890
0.890
=ZZZRT
00 +
11 =
ZZ =
=
+
+ωZ
ωZ
ωZ11=
=
=0.887
0.887
0.887+
+
+(0.012)(0.258)
(0.012)(0.258)
(0.012)(0.258)
=
=0.890
0.890
0.890
=
= ZZZ00+
PZZZ==
=
= 189.0(atm)
V
2
Z
RT
(0.890)(0.7302)(581.67)
RT = (0.890)(0.7302)(581.67)
(0.890)(0.7302)(581.67) = 189.0(atm)
ZZZZRT
P=
=
RT
RT factor
(0.890)(0.7302)(581.67)
P
= (0.890)(0.7302)(581.67)
= 189.0(atm)
189.0(atm) compressibilBecause the
acentric
is small, the
P
=
=
V
2 two- and three-parameter
PP =
=
= V
=
=
=
= 189.0(atm)
189.0(atm)
V
2222 Both the Redlich/Kwong
ity-factor correlations
are little different.
equation and
V
V
Because
the
acentric
factor
is
small,
the
twoand
three-parameter
compressibilBecause
the
acentric
factor
is small,
small,
the twotwo- and
and
three-parameter
compressibilthe
acentric
factor
is
the
three-parameter
compressibiltheBecause
generalized
compressibility-factor
correlation
give
answers
close
to the exper-con dos
Porque
el
factor
acéntrico
es
pequeño,
las
correlaciones
del
factor
de
compresibilidad
Because
Because
the
the acentric
acentric
factor
factor
isdifferent.
small,
small, the
theBoth
twotwo-the
and
andRedlich/Kwong
three-parameter
three-parameter compressibilcompressibility-factor
correlations
are
littleis
equation
and
ity-factor
correlations
are
little
different.
Both
the
Redlich/Kwong
equation
and
ity-factor
correlations
are
little
different.
Both
the
Redlich/Kwong
equation
and
imental
value
of
185(atm).
The
ideal-gas
equation
yields
a
result
that
is high
by
y tres parámetros
difieren
un
poco.
Tanto
la
ecuación
de
Redlich/Kwong
como
la
correlación
ity-factor
ity-factor
correlations
correlations
are
are little
little different.
different.
Both
Both the
the
Redlich/Kwong
Redlich/Kwong
equation
equation
and
and
the
generalized
compressibility-factor
correlation
give
answers
close
to
the
experthe
generalized
compressibility-factor
correlation respuestas
give answers
answers
close
to the
the experexperthe
generalized
compressibility-factor
correlation
give
close
to
14.6%.
generalizada
del
factor
de
compresibilidad
proporcionan
muy
próximas
al
valor
the
the generalized
generalized
compressibility-factor
correlation
correlation
give
give
answers
answers
close
close
to
tois
the
thehigh
experexperimental
value of
ofcompressibility-factor
185(atm). The
The ideal-gas
ideal-gas
equation
yields
result
that
by expeimental
value
185(atm).
equation
yields
aa result
result
that
is high
high
by
value
of
185(atm).
The
ideal-gas
equation
yields
a
that
is
by
rimentalimental
de
185(atm).
La
ecuación
del
gas
ideal
produce
un
resultado
que
es
mayor
en
14.6%.
imental
imental value
value of
of 185(atm).
185(atm). The
The ideal-gas
ideal-gas equation
equation yields
yields aa result
result that
that is
is high
high by
by
14.6%.
14.6%.
14.6%.
14.6%.
14.6%.
03-SmithVanNess.indd 107
8/1/07 13:00:06
108
108
108
108
108
CHAPTER 3. Volumetric Properties of Pure Fluids
CHAPTER 3. Volumetric Properties of Pure Fluids
CHAPTER 3. Volumetric Properties of Pure Fluids
CAPÍTULO
Propiedades
volumétricas
deFluids
fluidos puros
CHAPTER 3.3.Volumetric
Properties
of Pure
Example
3.12
Ejemplo
3.12 3.12
Example
Example
3.12
3 vessel immersed
A mass
mass
of g500
500
g of
of gaseous
gaseous
ammonia
is contained
contained
in aarecipiente
30,000-cm
3 sumergido en
Una masa
de 500
de g
amoniaco
gaseoso
está
contenida
en in
un
de33 30
000 cm
A
of
ammonia
is
30,000-cm
vessel
immersed
◦
Example
3.12
A
mass
of
500
g
of
gaseous
ammonia
is
contained
in
a
30,000-cm
vessel
immersed
in
a
constant-temperature
bath
at
65
C.
Calculate
the
pressure
of
the
gas
by:
◦
un baño a temperatura constante de 65 °C. Calcule la presión del gas mediante:
in a constant-temperature bath at 65◦ C. Calculate the pressure of the gas by:
3 vessel
in mass
a constant-temperature
bath
at 65 C.is Calculate
of the
gas by:
A
of 500 g of gaseous
ammonia
containedthe
in apressure
30,000-cm
immersed
(a)
The
ideal-gas
equation;
(b)
A
generalized
correlation.
◦
a) in
Laaecuación
del gas ideal;
b)
unaat(b)
correlación
generalizada.
(a)
The ideal-gas
equation;
correlation.
constant-temperature
bath
65 AC.generalized
Calculate
the
pressure of the gas by:
(a) The ideal-gas equation; (b) A generalized correlation.
(a) The ideal-gas equation; (b) A generalized correlation.
Solución
3.12 3.12
Solution
Solution
3.12
Solution
3.12
El volumen
molar
del
amoniaco
en el recipiente
es: is:
The molar
molar volume
volume
of ammonia
ammonia
in the
the vessel
vessel
The
of
in
is:
Solution
3.12
The molar volume oft ammonia
in
the
vessel
is:
t
V
30,000
V
t
t
−1
V
30,000
V
The molar volume
oft ammonia
vessel is:= 1,021.2 cm33 mol−1
= V t in
=the30,000
=V
=
=
= 1,021.2 cm3 mol−1
VV =
n
m/M
500/17.02
= 500/17.02 = 1,021.2 cm mol
V = nt = m/M
Vt
30,000
Vn
m/M
500/17.02
=
=
= 1,021.2 cm3 mol−1
V gas
= ideal,
(a)
By the
the ideal-gas
ideal-gas
equation,
a) Por la(a)
ecuación
del
By
nequation,
m/M
500/17.02
(a) By the ideal-gas equation,
(83.14)(65 +
+ 273.15)
273.15)
RT
RT
(a) By the ideal-gas
equation,
= (83.14)(65
= 27.53
27.53 bar
bar
=
=
=
PP =
(83.14)(65
+
273.15)
RT
1,021.2
1,021.2
= 27.53 bar
P = VV =
(83.14)(65
+ 273.15)
RT
V
1,021.2
=
= 27.53 bar
=
≈ 27.53/112.8
27.53/112.8
0.244), the
the genergener(b) Because
Because the
the P
reduced
pressure
is low
low (P
(Pr ≈
== 00.244),
(b)
reduced
pressure
is
V
1,021.2
r
b) Ya que
la
presión
reducida
es
menor
(P
≈
27.53/112.8
=
0.244),
la correlación
generalizada
del
1the
r
≈
27.53/112.8
=
0.244),
gener(b)
Because
the
reduced
pressure
is
low
(P
alized
virial-coefficient
correlation
should
suffice.
Values
of
B
and
B
are
given
0
1
r
0 y BValues
1 se conocen
alized
virial-coefficient
correlation
shouldde
suffice.
of B 0 por
and las
B 1ecuaciones
are given
coeficiente
virial
debe
ser
suficiente.
Los
valores
B
(3.65)
alized
virial-coefficient
correlation
should
Values
of B
and B the
aregenergiven
by Eqs.
Eqs.
(3.65)
and
(3.66),
with TTr is
=
338.15/405.7
= 0.834:
0.834:
= 0.244),
(b)
Because
theand
reduced
pressure
low
(Psuffice.
r ≈ 27.53/112.8
(3.66),
=
r = 338.15/405.7
y (3.66),by
Tr (3.65)
= 338.15/405.7
= with
0.834:
0 and B 1 are given
bycon
Eqs.
(3.65)
and (3.66),
with
Tr =should
338.15/405.7
= 0.834:
alized
virial-coefficient
correlation
suffice.
Values
of
B
0
=
−0.482
B11 ==
−0.232
0=
by Eqs. (3.65) and (3.66),
with
Tr = 338.15/405.7
0.834:
BB00 B=
−0.482
−0.232
–0.482 BB1 1= =
–0.232
B = −0.482
B = −0.232
0
Substitution
intoen
Eq.la(3.63)
(3.63)
with(3.63)
ω == 0.253
0.253
yields:
Becuación
= −0.482
B 1 = −0.232
Sustituyendo
w = 0.253
se obtiene:
Substitution
into
Eq.
with
ω
yields:
Substitution into Eq. (3.63) with ω = 0.253 yields:
= −0.482
−0.482
+
(0.253)(−0.232)
= −0.541
−0.541
Substitution into Eq.
with+
ω (0.253)(−0.232)
= 0.253 yields: =
B̂B̂ (3.63)
=
B̂ = −0.482 + (0.253)(−0.232) = −0.541
−(0.541)(83.14)(405.7)
B̂RT
RT
B̂cc = −0.482
+ (0.253)(−0.232) = −0.541 3
−1
= −161.8
−161.8 cm
cm3 mol
= −(0.541)(83.14)(405.7) =
mol−1
= B̂
BB =
B̂ RT
c = −(0.541)(83.14)(405.7)
P
112.8
3
−1
c
112.8
= −161.8 cm mol
B = Pc =
B̂ RT
Pfor
112.8
c c P: −(0.541)(83.14)(405.7)
Solve
Eq.
(3.38)
= −161.8 cm3 mol−1
=
= for P:
Solve Eq.B(3.38)
112.8
c P:
Solve Eq. (3.38)Pfor
(83.14)(338.15)
RT(3.38):(83.14)(338.15)
Resolviendo
P de lafor
ecuación
Solvepara
Eq. (3.38)
P:
= (83.14)(338.15) =
= 23.76 bar
bar
= RT
=
PP =
RT
− BB = 1,021.2
1,021.2 +
+ 161.8
161.8 = 23.76
23.76 bar
P = VV −
(83.14)(338.15)
VRT
−B
1,021.2 + 161.8
=
23.76 bar of pressure. The
P =is not necessary,
An iterative
iterative solution
solution
because BB is
is=independent
An
is Vnot−necessary,
because
B
1,021.2
+ 161.8 independent of pressure. The
An
iterative
solution
is
not
necessary,
because
B
is
independent
of pressure.
The
calculated
P
corresponds
to
a
reduced
pressure
of
P
= 23.76/112.8
23.76/112.8
= 0.211.
0.211.
r
calculated P corresponds to a reduced pressure of Pr =
=
calculated
P
corresponds
to
a
reduced
pressure
of
P
=
23.76/112.8
=
0.211.
Reference
to
Fig.
3.14
confirms
the
suitability
of
the
generalized
virial-coefficient
r
An
iterative
solution
is
not
necessary,
because
B
is
independent
of
pressure.
The
Referenceencontrar
to Fig. 3.14
suitability
of theBgeneralized
virial-coefficient
No es necesario
unaconfirms
solucióntheiterativa,
porque
es independiente
de la presión. El
Reference
3.14 confirms
the
suitability
of of
the
virial-coefficient
correlation.
calculated
PP Fig.
corresponds
to
a reduced
pressure
23.76/112.8
0.211.
r r=
correlation.
valor calculado
deto
corresponde
a una
presión
reducida
dePgeneralized
P
= 23.76/112.8
==0.211.
La figura
correlation.
Experimental
data
indicate
that
the
pressure
is
23.82
bar
at
the
given
condiReference
to
Fig.
3.14
confirms
the
suitability
of
the
generalized
virial-coefficient
3.14 confirma
que el uso de
la indicate
correlación
coeficiente
es apropiado.
Experimental
data
thatgeneralizada
the pressuredel
is 23.82
bar atvirial
the given
condiExperimental
data indica
indicate
that
the
pressure
23.82
barpara
at the
given
condi- conocitions.
Thus
the ideal-gas
ideal-gas
equation
yields
an es
answer
highbar
by
about
15%,
whereas
correlation.
Información
experimental
que
layields
presión
deis23.82
las15%,
condiciones
tions.
Thus
the
equation
an
answer
high
by
about
whereas
tions.
Thus
the
ideal-gas
equation
yields
an
answer
high
by
about
15%,
whereas
the
virial-coefficient
correlation
gives
an
answer
in
substantial
agreement
with
exExperimental
data
indicate
that
the
pressure
is
23.82
bar
at
the
given
condidas. Dethe
estevirial-coefficient
modo, la ecuación
del gas gives
ideal an
produce
respuesta que
es mayor
enexcasi 15%,
correlation
answeruna
in substantial
agreement
with
the
virial-coefficient
correlation
gives
an
answer
in
substantial
agreement
with
experiment,
even
though
ammonia
is
a
polar
molecule.
tions.
Thus
the
ideal-gas
equation
yields
an
answer
high
by
about
15%,
whereas
mientrasperiment,
que la correlación
delammonia
coeficiente
even though
is avirial
polarproporciona
molecule. una respuesta de conformidad conperiment,
even though
ammonia
is el
a polar
molecule.
the
virial-coefficient
correlation
gives
an answer
inuna
substantial
agreement
siderable
con
el experimento,
aun
cuando
amoniaco
es
molécula
polar. with experiment, even though ammonia is a polar molecule.
03-SmithVanNess.indd 108
8/1/07 13:00:23
3.7. Generalized
Generalized Correlations
Correlations for
for Liquids
Liquids
3.7.
3.7. Generalized
Generalized
Correlations
for Liquids
Liquids
3.7.
Correlations
3.7. Correlaciones
generalizadas
para for
líquidos
3.7
109
109
109 109
109
3.7
CORRELATIONS
FOR
3.7 GENERALIZED
GENERALIZED
CORRELATIONSPARA
FORLIQUIDS
LIQUIDS
CORRELACIONES
GENERALIZADAS
LÍQUIDOS
Although
the molar
molar
volumes
oflíquidos
liquids can
can
be
calculated
by means
meanspor
of medio
generalized
cubic equaequathe
volumes
liquids
be
calculated
by
of
generalized
cubic
AunqueAlthough
los
volúmenes
molares
de losof
son
posibles
de calcularse
de ecuaciones
cúbicas de
Although
the molar
molar
volumes
of
liquids
can
be
calculated
by means
means
of
generalized
cubic equaequaAlthough
the
volumes
of
liquids
can
be
calculated
by
of
generalized
cubic
tions
of
state,
the
results
are
often
not
of
high
accuracy.
However,
the
Lee/Kesler
correlation
tions
of
state,
the
results
are
often
not
of
high
accuracy.
However,
the
Lee/Kesler
correlation
estado, con
frecuencia
sonnot
muy
De cualquier
modo,
correlación
de Lee/Kesler
tions
of state,
state, los
the resultados
results are
areno
often
not
ofexactos.
high accuracy.
accuracy.
However,
the la
Lee/Kesler
correlation
tions
of
the
results
often
of
high
However,
the
Lee/Kesler
correlation
includes
data para
for subcooled
subcooledsubenfriados,
liquids, and
and Fig.
Fig.la3.13
3.13 illustrates
illustrates
curves
for both
both liquids
liquidslíquidos
and gases.
gases.
data
for
liquids,
curves
for
and
incluyeincludes
información
3.13 ilustra
curvas
como para
includes
data
forlíquidos
subcooled
liquids, and
andyFig.
Fig.figura
3.13 illustrates
illustrates
curves
fortanto
bothpara
liquids
and gases.
gases.
includes
data
for
subcooled
liquids,
3.13
curves
for
both
liquids
and
Values
for
both
phases
are
provided
in
Tables
E.1
through
E.4.
Recall,
however,
that
thisque esta
Values
for
both
phases
are
provided
in
Tables
E.1
through
E.4.
Recall,
however,
that
gases. Los
valores
ambas
fases
proporcionan
en lasE.1
tablas
E.1 aE.4.
E.4. Recall,
No obstante,
recuerde
Values
forde
both
phases
areseprovided
provided
in Tables
Tables
E.1
through
E.4.
Recall,
however,
thatthis
this
Values
for
both
phases
are
in
through
however,
that
this
correlation
is most
most suitable
suitable
for
nonpolar
and slightly
slightly
polar fluids.
fluids.
correlation
is
nonpolar
and
polar
correlación
es la más
parafor
fluidos
no polares
y ligeramente
polares.
correlation
isadecuada
most suitable
suitable
for
nonpolar
and slightly
slightly
polar fluids.
fluids.
correlation
is
most
for
nonpolar
and
polar
Inestán
addition,
generalized
equations
are available
available
forlathe
the estimation
estimation
of molar
molar volumes
volumes
ofde líquiIn
addition,
generalized
equations
are
for
of
of
Además,
disponibles
ecuaciones
generalizadas
para
de volúmenes
molares
In
addition,
generalized
equations
are available
available
forestimación
the22estimation
estimation
of molar
molar volumes
volumes
of
In
addition,
generalized
equations
are
for
the
of
of
22
saturated
liquids.
The
simplest
equation,
proposed
by
Rackett,
is
an
example:
22 es un ejemplo:
22
saturated
simplest
proposed
byby
Rackett,
is is
anan
example:
dos saturados.
Laliquids.
ecuación
más
simple,equation,
propuesta
por Rackett,
saturated
liquids.The
The
simplest
equation,
proposed
by
Rackett,22
is
an
example:
saturated
liquids.
The
simplest
equation,
proposed
Rackett,
example:
2/7
sat
(1−Trr ))2/72/7
=V
V Z
Z (1−T
(3.72)
V sat
sat
(1−Trr))2/7
=
(3.72)
V
V saturado
= ccV
VccccZZc(1−T
(3.72)(3.72)
V sat
=
(3.72)
V
c
An
alternative
form
of
this
equation
is
sometimes
useful:
An
alternative
form
ofof
this
equation
is is
sometimes
useful:
An
alternative
form
of
this
equation
sometimes
useful:
An
alternative
this
equation
sometimes
En ocasiones
es útil unaform
forma
alternativa
de is
esta
ecuación:useful:
P
2/7 ]
rP [1+(1−T
sat
)2/7
] ]]
2/7
rr ) ))2/7
Z
= PrP
Zrr cc[1+(1−T
(3.73)
sat
[1+(1−T
[1+(1−T
Z sat
=
Z
(3.73)
rr
=T
Z
(3.73)(3.73)
ZZsat
=
Z
(3.73)
Z saturado
Trr
cc
TTrr
The
only
data
required
are
the
critical
constants,
given
in
Table
B.1.
Results
are
usually
accuThe
only
data
required
are
the
critical
constants,
given
inin
Table
B.1.
Results
are
usually
accuLa única
información
que
se requiere
son
las constantes
críticas,
conocidas
en
la tabla
B.1.
Por lo
regular los
The
only
data
required
are
the
critical
constants,
given
in
Table
B.1.
Results
are
usually
accuThe
only
data
required
are
the
critical
constants,
given
Table
B.1.
Results
are
usually
accurate
to
1
or
2%.
rate
to
1
or
2%.
resultadosrate
tienen
exactitud de 1 o 2%.
rate
to 11una
or 2%.
2%.
to
or
23 developed a two-parameter corresponding-states
Lydersen,
Greenkorn,
and
Hougen23
23 desarrollaron
23
Lydersen,
Greenkorn,
and
developed
a aatwo-parameter
corresponding-states
Lydersen,
Greenkorn
y Hougen
una correlación
de estados
correspondientes con dos
Lydersen,
Greenkorn,
andHougen
Hougen23
developed
two-parameter
corresponding-states
Lydersen,
Greenkorn,
and
Hougen
developed
two-parameter
corresponding-states
correlation
for
estimation
of
liquid
volumes.
It
provides
a
correlation
of
reduced
density
ρrr as
as reducorrelation
for
estimation
of
liquid
volumes.
It
provides
a
correlation
of
reduced
density
ρ
parámetros
para la estimación
del volumen
líquidos.
Ésta proporciona
una of
correlación
de densidad
correlation
for estimation
estimation
of liquid
liquidde
volumes.
It provides
provides
correlation
of reduced
reduced
density
as
correlation
for
of
volumes.
It
aa correlation
density
ρρrr as
a
function
of
reduced
temperature
and
pressure.
By
definition,
a
function
of
reduced
temperature
and
pressure.
By
definition,
cida r r como
una función
de latemperature
temperaturaand
y lapressure.
presión reducidas.
Por definición,
function
of reduced
reduced
temperature
and
pressure.
By definition,
definition,
aa function
of
By
ρ
Vc
Vcc
ρrr ≡
≡ ρ ρρ=
= VcV
(3.74)
ρ
(3.74)
(3.74)(3.74)
≡ρ
=V
ρρrr ≡
(3.74)
ρcc =
V
ρ
V
ρcc
V
donde rwhere
en el punto
correlación
generalizada
se muestra
en laby
figura
is the
the density
density
at the
the crítico.
critical La
point.
The generalized
generalized
correlation
is shown
shown
by
Fig. 3.16.
3.16. Ésta es
c es laρ
where
ρdensidad
at
critical
point.
correlation
is
Fig.
3.16.
ccρ
where
ρis
is the
the density
density
at the
the
critical
point.The
The generalized
generalized
correlation
is shown
shown
by
Fig.
3.16.
where
at
critical
point.
The
correlation
is
by
Fig.
3.16.
cc is
posible This
usarla
de
manera
directa
con
la
ecuación
(3.74)
para
determinar
los
volúmenes
de
los
líquidos
This
figure
may
be
used
directly
with
Eq.
(3.74)
for
determination
of
liquid
volumes
if
the si se
figure
may
be
used
directly
with
Eq.
(3.74)
for
determination
of
liquid
volumes
if
the
This
figure
may
be
used
directly
with
Eq.
(3.74)
for
determination
of
liquid
volumes
if
the
This figure may be used directly with Eq. (3.74) for determination of liquid volumes if the
conoce value
el valor
crítico.is
mejor procedimiento
consiste
usar use
un volumen
líquido
conocido
value
of del
the volumen
critical volume
volume
isUn
known.
A better
better procedure
procedure
is to
toenmake
make
use
of aa single
single
known
of
the
critical
known.
A
is
of
known
value
of
the
critical
volume
is
known.
A
better
procedure
is
to
make
use
of
a
single
known
value of the critical volume is known. A better procedure is to make use of a single known
(estado liquid
1) y emplear
identidad,
liquid
volumela(state
(state
1) by
by the
the identity,
identity,
volume
1)
liquid
volume (state
(state
1) by
by the
the identity,
identity,
liquid
volume
1)
ρr11
(3.75)
=V
V ρrρρ
(3.75)
V22 =
rr11
(3.75)
V
= 11V
Vρ
(3.75)
V22 =
(3.75)
V
ρ11rr22
ρ
ρrr22
= required
required volume
volume
V
where
2V=
V
where
2
required
volume
volume
V22 == required
where
donde where
V2 = volumen
requerido
V =
= known
known volume
volume
V11 ==
known
volume
V
known
volume
conocido
V1 =V11volumen
ρ 11 ,, ρ
ρrr22 =
= reduced
reduced densities
densities read
read from
from Fig.
Fig. 3.16
3.16
= reduced
reduced
densitiesleídas
read from
from
Fig. 3.16
3.16
,, ρρ
densities
read
r r1,ρrrrρρ
densidades
reducidas
de
la Fig.
figura
3.16
rr22 =
rrr211=
This method
method gives
gives good
good results
results and
and requires
requires only
only experimental
experimental data
data that
that are
are usually
usually available.
available.
This
This method
method gives
gives good
good results
results and
and requires
requires only
only experimental
experimental data
data that
that are
are usually
usually available.
available.
This
Este método
da
buenos
resultados
y
sólo
requiere
de
información
experimental
que
usualmente
está
Figure
3.16
makes
clear
the
increasing
effects
of
both
temperature
and
pressure
on
liquid
den-disponiFigure
3.16
makes
clear
the
increasing
effects
ofof
both
temperature
and
pressure
onon
liquid
denFigure
3.16
makes
clear
the
increasing
effects
of
both
temperature
and
pressure
on
liquid
denFigure
3.16
makes
clear
the
increasing
effects
both
temperature
and
pressure
liquid
denble. La sity
figura
3.16
muestra
de
forma
clara
los
efectos
al
aumentar
la
temperatura
y
la
presión
sobre
la densisity
as
the
critical
point
is
approached.
asas
the
critical
point
is is
approached.
sity
as
the
critical
point
is
approached.
sity
the
critical
point
approached.
dad del líquido
conforme
se
acerca
al
punto
crítico.
Correlations
for
the
molar
densities
as
functions
of
temperature
are
given
for
many
pure
Correlations
forfor
the
molar
densities
asas
functions
ofof
temperature
are
given
forfor
many
pure
Correlations
for
the
molar
densities
as
functions
of
temperature
are
given
for
many
pure
Correlations
the
molar
densities
functions
temperature
are
given
many
pure
24
24
Las
correlaciones
para
lascoworkers.
densidades
molares
como funciones de la temperatura se conocen para varios
liquids
by Daubert
Daubert
and
coworkers.
24
24
liquids
by
and
liquids by
by Daubert
Daubert and
and coworkers.
coworkers. 24
liquids
líquidos puros por Daubert y colaboradores.
22 H. G. Rackett, J. Chem. Eng. Data, vol. 15, pp. 514–517, 1970; see also C. F. Spencer and S. B. Adler, ibid.,
22
H.
G.G.
Rackett,
J. J.
Chem.
Eng.
Data,
vol.
15,15,
pp.pp.
514–517,
1970;
seesee
also
C.C.
F. F.
Spencer
and
S. S.
B.B.
Adler,
ibid.,
22 H.
22
G.
Rackett,
J. Chem.
Chem.
Eng.
Data,
vol.
15,
pp.
514–517,
1970;
see
also
C.
F.
Spencer
and
S.
B.
Adler,
ibid.,
Rackett,
Eng.
Data,
vol.
514–517,
1970;
also
Spencer
and
Adler,
ibid.,
vol.
23,H.pp.
pp. 82–89,
82–89,
1978,
for aa review
review
of available
available
equations.
vol.
23,
1978,
for
of
equations.
vol.
23,
pp.
82–89,
1978,
for
a
review
of
available
equations.
vol. 23, pp. 82–89, 1978, for a review of available equations.
22 H. G. Rackett, J. Chem. Eng. Data, vol. 15, pp. 514-517, 1970; véase también C. F. Spencer y S. B. Adler, ibid., vol. 23, pp. 82-89,
23 A. L. Lydersen, R. A. Greenkorn, and O. A. Hougen, “Generalized Thermodynamic Properties of Pure Fluids,”
23
A.
L.L.
Lydersen,
R.R.
A.A.
Greenkorn,
and
O.O.
A.A.
Hougen,
“Generalized
Thermodynamic
Properties
of of
Pure
Fluids,”
23A.
23
A.
L.
Lydersen,
R.
A.
Greenkorn,
and
O.
A.
Hougen,
“Generalized
Thermodynamic
Properties
of
Pure
Fluids,”
Lydersen,
Greenkorn,
and
Hougen,
“Generalized
Thermodynamic
Properties
Pure
Fluids,”
1978, para
una Wisconsin,
revisión
deEng.
las ecuaciones
disponibles.
Univ.
Expt.
Sta.
Rept.
4, 1955.
1955.
Univ.
Wisconsin,
Eng.
Expt.
Sta.
Rept.
4,
Univ.
Wisconsin,
Eng.
Expt.
Sta.
Rept.
4,
1955.
Wisconsin,
Eng. Expt.ySta.
Rept.
4, 1955.
23 A. L. Univ.
Lydersen,
R. A. Greenkorn
O. A.
Hougen,
“Generalized Thermodynamic Properties of Pure Fluids”, Univ. Wisconsin, Eng.
24 T. 4,
24
E. 1955.
Daubert, R.
R. P.
P. Danner,
Danner, H.
H. M.
M. Sibul,
Sibul, and
and C.
C. C.
C. Stebbins,
Stebbins, Physical
Physical and
and Thermodynamic
Thermodynamic Properties
Properties of
of Pure
Pure
Expt. Sta. Rept.
T.
E.
Daubert,
24T.
24
T.
E.
Daubert, R.
R. P.
P. Danner,
Danner, H.
H. M.
M. Sibul,
Sibul, and
and C.
C. C.
C. Stebbins,
Stebbins, Physical
Physical and
and Thermodynamic
Thermodynamic Properties
Properties of
of Pure
Pure
E.
Daubert,
Chemicals:
Data
Compilation,
Taylor
&yFrancis,
Francis,
Bristol,
PA,
extantand
1995.
24 T. Chemicals:
Data
Compilation,
Taylor
&
Bristol,
PA,
extant
1995.
E.Chemicals:
Daubert,
R.
P.
Danner,
H.
M.
Sibul
C.
C.
Stebbins,
Physical
Thermodynamic
Properties
of
Pure
Chemicals:
Data
Chemicals: Data
Data Compilation,
Compilation, Taylor
Taylor &
& Francis,
Francis, Bristol,
Bristol, PA,
PA, extant
extant 1995.
1995.
Compilation, Taylor & Francis, Bristol, PA, existente 1995.
03-SmithVanNess.indd 109
8/1/07 13:00:32
110
CHAPTER 3.3.Volumetric
Properties
of Pure
CAPÍTULO
Propiedades
volumétricas
deFluids
fluidos puros
CHAPTER 3. Volumetric Properties of Pure Fluids
110
110
3.5
3.5
3.5
TrT�0.3
0.3
rr
3.0
3.0
3.0
2.5
2.5
 2.5
�r r
�r
2.0
2.0
2.0
Tr  0.95
Trr � 0.95
Tr � 0.95
1.0
0.97
0.97
0.99
0.97
0.99
0.99
1.5
1.5
1.5
1.0
1.0 0
0
1.0
0
Tr0.4
�
0.40.3
0.5
0.50.4
0.6
0.60.5
0.7
0.70.6
0.8 0.7
0.8
0.9 0.8
0.9
1.0 0.9
1.0
Líquido saturado
Saturated Liquid
Saturated Liquid
1
1
1
2
2
2
3
3
3
4
4
4
5
5
Pr
Pr 5
Pr
6
6
6
7
7
7
8
8
8
9
9
9
10
10
10
Figura 3.16: Correlación de densidad generalizada para líquidos.
Figure 3.16: Generalized density correlation for liquids.
Figure 3.16: Generalized density correlation for liquids.
Example
Ejemplo
3.13 3.13
Example
3.13
For ammonia at 310 K, estimate the density of:
Para el amoniaco a 310 K, calcule la densidad de:
For ammonia at 310 K, estimate the density of:
(a) The saturated liquid; (b) The liquid at 100 bar.
a) El líquido
saturado;
b) El liquid;
líquido (b)
a 100
(a) The
saturated
Thebar.
liquid at 100 bar.
Solution
Solución
3.13 3.13
Solution 3.13
(a) Apply Eq. (3.72) at the reduced temperature, Tr = 310/405.7 = 0.7641. With
(3.72)
atlathe
reduced
temperature,
= 310/405.7
= 0.7641.
With
a) Aplique
laApply
ecuación
temperatura
reducida,
= 0.7641.
Con V
Vc(a)
=
72.47 Eq.
and(3.72)
Z c =a0.242
(from Table
B.1), Tr T=r 310/405.7
C = 72.47
V
=
72.47
and
Z
=
0.242
(from
Table
B.1),
y ZC = 0.242
c (de la tabla B.1),
c
2/7
2/7
V sat = Vc Z c(1−Tr ) 2/7= (72.47)(0.242)(0.2359) 2/7= 28.33 cm3 mol−1
saturado
V V sat = Vc Z c(1−Tr ) = (72.47)(0.242)(0.2359) = 28.33 cm3 mol−1
For comparison, the experimental value is 29.14 cm3 mol−1 , a 2.7% difference.
