Transcript Notes 20

Lec 19: Entropy changes, relative
pressures and volumes, work
1
• For next time:
– Prepare for Midterm 2 on Thursday, November
6th
• Outline:
– Isentropic processes for ideal gases
– Internal reversible work
– Entropy balance equations
• Important points:
– Try to identify the “governing equations” and
not get bogged down in all the special cases
– Understand how to use the Tds relationships
– Don’t forget to apply the 2nd Law when
working problems with entropy or internal
reversible processes
2
Isentropic processes of ideal gases
with constant specific heats
• Before we had the situation where the
specific heat could be considered
constant, T2 dT
T
 cp
T1
T
 c p ln
2
T1
T2
p2
s 2  s1  c p ln  R ln
T1
p1
If this is zero, so that s2 = s1
T2
p2
c p ln
 R ln
T1
p1
3
Isentropic processes of ideal
gases with constant specific
heats: approximation
Then
 p2 
T2
ln
 ln 
T1
 p1 
R
cp
And
T2  p 2 
  
T1  p1 
R
cp
The value in the exponent is:
R k 1

cp
k
4
Isentropic processes of ideal
gases with constant specific
heats: approximation
So
T2  p 2 
  
T1  p1 
k 1
k
This is only applicable for an isentropic
process and cp, cv and k constant.
5
Isentropic processes of ideal
gases with constant specific
heats: approximation
From the other entropy change equation for
an ideal gas with constant specific heats
T2
v2
s 2  s1  0  c v ln  R ln
T1
v1
Simplifying, we get:
T2  v 1 
  
T1  v 2 
k 1
6
Isentropic processes of ideal
gases with constant specific
heats: approximation
We can combine the two T2/T1 equations
to get:
k
p2  v1 
  
p1  v 2 
Rearrange:
p2 v 2   p1 v1 
k
k
So the polytropic process with n=k is
an isentropic process.
7
TEAMPLAY
Assuming constant specific heats, find
the final pressure of air which undergoes
an isentropic process from P1 = 14.7
psia, 60°F to a final temperature of
620°F.
8
Isentropic processes
For ideal gases with variable specific heats,
the calculations are straightforward:
p2
s 2  s1  s (T2 )  s (T1 )  R ln
0
p1
o
o
Because it is isentropic, s2 – s1 = 0.
9
Isentropic processes
If T2 is needed and T1, p1, and p2 are known,
p2
s (T2 )  s (T1 )  Rln
p1
o
o
and one may need to interpolate to find T2.
10
Isentropic processes
If p1, T1, and T2 are known and p2 is needed,
 s o (T2 )  s o (T1 )  exp[s o (T2 ) / R ]
p2
 exp


o
p1
R
exp[
s
(T1 ) / R ]


Notice from this that p2 and p1 can be
given as functions of T, for isentropic
processes.
11
Isentropic processes
So, the relative “pressure” is defined and
tabulated:
pr (T )  exp[s (T )/R]
o
Note: pr is not a pressure in spite of
its name--it has no units. The right
hand side is also only dependent on
temperature.
12
Isentropic processes
We can use the pressure ratios directly
for determining pressure changes in
isentropic processes.
 p2

 p1

pr2
 
 s p r1
NOTE: Pr only works for isentropic
processes!!! Don’t try to use it for any
other process.
13
TEAMPLAY
Use relative pressures in your tables in your
books to find the final pressure of air which
undergoes an isentropic process from P1 =
14.7 psia, 60°F to 620°F.
14
Isentropic processes
There is a similar relationship for volumes:
 v2 
v r2
  
 v 1  s v r1
As with Pr, this only applies to isentropic
relationships and the right hand side only
depends on temperature.
15
Work in an internally reversible
flow system
• Earlier we had
w   pdv
• This was true for a quasiequilibrium
process.
16
Work for an internally reversible
flow system.
• Quasiequilibrium process--one for which
departures from equilibrium are
infinitesimally small.
• All states through which a system passes
in a quasiequilibrium process may be
considered to be themselves equilibrium
states.
17
Work for an internally reversible
flow system
• A reversible process must proceed through
a series of equilibrium states.
• Otherwise, there would be a tendency for
the system to change spontaneously,
which is irreversible.
• Therefore, a quasiequilibrium process is
an (internally) reversible process
18
Work for an internally reversible
flow system
• Consider an internally reversible steady
flow system:
• Second law
2
q int rev   T ds
1
19
Work for an internally reversible
flow system
• The first law (not limited to internally
reversible processes at this point) says

V2 

2
q  w  h 2  h1
2

V1 

2
2
 g( z 2  z 1 )
20
Work for an internally reversible
flow system
• If we do limit it to internally reversible
processes, which we will emphasize
with subscripts,
q int rev  w int rev  h 2  h1
2
2




V2
V1


2
2
 g(z 2  z1 )
• Then the heat transfer term can be
replaced
21
Work for an internally reversible
flow process
• Combining the laws yields
2
 T ds  w

V2 

2
int rev
 h 2  h1
1
2

V1 

2
2
 g( z 2  z 1 )
• Rearranging
w int rev
2
 V1 2 V2 2
   T ds  h 2  h1  

 g( z 1  z 2 )
2
2
1

22
Work for an internally reversible
flow process
• Now, use a Tds relation:
• Tds = dh – vdp
2
2
2
1
1
1
 T ds   dh   vdp
2
 Tds  ( h
1
2
2
 h 1 )    vdp
1
23
Work for an internally reversible
flow process
2
w int rev   
1

V1 
vdp 
2
2

V2 

2
2
 g( z 1  z 2 )
•An expression for internally reversible
work in a steady flow process.
24
Work for an internally reversible
flow process
• In the absence of KE and PE effects,
2
w int rev    vdp
1
• On a P-v diagram,
P
2
1
v
25
Work
• For open systems,
2
w int rev    vdp
1
• For closed systems
2
w int rev   pdv
1
26
Compressor work
• By using the relationship Pvn = constant
(or Pvk = constant) and solving for
v=
c / P 
1
n
2
, the expression w int rev    vdp
2
1
or w int rev    vdp as the book uses can
be integrated 1to get the expressions on
p. 310
27
Entropy change for a closed
system
 Q 
1  T 
2
S 2  S1
Entropy change of
system as it goes
from 1 to 2; can
be + or –
depending on the
other two terms.

