Entropy Change by Heat Transfer
• Define Thermal Energy Reservoir (TER)
– Constant mass, constant volume
– No work - Q only form of energy transfer
– T uniform and constant
S
U

V
dS
dU
dU   Q 
dS
TER

TER
1
T
dS
Q

T , dU
1
T
Q
 Q / T


Entropy Change by Heat Transfer
• Consider two TERs at different Ts, in
contact but isolated from surroundings
 PS  d ( S A  S B )   Q (

1
TA
Heat transfer between
TERs produces entropy
as long as TB>TA

1
TB
)0
TER
TA
Q
TER


TB
Second Law for Control Mass
• Mechanical Energy Reservoir (MER)
• CM interacts with a TER and an MER
• MER no disorder;
MER
W
provides only
reversible work
CM , T


Q
• Overall system isolated
rev
TER


2nd Law
 PS  dS system
 PS  d ( S TER  S MER  S CM )
 PS    Q
T
 dS CM
dS CM   Q
T
  PS  0
No entropy change could occur because:
- Isentropic process (Ps = 0)

- entropy production cancelled by heat loss
Ps - Q/T = 0
Alternative Approach to 2nd Law
Clausius
• It is impossible to design a cyclic device that
raises heat from a lower T to a higher T
without affecting its surroundings. (need
work)
Kelvin-Planck
• It is impossible to design a cyclic device that
takes heat from a reservoir and converts it to
work only (must have waste heat)
Carnot’s Propositions
•
Corollaries of Clausius and KelvinPlanck versions of 2nd Law:
1. It is impossible to construct a heat engine
that operates between two TERs that has
higher thermal efficiency than a reversible
heat engine. th,rev> th,irrev
2. Reversible engines operating between
the same TERs have the same th,rev
Carnot (Ideal) Cycle
• Internally reversible
• Interaction with environment reversible
Qh
T
Win
Wout
Reversible work
S - constant
Reversible heat transfer
T - constant
QL
S

Carnot efficiency
• Define efficiency:
W
 th 
PS 
net , out
QH
QH
TH


Q H  QL
QH
TH

Carnot
Ql
TL
TH
QL
QH
Ql
QH
TL

over a reversible cycle
QH
1

QH
QL
efficiency
 th ,Carnot  1 
PS  0

TH

W
TL
QL
:
TL
TH
TL
This is the best
one can do
Gibbs Equation
• State equations relate changes in T.D.
variables to each other: e.g.,
q - w = du
• If reversible and pdv work only
Tds = du + pdv
• In terms of enthalpy: dh = du + d(pv)
dh = du + pdv + vdp; Tds = dh -vdp-pdv+pdv
Tds = dh - vdp
Unique aspect of Thermodynamics
• The Gibbs Equations were derived
assuming a reversible process.
• However, it consists of state variables
only; i.e., changes are path
independent.
• Proven for reversible processes but
applicable to irreversible processes
also.
Enthalpy Relations for a Perfect Gas
du
Gibbs : ds =
Integrate
s 



:

2
1
T
2
1

ds =
cv
p
T

dv 
2
1
c v dT

T
c v dT 

R
2
1
v
R
dv
dv
v
  
 R ln  1 
T
 2 
dT
Show yourself:
s 

2
1
cp
 p 
 R ln  1 
T
 p 2 
dT
fn (T)
fn (p)
Calculating s
• Calculate temperature and pressure
effects separately
0
12
s

0
2
0
1
 s s
where s
0


T
T0
cp
dT
T
sO(T) values are tabulated for different gases
in Tables D
For a Calorically Perfect Gas
T 
 p 
T 
 
 s12  c p ln  2  R ln  2   c v ln  2  R ln  2 
 T1 
 p1 
 T1 
  1 
T 
 p 
T 
 p 


ln  2  ln  2  
ln  2  ln  2 
R
R  T1 
 p1    1  T1 
 p1 
 s12
cp

T  1
 p 
 ln  2   ln  2  
R
 T1 
 p1 
 s12
 

T2 T1  1 

ln
 p 2 p1 