−1 , a 2.7% difference.
For comparison,
experimental
is 29.14
cm3–1mol
Para comparación,
el valorthe
experimental
es value
de 29.14
cm3 mol
, y difiere
en 2.7%.
(b) The reduced conditions are:
b) Las condiciones reducidas son:
(b) The reduced conditions are:
100
Tr = 0.764
Pr = 100 = 0.887
Tr = 0.764
Pr =112.8 = 0.887
Sustituyendo
el valorthe
r rvalue,
= 2.38ρ(de
figura
3.16),
VC 112.8
enand
la ecuación
(3.74),
se gives:
obtiene:
Substituting
2.38
(from
Fig.y3.16),
Vc into Eq.
(3.74)
r =la
Substituting the value, ρr = 2.38 (from Fig. 3.16), and Vc into Eq. (3.74) gives:
Vc
72.47
V = Vc= 72.47= 30.45 cm3 mol−1
V =ρr =2.38 = 30.45 cm3 mol−1
ρr
2.38
03-SmithVanNess.indd 110
8/1/07 13:00:47
Problems
111
111
Problems
Problems
111
Problemas
111
Problems
111
3
−1
In comparison with the experimental value of 28.6 cm mol , this result is higher
3 mol−1 , this result is higher
In comparison
comparison
with
the
experimental
value
of 28.6
28.6
cm
3 mol
–1,−1
3 mol
by 6.5%. con
En In
comparación
elthe
valor
experimental
de of
28.6
cmcm
este
resultado
eshigher
mayor en 6.5%.
with
experimental
value
, this
result
3 mol
−1
In
comparison
with
the
experimental
value
of
28.6
cm
resultisis
higher
3 mol
−1 for
by
6.5%.
3
–1,,this
saturated
liquid a 310
If we startcon
with
the
experimental
value
of 29.14
cm
Si
comenzamos
el
valor
experimental
de
29.14
cm
mol
para
el
líquido
saturado
by
6.5%.
−1 for saturated liquid
by 6.5%.
mol
If
we
start
with
the experimental
experimental
value
ofthe
29.14
cm33 mol
−1=for
atwe
310
K, la
Eq.
(3.75)
may
be Para
used.
saturated
liquid
at
T
=
0.764,
ρ
K, se puede
usar
ecuación
(3.75).
elFor
líquido
saturado
a
T
0.764,
r
=
2.34
r
r(de
saturated
liquid
If
start
with
the
value
of
29.14
cm
3 mol−1
r for saturated
r1 liquid
1 =la figura
If we
start
with the
experimental
value
ofsaturated
29.14 cmliquid
at 310
310
K,
Eq.
(3.75)
may
be
used. For
For
theknown
atEq.
Tr (3.75)
= 0.764,
0.764,
ρr1 =
=
2.34
Fig. 3.16).
Substitution
of
intoat
gives:
3.16).
Con
la(from
sustitución
de los
valores
conocidos
envalues
la ecuación
(3.75)
se
obtiene:
at
K,
Eq.
(3.75)
may
be
used.
the
saturated
liquid
T
=
ρ
at 310
K, Eq.
be used. of
For
the saturated
liquid
Trr =gives:
0.764, ρrr11 =
2.34
(from
Fig.(3.75)
3.16).may
Substitution
known
values
into
Eq.at
(3.75)
2.34
(from
Fig.
3.16).
Substitution
of
known
values
into
Eq.
(3.75)
gives:
�
�
2.34 (from Fig. 3.16). Substitution
of known
values into Eq. (3.75) gives:
ρr1
2.34
�
�
�
�
V2 =ρV
= 28.65 cm3 mol−1
= (29.14)
1
2.34
r
�
�
1
−1
ρ
2.34
ρ
2.38
r
r
V
=
28.65
cm333 mol
=
V
=
(29.14)
mol−1
1
2
1 ρr1 = (29.14)
2.34 = 28.65 cm
VV22 =
V
1
ρ
2.38
r
= 28.65 cm mol−1
2 = V1ρr 2 = (29.14) 2.38
ρr22 agreement
En esencia,
resultado
concuerda
con el 2.38
valorthe
experimental.
This este
result
is in essential
with
experimental value.
1 interpolados
This
result
is in
indirecta
essential
agreement
with
thecorrelation
experimental
value.
La aplicación
de la
Lee/Kesler
con los
valores
de
and
Z 1 inDirect
application
ofcorrelación
the Lee/Kesler
with
valuesdeofZ 0Z y0 Z
This
result
is
essential
agreement
with
the
experimental
value.
0 and
1 inThis
result
is
in
essential
agreement
with
the
experimental
value.
3
–1
3
−1
Z
Direct
application
of
the
Lee/Kesler
correlation
with
values
of
Z
1which
las tablas
E.1 application
y E.2
conduce
aE.1
un
valor
deleads
33.87tocm
mol
,33.87
que tiene
error
importante,
sin
terpolated
from
Tables
and
E.2
a value
ofvalues
cmZ 0un
mol
,
is
and
Z
inDirect
of
the
Lee/Kesler
correlation
with
of
0 and Z 1 in3 of
Directafrom
application
of
the
Lee/Kesler
values
Z −1
terpolated
Tables
E.1
and
E.2owing
leads
to
value
ofwith
33.87
cm
mol
which is
is
3 mol
duda,
debido
la naturaleza
altamente
polar correlation
del
amoniaco.
significantly
in error,
noand
doubt
toaathe
highly
polar cm
nature
of−1
ammonia.
terpolated
from
Tables
E.1
E.2
leads
to
value
of
33.87
,,,which
3
−1
terpolated from
Tables
E.1
andowing
E.2 leads
to highly
a valuepolar
of 33.87
cmofmol
which is
significantly
in
error,
no
doubt
to
the
nature
ammonia.
significantly
significantlyin
inerror,
error,no
nodoubt
doubtowing
owingto
tothe
thehighly
highlypolar
polarnature
natureof
ofammonia.
ammonia.
PROBLEMAS
PROBLEMS
PROBLEMS
PROBLEMS
PROBLEMS
3.1. Exprese
la expansión
volumétrica
y la and
compresibilidad
isotérmica
como funciones
de laof
densidad
3.1. Express
the volume
expansivity
the isothermal
compressibility
as functions
–6
–1. ¿A
◦k
−6 presión
3.1.
Express
the
volume
expansivity
and
the
isothermal
compressibility
as
functions
of
r
y
sus
derivadas
parciales.
Para
el
agua
a
50
°C
y
1
bar,
=
44.18
×
10
bar
qué
density
ρ
and
its
partial
derivatives.
For
water
at
50
C
and
1
bar,
κ
=
44.18
×
10
3.1.
the
volume
expansivity
and
the
isothermal
compressibility as
functions
of
−6
3.1. Express
Express−1
the
volume
expansivity
andFor
thewater
isothermal
as44.18
functions
of
◦bar,
density
and
its
partial
derivatives.
at 50
50◦◦C
Ccompressibility
and
1 bar,
=
×que
10−6
debe
comprimirse
el
agua
a 50must
°C para
que
densidad
cambie
Suponga
k−6esbyindepenbar ρρ .and
To its
what
pressure
water
besucompressed
at 50
C1%?
toκκ change
its×
density
density
partial
derivatives.
For
water
at
and
1
=
44.18
10
◦
−1
◦
density
ρ
and
its
partial
derivatives.
For
water
at
50
C
and
1
bar,
κ
=
44.18
×
10
bar−1
To
what pressure
pressure
must
water be
beofcompressed
compressed
at 50◦ C
C to change
change its density
density by
diente
de
P.
1%?
Assume
that κ ismust
independent
P.
bar
... To
bar−1Assume
To what
what
pressure
must water
water
be
compressed atat 50
50◦ C to
to change its
its density by
by
1%?
that
κ
is
independent
of
P.
1%?
1%? Assume
Assumethat
thatκκ isisindependent
independentof
of P.
P.
3.2. Por
general, elvolume
coeficiente
de expansión
b y la compresibilidad
isotérmica
k depen3.2.loGenerally,
expansivity
β andvolumétrica
isothermal compressibility
κ depend
on T and
3.2.
Generally,
volume
expansivity
β
and
isothermal
compressibility
κ
depend
on
T
and
den
de
T
y
P.
Demuestre
que:
P.
Prove
that:
3.2.
volume
compressibility
κκ depend
� isothermal
�compressibility
�
�
3.2. Generally,
Generally,
volume expansivity
expansivity ββ and
and
isothermal
depend on
on TT and
and
P.
Prove that:
that:
∂κ
� �
�
�∂β
�
P.
Prove
�
�
�
=
−
P. Prove that:
∂β
∂κ
�∂β ∂�P
�
�
=T−
− ∂κ
∂β
∂κ ∂ T P
=
∂
P
∂
T
=
−
T
∂∂PP T
∂∂TT PP
T
P
3.3. The Tait equation for liquids is written
for an isotherm
as:
3.3.
The
Tait
equation
for
liquids
is
written
for
an
isotherm
as:
3.3.
Para
una
isoterma,
la
ecuación
de
Tait
para
líquidos
se
escribe
3.3.
as:� como:
� isotherm
3.3. The
TheTait
Taitequation
equationfor
forliquids
liquidsisiswritten
writtenfor
foran
an
isotherm
A�
P as:
�
��
V =��V0 1 −
AP
PB +
A
P
V
=
V
1
−
0
A
P
VV =
B+
+P
P
= VV00 11−
−B
P hypothetical molar or specific volwhere V is molar or specific volume, V0Bis+the
is
the
hypothetical
molaranor
orexpression
specific volvolwhere
V
is
molar
or
specific
volume,
V
0 is
ume
atel
pressure,
and
A and B V
are
positive
constants.
Find
for the
the
hypothetical
molar
where
V
isiszero
molar
specific
volume,
donde
V
es
volumen
o and
específico
yV0V
elconstants.
volumen
molar
o específico
hipotético
hypothetical
molar
or specific
specific
volwhere
V
molar or
ormolar
specific
volume,
0 0isesthe
ume
at
zero
pressure,
and
A
B
are
positive
Find
an
expression
for
the a presión
isothermal
compressibility
consistent
with
this
equation.
ume
at
zero
pressure,
and
A
and
B
are
positive
constants.
Find
an
expression
for
the
cero,
que A y B and
son constantes
positivas.
Encuentre una
para laforcompresibilidad
umemientras
at zerocompressibility
pressure,
A
and B are
positive
constants.
Findexpresión
an expression
the
isothermal
consistent
with
this equation.
equation.
isothermal
compressibility
consistent
with
this
isotérmica
que
sea
consistente
con
esta
ecuación.
isothermal compressibility consistent with this equation.
3.4. For liquid water the isothermal compressibility is given by:
3.4.
For liquid
liquid water
water the
the isothermal
isothermal compressibility
compressibility is
is given
given by:
by:
3.4.
For
3.4.
agua water
líquidathe
la isothermal
compresibilidad
isotérmicaisces
conocida
3.4.Para
Forelliquid
compressibility
given
by: por:
κ= c
cc(P + b)
V
=
κκκ =
V (P
(P +
+ b)
=V
V (P +b)
b)only. If 1 kg of water is compressed
where c and b are functions of temperature
where
c
and
b
are
functions
of
temperature
only.
If6011◦ C,
kghow
of water
water
iswork
compressed
muchis
is required?
isothermally
and
reversibly
from
1
to
500
bar
atIf
where
cyc band
bbúnicamente
are
functions
of
temperature
only.
◦
donde
c
son
funciones
de
la
temperatura.
Sikg
seof
comprime
kg
de agua de manera
where
and
are
functions
of
temperature
only.
If 1how
kg
of
water
is1compressed
compressed
◦
3
−1
C,
much
work
is
required?
isothermally
and
reversibly
from
1
to
500
bar
at
60
◦
At
60
C,
b
=
2,
700
bar
and
c
=
0.125
cm
g
.
how
much
work
isrequiere?
required?
isothermally
and
reversibly
from
11 to
500
bar
at
60
◦C,
◦
3
−1
isotérmica
y
reversible
desde
1
hasta
500
bar
a
60
°C,
¿cuánto
trabajo
se
A 60 °C, b =
C,
how
much
work
is
required?
isothermally
and
reversibly
from
to
500
bar
at
60
At 60
60◦ C,
C, bb =
= 2,
2, 700
700 bar
bar and
and cc =
= 0.125 cm
cm3 g−1 .
◦ C,
3 gand
–1. c = 0.125
2 At
700
bar
y bc ==0.125
At 60
2, 700cm
bar
0.125 cm3 gg−1..
03-SmithVanNess.indd 111
8/1/07 13:01:01
112112
CAPÍTULO
3. Propiedades
volumétricas
de fluidos puros
CHAPTER
3. Volumetric
Properties
of Pure Fluids
3.5. Calcule el trabajo reversible hecho al comprimir 1(pie)3 de mercurio
a una temperatura constante de
3 of mercury
3.5.32(°F),
Calculate
theuna
reversible
work
done
in compressing
1(ft)
atisotérmica
a constantdel mercurio
desde
presión
de
1(atm)
hasta
3
000
(atm).
La
compresibilidad
◦ F) from 1(atm) to 3,000(atm). The isothermal compressibility
temperature
of
32(
of
a 32(°F) es:
mercury at 32(◦ F) is:
κ/(atm)−1 = 3.9 × 10−6 − 0.1 × 10−9 P(atm)
3.6.
tetracloruro
carbono líquido
se asometen
a un cambio
de estado
3.6.Cinco
Five kilogramos
kilograms ofdeliquid
carbon de
tetrachloride
undergo
mechanically
reversible,
iso- isobárico,
◦
mecánicamente
reversible
a
1
bar,
durante
el
cual
la
temperatura
varía
de
0
a
20
°C.
Determine
DV t,
baric change of state at 1 bar during which the temperature changes from 0 C to 20◦ C.
t
t
t , and
t . The properties
W,Determine
Q, DH y �V
DUt ,. W
Suponga
las �U
siguientes
propiedades
tetracloruro
de carbono líquido a 1
, Q, �Hque
fordel
liquid
carbon tetrachloride
–3 K–1, C = 0.84−3
◦
−1–1, K–1, y r =
bar
y
0
°C
son
independientes
de
la
temperatura:
b
=
1.2
×
10
P
at 1 bar and 0 C may be assumed independent of temperature: β = 1.2 × 10 kJKkg
–3
−1 K−1 , and ρ = 1,590 kg m−3 .
1C
590
kg m .
P = 0.84 kJ kg
3.7.
sustancia for
para
la que
una constante,
se an
somete
a un proceso
isotérmico,
mecánicamente
3.7.Una
A substance
which
κ isk aes
constant
undergoes
isothermal,
mechanically
reversible
reversible
del
estado
inicial
(P
,
V
)
al
estado
final
(P
,
V
),
donde
V
es
el
volumen
1
1
2
2
process from initial state (P1 , V1 ) to final state (P2 , V2 ), where V is molar volume.molar.
the definition
κ, show que
that la
thetrayectoria
path of thedel
process
is described
by: por:
a) (a)A Starting
partir dewith
la definición
de k of
demuestre
proceso
está descrita
VV ==A(T)exp(-k
P) P)
A(T ) exp(−κ
b) (b)Determine
una
exacta which
que proporcione
el trabajo work
isotérmico
hecho
Determine
an expresión
exact expression
gives the isothermal
done on
1 molsobre
of 1 mol de
esta
sustancia
con
k
constante.
this constant-κ substance.
3.8.
molmole
de gas
congas
CP with
= (7/2)R
y C(7/2)R
de Pexpands
y T1 =
V = (5/2)R
1 = 8 barfrom
and se
C Vexpande
= (5/2)R
P1600
= K a P2 = 1
3.8.Un
One
of ideal
an ideal
CP =
bar
por
cada
una
de
las
trayectorias
siguientes:
8 bar and T1 = 600 K to P2 = 1 bar by each of the following paths:
Constant
volume;b)(b)
Constant temperature;
Adiabatically.
a)(a)
Volumen
constante;
temperatura
constante; c) (c)
en forma
adiabática.
Assuming mechanical
reversibility,
, Q,
, and
�Hcada
for proceso.
each process.
Suponiendo
reversibilidad
mecánica, calculate
calcule W,WQ,
DU�U
y DH
para
Dibuje cada traSketch
each
path
on
a
single
P
V
diagram.
yectoria en un solo diagrama PV.
3.9.Un
Angas
ideal
gasinicialmente
initially at 600
K and
undergoes
a four-step
reversible
3.9.
ideal,
a 600
K y1010bar
bar,
se somete
a un ciclomechanically
mecánicamente
reversible de cuacycle
in
a
closed
system.
In
step
12,
pressure
decreases
isothermally
to
3
bar;
in step
tro etapas en un sistema cerrado. En la etapa 12, la presión decrece isotérmicamente
a 3 bar; en la
23, pressure
decreases
at constant
volume
to 2 bar;
step
volume
decreases
etapa
23, la presión
disminuye
a volumen
constante
a 2 in
bar;
en 34,
la etapa
34, el
volumenatdisminuye a
constant
pressure;y en
andlainetapa
step 41,
41,elthe
gas
returnsenadiabatically
to itsainitial
state.inicial.
Take Considere:
presión
constante,
gas
regresa
forma adiabática
su estado
=
(7/2)R
and
C
=
(5/2)R.
C
P
V
CP = (7/2)R y CV = (5/2)R.
(a) Sketch the cycle on a P V diagram.
a) (b)Dibuje
el ciclo
en ununknown)
diagrama PV.
Determine
(where
both T and P for states 1, 2, 3, and 4.
b) Determine (donde no se tienen ) tanto a T como a P para los estados 1, 2, 3 y 4.
(c) Calculate Q, W , �U , and �H for each step of the cycle.
c) Calcule Q, W, DU y DH para cada etapa del ciclo.
3.10. An ideal gas, C P = (5/2)R and C V = (3/2)R, is changed from tP = 1 3bar and
3.10. Un gas
ideal,3con CP = (5/2)R y CV =
(3/2)R, cambia de P = 1 bar y V1 = 12 m a P2 = 12 bar y V2t
V1t =
12 m to P2 = 12 bar and V2t = 1 m3 by the following mechanically
reversible
3
= 1 m mediante los procesos mecánicamente reversibles siguientes:
processes:
a) (a)Compresión
Isothermalisotérmica.
compression.
b) (b)Compresión
adiabática seguida
porby
enfriamiento
a presiónpressure.
constante.
Adiabatic compression
followed
cooling at constant
c) (c)Compresión
adiabática
seguida
por
enfriamiento
a
volumen
constante.
Adiabatic compression followed by cooling at constant volume.
d) (d)Calentamiento
a
volumen
constante
seguido
por
enfriamiento
a presión constante.
Heating at constant volume followed by cooling at constant pressure.
03-SmithVanNess.indd 112
8/1/07 13:01:07
Problemas
113
Problems
113
113
Problems
e) Enfriamiento a presión constante seguido por calentamiento a volumen constante.
(e) Cooling at constant pressure followed by heating at constant volume. Problems
(e) Cooling
constant
followed
at constant
t y DHpressure
t para cada
Calcule
Q, W,atDU
uno deby
losheating
procesos,
y dibujevolume.
las trayectorias de todos los prot , and �H t for each of these processes, and sketch the paths of
Calculate
Q,
W
,
�U
cesos en un solo diagrama
PV.
t , and �H t for each of these processes, and sketch the paths of
Calculate
Q, Won, �U
all processes
a single
P V diagram.
(e) Cooling at constant pressur
all processes on a single P V diagram.
3.11. El gradiente térmico ambiental dT/dz caracteriza la variación local de la temperatura con la altura
Calculate Q, W , �U t , and �H
3.11.enThe
environmental
lapseLarate
dT /dz
characterizes
local
variation de
of acuerdo
temperature
la atmósfera
terrestre.
presión
atmosférica
varíathe
con
la elevación
con la fórmula
all processes on a single P V di
3.11. The
lapse
rate atmosphere.
dT /dz characterizes
the local
variation
temperature
withenvironmental
elevation in the
earth’s
Atmospheric
pressure
variesofwith
elevation
hidrostática.
with
elevation
in
the
earth’s
atmosphere.
Atmospheric
pressure
varies
with
elevation
according to the hydrostatic formula,
according to the hydrostatic formula,
3.11. The environmental lapse rate d
dP
with elevation in the earth’s atm
= −Mρg
dP
according to the hydrostatic for
dz = −Mρg
dz
where M is molar mass, ρ is molar density, and g is the local acceleration of gravity.
where
M isthat
mass,
ρ is molar
density,
gT
is related
the
oflocal
gravity.
donde
esmolar
lathe
masa
molar,
rises
densidad
molar
y glocal
estolaacceleration
de
Asssume
atmosphere
anlaideal
gas,and
with
Paceleración
by the polytropic
for-la gravedad.
Asssume
that
the
atmosphere
is
an
ideal
gas,
with
T
related
to
P
by
the
polytropic
forSuponga
que
la
atmósfera
es
un
gas
ideal,
con
T
relacionada
a
P
por
la
fórmula
politrópica,
ecuamula, Eq. (3.35c). Develop an expression for the environmental lapse rate in relation
mula,
Eq.
(3.35c).
Develop
an
expression
for
the
environmental
lapse
rate
in
relation
where
M
ción
(3.35c).
Desarrolle
una
expresión
para
el
gradiente
térmico
ambiental
en
relación
con
,isg,molar mass, ρ is m
to M, g, R, and δ.
Asssume that the atmosphere is
RtoyM,
d. g, R, and δ.
mula, Eq. (3.35c). Develop an
3.12. An evacuated tank is filled with gas from a constant-pressure line. Develop an expres�
to M,
g, R, and δ.
3.12.
Se
llena
con
gas
un
tanque
previamente
vacío
mediante
una
línea
a
presión
constante.
Desarrolle
3.12. An
evacuated
is filled withofgas
a constant-pressure
line. Develop
anthe
expresgas in
sion
relating tank
the temperature
thefrom
gas in
the tank to the temperature
T of
�
una
expresión
que
relacione
la
temperatura
del
gas
en
el
tanque
con
la
temperatura
T
′
del
gas
en la
the transfer
gas in
sion
the temperature
of the
gasconstant
in the tank
the temperature
T ofheat
the relating
line. Assume
the gas is ideal
with
heattocapacities,
and ignore
línea.
Suponga
que
el
gas
es
ideal
con
capacidades
caloríficas
constantes
e
ignore
la
transferencia
the
line.
Assume
the
gas
is
ideal
with
constant
heat
capacities,
and
ignore
heat
transfer
An evacuated tank is filled with
between the gas and the tank. Mass and energy balances for this problem are3.12.
treated
de
entre
el gas
el tanque.
Los and
balances
de balances
masa y energía
este problema
consideran
between
the gas
andythe
tank. Mass
energy
for thispara
problem
are treatedse
sion
relating the temperature of
incalor
Ex. 2.13.
en
in el
Ex.ejemplo
2.13. 2.13.
the line. Assume the gas is idea
between the gas and the tank. M
3.13. Show how Eqs. (3.36) and (3.37) reduce to the appropriate expressions for the four
3.13.
Demuestre
cómo
las
ecuaciones
(3.36)
y
(3.37)
se
reducen
a
las
expresiones
apropiadas
los
2.13.
3.13. Show
how values
Eqs. (3.36)
and (3.37)
reduce
the appropriate expressions for the fourin Ex.para
particular
of δ listed
following
Eq.to(3.37).
cuatro
valores
particulares
de
d
enumerados
después
de
la
ecuación
(3.37).
particular values of δ listed following Eq. (3.37).
3.13. Show how Eqs. (3.36) and (3.3
3.14. A tank of 0.1-m3 volume contains air at3 25◦ C and 101.33 kPa. The tank is connected
3 volume
3.14.
tanque
contiene
un volumen
de 0.1
m 25de◦ Caire
a 101.33
25 °C ykPa.
101.33
kPa.
Elistanque
estáparticular
conectadovalues
a
of δ listed follo
contains
air
at
and
The
tank
connected
3.14. Un
Atotank
of
0.1-m
a compressed-air line which supplies air at the constant conditions of 45◦ C and
◦
una
línea
de
aire
comprimido
que
proporciona
aire
en
condiciones
constantes
de
45
°C
y
1
500
kPa.
C
and
to1,500
a compressed-air
line
which
supplies
air
at
the
constant
conditions
of
45
kPa. A valve in the line is cracked so that air flows slowly into the tank until
Una
válvula
en valve
la línea se
que elso
aire fluye
bastante
lentitud
dentro
del tanque
1,500
kPa. A
theagrieta,
linepressure.
is así
cracked
air con
flows
slowly
intoenough
thehacia
tank
until
3.14.
A tank
of 0.1-m3 volume conta
the pressure
equalsinthe
line
If thethat
process
occurs
slowly
that
the
hasta
que
la
presión
se
equilibra
con
la
presión
de
la
línea.
Si
el
proceso
ocurre
muy
lentamente,
de
the
pressure equals
the line
pressure.
If how
the process
occurs
enough
theto a compressed-air
line which
temperature
in the tank
remains
at 25◦ C,
much heat
is lostslowly
from the
tank?that
Assume
◦
tal
manera
que
la
temperatura
se
mantiene
a
25
°C,
¿cuánto
calor
se
pierde
desde
el
tanque?
Supontemperature
in
the
tank
remains
at
25
C,
how
much
heat
is
lost
from
the
tank?
Assume
1,500
kPa.
A
valve
in
the line
air to be an ideal gas for which C P = (7/2)R and C V = (5/2)R.
ga
el an
aireideal
es un
gas
para
que
CP = (7/2)R
CV(5/2)R.
= (5/2)R.
(7/2)R
and C Vy =
airque
to be
gas
forideal
which
C Pel=
the pressure equals the line pre
in the tank remains
3.15. Gas at constant T and P is contained in a supply line connected through a valve to temperature
a
3.15.
aatTconstant
y P constantes
está
contenido
en
una
línea line
de suministro
conectada,
mediante
unatoválvula,
air
be
an
ideal
gas for which
3.15. Gas
Gas
T
and
P
is
contained
in
a
supply
connected
through
a
valve
to
a
closed tank containing the same gas at a lower pressure. The valve is opened to allow
aclosed
un tanque
cerrado
que the
contiene
el
mismo
gas apressure.
una presión
menor.
Laopened
válvulatoseallow
abre permitiendo
tank
containing
same
gas
at
a
lower
The
valve
is
flow of gas into the tank, and then is shut again.
que
dentro
tanque
después se cierra una vez más.
flowelofgas
gasfluya
into hacia
the tank,
anddel
then
is shuty again.
3.15. Gas at constant T and P is con
(a) Develop a general equation relating n 1 and n 2 , the moles (or mass) of gas in the
closed tank containing the same
una
ecuación
general
y nmoles
moles
(oUmasas)
de
gas
en eloftanque
a)(a) Desarrolle
1to
2, las
na2 ,nthe
(or mass)
of
gas
in
the
Develop
general
equation
n 1 and
tank at athe
beginning
and relating
endque
of relacione
the
process,
the
properties
and
U
,
the
flow
gas into the tank, and th
1
2
altank
principio
y
al
final
del
proceso,
con
las
propiedades
U
y
U
,
la
energía
interna
1 end2ofUthe
at the
beginning
the at
process,
to the properties
U2 , the
1 and
internal
energy
of theand
gas end
in theoftank
the beginning
and
process,
and del gas al
principio
yenthalpy
al final
proceso,
H ′,supply
laatentalpía
del gas
enthe
la
línea
de suministro
y(a)
a Q,Develop
el calor a general equation
internal
of del
thethe
gasgas
in in
theythe
tank
theline,
beginning
and
end
of the
process,toand
H � , theenergy
of
and
to Q,
heat
transferred
the
�
transferido
a
la
sustancia
en
el
tanque
durante
el
proceso.
Hmaterial
, the enthalpy
of the
gas inthe
theprocess.
supply line, and to Q, the heat transferred to the tank at the beginning and
in the tank
during
b) Reduzca
la
ecuación
generalthe
a su
forma más simple para el caso especial de un gas ideal conenergy of the gas in
material
in
the
tank
during
process.
(b) Reduce the general equation to its simplest form for the special case of an ideal internal
capacidades
caloríficas
constantes.
H � , the enthalpy of the gas
(b) Reduce
theconstant
generalheat
equation
to its simplest form for the special case of an ideal
gas with
capacities.
c) Además,
reduzca
la
ecuación
de
b)
para
el
caso
n
=
0.
1
material in the tank during
with constant
heat
capacities.
(c)gas
Further
reduce the
equation
of (b) for the case of
n 1 = 0.
d)(c) Después
reduzcathe
la equation
ecuación of
de(b)
c) para
el caso
en
que,
también
Q = 0.
(b)
Reduce
the general equatio
Further
reduce
for
the
case
of
n
=
0.
1
(d) Further reduce the equation of (c) for the case in which,
in addition, Q = 0.
gas
with
constant heat capa
(d) Further reduce the equation of (c) for the case in which, in addition, Q = 0.
(c) Further reduce the equation
(d) Further reduce the equation
03-SmithVanNess.indd 113
8/1/07 13:01:17
114
CAPÍTULO 3. Propiedades volumétricas de fluidos puros
e)
Considerando al nitrógeno como un gas ideal para el que CP = (7/2)R, aplique la ecuación apropiada para el que una alimentación estable de nitrógeno a 25 °C y 3 bar fluye hacia dentro de un
tanque vacío de 4 m3 de volumen, y calcule las moles de nitrógeno que fluyen hacia dentro del
tanque para igualar las presiones para dos casos:
1) Suponga que no fluye calor del gas al tanque o a través de las paredes del mismo.
2) El tanque pesa 400 kg, está perfectamente aislado, tiene una temperatura inicial de 25 °C,
cuenta con calor específico de 0.46 kJ kg–1 K–1 y es calentado por el gas, por lo que siempre
estará a la temperatura del gas en el tanque.
3.16. Desarrolle ecuaciones para determinar la temperatura final del gas que permanece en un tanque,
después de que se extrajo gas desde una presión inicial P1 hasta una presión final P2. Las cantidades
conocidas son la temperatura inicial, el volumen del tanque, la capacidad calorífica del gas, la capacidad calorífica total del contenido del tanque, P1 y P2. Suponga que el tanque siempre está a la
temperatura del gas que permanece en éste y que está perfectamente aislado.
3.17. Un tanque rígido no conductor, con un volumen de 4 m3, se divide en dos partes desiguales separadas por una membrana delgada. Un lado de la membrana, que representa 1/3 del tanque, contiene
gas nitrógeno a 6 bar y 100 °C, y el otro lado, que representa 2/3 del tanque, está vacío. La membrana se rompe y el gas llena el tanque.
a) ¿Cuál es la temperatura final del gas? ¿Cuánto trabajo se hace? ¿El proceso es reversible?
b) Describa un proceso reversible mediante el cual el gas pueda regresar a su estado inicial. ¿Cuánto trabajo se hace?
Suponga que el nitrógeno es un gas ideal para el que CP = (7/2)R y CV = (5/2)R.
3.18. Un gas ideal, inicialmente a 30 °C y 100 kPa, experimenta los siguientes procesos cíclicos en un
sistema cerrado:
a) En un proceso mecánicamente reversible, primero hay una compresión adiabática a 500 kPa, a
continuación un enfriamiento a presión constante de 500 kPa hasta 30 °C, y al final una expansión isotérmica hasta su estado original.
b) El ciclo experimenta los mismos cambios de estado, pero cada etapa es irreversible con una
eficiencia de 80% en comparación con la del correspondiente proceso mecánicamente reversible. Nota: la etapa inicial puede no ser más larga que el adiabático.
Calcule Q, W, DU y DH para cada etapa del proceso y para todo el ciclo. Considere CP = (7/2)R y
CV = (5/2)R.