Entropy transfer to the system via heat
transfer; can be
+ or –, depending on the
sign for Q.

S gen
Entropy production term: >0
when internal
irrerversibilities
are present; =0
when no int irr
are present; <0
never.
28
Entropy change of an internally
reversible process; heat transfer
2 V=c
T
P=c
1
s
The area under
either the red or
blue curves would
represent the heat
transfer for either
process:
2
q 12   Tds
1
29
TEAMPLAY
What does a reversible, adiabatic
process look like on a T-s diagram??
30
Entropy production and transfer
We can use entropy (the second law)
to find the minimum work (associated
with a reversible process).
31
Example Problem
Refrigerant 134a is compressed in a
piston-cylinder assembly from saturated
vapor at 10°F to a final pressure of 120
psia. Determine the minimum theoretical
work input required per unit mass of
refrigerant, in Btu/lb.
32
Look at possibilities on a Ts
diagram...
2?
P = 120 psia
T, F
2?
2?
P = 26.65 psia
10
1
S
Starting at 1, where is point 2?
What can we know?
33
Adiabatic Compression of R-134a
The first law gives us
q  w  u 2  u1  KE  PE
We know the process is adiabatic, so q = 0.
We ignore KE and PE.
Then w = u1 – u2
u1 = 94.68 Btu/lb from Table A-11E.
34
Adiabatic Compression of R-134a
We need to find u2 for minimum work on this
compressor.
A reversible system will have minimum work
done on it. We showed this in ch. 6 for
cycles. It is shown again for open systems
on pp. 308-309. That analysis can easily be
extended to closed systems by replacing h
with u.
w u u
1
2
Minimum work corresponds to the smallest
allowable value for u2, which we can
determine from the second law.
35
Adiabatic Compression of R-134a
q
s 2  s1  
 s gen
T
• q = 0 because it is adiabatic.
• sgen = 0 because the minimum work
occurs for a reversible process
(frictionless).
• So s2 = s1
• For irreversibilities, sgen > 0, which
implies that s2 > s1
36
Adiabatic Compression of R-134a
2a
P = 120 psia
T, F
2s
P = 26.65 psia
10
1
S
37
TEAMPLAY
The reversible situation is
s2 = s1 = 0.2214 Btu/(lbm-R)
For a simple, compressible system, you
now know enough to find u2s and w.
38
Entropy rate balance for control
volumes
For a closed system, we had (as one of
our forms of the second law)
Q
S 2  S1   S gen
Tb
For a closed system, a rate form can be
written:
Q j 
dS

 S gen
dt
j Tj
39
Entropy rate balance for control
volumes
Now, if entropy is transferred or convected,
into and out of the control volume, the
equation becomes, for an open system
Q j
dScv
    m i si   m e se  S gen
dt
j Tj
i inlet
e exit
40
Entropy rate balance for control
volumes
Q j
dScv
    m i si   m e se  S gen
dt
j Tj
i inlet
e exit
Rate of
change of
entropy
within the
control
volume
(CV).
Rate of
entropy
transfer
into the CV
as a result
of heat
transfer
Rate of
entropy
transfer
into the
CV with
the mass
flow.
Rate of
entropy
transfer
out of
the CV
with the
mass
flow.
Rate of
entropy
production due
to irreversibili
-ties in
the CV
41
Entropy rate balance for control
volumes
This can be simplified:
Q j
dScv
    m i si   m e se  S gen
dt
j Tj
i inlet
e exit
Assume steady flow, one heat transfer
term , and one exit and one inlet:
Q
0   m ( si  se )  S gen
T
42
Entropy rate balance for control
volumes

Now, if one divides through by m
we get
q
0   (si  se )  sgen
T
q
se  si  s2  s1   sgen
T
43
Entropy rate balance for control
volumes
q
s2  s1   sgen
T
•This equation is almost identical to the one
for closed systems. In that case, the whole
system goes between two states, 1 and 2.
•For an open system, the change is from
inlet to exit, and it is for the change of
condition of the mass in the system as it
moves from state 1 at the inlet to state 2 at
44
the exit.
Entropy rate balance for control
volumes
For sgen  0 (irreversibilities present):
q
s 2  s1 
T
If the process is also adiabatic, then
s2  s1
What can we conclude? An adiabatic,
irreversible process will always move to
the right on a Ts diagram.
45
Look at adiabatic compression
processes on a Ts diagram
T
2s
P2
2a
P2 > P 1
Forbidden
States
P1
1
s
46
Look at adiabatic expansion processes on a
Ts diagram
P1
T
1
P1 > P 2
Forbidden
States
P2
2s
2a
s
47
Isentropic Processes
(constant entropy)
For real substances, such as water, R-134a,
etc., use tabular entropy values or EES.
An isentropic compressor has R-134a
saturated vapor entering at 10°F and
leaving the compressor at 120 psia. What
does the process look like on a Ts diagram?
48
Adiabatic Compression of R-134a
P = 120 psia
T, F
2s
P = 26.65 psia
10
1
S
Pt 2 is at s2 = s1 and p = 120 lbf/in2.
49