Entropy Change by Heat Transfer
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Transcript Entropy Change by Heat Transfer
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
T0
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