3.19. Un metro cúbico de un gas ideal a 600 K y 1 000 kPa se expande hasta alcanzar cinco veces su
volumen inicial de la siguiente manera:
a) Mediante un proceso isotérmico mecánicamente reversible.
b) Mediante un proceso adiabático mecánicamente reversible.
03-SmithVanNess.indd 114
8/1/07 13:01:17
Problemas
115
c)
Mediante un proceso adiabático, irreversible, en el cual la expansión se hace contra una presión
restringente de 100 kPa.
Para cada caso calcule la temperatura y presión final y el trabajo realizado por el gas. CP = 21 J
mol–1 K–1.
3.20. Un mol de aire, inicialmente a 150 °C y 8 bar, se somete a los cambios mecánicamente reversibles
siguientes. Se expande isotérmicamente a una presión tal que cuando se enfría hasta 50 °C a volumen constante, su presión final es de 3 bar. Suponga que el aire es un gas ideal para el que CP =
(7/2)R y CV = (5/2)R. Calcule W, Q, DU y DH.
3.21. Un gas ideal fluye en estado estacionario por un tubo horizontal. No se añade calor ni se hace trabajo de flecha. El área de la sección transversal del tubo cambia con la longitud, lo cual hace que
cambie la velocidad. Deduzca una ecuación que relacione la temperatura con la velocidad del gas.
Si por una sección del tubo pasa nitrógeno a 150 °C a una velocidad de 2.5 m s–1, ¿cuál es su temperatura en otra sección donde su velocidad es de 50 m s–1? Sea CP = (7/2)R.
3.22. Un mol de gas ideal, inicialmente a 30 °C y 1 bar, cambia a 130 °C y 10 bar mediante tres distintos
procesos mecánicamente reversibles:
a) El gas se calienta primero a volumen constante hasta que su temperatura es de 130 °C; a continuación, se comprime isotérmicamente hasta que su presión es de 10 bar.
b) El gas se calienta primero a presión constante hasta que su temperatura es de 130 °C; a con­
tinuación, se comprime isotérmicamente hasta 10 bar.
c) El gas se comprime primero isotérmicamente hasta 10 bar; a continuación, se calienta a presión
constante hasta 130 °C.
Calcule Q, W, DU y DH en cada caso. Considere CP = (7/2)R y CV = (5/2)R. En otro caso, considere CP = (5/2)R y CV = (3/2)R.
3.23. Un mol de gas ideal, inicialmente a 30 °C y 1 bar, experimenta los cambios mecánicamente reversibles siguientes. Se comprime de manera isotérmica hasta un punto tal que cuando se calienta a
volumen constante hasta 120 °C, su presión final es de 12 bar. Calcule Q, W, DU y DH para el proceso. Tome CP = (7/2)R y CV = (5/2)R.
3.24. Un proceso consta de dos etapas: 1) un mol de aire a T = 800 K y P = 4 bar se enfría a volumen
constante hasta T = 350 K. 2) Después el aire se calienta a presión constante hasta que su temperatura llega a 800 K. Si este proceso de dos etapas se reemplaza por una sola expansión isotérmica del
aire desde 800 K y 4 bar hasta una presión final P, ¿cuál es el valor de P que hace que el trabajo de
los dos procesos sea el mismo? Suponga reversibilidad mecánica y considere el aire como un gas
ideal con CP = (7/2)R y CV = (5/2)R.
3.25. El siguiente es un esquema para encontrar el volumen interno VBt , de un cilindro de gas. El cilindro
se llena con un gas a baja presión P1, y se conecta mediante una línea y válvula pequeñas a un tanque de referencia evacuado y de volumen conocido VAt .
03-SmithVanNess.indd 115
8/1/07 13:01:18
116
116
116
116
CHAPTER 3. Volumetric Properties of Pure Fluids
CHAPTER 3. Volumetric Properties of Pure Fluids
CHAPTER 3. Volumetric Properties of Pure Fluids
Propiedades
de fluidos puros
is opened, and gas flows through the CAPÍTULO
line into the3.reference
tank. volumétricas
After the system
is opened,
and
gas flows
througha the
line into
the reference
tank.
After athe
system
returns
to its
initial
temperature,
sensitive
pressure
transducer
provides
value
for
t value
returns
toseits
initial
temperature,
alathe
sensitive
pressure
transducer
provides
for
isválvula
opened,
and
gas
line
into
reference
tank.
After
system
pressure
change
�P
inthrough
thepor
cylinder.
Determine
the
cylinder
volume
VaBthe
from
Lathe
abre
y elflows
gas fluye
línea
hacia
elthe
tanque
de referencia.
Después
dethe
que el sistema
the
pressure
change
�Pinicial,
in the un
cylinder.
Determine
the
volume
VaBt value
fromunthe
returns
its
initial temperature,
atransductor
sensitive
pressure
transducer
provides
for
following
regresa
a to
sudata:
temperatura
sensible
acylinder
la presión
proporciona
valor para el
tV t from the
following
data:
the pressure
change
�P
in
the
cylinder.
Determine
the
cylinder
volume
cambio
de
presión
DP
en
el
cilindro.
Determine
el
volumen
del
cilindro
V
a
partir
de
la informaB B
• V At =data:
256 cmt3 .
3; b) AP/P = –0.0639.
following
ción
siguiente:
a)
V
=
256
cm
t
3
1
• V A = 256 cmA .
• �P/P
1 = −0.0639.
t
3
V A = 256
cm .
•• �P/P
1 = −0.0639.
3.26. Un cilindro
horizontal,
no conductor y cerrado, contiene un pistón flotante no conductor y sin fric•que
�P/P
3.26.ción,
A closed,
nonconducting,
horizontal
cylinderA is
fitted
nonconducting,
friction1 =
divide
al−0.0639.
cilindro en
dos secciones,
y B.
Laswith
dos asecciones
contienen
masas iguales de
3.26.aire,
A closed,
nonconducting,
horizontal
fitted
with
frictionless,
floating
piston
which
divides
thecylinder
cylinder
A 1(atm).
and B. En
Thelatwo
sec- A se activa
inicialmente
en las
mismas
condiciones,
T1is=into
300Sections
K
y Pa1 nonconducting,
=
sección
less,
floating
whichhorizontal
divides
thecylinder
Sections
Aaumenta
and TB.
The
two
3.26.untions
Aelemento
closed,
nonconducting,
isinto
fitted
with
a nonconducting,
frictioncontain
equal
masses
ofeléctrico
air, initially
at the
same
conditions,
300
K secandTA en la sec1 =
depiston
calentamiento
ycylinder
la temperatura
del
aire
lentamente:
tions
contain
equal
masses
of air,
initially
the
same
conditions,
TB.
=
300two
Ktemand adiabática
less,
which divides
the
cylinder
into
Sections
A
and
The
sec1and
1(atm).apiston
An
electrical
heating
element
in
Section
A
is
activated,
the
air
P
1 =
ción
A floating
debido
la transferencia
de calor,
y Tat
en
la
sección
B
debido
a
la
compresión
B
=
1(atm).
An
electrical
heating
element
in
Section
A
is
activated,
and
the
air
temP
tions
contain
equal
masses
of
air,
initially
at
the
same
conditions,
T
=
300
K
and
1
peratures por
slowly
increase: T Alento
in Section
A because
of heat
and
Section
1unTgas
B inideal
provocada
el movimiento
del pistón.
Considere
el transfer,
aire como
con CP = R, y
peratures
slowly
increase:
T
in
Section
A
because
of
heat
transfer,
and
T
in
Section
=
1(atm).
An
electrical
heating
element
in
Section
A
is
activated,
and
the
air
temP
A
B
B 1nbecause
of
adiabatic
compression
by
the
slowly
moving
piston.
Treat
air
as
an
ideal
sea
el
número
de
moles
de
aire
contenidos
en
la
sección
A.
Para
el
proceso
descrito,
evalúe uno
A
7
Blos
because
of
adiabatic
compression
by number
the
as
ideal
peratures
T Ansiguientes:
in
Section
A slowly
because
of heat
transfer,
and air
TBA.
in an
Section
with
Cslowly
R, and
let
the
of moving
moles
ofpiston.
air in Treat
Section
For
the
P =
A be
degas
conjuntos
de27increase:
cantidades
gas
withasCdescribed,
let n one
of moving
moles
air in Treat
Section
Forideal
the
B because
of
adiabatic
compression
by
the
slowly
piston.
air A.
as an
P =
A beofthe
process
evaluate
thenumber
following
sets ofof
quantities:
2 R, and
7
process
as
described,
evaluate
one
of
the
following
sets
of
quantities:
gas
with
C
=
R,
and
let
n
be
the
number
of
moles
of
air
in
Section
A.
For
the
P A, si
A
2 P(final) = 1.25(atm).
a) (a)TAT, T,B Ty Q/n
, and
Q/n
, if P(final)
= 1.25(atm).
A asB described,
Aevaluate
process
one
of
the
following
sets
of
quantities:
b) (a)TBT, AQ/n
P(final),
= 425 K.
AB ,yand
Q/n Asi, ifTAP(final)
= 1.25(atm).
Q/n
if T A =
425 K.
(b) TB ,, T
A , and P(final),
c) (b)
T
,
Q/n
y
P(final),
si
T
=
325
K.=
ATB , T
A , and
BP(final)
Q/n
,
and
P(final),
if
T
425 K.
K.
(a)
Q/n
,
if
= 1.25(atm).
A
A
B A , and P(final),
A
if TB =
325
(c) T AA, Q/n
–1.
d) (c)
TAT
, TB, Q/n
y P(final),
siP(final),
Q/nA = if
3 kJ
mol
,
and
T
=
325
425
(b)
A P(final), if Q/n B
A = 3 kJ K.
mol−1 .
(d) T AAB, TB , and
A
,
T
,
and
P(final),
if
Q/n
=
3
kJ
mol−1 .
(d)
T
325 K.
(c) A Q/n
B A , and P(final), if TB
A
−1 . constantes se somete a un proceso arbitrario
3.27.
Un
mol
de
un
gas
ideal
con
capacidades
caloríficas
P(final),
if Q/n
kJ mol
(d) Tmole
3.27. One
of, and
an ideal
gas with
constant
capacities
undergoes an arbitrary mechanA , TB
A = 3heat
reversible.
Muestre
que:
3.27.mecánicamente
One
mole
of
an
ideal
gas
with
constant
heat
capacities
undergoes an arbitrary mechanically reversible process. Show that:
ically
reversible
process.
Show
that:
3.27. One mole of an ideal gas with constant heat capacities undergoes an arbitrary mechan1
ically reversible process. Show that:
�U = 1 �(P V )
γ
−
1 �(P V )
�U =
11
γ−
�U =
�(P V )
3.28.Deduzca
Derive an
theelwork
of mechanically
reversible, isothermal
γ −por
1 la compresión
3.28.
unaequation
ecuaciónfor
para
trabajo
hecho
isotérmica compression
mecánicamente reversi3.28.ble
Derive
anofequation
for an
theuna
work
of mechanically
reversible,
isothermal
of
1 mol
a gas
initial
pressure
P1 Pto1 ahasta
final
pressure
P2final
whenPcompression
equation
de
1 mol
de
gas,from
desde
presión
inicial
una
presión
cuando
la ecuación de
2the
of
1 mol
a virial
gas from
an
pressure
Ptruncated
to a final
P2 when compression
the equation
3.28.estado
Derive
equation
for
theinitial
work
of(3.11)]
mechanically
reversible,
isothermal
1 truncada
of
state
is
the
expansion
[Eq.
to:a:pressure
es an
laof
expansión
virial
[ecuación
(3.11)]
of state
thea virial
expansion
[Eq.
(3.11)] Ptruncated
to:pressure P2 when the equation
1 molisof
gas from
an initial
pressure
1 to a final
�
Z
=
1
+
B
P
of state is the virial expansion [Eq. (3.11)] truncated
to:
Z = 1 + B� P
How does the result compare with the
Z corresponding
= 1 + B � P equation for an ideal gas?
How does the result compare with the corresponding equation for an ideal gas?
se compara
este
resultado
con
la corresponding
ecuación
que corresponde
ideal?
How
does
theisresult
compare
with
the
equation foral
angas
ideal
gas?
3.29.¿Cómo
A
certain
gas
described
by the
equation
of state:
3.29. A certain gas is described by the equation of
� state: �
θ
3.29.
gas segas
describe
mediante
la ecuación
de
estado:
�bstate:
3.29.Cierto
A certain
is described
byPthe
equation
V =
RT + of
− θ �P
P V = RT + �b − RT � P
θ
RT
= RT of
+ T bonly.
− For this
P gas, determine expressions
Here, b is a constant and θ is P
aV
function
RT
Here,
is a constant
and θ is a function
T only.
Forpressure
this gas,coefficient
determine (∂
expressions
for
thebisothermal
compressibility
κ andofthe
thermal
P/∂ T )V .
for
thebexpressions
isothermal
compressibility
κ and
the
thermal
pressure
coefficient
(∂
P/∂ T )V .
Here,
is
a
constant
and
θ
is
a
function
of
T
only.
For
this
gas,
determine
expressions
These
should
contain
only
T
,
P,
θ
,
dθ/dT
,
and
constants.
Aquí, b es una constante y q es una función sólo de T. Para este gas, determine las expresiones
para
These
expressions
should
contain only
T ,the
P, thermal
θ , dθ/dTpressure
, and constants.
for
the
isothermal
compressibility
κ
and
coefficient
(∂
P/∂
T )V .expresiones
la compresibilidad isotérmica
k
y
el
coeficiente
de
presión
térmica
(∂P/∂T)
.
Estas
V
◦ C the second
These
expressions
should
T and
, P, third
θ, dθ/dT
and constants.
3.30.deben
For
methyl
chloride
atP,100
virial, coefficients
are:
contener
sólo T,
q,contain
dq/dT yonly
constantes.
3.30. For methyl chloride at 100◦ C the second and third virial coefficients are:
3
−1
◦C
B = −242.5
cmthe
mol
C =virial
25,200
cm6 mol−2
3.30.Para
Forelmethyl
at 100
second
and ythird
coefficients
are:
3 el
−1
−2 son:
3.30.
clorurochloride
deB metilo
a
100
°C,
segundo
tercer
coeficientes
viriales
= −242.5 cm mol
C = 25,200 cm6 mol
B = −242.5 cm3 3mol−1
C = 25,200 cm6 mol−2
B = –242.5 cm mol–1 C = 25 200 cm6 mol–2
03-SmithVanNess.indd 116
8/1/07 13:01:33
Problems
117
Problems
117
117
Problems
Problemas
117
117
Problems
Calculate the work of mechanically reversible, isothermal compression of 1 mol of
Calculate
the work
of mechanically
reversible,
isothermal
compression
of
1 de
mol of de metimethyl
chloride
1 bar toisotérmica,
55 bar
at mecánicamente
100◦ C.
Base calculations
the
Calcule
el trabajo
defrom
compresión
reversible
deon1of
mol
Calculate
the work
of mechanically
reversible,
compression
1 following
molcloruro
of
◦ C.isothermal
Calculate
the
work
of
mechanically
reversible,
isothermal
compression
of
1
mol of
methyl
chloride
from
1
bar
to
55
bar
at
100
Base
calculations
on
the
following
◦ C. Base
forms1chloride
of
the virial
lomethyl
desde
hasta
55from
barequation:
a1100
Apoye
sus100
cálculos
en
las
formas
de
la
ecuación
virial
siguientes:
bar °C.
to 55
bar at
calculations
on
the
following
methylofchloride
1 bar to 55 bar at 100◦ C. Base calculations on the following
forms
the virialfrom
equation:
forms of the virial equation:
B
C
forms of the virial equation:
Z = 1 +B +C 2
a) (a)
(a)
Z = 1 + BBV+ CC2V
(a)(b)
ZZ
= 1 + V B+� PV+
� 2
(a)
Z ==1 1++
V +V2 C P
(b)
ZB = 1 + BV�� P +VC2�� P22 C − B 2
b)(b)
Z = 1 + B �P + C �P� 2
B � = B Z = 1 +and
(b)where
B P + C�CP=C − B22 2
�
B
B
RT
(RT
B� = B
and
C = CC −
where
− )B22)
B � = RT and
donde
y CC�� =
where
(RT
B
=
=
and
where
RT exactly the same result?
(RT )22
Why don’t both equations give
RT
(RT )
Why don’t both equations give exactly the same result?
Whyqué
don’t
both
equations
give exactly
thecon
same
result?
¿Por
no
se
obtiene
el
mismo
resultado
ambas
ecuaciones?
Why don’t both equations give exactly the same result?
3.31. Any equation of state valid for gases in the zero-pessure limit implies a full set of virial
3.31. Any
equation ofShow
state that
validthe
forsecond
gases inand
the zero-pessure
limit implies
a full setthe
of generic
virial
coefficients.
virial
implied
3.31.Cualquier
Any
equation
of state
forque
gases
in thethird
zero-pessure
implies
a fullby
setcero,
of virial
3.31.
ecuación
de valid
estado,
es válida
para
gasescoefficients
enlimit
el límite
de presión
implica un con3.31. coefficients.
Any
equation
of
state
valid
for
gases
in
the
zero-pessure
limit
implies
a
full
set
of
virial
Show
that
the
second
and
third
virial
coefficients
implied
by
the
generic
cubic
equation
of
state,
(3.42),
are:
coefficients.
Show
that
theEq.
second
and
third
virial coefficients
implied
by thecoeficientes
generic
junto
completo
de
coeficientes
viriales.
Demuestre
que
el
segundo
y
tercero
viriales
coefficients.
Show
that Eq.
the (3.42),
second are:
and third virial coefficients implied by the generic
cubic
equation
of state,
cubic equation
of state, Eq.
(3.42),cúbica
are: genérica, ecuación (3.42), son:
implícitos
en
la
ecuación
de
estado
cubic equation of state, Eq.a(T
(3.42),
(� + σ )ba(T )
) are:
C =2b2 +(� + σ )ba(T )
B = b −a(T )
)
RT ))
C = b + (� + σσ)ba(T
B = b − a(TRT
)ba(T
C = b22 + (� + RT
B = b − a(T
RT )
C =b +
B = b − RT
RT
Specialize the result forRT
B to the Redlich/Kwong equation RT
of state, express it in reSpecialize
the result
for B to the
Redlich/Kwong
equation
of state,
express it in reduced form,
and compare
numerically
with the
generalized
correlation
Specialize
the result
for B toitthe
Redlich/Kwong
equation
of state,
express itforin Bre-for
Specialize
the
result
for
B
to
the
Redlich/Kwong
equation
of
state,
express
in for
reduced
form,
and
compare
it
numerically
with
the
generalized
correlation
forit B
Adapte
el
resultado
para
B aitDiscuss
lanumerically
ecuación
estado
Redlich/Kwong,
expréselo
simple
fluids,
(3.65).
whatde
you
find.
duced
form,
andEq.
compare
with
the de
generalized
correlation
for B en
forsu forma re­
duced
form,
and
compare
it
numerically
with
the
generalized
correlation
for
B
for
simple
fluids,
Eq.
(3.65).
Discuss
what
you
find.
ducida
compárelo
numéricamente
conyou
la correlación
generalizada para B para fluidos simples,
simpleyfluids,
Eq. (3.65).
Discuss what
find.
simple fluids,
Eq.
(3.65).
Discuss
what you find.
ecuación
(3.65).
Analice
resultados.
3.32.
Calculate
Z and
V forsus
ethylene
at 25◦ C and 12 bar by the following equations:
3.32. Calculate Z and V for ethylene at 25◦ C and 12 bar by the following equations:
3.32. Calculate Z and V for ethylene at 25◦◦C and 12 bar by the following equations:
C and
12 bar
thefollowing
followingexperimental
equations: values
3.32. Calculate
Ztruncated
and V for
ethylene
at 25[Eq.
(a) The
virial
equation
(3.40)]
withbythe
3.32. Calcule
Z truncated
yV
para elvirial
etileno
a 25 ºC[Eq.
y 12(3.40)]
bar
mediante
ecuaciones
siguientes:values
(a) The
equation
with thelas
following
experimental
of truncated
virial coefficients:
(a) The
virial equation [Eq. (3.40)] with the following experimental values
(a) of
The
truncated
virial equation [Eq. (3.40)] with the following experimental values
virial
coefficients:
virial coefficients:
a) Laof
virial truncada [ecuación
(3.40)], con los valores experimentales
siguientes de los
3
6
ofecuación
virial coefficients:
mol−1
C = 7,200 cm
mol−2
B = −140 cm
3
6
−1
−2
coeficientes viriales:
C = 7,200 cm mol
B = −140 cm mol
C = 7,200 cm66 mol−2
B = −140 cm33 mol−1
−1
C = a7,200
cm
mol−2
B virial
= −140
cm3 mol
(b) The truncated
equation
[Eq.
(3.38)],
with
value
of B
–1 6 mol
–2 from the generalized
B
=
–140
cm
mol
C
=
7
200
cm
(b) The
truncated
virial equation
[Eq. (3.38)], with a value of B from the generalized
Pitzer
correlation
(3.63)].
(b) The
truncated
virial [Eq.
equation
[Eq. (3.38)], with a value of B from the generalized
(b) Pitzer
The truncated
virial
equation
[Eq. (3.38)], with a value of B from the generalized
correlation
[Eq.
(3.63)].
The correlation
Redlich/Kwong
equation.
b) (c)
LaPitzer
ecuación
virial truncada
[ecuación (3.38)], con un valor de B obtenido de la correlación ge[Eq. (3.63)].
Pitzer
correlation
[Eq.
(3.63)].
(c)neralizada
The
Redlich/Kwong
equation.equation.
de Pitzer [ecuación
The
Soave/Redlich/Kwong
(c)(d)
The
Redlich/Kwong
equation.(3.63)].
(c) The
The Soave/Redlich/Kwong
Redlich/Kwong equation.
(d)
equation.
c)(d)(e)
LaThe
ecuación
de Redlich/Kwong.
The
Peng/Robinson
equation.
Soave/Redlich/Kwong
equation.
(d)
The
Soave/Redlich/Kwong
equation.
(e)
The
Peng/Robinson
equation.
d)(e)LaThe
ecuación
de Soave/Redlich/Kwong.
Peng/Robinson
equation.
Peng/Robinson
equation.
e) (e)Calculate
LaThe
ecuación
de Peng/Robinson.
3.33.
Z and
V for ethane at 50◦ C and 15 bar by the following equations:
3.33. Calculate Z and V for ethane at 50◦◦ C and 15 bar by the following equations:
3.33. Calculate Z and V for ethane at 50 ◦C and 15 bar by the following equations:
C
15 bar
by
the
equations:
3.33.Calcule
Calculate
and
ethane
(a) The
truncated
virial
with
thefollowing
following
experimental values
3.33.
Z y ZV
paraVelfor
etano
aequation
50at°C50y [Eq.
l5 and
bar(3.40)]
con
las
ecuaciones
siguientes:
(a) The
truncated
virial equation [Eq. (3.40)] with the following experimental values
of truncated
virial coefficients:
(a) The
virial equation [Eq. (3.40)] with the following experimental values
(a) of
The
truncated
virial equation [Eq. (3.40)] with the following experimental values
virial
coefficients:
virial coefficients:
a) Laof
virial truncada [ecuación
(3.40)], con los valores experimentales
siguientes de los
3
6
ofecuación
virial coefficients:
mol−1
C = 9,650 cm
mol−2
B = −156.7 cm
3
6
coeficientes viriales:
−1
−2
C = 9,650 cm mol
B = −156.7 cm mol
C = 9,650 cm66 mol−2
B = −156.7 cm33 mol−1
−1
−2
C =a 9,650
cmB–2mol
B=
−156.7
cm3 mol
–1 (b) The truncated
virial
equation
[Eq.
(3.38)],
with
value
B = –156.7 cm mol
C = 9 650
cm6 of
mol from the generalized
(b) The
truncated
virial equation
[Eq. (3.38)], with a value of B from the generalized
Pitzer
correlation
(3.63)].
(b) The
truncated
virial [Eq.
equation
[Eq. (3.38)], with a value of B from the generalized
(b)
The
truncated
virial
equation
[Eq. (3.38)],
value
ofde
B Bfrom
the generalized
Pitzer
correlation
[Eq.
(3.63)].
b) (c)
LaPitzer
ecuación
virial truncada
[ecuación
(3.38)],with
conaun
valor
obtenido
a partir de la correlaThe correlation
Redlich/Kwong
equation.
[Eq. (3.63)].
Pitzer
correlation
[Eq.
(3.63)].
(c)(d)
The
Redlich/Kwong
equation.
ción
generalizada
de
Pitzer
[ecuación
(3.63)].
The
Soave/Redlich/Kwong
(c) The
Redlich/Kwong
equation.equation.
(c)LaThe
The
Redlich/Kwong
equation.
(d)
Soave/Redlich/Kwong
equation.
c)(d)
ecuación
de Redlich/Kwong.
(e)The
The
Peng/Robinson
equation.
Soave/Redlich/Kwong
equation.
(d)
The
Soave/Redlich/Kwong
equation.
Peng/Robinson
equation.
d) (e)LaThe
ecuación
de Soave/Redlich/Kwong.
(e) The Peng/Robinson equation.
Peng/Robinson
equation.
e) (e)LaThe
ecuación
de Peng/Robinson.
03-SmithVanNess.indd 117
8/1/07 13:01:48
CHAPTER
CHAPTER
3. 3.
Volumetric
Volumetric
Properties
Properties
of of
Pure
Pure
Fluids
Fluids
118118
118
CAPÍTULO 3. Propiedades volumétricas de fluidos puros
3.34.
3.34.
Calculate
Calculate
Z and
Z and
V V
forfor
sulfur
sulfur
hexafluoride
hexafluoride
at 75
at ◦75
C◦and
C and
1515
barbar
byby
thethe
following
following
equaequa3.34. Calcule
Z
y
V
para
el
hexafluoruro
de
azufre
a
75
°C y 15 bar mediante las ecuaciones siguientes:
tions:
tions:
The
The
truncated
truncated
virial
virial
equation
equation
[Eq.
[Eq.
(3.40)]
(3.40)]
with
with
thethe
following
following
experimental
values
values
a)(a)(a)
La
ecuación
virial
truncada
[ecuación
(3.40)],
con
los
valores experimental
experimentales
siguientes
de los
of of
virial
virial
coefficients:
coefficients:
coeficientes
viriales:
3 3 −1 −1
6 6 −2 −2
B ==
B–194
=
−194
−194
cm
mol
mol
C =C
=15
=
15,300
15,300
cm6cm
mol
mol
3 cm
–1
–2
B
cm
mol
C
300
cm
mol
The
The
truncated
truncated
virial
virial
equation
equation
[Eq.
[Eq.
(3.38)],
(3.38)],
with
with
a value
a value
of of
Bdefrom
BBfrom
thethe
generalized
generalized
b)(b)(b)
La
ecuación
virial
truncada
[ecuación
(3.38)],
con
un
valor
obtenido
de la correlación gePitzer
Pitzer
correlation
correlation
[Eq.
[Eq.
(3.63)].
(3.63)].
neralizada de Pitzer [ecuación (3.63)].
The
The
Redlich/Kwong
Redlich/Kwong
equation.
equation.
c)(c)(c)
La
ecuación
de Redlich/Kwong.
The
The
Soave/Redlich/Kwong
Soave/Redlich/Kwong
equation.
equation.
La
ecuación
de Soave/Redlich/Kwong.
d)(d)(d)
The
The
Peng/Robinson
Peng/Robinson
equation.
equation.
e)(e)(e)
La
ecuación
de Peng/Robinson.
3 mol
3 mol
−1 ,−1and
ForFor
sulfur
sulfur
hexafluoride,
hexafluoride,
Tc T=
318.7
318.7
K, K,
Pc P=
=
37.6
37.6
bar,bar,
V V= =
198198
cmcm
, and
c =
Para el hexafluoruro de azufre,
Tc = 318.7 K,
Pc c = 37.6 bar, cVc c= 198 cm3 mol–1, y w = 0.286.
ω=
ω=
0.286.
0.286.
3.35. Determine Z y V para el vapor a 250 °C
1 800 kPa a partir de lo siguiente:
◦ C◦y
3.35.
3.35.
Determine
Determine
Z and
Z and
V for
V for
steam
steam
at 250
at 250
and
C and
1,800
1,800
kPakPa
byby
thethe
following:
following:
a)(a)(a)
Mediante
la ecuación
virial
truncada
[ecuación
(3.40)]
con losexperimental
valores
experimentales
The
The
truncated
truncated
virial
virial
equation
equation
[Eq.
[Eq.
(3.40)]
(3.40)]
with
with
thethe
following
following
experimental
values
values siguientes
de
los
coeficientes
viriales:
of of
virial
virial
coefficients:
coefficients:
3 3mol
6 6mol
3 –1−1
6 –2−2 −2
−1
=
cm
C
800
cm
mol
mol
C =C
=–5
=
−5,800
−5,800
cmcm
mol
mol
BB =
B –152.5
=
−152.5
−152.5
cmcm
b)(b)(b)
Con
latruncated
ecuación
virial
truncada
[ecuación
(3.38)],
con
unofvalor
Bthe
obtenido
de la correlación
The
The
truncated
virial
virial
equation
equation
[Eq.
[Eq.
(3.38)],
(3.38)],
with
with
a value
a value
of
B from
B de
from
the
generalized
generalized
generalizada
de
Pitzer
[ecuación
(3.63)].
Pitzer
Pitzer
correlation
correlation
[Eq.
[Eq.
(3.63)].
(3.63)].
c) Mediante las tablas de vapor (apéndice F).
(c)(c)
The
The
steam
steam
tables
tables
(App.
(App.
F).F).
3.36. Con respecto a las expansiones viriales, ecuaciones (3.11) y (3.12), demuestre que:
3.36.
3.36.
With
With
respect
respect
to to
thethe
virial
virial
expansions,
expansions,
Eqs.
Eqs.
(3.11)
(3.11)
andand
(3.12),
(3.12),
show
show
that:
that:
�� ��
�� ��
∂ Z∂ Z
∂ Z∂ Z
B � B=� =
and B =
B=
and
y ∂ P∂ PT,P=0
∂ρ∂ρT,ρ=0
T,P=0
T,ρ=0
where
where
ρ1/V.
≡
1/V
1/V
. .
donde
rρ ≡≡
3.37.
3.37.
Equation
Equation
(3.12)
when
when
truncated
truncated
to four
toafour
terms
terms
accurately
accurately
represents
represents
the
volumetric
volumetric
data
data
3.37.
Cuando
la(3.12)
ecuación
(3.12)
se trunca
cuatro
términos,
describe
conthe
exactitud
la información
volu◦C
◦with:
for
for
methane
methane
gas
gas
at
0
at
0
C
with:
métrica para el gas metano a 0 °C con:
3 3 −1 −1
6 6 −2 −2
9 9 −3 −3
mol
mol–1 CCC
===
2,620
mol
D =D
=5=
5,000
5,000
cm9cm
mol
mol
B=
BB=
−53.4
cmcm
cm
3 mol
6mol
–2 D
–3
= −53.4
–53.4
2 2,620
620cm
cmcm
mol
000
cm
mol
Use
Use
these
these
data
data
to to
prepare
prepare
plot
a plot
of of
Z
vs.
Z gráfica
vs.
P for
P for
methane
at 0at◦ C
0◦from
C
0 to
0 to
200
200
bar.bar.
a)(a)(a)
Use
esta
información
para apreparar
una
de methane
Z en función
defrom
P para
metano
a 0 °C de 0 a
(b)(b)
To
To
what
what
pressures
pressures
do
do
Eqs.
Eqs.
(3.38)
(3.38)
and
and
(3.39)
(3.39)
provide
provide
good
good
approximations?
approximations?
200 bar.
b) ¿Para qué presiones las ecuaciones (3.38) y (3.39) proporcionan buenas aproximaciones?
