ChemE 260
Entropy Balances
On Open and Closed Systems
Dr. William Baratuci
Senior Lecturer
Chemical Engineering Department
University of Washington
TCD 8: A & B
CB 6: 10 + Supplement
May 10, 2005
Entropy Balance: Closed System
•
1st Law:
dU  Q  W
•
2nd Law, Internally
Reversible Processes:
Q  TdS
Boundary Work, Internally
Reversible Processes:
Wb  PdV
•
Gibbs 1st Equation:
dU  TdS  pdV
•
Entropy Balance Equation
•
– Integral Form:
– Differential Form:
– Rate Form:
Baratuci
ChemE 260
May 10, 2005
(Usually assume Wtot = Wb.)
Q
S 2  S1  
 Sgen
T
1
Q
dS 
  Sgen
T
dS Q
  S gen
dt T
2
Entropy Balance: Open System
•
General:
•
Steady-state,
SISO:
•
Entropy generation
within the system:
Baratuci
ChemE 260
May 10, 2005
dSsys
inlets
dt
  m in Sˆ in 


outlets

Q
mout Sˆ out  
 S gen
Tsys
Q
m Sˆ out  Sˆ in  
 S gen
Tsys


Q
S gen  m Sˆ out  Sˆ in  
Tsys
The
•
st
1
Law and Entropy
1st Law, Steady-state, Internally
Reversible, SISO:
Q  Wnot b

•
From the definition of
entropy:
Wnot b
m
Q
ˆ
Q 
m


v 2
g
 m  H 

z 
2g
g
C
C




v 2
g
    H 

z 
2g
g
m 
C
C

Q
out
 TdSˆ
in
•
Gibbs 2nd Equation:
ˆ  VdP
ˆ
TdSˆ  dH
•
Integrating for an
open system:
out
out
in
in
Baratuci
ChemE 260
May 10, 2005
 TdSˆ  Hˆ   Vˆ dP
Mechanical Energy Balance Equation
•
Combine all the equations from the previous slide:

•
The MEBE:
Wnot b
m
 ˆ out ˆ
 

v 2
g
   H   V dP    H 

z 
2g
g
C
C

in

 

out
Wnot b

m
in
out
•
If Wnot b = 0 :
0


in
v 2
g
V̂ dP 

z
2gC gC
2

v
g
ˆ dP 
V

z
2gC gC
P v 2
g
0


z
 2 gC gC
•
If Ekin = Epot = 0 :
Baratuci
ChemE 260
May 10, 2005

Wnot b
m
ˆ
 W
not b 
out

in
Bernoulli
Equation
Bernoulli Equation
(incompressible fluid)
ˆ dP
V
Usually,
Wnot b = Wsh
Shaft Work & PV Diagrams
Wsh

m
ˆ 
 W
sh
out

ˆ dP
V
in
• Polytropic Processes
ˆ C
PV
C
Ŵsh     
P
in 
out
or:
C
V̂   
P
1/ 
out
dP   C1 / 

in
1/ 
dP
P1 / 

ˆ P V
ˆ
P2 V
2
1 1
 1

•
 1 :
ˆ 
W
sh
•
=1:
 P2 
ˆ
ˆ
Wsh   P1 V1 Ln  
 P1 
Baratuci
ChemE 260
May 10, 2005

PV Diagram: Polytropic
Baratuci
ChemE 260
May 10, 2005
Summary: Wsh Polytropic Processes
Process Type
Real Fluids
Ideal Gases
Ŵsh  0
Ŵsh  0
=0
=1
Isothermal
ˆ P V
ˆ Ln  P2 
W
sh
1 1
 P1 
Ŵsh  
R T1
P
Ln  2 
MW
 P1 
  1: Polytropic
ˆ 
W
sh

ˆ P V
ˆ
P2 V
2
1 1
 1

Ŵsh  
 =  = constant
ˆ 
W
sh

ˆ P V
ˆ
P2 V
2
1 1
 1


Isentropic
Ŵ  


ˆ V
ˆ P  P 
W
sh
2
1
 = : Isochoric
Baratuci
ChemE 260
May 10, 2005
•
sh

R
 T2  T1 
  1 MW
R
 T2  T1 
  1 MW
Ŵsh  
R
 T2  T1 
MW
Wsh is actially all work other than flow work.
Next Class …
• Problem Session
• After that…
– Isentropic Efficiency
• Define an efficiency for a process by comparing actual
performance to the performance of an isentropic process
• Nozzles, compressors and turbines
• New diagram: HS Diagram
– Multi-Stage Compressors
• Intercooler HEX’s reduce work input requirement
Baratuci
ChemE 260
May 10, 2005
Example #1
• Air is compressed from 1 bar and 310 K to
8 bar. Calculate the specific work and heat
transfer if the air follows a polytropic
process path with  = 1.32. Assume air is
an ideal gas in this process.
Baratuci
ChemE 260
May 10, 2005
Example #2
• A turbine lets down steam from 5 MPa and
o
500 C to saturated vapor at 100 kPa while
producing 720 kJ/kg of shaft work. The
outer surface of the turbine is at an average
o
temperature of 200 C. Determine the heat
losses from the turbine and the entropy
generation in the turbine in kJ/kg-K.
Baratuci
ChemE 260
May 10, 2005