1.5 Las Relaciones T ds: Sistema cerrado

24/02/2014

1.5 ��

 δ Q    T  int,rev

2  δ Q  ∆S = ∫   1  T int,rev int, rev

dS = 

��

δ W int,rev, sale

= PdV esp

��

∫

W int . rev . = PdV esp

��

��

δ Qint, rev

δ Q = TdS

��:

��

− δ W int, rev, sale = dU ��

/ δ W int . rev. = PdV esp

Tds = du + PdV esp

TdS = dU + PdV esp

δ Q

��

��

δ Q

δ Qint, rev

��

��

− δ W int,rev , sale = dH

��

Tds = dh − V esp dP

H = U + PV → dh = du + PdV esp + V esp dP

��

��

��

TERMODINÁMICA DE GASES Y VAPORES

DR. ARTURO MORALES


DR. ARTURO MORALES

1.6 ��

��

ds =

du T

+

��

PdV esp

ds =

T

dh T

−

�� :

ds =

V espdP

du = CvdT y

dh = CpdT

� ��

du T

T ds =

�� , ��

dV esp ≅ 0

Cp = Cv = C

du T

=

CdT T

+

PdV esp T 2

∫

s2 − s1 = C (T ) 1

dT T

≅ C prom In

T 2 T 1

PV = = RT

�� s2 − s1 = C prom In

T 2 = 0 → T 2 = T 1 T 1

�� , �� (� 2��1�0), �� TERMODINÁMICA DE GASES Y VAPORES

DR. ARTURO MORALES


DR. ARTURO MORALES

��

�� D�� :

E� �� 18 �C, �� 25 �� 350 �C. D�� .

B�� P 2 = 5 MPa

Cpagua = 4 .18kJ / kgK

�� ∆� � �2 � �1 � 0.2�0 ��/��

∆U hierro + ∆U agua = 0

T 2 = 120 K Cp2 = 3.486 kJ / kgK

Bomba

E entra − E sale = ∆E sistema

[mC (T 2 − T 1 ) ]hierro + [mC (T 2 − T 1 ) ]agua = 0

�� Cphierro = 0.45kJ / kgK

T 2 = 26.7 C o

��

��

S gen = ∆S hierro + ∆S agua

∆S hiero =

P 1 = 1 MPa T 1 = 110 K Cp1 = 3.471 kJ / kgK

Error =

∆ sactual − ∆ sideal 0.27 − 0.303 x100 = = 12.2% ∆ s actual 0.27


DR. ARTURO MORALES

∆S agua =


DR. ARTURO MORALES

1

24/02/2014

1.� ��

P = RT / V esp

��

du = CvdT

��

dh = CpdT

�. �� 2

∫

s2 − s1 = Cv (T )

ds =

du T

+

PdV esp T

dT T

1

1. �� 2

→ s2 − s1 = ∫ Cv (T ) 1

dT V + RIn 2 T V 1

V 2

+ RIn

s2 − s1 = Cvavg In

V 1 T 2 V + RIn 2 T 1 V 1

 kJ   kg − K   

2. ��

ds =

dh T

−

V esp dP T

2

→ s2 − s1 = ∫ Cp(T ) 1

dT P − RIn 2 T P 1

2

∫

s2 − s1 = Cp(T )

�� C� � C� ��

1

�. ��


s2 − s1 = Cpavg In

��

T 2 P − RIn 2 T 1 P 1

 kJ   kg − K   

�. �� TERMODINÁMICA DE GASES Y VAPORES

DR. ARTURO MORALES


DR. ARTURO MORALES

��

�. �� 2

∫

� �� s2 − s1 = Cp(T ) 1


�� ) ��

��

�� 1� T

∫

s = Cp(T ) o

0

dT T

s2 − s1 = s 0 2 − s 01 − RIn

�0 �� .