3.38.
3.38.
Calculate
Calculate
thethe
molar
molar
volume
volume
of of
saturated
saturated
liquid
liquid
andand
thethe
molar
molar
volume
volume
of of
saturated
saturated
3.38. Calcule
elby
volumen
molar del líquido
saturado
y of
delof
volumen
molar
vapor
saturado
mediante la
vapor
vapor
by
the
the
Redlich/Kwong
Redlich/Kwong
equation
equation
forfor
oneone
the
the
following
following
anddel
and
compare
compare
results
results
ecuación
de Redlich/Kwong
para
una de lascorrelations.
condiciones
y sustancias siguientes y compare los resulwith
with
values
values
found
found
byby
suitable
suitable
generalized
generalized
correlations.
tados con los valores encontrados mediante el empleo de las correlaciones generalizadas adecuadas.
(a)(a)
Propane
Propane
at 40
at ◦40
C◦where
C where
P sat
P sat
= 13.71
= 13.71
bar.bar.
saturado
a) Propano a 40 °C donde P
= 13.71 bar.
03-SmithVanNess.indd 118
8/1/07 13:01:57
119
Problemas
b)
c)
d)
e)
f)
g)
h)
i)
j)
k)
l)
m)
n)
o)
p)
q)
r)
s)
t)
Propano a 50 °C donde P saturado = 17.16 bar.
Propano a 60 °C donde P saturado = 21.22 bar.
Propano a 70 °C donde P saturado = 25.94 bar.
n-butano a 100 °C donde P saturado = 15.41 bar.
n-butano a 110 °C donde P saturado = 18.66 bar.
n-butano a 120 °C donde P saturado = 22.38 bar.
n-butano a 130 °C donde P saturado = 26.59 bar.
Isobutano a 90 °C donde P saturado = 16.54 bar.
Isobutano a 100 °C donde P saturado = 20.03 bar.
Isobutano a 110 °C donde P saturado = 24.01 bar.
Isobutano a 120 °C donde P saturado = 28.53 bar.
Cloro a 60 °C donde P saturado = 18.21 bar.
Cloro a 70 °C donde P saturado = 22.49 bar.
Cloro a 80 °C donde P saturado = 27.43 bar.
Cloro a 90 °C donde P saturado = 33.08 bar.
Bióxido de azufre a 80 °C donde P saturado = 18.66 bar.
Dióxido de azufre a 90 °C donde P saturado = 23.31 bar.
Dióxido de azufre a 100 °C donde P saturado = 28.74 bar.
Dióxido de azufre a 110 °C donde P saturado = 35.01 bar.
3.39. Use la ecuación de Soave/Redlich/Kwong para calcular el volumen molar del líquido y del vapor
saturados, para la sustancia y condiciones conocidas por uno de los incisos del problema 3.38, y
compare sus resultados con los valores que se encuentran mediante las correlaciones generalizadas
adecuadas.
3.40. Use la ecuación de Peng/Robinson para calcular los volúmenes molares del líquido y del vapor saturados para la sustancia y condiciones dadas por uno de los incisos del problema 3.38, y compare sus
resultados con los valores que se encuentran mediante las correlaciones generalizadas adecuadas.
3.41. Calcule lo siguiente:
a) El volumen ocupado por 18 kg de etileno a 55 °C y 35 bar.
b) La masa de etileno contenida en un cilindro de 0.25 m3 a 50 °C y 115 bar.
3.42. El volumen molar de la fase vapor de un compuesto particular se reporta como
23 000 cm3 mol–1 a 300 K y 1 bar. Ninguna otra información está disponible. Sin que suponga comportamiento de un gas ideal determine una estimación razonable del volumen molar del vapor a 300
K y 5 bar.
3.43. Con una buena aproximación, ¿cuál es el volumen molar del vapor de etanol a 480 °C y 6 000 kPa?
¿Cómo se compara este resultado con el valor que corresponde a un gas ideal?
3.44. Se utiliza un recipiente de 0.35 m3 para almacenar propano líquido a su presión de vapor. Las consideraciones de seguridad dictan que a una temperatura de 320 K el líquido no debe ocupar más de
80% del volumen total del recipiente. Bajo estas condiciones determine las masas de vapor y de líquido dentro del recipiente. A 320 K la presión de vapor del propano es 16.0 bar.
03-SmithVanNess.indd 119
8/1/07 13:01:58
120
CAPÍTULO 3. Propiedades volumétricas de fluidos puros
3.45. Un tanque de 30 m3 contiene 14 m3 de n-butano líquido en equilibrio con su vapor a 25 °C. Estime
la masa de vapor del n-butano contenida en el tanque. La presión de vapor del n-butano a la temperatura dada es 2.43 bar.
3.46. Calcule:
a) La masa de etano contenida en un recipiente de 0.15 m3 a 60 °C y 14 000 kPa.
b) La temperatura a la que 40 kg de etano, almacenados en un recipiente de 0.15 m3, ejercen una
presión de 20 000 kPa.
3.47. ¿A qué presión debe llenarse un recipiente de 0.15 m3 a 25 °C para guardar 40 kg de etileno?
3.48. Si 15 kg de H2o se calientan a 400 °C en un recipiente de 0.4 m3, ¿que presión se genera?
3.49. A un recipiente de 0.35 m3 contiene vapor de etano a 25 °C y 2 200 kPa. Si el recipiente se calienta
a 220 °C, ¿qué presión se genera dentro de él?
3.50. ¿Cuál es la presión en un recipiente de 0.5 m3 cuando se carga con 10 kg de dióxido de carbono a
30 °C?
3.51. A un recipiente rígido, lleno a la mitad de su volumen con nitrógeno líquido en su punto normal de
ebullición, se le permite calentarse a 25 °C. ¿Qué presión se desarrolla? El volumen molar del nitrógeno líquido en su punto normal de ebullición es 34.7 cm3 mol–1.
3.52. El volumen específico del isobutano líquido a 300 K y 4 bar es 1.824 cm3 g–1. Calcule el volumen
específico a 415 K y 75 bar.
3.53. La densidad del n-pentano líquido es 0.630 g cm–3 a 18 °C y 1 bar. Estime su densidad a 140 °C y
120 bar.
3.54. Calcule la densidad del etanol líquido a 180 °C y 200 bar.
3.55. Estime el cambio de volumen en la vaporización del amoniaco a 20 °C. A esta temperatura, la presión de vapor del amoniaco es 857 kPa.
3.56. La información PVT es posible adquirirse mediante el procedimiento siguiente. La masa m de una
sustancia de masa molar M se introduce en un recipiente de volumen total V t conocido y térmicamente controlado. Al sistema se le permite alcanzar el equilibrio, y se miden la temperatura T y la
presión P.
a) Aproximadamente, ¿qué porcentajes de error se pueden permitir en las variables observadas (m,
M, V t y P) si el error máximo permisible en el factor de compresibilidad Z es ±1%?
b) Aproximadamente, ¿qué porcentajes de error se permiten en las variables observadas si el error
máximo permisible en los valores calculados del segundo coeficiente virial B es ±1%? Suponga
que Z 0.9 y que los valores de B se calculan mediante la ecuación (3.39).
03-SmithVanNess.indd 120
8/1/07 13:01:58
Problems
Problems
Problemas
121
121
121
3.57.
gasdescribed
descrito por
de Redlich/Kwong
y para
temperatura
mayor
que Tc, desa3.57.
For
aa gas
by
the
Redlich/Kwong
equation
for
temperature
greater
than
3.57.Para
Forun
gas
described
byla
theecuación
Redlich/Kwong
equation and
and
for aauna
temperature
greater
than
TTcc,, develop
expressions
two
slopes,
rrolle
expresiones
para lasfor
dosthe
pendientes
límite,
develop
expressions
for
the
two limiting
limiting
slopes,
�
�
� �
� �
�
�
∂∂ZZ
∂∂ZZ
lim
lim
lím
lím
lim
lim
P→∞
P→0
P→∞ ∂∂ P
P→0 ∂∂ P
P TT
P TT
Observe
quein
ellimit
límiteas
0,Vand
→ ∞,
el límite
conforme
P→
Note
the
PP →
that
the
limit
as
∞,
b.
Note that
that
inen
the
limit
asconforme
→ 0,
0, VVP →
→ ∞,
∞,
and
thatyin
inque
theen
limit
as PP →
→
∞, VV →
→
b. ∞, V → b.
metano
60(°F)
yF)1(atm)
es equivalente
a 1(gal)
de gasolina
comoas
3.58.
3.58.
140(ft)
of
methane
gas
and
is
to
of
3.58.SiIf
If140(pie)
140(ft)333 de
of gas
methane
gasa at
at 60(
60(◦◦F)
and 1(atm)
1(atm)
is equivalent
equivalent
to 1(gal)
1(gal)
of gasoline
gasoline
ascombustible
para
de automóvil,
¿cuál
sería
el volumen
del tanque
necesitato
fuel
an
engine,
what
would
be
of
tank
hold
fuelunfor
formotor
an automobile
automobile
engine,
what
would
be the
the volume
volume
of the
theque
tankserequired
required
topara
holdmantener el
methane
3,000(psia)
and
F)
an
equivalent
to
metano
a 3at
000(psia)
y 60(°F)
en◦◦una
equivalente
a 10(gal)
de of
gasolina?
F) in
incantidad
an amount
amount
equivalent
to 10(gal)
10(gal)
of gasoline?
gasoline?
methane
at
3,000(psia)
and 60(
60(
121
Problems
3.59.
Determine
aa good
for
factor
hydrogen
va3.59.Determine
Determine
good
estimate
for the
the
compressibility
factor ZZ of
of saturated
saturated
hydrogen
va-de hidróge3.59.
una
buenaestimate
estimación
paracompressibility
el factor de compresibilidad
Z de vapor
saturado
at
25
and
bar.
For
an
ZZ =
por
atK
25yK
K3.213
and 3.213
3.213
bar.
For comparison,
comparison,
an experimental
experimental
value
is 0.7757.
= 0.7757.
0.7757.
nopor
a 25
bar. Por
comparación,
un valor
experimentalvalue
es Z is
=
3.57. For a gas described by the Redlich/Kwong equation and for a temperature greater than
3.60.
Boyle
is
temperature
for
3.60.LaThe
The
BoyleTtemperature
temperature
is the
the
temperature
for which:
which: slopes,
for
limiting
3.60.
temperatura
de Boyleexpressions
es
aquella
parathe
la two
que:
c , develop
�
�
� �
�
�
�
�
∂∂∂ZZZ
∂Z
=
lim
= 00
lim
lim
lim
lím
P→0
PP TTT
P→∞ ∂ P T
P→0 ∂∂∂P
P→0
Show
that
second
virial
coefficient
is
at
Boyle
temperature.
a) (a)
Muestre
que
el
segundo
coeficiente
virial
eszero
ceroand
la
temperatura
deas
Boyle.
(a)
ShowNote
that the
the
second
virial
coefficient
is
zero
ata the
the
Boyle
temperature.
that
in the limit
as
P→
0, VBBB→
∞,
that
in the
limit
P → ∞, V → b.
la the
correlación
generalizada
para
(3.63),
para estimar
la temperatura
b) (b)
Use
Use
correlation
for
Eq.
to
the
Boyle
(b)
Use
the generalized
generalized
correlation
forB,B,
B,ecuación
Eq. (3.63),
(3.63),
to estimate
estimate
the reduced
reduced
Boyle de Boyle
3.58.
Ifpara
140(ft)
of simples.
methane
reducida
fluidos
temperature
for
fluids.
temperature
for3simple
simple
fluids.gas at 60(◦ F) and 1(atm) is equivalent to 1(gal) of gasoline as
fuel for an automobile engine, what would be the volume of the tank required to hold
◦ F) in an amount equivalent to 10(gal) of gasoline?
methane
3,000(psia)
andis
3.61.
entregagas
gas
naturalat(suponga
metano
puro)
a unato
ciudad
través
de unat
conducto
con una relación
3.61.
(assume
pure
delivered
aa city
pipeline
aa volumetric
3.61.SeNatural
Natural
gas
(assume
pure methane)
methane)
is60(
delivered
to
cityavia
via
pipeline
at
volumetric
◦◦F) promedio
rate
million
standard
cubic
per
Average
delivery
conditions
volumétrica
150 millones
piesfeet
cúbicos
estándar
por día.
Las condiciones
de50(
entrega
rate of
of 150
150de
million
standardde
cubic
feet
per day.
day.
Average
delivery
conditions are
are
50(
F)
3.59.
aDetermine:
good estimate for the compressibility factor Z of saturated hydrogen vaand
300(psia).
Determine:
and
300(psia).
Determine:
son
50(°F)
yDetermine
300(psia).
por at 25 K and 3.213 bar. For comparison, an experimental value is Z = 0.7757.
volumetric
delivery
rate
cubic
feet
day.
(a)
The
volumetric
delivery
rate in
in actual
actual
cubic
feet per
per
day.por día.
a) (a)
LaThe
relación
de entrega
volumétrica
en pies
cúbicos
reales
(b)
The
molar
delivery
rate
in
kmol
per
hour.
(b)
The
molar
delivery
rate
in
kmol
per
hour.
Thede
Boyle
temperature
the temperature
b) La3.60.
relación
entrega
molar eniskmol
por hora. for which:
−1..
gas
at
conditions
m
ss−1
(c)
The
gas velocity
velocity
atadelivery
delivery
conditions
in
m�
c) (c)
LaThe
velocidad
del gas
las condiciones
de in
entrega
en�m s–1.
∂Z
= 0of
lim
The
inside
diameter
The pipe
pipe is
is 24(in)
24(in) schedule-40
schedule-40 steel
steel with
with an
an
inside∂ P
diameter
of 22.624(in).
22.624(in). Standard
Standard
P→0
T interior de 22.624(in). Las condiciones
Laconditions
tubería
es
de
acero
de
24(in),
cédula
40,
con
un
diámetro
◦
◦
F) and
and 1(atm).
1(atm).
conditions are
are 60(
60( F)
estándar son 60(°F) y 1 atm.
(a) Show that the second virial coefficient B is zero at the Boyle temperature.
3.62.
correlations
use
critical
compressibility
factor
rather
3.62. Some
Some corresponding-states
corresponding-states
correlations
use the
thefor
critical
compressibility
factor ZZ
cc,, rather
(b) Use the generalized
correlation
B, Eq.
(3.63), to estimate
the
reduced Boyle
deω,
estados
correspondientes
usan
eltypes
factor
de
compresibilidad
crítica Zc, más
3.62. Algunas
than
acentric
factor
as
a
third
parameter.
The
two
of
correlation
(one
based
than the
thecorrelaciones
acentric
factor
ω,
as
a
third
parameter.
The
two
types
of
correlation
(one
based
temperature for simple fluids.
que
el
factor
acéntrico
w,
como
un
tercer
parámetro.
Las
dos
clases
de
correlación
(uno
on
on TTcc,, PPcc,, and
and ZZcc,, the
the other
other on
on TTcc,, PPcc,, and
and ω)
ω) would
would be
be equivalent
equivalent were
were there
there aa oneone- basado en
,to-one
Pc y Zcorrespondence
,
el
otro
en
T
,
P
y
w)
serían
equivalentes
donde
exista
una
correspondencia
uno a uno
Tcto-one
c
c
and
data
B
aa test
between
ZZcc methane)
and ω.
ω. The
The
data of
of App.
App.
B allow
allow
test of
ofatthis
this
between
3.61.correspondence
Natural gas c(assume
pure
is delivered
to a city
via pipeline
a volumetric
y
w.
La
información
del
apéndice
B
permite
una
prueba
de
esta
correspondencia.
Prepare
entre
Z
c
◦ F)
correspondence.
Prepare
aa plot
of
ω
see
how
well
with
correspondence.
Prepare
plot
of ZZccubic
vs. feet
ω to
toper
seeday.
how
well ZZcdelivery
correlates
with ω.
ω.are 50(
rate of 150
million
standard
Average
conditions
c vs.
c correlates
en
función
de
w
para
ver
qué
tan
bien
Z
se
correlaciona
con
w.
Desarrolle
una
una
gráfica
de
Z
c
c
Develop
linear
correlation
(Z
=
a
+
bω)
for
nonpolar
substances.
Develop aaand
linear
correlation
(Z
=
a
+
bω)
for
nonpolar
substances.
300(psia). Determine:
cc
correlación lineal (Zc = a + bw) para sustancias no polares.
The volumetric
delivery rate
in actual
cubic volume)
feet
per day.
3.63.
suggests
that
(paths
of
are
3.63. Figure
Figure 3.3
3.3(a)
suggests
that the
the isochores
isochores
(paths
of constant
constant
volume)
are approximately
approximately
figura
3.3
sugiere
que
las
isocoras
(trayectorias
de
volumen
constante)
sonisochores.
líneas casi rectas en
3.63. Lastraight
(b) on
The
delivery
rate that
in kmol
per hour. models
aa PPmolar
TT diagram.
Show
the
straight lines
lines
on
diagram.
Show
that
the following
following
models imply
imply linear
linear
isochores.
−1
un diagrama(c)
PT.The
Muestre
que
los
siguientes
modelos
implican
isocoras
lineales.
gas velocity at delivery conditions in m s .
(a)
(a) Constant-β,
Constant-β, κκ equation
equation for
for liquids.
liquids. (b)
(b) Ideal-gas
Ideal-gas equation.
equation. (c)
(c) Van
Van der
der Waals
Waals
The
pipe
is 24(in)
schedule-40
steel with
an inside
of 22.624(in).
Standard
equation.
equation.
para
líquidos
k, –b
constante. b) Ecuación
dediameter
gas ideal. c) Ecuación
de van
a) Ecuación
conditions are 60(◦ F) and 1(atm).
der Waals.
03-SmithVanNess.indd 121
3.62. Some corresponding-states correlations use the critical compressibility factor Z c , rather
than the acentric factor ω, as a third parameter. The two types of correlation (one based
on Tc , Pc , and Z c , the other on Tc , Pc , and ω) would be equivalent were there a oneto-one correspondence between Z c and ω. The data of App. B allow a test of this
10/1/07
correspondence. Prepare a plot of Z c vs. ω to see how well Z c correlates with ω.
14:08:10
122122
122
122
CAPÍTULO
3. Propiedades
volumétricas
de
fluidos puros
CHAPTER
Volumetric
Properties
Fluids
CHAPTER
CHAPTER
3.3.3.
Volumetric
Volumetric
Properties
Properties
ofofPure
ofPure
Pure
Fluids
Fluids
3.64. A continuación se muestra el árbol de decisión de la ecuación de estado. Para cada elemento discu3.64.
Shown
below
Equation
Decision
Tree.
For
each
item
discuss
the
condi3.64.
3.64.
Shown
Shown
below
below
isisthe
isthe
the
Equation
Equation
ofofState
ofState
State
Decision
Decision
Tree.
Tree.
For
For
each
each
item
item
discuss
discuss
the
the
condicondita tions
las condiciones
másits
apropiadas
para su uso.
appropriate
use.
tions
tions
appropriate
appropriate
totoits
to its
use.
use.
��
�
(a)(a)
Ideal
gas
(a)
Ideal
Ideal
gas
gas ideal
a)
Gas
Gas�
Gas
Gas
Gas
��
��
�
b) virial
Ecuación
virial de dos términos
(b)
2-term
virial
equation
(b)
(b)
2-term
2-term
virial
equation
equation
��
�
c)equation
Ecuación
cúbica
de estado
(c)(c)
Cubic
equation
(c)
Cubic
Cubic
equation
ofof
state
ofstate
state
d) Tablas de Lee/Kesler, apéndice E ��
�
(d)
Lee/Kesler
tables,
Appendix
(d)
(d)
Lee/Kesler
Lee/Kesler
tables,
tables,
Appendix
Appendix
EE E
Gas
Gas
Gas
ororor �
¿Gas o
��
liquid?
liquid?
liquid?
líquido? ��
�
(e)(e)
Incompressible
liquid
(e)
Incompressible
Incompressible
liquid
liquid
e) Líquido incompresible
��
�
(f)
Rackett
equation,
Eq.
(3.72)
(f) (f)
Rackett
Rackett
equation,
equation,
Eq.
Eq.
(3.72)
(3.72)
Liquid
Liquid
Liquid
f) Ecuación de Rackett, ecuación (3.72)
��
�
Líquido
��
�
(g)
Constant
(g)
(g)
Constant
Constant
ββand
βand
and
κκ κ
g) b y k constantes
��
�
(h)
Lydersen
chart,
Fig.
3.16
(h)
(h)
Lydersen
Lydersen
etetal.
etal.al.
chart,
chart,
Fig.
Fig.
3.16
3.16
h) Diagrama
de
Lydersen,
et al., figura 3.16
3.65.
Un
gas
ideal,
inicialmente
a◦25
°C
1 bar,
bar,
se somete
alfollowing
siguiente
proceso
cíclicoininen
◦25
3.65.
An
ideal
gas,
initially
C◦and
undergoes
the
cyclic
processes
3.65.
3.65.
An
An
ideal
ideal
gas,
gas,
initially
initially
atat25
at25
C
Cand
and
1y1bar,
1bar,
undergoes
undergoes
the
the
following
following
cyclic
cyclic
processes
processes
aina un
a sistema
cerrado:
closed
system:
closed
closed
system:
system:
a)(a)
En
mecánicamente
reversibles
se
comprime
de
manera to
adiabática
(a)
reversible
processes,
compressed
adiabatically
to5to5bar,
(a)
InInprocesos
mechanically
Inmechanically
mechanically
reversible
reversible
processes,
processes,
ititprimero
isitisfirst
isfirst
first
compressed
compressed
adiabatically
adiabatically
5bar,
bar,a 5 bar, a
◦hasta
◦ C,
continuación
seataenfría
a presión
constante
de
525
bar
25
°C
yexpanded
por
últimoisotherseisotherexpande isotérC,
and
finally
expanded
isotherthen
cooled
pressure
25◦25
C,
and
and
finally
finally
expanded
then
then
cooled
cooled
at
ataconstant
aconstant
constant
pressure
pressure
ofof5of5bar
5bar
bar
toto
to
micamente
a its
suoriginal
presión
original.
mally
pressure.
mally
mally
totoits
toits
original
original
pressure.
pressure.
b)(b)
El
ciclo
es
irreversible
y
cada
etapa
tiene
una
eficiencia
de of
80%
en
comparación
con el corres(b)
The
cycle
and
each
step
has
anan
efficiency
ofof
80%
compared
with
(b)
The
The
cycle
cycle
isisirreversible
isirreversible
irreversible
and
and
each
each
step
step
has
has
an
efficiency
efficiency
80%
80%
compared
compared
with
with
pondiente
proceso
que
es
mecánicamente
reversible.
El
ciclo
completo
consiste
en
una
the
corresponding
mechanically
reversible
process.
The
cycle
still
consists
the
the
corresponding
corresponding
mechanically
mechanically
reversible
reversible
process.
process.
The
The
cycle
cycle
still
still
consists
consists
ofofan
ofan
an etapa de
compresión
adiabática,
una
etapa
de
enfriamiento
isobárico
y
una
expansión
isotérmica.
adiabatic
compression
step,
isobaric
cooling
step,
and
isothermal
expansion.
adiabatic
adiabatic
compression
compression
step,
step,
ananan
isobaric
isobaric
cooling
cooling
step,
step,
and
and
ananan
isothermal
isothermal
expansion.
expansion.
Calculate
Q,Q,
�U
and
�H
for
each
step
the
process
and
for
the
cycle.
Calculate
Calculate
Q,
WW,W
,∆H
�U
, ,para
and
, and
�H
�H
for
for
each
each
step
step
ofofof
the
process
process
and
and
for
for
the
the
cycle.
cycle.
Calcule
Q, W,
∆U
y,�U
cada
etapa
del
proceso
y the
para
el ciclo.
Considere
que Take
CTake
= (7/2)R y CV
PTake
and
CPC
=(7/2)R
(7/2)R
and
and
CCVC
=(5/2)R.
(5/2)R.
P=
V=
P=(7/2)R
V=(5/2)R.
=C(5/2)R.
3.66.
Show
that
the
density-series
second
virial
coefficients
can
derived
from
isothermal
3.66.
3.66.
Show
Show
that
that
the
the
density-series
density-series
second
second
virial
virial
coefficients
coefficients
can
can
be
derived
derived
from
from
isothermal
isothermala partir de
3.66.
Demuestre
que
los
segundos coeficientes
viriales
en serie
debebe
densidad
pueden
obtenerse
volumetric
data
via
the
expression:
volumetric
volumetric
data
data
via
via
the
the
expression:
expression:
información
volumétrica
isotérmica
mediante la expresión:
(Z(Z
density)
lim
(Z
−−1)/ρ
−1)/ρ
1)/ρ ρρr(densidad
(molar
ρ(molar
(molar
density)
density)
≡≡1/V
≡1/V
1/V
BB=
B==
lim
lim
lím
molar)
1/V)
ρ→0
ρ→0
ρ→0
3.67.
Haga
uso
de
la ecuación
del
problema
anterior
yand
ladata
información
deF.2
laF.2
tabla
F.2
para
obtener
3.67.
Use
the
equation
preceding
problem
and
data
from
Table
3.67.
3.67.
Use
Use
the
the
equation
equation
ofofthe
ofthe
the
preceding
preceding
problem
problem
and
data
from
from
Table
Table
F.2
totoobtain
toobtain
obtain
a avalue
avalue
value un valor
de
B
para
el
agua
a
una
de
las
siguientes
temperaturas:
water
following
temperatures:
ofofof
BBfor
Bfor
for
water
water
atatone
atone
one
ofofthe
ofthe
the
following
following
temperatures:
temperatures:
◦C
◦C
◦C
◦C
◦ C(b)
◦C
◦ Cb)
◦C
◦C
a)(a)
300
°C
350
°C
(b)
350
(c)
400
(a)
300
(b)
350
350
(c)
(c)
400
400
(a)
300
300
03-SmithVanNess.indd 122
c) 400 °C
10/1/07 14:08:52
Problemas
Problems
Problems
Problems
123
123
123
123
3.68.
Obtenga
los
valores
de
W,
ψ
yand
ZZcZcdados
en
la
tabla
3.1
para:
3.68.
3.68.
3.68.
Derive
Derive
Derive
the
the
the
values
values
values
ofof
of
�,
�,
�,
�,
�,
�,
and
and
Zcgiven
given
ininin
Table
Table
Table
3.1
3.1
3.1
for:
for:
for:
cgiven
a)(a)
La
ecuación
de estado deequation
Redlich/Kwong.
(a)
(a)The
The
The
Redlich/Kwong
Redlich/Kwong
Redlich/Kwong
equation
equation
ofofof
state.
state.
state.
b)(b)
La
ecuación
de estado de Soave/Redlich/Kwong.
(b)
(b)The
The
The
Soave/Redlich/Kwong
Soave/Redlich/Kwong
Soave/Redlich/Kwong
equation
equation
equation
ofofof
state.
state.
state.
c)(c)
La
ecuación
de
estado
de
Peng/Robinson.
(c)
(c)The
The
The
Peng/Robinson
Peng/Robinson
Peng/Robinson
equation
equation
equation
ofofof
state.
state.
state.
3.69. Suponga que se tienen disponibles datos de Z en función de Pr a Tr constante. Demuestre que el
data
data
are
are
are
available
available
available
at
constant
constant
constant
TTr rT. .r Show
.Show
Show
that
that
that
the
the
the
reduced
reduced
reduced
densitydensitydensity3.69.
3.69.
3.69.
Suppose
Suppose
Suppose
ZZZvs.
vs.
vs.PPr P
r data
rvirial
segundo
coeficiente
reducido
enatat
serie
de densidad
puede
obtenerse
a partir
de tales datos
series
series
series
second
second
second
virial
virial
virial
coefficient
coefficient
coefficient
can
can
can
be
be
be
derived
derived
derived
from
from
from
such
such
such
data
data
data
via
via
via
the
the
the
expression:
expression:
expression:
mediante la expresión:
B̂B̂B̂
===lim
lim
lim
(Z
(Z
(Z
−−−
1)Z
1)Z
1)Z
TTr rT/P
/P
lím
r /P
rr r
PP
r →0
rP→0
r →0
Sugerencia:
fundamente
el desarrolloon
en
la
expansión
virial
en densidad
completa,
ecuación (3.12).
Suggestion:
Suggestion:
Suggestion:
Base
Base
Base
the
the
the
development
development
development
on
on
the
the
the
full
full
full
virial
virial
virial
expansion
expansion
expansion
ininin
density,
density,
density,
Eq.
Eq.
Eq.
(3.12)
(3.12)
(3.12)
Utilice
el
resultado
del
problema
anterior
yand
los
datos
de
laTable
tabla
E.1
obtener
un valor
3.70.
3.70.
3.70.
3.70.
Use
Use
Use
the
the
the
result
result
result
ofofof
the
the
the
preceding
preceding
preceding
problem
problem
problem
and
and
data
data
data
from
from
from
Table
Table
E.1
E.1
E.1
topara
toto
obtain
obtain
obtain
aavalue
avalue
value
ofofof de B̂ para
fluidos
simples
con
T
=
1.
Compare
el
resultado
con
el
valor
dado
por
la
ecuación
(3.65).
r at
B̂B̂B̂
for
for
for
simple
simple
simple
fluids
fluids
fluids
atat
TTr rT==r 1.
=1.1.
Compare
Compare
Compare
the
the
the
result
result
result
with
with
with
the
the
the
value
value
value
implied
implied
implied
by
byby
Eq.
Eq.
Eq.
(3.65).
(3.65).
(3.65).
3.71.
La
siguiente
conversación
fuewas
escuchada
en in
los
pasillos
de una
compañía
de ingeniería. 3.71.
3.71.
3.71.
The
The
The
following
following
following
conversation
conversation
conversation
was
was
overheard
overheard
overheard
inin
the
the
the
corridors
corridors
corridors
ofofof
aimportante
alarge
alarge
large
engineering
engineering
engineering
firm.
firm.
firm.
New
New
New
engineer:
engineer:
engineer:
”Hi,
”Hi,
”Hi,
boss.
boss.
boss.
Why
Why
Why
the
the
the
big
big
big
smile?”
smile?”
smile?”
Ingeniero
novato:
“Hola,
jefe.
¿Por
qué
tan
contento?
Old-timer:
Old-timer:
Old-timer:
”I
”I
”I
finally
finally
finally
won
won
won
a
a
wager
a
wager
wager
with
with
with
Harry
Harry
Harry
Carey,
Carey,
Carey,
from
from
from
Research.
Research.
Research.He
He
He
bet
bet
bet
me
me
me
that
that
that Me aposVeterano: “Por fin le gané una apuesta a Uriel Ortega,
del
departamento
de
investigación.
I
I
couldn’t
I
couldn’t
couldn’t
come
come
come
up
up
up
with
with
with
a
a
quick
a
quick
quick
but
but
but
accurate
accurate
accurate
estimate
estimate
estimate
for
for
for
the
the
the
molar
molar
molar
volume
volume
volume
of
of
of
ararar- del argón
tó que no se
me podría ocurrir una estimación rápida pero precisa para el volumen molar
◦◦CC
◦C
gon
gon
gon
at
at
at
30
30
30
and
and
and
300
300
300
bar.
bar.
bar.
Nothing
Nothing
Nothing
to
to
to
it;
it;
it;
I
I
used
I
used
used
the
the
the
ideal-gas
ideal-gas
ideal-gas
equation,
equation,
equation,
and
and
and
got
got
got
about
about
about
a 30 °C33y3 300
bar.
Nada más fácil: utilicé la ecuación del gas ideal y obtuve aproximadamente 83
−1
−1
−1
83
83
cm
cm
cm–1
mol
mol
. .Harry
.negó
Harry
Harry
shook
shook
shook
his
his
his
head,
head,
head,
but
but
but
paid
paid
paid
up.
up.
up.
What
What
What
do
dodo
you
you
you
think
think
think
about
about
that?”
that?”
that?”
383
cm
mol
.mol
Uriel
con
la cabeza,
pero
pagó.
¿Qué
piensas
acerca
deabout
esto?
New
New
New
engineer
engineer
engineer
(consulting
(consulting
(consulting
his
his
his
thermo
thermo
thermo
text):
text):
text):
”I”I”I
think
think
think
you
you
you
must
must
must
be
bebe
living
living
living
right.”
right.”
right.”
Ingeniero novato (consultando su texto de termodinámica): “Pienso que debe de estar en lo correcto”.
Argon
Argon
Argon
atatat
the
the
the
stated
stated
stated
conditions
conditions
conditions
isisis
not
not
not
an
anan
ideal
ideal
ideal
gas.
gas.
gas.Demonstrate
Demonstrate
Demonstrate
numerically
numerically
numerically
why
why
why
the
the
the
Elold-timer
argón
a won
las
condiciones
old-timer
old-timer
won
won
his
his
his
wager.
wager.
wager.establecidas no es un gas ideal. Demuestre numéricamente por qué el
veterano ganó la apuesta.
3.72.
3.72.
3.72.
Five
Five
Fivemol
mol
molofofof
calcium
calcium
calcium
carbide
carbide
are
are
are
combined
combinedwith
with
with10
1010mol
mol
molofofde
ofwater
water
waterinen
ininaaun
closed,
aclosed,
closed,rigid,
rigid,
rigid,
3.72.
Cinco
moles
de
carburocarbide
de
calcio
secombined
combinan
con
10
moles
agua
recipiente
cerrado, rígi33internal
3internal
3
high-pressure
high-pressure
high-pressure
vessel
vessel
vessel
of
of
of
1800
1800
1800
cm
cm
cm
internal
empty
empty
empty
volume.
volume.
volume.
Acetylene
Acetylene
Acetylene
gas
gas
gas
is
is
is
produced
produced
produced
do, de alta presión y con un volumen vacío interno de 1 800 cm . Se produce gas de acetileno meby
byby
the
the
the
reaction:
reaction:
diante
lareaction:
reacción:
(s)
++
2H
2H
2H
→
→
CC2C
(g)
+
Ca(OH)
Ca(OH)
Ca(OH)
(s)
CaC
CaC
CaC
CaC
++
O(l)
→
H
(g)
+++
22(s)
2(s)
22O(l)
2 O(l)
22H
2 (g)
22(s)
2 (s)
2(s)
2H
22(g)
ElThe
recipiente
contiene
un
embalaje
con
una porosidad
de
40%
para
evitar ladecomposidescomposición
exploThe
The
vessel
vessel
vessel
contains
contains
contains
packing
packing
packing
with
with
with
aaporosity
aporosity
porosity
ofofof
40%
40%
40%
tototo
prevent
prevent
prevent
explosive
explosive
explosive
decomposidecomposi◦de
◦CC
◦C
siva
del
acetileno.
Las condiciones
iniciales
son
25
°C
1bar,
bar,
ythe
la
reacción
llega
atoto
completarse.
and
and
and
11ybar,
1bar,
and
and
and
the
the
reaction
reaction
reaction
goes
goes
goes
to
tion
tion
tion
ofofof
the
the
the
acetylene.
acetylene.
acetylene.
Initial
Initial
Initial
conditions
conditions
conditions
are
are
are
25
2525
La
reacción esThe
exotérmica,
pero
debido a but
la
transferencia
detransfer,
calor,
lathe
temperatura
final es de sólo
completion.
completion.
completion.
The
The
reaction
reaction
reaction
isis
is
exothermic,
exothermic,
exothermic,
but
but
owing
owing
owing
tototo
heat
heat
heat
transfer,
transfer,
the
the
final
final
final
temperatemperatempera◦◦C.
◦la
125
°C.
Determine
presión
final
en
el recipiente.
ture
ture
ture
isisis
only
only
only
125
125
125
C.
C.
Determine
Determine
Determine
the
the
the
final
final
final
pressure
pressure
pressure
ininin
the
the
the
vessel.
vessel.
vessel.
–1
◦◦C,
◦ C,
333mol
3mol
−1
−1
−1
Nota:
aAt
125
°C
elC,
volumen
molar
delof
Ca(OH)
de
33.0
cm
mol
.. .Ignore
los
efectos
de cualquier
the
the
the
molar
molar
molar
volume
volume
volume
ofof
Ca(OH)
Ca(OH)
Ca(OH)
33.0
33.0
33.0
cm
cm
cm
mol
Ignore
.Ignore
Ignore
the
the
the
effects
effects
effects
Note:
Note:
Note:
AtAt
125
125
125
2 es
22is
2isis
gas
(por
ejemplo,
aire)
que
de manera
inicial
haya
estado presente en dicho recipiente.
ofofof
any
any
any
gases
gases
gases
(e.g.,
(e.g.,
(e.g.,
air)
air)
air)
initially
initially
initially
present
present
present
ininin
the
the
the
vessel.
vessel.
vessel.
◦◦CC
◦C
3.73.
Se
requiere
almacenar
3535,000
000
kg
deofofpropano,
que
es recibido
como
un
gas
a 1(atm).
10
°C y 1(atm). Se
3.73.
3.73.
3.73.
Storage
Storage
Storage
isisis
required
required
required
for
for
for
35,000
35,000
kg
kgkg
of
propane,
propane,
propane,
received
received
received
asasas
aagas
agas
gas
atatat
10
10
10
and
and
and
1(atm).
1(atm).
proponen
dos opciones:
Two
Two
Two
proposals
proposals
proposals
have
have
have
been
been
been
made:
made:
made:
◦◦CC
◦ and
(a)
(a)
Store
Store
Store
ititas
itasas
aacomo
gas
agas
gas
atatun
at
10
1010
and
1(atm).
1(atm).
a)(a)
Almacenarlo
gasC
a and
101(atm).
°C
y 1(atm).
◦◦CC
◦C
and
and
6.294(atm).
6.294(atm).
6.294(atm).
For
For
For
this
this
thiseste modo
(b)
(b)
(b)
Store
Store
Store
it
it
as
it
as
as
a
a
liquid
a
liquid
liquid
in
in
in
equilibrium
equilibrium
equilibrium
with
with
with
its
itsits
vapor
vapor
vapor
at
10
1010
b) Almacenarlo como un líquido en equilibrio
con
suatat
vapor
aand
10
°C
y 6.294(atm).
Para
mode
mode
mode
of
of
of
storage,
storage,
storage,
90%
90%
90%
of
of
of
the
the
the
tank
tank
tank
volume
volume
volume
is
is
is
occupied
occupied
occupied
by
by
by
liquid.
liquid.
liquid.
de almacenamiento, 90% del volumen del tanque es ocupado por líquido.
03-SmithVanNess.indd 123
10/1/07 14:09:43
124
124124
CAPÍTULO
3. Propiedades
volumétricas
de fluidos puros
CHAPTER
Volumetric
Properties
of Pure
Fluids
CHAPTER
3. 3.
Volumetric
Properties
of Pure
Fluids
Compare
las dostwo
opciones,
discutiendo
lospros
puntos
acons
favorofy each.
en contraquantitative
de cada una.where
Trate de ofrecer
Compare
proposals,
discussing
Compare
thethe
two proposals,
discussing
pros
andand
cons
of each. BeBe
quantitative where
argumentos
cuantitativos
siempre
que
sea
posible.
possible.
possible.
3.74. La definición del factor de compresibilidad Z, ecuación (3.10), puede escribirse de la forma más
3.74.
The
definition
compressibility
factor
, Eq.
(3.10),
may
written
more
3.74.
The
definition
of of
compressibility
factor
Z , ZEq.
(3.10),
may
be be
written
in in
thethe
more
intuitiva:
intuitive
form:
intuitive
form:
V V
Z≡
Z≡
V (gas
(idealgas)
ideal)
V (idealgas)
where
both
volumes
are
at
the
same
T
and
P.
Recall
ideal
aunmodel
sub-es una suswhere
both
volumes
are
at
the
same
T
and
P.
Recall
that
ideal
gasgas
isque
aismodel
donde ambos volúmenes se encuentran a las mismas
T that
y an
P. an
Recuerde
gassubideal
stance
comprising
particles
with
no
intermolecular
forces.
Use
the
intuitive
definition
stance
comprising
particles
with
no
intermolecular
forces.
Use
the
intuitive
definition
tancia modelo que tiene partículas sin fuerzas intermoleculares. Utilice la definición intuitiva de Z
of
Z argue
to argue
that:
of
Z to
that:
para
sostener
que:
attractions
promote
values
Z 1.
< 1.
(a)(a)
Intermolecular
promote
values
of of
Zvalores
<
a)
LasIntermolecular
atraccionesattractions
intermoleculares
promueven
de Z < 1.
repulsions
promote
values
Z 1.
> 1.de Z > 1.
(b)(b)
Intermolecular
promote
values
of of
Zvalores
>
b)
LasIntermolecular
repulsionesrepulsions
intermoleculares
promueven
Aequilibrio
balance
of
repulsions
implies
Z
=1.(Note
1.(Nótese
(Note
gas es un caso
(c)(c)
A
of attractions
andand
repulsions
implies
thatthat
Z=
1.
thatthat
an an
ideal
gas
c)
Unbalance
deattractions
atracciones
y repulsiones
implica
que
Z=
que
unideal
gas
ideal
a special
case
there
or repulsions.)
is
aisspecial
case
forfor
which
areare
no no
attractions
or repulsions.)
especial
para
el
que
nowhich
haythere
atracciones
ni attractions
repulsiones).
3.75.
Escriba
forma
general
ecuación
estado
3.75.
Write
the
general
form
of una
an
equation
ofdestate
3.75.
Write
thela
general
form
of de
an
equation
of state
as:as: como:
=+11 Z
+rep
ZZrep
ZZatr(T,
(T,
(ρ)
−attr
(T,
ZZ 1=
+
rep(r)
attr
(ρ)
− –Z
ρ)r)ρ)
Z=
where
Z (r)
represents
from
representsla contriburep (ρ)
attr
ZZrep
represents
contributions
from
repulsions,
andand
ρ)
represents
where
donde
representa
la contributions
contribución
de
lasrepulsions,
repulsiones
yZ attr
ZZatr(T,
(T,(T,
r)ρ)
representa
rep(ρ)
contributions
from
attractions.
What
repulsive
and
attractive
contributions
contributions
from
attractions.
areare
thethe
repulsive
and
attractive
contributions
ción
de las atracciones.
¿CuálesWhat
son
las
contribuciones
repulsivas
y de
atracción
detolatoecuación de
the
Waals
equation
of state?
the
vanvan
derder
Waals
of state?
estado
de
van
der equation
Waals?
3.76.
below
four
proposed
modifications
Waals
equation
of
state.
3.76.
A Given
continuación
se
proponen
cuatro
modificaciones
avan
la van
ecuación
deequation
estado
deofvan
der Waals. ¿Son
3.76.
Given
below
areare
four
proposed
modifications
of of
thethe
derder
Waals
state.
Are
any
of
these
modifications
reasonable?
Explain
carefully;
statements
such
as,
“It afirmaciorazonables
algunas
de estas modificaciones?
Explique
de manera
cuidadosa
poras,qué
Are
any of these
modifications
reasonable? Explain
carefully;
statements
such
“It las
isn’t
cubic
in
volume.”
do
not
qualify.
nes del
tipoin“no
es cúbica
en el
volumen”, no son válidas.
isn’t
cubic
volume.”
do not
qualify.
RTRT a a
P=
(a)(a)
−−
P=
V b− b V V
V−
RTRT
a a
P=
(b)(b)
P=
−2 −
2
(V
−
b)
V V
(V − b)
RTRT
a a
P=
(c)(c)
P=
− −2 2
V (V
− b) V V
V (V
− b)
RTRT a a
P= −− 2
(d)(d)
P=
V V V 2V
3.77.
With
reference
to Pb.
2.43,
assume
to be
an
ideal
develop
expression
3.77.
With
reference
Pb.
2.43,
assume
air air
toque
be
an
ideal
andand
develop
an an
expression
3.77.
Con
respecto
altoproblema
2.43,
suponga
el aire
esgas,
ungas,
gas
ideal
y desarrolle
una expresión en la
giving
the
household
air
temperature
as
a
function
of
time.
giving
the
household
air
temperature
as
a
function
of
time.
que proporcione la temperatura del aire doméstico como una función del tiempo.
3.78.
A garden
hose
water
valve
shut
closed
insethe
sun,
of
3.78.
AUna
garden
hose
with
thethe
water
valve
shut
andand
thethe
nozzle
closed
sitssits
in the
sun,
fullfull
ofexpuesta
3.78.
manguera
dewith
jardín
con
la
válvula
del
agua
y lanozzle
boquilla
cerradas
encuentra
a los
◦ C and 6 bar. After some time the temperature
◦10
liquid
water.
Initially,
the
water
is
at
C
and
6
bar.
After
some
time
the
temperature
liquid
water.
Initially,
the
water
is
at
10
rayos del sol y llena de agua
líquida.
Al
principio,
el
agua
tiene
una
temperatura
de
10
°C
y
6
bar.
◦ C. Owing to the increase in temperature and pressure and the
of the
water
rises
to ◦40
of
the
water
totiempo
40
C. la
Owing
to the increase
temperature
pressure and the
Después
de rises
cierto
temperatura
del aguainasciende
a 40 and
°C. Obedeciendo
al incremento de
elasticity
of
the
hose,
the
internal
of the
hose
increases
0.35%.
Estimate
elasticity
of the
hose,
theasí
internal
of the
hose
increases
by
0.35%.
Estimate
temperatura
y presión,
como
adiameter
ladiameter
elasticidad
de
la
manguera,
el by
diámetro
interno
de ésta se inthe
final
pressure
of
the
water
in
the
hose.
the
final
pressure
of
the
water
in
the
hose.
crementa 0.35%. Calcule la presión final del agua en la manguera.
−1 −1
Data: β(ave)
= 250
× −6
10−6−1
K−1 ; κ(ave)
= 45
× −6
10−6 bar
= 250
× 10
= 45
× ×10
Data:
Datos:β(ave)
b(prom)
= 250
× 10–6KK–1; ;κ(ave)
k(prom)
= 45
10–6bar
bar–1
03-SmithVanNess.indd 124
8/1/07 13:02:44
Capítulo 4
Efectos térmicos
La transferencia de calor es una de las operaciones más comunes en la industria química. Considere, por
ejemplo, la fabricación del etilenglicol (un agente anticongelante) mediante la oxidación de etileno a óxido de
etileno y de su hidratación posterior a glicol. La reacción de oxidación catalítica resulta más efectiva cuando
se realiza a temperaturas cercanas a 250 °C. Por lo tanto, los reactivos etileno y aire se calientan a esta temperatura antes de que entren al reactor. Para diseñar el precalentador es necesario saber cuánto calor se transfiere. Las reacciones de combustión del etileno con el oxígeno en la cama catalítica tienden a incrementar la
temperatura. Sin embargo, si se elimina el calor del reactor, la temperatura no aumenta más de 250 °C. Las
temperaturas superiores promueven la formación de CO2, que es un producto no deseado. Para el diseño del
reactor se requiere conocer la rapidez de transferencia de calor, y ésta depende de los efectos térmicos asociados con las reacciones químicas. El producto óxido de etileno se hidrata a glicol por la absorción en agua. El
calor se desprende debido no sólo al cambio de fase y al proceso de disolución, sino también por una reacción
de hidratación entre el óxido de etileno disuelto y el agua. Al final, el glicol se recupera del agua a través de
la destilación, un proceso de vaporización y condensación que origina la separación de una solución en sus
componentes.
Todos los efectos térmicos importantes se ilustran por medio del proceso, relativamente simple, de la
fabricación de una sustancia química. A diferencia de los efectos del calor sensible, caracterizados por los
cambios de temperatura, los efectos térmicos de una reacción química, la transición de fase, así como la formación y la separación de soluciones se determinan a partir de medidas experimentales hechas a temperatura
constante. En este capítulo, la termodinámica se aplica a la evaluación de la mayoría de los efectos térmicos
que acompañan a las operaciones físicas y químicas. No obstante, los efectos térmicos de los procesos de
mezclado, que dependen de las propiedades termodinámicas de las mezclas, se tratan en el capítulo 12.
4.1
EFECTOS DEL CALOR SENSIBLE
La transferencia de calor a un sistema en el que no hay transiciones de fase, ni reacciones químicas o algún
cambio en la composición, provoca que cambie la temperatura del sistema. Nuestro propósito es desarrollar
relaciones entre la cantidad de calor transferida y el cambio de temperatura resultante.
Cuando el sistema es una sustancia homogénea de composición constante, la regla de la fase indica que
al fijar los valores de dos propiedades intensivas se establece su estado. Por lo tanto, la energía interna molar
o específica de una sustancia se puede expresar como una función de las otras dos variables de estado. Cuando
125
04-SmithVanNess.indd 125
8/1/07 13:07:19
126
CAPÍTULO
4. Efectos
CHAPTER
4. Heat
Heat
Effectstérmicos
126
126
CHAPTER 4.
4.
Effects
CHAPTER
Heat Effects
126
126
CHAPTER 4. Heat Effects
éstas seother
seleccionan
en forma
arbitraria
como
temperatura
y volumen
molar and
o específico,
U = U(T, V).
state
variables.
When
these
are
arbitrarily
selected
as
temperature
molar or
or specific
specific
other
state
variables.
When
these
are
arbitrarily
selected
as
temperature
and
molar
De donde,
other
state
When
these are arbitrarily selected as temperature and molar or specific
volume,
U variables.
=U
U (T,
(T, V
V ).
).
Whence,
other state
variables.
When
these are arbitrarily selected as temperature and molar or specific
volume,
U
=
Whence,
volume,
U
=
U (T,
V ).
Whence,
�
�
�
�
volume, U = U (T, V ). Whence, �
�
�
�
∂U �
∂U �
� ∂U
�
∂U
dU
=
dT
+
V
� dT + � ∂U � dd V
dU =
= � ∂U
∂T
T V dT
∂V
V T dV
dU
+ ∂∂U
∂U
∂
dU = ∂ T VVV dT + ∂ V TTT d V
∂T V
∂V T
As aa result
result
of
Eq.
(2.16)(2.16),
this becomes:
becomes:
Como resultado
de
la
ecuación
ésta
será:
As
of
Eq.
(2.16)
this
As a result of Eq. (2.16) this becomes:
�
�
As a result of Eq. (2.16) this becomes:
�
�
∂U �
�
∂U
dT +
+ � ∂U � dd V
V
dU =
=C
C V dT
dU
V dT +
∂V
V T dV
dU
=C
∂U
V
V
∂
T
dU = C V dT + ∂ V TT d V
∂V T
The
final
term
may
be
set
equal
to
zero in
in two
two circumstances:
circumstances:
The final
final term
term may
may be
be set
set equal
equal to
to zero
zero
The
in
two
circumstances:
Existe laThe
posibilidad
quebeelset
término
seainigual
cero bajo dos circunstancias:
final termdemay
equal final
to zero
two acircumstances:
• For
For any
any constant-volume
constant-volume process,
process, regardless
regardless of
of substance.
substance.
•
• For anyproceso
constant-volume
regardless
of substance.
• Para cualquier
a volumenprocess,
constante
sin importar
la sustancia.
• For any constant-volume process, regardless of substance.
•
Whenever
the
internal
energy
is
independent
of
volume,
regardless of
of the
the process.
process. This
This
Whenever
the internal
internal
energy
is independent
independent
of volume,
volume,sinregardless
regardless
• Siempre
que la energía
interna energy
sea independiente
del volumen,
considerar
proceso.
Esto
es verda•• Whenever
the
is
of
ofelthe
process.
This
is
exactly
true
for
ideal
gases
and
incompressible
fluids
and
approximately
true
for
low• is
Whenever
the
internal
energy
is
independent
of
volume,
regardless
of
the
process.
This
is
exactly
true
for
ideal
gases
and
incompressible
fluids
and
approximately
true
for
lowdero para
gases
ideales
y
fluidos
incompresibles,
y
aproximadamente
cierto
para
gases
a
baja
exactlygases.
true for ideal gases and incompressible fluids and approximately true for low-presión.
pressure
is
exactly
true for ideal gases and incompressible fluids and approximately true for lowpressure
gases.
pressure gases.
pressure gases.
En cualquier
caso,
dU =
=C
In either
either
case,
dU
CVV dT
dT
In
case,
dU
=C
C
dT
V dT
In
either case,
dU
=
V
V
In either case,
dU =
� C
T V dT
�
T22
�
T
y
C V dT
dT
(4.1) (4.1)
and
�U =
= � T22 C
(4.1)
and
�U
V dT
(4.1)
and
�U
= TT1T2 C
V
V
1
(4.1)
and
�U = TT11 C V dT
T1
For
a
mechanically
reversible
constant-volume
process,
Q
=
�U
,
and
Eq.
(2.19)
may
be escriPara unFor
proceso
a volumen constante
reversible,
∆U,�U
y la and
ecuación
(2.19) may
se puede
mechanically
reversiblemecánicamente
constant-volume
process, QQ
Q= =
=
Eq. (2.19)
(2.19)
be
For
aa mechanically
reversible
constant-volume
process,
�U ,, and
Eq.
may be
written
for
a
unit
mass
or
a
mole:
bir parawritten
unaamol
For
mechanically
reversible
constant-volume process, Q = �U , and Eq. (2.19) may be
foro aamasa
unit unitaria:
mass
or aa mole:
mole:
written
for
unit
mass
or
� T2
written for a unit mass or a mole:
�
T
�
TT22 C V dT
Q
=
�U
=
Q=
= �U
�U =
= �T T22 C
C V dT
dT
Q
1
V
V
Q = �U = TTT111 C V dT
T1
Similarly, the
the molar
molar or
or specific
specific enthalpy
enthalpy may
may
be expressed
expressed as
as a function
function of
of temperature
temperature
Similarly,
be
De manera
similar,
la
entalpía
molar
o
específica
se expresa
en función
de la temperatura
y de la preSimilarly,
the H
molar
or(T,
specific
enthalpy may
be
expressed
as aa function
of temperature
and
pressure.
Then
=
H
P),
and
Similarly,
the
molar
or
specific
enthalpy
may
be
expressed
as
a
function
of
temperature
and
pressure.
Then
H
=
H
(T,
P),
and
sión. En
tal
caso,
H
=
H(T,
P),
y
and pressure. Then H = H (T, P), and
�
�
�
�
and pressure. Then H = H (T, P), and
�
�
�
�
∂H
H�
∂H
H�
�
�
∂
∂
dT +
+ � ∂ H � dd P
P
dH
H=
= � ∂ H � dT
T P dT +
∂H
P T dP
dd H
= ∂∂ T
H
∂
P
P
T
d H = ∂ T PP dT + ∂ P TT d P
∂T P
∂P T
As
a
result
of
Eq.
(2.20)
this
becomes:
As
a
result
of
Eq.
(2.20)
this
becomes:
As a result of Eq. (2.20) this becomes:
Como resultado
deof
la Eq.
ecuación
ésta será:
�
�
As a result
(2.20)(2.20),
this becomes:
�
�
∂H
H�
�
∂
dT +
+ � ∂ H � dd P
P
dH
H=
=C
C P dT
P dT +
P T dP
dd H
=C
∂∂ H
P
P
P
T
d H = C P dT + ∂ P TT d P
∂P T
Again,
two
circumstances
allow
the final
final term
term to
to be
be set
set equal
equal to
to zero:
zero:
Again, two
two circumstances
circumstances allow
allow the
the
Again,
final term
to be
set equal
to zero:
twocircunstancias
circumstancespermiten
allow theque
final term
to befinal
set equal
to zero:
Una vezAgain,
más,
dos
término
seasubstance.
igual
a cero:
• For
For
any constant-pressure
constant-pressure process,
process,elregardless
regardless
of the
the
•
any
of
substance.
• For anyproceso
constant-pressure
process, regardless
of the
substance.
• Para cualquier
a presión constante,
sin importar
la sustancia.
• For any constant-pressure
process, regardless
of the
substance.
•
Whenever
the
enthalpy
of
the
substance
is
independent
of pressure,
pressure, regardless
regardless of
of the
the
•
Whenever
the
enthalpy
of
the
substance
is
independent
of
Whenever
the enthalpy
oftrue
thefor
substance
is independent
of
pressure,
regardless
the Esto
que la entalpía
de
la sustancia
sea
independiente
de
la presión,
sin
considerar
el of
proceso.
• Cada•vez
process.
This
is
exactly
ideal
gases
and
approximately
true
for
low-pressure
• process.
WheneverThis
the is
enthalpy
thefor
substance
is independent
of pressure,
of the
process.
This
exactlyoftrue
true
for
ideal gases
gases
and approximately
approximately
true regardless
for low-pressure
low-pressure
exactly
ideal
and
for
es precisamente
cierto is
para
gases ideales,
y aproximadamente
válido paratrue
gases
a baja presión.
gases.
process.
This
is
exactly
true
for
ideal
gases
and
approximately
true
for
low-pressure
gases.
gases.
gases.
04-SmithVanNess.indd 126
8/1/07 13:07:21
4.1. Sensible Heat Effects
4.1.
Sensible
Heat
Effects
4.1. Sensible
Sensible
Heat
Effects
4.1.
4.1.
Sensible
Heat
Effects
Effects
4.1. Efectos
del
calorHeat
sensible
In either case,
In
either
case,
Ineither
either
case,
En cualquier
caso,
In
In
either
case,
case,
y
and
and
and
and
and
127
127
127
127
127
d H = C P dT
H
=
C
dT
H=
=
CPPPPdT
dT
dH
=C
C
dT
ddddH
H
=
C
dT
� T2P
��
�H = �� TTTT2T2222C P dT
CC
dT
�H
=
CPPPPdT
dT
�H=
= T1 C
dT
�H
�H
=
T
TTTT
11111
P
127
(4.2)
(4.2)
(4.2) (4.2)
(4.2)
(4.2)
Q ∆H
= �H
mechanically
reversible,reversibles
constant-pressure,
closed-system
processes
Por otraMoreover,
parte, Q =
parafor
procesos
mecánicamente
de sistema
cerrado, a presión
constante
Moreover,
Q
=
�H
for
mechanically
reversible,
constant-pressure,
closed-system
processes
Moreover,
Qand
=�H
�Hthe
fortransfer
mechanically
reversible,
constant-pressure,
closed-system
processes
Moreover,
Moreover,
Q
Q
=
=
�H
for
for
mechanically
mechanically
reversible,
reversible,
constant-pressure,
constant-pressure,
closed-system
closed-system
processes
processes
and
[Eq.
(2.23)]
for
of
heat
in
steady-flow
exchangers
where
�E
[ecuación (2.23)] y para la transferencia de calor en intercambiadores de flujo estableP donde�E
∆EKP are
y ∆EK son
and
are
[Eq.
(2.23)]
and
for
the
transfer
of
heat
in
steady-flow
exchangers
where
�E
and�E
�EKKKKare
are
[Eq.(2.23)]
(2.23)]
andfor
for
the
transfer
ofheat
heat
insteady-flow
steady-flow
exchangerswhere
where�E
�EPPPPand
�E
[Eq.
[Eq.
(2.23)]
and
and
for
the
transfer
transfer
of
of
heat
in
in
steady-flow
exchangers
where
�E
=the
0 [Eq.
(2.33)].
In cualquier
either
case,
negligible
P and�E
K are
insignificantes
y Wand
0 s[ecuación
(2.33)].
En
caso, exchangers
s =W
=
0
[Eq.
(2.33)].
In
either
case,
negligible
and
W
=
0
[Eq.
(2.33)].
In
either
case,
negligible
and
W
s
[Eq.(2.33)].
(2.33)].InIneither
eithercase,
case,
negligible
negligibleand
andWWssss==00[Eq.
�
��� TTT222
�
(4.3) (4.3)
Q = �H = TTT222 C P dT
CC
dT
(4.3)
Q
=
�H
=
CPPPPdT
dT
(4.3)
Q=
=�H
�H=
= T1 C
dT
(4.3)
(4.3)
Q
Q
=
�H
=
P
T
TTTT
11111
La aplicación
más común
de esta application
ecuación enofingeniería
es enislatotransferencia
de calor
en flujo estable.
The common
engineering
this equation
steady-flow heat
transfer.
The
common
engineering
application
of
this
equation
isis
to
steady-flow
heat
transfer.
Thecommon
commonengineering
engineeringapplication
applicationof
ofthis
thisequation
equationis
isto
tosteady-flow
steady-flowheat
heattransfer.
transfer.
The
The
common
engineering
application
of
this
equation
to
steady-flow
heat
transfer.
Temperature
Dependence
of
the
Heat Capacity
Dependencia
de laDependence
temperaturaof
de
la capacidad
calorífica
Temperature
Dependence
of
the
Heat
Capacity
Temperature
Dependence
ofthe
the
HeatCapacity
Capacity
Temperature
Heat
Evaluation of the integral in Eq. (4.3) requires knowledge of the temperature dependence of
Evaluation
of
the
integral
in
Eq.
(4.3)
requires
knowledge
of
the
temperature
dependence
of
Evaluation
of
theintegral
integral
inEq.
Eq.given
(4.3)
requires
knowledge
ofthe
thethe
temperature
dependence
of
Evaluation
Evaluation
of
the
the
integral
in
in
Eq.
(4.3)
(4.3)
requires
requires
knowledge
knowledge
of
of
the
temperature
temperature
dependence
dependence
of
of
La evaluación
laof
integral
deislausually
ecuación
(4.3)
requiere
el conocimiento
de
la dependencia
de la temperatuthe
heatdecapacity.
This
by
an empirical
equation;
two
simplest
expressions
the
heat
capacity.
This
is
usually
given
by
an
empirical
equation;
the
two
simplest
expressions
the
heat
capacity.
This
is
usually
given
by
an
empirical
equation;
the
two
simplest
expressions
the
the
heat
heat
capacity.
capacity.
This
This
is
is
usually
usually
given
given
by
by
an
an
empirical
empirical
equation;
equation;
the
the
two
two
simplest
simplest
expressions
expressions
ra de la of
capacidad
practicalcalorífica.
value are:Ésta, por lo general, se proporciona por una ecuación empírica; las dos expresioof
practical
value
are:
of
practical
valueare:
are:
of
practical
practical
value
are:
nes másof
simples
devalue
valor
C P práctico son:
CP
−2
CC
CC
and
CPPPP = α + βT + γ T 222
CPPPP = a + bT + cT −2
C
C
−2
2
2
−2
and
=
α
+
βT
+
γ
T
=
+
bT
+
cT
RP =
RP =
and
=ααα++
+βT
βT++
+γγγTTT2
=aaaa+
+bT
bT+
+cT
cT−2
−2
y
and
and
=
βT
=
+
bT
+
cT
RRR
RRR
R
R
where α, β, and γ and a, b, and c are constants characteristic of the particular substance. With
where
α,
β,
and
γ and
and
a,
b,
and
are
constants
characteristic
of
the
particular
substance.
With
where
α,así
β,and
and
and
a,cb,
b,
and
areconstants
constants
characteristic
ofthe
theparticular
particular
substance.
With del
donde a,
b exception
y α,
g,
como
a,
blast
ya,
son
constantes
características
laof
sustancia
particular.
Con
excepción
where
where
α,
β,
β,
and
and
a,
b,
and
and
ccccare
are
constants
characteristic
characteristic
of
the
particular
substance.
substance.
With
With
the
ofγγγthe
term,
these
equations
are
of thedesame
form.
We
therefore
combine
the
exception
of
the
last
term,
these
equations
are
of
the
same
form.
We
therefore
combine
the
exception
of
the
last
term,
these
equations
are
of
the
same
form.