P 2 P 1

�) �� T avg = 37 C → Cpavg = 1.006 o

�� , �� . �� .

kJ kgK

� �� 0, �� TERMODINÁMICA DE GASES Y VAPORES

DR. ARTURO MORALES


��

�� . C�� )�� , �)�� ) �� P 2 = 15 MPa

�) ��

s2 = s1

Bomba

��

s2 − s1 = Cv prom In In

T 2 T 1

=−

In

�) �� Cp = R + Cv

x1 = 0

T 2 T 1

+ RIn

R Cv prom

In

V 2

s 2 − s1 = Cp prom In

V 1

V 2

In

V 1

 V  Cv = In  1  T 1  V 2 

T 2

k =

 T 2     T 1  isoentrópi TV esp

��


DR. ARTURO MORALES

T 2 R P In 2 = T 1 Cp prom P 1

Cp

R

Cv

Cv

k −1

co

 V  =  1   V 2 

R

prom

= k − 1

1� ��

�) ��

T 2 P − RIn 2 T 1 P 1

R

Propiedad de logaritmos In ( a x ) = xIn ( a )

P 1 = 10 kPa

DR. ARTURO MORALES

k −1

= cte.

In

 P  Cp prom T 2 = In  1  T 1  P 2  2��

 T 2   P  =  1   T   1  isoentrópi co  P 2 

(1 − k ) / k

TP (1− k )/ k = cte.

3��

 P 2   V  =  1   P   1  isoentrópi co  V 2 

k

PV esp = cte. k

�� .


DR. ARTURO MORALES

2

24/02/2014

��

�� 0 0 �� s 2 − s1 = s 2 − s 1 − RIn

s 0 2 = s 01 + RIn

P 2 P 1

�� .

P 2 P 1

D�� :

�� : �� :

V esp 2 V esp1

=

P 1V esp1 T 1

=

s2 − s1 = Cv prom In

P relativa �(�)

T 2 T 1

+ RIn

1� ��

P 2 s 0 2 − s 01 exp( s 0 2 / R ) P r 2 = exp = = P 1 R exp( s 01 / R ) P r 1

�� :

�� 14 � �� 50�F �� 320�F �� . D�� =1.667

 T 2   V  =  2   T   1  isoentrópi co  V 1 

k −1

V 2

s 2 − s1 = Cp prom In

V 1

2�� .

 T 2   P  =  1   T   1  isoentrópi co  P 2 

(1 − k ) / k

T 2 T 1

− RIn

P 2 P 1

3�� .

 P 2   V  =  1   P   1  isoentrópi co  V 2 

k

P 2V esp2 T 2

�� (�)

T 2 P 1 T 2 exp( s10 / R ) T 2 / P relativa 2 V r 2 = = = T 1 P 2 T 1 ( s20 / R ) T 1 / P relativa1 V r 1


DR. ARTURO MORALES


DR. ARTURO MORALES

��

�� 22�C � 95 �� . �� 1/�2=8, �� . k −1

TV esp ��

= cte.

��

TP (1−k ) / k = cte.

�� 1.5 � 3 �� C�2 � �� 100 ��. �� 150 �� . D�� .

k

PV esp = cte.

 T 2   V  =  1   T   1  isoentrópi co  V 2 

s2 − s1 = Cv prom In

k −1

CvCO2 = 0.657kJ / kgK

T 2 T 1

+ RIn

V 2 V 1

s2 − s1 = Cp prom In

T 2 T 1

− RIn

P 2 P 1

C��

1. �� 2 , �� . 2. �� 2 �� .

D��

�� C��

��

T 1 → V r 1 → V r 2 → T 2

V r 1 @ 295 K = 647.9 �� A�17 ��


V r 2 = V r 1

V esp 2 V esp1

T 2 = 662.7 K

1  = 647 .7   = 80.99  8  A �� 2 �� 2

DR. ARTURO MORALES


DR. ARTURO MORALES

�� 6��. E�� A. C�� & �� A. B�� : 7�70 7�83 7�93 7�95 7�99


DR. ARTURO MORALES

3

1.5 Las Relaciones T ds: Sistema cerrado

Recommend Documents