We
therefore
combine
último the
término,
estas
ecuaciones
tienen
la
misma
forma.
Por
lo
tanto,
las
combinamos
para
proporcionar
una
the
exception
exception
of
of
the
the
last
last
term,
term,
these
these
equations
equations
are
are
of
of
the
the
same
same
form.
form.
We
We
therefore
therefore
combine
combine
them to provide a single expression:
them
to
provide
single
expression:
themto
toprovide
provideaaaasingle
singleexpression:
expression:
sola expresión:
them
them
to
provide
single
expression:
CP
−2
CC
(4.4)
CPPPP = A + BT + C T 222+ DT −2
C
−2
−2
=
+
BT
+
CC
+
DT
(4.4)
RP =
=AAAA+
+BT
BT+
+C
CTTTT222+
+DT
DT−2
(4.4) (4.4)
−2
=
+
BT
+
+
DT
(4.4)
(4.4)
R
R
RR
where either C or D is zero, depending
on the substance considered. Because the ratio C P /R
Donde where
ya
seaeither
C
o DC
esor
cero
dependiendo
de laon
sustancia
considerada.
Puesto
que la the
relación
where
either
CC
or
D
isis
zero,
depending
on
the
substance
considered.
Because
the
ratio
CC
/Rno tiene
where
either
C
or
Dis
iszero,
zero,
depending
on
thesubstance
substance
considered.
Because
theratio
ratioC
CPPP/R
P/R
/R
either
or
D
D
zero,
depending
on
the
the
substance
considered.
considered.
Because
Because
the
ratio
C
iswhere
dimensionless,
the
units
ofdepending
C
PP/R
P are governed by the choice of R.
dimensiones,
las
unidades
de
C
dependen
de
la
elección
de
R.
are
governed
by
the
choice
of
R.
is
dimensionless,
the
units
of
C
are
governed
by
the
choice
of
R.
is
dimensionless,
the
units
of
C
P
P
P
aregoverned
governed
by
bythe
thechoice
choice
ofofR.
R.
isisdimensionless,
dimensionless,
the
units
units
CCPgases
As shown inthe
Chap.
6,ofof
for
the
ideal-gas
heat
capacity,
rather than the actual heat caPPare
Como se
mostrará
en
el capítulo
6,
para
los
gases seheat
usa capacity,
la
capacidad
calorífica
del
gasheat
ideal,
en lugar
As
shown
inin
Chap.
6,
for
gases
the
ideal-gas
heat
capacity,
rather
than
the
actual
caAs
shown
Chap.
6,for
forgases
gases
the
ideal-gas
heat
capacity,
rather
than
theactual
actual
heat
caAs
shown
Chap.
Chap.
6,
6,
for
gases
the
the
ideal-gas
ideal-gas
heat
capacity,
rather
rather
than
than
the
the
actual
heat
cacapacityAs
is shown
used
ininin
the
evaluation
of
such
thermodynamic
properties
as the
enthalpy.
Theheat
reason
de la capacidad
calorífica
real
en
la
evaluación
de
las
propiedades
termodinámicas
como
la
entalpía.
La
razón
pacity
is
used
in
the
evaluation
of
such
thermodynamic
properties
as
the
enthalpy.
The
reason
pacity
is
used
in
the
evaluation
of
such
thermodynamic
properties
as
the
enthalpy.
The
reason
pacity
isused
usedininthe
theevaluation
evaluationofofevaluation
such
suchthermodynamic
thermodynamic
properties
propertiesas
asthe
theenthalpy.
enthalpy.
The
reason
ispacity
thatisthermodynamic-property
is most conveniently
accomplished
inThe
tworeason
steps:
es que is
la
evaluación
termodinámica
de
la
propiedad
se
consigue
en
forma
más
conveniente
siguiendo
is
that
thermodynamic-property
evaluation
is
most
conveniently
accomplished
in
two
steps:
is
that
thermodynamic-property
evaluation
is
most
conveniently
accomplished
in
two
steps:
isthat
that
thermodynamic-property
thermodynamic-property
evaluation
evaluationisisideal-gas
most
mostconveniently
conveniently
accomplished
accomplished
inintwo
twosteps:
steps: dos
first,
calculation
of values for a hypothetical
state wherein
ideal-gas heat
capacities
pasos: primero,
se calculan
los
valores
un hipotético
estado
delwherein
gas
ideal,
en donde
se usan
las capacidafirst,
calculation
of
values
for
aaapara
hypothetical
ideal-gas
state
wherein
ideal-gas
heat
capacities
first,
calculation
of
values
for
hypothetical
ideal-gas
state
wherein
ideal-gas
heat
capacities
first,
first,
calculation
calculation
of
of
values
values
for
for
a
hypothetical
hypothetical
ideal-gas
ideal-gas
state
state
wherein
ideal-gas
ideal-gas
heat
heat
capacities
capacities
are used; second, correction of the ideal-gas-state values to the real-gas values. A real gas
des caloríficas
delsecond,
gas ideal;
segundo,of
hace
una corrección
de los
del estado
de A
gas
ideal
para que
are
used;
second,
correction
the
ideal-gas-state
values
to
the
real-gas
values.
A
real
gas
areused;
used;
second,
correction
ofsethe
the
ideal-gas-state
values
tovalores
thewhen
real-gas
values.
A
real
gas
are
are
used;
second,
correction
correction
of
the
ideal-gas-state
ideal-gas-state
values
to
to
the
the
real-gas
real-gas
values.
values.
A
real
gas
gas
becomes
ideal
in the
limit as ofP
→
0;
if it were to values
remain
ideal
compressed
toreal
a finite
reflejenbecomes
los valores
delin
gas
real.
Unas
gas
será
en to
el
límite
conforme
P→
0; si continuara
siendo
ideal
becomes
ideal
in
the
limit
as
PPPreal
→
0;
ifififideal
itititwere
were
to
remain
ideal
when
compressed
to
aaafinite
finite
becomes
ideal
in
the
limit
as
→
0;
were
to
remain
ideal
when
compressed
to
finite
becomes
ideal
ideal
in
the
the
limit
limit
as
P
→
→
0;
0;
if
it
were
to
remain
remain
ideal
ideal
when
when
compressed
compressed
to
to
a
finite
pressure, it would exist in a hypothetical ideal-gas state. In such states gases have propercuandopressure,
se
comprime
a presiones
finitas,
existiría en
un estado
de gas
ideal
hipotético.
Los
gases
en sus estapressure,
itititwould
would
exist
in
aaahypothetical
hypothetical
ideal-gas
state.
In
such
states
gases
have
properpressure,
would
exist
in
hypothetical
ideal-gas
state.
In
such
states
gases
have
properpressure,
it
would
exist
exist
in
in
a
hypothetical
ideal-gas
ideal-gas
state.
state.
In
In
such
such
states
states
gases
gases
have
have
properproperties that reflect their molecular structure just as do real gases. Ideal-gas-state heat capacities
dos de ties
gas
tienen
propiedades
reflejan just
su
estructura
molecular
al igual que en
loscapacities
gases
reales. Por
ties
that
reflect
their
molecular
structure
just
as
do
real
gases.
Ideal-gas-state
heat
capacities
tiesideal
thatreflect
reflecttheir
their
molecular
structure
just
asdo
doreal
real
gases.Ideal-gas-state
Ideal-gas-state
heat
capacities
ig molecular
ig que
ties
that
that
reflect
molecular
structure
structure
as
as
do
real
gases.
Ideal-gas-state
heat
heat
capacities
(designated
by Ctheir
thereforejust
different
for gases.
different
although
functions
of
ig
ig
ig
ig )gases;
ig and C ig
ig) are
P
V
ig
ig
ig
lo tanto,
las
capacidades
térmicas
del
gas
ideal
(denotadas
por
C
y
C
son
diferentes
para
gases
distintos;
ig
ig
(designated
by
C
and
C
)
are
therefore
different
for
different
gases;
although
functions
of
and
C
)
are
therefore
different
for
different
gases;
although
functions
of
(designated
by
C
P
V
and
and
CCVVVVV))are
aretherefore
therefore
different
differentfor
fordifferent
differentgases;
gases;although
althoughfunctions
functionsof
of
(designated
(designatedby
by
CCPPPPPare
temperature,
they
independent
of pressure.
aunquetemperature,
son
funciones
de
la
temperatura,
son
independientes
de
la
presión.
temperature,
they
are
independent
of
pressure.
temperature,
they
are
independent
of
pressure.
temperature,
they
they
are
arecapacities
independent
independent
ofofpressure.
pressure.
Ideal-gas
heat
increase
smoothly with increasing temperature toward an upper
Las capacidades
caloríficas
delincrease
gas
ideal
aumentan
deincreasing
manera uniforme
conforme
se
la
Ideal-gas
heat
capacities
increase
smoothly
with
increasing
temperature
toward
an
upper
Ideal-gas
heatcapacities
capacities
increase
smoothly
with
increasing
temperature
toward
anincrementa
upper
Ideal-gas
heat
capacities
increase
smoothly
smoothly
with
with
increasing
temperature
temperature
toward
toward
an
an
upper
upper
limit, Ideal-gas
which
is heat
reached
when all
translational,
rotational,
and vibrational
modes
of molecular
temperatura
hacia
un
límite
superior,
el
cual
se
alcanza
cuando
se
excitan
completamente
todos
los
modos
de
ig
limit,
which
is
reached
when
all
translational,
rotational,
and
vibrational
modes
of
molecular
limit,
which
is
reached
when
all
translational,
rotational,
and
vibrational
modes
of
molecular
limit,
limit,which
which
isisreached
reached
when
alltranslational,
translational,
rotational,
rotational,
and
and
vibrational
modes
modes
of
molecular
molecular
motion
are fully
excitedwhen
[seeall
Eq.
(16.18)]. The
influence
ofvibrational
temperature
on Cof
for
argon,
ig
ig
P
ig
movimiento
molecular
traslación,
rotación
y vibración
[véase of
la
ecuación
(16.18)].
En
laargon,
figura 4.1 se
ig
motion
are
fully
excited
[see
Eq.
(16.18)].
The
influence
of
temperature
on
CC
for
argon,
motion
are
fullyde
excited
[see
Eq.(16.18)].
(16.18)].
Theinfluence
influence
oftemperature
temperature
onC
Cig
for
argon,
motion
motion
are
are
fully
fully
excited
excited
[see
[see
Eq.
Eq.
(16.18)].
The
The
influence
of
temperature
on
on
for
for
argon,
P
P
nitrogen,
water,
and
carbon
dioxide
is
illustrated
in
Fig.
4.1.
Temperature
dependence
is
exPPP
ig
muestranitrogen,
la
influencia
de
la
temperatura
en
C
para
el
argón,
el
nitrógeno,
el
agua
y
el
dióxido
de
La
nitrogen,
water,
and
carbon
dioxide
is
illustrated
in
Fig.
4.1.
Temperature
dependence
is
exnitrogen,
water,
and
carbon
dioxide
is
illustrated
in
Fig.
4.1.
Temperature
dependence
iscarbono.
exP
nitrogen,
water,
water,and
andby
carbon
carbon
dioxide
dioxide
isisillustrated
illustrated
ininhere
Fig.
Fig.written:
4.1.
4.1. Temperature
Temperaturedependence
dependenceisis
exexpressed
analytically
equations
such
as
Eq. (4.4),
dependencia
de
la
temperatura
se
expresa
en
forma
analítica
mediante
ecuaciones
como
la
(4.4),
que
aquí
se
pressed
analytically
by
equations
such
as
Eq.
(4.4),
here
written:
pressed
analytically
by
equations
such
as
Eq.
(4.4),
here
written:
pressed
pressedanalytically
analyticallyby
byequations
equationssuch
suchasasEq.
Eq.(4.4),
(4.4),here
herewritten:
written:
ig
escribe como:
C ig
igig
−2
CC
CPPigPig = A + BT + C T 222+ DT −2
C
−2
222 +
PPP =
−2
DT
A
+
BT
+
C
T
R
+
DT
=
A
+
BT
+
C
T
−2
DT −2
==AA++BT
BT++CCTT ++DT
RRRR
04-SmithVanNess.indd 127
8/1/07 13:07:25
128
CAPÍTULO 4. Efectos térmicos
CHAPTER 4. Heat Effects
128
Los valores de los parámetros se dan en la tabla C.1 del apéndice C para varios gases orgánicos e inorgánicos
1
comunes.
En laofliteratura
es posible
ecuaciones
exactas
más complejas.
Values
the parameters
areencontrar
given in Table
C.1 ofmás
App.
C for pero
a number
of common organic
and inorganic gases. More accurate but more complex equations are found in the literature.1
7
CO 2
7
CO 2
6
6
H 2O
H 2O
5
5
Figura 4.1: Capacidades caloríficas de
4.1: nitrógeno,
Ideal-gas heat
gas idealFigure
del argón,
agua y
capacities
of
bióxido de carbono. argon, nitrogen,
water, and carbon dioxide.
CPig
R
CPig
N2
4
R
N2
4
3
3
Ar
Ar
2
2
500
500
1 000
T/K
1500
1000
T/K
1 500
2 000
2000
Como resultado
deof
la Eq.
ecuación
lasideal-gas
dos capacidades
caloríficas
del gas ideal están relacionadas:
As a result
(3.19),(3.19),
the two
heat capacities
are related:
ig
ig
CV
C
= P −1
R
R
ig
(4.5) (4.5)
ig
The temperature dependence of C V /R follows from the temperature dependence of C P /R. ig
La dependencia de la temperatura de CVig/R
es consecuencia
de la dependencia de la temperatura de CP /R.
ig ig
ig
The de
effects
of temperature
C Vobtienen
are determined
experiment, most
often
Los efectos
la temperatura
sobreon
CPigCoP Cor
se
de manerabyexperimental;
con frecuencia,
se
V
from
spectroscopic
data
and
knowledge
of
molecular
structure
through
calculations
based
on
calculan mediante métodos de la mecánica estadística a partir de datos espectroscópicos y del conocimiento
statisticalmolecular.
mechanics.
Where
data areexperimental
not available,semethods
estimation
are
de la estructura
Cuando
noexperimental
se tiene información
empleanofmétodos
de estimación,
2
2
employed,
as
described
by
Reid,
Prausnitz,
and
Poling.
como los descritos por Reid, Prausnitz, Poling y O’Connell.
ideal-gas
heat capacities
exactly
forpara
real gases
gases reales
only atsólo
zeroapressure,
AunqueAlthough
las capacidades
caloríficas
del gasare
ideal
son correct
correctas
presión cero, la
the
departure
of
real
gases
from
ideality
is
seldom
significant
at
pressures
below
severaldebars,
desviación de la idealidad de los gases reales rara vez es significativa en presiones por debajo
varios bar,
ig ig
ig
here
Cy
C
are
usually
good
approximations
to
their
true
heat
capacities.
y en esteand
caso
CPig
por
lo
general
son
buenas
aproximaciones
de
sus
capacidades
térmicas
reales.
P Cand
V
V
1 Véase F.
1 See
A. Aly
y L.
L. and
Lee,L.
Fluid
Phase
Equilibria,
vol. 6, pp. 169-179,
y su bibliografía;
véase
también T.see
E. Daubert,
R. P.
F. A.
Aly
L. Lee,
Fluid
Phase Equilibria,
vol. 6, pp.1981,
169–179,
1981, and its
bibliography;
also
Danner, H.
M.
y C. R.
C. Stebbins,
Thermodynamic
Properties
of Pureand
Chemicals:
Data Compilation,
& Francis,
T. E. Sibul
Daubert,
P. Danner,Physical
H. M. and
Sibul,
and C. C. Stebbins,
Physical
Thermodynamic
Properties Taylor
of Pure
Chemicals:
Bristol, PA,
existente Data
1995.Compilation, Taylor & Francis, Bristol, PA, extant 1995.
2 J. M. Prausnitz, B. E. Poling y J. P. O’Connell, The Properties of Gases and Liquids, 5a. edición, capítulo 3, McGraw-Hill, Nueva
2
York, 2001. J. M. Prausnitz, B. E. Poling, and J. P. O’Connell, The Properties of Gases and Liquids, 5th ed., chap. 3, McGrawHill, New York, 2001.
04-SmithVanNess.indd 128
8/1/07 13:07:27
4.1. Sensible Heat Effects
4.1. Sensible Heat Effects
4.1. Sensible Heat Effects
Example
4.1
4.1. Efectos
del calor sensible
129
129
129
Example
4.1
The parameters listed in Table C.1 require use of Kelvin temperatures in Eq. (4.4).
Example
4.1
The
parameters
listed
Table
C.1
require
use of Kelvin
temperatures
in Eq. (4.4).
Equations
sameinform
may
also
be developed
for use
with temperatures
in ◦◦C,
Ejemplo
4.1◦◦of the
◦
129
C,
Equations
thebut
same
mayC.1
also
be developed
for use
with
temperatures
(R), parameters
and (ofF),
the form
parameter
values
are
different.
The
molar
heatincapacity
of
The
listed
in
Table
require
use
of Kelvin
temperatures
Eq.in(4.4).
◦ F), but theen
Los parámetros
lastate
tabla
requieren
del usoof
de
temperaturas
Kelvin
en
ecuación
(R),
and enumerados
(in
values
different.
The
molar
heat
capacity
methane
isC.1
given
asare
a function
temperature
in kelvins
by:
C,
Equations
ofthe
theideal-gas
same parameter
form
may
also
be
developed
for
use
with
temperatures
inla◦ of
(4.4). De
esta
forma,
pueden
desarrollar
ecuaciones
para
con
temperaturas
methane
in
thetambién
ideal-gas
state
is given
as aare
function
of temperature
kelvins
by: of en °C,
(R),
and
(◦ F),
but
these
parameter
values
different.
Theusarse
molar in
heat
capacity
ig
ig
(R) y (°F),
aunque
los valores
de as
losaparámetros.
La capacidadincalorífica
molar del meC PP state
methane
inson
the diferentes
ideal-gas
is given
function
of temperature
kelvins by:
−3
−6 22
−3
−6
ig =dada
1.702
+
9.081
×
10
T
−
2.164
×
10
T
tano en estado de gas ideal
está
como
una
función
de
la
temperatura
en
kelvins
por:
C
P
R
−3
−6 2
ig = 1.702 + 9.081 × 10 T − 2.164 × 10 T
CRP
−3
−6 T 2
ig
ig
= 1.702
9.081Table
× 10 C.1.
T −Develop
2.164 × 10
where the parameterR values
are+from
an equation
for C PP /R for
ig
◦
where
the parameter
temperatures
in ◦ C. values are from Table C.1. Develop an equation for C Pig/R for
◦ C.
donde los
valores
de
los
parámetros
se toman de la tabla C.1. Desarrolle una ecuación para Cig /R con tempetemperatures
in
where the parameter
values are from Table C.1. Develop an equation for C PP /R for
raturas en
°C.
temperatures
in ◦ C.
Solution 4.1
Solution
4.1
Solución 4.1
The relation
between the two temperature scales is:
Solution
4.1
La relación
entre las
dos escalas
de latemperature
temperaturascales
es: is:
The relation
between
the two
T K = t ◦◦C scales
+ 273.15
The relation between the two temperature
is:
= tt ◦°C
TTKK=
C ++273.15
273.15
Therefore,
functionde
of t,t, T K = t ◦ C + 273.15
Por lo tanto,
como as
unaa función
Therefore, as a function of t,
ig
CPPig as a function of t, −3
Therefore,
−6 (t + 273.15)22
ig = 1.702 + 9.081 × 10−3 (t + 273.15) − 2.164 × 10−6
CR
2
P
−3
−6
ig = 1.702 + 9.081 × 10 (t + 273.15) − 2.164 × 10 (t + 273.15)
CRP
−3
ig
ig
= 1.702
− 2.164 × 10−6 (t + 273.15)2
C PP+ 9.081 × 10 (t + 273.15)
R
−3
−6 t22
−3
ig
=
4.021
+
7.899
×
10
or
t
− 2.164 × 10−6
o
CR
P
−3
−6 2
or
ig = 4.021 + 7.899 × 10 t − 2.164 × 10 t
CRP
= 4.021 + 7.899 × 10−3 t − 2.164 × 10−6 t 2
or
R
Gas mixtures of constant composition behave exactly as do pure gases. In an ideal-gas
Las
mezclas
de
gasesofdeconstant
composición
constante
seanother,
comportan
como
losideal-gas
gases puros.
En
Gasthe
mixtures
exactlyand
asexactamente
do pure
In
an
mixture
molecules
have
no composition
influence
onbehave
one
each
gasgases.
exists
independent
of
una mezcla
de
gases
ideales
las
moléculas
carecen
de
alguna
influencia
entre
sí,
y
cada
gas
existe
de
manera
mixture
themixtures
molecules
no influence
onaone
another,
andaseach
gasmole-fraction-weighted
exists independent
of
the others.
The ideal-gas
heat
capacity
of
mixture
is therefore
Gas
of have
constant
composition
behave
exactly
do the
pure
gases.
In an ideal-gas
independiente
de
los
otros.
Debidoheat
a eso,
la capacidad
calorífica
del gases
gas ideal
de
una
mezcla
es la suma
de las
the
others.
The
ideal-gas
capacity
of
a
mixture
is
therefore
the
mole-fraction-weighted
sum
of
the
heat
capacities
of
the
individual
gases.
Thus
for
A,
B,
and
C
the
molar
heat
mixture the molecules have no influence on one another, and each gas exists independent of
capacidades
caloríficas
ponderadas
con
la
fracción
mol
de
los
gases
individuales.
De
este
modo,
para
los
sum
of theofheat
capacities
of
thecapacity
individual
gases.
Thus
gases A,
and C the molar heat
capacity
aThe
mixture
in the
ideal-gas
state
the
others.
ideal-gas
heat
of is:
a mixture
is for
therefore
theB,mole-fraction-weighted
gases A,capacity
B
y
C
la
capacidad
calorífica
molar
de
una
mezcla
en
el
estado
de
gas
ideal
es:
of aheat
mixture
in theofideal-gas
state is:gases. Thus for gases A, B, and C the molar heat
sum of the
capacities
the individual
ig
ig
ig
ig
ig
ig
ig
ig
=
y AACis:
(4.6)
capacity of a mixture in the C
ideal-gas
state
BC P
CC P
Pmixture
PAA + y B
PBB + yC
PCC
P
P
mixture
ig
ig
ig
ig
C Pmixture = y A C PA + y B C PB + yC C PC
(4.6) (4.6)
mezcla
ig
ig
ig
ig
ig
ig
ig
ig
ig
ig
C Pmixture
= yAC
y B C PB +
(4.6)
C C PCA, B, and C in the ideal-gas
where C PPAA , C PPBB , and C PPCC are
the molar
heat
capacities
ofypure
PA +
ig
ig
ig
ig
ig
ig
donde Cwhere
,
C
y
C
son
las
capacidades
caloríficas
molares
de
A,
B
y
C
puras
en
estado
de
gas
ideal y yA,
C
,
C
,
and
C
are
the
molar
heat
capacities
of
pure
A,
B,
and
C
in
the
ideal-gas
state,
and
y
,
y
,
and
y
are
mole
fractions.
A
B
C
A
B
C
PB
PC
PA PBPA PC
igy las
igfracciones
yB y yCstate,
representan
molares.
and
,Cygases,
yCC ig
are
mole
fractions.
A
B ,, and
As
the
heat
capacities
of
solids
and
liquids
are
found
by
experiment.
Paramwhere
C Pwith
,
and
are
the
molar
heat
capacities
of
pure
A,
B,
and
C
in
the
ideal-gas
PB
PC
A
Alstate,
igualfor
que
conthe
los
gases,
lasfractions.
capacidades
caloríficas
sólidos
y líquidos
seParamdeterminan
de
As
with
capacities
and liquids
arelos
found
bygiven
experiment.
eters
the
dependence
of of
C PPsolids
as expressed
by de
Eq.
(4.4)
are
for a few
solids
and
yocurre
, gases,
y B , and
yCheat
are
mole
Atemperature
maneraeters
experimental.
Los parámetros
para
de liquids
la temperatura
de
Cgiven
como
expresan
for
temperature
ofdependencia
CofPC.solids
asCorrelations
expressed
by
are
for
amany
few
solids
Pexperiment.
and
liquids
in
Tables
C.2
and C.3
oflaApp.
forEq.
the
heat
capacities
ofse
solidsmedianAsthe
with
gases,
the dependence
heat
capacities
and
are(4.4)
found
by
Paramte la ecuación
están
dados
paraC.3
algunos
y líquidos
en
las33(4.4)
tablascapacities
C.2
y C.3
apéndice
and
Tables
C.2
ofGreen
App.
Correlations
for
the
heat
ofdel
many
solids
andliquids
liquids
given
by and
Perry
and
and
by
Daubertby
et Eq.
al.
eters
for(4.4)
theinare
temperature
dependence
ofsólidos
CC.
expressed
are
given
for
a few
solids C. Las
P as
3
correlaciones
para are
lasTables
capacidades
deand
diversos
sólidoset
líquidos
están dadas
por Perry
y Green,
and
given
by Perry
andofGreen
by Daubert
and liquids
liquids
in
C.2
andcaloríficas
C.3
App.
C.
Correlations
foryal.
the
heat capacities
of many
solids
3
33R. Daubert
así como
por
ygiven
colaboradores.
H. Perry
D. Green,
Perry’s
Chemical
Handbook,
and
liquids
areand
by Perry
and
GreenEngineers’
and by Daubert
et 7th
al.3ed., Sec. 2, McGraw-Hill, New York,
3 R. T.
1997;
Daubert
op. cit.
H.E.Perry
and etD.al.,
Green,
Perry’s Chemical Engineers’ Handbook, 7th ed., Sec. 2, McGraw-Hill, New York,
1997;
E. Daubert et al., op. cit.
3 R.T.H.
Perry and D. Green, Perry’s Chemical Engineers’ Handbook, 7th ed., Sec. 2, McGraw-Hill, New York,
3 R. H.
1997;
T. yE.D.Daubert
al., op.Chemical
cit.
Perry
Green,etPerry’s
Engineer’s Handbook, 7a. edición, sección 2, McGraw-Hill, Nueva York, 1997; T. E.
Daubert et al., op. cit.
04-SmithVanNess.indd 129
8/1/07 13:07:30
130
130
130
130
130
130
130
130
CHAPTER
4.
Heat
Effects
CHAPTER
4. 4.
Heat
Effects
CHAPTER
Effects
CAPÍTULO
4. Heat
Efectos
térmicos
CHAPTER
4.4. Heat
Effects
CHAPTER
Heat
Effects
CHAPTER 4.
4. Heat
Heat Effects
Effects
CHAPTER
Evaluation
the
Sensible-Heat
Integral
Evaluation
of
the
Sensible-Heat
Integral
Evaluación
de laof
integral
del calor sensible
Evaluation
of
the
Sensible-Heat
Integral
Evaluation
of
the
Sensible-Heat
Integral
�
�
Evaluation
of
the
Sensible-Heat
Integral
�
Evaluation
of
the
Sensible-Heat
Integral
Evaluation
of
the
Sensible-Heat
Integral
Evaluation
of
the
integral
isisaccomplished
by
substitution
for
CCPPCas
aafunction
of
TT,,T ,
��CCPPCdT
Evaluation
ofof
thethe
integral
dT
accomplished
byby
substitution
forfor
as
function
ofof
Evaluation
integral
is
accomplished
substitution
a function
P dT
P as
� Cse
La evaluación
deby
laofformal
integral
∫ CP �dT
consigue
mediante
labysustitución
para
CCPresult
función
de
Evaluation
the
integral
dT
istemperature
accomplished
substitution
for
asasaisis
function
ofofT,
TT,seguida
Pcomo
followed
integration.
For
limits
of
TT00Tand
TTfor
the
conveniently
Evaluation
of
theintegral
integral
CPPPPdT
dTFor
is
accomplished
bysubstitution
substitution
forthe
C
function
followed
byof
formal
integration.
For
temperature
limits
of
and
result
conveniently
followed
by
formal
integration.
temperature
limits
of
Tthe
is
conveniently
Evaluation
of
the
integral
dT
accomplished
by
substitution
for
as aaafunction
function
of TTen
Evaluation
the
CC
isis
accomplished
by
CCresult
as
of
,, , forma
0yand
PPPresult
por la integración
formal.
Para
los
límites
de
las
temperaturas
de
T
T,
el
resultado
se
expresa
followed
by
formal
integration.
For
temperature
limits
of
T
and
T
the
is
conveniently
0
0
expressed
as:
followed
by
formal
integration.
For
temperature
limits
of
T
and
T
the
result
is
conveniently
expressed
as:
0
expressed
as:
followed
by formal
formal integration.
integration. For
For temperature
temperature limits
limits of
of TT00 and
and TT the
the result
result isis conveniently
conveniently
followed
by
conveniente
como:
expressed
as:
expressed
as:
expressed
as:
�� � �� �
expressed
�� T�T as:
BB B22 22
CC C33 33
DD D
−
1 �1�
��ττ −
� CTCPPC P
τ 1−
2 −
2 1)
3 −
3 1)
−
+
+
+
=
AT
(4.7)
�ττ−−11�
�
dTdT
==
ATAT
(4.7)
(τ0 (τ
−1)
1)1)
++
(τ
−
1)1)
++
(τ
−
1)1)
+ +DD�
BBTT002T(τ
CCTT003T(τ
���TTTTCCP dT
00(τ
−
(τ
−
(τ
−
(4.7) (4.7)
2
3
0
0
RRPPPRdT
2
3
T
τ
C
τ
−
1
B
C
D
2
2
3
3
τ
−
1
2
3
T
τ
=
AT
T
T
(4.7)
(τ
−
1)
+
(τ
−
1)
+
(τ
−
1)
+
B
C
D
0
TT00 TC
2
3
T
τ
0 0 (τ − 1) +
2 (τ
3 (τ
22 − 1) +
33 − 1) + 0 0
0
0
2
3
dT
=
AT
T
T
(4.7)
0 R dT
(τ
−
1)
+
(τ
−
1)
+
(τ
−
1)
+
dT
=
AT
T
T
(4.7)
0
(τ
T
T
(4.7)
0
2
3
T
τ
(τ
−
1)
+
−
1)
+
(τ
−
1)
+
=
AT
0
0
0
T0T
0
RR
222 0
333 00
τττ
TTT
000
TT00 0 R
TT T
ττ ≡
donde where
where
≡
where
τ ≡T
where
ττ≡≡TTT0T
0T
where
where
τ
≡ TT0 0
where
τ≡
0 is straightforward.
T
T
00�H
Given
and
TT,,Tthe
of
QQ Q
or
�H
isisTthe
calDados T
y T,Tel
de
Qcalculation
ocalculation
∆H es directo.
Esor
menos
directo
el cálculoLess
deLess
T,direct
dados
ythe
Q
ocal∆H. En
Given
T00Tcálculo
and
the
calculation
ofof
or
�H
is is
straightforward.
Less
direct
cal0 Given
0the
, the
straightforward.
direct
is
0 and
Given
T
and
T
,
the
calculation
of
Q
or
�H
isisstraightforward.
Less
direct
is
the
cal0
and
Q
or
�H
.
Here,
an
iteration
scheme
may
be
useful.
Factoring
culation
of
TTTT
,,T
given
T
Given
and
T
,
the
calculation
of
Q
or
�H
straightforward.
Less
direct
is
the
caleste caso,
un Given
planteamiento
iterativo
es
de
utilidad.
Al
factorizar
(τ
–
1)
de
cada
término
del
lado
derecho
de
culation
ofof
given
and
Q
or
�H
.
Here,
an
iteration
scheme
may
be
useful.
Factoring
0
0
and
Q
or
�H
.
Here,
an
iteration
scheme
may
be
useful.
Factoring
culation
T
,
given
T
0
Given
and
T
,
the
calculation
of
Q
or
�H
is
straightforward.
Less
direct
is
the
caland TT, the
0 calculation of Q or �H is straightforward. Less direct is the calculation
ofof
TTeach
,00,given
and
QQ
oror�H
. . Here,
anan
iteration
scheme
may
bebeuseful.
Factoring
0 on
(τ
−
1)
from
term
the
right-hand
side
of
Eq.
(4.7)
gives:
culation
given
and
�H
Here,
iteration
scheme
may
useful.
Factoring
la ecuación
(τ
−(4.7)
1)1)
from
term
on
the
right-hand
side
of
Eq.
(4.7)
gives:
0
(τ
−
from
each
term
on
the
right-hand
side
of
Eq.
(4.7)
gives:
culation
ofse
, given
given
and
Q
or
�H
.
Here,
an
iteration
scheme
may
be
useful.
Factoring
culation
of
TTobtiene:
,each
TTT
and
Q
or
�H
.
Here,
an
iteration
scheme
may
be
useful.
Factoring
00
(τ(τ−−1)1)from
each
term
on
the
right-hand
side
ofofEq.
(4.7)
gives:
eachterm
termon
theright-hand
right-handside
sideof
Eq.(4.7)
(4.7)gives:
gives:
(τ −
− 1)
1) from
from
each
term
the
right-hand
side
of Eq.
Eq.
(4.7)
gives:
��
�on
(τ
�from
�on�the
� T�T each
T
C
B
C
DD D� �
B B22 2
C C33 322 2
��
��T CPPC P
−
+
+
+ ττ +
+
=
��AT
(τ (τ
−1)
1)1)
+
(τ
+1)
1)1)
++
(τ
+
1)1)
+ +DD���(τ
dTdT
==
ATAT
BBTT002T(τ
CCTT003T(τ
�� TTTCCP dT
00 +
−
+
τ 1)
+
0+
2 2+ +
0 (τ
0 (τ
RRPPPRdT
2
3
TD
B
C
C
2
3
B
C
2
3
Tτ00T0(τ(τ−−1)1)
+
(τ
+
1)
+
(τ
+
τ
+
1)
+
=
AT
T
T
TT00 TC
2
3
0
2 + τ + 1) +ττD
22 (τ + 1) +
33 (τ
0
0
2
+
dT
=
AT
T
T
0 R dT
0
+ 22TT000(τ
(τ +
+ 1)
1) +
+ 33TT000(τ
(τ +
+ ττ +
+ 1)
1) +
+ ττTT0 (τ
dT =
= AT
AT00 +
(τ −
− 1)
1)
T0
RR
22
33
ττTT000
TT0T0 0 R
TT −
T
−
T
0
T −0T
Because
ττ −
Because
−
1=
=
TT−−TT0 0
Because
τ 1−
1=
T
T
−
T00TT00
τ
−
1
=
T
−
Ya que Because
Because
Because
−−111=
== TT0 0
Because
τττ−
this
may
be
written:
TT000
this
may
bebe
written:
this
may
written:
this
may
be
written:
this
may
be
written:
ésta se puede
escribir
como: �� �
�� �
this may
may
this
be
�be
� Twritten:
�written:
��
��
CC C22 22
DD D�
BB B
��TT CTCPPC P
�(T
�
�
2
2
+
+
+
1)
+
−
=
(τ0 (τ
+1)
1)1)
++
(τ
+τ+
τ+
+
1)1)
+ +DD
(T(T
−T
T00)T) )
dTdT
= =AA+
++
CCTT002T(τ
BBTT00(τ
�� TTTCCP dT
+
(τ
τ
+
−
A
T
2
2
0
2
2
RRPPPRdT
2
3
C
C
D
B
2
2
τ
T
C
B
2
3
(τ
+
1)
+
(τ
+
τ
+
1)
+
−−TT0 ) )0
=
A
+
T
T
D
TT00 TC
τ
T
2
3
0
τ
T
22 (τ
22 + τ + 1) + 002 0 (T
0
(τ
+
1)
+
(T
dT
=
A
+
T
T
0 R dT
0
(τ +
+ 1)
1) +
+ 33TT000(τ
(τ +
+ ττ +
+ 1)
1) +
+ τ T0222 (T
(T −
− TT00)0)
dT =
= AA +
+ 22TT00(τ
T0
RR
22
33
τττTTT
TT0T0 0 R
000
We
identify
the
quantity
in
square
brackets
as
�C
/R,/R,
where
�C
isis defined
as
aa mean
WeWe
identify
thethe
quantity
in in
square
brackets
asas
�C�C
where
�C�C
defined
asas
mean
PP��HP
PP��HP
identify
quantity
square
brackets
where
is
defined
a mean
H�/R,
H�H
H where
We
identify
the
quantity
in
square
brackets
as
�C
�
/R,
�C
�
is
defined
as
aamean
P
P
H
H
heat
capacity:
We
identify
the
quantity
in
square
brackets
as
�C
�
/R,
where
�C
�
is
defined
as
mean
Identificamos
la
cantidad
entre los
rectangulares
⟨C
⟩
se
define
como una
heat
capacity:
P
P⟩ H /R,
P
H
heat
capacity:
We
identify
the quantity
quantity
inparéntesis
square brackets
brackets
as �C
�CPPP�como
�HHH/R,
/R,⟨C
where
�CPdonde
�
is
defined
as
a
mean
�
is
defined
as
a
mean
H
We
identify
the
in
square
as
where
�C
P HH
heat
capacity:
heat
capacity:
capacidad
heatcalorífica
capacity:media:
heat
capacity:
�C
CC C22 22
DD D
BB B
�C�C
PP��HP
H�H
2 +
2 ττ +
+
+
1)
+
(4.8)
=
AA+
�C
(τ0 (τ
+1)
1)1)
++
(τ
+
+
1)
+
(4.8)
==
++
CCTT002T(τ
DD
BBTT00(τ
+
(τ
+
τ
+
1)
+
(4.8)
A
T
PP�H
2
22 2
�
�C
3C
�CRPRP��H
�C
D
20 222+ τ + 1) + ττT
D
2 T2T0 (τ(τ++1)1)++C
3 T3T022(τ
(4.8)
T
RHH ==AA++2BB
τ
T
0
(τ
+
τ
+
1)
+
(4.8) (4.8)
0
2
0(τ +
+ 1)
1) +
+ 33TT000(τ
(τ +
+ ττ +
+ 1)
1) +
+ τ T02220
(4.8)
= AA +
+ 22TT00(τ
RR =
(4.8)
RR
22
33
τττTTT
000
Equation
(4.2)
may
therefore
be
written:
Equation
(4.2)
may
therefore
bebe
written:
Equation
(4.2)
may
therefore
written:
Equation
(4.2)
may
therefore
bebewritten:
(4.2)may
may
therefore
written:
Equation
(4.2)
may
therefore
be written:
written:
(4.2)
therefore
be
DebidoEquation
aEquation
eso, la ecuación
(4.2)
se escribe
como:
(T (T
−
(4.9)
�H
=
−TT00)T) 0 )
(4.9)
�H
= �C
�C�C
PP��HP
(4.9)
�H
=
H�(T
H −−
(T
TT0 ) )
(4.9)
�H
==�C
PP�H
�
(T
−
(4.9) (4.9)
�H
�C
0
(T −
− TT00))
(4.9)
�H =
= �C
�CPP��HHH(T
(4.9)
�H
The
angular
brackets
enclosing
CCPPCidentify
itit as
aa mean
value;
subscript
“H
”” denotes
aa mean
The
angular
brackets
enclosing
identify
as
mean
value;
subscript
“H“H
denotes
mean
The
angular
brackets
enclosing
it as
a mean
value;
subscript
” denotes
a mean
P identify
The
angular
brackets
enclosing
C
identify
it
as
a
mean
value;
subscript
“H
”
denotes
asimilar
mean
P
value
specific
to
enthalpy
calculations,
and
distinguishes
this
mean
heat
capacity
from
a
similar
The
angular
brackets
enclosing
C
identify
it
as
a
mean
value;
subscript
“H
”
denotes
a
mean “H”
value
specific
to to
enthalpy
calculations,
and
distinguishes
this
mean
heat
capacity
from
a
P
value
specific
enthalpy
calculations,
and
distinguishes
this
mean
heat
capacity
from
a
similar
Los paréntesis
angulares
que
contienen
se
identifican
como
su
valor
medio;
en
tanto,
el
subíndice
The
angular
brackets
enclosing
identify
it
as
a
mean
value;
subscript
“H
”
denotes
a
mean
The
angular
brackets
enclosing
CCC
identify
it
as
a
mean
value;
subscript
“H
”
denotes
a
mean
P
PP
value
specific
to
enthalpy
calculations,
and
distinguishes
this
mean
heat
capacity
from
a
similar
quantity
introduced
in
the
next
chapter.
value
specific
to
enthalpy
calculations,
and
distinguishes
this
mean
heat
capacity
from
a
similar
quantity
introduced
in in
thethe
next
chapter.
quantity
introduced
next
chapter.
denota value
un
valor
medio
para
cálculos
dedistinguishes
la entalpía ythis
distingue
esta capacity
capacidad
calorífica
media de
value
specific
toespecífico
enthalpy
calculations,
and
distinguishes
this
mean heat
heat
capacity
from
similar
specific
to
enthalpy
calculations,
and
mean
from
aa similar
quantity
introduced
inin
the
next
chapter.
Solution
of
Eq.
for
TT
gives:
quantity
introduced
the
next
chapter.
Solution
ofof
Eq.
(4.9)
for
Solution
Eq.
(4.9)
for
Tgives:
gives:
una cantidad
similar
que
será
introducida
en
el siguiente capítulo.
quantity
introduced
in(4.9)
the
next
chapter.
quantity
introduced
in
the
next
chapter.
Solution
ofofEq.
(4.9)
for
TTgives:
Solution
Eq.(4.9)
(4.9)
forT
gives:
La solución
de laof
(4.9)
T da:
Solution
ofecuación
Eq.
(4.9)
for
Tpara
gives:
Solution
Eq.
for
gives:
�H
�H
�H+ T
TT =
(4.10)
++
T00T0
(4.10)
�H
T= =
(4.10)
�H
�H
�C
TT==�C
(4.10)
�H
PP��HP
�C
�H++TT0 0
H
(4.10) (4.10)
= �C
+
T
(4.10)
+
T
(4.10)
�
TT =
0
P
0
�CPP��H�H
�C
�C
HH) allows evaluation of �C � by Eq. (4.8).
AA starting
value
for
TT (and
hence
for
ττ =
TTP/T
evaluation
ofof
�C�C
Eq.
(4.8).
starting
value
forfor
hence
forfor
/T
PP �HP
00 ) 0allows
) allows
evaluation
� byby
Eq.
(4.8).
A
starting
value
T(and
(and
hence
τ= =
T /T
allows
evaluation
ofof�C
�HH� H
by
Eq.
(4.8).
A
starting
value
for
T
(and
hence
for
τ
=
T
/T
P
0 )a
Substitution
of
this
value
into
Eq.
(4.10)
provides
new
value
of
T
from
which
to
reevaluate
)
allows
evaluation
�C
by
Eq.
(4.8). (4.8).
A
starting
value
for
T
(and
hence
for
τ
=
T
/T
Substitution
of
this
value
into
Eq.
(4.10)
provides
a
new
value
of
T
from
which
to
reevaluate
P
0) allows
Un valor
para
Tof(yfor
por
tanto
de
τ Eq.
= T/T
evaluación
deof⟨CTPof
⟩of
mediante
la
ecuación
Substitution
this
(4.10)
a new evaluation
value
from
which
reevaluate
allows
evaluation
�C
�HHHby
byto
Eq.
(4.8).
Ainicial
starting
value
for
(andinto
hence
for
=provides
/T00la
A
starting
value
TTvalue
(and
hence
for
=
TT/T
0ττ) permite
H �C
)
�
Eq.
(4.8).
P
P
Substitution
ofofthis
value
into
Eq.
(4.10)
provides
a new
value
of
TTfrom
which
totoreevaluate
�C
continues
to
convergence
on
aun
value
of
Tde
Substitution
this
valueinto
into
Eq.
(4.10)
provides
new
value
fromwhich
which
reevaluate
�C
�HPH.�.HIteration
Iteration
continues
to(4.10)
convergence
onon
afinal
final
value
ofof
T.of
.Tof
PP�este
Al sustituir
valor
la
ecuación
se
obtiene
valor
T.TT
a partir
del
cualto
vuelve a evaluar
�C
. Iteration
continues
to
convergence
anuevo
final
value
Substitution
ofen
this
value
into
Eq.
(4.10)
provides
new
value
of
from
which
tosereevaluate
reevaluate
Substitution
of
this
value
Eq.
(4.10)
provides
aaanew
value
from
�C
�
.
Iteration
continues
to
convergence
on
a
final
value
of
T
.
PPH
�C
�
.
Iteration
continues
to
convergence
on
a
final
value
of
T
.
⟨CP⟩ H. �C
La
continúa
hastato
la convergence
convergenciaon
para
un valor
�CPPiteración
Iteration
continues
to
convergence
on
final
valuefinal
of TTde
��HHH.. Iteration
continues
aa final
value
of
.. T.
04-SmithVanNess.indd 130
8/1/07 13:07:37
4.1.
4.1. Sensible
Sensible Heat
Heat Effects
Effects
4.1. Efectos
del calorHeat
sensible
4.1. Sensible
Effects
4.1. Sensible
Sensible Heat
Heat Effects
Effects
4.1.
131
131
131
131
131
131
Example 4.2
Ejemplo
4.2 4.2
Example
Example
4.2 required
Example
Calculate
heat
Calculate the
the 4.2
heat
required to
to raise
raise the
the temperature
temperature of
of 1
1 mol
mol of
of methane
methane from
from 260
260
◦◦Cnecesario
Calculate
the
heat para
required
to raise
the
temperature
of 1de
mol
of
methane
Calculeto
calor
aumentar
la
de 1 mol
metano
de 260from
amay
600260
°C en un
a
process
at
a
pressure
sufficiently
low
that
methane
be
toel600
600
C in
in
a steady-flow
steady-flow
process
attemperatura
a
pressure
sufficiently
low
that
methane
may
be
Calculate
the
heat required
required
to raise
raise
the
temperature
of 11 mol
mol
of
methane
from
260
Calculate
the
heat
to
the
temperature
of
of
methane
from
260
◦C
to
600
in
a
steady-flow
process
at
a
pressure
sufficiently
low
that
methane
may
be un gas
procesoconsidered
de
flujo
estable
a
una
presión
lo
suficientemente
baja
para
que
el
metano
se
considere
◦
◦
an
ideal
gas.
considered
ideal gas. process
to
600 C
C in
in aan
a steady-flow
steady-flow
process at
at aa pressure
pressure sufficiently
sufficiently low
low that
that methane
methane may
may be
be
to
600
ideal. considered an ideal gas.
considered an
an ideal
ideal gas.
gas.
considered
Solution
4.2
Solution
4.2
Solución
4.2
Solution
4.2
Solution
4.2
Solution
4.2
Equation
in
with
Eq.
(4.7)
the
result.
ParamLa ecuación
(4.3) (4.3)
en
combinación
con la
(4.7)
proporciona
el resultado
requerido.
parámetros
Equation
(4.3)
in combination
combination
with
Eq.
(4.7) provides
provides
the required
required
result.Los
Paramig
igEquation (4.3)
combination
with
Eq.temperatures
(4.7) provides
the required result. Paramig /Rin
eters
for
C
come
from
Table
C.1;
are:
para CPEquation
/R
se
obtienen
de
la
tabla
C.1;
las
temperaturas
son:
Equation
(4.3)
income
combination
withC.1;
Eq.temperatures
(4.7) provides
provides
the required
required result.
result. ParamParam(4.3)
combination
with
Eq.
(4.7)
the
eters for C
/Rin
from Table
are:
P
Pig /R
eters for Cig
come from Table C.1; temperatures are:
ig
eters for
for CCPPP /R
/R come
come from
from Table
Table C.1;
C.1; temperatures
temperatures are:
are:
eters
873.15
873.15 1.6377
T
T
ττ =
873.15 =
T00 =
= 533.15
533.15 K
K
T=
= 873.15
873.15 K
K
= 533.15
= 1.6377
873.15 = 1.6377
T0 = 533.15 K
T = 873.15 K
τ = 873.15
533.15
= 533.15
533.15 K
K
= 873.15
873.15 K
K
= 533.15 =
= 1.6377
1.6377
TT00 =
TT =
ττ =
�� 873.15 ig
533.15
533.15
ig
873.15 C
� 873.15 C PPig dT
Whence,
Q
Whence,
Q=
= �H
�H =
=R
R�� 873.15
ig
873.15 C
Pig dT
CR
Whence,
Q = �H = R 533.15
R
PP dT
533.15 C
De donde,
Whence,
Q=
= �H
�H =
= RR 533.15
dT
Whence,
Q
R dT
R
533.15 R
533.15
��
−3
−3
9.081
�
9.081 ×
× 10
10−3
2 22
T
Q
T02(τ
(τ −
− 1)
1) +
+ 9.081 2× 10−3
(τ −
− 1)
1)
Q=
= (8.314)
(8.314)��1.702
1.702 T
T00(τ
9.081 ×
×
10−3 T02022 (τ222 − 1)
Q = (8.314) 1.702 T0 (τ − 1) + 9.081
2 10
T
(τ −
− 1)
1) +
+
(τ
−
1) �
Q=
= (8.314)
(8.314) 1.702
1.702TT00(τ
T
(τ
−
1)
Q
2
00
�
22
−6
−6
�
2.164
×
10
2.164
×
10
3
3
3
3
�
−6 T
−
(τ −
− 1)
1)� =
= 19,778
19,778 JJ
T0 (τ
− 2.164 3× 10−6
2.164 ×
×
10−6 T03033 (τ333 − 1) = 19,778 J
− 2.164
3 10
(τ −
− 1)
1) =
= 19,778
19,778 JJ
−
TT00 (τ
−
3
33
Uso de las funciones definidas
Use
Use of
of Defined
Defined Functions
Functions
Use
of
Defined
Functions
��
Use
of
Defined
Functions
Use
of
Defined
Functions
La integral
a menudo
en los cálculos
termodinámicos.
Por lo As
tanto,
por razones
de conveThe
integral
often
in
calculations.
aa matter
of
P/R)dT
� (C
P /R)
The∫(C
integral
(C
/R) dT
dTaparece
often appears
appears
in thermodynamic
thermodynamic
calculations.
As
matter
of conveconveP
�� (Cde
The
integral
/R)
dT often
appears
inofthermodynamic
calculations.
As a matter of
conveniencia,The
el lado
derecho
la define
ecuación
se
define
como
la función
ICPH(T0,T;A,B,C,D)
y convese
escribe una
nience,
we
the
right
side
Eq.
(4.7)
as
function,
ICPH(T0,T;A,B,C,D),
nience,
we therefore
therefore
define
the(4.7)
right
side
Eq.
(4.7)
as the
the
function,
ICPH(T0,T;A,B,C,D),
The
integral
(CPPP/R)
/R)
dT often
often
appears
inof
thermodynamic
calculations.
As aa matter
matter of
of
conveintegral
(C
dT
appears
in
thermodynamic
calculations.
As
nience,
we
therefore
define
the
right
side
of
Eq.
(4.7)
as
the
function,
ICPH(T0,T;A,B,C,D),
rutina de
computadora
para
su
evaluación.
En
tal
caso
la
ecuación
(4.7)
será
entonces:
and
write
a
computer
routine
for
its
evaluation.
Equation
(4.7)
then
becomes:
and
write
a
computer
routine
for
its
evaluation.
Equation
(4.7)
then
becomes:
nience, we
we therefore
therefore define
define the
the right
right side
side of
of Eq.
Eq. (4.7)
(4.7) as
as the
the function,
function, ICPH(T0,T;A,B,C,D),
ICPH(T0,T;A,B,C,D),
nience,
and write a computer routine for its evaluation. Equation (4.7) then becomes:
and write
write aa computer
computer routine
routine��for
for
its evaluation.
evaluation. Equation
Equation (4.7)
(4.7) then
then becomes:
becomes:
and
TT its
CP
� C
ICPH(T0,T;A,B,C,D)
�� TTT C PP dT
dT ≡
≡ ICPH(T0,T;A,B,C,D)
ICPH(T0,T;A,B,C,D)
CR
TT0 C
RPP dT ≡ ICPH(T0,T;A,B,C,D)
0
dT ≡
≡ ICPH(T0,T;A,B,C,D)
ICPH(T0,T;A,B,C,D)
T0 R dT
R
TT00 R
The
function
name
is
ICPH,
and
the
quantities
in
are
the
and
El nombre
la función
es ICPH,
y lasand
cantidades
entre paréntesis
son las
T0 y T,T
Thedefunction
name
is ICPH,
the quantities
in parentheses
parentheses
arevariables
the variables
variables
T00seguidas
and T
T ,, por los
The
function
name
is
ICPH,
and
the
quantities
in
parentheses
are
the
variables
T
and
T
,
0
followed
by
parameters
A,
B,
C,
and
D.
When
these
quantities
are
assigned
numerical
values,
parámetros
B, by
C yparameters
D. Cuando
estas
se les
valores
numéricos,
la notación
representa
followed
A,a B,
C,
and
When
these
quantities
are
assigned
numerical
values,
The A,
function
name
ICPH,
andcantidades
theD.
quantities
inasignan
parentheses
are
the variables
variables
and
The
function
name
isis ICPH,
and
the
quantities
in
parentheses
are
the
TT00 and
TT,,
followed
by
parameters
A,
B,
C,
and
D.
When
these
quantities
are
assigned
numerical
values,
the
notation
represents
aaA,
value
for
the
integral.
Thus,
for
evaluation
of
in
un valorfollowed
para
la integral.
Así, para
la
de
Q en
el ejemplo
the
notation
represents
value
forand
theD.
integral.
Thus,
for the
the 4.2:
evaluation
of Q
Qnumerical
in Ex.
Ex. 4.2:
4.2:values,
followed
by parameters
parameters
A,
B,evaluación
C,
and
D.
When
these
quantities
are assigned
assigned
numerical
values,
by
B,
C,
When
these
quantities
are
the notation represents a value for the integral. Thus, for the evaluation of Q in Ex. 4.2:
the notation
notation represents
represents aa value
value for
for the
the integral.
integral. Thus,
Thus, for
for the
the evaluation
evaluation of
of Q
Q in
in Ex.
Ex. 4.2:
4.2:
the
Q
8.314
ICPH(533.15,873.15;1.702,9.081E-3,-2.164E-6,0.0)
19,778
Q
=
8.314 ××
×ICPH(533.15,873.15;1.702,9.081E-3,-2.164E-6,0.0)
ICPH(533.15,873.15;1.702,9.081E-3,-2.164E-6,0.0)==
=19
19,778
Q=
= 8.314
778 J JJ
Q = 8.314 × ICPH(533.15,873.15;1.702,9.081E-3,-2.164E-6,0.0) = 19,778 J
Q=
= 8.314
8.314 ×
× ICPH(533.15,873.15;1.702,9.081E-3,-2.164E-6,0.0)
ICPH(533.15,873.15;1.702,9.081E-3,-2.164E-6,0.0) =
= 19,778
19,778 JJ
Q
Representative
computer
programs
for
evaluation
of
the
integral
are
given
in
D.
Representative
computer
programs
for
evaluation
of
the
integral
are
given
in App.
App.
D.integral.
En el apéndice
D se dan
programas
representativos
de computadora
para are
la evaluación
de laD.
Representative
computer
programs
for evaluation
of the integral
given
in as
App.
For
added
flexibility
the
programs
also
evaluate
the
dimensionless
quantity
�C
�
/R
given
P
For
added
flexibility
the
programs
also
evaluate
the
dimensionless
quantity
�C
�
/R
as
given
H
Representative
computer
programs
for
evaluation
of
the
integral
are
given
in
App.
D.expresa
Representative
computer
programs
for
evaluation
of
the
integral
are
given
in
App.
D.
P
H
Para mayor
flexibilidad,
los programas
también
evalúanthe
la cantidad
adimensional
⟨C
se
ForEq.
added
flexibility
the programs
also
evaluate
dimensionless
quantity
�CPP⟩ �HH/R
/Rcomo
as given
by
(4.8).
The
right-hand
side
of
this
equation
is
another
function,
MCPH(T0,T;A,B,C,D).
by
Eq.
(4.8).
The
right-hand
side
of
this
equation
is
another
function,
MCPH(T0,T;A,B,C,D).
For
added
flexibility
the
programs
also
evaluate
the
dimensionless
quantity
�C
�
/R
as
given
For
added
flexibility
the
programs
also
evaluate
the
dimensionless
quantity
�C
�
/R
as
given
P
P
mediante
ecuación
(4.8).
El lado side
derecho
de equation
esta ecuación
es otra
función,
MCPH(T0,T;A,B,C,D).
Con
HH
bylaEq.
(4.8).
The
right-hand
of this
is another
function,
MCPH(T0,T;A,B,C,D).
With
definition,
Eq.
becomes:
With
this
definition,
Eq. (4.8)
(4.8)
becomes:
by
Eq.this
(4.8).
The right-hand
right-hand
side
of this
this equation
equation isis another
another function,
function, MCPH(T0,T;A,B,C,D).
MCPH(T0,T;A,B,C,D).
by
Eq.
(4.8).
The
side
of
esta definición,
ecuación
(4.8)
será:
With thisladefinition,
Eq. (4.8)
becomes:
With this
this definition,
definition, Eq.
Eq. (4.8)
(4.8) becomes:
becomes:
With
�C
�C PP ��H
H = MCPH(T0,T;A,B,C,D)
�CRP �H
= MCPH(T0,T;A,B,C,D)
�CPR
�C
MCPH(T0,T;A,B,C,D)
P��HH = MCPH(T0,T;A,B,C,D)
= MCPH(T0,T;A,B,C,D)
MCPH(T0,T;A,B,C,D)
R =
RR
04-SmithVanNess.indd 131
8/1/07 13:07:42
CHAPTER 4. Heat Effects
132
CHAPTER 4. Heat Effects
132
CHAPTER
4. Heat
Effectstérmicos
132
132
CAPÍTULO
4. Efectos
A specific numerical value of this function is:
A specific numerical value of this function is:
A numérico
specific numerical
value
of this
function
Un valor
específico
de esta
función
es: is:
MCPH(533.15,873.15;1.702,9.081E-3,-2.164E-6,0.0)
= 6.9965
MCPH(533.15,873.15;1.702,9.081E-3,-2.164E-6,0.0) = 6.9965
6.9965
representingMCPH(533.15,873.15;1.702,9.081E-3,-2.164E-6,0.0)
�C P � /R for methane in the calculation of Ex. 4.2. By Eq.=(4.9),
representing �C P �HH /R for methane in the calculation of Ex. 4.2. By Eq. (4.9),
representando
a ⟨CP⟩�C
el metano
eninelthe
cálculo
del ejemplo
la ecuación (4.9),
representing
/R for
methane
calculation
of Ex.4.2.
4.2.Mediante
By Eq. (4.9),
H /R
P �Hpara
�H = (8.314)(6.9965)(873.15 − 533.15) = 19,778 J
�H = (8.314)(6.9965)(873.15 − 533.15) = 19,778 J
∆H== (8.314)(6.9965)(873.15
(8.314)(6.9965)(873.15 –−533.15)
778 J J
�H
533.15)==1919,778
Example 4.3
Example
4.3
Ejemplo
4.3
What is
the final
temperature when heat in the amount of 0.4 × 1066(Btu) is added to
Example
4.3
What is the final temperature when heat
in the amount of 0.4 × 10 (Btu) is added to
◦
25(lblamol)
of ammonia
in
a steady-flow
process
at
1(atm)?
¿Cuál será
temperatura
final initially
cuandoat
se500(
agrega
una
× 106(Btu) a 25(lb mol)
◦F)
6 (Btu)
What
theoffinal
temperature
when
heat
amountdeofcalor
0.4 ×de100.4
is added to
25(lb is
mol)
ammonia
initially
at
500(
F) in
in the
a cantidad
steady-flow
process
at
1(atm)?
de amoniaco inicialmente a 500(°F) en un proceso
de flujo estable a 1(atm)?
◦
25(lb mol) of ammonia initially at 500( F) in a steady-flow process at 1(atm)?
Solution 4.3
Solución
4.3
Solution
4.3
4.3
Si ∆HSolution
esIfel�H
cambio
de
la entalpía
parafor
1(lb
mol),
Q=
y , and
is the
enthalpy
change
1(lb
mol),
Q n=∆H
n �H
If �H is the enthalpy change for 1(lb mol), Q = n �H , and
If �H is the enthalpy change
for ×
1(lb
Q = n �H , and
6
Q
0.4
10mol),
�H = Q = 0.4 × 106 = 16,000(Btu)(lb mol)−1
25 6 = 16,000(Btu)(lb mol)−1
�H = n =
Qn
0.4 ×
2510
�H =
=
= 16,000(Btu)(lb mol)−1
n requires
25 temperatures in kelvins; therefore, conversion
The
heat-capacity
equation
La ecuación
para la capacidad
calorífica
requiere
que lasintemperaturas
estén en
kelvins; por lo
The
requires
temperatures
kelvins;
therefore,
conversion
−1 is equivalent
of
allheat-capacity
units
to the equation
SI
system
is indicated.
Because
1 J mol
to
–1 es equivalente
−11 is
tanto, seThe
necesita
convertir
todas
lasrequires
unidades
al sistema
SI.kelvins;
Puesto
que
J mol
of
all
units
to
the
SI
system
is
indicated.
Because
1
J
mol
equivalent
to
heat-capacity
equation
temperatures
in
therefore,
conversion
−1
,
divide
the
preceding
result
by
0.4299:
0.4299(Btu)(lb
mol)
–1, el resultado
a 0.4299(Btu)(lb
mol)
se divide
entreby0.4299:
, divideanterior
preceding
result
0.4299(Btu)(lb
mol)
of
all units
to the
SI−1system
isthe
indicated.
Because
10.4299:
J mol−1 is equivalent to
−1 , divide the preceding result by 0.4299:
0.4299(Btu)(lb mol)�H
37,218
mol–1−1
∆H=
= 16,000/0.4299
16
000/0.4299 ==
218 JJJmol
�H
=
16,000/0.4299
=3737,218
mol−1
−1
�H = 16,000/0.4299
500 + 459.67= 37,218 J mol
With
T0 = 500 + 459.67 = 533.15 K
1.8
With
T0 =
= 533.15 K
Con
500 +1.8
459.67
With
T0 =
= 533.15 K
1.8
then for any value of T :
then for any value of T :
A continuación,
paravalue
cualquier valor de T:
then for any
�C P �H of T :
�C P �H = MCPH(533.15,T;3.578,3.020E-3,0.0,-0.186E+5)
R
�C PR�H = MCPH(533.15,T;3.578,3.020E-3,0.0,-0.186E+5)
= MCPH(533.15,T;3.578,3.020E-3,0.0,-0.186E+5)
R
Iteration between
this equation and Eq. (4.10) starts with a value T ≥ T0 , and
Iteration between
thisvalue:
equation and Eq. (4.10) starts with a value T ≥ T0 , and
converges
on the final
convergesbetween
on the final
Iteration
this value:
equation and Eq. (4.10) starts with a value T ≥ T0 , and
La iteración entre esta ecuación y la (4.10) comienza con un valor T ≥ T0, y converge al valor
converges on the final
T =value:
1,250 K
or
1,790(◦◦F)
final:
T = 1,250 K
or
1,790( F)
◦
T =
1,250
K
or
1,790(
T = 1 250 K
o
1 790 (°F) F)
04-SmithVanNess.indd 132
8/1/07 13:07:46
4.2. Latent
of Pure Substances
4.2. Calores
latentesHeats
de sustancias
puras
4.2
133
133
4.2 LATENT
HEATS OF
SUBSTANCES
CALORES
LATENTES
DE PURE
SUSTANCIAS
PURAS
Cuando When
una sustancia
pura se convierte
al estado
líquido
desdeorelvaporized
estado sólido
seliquid
vaporiza
a partir de un
a pure substance
is liquefied
from the
solid state
from othe
at constant
estado líquido
a presión
constante,
no hay cambio
en lathe
temperatura;
de cualquier
modo, of
el proceso
pressure,
no change
in temperature
occurs;alguno
however,
process requires
the transfer
a
requierefinite
la transferencia
finita de
calorheat
a laeffects
sustancia.
Estos the
efectos
llamados
amount of de
heatuna
to cantidad
the substance.
These
are called
latenttérmicos
heat of son
fusion
calor latente
de fusión
y calor
latente de vaporización.
De manera
similar,
existen calores
de transición
and the
latent heat
of vaporization.
Similarly, there
are heats
of transition
accompanying
the que
acompañan
el cambio
de una sustancia
desolid
un estado
a otro;forpor
ejemplo,the
el heat
calorabsorbed
absorbidowhen
cuando los
change
of a substance
from one
state sólido
to another;
example,
cristalesrhombic
rómbicoscrystalline
de azufre cambian
a una estructura
monoclínica
a 95 °C
es de
360isJ por
sulfur changes
to the monoclinic
structure
at y951◦bar
C and
1 bar
360cada
J forátomogramo. each gram-atom.
La principal
estos of
procesos
es processes
la coexistencia
dos fases. De
acuerdo
con la
regla de
The característica
characteristic de
feature
all these
is thede
coexistence
of two
phases.
Acfase, un cording
sistema to
dethe
dosphase
fases rule,
que está
formadosystem
por unaconsisting
sola especie
univariante,
estado intensivo
se
a two-phase
of aessingle
species yis su
univariant,
and
determina
la especificación
de una sola
intensiva.
calor
latenteproperty.
que acompaña
un camitspor
intensive
state is determined
bypropiedad
the specification
of Así,
just el
one
intensive
Thus athe
bio de fase
es heat
una accompanying
función sólo dea la
temperatura,
está relacionada
con otras
deltosistema
latent
phase
change isque
a function
of temperature
only,propiedades
and is related
medianteother
una system
ecuación
termodinámica
exacta:
properties
by an exact
thermodynamic equation:
�H = T �V
sat
d P saturado
dT
(4.11) (4.11)
where for a pure species at temperature T ,
donde para una especie pura a temperatura T,
�H = latent heat
∆H = calor
latente
�V =
volume change accompanying the phase change
∆V = cambio
de volumen
que acompaña el cambio de fase
sat
P
=
saturation
pressure
Psaturado = presión de saturación
The derivation of this equation, known as the Clapeyron equation, is given in Chap. 6.
satcapítulo
La deducción de
estaEq.
ecuación,
ecuación de Clapeyron,
se da denP el
6. slope of
/dT is the
When
(4.11) isconocida
applied como
to the vaporization
of a pure liquid,
Cuando
se aplica
la ecuación (4.11) a curve
la vaporización
de un líquido
puro, d�V
Psaturado
/dTdifference
es la pendiente
the vapor
pressure-vs.-temperature
at the temperature
of interest,
is the
de la curva
de presión
vapor en
la temperatura
a la liquid,
temperatura
de interés,
∆V esheat
la diferencia
between
molarde
volumes
of función
saturateddevapor
and saturated
and �H
is the latent
of
entre losvaporization.
volúmenes molares
de vapor
saturado
líquido saturado,
y ∆H es el calor
de vaporización.
Thus values
of �H
may ybedecalculated
from vapor-pressure
andlatente
volumetric
data.
Así, los valores
de ∆Hheats
se pueden
calcular
a partir de
la presión de vapor
y de información
volumétrica.
Latent
may also
be measured
calorimetrically.
Experimental
values are
available
4 Correlations
Los
también
miden
de manera
calorimétrica.
dispone
de valores
experimentales
at calores
selectedlatentes
temperatures
forsemany
substances.
forSethe
latent heats
of many
com4 Las correlaciones para
para muchas
sustancias
a diferentes
temperaturas.
los calores latentes
data aredenotmuchos
pounds
as a function
of temperature
are given
by Daubert et al.5 Nevertheless,
5 Sin embargo, no siempre
compuestos
en función
deatlathe
temperatura
están
porand
Daubert
y colaboradores.
always
available
temperature
of dados
interest,
in many
cases the data
necessary for aphay información
a laare
temperatura
de interés,
y enevent
muchos
casos tampoco
se conoce
información
plication disponible
of Eq. (4.11)
also not known.
In this
approximate
methods
are used
for
necesariaestimates
para aplicar
ecuación
En esta asituación
se usanBecause
métodos
aproximados
para estimar
el
of thelaheat
effect (4.11).
accompanying
phase change.
heats
of vaporization
are
efecto térmico
que most
acompaña
un cambio
fase. Debido
a que
losthey
calores
vaporización
los más imby far the
important
from adepractical
point of
view,
havedereceived
most son
attention.
6 Alternative
portantesOne
desde
un puntois de
han recibido
mayor known
atención.
Un procedimiento
de uso
común es
procedure
usevista
of apráctico,
group-contribution
method,
as UNIVAP.
meth6
el método
la contribución
de grupo, conocido como UNIVAP. Los métodos alternativos sirven para uno
odsdeserve
one of two purposes:
de dos propósitos:
• Prediction of the heat of vaporization at the normal boiling point, i.e., at a pressure of
• Predicción
del caloratmosphere,
de vaporización
en as
el 101,325
punto dePa.
ebullición normal, es decir, a una presión de 1 at1 standard
defined
mósfera estándar, definida como 101 325 Pa.
4 V. Majer and V. Svoboda, IUPAC Chemical Data Series No. 32, Blackwell, Oxford, 1985; R. H. Perry and
D. Green, op. cit., Sec. 2.
4
5
Daubert et
al., op.
cit.
V. Majer yT.V.E.Svoboda,
IUPAC
Chemical
Data Series No. 32, Blackwell, Oxford, 1985; R. H. Perry y D. Green, op. cit., sección 2.
5 T. E. Daubert et al., op. cit.
6 M. Klüppel, S. Schulz, and P. Ulbig, Fluid Phase Equilibria, vol. 102, pp. 1–15, 1994.
6 M. Klüppel, S. Schultz y P. Ulbig, Fluid Phase Equilibria, vol. 102, pp. 1-15, 1994.
04-SmithVanNess.indd 133
8/1/07 13:07:47
CHAPTER 4.
4. Heat
Heat Effects
Effects
134
CHAPTER
134
CHAPTER 4. Heat Effects
134
134
CHAPTER
4. 4.
Heat
Effects
CAPÍTULO
Efectos
térmicos
CHAPTER 4. Heat Effects
134
Estimation of
of the
the heat
heat of
of vaporization
vaporization at
at any
any temperature
temperature from
from the
the known
known value
value at
at aa
•• Estimation
•• Estimation
of
the
heat
of
vaporization
at
any
temperature
from
the
known
value
at
• Estimación
del
calor
de
vaporización
a
cualquier
temperatura
a
partir
del
valor
conocido
a
una
Estimation
of
the
heat
of
vaporization
at
any
temperature
from
the
known
value
at aatempesingle
temperature.
single temperature.
single
temperature.
•
Estimation
of
the
heat
of
vaporization
at
any
temperature
from
the
known
value
at
a
ratura particular.
single temperature.
single
temperature.
Rough estimates
estimates of
of latent
latent heats
heats of
of vaporization
vaporization for
for pure
pure liquids
liquids at
at their
their normal
normal boiling
boiling
Rough
Rough
estimates
of
heats
of
for
their
boiling
Estimaciones
burdas
de los
calores
latentes
de vaporización
paraliquids
líquidosat
en
sus puntos
Rough
estimates
of latent
latent
heats
of vaporization
vaporization
for pure
pure
liquids
atpuros
their normal
normal
boilingde ebupoints are
are given
given
by Trouton’s
Trouton’s
rule:
points
by
rule:
points
are
given
by
Trouton’s
rule:
estimates
of regla
latent
heats
of �H
vaporization
for pure liquids at their normal boiling
llición normal
están
dadas
la
de
Trouton:
pointsRough
are given
bypor
Trouton’s
rule:
�Hnn
�H
∼ 10
10
points are given by Trouton’s rule:
n∼
�H
10
RTnnn ∼
RT
∼
10
�H
RT
RTnn ∼ 10
is the
the absolute
absolute temperature
temperature of
of the
theRT
normal
boiling point.
point. The
The units
units of
of �H
�Hnn,, R,
R, and
and TTnn
where TTnn is
normal
boiling
where
n
the absolute
temperature
of
point. The
units of
�H
and
where
T is
n ,, R,
absolute
temperature
of the
the normal
normal boiling
boiling
The
�H
and TTnn
where
must
beTnchosen
chosen
so
that �H
�H
/RTnn is
is dimensionless.
dimensionless.
Datingpoint.
from 1884,
1884,units
thisof
rule
still
provides
must
be
that
Dating
from
this
rule
still
provides
n is the so
n R,
nn/RT
must
be
chosen
so
that
�H
/RT
is
dimensionless.
Dating
from
1884,
this
rule
still
provides
is
the
absolute
temperature
of
the
normal
boiling
point.
The
units
of
�H
,
R,
and
where
T
n
n
ncheck
n n, Rprovides
n deben
donde Taamust
es labetemperatura
absoluta
del punto
de ebullición
normal.
Las
unidades
∆H
y TnTse
chosen
sowhether
that
�Hvalues
is dimensionless.
from
thisde
rule
still
nsimple
simple
check on
on
whether
values
calculated
by other
otherDating
methods
are1884,
reasonable.
Representative
by
methods
are
reasonable.
Representative
n /RT
ncalculated
a
simple
check
on
whether
values
calculated
by
other
methods
are
reasonable.
Representative
must
be
chosen
so
that
�H
/RT
is
dimensionless.
Dating
from
1884,
this
rule
still
provides
n
n
elegir de
tal
forma
que
∆H
/RT
no
tenga
dimensiones.
Desde
1884,
esta
regla
proporciona
una
comprobaa
simple
check
on
whether
values
calculated
by
other
methods
are
reasonable.
Representative
n
n
experimental
values
for
this
ratio
are
Ar,
8.0;
N
,
8.7;
O
,
9.1;
HCl,
10.4;
C
H
,
10.5;
H
S,
experimental values for this ratio are Ar, 8.0; N22, 8.7; O22, 9.1; HCl, 10.4; C66H66, 10.5; H22S,
values
for
this
ratio
are
Ar,
O
,, 9.1;
HCl,
10.4;
C
S,
aexperimental
simple
on whether
values
calculated
byN
methods
reasonable.
Representative
2 ,, 8.7;
ción simple
de
sicheck
los
valores
calculados
por
otros
Valores
experimental
values
this
ratioson
arerazonables
Ar, 8.0;
8.0;
Nother
O22métodos.
9.1;are
HCl,
10.4; experimentales
C66H
H66,, 10.5;
10.5; H
H22represenS,
O,
13.1. for
10.6;
and
H
13.1.
10.6;
and
H
2 8.7;
22O,
O,
13.1.
10.6;
and
H
experimental
values
for
this
ratio
are
Ar,
8.0;
N
,
8.7;
O
,
9.1;
HCl,
10.4;
C
H
,
10.5;
H
S,
2
2 10.4; C26H6, 10.5; H2S, 10.6,
6 y6 H2O,
2
7
7
tativos de
estaOf
relación
son
Ar,
8.0;
N
,
8.7;
O
,
9.1;
HCl,
13.1.
10.6;
and
H
O,
13.1.
2
2
Of the
the2same
same nature,
nature, but
but not
not quite
quite so
so simple,
simple, is
is the
the equation
equation proposed
proposed by
by Riedel:
Riedel: 7
7
Of
same
nature,
but
so
is
proposed
by
O,
13.1.
10.6;
and
H2naturaleza,
7
De
la misma
pero
nonot
tanquite
simple,
es la ecuación
propuesta
por Riedel:
Of the
the
same
nature,
but
not
quite
so simple,
simple,
is the
the equation
equation
proposed
by Riedel:
Riedel:
7
Of the same nature, but not
quite
so
simple,
is
the
equation
proposed
by
Riedel:
�Hnn
− 1.013)
1.013)
1.092(ln PPcc −
�H
1.092(ln
�H
1.013)
PPc −
= 1.092(ln
(4.12)
(4.12)
n=
1.092(ln
�H
−
1.013)
n
c
=
(4.12)
RT
0.930
−
T
RT
0.930
−
T
n
r
n
r
nn
=
(4.12) (4.12)
�H
1.092(ln
Pc−−T1.013)
RT
0.930
n
r
RTn =
0.930 − Trnn
(4.12)
n
where PPcc is
is the
the critical
critical pressure
pressureRT
innbars
bars and
and 0.930
is −
theTrreduced
reduced
temperature at
at TTnn.. Equation
Equation
where
in
TTrrnn is
the
temperature
P
is
the
critical
pressure
in
bars
and
T
is
at
donde Pwhere
es
la
presión
crítica
en
bar
y
T
es
la
temperatura
reducida
a Tntemperature
. Resulta sorprendente
que, para ser
c
r
c
r
n
n
where is
is the critical
pressure
bars
and Trn expression;
is the
the reduced
reduced
temperature
at TTnn.5.5 Equation
Equation
(4.12)
isPcsurprisingly
surprisingly
accurate
forinan
an
empirical
expression;
errors
rarely exceed
exceed
percent.
(4.12)
accurate
for
empirical
errors
rarely
percent.
(4.12)
is
surprisingly
accurate
for
an
empirical
expression;
errors
rarely
exceed
5
percent.
where
P
is
the
critical
pressure
in
bars
and
T
is
the
reduced
temperature
at
T
.
Equation
una expresión
empírica,
la
ecuación
(4.12)
es
exacta;
los
errores
rara
vez
exceden
a
5%.
Aplicándola
c
r
n
n
(4.12) isto
accurate for an empirical expression; errors rarely exceed 5 percent.al agua
Applied
tosurprisingly
water itit gives:
gives:
Applied
water
Applied
to
water
it
gives:
(4.12)
is
surprisingly
accurate
for an empirical expression; errors rarely exceed 5 percent.
se obtiene:
Applied to water it gives:
Applied to water it gives:�H
1.092(ln 220.55
220.55 −
− 1.013)
1.013)
�Hnn
1.092(ln
220.55
−
1.013)
�H
= 1.092(ln
= 13.56
13.56
n=
1.092(ln
220.55
−
1.013) =
�H
n
=
= 13.56
RTnn = 1.092(ln
0.930
− 0.577
0.577
RT
0.930
−
�H
220.55
− 1.013) = 13.56
RT
0.930
−
0.577
n
RTn =
0.930 − 0.577
= 13.56
0.930 − 0.577=
−1
=n(13.56)(8.314)(373.15)
(13.56)(8.314)(373.15)
= 42,065
42,065 JJ mol
mol−1
Whence,
�HnnRT
=
Whence,
�H
De donde,
−1
=
(13.56)(8.314)(373.15)
=
42,065
J
mol
Whence,
�H
−1
n
(13.56)(8.314)(373.15) = 42,065 J mol
Whence,
�Hn =−1
−1
−1
−1
−1
This
corresponds
to
2,334
J
g
;
the
steam-table
value
of
2,257
J
g
is
loweren
by3.4%.
3.4 percent.
percent.
–1
–1
This
corresponds
to
2,334
J
g
;
the
steam-table
value
of
2,257
J
g
is
lower
by
3.4
=
(13.56)(8.314)(373.15)
=
42,065
mol
Whence,
�H
n
−1
−1
Esto corresponde
a 2 334 to
J g2,334
; el Jvalor
lasteam-table
tabla de vapor
deof
2 257
J gJ g −1
es is
menor
This
corresponds
gg−1;;de
the
value
2,257
lower
by
3.4
percent.
This
corresponds
to
2,334
J
the
steam-table
value
of
2,257
J
g
is
lower
by
3.4
percent.
Estimates
of
the
latent
heat
of
vaporization
of
a
pure
liquid
at
any
temperature
from
the a partir
Estimates
of
the
latent
heat
of
vaporization
of
a
pure
liquid
at
any
temperature
from
the
−1
−1
Las
estimaciones
del
calor
latente
de
vaporización
de aunpure
líquido
puro
aiscualquier
temperatura
Estimates
of
the
latent
heat
of
vaporization
of
liquid
at
any
temperature
from
the
This
corresponds
to
2,334
J
g
;
the
steam-table
value
of
2,257
J
g
lower
by
3.4
percent.
Estimates
of
the
latent
heat
of
vaporization
of
pure
liquid
at
any
temperature
from
the conoknown
value
at
a
single
temperature
may
be
based
on
a
known
experimental
value
or
on
a
value
known
value
at
a
single
temperature
may
be
based
on
a
known
experimental
value
or
on
a
value
de un valor
conocido
a auna
temperatura
particular,
se
pueden
hacer
con
base
en temperature
un value
valor or
experimental
known
value
at
single
temperature
may
be
based
on
a
known
experimental
on
a
value
Estimates
of
the
latent
heat
of
vaporization
of
a
pure
liquid
at
any
from
the
88 has
known
value
at
a
single
temperature
may
be
based
on
a
known
experimental
value
or
on
a
value
8
estimated
by
Eq.
(4.12).
The
method
proposed
by
Watson
has
found
wide
acceptance:
estimated
by
Eq.
(4.12).
The
method
proposed
by
Watson
found
wide
acceptance:
8
cido o sobre
unvalue
valor
estimado
por
la ecuación
(4.12).
El
método
propuesto
por Watson
goza
una amplia
estimated
by
Eq.
(4.12).
The
proposed
by
Watson
found
acceptance:
known
a single
may
be based
a known
experimental
or
on adevalue
8 has
estimated
byat
Eq.
(4.12).temperature
The method
method
proposed
byon
Watson
has
found wide
widevalue
acceptance:
aceptación:
8 has found wide acceptance:
�0.38
� by Watson
�
�
estimated by Eq. (4.12). The method proposed
0.38
�
− TTrr22 �
�H22
�H
�0.38
�111−
−
T
�H
0.38
=
(4.13)
=
(4.13)
r
2
2
−TTTrrr2�0.38
�H112 = � 111−
(4.13)
�H
−
�H
11
=
(4.13) (4.13)
1
−
T
�H
�H
�H211 = 11 −
− TTrrr211
(4.13)
�H1 its
− Tis
This equation
equation is
is simple
simple and
and fairly
fairly accurate;
accurate;
its1use
use
isr1illustrated
illustrated in
in the
the following
following example.
example.
This
This
is
and
fairly
use
in
following
Esta ecuación
es simple
ysimple
bastante
su uso seits
ilustra
el siguiente
ejemplo.
This equation
equation
is simple
andexacta;
fairly accurate;
accurate;
its
use is
isenillustrated
illustrated
in the
the
following example.
example.
This equation is simple and fairly accurate; its use is illustrated in the following example.
134
Example
4.4
Example
Ejemplo
4.4 4.4
Example
4.4
−1,, estimate
Given that
that the
the latent
latent heat
heat of
of vaporization
vaporization of
of water
water at
at 100
100◦◦C
C is
is 2,257
2,257 JJ gg−1
estimate
Given
Example
Given that the4.4
latent ◦◦heat of vaporization of water at 100◦◦C is 2,257
J g−1
−1, estimate
–1
Dado que
el
calor
latente
de vaporización
del aguaof
a water
100 °Cates100
2 ◦257
, calcule
calor latente a
Given
that
theat
latent
C isJ g2,257
J g−1 el
, estimate
the
latent
heat
at
300
C. of vaporization
the
latent
heat
300
C.
◦heat
the
at
thatheat
the latent
◦C.
300 °C.Given
the latent
latent
heat
at 300
300◦heat
C. of vaporization of water at 100 C is 2,257 J g , estimate
the latent heat at 300 C.
77L
Riedel, Chem.
Chem. Ing.
Ing. Tech.,
Tech., vol.
vol. 26,
26, pp.
pp. 679–683,
679–683, 1954.
1954.
Riedel,
7L
Riedel, Chem. Ing. Tech., vol. 26, pp. 679–683, 1954.
7L
L Riedel, Chem. Ing. Tech., vol. 26, pp. 679–683, 1954.
878K.
K. Riedel,
M. Watson,
Watson,
Ind.
Eng.
Chem.,
vol.
35, 679–683,
pp. 398–406,
398–406,
1943.
Chem.
Ing.
Tech.,
vol.vol.
26, pp.
1954.
M.
Ind.
Eng.
Chem.,
35,
pp.
1943.
8L
7 L. Riedel,
M. Watson,
Ind. Eng.
Chem.,
vol. 35, pp.
398–406, 1943.
Chem.
Ing. Tech.,
vol. 26,
pp. 679-683,
1954.
8K.
K. M. Watson, Ind. Eng. Chem., vol. 35, pp. 398–406, 1943.
8 K. M. Watson,
8 K. M. Watson,
Ind. Eng.Ind.
Chem.,
35, pp.
1943. 1943.
Eng.vol.
Chem.,
vol.398-406,
35, pp. 398–406,
04-SmithVanNess.indd 134
8/1/07 13:07:52
4.3. Standard Heat of Reaction
4.3. Calor estándar de reacción
135
135
Solución 4.4
Solution 4.4
calor latente a 100 °C = 2◦ 257 J g–1
�H = latent heat at 100 C = 2,257 J g−1
calor1 latente a 300 °C ◦
�H2 = latent heat at 300 C
373.15/647.1
= 0.577
Tr1 = 373.15/647.1 = 0.577
573.15/647.1
= 0.886
Tr2 = 573.15/647.1 = 0.886
En seguida,
ecuación
Thenpor
by la
Eq.
(4.13), (4.13),
Sean ∆H1 =
Let
∆H2 =
Tr1 =
Tr2 =
�H2 = (2, 257)
�
1 − 0.886
1 − 0.577
�0.38
= (2, 257)(0.270)0.38 = 1,371 J g−1
value given en
in las
the tablas
steam de
tables
is 1,406
J g−1
El valorThe
proporcionado
vapor
es 1 406
J g.–1.
4.3
4.3 STANDARD HEAT OF REACTION
CALOR ESTÁNDAR DE REACCIÓN
Heat térmicos
effects discussed
so hasta
far have
been
physical
processes.
Chemical
also are
Los efectos
analizados
ahora
hanforsido
sólo para
procesos
físicos. reactions
Las reacciones
químicas
accompanied
either
by
the
transfer
of
heat
or
by
temperature
changes
during
the
course
of
también se acompañan, ya sea por una transferencia de calor o por variaciones de temperatura durante
el
reaction—in
some cases
both. These
effects
are manifestations
of the
differences in de
molectranscurso
de una reacción,
y enby
algunos
casos por
ambas.
Estos efectos son
manifestaciones
las diferenstructure, molecular
and therefore
in energy,
of the products
and reactants.
For example,
cias en ular
la estructura
y, en
consecuencia,
en la energía
de los productos
y dethelosreactants
reactivos. Por
in
a
combustion
reaction
possess
greater
energy
on
account
of
their
structure
than
do the prodejemplo, los reactivos en una reacción de combustión poseen una energía mayor que los productos
a causa de
ucts, and
this energía
energy must
be transferred
the surroundings
as heato or
produce obtenidos
products a una
su estructura,
y esta
debeeither
ser transferida
a lostoalrededores
como calor
productos
at an elevada.
elevated temperature.
temperatura
Each
vast number
of possible
may be
outmaneras,
in many y cada
Cada una
de of
lasthe
muchas
reacciones
químicaschemical
se puedereactions
llevar a cabo
de carried
diferentes
different
ways,
and
each
reaction
carried
out
in
a
particular
way
is
accompanied
by
a
particular
reacción que se realiza en un procedimiento particular se acompaña por un efecto térmico propio. La tabuheat
effect.
Tabulation
of all possible
forreacciones
all possiblerealizables
reactions is
WeDebido
lación de
todos
los efectos
térmicos
posibles heat
para effects
todas las
es impossible.
poco práctica.
therefore
calculate
the
heat
effects
for
reactions
carried
out
in
diverse
ways
from
data
a eso, calculamos los efectos térmicos para las reacciones que se realizan de diversas maneras, afor
partir de
reactions
out inque
a standard
way.deThis
reduces
the required
data to
minimum.
información
paracarried
reacciones
se realizan
modo
estándar.
Esto reduce
al amínimo
la información reThe heat associated with a specific chemical reaction depends on the temperatures of
querida.
both the reactants and products. A consistent (standard) basis for treatment of reaction heat
El calor asociado con una reacción química determinada depende de la temperatura de los reactivos y
effects results when the products of reaction and the reactants are all at the same temperature.
de los productos. Una base consistente (estándar) para el tratamiento de los efectos térmicos de una reacción
Consider the flow-calorimeter method for measurement of heats of combustion of fuel
se obtiene cuando los productos de una reacción y los reactivos, todos, están a la misma temperatura.
gases. The fuel is mixed with air at ambient temperature and the mixture flows into a combusConsidere el método del calorímetro de flujo para medir los calores de combustión de los gases comtion chamber where reaction occurs. The combustion products enter a water-jacketed section
bustibles. El combustible se mezcla con aire a temperatura ambiente y la mezcla fluye hacia una cámara de
in which they are cooled to the temperature of the reactants. Because no shaft work is produced
combustión donde ocurre la reacción. Los productos de la combustión entran a una sección de enfriamiento
by the process, and the calorimeter is built so that changes in potential and kinetic energy are
con agua, en la que se enfrían a la temperatura de los reactivos. Puesto que no se produce trabajo de flecha
negligible, the overall energy balance, Eq. (2.32), reduces to
debido al proceso, y el calorímetro está construido de tal forma que son insignificantes los cambios en las
energías potencial y cinética, el balance de energíaQtotal,
la ecuación (2.32), se reduce a
= �H
= ∆H
Thus the heat flowing from the calorimeter Q
and
absorbed by the water is equal in magnitude to
the enthalpy change caused by the combustion reaction. The enthalpy change of reaction �H
Así, la transmisión del calor desde el calorímetro y que es absorbido por el agua es idéntico en magnitud al
is called the heat of reaction.
cambio de la entalpía causado por una reacción de combustión. El cambio de la entalpía de reacción ∆H se
conoce como el calor de reacción.
04-SmithVanNess.indd 135
8/1/07 13:07:53
136
136
CHAPTER 4. Heat Effects
CHAPTER 4. Heat Effects
136
CAPÍTULO 4. Efectos térmicos
For purposes of data tabulation with respect to the reaction,
For purposes of data tabulation with respect to the reaction,
Con el propósito de tabular la información
con→
respecto
la reacción,
a A + bB
l L + ma M
a A + bB → l L + m M
aA + bB → lL + mM
the standard heat of reaction is defined as the enthalpy change when a moles of A and b moles
the
standard
heat
of
reaction
is
defined
as theTenthalpy
change
when of
a moles
of
and bofmoles
of
B in their
standard se
states
at como
temperature
react
to form
l moles
L and
mAA
moles
M in
el calorof
estándar
de standard
reacción
define
el cambio
de entalpía
cuando
aofmoles
dem
y b moles
B
in
their
states
at
temperature
T
react
to
form
l
moles
L
and
moles
of MdeinB en sus
their
standard
states
at
the
same
temperature
T
.
estadostheir
estándar
a temperatura
T same
reaccionan
para formar
standard
states at the
temperature
T . l moles de L y m moles de M en sus estados estándar a la misma
temperatura
T.
A standard state is a particular state of a species at temperature T
A
standard
state conditions
is a particular
state of acomposition,
species at temperature
T
and
at specified
of pressure,
and physical
Un estado
estándar
es un conditions
estado particular
de una especie
a temperatura
T y en condicioand
at
specified
of
pressure,
composition,
and
physical
condition as,
e.g., gas,
liquid, or solid.
nes determinadas
de presión,
composición
y condición física, por ejemplo, gas, líquido o
condition as,
e.g., gas,
liquid, or solid.
sólido. A standard-state pressure of 1 standard atmosphere (101,325 Pa) was in use for many
standard-state
pressure of
standard
atmosphere
Pa) wasisin1 use
5 Pa),
years,Aand
older data tabulations
are1 for
this pressure.
The (101,325
present standard
bar for
(105many
Durante
muchos
años
se
utilizó como
presión
de estadoThe
estándar
a 1standard
atmósfera
estándar
(101
325 Pa)
years,
and
older
data
tabulations
are
for
this
pressure.
present
is
1
bar
(10
Pa),
but for purposes of this chapter, the difference is of negligible consequence. With5 respect
to
y las antiguas
tabulaciones
de
datos
son
para
esta
presión.
Ahora
el
estándar
es
1
bar
(10
Pa);
de
cualquier
but
for purposes
this chapter,
the difference
is of negligible
consequence.
With For
respect
to
composition,
the of
standard
states used
in this chapter
are states of
the pure species.
gases,
modo, composition,
para los finesthe
de standard
este capítulo,
la
diferencia
tiene consecuencias
insignificantes.
Con
respecto a la
states
used
in
this
chapter
are
states
of
the
pure
species.
For
gases,
the physical state is the ideal-gas state and for liquids and solids, the real state at the standardcomposición,
en estestate
capítulo
los
estados estándar
queliquids
se usanand
sonsolids,
los estados
destate
las especies
puras. Para los
the
ideal-gas
and for
real
at the
standardstatephysical
pressure
andisatthethe
system state
temperature.
In summary,
thethe
standard
states
used
in this
gases, el
estado
físico
es
el
estado
de
gas
ideal;
y
para
líquidos
y
sólidos,
el
estado
real
a
la
presión
del estado
state
pressure
chapter
are: and at the system temperature. In summary, the standard states used in this
estándarchapter
y a la temperatura
del
sistema.
En
resumen,
en
este
capítulo
los
estados
estándar
usados
son:
are:
•
Gases:
The
pure
substance
in the
ideal-gas
state at 1 bar.
• Gases:
sustancia
pura
ensubstance
el estado de
gasideal-gas
ideal a 1 state
bar.
• Gases:
The
pure
in the
at 1 bar.
• Liquids
and líquido
solids: oThe
real puro
pure real
liquid
solid at 1 bar.
• Líquidos
y sólidos:
sólido
a 1orbar.
• Liquids and solids: The real pure liquid or solid at 1 bar.
Property
values
in theen
standard
state
are denoted
by the
symbol.
For example,
Los valores
de una
propiedad
el estado
estándar
se indican
condegree
el símbolo
de grado.
Por ejemplo, CP°
values
in estado
the
state
are denoted
by theestándar
degree
C P◦◦ isProperty
thecalorífica
standard-state
heatstandard
capacity.
Because
the elstandard
state
forsymbol.
gaseslos
is For
the example,
ideal-gas
es la capacidad
en el
estándar.
Puesto
que
estado
para
gases
es el estado de
C Ppara
is the
standard-state
heat capacity.
the standard
state
for
isal the
ideal-gas
◦ for
gas ideal,
gases
CP°isesidentical
idéntica
a CPigC
, yigla,Because
información
deTable
la tabla
C.1
se gases
aplica
estado
estándar
gases
with
and
the
data
of
C.1
apply
to
the
standard
state para
state, Clos
ig
P
P
for
gases
is identical
with
C P , estándar
and
of
Table
C.1
apply
to the
standard
state es la
state,
C P◦ las
los gases.
condiciones
para
estado
son
fijas
excepto
la temperatura,
que
siempre
forTodas
gases.
All
conditions
for aun
standard
statethe
aredata
fixed
except
temperature,
which
is always
for
All
conditions
for aStandard-state
standard
stateproperties
are fixedson
except
temperature,
is always
del sistema.
Debido
a eso,
las system.
propiedades
del estado
estándar
funciones
sólo
de lawhich
temperatura.
El estado
the gases